tag:blogger.com,1999:blog-37486255103489613422021-03-03T22:01:09.953-08:00solidangl.esBill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.comBlogger29125tag:blogger.com,1999:blog-3748625510348961342.post-42853099146130335702021-03-03T14:39:00.053-08:002021-03-03T22:01:09.711-08:00Inverse Circular Functions<!--For permalink, put "spr21-mat130-" and then lesson name with hyphens.--> <p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p><p><a href="#previewactivity">Go to Preview Activity</a></p><hr /><p> Textbook Reference: <a href="https://openstax.org/books/precalculus/pages/6-3-inverse-trigonometric-functions" target="_blank">Section 6.3 <b>Inverse Trigonometric Functions</b></a></p><hr /> <p>As we've been solving equations this week, perhaps you've noticed that something seems a little off...</p><p><b>All of the equations have been too "nice."</b></p><p>What do we do with an equation like this? \[\sin(\theta)=\dfrac13\] This equation <i>should</i> have a solution, since \(\dfrac13\) is well within the range of <b style="color: #2980b9;">heights</b> on the unit circle — it's between \(-1\) and \(1\). But there's no "nice" angle with a <b style="color: #2980b9;">height</b> of \(\dfrac13\). Are we just out of luck? </p><p>Of course not.</p><p>We need to dive a little deeper and figure out <i>how to "undo" a circular function</i> ... using <b>inverse functions</b>.</p><h2>Inverse functions</h2><p>If a function is a "machine" that turns inputs into outputs, then an <b><u>inverse function</u></b> is a corresponding "machine" that turns the outputs back into inputs.</p><p>For example, if \(f(x)=3x+1\), then \(f\) is a function that first triples a number and then adds \(1\). Then, its inverse function \(f^{-1}\) needs to "undo" these steps — first <i>subtract</i> \(1\), and then <i>divide</i> by \(3\) — so \(f^{-1}(x)=\dfrac{x-1}3\).</p><p>Many of the functions you've learned about so far are inverses of each other:</p> <style>table, th, td { border: none; border-collapse: collapse; } </style><table class="center"> <tbody> <tr> <td>Addition</td> <td>\(\leftrightarrow\)</td> <td>Subtraction</td> </tr> <tr> <td>Multiplication</td> <td>\(\leftrightarrow\)</td> <td>Division</td> </tr> <tr> <td>Powers*</td> <td>\(\leftrightarrow\)</td> <td>Roots</td> </tr> <tr> <td>Exponentials</td> <td>\(\leftrightarrow\)</td> <td>Logarithms</td> </tr> </tbody> </table><p>There's a catch though.</p><p>It becomes much more difficult to specify an inverse when your original function isn't <b>one-to-one</b>.</p><p>For example, did you notice that asterisk on "Powers" above? Say we have \(f(x)=x^2\), and I tell you that \(f(a)=4\). Then you don't have quite enough information to figure out what \(a\) is, because there are two possibilities: \(a\) could be \(2\), or it could be \(-2\). This happens because \(f\) isn't a <b>one-to-one</b> function — it doesn't pass the <b>horizontal line test</b> — so, given an output, you can't always <i>uniquely</i> determine the original input.</p><p style="text-align: center;"><iframe frameborder="0" height="500px" src="https://www.desmos.com/calculator/lvnvpmg7ic?embed" style="border: 1px solid #ccc;" width="500px"></iframe><br /><i>Drag the black point to move the horizontal line.<br />Notice how most outputs have two inputs.</i></p><p>We do have a trick up our sleeve though: we can <b>restrict</b> ourselves to only looking at <i>part</i> of the original graph of \(f(x)\), and decide that our inverse function should always give values on that branch.</p><p>That's what happens when we define \(\sqrt{x}\) to always give the <i>positive</i> square root of \(x\), so that \(\sqrt{4}=2\) rather than being undecided between \(2\) and \(-2\).</p><p>What's more, if you look at the graph of our newly-defined inverse function, you should notice that it's a <b>reflection</b> of our original graph, over the line \(y=x\) ... which makes sense, because we're swapping our inputs and outputs, that is, \(x\) and \(y\) coordinates.</p><p style="text-align: center;"><iframe frameborder="0" height="500px" src="https://www.desmos.com/calculator/kidt2we4j4?embed" style="border: 1px solid #ccc;" width="500px"></iframe><br /><i>Drag either the red or blue points.<br /> What do you notice about the coordinates?</i></p> <h2>Creating inverses of circular functions</h2><p>To solve our equation \(\sin(\theta)=\dfrac13\) above, we'd like to create an inverse of the sine function, but again we have the problem that it's not one-to-one. In fact, it's <i>very</i> not one-to-one.\( \DeclareMathOperator{\Sin}{Sin} \DeclareMathOperator{\Cos}{Cos} \DeclareMathOperator{\Tan}{Tan} \DeclareMathOperator{\Csc}{Csc} \DeclareMathOperator{\Sec}{Sec} \DeclareMathOperator{\Cot}{Cot} \DeclareMathOperator{\Arcsin}{Arcsin} \DeclareMathOperator{\Arccos}{Arccos} \DeclareMathOperator{\Arctan}{Arctan} \DeclareMathOperator{\Arccsc}{Arccsc} \DeclareMathOperator{\Arcsec}{Arcsec} \DeclareMathOperator{\Arccot}{Arccot} \)</p><p style="text-align: center;"><iframe src="https://www.desmos.com/calculator/q8in1jojpy?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe><br/><i>Drag the black point. Notice how many intersections there are!</i></p><p>But we can use the same trick from earlier, where we only focus on <i>part</i> of the graph of \(y=\sin(x)\). We could use any part we want as long as it passes the horizontal line test, but for convenience's sake we'll choose the "simplest" part of the graph: the part containing the origin, from \(-\dfrac\pi2\) to \(\dfrac\pi2\).</p><p style="text-align: center;"><iframe src="https://www.desmos.com/calculator/hvigirx52c?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe><br/><i>Drag the black point, and notice that this part of the graph passes the horizontal line test.</i></p><p>If we reflect this graph over the line \(y=x\), swapping all the input and outputs, we get a new graph that should serve as our inverse function:</p><p style="text-align: center;"><iframe src="https://www.desmos.com/calculator/wc6xzllbju?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe></p><p>We'll call this the <b style="color: #8e44ad;"><u>inverse sine function</u></b>, and we'll write it as \(\color{flatpurple}\Sin^{-1}(x)\).</p><p><i>(By the way, I usually write inverse circular functions with a capital letter at the beginning. You'll also see them with a lowercase letter instead; for example, Desmos uses all-lowercase.)</i></p><p>To put it more succinctly:</p><p style="text-align: center; font-size: large;"><b>\(\color{flatpurple}\Sin^{-1}(x)\) is the <b style="color: #8e44ad;">angle</b><br />between \(-\dfrac\pi2\) and \(\dfrac\pi2\)<br />whose <b style="color: #2980b9;">sine</b> is \(x\).</b></p><p>For example, notice that \(\color{flatblue}\sin\left(\dfrac\pi6\right)=\sin\left(\dfrac{5\pi}6\right)=\dfrac12\), so \(\color{flatpurple}\Sin^{-1}\left(\dfrac12\right)=\dfrac\pi6\), since \(\color{flatpurple}\dfrac\pi6\) is the unique angle that falls within the range from \(-\dfrac\pi2\) to \(\dfrac\pi2\).</p><p>By restricting the other circular functions similarly by defining the ranges we want for their <b style="color: #8e44ad;">inverses</b>:</p><ul> <li>\(\color{flatpurple}\Sin^{-1}(x)\), \(\color{flatpurple}\Csc^{-1}(x)\), and \(\color{flatpurple}\Tan^{-1}(x)\) always give angles between \(-\dfrac\pi2\) and \(\dfrac\pi2\).<br/> <div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-IZSMbj6TC-c/YEBPoKdDGLI/AAAAAAAAeao/UX5hx2wMV6APyNibBfIJXWfYXAr-fsv4wCLcBGAsYHQ/s1475/invright.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="60%" data-original-height="1475" data-original-width="1387" src="https://1.bp.blogspot.com/-IZSMbj6TC-c/YEBPoKdDGLI/AAAAAAAAeao/UX5hx2wMV6APyNibBfIJXWfYXAr-fsv4wCLcBGAsYHQ/s1475/invright.png"/></a></div> </li> <li>\(\color{flatpurple}\Cos^{-1}(x)\), \(\color{flatpurple}\Sec^{-1}(x)\), and \(\color{flatpurple}\Cot^{-1}(x)\) always give angles between \(0\) and \(\pi\).<br/> <div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-VNCzix1HGLo/YEBP33CXUfI/AAAAAAAAeaw/jQXz-UUfikQ8FTOvI7KvtbkmFEEe743UwCLcBGAsYHQ/s1382/invtop.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="60%" data-original-height="1341" data-original-width="1382" src="https://1.bp.blogspot.com/-VNCzix1HGLo/YEBP33CXUfI/AAAAAAAAeaw/jQXz-UUfikQ8FTOvI7KvtbkmFEEe743UwCLcBGAsYHQ/s1382/invtop.png"/></a></div> </li></ul><p>Here's the big take-away:</p><p style="text-align: center; font-size: large;"><b><span style="color: #8e44ad;">Inverse circular functions</span> tell us <span style="color: #8e44ad;">angles</span>.</b></p><hr /><a id="previewactivity"></a><h1 style="text-align: left;">Preview Activity 10</h1><p> <i>Answer these questions and submit your answers as a document <b>on Moodle</b>. (Please submit as .docx or .pdf if possible.)</i></p><style>table, th, td { border: 1px solid black; border-collapse: collapse; } </style><ol> <li style="padding-bottom: 1.45em; text-align: left;">Fill in the following table of values for \(\color{flatpurple}\Sin^{-1}(x)\).<br/> <table class="center"> <tbody> <tr> <td>\(x\)</td> <td style='width: 50px'>\(-1\)</td> <td style='width: 50px'>\(-\dfrac{\sqrt3}2\)</td> <td style='width: 50px'>\(-\dfrac{\sqrt2}2\)</td> <td style='width: 50px'>\(-\dfrac12\)</td> <td style='width: 50px'>\(0\)</td> <td style='width: 50px'>\(\dfrac12\)</td> <td style='width: 50px'>\(\dfrac{\sqrt2}2\)</td> <td style='width: 50px'>\(\dfrac{\sqrt3}2\)</td> <td style='width: 50px'>\(1\)</td> </tr> <tr> <td>\(\color{flatpurple}\Sin^{-1}(x)\)</td> <td></td> <td></td> <td></td> <td></td> <td></td> <td></td> <td></td> <td></td> <td></td> </tr> </tbody> </table> <br/> Make sure all values are between \(-\dfrac\pi2\) and \(\dfrac\pi2\). <li style="padding-bottom: 1.45em; text-align: left;">Fill in the following table of values for \(\color{flatpurple}\Cos^{-1}(x)\).<br/> <table class="center"> <tbody> <tr> <td>\(x\)</td> <td style='width: 50px'>\(-1\)</td> <td style='width: 50px'>\(-\dfrac{\sqrt3}2\)</td> <td style='width: 50px'>\(-\dfrac{\sqrt2}2\)</td> <td style='width: 50px'>\(-\dfrac12\)</td> <td style='width: 50px'>\(0\)</td> <td style='width: 50px'>\(\dfrac12\)</td> <td style='width: 50px'>\(\dfrac{\sqrt2}2\)</td> <td style='width: 50px'>\(\dfrac{\sqrt3}2\)</td> <td style='width: 50px'>\(1\)</td> </tr> <tr> <td>\(\color{flatpurple}\Cos^{-1}(x)\)</td> <td></td> <td></td> <td></td> <td></td> <td></td> <td></td> <td></td> <td></td> <td></td> </tr> </tbody> </table> <br/> Make sure all values are between \(0\) and \(\pi\). </li> <li style="padding-bottom: 1.45em; text-align: left;">What's the difference between \(\color{flatpurple}\Sin^{-1}(x)\) and \(\color{flatlightblue}\csc(x)\)? Explain the best you can.</li> <li style="padding-bottom: 1.45em; text-align: left;">Use your calculator or Desmos to solve the equation from the beginning of the section: \(\sin(x)=\dfrac13\). Are there other solutions?</li> <li> Answer <b>AT LEAST</b> one of the following questions: <ol style="list-style-type: lower-alpha;"> <li>What was something you found interesting about this reading?</li> <li>What was an "a-ha" moment you had while doing this reading?</li> <li>What was the muddiest point of this reading for you?</li> <li>What question(s) do you have about anything you've read?</li> </ol> </li></ol><p><a href="https://drive.google.com/file/d/1v4TqfkoG9fziftUoWlawxCgqp1dM07an/view?usp=sharing" target="_blank"><b>Answers</b></a></p><hr /><p> <a id="previewactivity"></a><a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-88086380167287369232021-03-01T16:12:00.019-08:002021-03-02T17:40:29.405-08:00Advanced Equation Techniques<!--For permalink, put "spr21-mat130-" and then lesson name with hyphens.--> <p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p><p><a href="#previewactivity">Go to Preview Activity</a></p><hr /><p> Textbook Reference: <a href="https://openstax.org/books/precalculus/pages/7-5-solving-trigonometric-equations" target="_blank">Section 7.5 <b>Solving Trigonometric Equations</b></a></p><hr /> <p>Let's take a closer look at "splitting", in the context where it's most useful: factoring.</p><p><i>(Again, if you remember this from last semester, you can actually skip right to the <a href="#previewactivity">Preview Activity</a>, but if you need the refresher, read on.)</i></p><h2>Splitting polynomials</h2><p>Say we've got the following quadratic equation. \[2x^2+6x-20=0\] You have a couple of ways to solve this at your disposal: </p><ul> <li>On one hand, you could complete the square or use the Quadratic Formula. These are guaranteed to work (in fact they're actually the same thing), but they can be a little long-winded and tedious.</li> <li>On the other hand, you could try to factor the left side. This <i>isn't</i> guaranteed to always work, but if it does, it'll make it much easier.</li></ul><p>By now, you've probably already noticed that all the coefficients are divisible by \(2\), so at least we can factor that out: \[2(x^2+3x-10)=0\] That new quadratic in the parentheses is relatively easy to factor, since its leading coefficient is \(1\). We need to find two numbers whose <b>product</b> is \(-10\) and whose <b>sum</b> is \(-7\); after some thought, you should see that \(5\) and \(-2\) will do the trick. This means we can now factor the expression as follows: \[2(x+5)(x-2)=0\] Cool, now it's factored. But how is this supposed to help us? </p><p>Here's the key idea behind <i>why</i> we factor:</p><p style="text-align:center;"><b>If a <i>product</i> is zero,<br>then one of the <i>factors</i> must be zero.</b></p><p>Since we have three things multiplied together, we now have three possibilities:</p><ul> <li>We could have \(2=0\). That leads nowhere pretty fast, though, so we'll ignore it.</li> <li>We could have \(x+5=0\), which would mean \(x=-5\).</li> <li>We could have \(x-2=0\), which would mean \(x=2\).</li></ul><p>Thus we end up with two solutions: \(x=-5\), or \(x=2\).</p><h2>Taking it up a notch</h2><p>Let's do a harder one: \[6x^2-x-2=0\] This one doesn't have that common factor of \(2\), and even if it did we still wouldn't have a leading coefficient of \(1\). </p><p>We can still do it with a bit of guess-and-check, but personally I'm a fan of the so-called "area model" or "<a href="https://www.youtube.com/watch?v=OKkg2kUsDq4">box method</a>". Here's a video if you've never seen the method before (done on a different problem):</p><div class="separator" style="clear: both; text-align: center;"><iframe class="BLOG_video_class" allowfullscreen="" youtube-src-id="OKkg2kUsDq4" width="640" height="366" src="https://www.youtube.com/embed/OKkg2kUsDq4"></iframe></div><p>So here's how we'd set it up for our problem:</p><div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-nib9AJLqpBo/YD2aulXbV4I/AAAAAAAAeaA/avemrrMQCps2EQ6skwDe_LHnCMa-9O7rwCLcBGAsYHQ/s927/boxmethod1.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="35%" data-original-height="594" data-original-width="927" src="https://1.bp.blogspot.com/-nib9AJLqpBo/YD2aulXbV4I/AAAAAAAAeaA/avemrrMQCps2EQ6skwDe_LHnCMa-9O7rwCLcBGAsYHQ/s927/boxmethod1.png"/></a></div><p>To fill in the other two boxes, we need two numbers whose product is \(6\cdot-2=-12\) and whose sum is \(1\). This leads us to \(-4\) and \(3\), so we'll use those numbers to split up the middle term:</p><div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-NimcAnyX3n8/YD2auvrpxEI/AAAAAAAAeZ4/gtuacDz8CoIFhv_BT76LYMJUaIJFyFsgwCLcBGAsYHQ/s941/boxmethod2.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="35%" data-original-height="586" data-original-width="941" src="https://1.bp.blogspot.com/-NimcAnyX3n8/YD2auvrpxEI/AAAAAAAAeZ4/gtuacDz8CoIFhv_BT76LYMJUaIJFyFsgwCLcBGAsYHQ/s941/boxmethod2.png"/></a></div><p>The greatest common factor of the top row is \(2x\), so we put that to the left, and after that the rest of the table fills itself in:</p><div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-pKZkcStUZAQ/YD2aukl1l6I/AAAAAAAAeZ8/auhJ7iifR_sostOziFrNdcxPhSy_gfLkACLcBGAsYHQ/s961/boxmethod3.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="35%" data-original-height="577" data-original-width="961" src="https://1.bp.blogspot.com/-pKZkcStUZAQ/YD2aukl1l6I/AAAAAAAAeZ8/auhJ7iifR_sostOziFrNdcxPhSy_gfLkACLcBGAsYHQ/s961/boxmethod3.png"/></a></div><p>Essentially we've reconstructed the original multiplication problem: \[(2x+1)(3x-2)=0\] Since we have a product of two things equaling zero, this splits into two possibilities: </p><ul> <li>We could have \(2x+1=0\), which would mean \(x=-\dfrac12\).</li> <li>We could have \(3x-2=0\), which would mean \(x=\dfrac23\).</li></ul><p>Thus we end up with two solutions: \(x=-\dfrac12\), or \(x=\dfrac23\).<br></p><p>These aren't the only ways to do factoring, but at least it should jog your memory. You'll need these techniques to be able to tackle some of the new equations we'll solve in class with circular functions.</p> <hr /><a id="previewactivity"></a><h1 style="text-align: left;">Preview Activity 9</h1><p> <i>Answer these questions and submit your answers as a document <b>on Moodle</b>. (Please submit as .docx or .pdf if possible.)</i></p><style>table, th, td { border: 1px solid black; border-collapse: collapse; } </style><ol> <li style="padding-bottom: 1.45em; text-align: left;">Consider this equation: \[x^2=x\] A student solves the equation by dividing both sides by \(x\) and leaving \(x=1\).<br> Show that there is another solution. Why did the student miss this solution? What should they have done instead? </li> <li style="padding-bottom: 1.45em; text-align: left;">Consider this equation: \[x^2-6x+8=1\] A student solves the equation by factoring the left-hand side, which gives \((x-2)(x-4)=1\). They split the equation into two equations, \(x-2=1\) and \(x-4=1\), and conclude that solutions \(x=3\) or \(x=5\). <ol style="list-style-type: lower-alpha;"> <li>How can you show the student that these are not solutions of the original problem?</li> <li>Explain the student's mistake in reasoning.</li> <li>Correctly solve the equation.</li> </ol> </li> <li style="padding-bottom: 1.45em; text-align: left;">Solve this equation: \[3x^2+10x-8=0\] If you really want to, you can just complete the square or use the Quadratic Formula to solve it, but the left-hand side is factorable, so you should try to factor it. It'll be good practice.<br>(If you need a bunch more practice because your factoring is rusty, here's some <a href="https://cdn.kutasoftware.com/Worksheets/Alg2/Quadratic%20Equations%20By%20Factoring.pdf">optional extra practice</a> with answers provided. </li> <li> Answer <b>AT LEAST</b> one of the following questions: <ol style="list-style-type: lower-alpha;"> <li>What was something you found interesting about this reading?</li> <li>What was an "a-ha" moment you had while doing this reading?</li> <li>What was the muddiest point of this reading for you?</li> <li>What question(s) do you have about anything you've read?</li> </ol> </li></ol><p><a href="https://drive.google.com/file/d/1v4TqfkoG9fziftUoWlawxCgqp1dM07an/view?usp=sharing" target="_blank"><b>Answers</b></a></p><hr /><p> <a id="previewactivity"></a><a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-51655108720748326172021-02-27T19:44:00.008-08:002021-02-27T22:09:19.214-08:00Solving Equations with Circular Functions<!--For permalink, put "spr21-mat130-" and then lesson name with hyphens.--> <p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p><p><a href="#previewactivity">Go to Preview Activity</a></p><hr /><p> Textbook Reference: <a href="https://openstax.org/books/precalculus/pages/7-5-solving-trigonometric-equations" target="_blank">Section 7.5 <b>Solving Trigonometric Equations</b></a></p><p><i>At this point, we're going to start jumping around a bit and doing things in a slightly different order than the textbook, to make the material "flow" a bit better. If you want to get practice by doing exercises, try to look for exercises that look like what we're doing in and outside of class; if something looks unfamiliar, it might be something we'll get to later in the semester.</i></p><hr /> <p>Ahh, solving equations ... the bread and butter of algebra. Brings back good memories, doesn't it?</p><p>Well ... it brings back memories, at least. Sadly, how good those memories are might vary.</p><p>In previous classes, you likely learned a whole ton of different techniques for solving equations, with different steps for different kinds of functions — linear functions, quadratic functions, polynomial functions, exponential functions, logarithmic functions, and so on. It might have felt overwhelming at times.</p><p>But I'm going to let you in on a little secret.</p><p>It turns out that there are really only two algebraic techniques for solving equations:</p><p style="text-align:center;">"<b><u>Undoing</u></b>" and "<b><u>Splitting</u></b>".</p><p><i>If you had me last semester, you already know this, and you can <a href="#skip">skip this section</a> if you'd like. If not, or if you just want a quick refresher, read on.</i></p><h2>Undoing and splitting</h2><p>Suppose you I tell you I'm thinking of a number.</p><p>I can take my number, and do the following things in order:</p><ul> <li>Cube it.</li> <li>Double it.</li> <li>Add seven.</li> <li>Divide it by five.</li></ul><p>When I do these things, the answer I end up with is \(27\).</p><p>How can you figure out my original number?</p><p>You could just guess a bunch of different numbers and check whether you end up with \(27\). But that might take a while, so a more efficient and clever solution would be to roll the clock back and "<b>undo</b>" each of my steps:</p><ul> <li>Start off with the answer, \(27\).</li> <li>Multiply it by \(5\), which gives \(135\).</li> <li>Subtract \(7\), which gives \(128\).</li> <li>Cut it in half, which gives \(64\).</li> <li>Finally, take the cube root, which gives \(4\).</li></ul><p>Phrased algebraically, what you just did might look like this: \[ \begin{align*} \dfrac{2x^3+7}{5} &= 27\\ 2x^3+7 &= 135 & \text{Multiply both sides by }5\\ 2x^3 &= 128 & \text{Subtract }7\text{ from both sides}\\ x^3 &= 64 & \text{Divide both sides by }2\\ x &= 4 & \text{Take cube root of both sides} \end{align*} \] </p><p>What we're essentially doing is using the idea of <b>inverse functions</b> to get our variable by itself — each thing we did to \(x\) is like a layer, and by "undoing" those things, we're peeling back the layers.</p><p>We can run into some problems though. Take a look at this equation now: \[2|x+3|+1 = 7\] We can subtract \(1\) from both sides and then divide both sides by \(2\), but then we're left with: \[|x+3|=3\] It's more difficult to "peel back" the absolute value function, since it's not <b>one-to-one</b> — there are multiple inputs that could give us the same output. </p><p>But if we think about how absolute values work, we can realize that there are two numbers with an absolute value of \(3\): namely, \(3\) and \(-3\). This allows us to "<b>split</b>" our complicated equation into two easier-to-solve equations: \[x+3=3\qquad\qquad x+3=-3\] Solving these equations individually gives us our two solutions: \(x=0\) or \(x=-6\). </p><a id="skip"></a><h2>Using circular functions</h2><p>Now that you're familiar with the unit circle, you already can solve some basic equations involving the circular functions.</p><p>...wait you <i>are</i> familiar with the unit circle, right?</p><p>If you haven't spent some time with your best-friend-the-unit-circle in some time, you might need to go back and review it. Sit down with it. Have some deep conversations. Get to where you can <i>visualize</i> the <b style="color: #8e44ad;">angle</b> measures (in both degrees <i>and</i> radians, as well as the <b style="color: #2980b9;">height</b>, <b style="color: #c0392b;">overness</b>, and <b style="color: #27ae60;">slope</b> of those important points on the rim. Doing so will mean you spend less time going back and forth looking things up, and more time actually solving problems.</p><p>Consider the following equation: \[\sin(x)=\dfrac12\] What this is asking is, "what are all angles on the unit circle that have a <b style="color: #2980b9;">height</b> of \(\dfrac12\)? </p><p>If you look back at your unit circle — or, even better, <i>if you can see it in your mind's eye</i> — you should see that there are two solutions: \(x=\dfrac\pi6\), and \(x=\dfrac{5\pi}6\).</p><div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-38TdCC9-QjY/YDsLFXlym2I/AAAAAAAAeZY/Dj5bGAEGxV4VMh01Nw2sRj3N_B4IXYIgwCLcBGAsYHQ/s1293/sinpi%253D12.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="75%" data-original-height="1273" data-original-width="1293" src="https://1.bp.blogspot.com/-38TdCC9-QjY/YDsLFXlym2I/AAAAAAAAeZY/Dj5bGAEGxV4VMh01Nw2sRj3N_B4IXYIgwCLcBGAsYHQ/s1293/sinpi%253D12.png"/></a></div><p>But is that <i>all</i> the solutions?</p><p>Remember, the circular functions are <b>periodic</b>, so they repeat forever for infinitely many <b>coterminal</b> angles. So, for example, \(x=\dfrac{13\pi}{6}\) would be a solution, as would \(x=-\dfrac{7\pi}{6}\).</p><p>We can see this even more clearly if we <b>graph</b> both sides of our equation:</p><p style="text-align:center;"><iframe src="https://www.desmos.com/calculator/etyyzbbuwb?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe><br /><i>Click the points of intersection to see their coordinates.</i></p><p>Each \(x\)-value where the two graphs intersect is a solution. (As you can see, there's quite a few.)</p><p>But I have some good news:</p><p>Once you can visualize angles on the unit circle, <b>everything else is just applying what you already know about solving equations.</b> So, most of the time, all you really have to do is attempt to get your circular function by itself (through one means or another), and then call up your best friend the unit circle to help you out!</p> <hr /><a id="previewactivity"></a><h1 style="text-align: left;">Preview Activity 8</h1><p> <i>Answer these questions and submit your answers as a document <b>on Moodle</b>. (Please submit as .docx or .pdf if possible.)</i></p><style>table, th, td { border: 1px solid black; border-collapse: collapse; } </style><ol> <!-- <li style="padding-bottom: 1.45em; text-align: left;">Complete the following table for \(\color{flatgreen}y=\tan(x)\):<br/> <table class="center"> <tbody> <tr> <td>\(x\)</td> <td style='width: 40px'>\(-\tfrac\pi2\)</td> <td style='width: 40px'>\(-\tfrac\pi4\)</td> <td style='width: 40px'>\(0\)</td> <td style='width: 40px'>\(\tfrac\pi4\)</td> <td style='width: 40px'>\(\tfrac\pi2\)</td> </tr> <tr> <td>\(\color{flatgreen}\tan(x)\)</td> <td></td> <td></td> <td></td> <td></td> <td></td> </tr> </tbody> </table> <br/> Then sketch a graph of \(\color{flatgreen}y=\tan(x)\) on the interval \([-\tfrac\pi2,\tfrac\pi2]\) as accurately as possible <b>by hand</b>, including these points (and the two poles). </li> --> <li style="padding-bottom: 1.45em; text-align: left;">Find both solutions to the equation \(\tan(\theta)=-1\) in the interval \(0\le\theta\le 2\pi\). Then find at least two more solutions outside this interval. Include a unit circle sketch to support your answer.</li> <li style="padding-bottom: 1.45em; text-align: left;">Solve the equation \(2\cos(x)+1=0\). Explain the steps you take to solve it. Include a unit circle sketch to support your answer. </li> <li style="padding-bottom: 1.45em; text-align: left;">Come up with an equation (using a circular function) that has only <b>one</b> solution on the interval \([0,2\pi]\). Explain why it only has one solution. Include a unit circle sketch to support your answer. </li> <li> Answer <b>AT LEAST</b> one of the following questions: <ol style="list-style-type: lower-alpha;"> <li>What was something you found interesting about this reading?</li> <li>What was an "a-ha" moment you had while doing this reading?</li> <li>What was the muddiest point of this reading for you?</li> <li>What question(s) do you have about anything you've read?</li> </ol> </li></ol><p><a href="https://drive.google.com/file/d/1v4TqfkoG9fziftUoWlawxCgqp1dM07an/view?usp=sharing" target="_blank"><b>Answers</b></a></p><hr /><p> <a id="previewactivity"></a><a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-37169988356894318842021-02-24T21:12:00.027-08:002021-02-25T08:06:22.938-08:00Graphs of Other Circular Functions<!--For permalink, put "spr21-mat130-" and then lesson name with hyphens.--> <p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p><p><a href="#previewactivity">Go to Preview Activity</a></p><hr /><p> Textbook Reference: <a href="https://openstax.org/books/precalculus/pages/6-2-graphs-of-the-other-trigonometric-functions" target="_blank">Section 6.2 <b>Graphs of the Other Trigonometric Functions</b></a></p><hr /> <p>Now that you've got a feel for the graphs of the <b style="color: #2980b9;">sine</b> and <b style="color: #c0392b;">cosine</b> functions, let's take a look at the graphs of the other four circular functions. While these graphs aren't usually used to represent <i>physical</i> things, they do give us a good visual idea of how the functions behave at a glance — especially what happens as the values get very large.</p><h2>The tangent and cotangent functions</h2><p>Let's start with the <b style="color: #27ae60;">tangent</b> function.</p><p style="text-align:center;"><iframe src="https://www.desmos.com/calculator/ilj6kjzkb4?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe></p><p>...whoa.</p><p>That's a pretty wild-looking graph! What the heck is going on?</p><p>Let's break it down.</p><p> Remember our definition of the <b style="color: #27ae60;">tangent</b> function: \[ {\color{flatgreen}\tan(\theta)}=\dfrac{\color{flatblue}\sin(\theta)}{\color{flatred}\cos(\theta)} \] This means we can understand the <b style="color: #27ae60;">tangent</b> by thinking of what happens to the <b style="color: #2980b9;">sine</b> and <b style="color: #c0392b;">cosine</b>. </p><ul> <li>When \(\color{flatblue}\sin(\theta)=0\), the numerator is zero, so we'll have \(\color{flatgreen}\tan(\theta)=0\). This happens at \(0\), \(\pm\pi\), \(\pm2\pi\), and so on.</li> <li>When \({\color{flatblue}\sin(\theta)}={\color{flatred}\cos(\theta)}\), the numerator and denominator are the same, so we'll have \(\color{flatgreen}\tan(\theta)=1\). Similarly, when \({\color{flatblue}\sin(\theta)}={-\color{flatred}\cos(\theta)}\), the numerator and denominator are opposites, so we'll have \(\color{flatgreen}\tan(\theta)=-1\). These happen at \(\pm\tfrac\pi4\), \(\pm\tfrac{3\pi}4\), \(\pm\tfrac{5\pi}4\), and so on.</li> <li>When \(\color{flatred}\cos(\theta)=0\), the denominator is zero, so \(\color{flatgreen}\tan(\theta)\) will be <b style="color: #27ae60;">undefined</b>. Remember, however, that as a number shrinks to zero, its reciprocal blows up to infinity (in either direction), so we can also think of it as \(\color{flatgreen}\tan(\theta)=\overset\sim\infty\). This shows up on the graph as a bunch of <b>vertical asymptotes</b>, at \(\pm\tfrac\pi2\), \(\pm\tfrac{3\pi}2\), \(\pm\tfrac{5\pi}2\), and so on. <ul> <li>I prefer to call these vertical asymptotes "<b><u>poles</u></b>" instead. First of all, it's more consistent with what they're called in the rest of math when functions blow up to infinity. Second of all, it's shorter and easier to spell.</li> </ul> </li></ul><p>It also helps to remember that the <b style="color: #27ae60;">tangent</b> function tells us the <b style="color: #27ae60;">slope</b> of an angle! As we go around the unit circle, the slope increases rapidly, until we have an undefined (or infinite) slope at \(\tfrac\pi2\). After this, the slope is steep but negative, which makes the graph appear to "wrap around" infinity and come out the other side on the bottom of the graph! Eventually we hit zero again and the cycle repeats all over again.</p><p>By the way, notice that the period of the <b style="color: #27ae60;">tangent</b> function is actually \(\pi\). That's because it only takes half of a circle to get to another point with the same <b style="color: #27ae60;">slope</b> — for example, the <b style="color: #27ae60;">slope</b> is \(1\) at both \(\tfrac\pi4\) and \(\tfrac{5\pi}4\).</p><p>The <b style="color: #2ecc71;">cotangent</b> function looks similar:</p><p style="text-align:center;"><iframe src="https://www.desmos.com/calculator/zofzbxil2k?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe></p><p>You'll compare the <b style="color: #2ecc71;">cotangent</b> function to the <b style="color: #27ae60;">tangent</b> function in the <a href="#previewactivity">Preview Activity</a> below.</p><h2>The secant and cosecant functions</h2><p>Take a look at the graph of the <b style="color: #3498db;">cosecant</b> function.</p><p style="text-align:center;"><iframe src="https://www.desmos.com/calculator/g3qg5f3muj?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe></p><p>...whoa.</p><p>This graph is even weirder than the last ones. How can we break this one down?</p><p>Well, remember that the <b style="color: #3498db;">cosecant</b> function is the <i>reciprocal</i> of the <b style="color: #2980b9;">sine</b> function: \[ {\color{flatlightblue}\csc(\theta)}=\dfrac{1}{\color{flatblue}\sin(\theta)} \] So, maybe we can understand it by comparing those two graphs. Let's put them both on the same graph:</p><p style="text-align:center;"><iframe src="https://www.desmos.com/calculator/nc4a3avmnt?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe></p><p>Some things to notice here:</p><ul> <li>The <b style="color: #3498db;">cosecant</b> and <b style="color: #2980b9;">sine</b> graphs touch when they have values of \(1\) or \(-1\).<br> This makes sense, because \(\dfrac{1}{\color{flatblue}1}=\color{flatlightblue}1\) and \(\dfrac{1}{\color{flatblue}-1}=\color{flatlightblue}-1.\)</li> <li>As the <b style="color: #2980b9;">sine</b> gets smaller (that is, closer to zero), the <b style="color: #3498db;">cosecant</b> gets larger (that is, closer to infinity). As a result, when \(\color{flatblue}\sin(\theta)=0\), the value of \(\color{flatlightblue}\csc(\theta)\) is <b style="color: #3498db;">undefined</b> — or, once again, you can often think of it as \(\color{flatlightblue}\overset\sim\infty\). Graphically, this means that wherever \(\color{flatblue}y=\sin(x)\) has an <b style="color: #2980b9;">\(x\)-intercept</b>, \(\color{flatlightblue}y=\csc(x)\) has a <b style="color: #3498db;">pole</b>.</li></ul><p>And, as you might imagine, the graph of <b style="color: #e74c3c;">secant</b> has a similar shape:</p><p style="text-align:center"><iframe src="https://www.desmos.com/calculator/jk6zkggh01?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe></p><p>And we have a similar relationship to the <b style="color: #c0392b;">cosine</b>:</p><p style="text-align:center"><iframe src="https://www.desmos.com/calculator/sftkmaxwsc?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe></p><p>Take a careful look at these graphs to see how they relate to each other.</p><p>By the way, here's a useful intuition to have for these reciprocal graphs:</p><p style="text-align:center;font-size: large;">Reciprocals swap <b>large</b> values with <b>small</b> values.</p><p>If \(\color{flatred}\cos(\theta)\) is small, you can expect \(\color{flatlightred}\sec(\theta)\) to be large. Likewise, if \(\color{flatgreen}\tan(\theta)\) is large, you can expect \(\color{flatlightgreen}\cot(\theta)\) to be small.</p> <h2>Putting it all together</h2><p>If you want to see all the graphs together, you can take a look at <a href="https://www.desmos.com/calculator/pj3vyzd5bb">this Desmos link</a>.</p><p>One side note about using Desmos by the way — <i>it won't actually draw asymptotes by itself.</i> In order to get them to appear on the graphs above, I actually had to tell Desmos where to put them (which you will see if you open them up yourself and look at how they're set up.</p><p><br></p><p>By the way, you might understandbly be feeling a bit overwhelmed because of how many different graphs we've looked at.</p><p>If so, let me put this in a broader perspective.</p><p>It's essential to be able to sketch out the basic graphs of \(\color{flatblue}y=\sin(x)\) and \(\color{flatred}y=\cos(x)\), since they're arguably the two most important circular functions and you want to have a good intuition for how they work.</p><p>However, producing the graphs of the other four functions from memory isn't nearly as important — it's enough to have an overall idea of what they look like, and let technology do the rest if necessary. What's <i>far</i> more important is that you <b>understand</b> where the graphs come from, why they look like they do, and how they're related.</p> <hr /><a id="previewactivity"></a><h1 style="text-align: left;">Preview Activity 7</h1><p> <i>Answer these questions and submit your answers as a document <b>on Moodle</b>. (Please submit as .docx or .pdf if possible.)</i></p><style>table, th, td { border: 1px solid black; border-collapse: collapse; } </style><ol> <!-- <li style="padding-bottom: 1.45em; text-align: left;">Complete the following table for \(\color{flatgreen}y=\tan(x)\):<br/> <table class="center"> <tbody> <tr> <td>\(x\)</td> <td style='width: 40px'>\(-\tfrac\pi2\)</td> <td style='width: 40px'>\(-\tfrac\pi4\)</td> <td style='width: 40px'>\(0\)</td> <td style='width: 40px'>\(\tfrac\pi4\)</td> <td style='width: 40px'>\(\tfrac\pi2\)</td> </tr> <tr> <td>\(\color{flatgreen}\tan(x)\)</td> <td></td> <td></td> <td></td> <td></td> <td></td> </tr> </tbody> </table> <br/> Then sketch a graph of \(\color{flatgreen}y=\tan(x)\) on the interval \([-\tfrac\pi2,\tfrac\pi2]\) as accurately as possible <b>by hand</b>, including these points (and the two poles). </li> --> <li style="padding-bottom: 1.45em; text-align: left;">Compare and contrast the graphs of \(\color{flatgreen}y=\tan(x)\) and \(\color{flatlightgreen}y=\cot(x)\). Pay special attention to where the zeros and poles are.</li> <li style="padding-bottom: 1.45em; text-align: left;">The value of \(\tfrac\pi2\) is about \(1.5707963268\). Complete the following table.<br/> <table class="center"> <tbody> <tr> <td>\(x\)</td> <td style='width: 40px'>1.5</td> <td style='width: 50px'>1.57</td> <td style='width: 70px'>1.5707</td> <td style='width: 80px'>1.57079</td> <td style='width: 90px'>1.570796</td> </tr> <tr> <td>\(\color{flatlightred}\sec(x)\)</td> <td></td> <td></td> <td></td> <td></td> <td></td> </tr> </tbody> </table> <br/> How does this support the behavior of \(\color{flatlightred}\sec(x)\) as \(x\) gets close to \(\tfrac\pi2\)? </li> <li style="padding-bottom: 1.45em; text-align: left;">Look at the graphs of all six circular functions. What kinds of <b>symmetry</b> do you notice in the various graphs? </li> <li> Answer <b>AT LEAST</b> one of the following questions: <ol style="list-style-type: lower-alpha;"> <li>What was something you found interesting about this reading?</li> <li>What was an "a-ha" moment you had while doing this reading?</li> <li>What was the muddiest point of this reading for you?</li> <li>What question(s) do you have about anything you've read?</li> </ol> </li></ol><p><a href="https://drive.google.com/file/d/1v4TqfkoG9fziftUoWlawxCgqp1dM07an/view?usp=sharing" target="_blank"><b>Answers</b></a></p><hr /><p> <a id="previewactivity"></a><a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-28891601368711328142021-02-22T22:32:00.010-08:002021-02-23T05:50:58.132-08:00Transforming Sinusoids - Exploration<p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p><hr /><p>I noticed that a lot of people struggled with transforming sinusoids on the Desmos activity in class. I think part of this was that I left it so open-ended — it was difficult to really get a grasp of how all the different values of \(a\), \(b\), \(c\), and \(d\) affected the graphs of \(\color{flatblue}y=a\sin\big(b(x-c)\big)+d\) and \(\color{flatred}y=a\cos\big(b(x-c)\big)+d\).</p><p>So, in response, I've programmed a new Desmos activity that should make it easier to see how to transform the graphs as well as how to match a given sinusoidal curve.</p><div class="separator" style="clear: both; text-align:center;"><a href="https://www.desmos.com/calculator/bucdgp916o" target="_blank" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="100%" data-original-height="1051" data-original-width="1948" src="https://1.bp.blogspot.com/-ewkhTL6G07Y/YDSh81B5SfI/AAAAAAAAeWc/AG1ZlaSrV6cr5QBBN05ix7-UB12VHJw7wCLcBGAsYHQ/s1948/sinusoiddesmos.png"/></a><a href="https://www.desmos.com/calculator/bucdgp916o" target="_blank"><b>View the Activity</b></a></div> <h1 style="text-align: left;">Preview Activity 6</h1><p> <i>Answer these questions and submit your answers as a document <b>on Moodle</b>. (Please submit as .docx or .pdf if possible.)</i></p><p>On the Desmos activity above, turn the folders on for Curve 1, Curve 2, and Curve 3 by clicking the <b>folder icons</b>:</p><div class="separator" style="clear: both; text-align:center;"><img alt="" border="0" data-original-height="189" data-original-width="65" src="https://1.bp.blogspot.com/-UwjIbBBaorI/YDSi6u9lVFI/AAAAAAAAeWk/ltwDQQ78tNIZoEJEZJm9U-OtV5v_ngG1wCLcBGAsYHQ/s0/folders.png"/></div><p>For <b>each</b> curve, find <b>three different graphs</b> that match it, including at least one <b style="color: #2980b9;">sine</b> and at least one <b style="color: #c0392b;">cosine</b> each. (So you should end up with <b>nine equations total</b>.)</p><ul><li><i>Tip: Try letting \(a\) be negative sometimes to get more equations for the same curve.</i></li></ul><p>Submit your nine equations, and then let me know if there was anything about this particular activity that made more sense than the one in class, or if there are still questions you have after doing it.</p><hr /><p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-2891241704923935602021-02-20T20:13:00.036-08:002021-02-22T22:57:35.124-08:00Graphs of Sine and Cosine<!--For permalink, put "spr21-mat130-" and then lesson name with hyphens.--><!--<style>/* Tooltip container */ .tooltip { position: relative; display: inline-block; border-bottom: 1px dotted black; /* If you want dots under the hoverable text */ } /* Tooltip text */ .tooltip .tooltiptext { visibility: hidden; width: 360px; background-color: black; color: #fff; text-align: center; padding: 5px 0; border-radius: 6px; /* Position the tooltip text - see examples below! */ position: absolute; z-index: 1; } /* Show the tooltip text when you mouse over the tooltip container */ .tooltip:hover .tooltiptext { visibility: visible; } </style>--><p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p><p><a href="#previewactivity">Go to Preview Activity</a></p><hr /><p> Textbook Reference: <a href="https://openstax.org/books/precalculus/pages/6-1-graphs-of-the-sine-and-cosine-functions" target="_blank">Section 6.1 <b>Graphs of the Sine and Cosine Functions</b></a></p><hr /> <p>When you're getting to know someone, it's worth it to learn different things about them. You might know someone's favorite pizza topping or what kind of movies they like, but once you start talking about things like their family or where they grew up, you start to get a better understanding of who they are, and your friendship gets stronger.</p><p>...wait, what does this have to do with math?</p><p>Well, for the past few lessons, we've been primarily focusing on the <b>geometric</b> properties of the circular functions — the way that they tell us things like <b style="color: #2980b9;">height</b>, <b style="color: #c0392b;">overness</b>, and <b style="color: #27ae60;">slope</b> of an angle on the unit circle. These are great intuitions to have, but they're really only part of their story — we need to get to know them better. For the next few lessons, we'll be focusing on how they behave as <b>functions</b> in their own right — much like how you studied functions like polynomials and logarithms.</p><h2>Getting a better picture</h2><p>One of the best ways to really get to know a function is to look at its graph. They do say, after all, that a picture is worth a thousand words.</p><p>If that's the case, then a GIF is worth a thousand pictures:</p><div class="separator" style="clear: both;"><a href="https://www.mathwarehouse.com/animated-gifs/images/sine-cosine-unit-circle-animation.gif" style="display: block; padding: 1em 0px; text-align: center;"><img alt="" border="0" data-original-height="325" data-original-width="500" src="https://www.mathwarehouse.com/animated-gifs/images/sine-cosine-unit-circle-animation.gif" /></a></div><p>As our point moves around the circle, its <b style="color: #2980b9;">height</b> and <b style="color: #c0392b;">overness</b> fluctuate, back and forth between \(-1\) and \(1\). The graph we get as a result is shaped like a wave — a shape we call a <b><u>sinusoid</u></b>.</p><p>Here's a portion of the graph of \(\color{flatblue}y=\sin(x)\), plotted on the interval \(0\le x\le 2\pi\):</p><p style="text-align: center;"><iframe frameborder="0" height="500px" src="https://www.desmos.com/calculator/ps0k69ksgg?embed" style="border: 1px solid #ccc;" width="500px"></iframe></p><p>And then the same portion of the graph of \(\color{flatred}y=\cos(x)\), over the same interval:</p><p style="text-align: center;"><iframe frameborder="0" height="500px" src="https://www.desmos.com/calculator/fbtc55wcja?embed" style="border: 1px solid #ccc;" width="500px"></iframe></p><p>Note that I said these are only a <i>portion</i> of the graph. The full graphs are more like this:</p><p style="text-align: center;"><iframe frameborder="0" height="500px" src="https://www.desmos.com/calculator/edpyvuqnb3?embed" style="border: 1px solid #ccc;" width="500px"></iframe></p><p>Notice how the basic shape of each (which is a little darker in the above graph) repeats forever in either direction. This is what we mean we say that the <b style="color: #2980b9;">sine</b> and <b style="color: #c0392b;">cosine</b> are <b><u>periodic</u></b> functions, and we call one full cycle a <b><u>period</u></b>.</p><p>By the way, did you notice that I wrote \(\color{flatblue}\sin(x)\) above instead of \(\color{flatblue}\sin(\theta)\)? What's up with that?</p><p>Once we recognize the sine as a <b>function</b>, we can use whatever variables we want. Our input <i>might</i> be the angle of a unit circle, so we might use \(\theta\). But it also might be the time that we've been on a Ferris wheel, so we might end up using something else like \(t\).</p><p>Remember the immortal words of Shakespeare:</p><p style="text-align: center;"><i>"A variable by any other name would smell as sweet."</i></p><p>(Okay, maybe I paraphrased a bit.)</p><h2>More about sinusoids</h2><p>We really unlock the true potential of sinusoidal graphs when we <b>transform</b> them:</p><p style="text-align: center;"><iframe frameborder="0" height="500px" src="https://www.desmos.com/calculator/qhgxs2ddka?embed" style="border: 1px solid #ccc;" width="500px"></iframe></p><p>To understand what we're seeing, it's helpful to define a few important terms:</p><ul> <li>The <b style="color: #e67e22"><u>midline</u></b> is the horizontal line in the center of the wave, halfway between the maximum and the minimum height.</li> <li>The <b style="color: #8e44ad"><u>amplitude</u></b> is the distance from the midline to either the maximum or the minimum; in other words, it's the maximum displacement from the midline.</li> <ul><li>Amplitude is always given as a <b>positive</b> number.</li></ul> <li>The <b style="color: #27ae60"><u>period</u></b>, as mentioned before, is how the length of one full cycle of the wave.</li></ul><p style="text-align:center;"><a href="https://1.bp.blogspot.com/-vOVd7qwBzWs/YDHvYfz6gcI/AAAAAAAAeV4/3uHTTn-2JSs1yAnVURc-JFwwePY0YI_TwCLcBGAsYHQ/s2322/waveparts.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="100%" data-original-height="958" data-original-width="2322" src="https://1.bp.blogspot.com/-vOVd7qwBzWs/YDHvYfz6gcI/AAAAAAAAeV4/3uHTTn-2JSs1yAnVURc-JFwwePY0YI_TwCLcBGAsYHQ/s2322/waveparts.png"/></a><br/><i>Click the picture to enlarge it.</i></p><p>For example, look at the following graph:</p><p style="text-align: center;"><iframe frameborder="0" height="500px" src="https://www.desmos.com/calculator/q1m2ndvjg6?embed" style="border: 1px solid #ccc;" width="500px"></iframe></p><ul> <li>The <b style="color: #e67e22">midline</b> is the line \(y=1\).</li> <li>The <b style="color: #8e44ad">amplitude</b> is \(3\).</li> <li>The <b style="color: #27ae60">period</b> is \(4\).</li></ul><p>In class you'll explore how to produce a sinusoidal wave with any <b style="color: #e67e22">midline</b>, <b style="color: #8e44ad">amplitude</b>, and <b style="color: #27ae60">period</b> you'd like — which will be essential to using these waves in all sorts of applications!</p> <hr /><a id="previewactivity"></a><h1 style="text-align: left;">Preview Activity 5</h1><p> <i>Answer these questions and submit your answers as a document <b>on Moodle</b>. (Please submit as .docx or .pdf if possible.)</i></p><style>table, th, td { border: 1px solid black; border-collapse: collapse; } </style><ol> <li style="padding-bottom: 1.45em; text-align: left;">Complete the following table for \(\color{flatblue}y=\sin(x)\):<br/> <table class="center"> <tbody> <tr> <td>\(x\)</td> <td style='width: 40px'>\(0\)</td> <td style='width: 40px'>\(\tfrac\pi2\)</td> <td style='width: 40px'>\(\pi\)</td> <td style='width: 40px'>\(\tfrac{3\pi}2\)</td> <td style='width: 40px'>\(2\pi\)</td> </tr> <tr> <td>\(\color{flatblue}\sin(x)\)</td> <td></td> <td></td> <td></td> <td></td> <td></td> </tr> </tbody> </table> <br/> Then sketch a graph of \(\color{flatblue}y=\sin(x)\) on the interval \([0,2\pi)\) as accurately as possible <b>by hand</b>, including these five points. </li> <li style="padding-bottom: 1.45em; text-align: left;">Repeat the above exercise for \(\color{flatred}y=\cos(x)\). Then compare and contrast the two graphs. <ul><li>By the way, you should practice sketching the graphs of \(\color{flatblue}y=\sin(x)\) and \(\color{flatred}y=\cos(x)\) from memory a few times. It <i>really</i> helps to be familiar with the basic shapes of these two functions!</li></ul> </li> <li style="padding-bottom: 1.45em; text-align: left;">Refer back to the graphs of \(\color{flatblue}y=\sin(x)\) and \(\color{flatred}y=\cos(x)\). <ol style="list-style-type: lower-alpha;"> <li>What is the <b style="color: #e67e22">midline</b> of each of these graphs?</li> <li>What is the <b style="color: #8e44ad">amplitude</b>?</li> <li>What is the <b style="color: #27ae60">period</b>?</li> </ol> </li> <li style="padding-bottom: 1.45em; text-align: left;">Explain in your own words why it should make sense that the <b style="color: #2980b9;">sine</b> and <b style="color: #c0392b;">cosine</b> functions are <b>periodic</b> — that is, why their graphs repeat forever. (It might help to remember what they represent on the unit circle!) </li> <li> Answer <b>AT LEAST</b> one of the following questions: <ol style="list-style-type: lower-alpha;"> <li>What was something you found interesting about this reading?</li> <li>What was an "a-ha" moment you had while doing this reading?</li> <li>What was the muddiest point of this reading for you?</li> <li>What question(s) do you have about anything you've read?</li> </ol> </li></ol><p><a href="https://drive.google.com/file/d/1v4TqfkoG9fziftUoWlawxCgqp1dM07an/view?usp=sharing" target="_blank"><b>Answers</b></a></p><hr /><p> <a id="previewactivity"></a><a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-27580351026121184732021-02-19T18:26:00.001-08:002021-02-22T23:14:09.580-08:00Right Triangle Trigonometry (Wrap-Up)<p>Since most of class was just spent working on the specific problems, this wrap-up will be short ... just a couple of tips:</p><ul> <li>When finding the values of circular functions based on some initial information (like a point or some given values), drawing the respective "bowtie triangle" we saw in class helps make sure everything ends up in the right place.</li> <li>When solving an application problem involving trigonometry, there are many possible paths to solve a given problem! Just be careful not to round things off too early or your answers might be a little bit off.</li></ul><p>I've added the cofunction identities to our <a href="https://www.solidangl.es/2021/02/spr21-mat130-list-of-identities.html">list of identities</a>. Check back with that list every so often as we continue to work on it.</p><hr><p><a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-25989744781736359362021-02-19T16:59:00.017-08:002021-02-22T23:01:12.683-08:00Fundamental Identities (Wrap-Up)<p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p><p>Here's the full set of tangents from the unit circle:</p><div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-C4VjLLzt_z0/YDBSwyGlaBI/AAAAAAAAeVc/1s_audJkDgUIlmD6pSkBsfytnIud-280QCLcBGAsYHQ/s1494/tangents.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="90%" data-original-height="1490" data-original-width="1494" src="https://1.bp.blogspot.com/-C4VjLLzt_z0/YDBSwyGlaBI/AAAAAAAAeVc/1s_audJkDgUIlmD6pSkBsfytnIud-280QCLcBGAsYHQ/s1490/tangents.png"/></a></div><p>Remember that most references will say that \(\color{flatgreen}\tan(\tfrac\pi2)\) as <b style="color: #27ae60;">undefined</b> — and it is, if you insist that your answer has to be a "real" number. But it's also helpful to think of the answer as \(\color{flatgreen} \overset\sim\infty\), since this also lets you work with cotangents nicely, as shown in the <a href="https://drive.google.com/file/d/14OrifZdksqA4TvOIY7Uo6rHRhJNLs6IH/view?usp=sharing">answers</a> to the Preview Activity. Just be careful — <b>infinity is not a "real" number</b>, so you can't expect it to always behave "nicely" in computations. (That's actually what you'll learn about in calculus — how to properly handle infinity!)</p><p>We also started establishing a list of identities — fundamental relationships between the various circular functions. I'll be keeping track of the identities we've learned on <a href="https://www.solidangl.es/2021/02/spr21-mat130-list-of-identities.html">a single page</a> to keep them all in one place!</p><p>The ones I'd recommend knowing cold are the top three reciprocal identities, the two quotient identities, and the first of the Pythagorean Identities (\({\color{flatblue}\sin^2(\theta)}+{\color{flatred}\cos^2(\theta)} = 1\)). You can always rederive the other ones if you need them (for example by dividing the Pythagorean identities through by \(\color{flatblue}\sin^2(\theta)\) or \(\color{flatred}\cos^2(\theta)\), though if you use them enough, you'll find that you'll end up knowing them without having to actively memorize them!</p><hr><p><a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-14283533677862512932021-02-19T16:55:00.016-08:002021-02-19T18:30:44.534-08:00List of Identities<p><a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p><hr /><h3>Reciprocal Identities</h3><p> \[ \begin{align*} {\color{flatlightblue}\csc(\theta)} &= \dfrac{1}{\color{flatblue}\sin(\theta)} & {\color{flatlightred}\sec(\theta)} &= \dfrac{1}{\color{flatred}\cos(\theta)} & {\color{flatlightgreen}\cot(\theta)} &= \dfrac{1}{\color{flatgreen}\tan(\theta)} \\ {\color{flatblue}\sin(\theta)} &= \dfrac{1}{\color{flatlightblue}\csc(\theta)} & {\color{flatred}\cos(\theta)} &= \dfrac{1}{\color{flatlightred}\sec(\theta)} & {\color{flatgreen}\tan(\theta)} &= \dfrac{1}{\color{flatlightgreen}\cot(\theta)} \end{align*} \] </p><h3>Quotient Identities</h3><p> \[ \begin{align*} {\color{flatgreen}\tan(\theta)} &= \dfrac{\color{flatblue}\sin(\theta)}{\color{flatred}\cos(\theta)} & {\color{flatlightgreen}\cot(\theta)} &= \dfrac{\color{flatred}\cos(\theta)}{\color{flatblue}\sin(\theta)} & \end{align*} \] </p><h3>Pythagorean Identities</h3><p> \[ \begin{align*} {\color{flatblue}\sin^2(\theta)}+{\color{flatred}\cos^2(\theta)} &= {1} \\ {\color{flatgreen}\tan^2(\theta)}\ \ \ +\ \ \ {1}\ \ \ &= {\color{flatlightred}\sec^2(\theta)} \\ {1}\ \ \ +\ \ \ {\color{flatlightgreen}\cot^2(\theta)} &= {\color{flatlightblue}\csc^2(\theta)} \end{align*} \] </p><h3>Cofunction Identities</h3><p> \[ \begin{align*} {\sin({\color{flatpurple}\theta})} &= {\cos({\color{flatpink}\tfrac\pi2-\theta})} & {\tan({\color{flatpurple}\theta})} &= {\cot({\color{flatpink}\tfrac\pi2-\theta})} & {\sec({\color{flatpurple}\theta})} &= {\csc({\color{flatpink}\tfrac\pi2-\theta})} \\ {\cos({\color{flatpurple}\theta})} &= {\sin({\color{flatpink}\tfrac\pi2-\theta})} & {\cot({\color{flatpurple}\theta})} &= {\tan({\color{flatpink}\tfrac\pi2-\theta})} & {\csc({\color{flatpurple}\theta})} &= {\sec({\color{flatpink}\tfrac\pi2-\theta})} \end{align*} \] </p><hr /><p><a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-18013639508740056252021-02-15T15:30:00.018-08:002021-02-22T23:01:26.327-08:00The Circular Functions (Wrap-Up)<style> img { image-rendering: -moz-crisp-edges; /* Firefox */ image-rendering: -o-crisp-edges; /* Opera */ image-rendering: -webkit-optimize-contrast;/* Webkit (non-standard naming) */ image-rendering: crisp-edges; -ms-interpolation-mode: nearest-neighbor; /* IE (non-standard property) */ } </style><p>Here's a "cleaned up" copy of the complete unit circle whose points you filled out today:</p><div class="separator" style="clear: both; text-align:center;"><a href="https://1.bp.blogspot.com/-6zFQs8PjNQ8/YCsB5gxLJXI/AAAAAAAAeSM/DeZKbSFN3TcdGa_xjjJkrNyHCQdTipnQQCLcBGAsYHQ/s1779/uccompletepoints.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="100%" data-original-height="1511" data-original-width="1779" src="https://1.bp.blogspot.com/-6zFQs8PjNQ8/YCsB5gxLJXI/AAAAAAAAeSM/DeZKbSFN3TcdGa_xjjJkrNyHCQdTipnQQCLcBGAsYHQ/s1779/uccompletepoints.png"/></a><br><i>Click the picture to enlarge it, especially if it's blurry.</i></div><p>A few important things to remember:</p><ul> <li>The only five magnitudes that ever show up as coordinates for these "nice" points are (in order from least to greatest): \[0,\ \dfrac12,\ \dfrac{\sqrt2}2,\ \dfrac{\sqrt3}2,\ 1\] Use negative signs where needed.</li> <li>The numbers \(0\) and \(1\) are always paired up, as are \(\dfrac12\) and \(\dfrac{\sqrt3}2\), while \(\dfrac{\sqrt2}2\) is always paired with itself.</li> <li>Instead of \(\dfrac{\sqrt{2}}{2}\), you can just as easily write \(\dfrac1{\sqrt2}\) instead — either one is as good as the other. There's no hard-and-fast "rule" that says you can <b>never</b> have a radical in the denominator! It all depends on what you want to do with it. Choose the form that lends itself best to the situation. <ul> <li>If you're curious, one reason that <b>historically</b> we used to rationalize the denominator has to do with by-hand calculations. If you want to calculate \(\dfrac1{\sqrt2}\) by hand, that's about \(1\div 1.414\) — set up that division problem and before long you'll see why it's unwieldy. But if you use the equivalent form \(\dfrac{\sqrt2}2\), you can see it's much easier to calculate as about \(0.707\).</li> </ul> </li> <li>Don't bother using silly mnemonics to memorize the quadrants in which <b style="color: #2980b9;">sine</b> and <b style="color: #c0392b;">cosine</b> are positive or negative. Instead, remember what these things <b>mean</b> (<b style="color: #2980b9;">height</b> and <b style="color: #c0392b;">overness</b> respectively) and use your eyes (or your mind's eye) to look for relationships. <b>There's no need to <i>memorize</i> it if you can <i>visualize</i> it.</b></li></ul><hr><p><a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-42689618330152893792021-02-13T14:57:00.062-08:002021-02-22T23:02:13.436-08:00Right Triangle Trigonometry<!--For permalink, put "spr21-mat130-" and then lesson name with hyphens.--><!--<style>/* Tooltip container */ .tooltip { position: relative; display: inline-block; border-bottom: 1px dotted black; /* If you want dots under the hoverable text */ } /* Tooltip text */ .tooltip .tooltiptext { visibility: hidden; width: 360px; background-color: black; color: #fff; text-align: center; padding: 5px 0; border-radius: 6px; /* Position the tooltip text - see examples below! */ position: absolute; z-index: 1; } /* Show the tooltip text when you mouse over the tooltip container */ .tooltip:hover .tooltiptext { visibility: visible; } </style> --><p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p><p><a href="#previewactivity">Go to Preview Activity</a></p><hr /><p> Textbook Reference: <a href="https://openstax.org/books/precalculus/pages/5-4-right-triangle-trigonometry" target="_blank">Section 5.4 <b>Right Triangle Trigonometry</b></a></p><hr /><script type="text/x-mathjax-config"> MathJax.Hub.Config({ TeX: { extensions: ["color.js"] }}); </script> <p>You've probably noticed by now that there have been an awful lot of <b>right triangles</b> popping up in our study of the circular functions. In fact, I'm willing to bet that the first time you were introduced to the <b style="color: #2980b9;">sine</b>, <b style="color: #c0392b;">cosine</b>, and <b style="color: #27ae60;">tangent</b> functions in high school, they didn't have anything to do with circles at all! Instead, these three functions also come up as certain <b>ratios</b> between the sides of right triangles.</p><p>You even learned a word for the study of measuring triangles: <b><u>trigonometry</u></b>.</p><p>Yup, that's all "trigonometry" means — "triangle-measuring." That's it.</p><p>Anyway, this explains why the six <b>circular functions</b> we've studied so far are often instead called <b><u>trigonometric functions</u></b>.</p><p>We'll continue to call them <b>circular functions</b> when possible, because they're useful for a whole lot more than measuring triangles — you'll see soon that they can model all sorts of other "real-world" <a href="#asterisk">*</a> phenomena like waves. Nevertheless, it's still useful to study what triangle have to do with circles in the first place.</p><h2>Trigonometric ratios</h2><p>Take a look at the right triangle in the figure below.</p><div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-GUmG0otYywo/YCyLp_R8thI/AAAAAAAAeTA/L5q6sOocHgQyvxel_M8cchIW9qlPahF8QCLcBGAsYHQ/s923/rt1.png" style="display: block; padding: 1em 0px; text-align: center;"><img alt="" border="0" data-original-height="869" data-original-width="923" src="https://1.bp.blogspot.com/-GUmG0otYywo/YCyLp_R8thI/AAAAAAAAeTA/L5q6sOocHgQyvxel_M8cchIW9qlPahF8QCLcBGAsYHQ/s923/rt1.png" width="50%" /></a></div><p>Imagine you're standing at the <b style="color: #8e44ad;">angle</b> marked \(\color{flatpurple}\theta\). Then one of the <b><u>legs</u></b> is across the triangle from you — we'll call it the <b style="color: #2980b9;"><u>opposite</u></b> leg — while the other leg is next to you — we'll call it the <b style="color: #c0392b;"><u>adjacent</u></b> leg. There's also the <b><u>hypotenuse</u></b>, which is always the longest side of the right triangle, sitting across from the right angle.</p><p>An interesting thing to notice is that as soon as we know the measure of the <b style="color: #8e44ad;">angle</b> \(\color{flatpurple}\theta\), we automatically know that our triangle is similar to <b>all other right triangles</b> with an <b style="color: #8e44ad;">angle</b> \(\color{flatpurple}\theta\). In some sense, the ratios are "locked in."</p><div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-xcNrf3e2IRw/YCyLscHd0OI/AAAAAAAAeTE/SyVtLorInNgyNPHaZTWIK794_lz6QMqVACLcBGAsYHQ/s1608/rt2.png" style="display: block; padding: 1em 0px; text-align: center;"><img alt="" border="0" data-original-height="1278" data-original-width="1608" src="https://1.bp.blogspot.com/-xcNrf3e2IRw/YCyLscHd0OI/AAAAAAAAeTE/SyVtLorInNgyNPHaZTWIK794_lz6QMqVACLcBGAsYHQ/s1608/rt2.png" width="70%" /></a></div><p>To make things simple, we could imagine <b>rescaling</b> our triangle so that the <b>hypotenuse</b> is \(1\) — which means that our triangle would fit nicely in the first quadrant of the unit circle (though we might have to flip or rotate it to get there).</p><p>Once we do that, the <b style="color: #2980b9;">opposite</b> side would line up nicely with the <b style="color: #2980b9;">sine</b>, and the <b style="color: #c0392b;">adjacent</b> side would line up nicely with the <b style="color: #c0392b;">cosine</b>. But in order to get to this triangle, we had to divide through by the length of the <b>hypotenuse</b>.</p><div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-xWj2Mjjxeg0/YCyLsmARYLI/AAAAAAAAeTI/EBvBplgs8jQhi6EfOTaVtmgxMa9bfO4-gCLcBGAsYHQ/s1515/rt3.png" style="display: block; padding: 1em 0px; text-align: center;"><img alt="" border="0" data-original-height="1474" data-original-width="1515" src="https://1.bp.blogspot.com/-xWj2Mjjxeg0/YCyLsmARYLI/AAAAAAAAeTI/EBvBplgs8jQhi6EfOTaVtmgxMa9bfO4-gCLcBGAsYHQ/s1515/rt3.png" width="90%" /></a></div><p>That leads us to two useful ratio definitions: \[ {\color{flatblue}\sin(\theta)}=\dfrac{\color{flatblue}\text{opposite}}{\text{hypotenuse}} \qquad {\color{flatred}\cos(\theta)}=\dfrac{\color{flatred}\text{adjacent}}{\text{hypotenuse}} \] Dividing the two ratios, we end up with one more: \[ {\color{flatgreen}\tan(\theta)}=\dfrac{\color{flatblue}\sin(\theta)}{\color{flatred}\cos(\theta)}=\dfrac{{\color{flatblue}\text{opp}}/\text{hyp}}{{\color{flatred}\text{adj}}/\text{hyp}}=\dfrac{\color{flatblue}\text{opp}}{\text{hyp}}\cdot\dfrac{\text{hyp}}{\color{flatred}\text{adj}}=\dfrac{\color{flatblue}\text{opposite}}{\color{flatred}\text{adjacent}} \] </p><p>You might have learned the mnemonic <b>"<span style="color: #2980b9;">SO</span>H-<span style="color: #c0392b;">CA</span>H-<span style="color: #27ae60;">T</span><span style="color: #2980b9;">O</span><span style="color: #c0392b;">A</span>"</b> as a way to remember these relationships:</p><ul> <li><span style="color: #2980b9;"><b>S</b>ine</span> is <span style="color: #2980b9;"><b>O</b>pposite</span> over <b>H</b>ypotenuse</li> <li><span style="color: #c0392b;"><b>C</b>osine</span> is <span style="color: #c0392b;"><b>A</b>djacent</span> over <b>H</b>ypotenuse</li> <li><span style="color: #27ae60;"><b>T</b>angent</span> is <span style="color: #2980b9;"><b>O</b>pposite</span> over <span style="color: #c0392b;"><b>A</b>djacent</span></li></ul><p>This can help if you need to quickly recall which is which, but now you can also remember them by imagining you've moved and scaled your triangle so the <b style="color: #8e44ad;">angle</b> in question is in standard position on the unit circle!</p><h2>Solving triangles</h2><p>Let's look at a "real-world" application. <a href="#asterisk">*</a></p><p>Suppose you're looking in front of a really tall building. You have no way of measuring the actual height of the building directly, but you know that when you stand \(300\) feet away from it (about one city block), you have to look up at an <b style="color: #8e44ad;">angle of elevation</b> of \(62^\circ\) to see the top. (You might measure this with an <a href="https://en.wikipedia.org/wiki/Inclinometer">inclinometer</a>.)</p><div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-dkkkWF9jIiQ/YCyuWDnQZVI/AAAAAAAAeUo/I6hLrRVdy2U2R4LZTdD_mUWKouF8Ir-dACLcBGAsYHQ/s1325/building.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="35%" data-original-height="1325" data-original-width="934" src="https://1.bp.blogspot.com/-dkkkWF9jIiQ/YCyuWDnQZVI/AAAAAAAAeUo/I6hLrRVdy2U2R4LZTdD_mUWKouF8Ir-dACLcBGAsYHQ/s1325/building.png"/></a></div><p>How could you figure out the <b style="color: #2980b9;">height</b> of the building?</p><p>If you draw a triangle to model the situation, you can notice that the \(300\)-foot leg is <b style="color: #c0392b;">adjacent</b> to your <b style="color: #8e44ad;">angle of elevation</b>, and you're looking for the length of the <b style="color: #2980b9;">opposite</b> leg. These two lengths are related by the <b style="color: #27ae60;">tangent</b> ratio, so you can set up an equation: \[ \begin{align*} {\color{flatgreen}\tan(62^\circ)}&=\dfrac{\color{flatblue}x}{\color{flatred}300}\\ 300\cdot\tan(62^\circ)&=x\\ x&\approx 564.21793 \end{align*} \] Thus the <b style="color: #2980b9;">height</b> of the building is about <b style="color: #2980b9;">\(564\) feet</b>.</p><p>Note that you'd need to use a calculator to find \(\color{flatgreen}\tan(62^\circ)\), since that's not one of our "nice" angles, but let's be real — you can always pull out your cell phone for that. Just make sure your calculator is expecting angles in degrees ... after all, \(62\) radians is a VERY different angle!</p><h2>A word of caution</h2><p>It's important to remember that terms like "<b style="color: #2980b9;">opposite</b>" and "<b style="color: #c0392b;">adjacent</b>" are relative to what <b style="color: #8e44ad;">angle</b> you're using as your reference point. For example, look at the following triangle:</p><div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-OsZWSgqpr4E/YCyLsoqqfAI/AAAAAAAAeTM/JvQFTViryocMusYTx4k9oylEbtgtyueCgCLcBGAsYHQ/s1025/rt4.png" style="display: block; padding: 1em 0px; text-align: center;"><img alt="" border="0" data-original-height="783" data-original-width="1025" src="https://1.bp.blogspot.com/-OsZWSgqpr4E/YCyLsoqqfAI/AAAAAAAAeTM/JvQFTViryocMusYTx4k9oylEbtgtyueCgCLcBGAsYHQ/s1025/rt4.png" width="40%" /></a></div><p>If you're standing at the <b style="color: #8e44ad;">angle</b> marked \(\color{flatpurple}\alpha\), then \(x\) is the <b style="color: #2980b9;">opposite</b> leg and \(y\) is the <b style="color: #c0392b;">adjacent</b> leg. But if your friend is standing at the <b style="color: #f368e0;">angle</b> marked \(\color{flatpink}\beta\) instead, then \(x\) is now the <b style="color: #c0392b;">adjacent</b> leg to them while \(y\) is now the <b style="color: #2980b9;">opposite</b> leg from them.</p><p>In other words, <b>perspective matters</b>.</p><p>But the nice thing is, if you both perform calculations using your respective angles, you'll end up getting the same answers as long as you're consistent!</p> <p><a id="asterisk"></a>* <i>To be honest, I'm not really a fan of the phrase "real-world" in relation to math. It sends the message that thinking through puzzles and problem-solving is "fake." Don't get me wrong, it's really cool that mathematics can model so many things about the world around us! But even if you're just playing with numbers and pictures and finding patterns, that's just as "real" because <b>you</b> are the one doing it.</i></p> <hr /><a id="previewactivity"></a><h1 style="text-align: left;">Preview Activity 4</h1><p> <i>Answer these questions and submit your answers as a document <b>on Moodle</b>. (Please submit as .docx or .pdf if possible.)</i></p><ol> <li style="padding-bottom: 1.45em; text-align: left;"> You'll need to use a calculator to answer these questions. (Make sure it's in degree mode!) <ol style="list-style-type: lower-alpha;"> <li>What is \(\sin 20^\circ\)?</li> <li>What is \(\cos 70^\circ\)?</li> <li>What is \(\cos 10^\circ\)?</li> <li>What is \(\sin 80^\circ\)?</li> <li>What relationship are you seeing in the previous exercises? Make a conjecture (a guess as to what you think is going on), and try it out on a new set of numbers to see if it does in fact work.</li> </ol> </li> <li style="padding-bottom: 1.45em; text-align: left;"> Consider the triangle below.<br> <div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-ky0vQVFU5UU/YCyR4JwbfPI/AAAAAAAAeUM/vyLwpbNF7tQTLJoK0jugwxeWJuLQvaX6wCLcBGAsYHQ/s962/rt345.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="40%" data-original-height="962" data-original-width="761" src="https://1.bp.blogspot.com/-ky0vQVFU5UU/YCyR4JwbfPI/AAAAAAAAeUM/vyLwpbNF7tQTLJoK0jugwxeWJuLQvaX6wCLcBGAsYHQ/s962/rt345.png"/></a></div> <ol style="list-style-type: lower-alpha;"> <li>Calculate \(\sin({\color{flatpurple}\alpha})\), \(\cos({\color{flatpurple}\alpha})\), and \(\tan({\color{flatpurple}\alpha})\).</li> <li>Calculate \(\sin({\color{flatpink}\beta})\), \(\cos({\color{flatpink}\beta})\), and \(\tan({\color{flatpink}\beta})\).</li> <li>How do the results of the previous two questions compare?</li> </ol> </li> <li> Answer <b>AT LEAST</b> one of the following questions: <ol style="list-style-type: lower-alpha;"> <li>What was something you found interesting about this reading?</li> <li>What was an "a-ha" moment you had while doing this reading?</li> <li>What was the muddiest point of this reading for you?</li> <li>What question(s) do you have about anything you've read?</li> </ol> </li></ol><p><a href="https://drive.google.com/file/d/1v4TqfkoG9fziftUoWlawxCgqp1dM07an/view?usp=sharing" target="_blank"><b>Answers</b></a></p><hr /><p> <a id="previewactivity"></a><a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-91210787973042848332021-02-13T14:56:00.055-08:002021-02-19T18:12:44.369-08:00Fundamental Identities<!--For permalink, put "spr21-mat130-" and then lesson name with hyphens.--><p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p><p><a href="#previewactivity">Go to Preview Activity</a></p><hr /><p> Textbook Reference: <a href="https://openstax.org/books/precalculus/pages/5-2-unit-circle-sine-and-cosine-functions" target="_blank">Section 5.3 <b>The Other Trigonometric Functions</b></a> </p><hr /><script type="text/x-mathjax-config"> MathJax.Hub.Config({ TeX: { extensions: ["color.js"] }}); </script> <p> So far you've been introduced to the two most famous circular functions, the <b style="color: #2980b9;">sine</b> and <b style="color: #c0392b;">cosine</b>, which respectively represent the <b style="color: #2980b9;">height</b> and <b style="color: #c0392b;">overness</b> of an angle projected onto the unit circle. These two functions are so useful in math as a whole that, if we really wanted, we could do everything for the semester in terms of these two functions. However, there are four useful ratios we can build out of these two functions that are so commonly encountered that we'll give them their own names: the <b><u>tangent</u></b>, the <b><u>cotangent</u></b>, the <b><u>secant</u></b>, and the <b><u>cosecant</u></b>. </p><h2>The tangent function</h2><p> By far the most useful of these new functions is the <b style="color: #27ae60;"><u>tangent</u></b> function, which is written \(\color{flatgreen}\tan(\theta)\). The most useful intuition to motivate the <b style="color: #27ae60;">tangent</b> is this: </p><p style="text-align: center;"> <b><span style="font-size: large;">The <span style="color: #27ae60;">tangent</span> of an angle<br />is the <span style="color: #27ae60;">slope</span> of the terminal side.</span></b></p><p style="text-align:center;"><iframe src="https://www.desmos.com/calculator/wbpkcgstlu?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe> <br><i>Drag the point to change the angle.</i></p><p> Yes, this is the exact same "rise-over-run" <b style="color: #27ae60;">slope</b> you've encountered in your previous algebra classes! Remeber that the <b style="color: #27ae60;">slope</b> of a line is the <b>ratio</b> between the <b style="color: #2980b9;">vertical change</b> (the "<b style="color: #2980b9;">rise</b>") and the <b style="color: #c0392b;">horizontal change</b> (the "<b style="color: #c0392b;">run</b>"): \[{\color{flatgreen}\text{slope}}=\dfrac{\color{flatblue}\Delta y}{\color{flatred}\Delta x}=\dfrac{\color{flatblue}y_2-y_1}{\color{flatred}x_1-x_1}\] Thinking about this in terms of our unit circle, you might realize that this is just the <b style="color: #2980b9;">height</b> divided by the <b style="color: #c0392b;">overness</b>! This gives us a really nice way to calculate the <b style="color: #27ae60;">tangent</b> of any* angle: \[{\color{flatgreen}\tan(\theta)}=\dfrac{\color{flatblue}\sin(\theta)}{\color{flatred}\cos(\theta)}\] </p><p><i>*...well, almost any angle. You'll see what I mean later.</i></p><h2>The reciprocal functions</h2><p> The <b style="color: #2980b9;">sine</b>, <b style="color: #c0392b;">cosine</b>, and <b style="color: #27ae60;">tangent</b> functions are the most well-known circular functions. However, in certain applications, it's sometimes easier to work with their reciprocals, which we'll define as follows: </p><ul> <li>The <b style="color: #3498db;"><u>cosecant</u></b> function is the reciprocal of the <b style="color: #2980b9;">sine</b> function. \[{\color{flatlightblue}\csc(\theta)}=\dfrac1{\color{flatblue}\sin(\theta)}\] </li> <li>The <b style="color: #e74c3c;"><u>secant</u></b> function is the reciprocal of the <b style="color: #c0392b;">cosine</b> function. \[{\color{flatlightred}\sec(\theta)}=\dfrac1{\color{flatred}\cos(\theta)}\] </li> <li>The <b style="color: #2ecc71;"><u>cotangent</u></b> function is the reciprocal of the <b style="color: #27ae60;">tangent</b> function. \[{\color{flatlightgreen}\cot(\theta)}=\dfrac1{\color{flatgreen}\tan(\theta)}\] </li></ul><blockquote> <h3>CAUTION!</h3> <p>Be careful not to get these mixed up! It's a common mistake to think that \({\color{flatlightred}\sec(\theta)}=\dfrac1{\color{flatblue}\sin(\theta)}\) and \({\color{flatlightblue}\csc(\theta)}=\dfrac1{\color{flatred}\cos(\theta)}\). A mnemonic that may help is to look at the third letter of each reciprocal function's name:</p> <ul> <li><b>SE<span style="font-size: x-large; color: #e74c3c;">C</span>ANT</b> is the reciprocal of <b><span style="font-size: x-large; color: #c0392b;">C</span>OSINE</b>.</li> <li><b>CO<span style="font-size: x-large; color: #3498db;">S</span>ECANT</b> is the reciprocal of <b><span style="font-size: x-large; color: #2980b9;">S</span>INE</b>.</li> <li><b>CO<span style="font-size: x-large; color: #2ecc71;">T</span>ANGENT</b> is the reciprocal of <b><span style="font-size: x-large; color: #27ae60;">T</span>ANGENT</b>.</li> </ul> <p><i>(By the way, if you want to know where the names "<b style="color: #27ae60;">tangent</b>" and "<b style="color: #e74c3c;">secant</b>" <b>actually</b> come from, you might want to look at Extension Problem 2!)</i></p></blockquote><h2>Example</h2><p>Suppose \(\color{flatpurple}\phi=\dfrac\pi6\). We can see from the unit circle that \(\color{flatblue}\sin(\phi)=\dfrac12\) and \(\color{flatred}\cos(\phi)=\dfrac{\sqrt{3}}2\).</p><ul> <li>The tangent can be found by dividing the two: \[{\color{flatgreen}\tan(\phi)}=\dfrac{\color{flatblue}\sin(\phi)}{\color{flatred}\cos(\phi)}=\dfrac{\color{flatblue}1/2}{\color{flatred}\sqrt{3}/2}=\dfrac12\cdot\dfrac2{\sqrt{3}}=\color{flatgreen}\dfrac1{\sqrt{3}}\]</li> <li>The other functions are just reciprocals of our other three: \[ \begin{align*} {\color{flatlightblue}\csc(\phi)}&=\dfrac{1}{\color{flatblue}\sin(\phi)}=\dfrac{1}{\color{flatblue}1/2}={\color{flatlightblue}2}\\ {\color{flatlightred}\sec(\phi)}&=\dfrac{1}{\color{flatred}\cos(\phi)}=\dfrac{1}{\color{flatred}\sqrt{3}/2}={\color{flatlightred}\dfrac2{\sqrt{3}}}\\ {\color{flatlightgreen}\cot(\phi)}&=\dfrac{1}{\color{flatgreen}\tan(\phi)}=\dfrac{1}{\color{flatgreen}1/\sqrt{3}}=\color{flatlightgreen}\sqrt{3}\\ \end{align*} \] </ul><h2>Identities</h2><p>Once you get a feel for the circular functions, you start noticing a whole bunch of relationships between them — like the same numbers getting passed around, or certain ways that they complement each other. The equation \[{\color{flatgreen}\tan(\theta)}=\dfrac{\color{flatblue}\sin(\theta)}{\color{flatred}\cos(\theta)}\] is an example of such a relationship — given the values of any two of these basic functions, you can find the third.</p><p>This is an example of what's called an <b><u>identity</u></b> in mathematics: an equation that's always true for any valid numbers that get plugged into it. Pick any value of \(\theta\) you want, and as long as \(\color{flatblue}\sin(\theta)\), \(\color{flatred}\cos(\theta)\), and \(\color{flatgreen}\tan(\theta)\) are all defined, you can be sure that the above equation will hold true.</p><p>We'll see many examples of identities as you learn more about the circular functions. It's important not to think of them as just a bunch of things to memorize — after all, in the real world, you can always just <a href="https://www.google.com/search?q=trig+identities">look them up</a> if you need them. What's important is <b>understanding</b> them and seeing where they come from.</p><h2>A caveat</h2><p>Remember how I said we can find the tangent of "almost" any angle, and put a note saying "<i>*...well, almost any angle. You'll see what I mean later.</i>"?</p><p>It's later.</p><p>What happens if we try to find the <b style="color: #27ae60;">tangent</b> of a <b>right angle</b> (\(\pi/2\) radians)?</p><p>Look what happens if we try to use our formula from earlier: \[ {\color{flatgreen}\tan(\pi/2)}=\dfrac{\color{flatblue}\sin(\pi/2)}{\color{flatred}\cos(\pi/2)}=\dfrac{\color{flatblue}1}{\color{flatred}0} \] </p><p>Uh-oh. We just divided by zero.</p><p>That's a bad thing, right? That's why most people say that the <b style="color: #27ae60;">tangent</b> of a right angle is "<b>undefined</b>." They also say that the <b style="color: #27ae60;">slope</b> of a vertical line is undefined for the same reason, so that makes sense, since we just said earlier that the <b style="color: #27ae60;">tangent</b> and the <b style="color: #27ae60;">slope</b> are related. From a function point of view, we would say that \(\pi/2\) isn't in the <b>domain</b> of the tangent function — it's an invalid input.</p><p>So when you're calculating these functions, if you run into an angle that makes you divide by zero, you know you've found an angle for which that function is undefined. And you just have to accept that sometimes these functions just don't work.</p><p><i>But I'm going to let you in on a little secret.</i></p><p>If you want, instead of thinking of \(1/0\) as "undefined", you can actually think about it as "infinity", if you're <b>really</b> careful about what that means. Going into how to do that is a bit of a rabbit hole, and you can regard it as optional, but if you want to know how to break the rules and <b>actually divide by zero</b>, you might want to <a href="https://www.1dividedby0.com" target="_blank">look into it sometime</a>. 😉</p><hr /><a id="previewactivity"></a><h1 style="text-align: left;">Preview Activity 3</h1><p> <i>Answer these questions and submit your answers as a document <b>on Moodle</b>. (Please submit as .docx or .pdf if possible.)</i></p><ol> <li style="padding-bottom: 1.45em; text-align: left;">Find the value of all six circular functions of \(\theta=\dfrac\pi4\).</li> <li style="padding-bottom: 1.45em; text-align: left;">If \({\color{flatlightblue}\csc(\theta)}=\dfrac1{\color{flatblue}\sin(\theta)}\), what do you think \(\dfrac1{\color{flatlightblue}\csc(\theta)}\) should be, and why?<br> What about \(\dfrac1{\color{flatlightred}\sec(\theta)}\) and \(\dfrac1{\color{flatlightgreen}\cot(\theta)}\)?</li> <li style="padding-bottom: 1.45em; text-align: left;">We saw earlier that \({\color{flatgreen}\tan(\theta)}=\dfrac{\color{flatblue}\sin(\theta)}{\color{flatred}\cos(\theta)}\). What do you think \(\dfrac{\color{flatred}\cos(\theta)}{\color{flatblue}\sin(\theta)}\) should be, and why?</li> <li style="padding-bottom: 1.45em; text-align: left;">What do you think should be the value of \(\color{flatlightgreen}\cot(\pi/2)\)? Why?</li> <li> Answer <b>AT LEAST</b> one of the following questions: <ol style="list-style-type: lower-alpha;"> <li>What was something you found interesting about this reading?</li> <li>What was an "a-ha" moment you had while doing this reading?</li> <li>What was the muddiest point of this reading for you?</li> <li>What question(s) do you have about anything you've read?</li> </ol> </li></ol><p><a href="https://drive.google.com/file/d/14OrifZdksqA4TvOIY7Uo6rHRhJNLs6IH/view?usp=sharing" target="_blank"><b>Answers</b></a></p> <hr /><p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-41274129737744083312021-02-12T21:11:00.103-08:002021-02-22T23:03:19.019-08:00The Circular Functions<!-- For permalink, put "spr21-mat130-" and then lesson name with hyphens. --><p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html" >Return to MAT 130 main page</a ></p><p><a href="#previewactivity">Go to Preview Activity</a></p><hr /><p> Textbook Reference: <a href="https://openstax.org/books/precalculus/pages/5-2-unit-circle-sine-and-cosine-functions" target="_blank" >Section 5.2 <b>Unit Circle: Sine and Cosine Functions</b></a ></p><br /><p> <i >I first saw this brilliant way of introducing the circular functions because of <a href="https://twitter.com/jamestanton">James Tanton</a>, the "Mathematician-at-Large" for the Mathematical Association of America, so a big thanks goes to him. You're welcome to watch <a href="https://www.youtube.com/watch?v=Lr9Epp5eKUQ" >his video on YouTube</a > to see his own presentation of it.</i ></p><hr /><script type="text/x-mathjax-config"> MathJax.Hub.Config({ TeX: { extensions: ["color.js"] }}); </script> <p> Imagine you're in 5th century India. Back then you wouldn't have had a cell phone and Facebook, so you might have had a lot of time to just sit around and ponder the wonders of the universe we live in — wonders such as the sun. You might find yourself wondering: </p><p style="text-align: center;"> <i ><b >Just what is that big beautiful ball in the sky we call the "sun"?</b ></i ></p><p style="text-align: center;"> <i><b>How far away is it?</b></i></p><p style="text-align: center;"> <i><b>How high up is it?</b></i></p><div class="separator" style="clear: both;"> <a href="https://1.bp.blogspot.com/-idiekdJRFKs/YCd2n8n8NhI/AAAAAAAAeQs/PiO3YZgVUrcjejA1Ln5C89Ail2a-v-0PACLcBGAsYHQ/s1628/sun1.png" style="display: block; padding: 1em 0px; text-align: center;" ><img alt="" border="0" data-original-height="1099" data-original-width="1628" src="https://1.bp.blogspot.com/-idiekdJRFKs/YCd2n8n8NhI/AAAAAAAAeQs/PiO3YZgVUrcjejA1Ln5C89Ail2a-v-0PACLcBGAsYHQ/s1628/sun1.png" width="100%" /></a></div><p> It's worth noting that a lot of the sort of math we're learning this semester was motivated by astronomy. People from many different cultures kept track of the motion of the sun, moon, stars, and planets, for both practical and philosophical reasons. We're concentrating on India in particular because the math they figured out is what leads to the idea of circular functions as we know them today. </p><p> Anyway, let's get back to figuring out where the sun is. Assuming the sun goes in a (roughly) circular path, it would always be the same <b>radius</b> away from the center of the earth. If you could figure out that radius, then you could do a little geometry to figure out how high the sun is based on your angle. To make things simple, you decide that the distance from the earth to the sun should always be \(1\). That way, if years down the line scientists were ever able to actually figure out that distance, all it takes is a quick multiplication and all your calculations will be accurate! (In fact, scientists today do call this distance \(1\) AU, which is short for <u style="font-weight: bold;">astronomical unit</u>.) </p><p> Likewise, to figure out how high up the sun is, you can't just go up there and drop a really long rope down as an altitude. The best you can do from your safe vantage point on the earth is measure the <b style="color: #8e44ad;">angle of elevation</b> as you look up at it (not too directly, of course). So, at this point you want to find out a relationship between the <b style="color: #8e44ad;">angle</b> and the <b style="color: #2980b9;">height</b>, watching how those quantities change as the sun rises in the east and sets in the west. While you're at it, you might as well also try to figure out how far left or right the sun is — its "<b style="color: #c0392b;" >overness</b >", so to speak. </p><div class="separator" style="clear: both;"> <a href="https://1.bp.blogspot.com/-H52ktnMudcY/YCd2n3SRKqI/AAAAAAAAeQw/GR4oTCKvckU5qWJlSj43XXhCLE-ILKgOQCLcBGAsYHQ/s1809/sun2.png" style="display: block; padding: 1em 0px; text-align: center;" ><img alt="" border="0" data-original-height="1014" data-original-width="1809" src="https://1.bp.blogspot.com/-H52ktnMudcY/YCd2n3SRKqI/AAAAAAAAeQw/GR4oTCKvckU5qWJlSj43XXhCLE-ILKgOQCLcBGAsYHQ/s1809/sun2.png" width="100%" /></a></div><p> And that's exactly what they did in India — they developed a whole bunch of mathematics to figure out the <b style="color: #2980b9;">height</b> and <b style="color: #c0392b;">overness</b> of various angles, writing all sorts of manuscripts in Sanskrit and pushing the boundaries of mathematical knowledge at the time. </p><h2>Mistranslations</h2><p> From here, the story gets pretty interesting as different cultures passed around this knowledge. </p><p> In the 9th and 10th centuries, Persian scholars came along these Sanskrit manuscripts with all sorts of interesting techniques for analyzing circles, and they decided to translate the manuscripts into Arabic. The Sanskrit word used for the <b style="color: #2980b9;">height</b> was "<b style="color: #2980b9;" >jya</b >" <i >(which comes from the word for a bow-and-arrow, which you might to see if you extend the height downward to the other side of the circle)</i > or "<b style="color: #2980b9;">jiva</b>". The Persian scholars liked the second word, so they didn't try to translate it, instead deciding to just transcribe it as closely as possible from the Sanskrit (much like how when we talk about the famous Japanese food "寿司", we don't translate it as "vinegared rice" but just call it "sushi.") This gave them the word "<b style="color: #2980b9;" >jiba</b >" — except that in Semitic languages like Arabic and Hebrew, writing vowels is optional, so instead of "<b style="color: #2980b9;">jiba</b>" they really just wrote "<b style="color: #2980b9;">jb</b>". </p><p> Now fast-forward to the 13th century, when European scholars came across these Arabic manuscripts and started to translate them into Latin. They got really confused when they saw "<b style="color: #2980b9;">jb</b>", because the only Arabic word they knew with those consonants was "<b style="color: #2980b9;" >jaib</b >," which means a cavity or a bay — kind of a strange word to use for the height of the sun. Nevertheless, they went with it, and translated it to the Latin word for a cavity or bay: "<b style="color: #2980b9;">sinus</b>." Many countries today still call this height the <b style="color: #2980b9;">sinus</b>, but in English it got modified just a bit more to become what we know today as the <b style="color: #2980b9;"><u>sine</u></b> of an angle. </p><p> In other words, our word <b style="color: #2980b9;">sine</b> has been translated from Sanskrit, to Arabic, to bad Arabic, to Latin, and finally to English. Yikes. </p><h2>Height and Overness</h2><p style="text-align: center;"> <i>"Wait, so what's the point of this history lesson?"</i></p><p> Because now we know what the <b style="color: #2980b9;">sine</b> is supposed to mean: </p><p style="text-align: center;"> <b ><span style="font-size: large;" >The <span style="color: #2980b9;"><u>sine</u></span> of an <span style="color: #8e44ad;">angle</span><br />is its <span style="color: #2980b9;">height</span> on the unit circle.</span ></b ></p><p> And you can remember the mental image of finding the <b style="color: #2980b9;">height</b> of the sun. </p><p>Now, what about the <b style="color: #c0392b;">overness</b>?</p><p> Well, the <b style="color: #c0392b;">overness</b> of a particular angle can be drawn with a very similar picture to the <b style="color: #2980b9;">height</b>, but instead using the <b style="color: #16a085;">complementary angle</b>: </p><div class="separator" style="clear: both;"> <a href="https://1.bp.blogspot.com/-rRJiXkx6BTY/YCd2n3QQzbI/AAAAAAAAeQo/fXlL0aKlKmompl0W5YR7qPExAe-vrBrtQCLcBGAsYHQ/s1774/sun3.png" style="display: block; padding: 1em 0px; text-align: center;" ><img alt="" border="0" data-original-height="911" data-original-width="1774" src="https://1.bp.blogspot.com/-rRJiXkx6BTY/YCd2n3QQzbI/AAAAAAAAeQo/fXlL0aKlKmompl0W5YR7qPExAe-vrBrtQCLcBGAsYHQ/s1774/sun3.png" width="100%" /></a></div><p> In Latin this was written as "<b style="color: #16a085;">complementi</b> <b style="color: #2980b9;">sinus</b>" (meaning "<b style="color: #2980b9;" >sine</b > of the <b style="color: #16a085;">complement</b>"), and was eventually abbreviated to "<b style="color: #16a085;">co.</b ><b style="color: #2980b9;">sinus</b>." </p><p> As you might guess, nowadays we call it the "<b style="color: #c0392b;" ><u>cosine</u></b >". </p><p>So now we have another useful intuition:</p><p style="text-align: center;"> <b ><span style="font-size: large;" >The <span style="color: #c0392b;"><u>cosine</u></span> of an <span style="color: #8e44ad;">angle</span><br />is its <span style="color: #c0392b;">overness</span> on the unit circle.</span ></b ></p><p> And you can remember the mental image of finding the <b style="color: #c0392b;">overness</b> of the sun. </p><p> What we've just introduced here are the primary object of study this semester, the <b><u>circular functions</u></b >, the <b style="color: #2980b9;">sine</b> (written \(\color{flatblue}\sin(\theta)\)) and the <b style="color: #c0392b;">cosine</b> (written \(\color{flatred}\cos(\theta)\)). Essentially, if you input any <b style="color: #8e44ad;">angle</b>, they will tell you the <b style="color: #2980b9;">height</b> and <b style="color: #c0392b;">overness</b> respectively. </p><h2>Example</h2><p> Let's look what happens when the sun first rises in the east (at \(0^\circ\) or \(0\) radians). </p><div class="separator" style="clear: both;"> <a href="https://1.bp.blogspot.com/-AHXFa2dRtPY/YCd4EzeGf2I/AAAAAAAAeRU/KZRsOa7lV6866gl9qGk0HjHqJqp-dh7pQCLcBGAsYHQ/s1958/sun4.png" style="display: block; padding: 1em 0px; text-align: center;" ><img alt="" border="0" data-original-height="996" data-original-width="1958" src="https://1.bp.blogspot.com/-AHXFa2dRtPY/YCd4EzeGf2I/AAAAAAAAeRU/KZRsOa7lV6866gl9qGk0HjHqJqp-dh7pQCLcBGAsYHQ/s1958/sun4.png" width="100%" /></a></div><p> At this point, the sun is all the way over to the right, so the <b style="color: #c0392b;">overness</b> is <span style="color: #c0392b;">\(1\)</span>. It hasn't risen at all in the sky, so its <b style="color: #2980b9;">height</b> is <span style="color: #2980b9;">\(0\)</span>. In modern mathematical notation, we would say that \(\color{flatred}\cos(0^\circ)=1\) and \(\color{flatblue}\sin(0^\circ)=0\), and so the \(({\color{flatred}x},{\color{flatblue}y})\) coordinates of the point corresponding to \(0^\circ\) are \(({\color{flatred}1},{\color{flatblue}0})\). </p><h2>Conclusion</h2><p>Here's the big take-away:</p><p style="text-align: center;"> <b ><span style="font-size: large;" >For any <span style="color: #8e44ad;">angle</span> \(\color{flatpurple}\theta\),<br /> the corresponding point<br /> on the unit circle is<br /> \(({\color{flatred}\cos(\theta)},{\color{flatblue}\sin(\theta)})\).</span ></b ></p><p style="text-align: center;"> <iframe frameborder="0" height="500px" src="https://www.desmos.com/calculator/a9yv5f26za?embed" style="border: 1px solid #ccc;" width="500px" ></iframe ><br /> <i >Drag the sun around to change the diagram.<br />Click on the sun to turn its coordinates on or off.</i ></p><p> In class, you'll be working to figure out the <b style="color: #2980b9;">sine</b> and <b style="color: #c0392b;">cosine</b> of various nice <b style="color: #8e44ad;">angles</b>, based on the properties of particularly "nice" right triangles. </p> <hr /><a id="previewactivity"></a><h1 style="text-align: left;">Preview Activity 2</h1><p> <i >Answer these questions and submit your answers as a document <b>on Moodle</b>. (Please submit as .docx or .pdf if possible.)</i ></p><ol> <li style="padding-bottom: 1.45em; text-align: left;"> What are the <b style="color: #2980b9;">height</b> and <b style="color: #c0392b;">overness</b> of the sun when it's directly overhead at noon?<br />How would you write this in modern mathematical notation?<br /> </li> <li style="padding-bottom: 1.45em; text-align: left;"> What are the <b style="color: #2980b9;">height</b> and <b style="color: #c0392b;">overness</b> of the sun when it sets in the west?<br />How would you write this in modern mathematical notation?<br /> </li> <li style="padding-bottom: 1.45em; text-align: left;"> What are the <b style="color: #2980b9;">height</b> and <b style="color: #c0392b;">overness</b> of the sun at midnight (assuming it's directly under where you are on the earth)?<br />How would you write this in modern mathematical notation?<br /> </li> <li style="padding-bottom: 1.45em; text-align: left;"> What are the <b style="color: #2980b9;">height</b> and <b style="color: #c0392b;">overness</b> of the sun when it makes a \(\color{flatpurple}32^\circ\) <b style="color: #8e44ad;">angle of elevation</b>?<br />How would you write this in modern mathematical notation?<br /><i >(You'll need to use a calculator, or you can use the interactive diagram above.)</i > </li> <li style="padding-bottom: 1.45em; text-align: left;"> At how many times during the day is the <b style="color: #2980b9;">height</b> of the sun equal to \(\color{flatblue}\dfrac12\)? What are the <b style="color: #8e44ad;">angles</b> at which this occurs?<br /> </li> <li> Answer <b>AT LEAST</b> one of the following questions: <ol style="list-style-type: lower-alpha;"> <li>What was something you found interesting about this reading?</li> <li>What was an "a-ha" moment you had while doing this reading?</li> <li>What was the muddiest point of this reading for you?</li> <li>What question(s) do you have about anything you've read?</li> </ol> </li></ol><p><a href="https://drive.google.com/file/d/1e8aJxMRiratrjjeAhSix8cuxRbZTZQWE/view?usp=sharing" target="_blank"><b>Answers</b></a></p><hr /><p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-80415046492566473152021-02-12T20:44:00.015-08:002021-02-22T23:03:27.804-08:00Clock angles solution<p><a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p><hr /><p>This is an example of what I would call a thorough written solution to the clock angle problem today.</p><blockquote> <p><b>Problem:</b><br>When an analog clock reads 5:45, what is the radian measure of the angle between the hour hand and the minute hand?</p> <p><b>Solution:</b></p> <blockquote> <p>At 5:45, the minute hand will be exactly at the \(9\). In addition, since \(\dfrac34\) of an hour has passed, the hour hand will have moved \(\dfrac34\) of the way from the \(5\) to the \(6\), as in the diagram below.</p> <div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-_8hMBiaxLvI/YCdYZexubGI/AAAAAAAAePI/dHEcTFxtI-AvYm4QpVNfBcnT8hDwRfzwACLcBGAsYHQ/s1119/clocksoln.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="75%" data-original-height="1084" data-original-width="1119" src="https://1.bp.blogspot.com/-_8hMBiaxLvI/YCdYZexubGI/AAAAAAAAePI/dHEcTFxtI-AvYm4QpVNfBcnT8hDwRfzwACLcBGAsYHQ/s1119/clocksoln.png"/></a></div> <p>Now, the angle between the \(6\) and the \(9\) (in blue) is \(\dfrac14\) of a circle, and the angle between the \(6\) and the hour hand (in purple) is \(\dfrac14\) of \(\dfrac1{12}\) of a circle, which equals \(\dfrac1{48}\). If we add these fractions, we find that the total shaded angle is \[\dfrac14+\dfrac1{48}=\dfrac{12}{48}+\dfrac1{48}=\dfrac{13}{48}.\]Since a full circle is \(2\pi\) radians, we can multiply this fraction by \(2\pi\), thus giving us an angle of \(\dfrac{13}{48}(2\pi)=\dfrac{13\pi}{24}\) radians. \(\blacksquare\)<br /> </blockquote></blockquote><p>Notice that <b>at no point did we need to use degrees</b>! In order to "speak radians fluently," you need to remember the key point from the previous reading:</p><p style="text-align:center;"><b>Radians are angles measured as fractions of \(2\pi\).</b></p><p>Remembering this key piece of intuition is what will help you become "fluent" in radians as we go throughout the semester.</p><hr /><p><a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-17245806361720625372021-02-10T09:09:00.022-08:002021-02-22T23:04:19.028-08:00The Unit Circle (wrap-up)<p>After each class, I'm going to write up brief summaries and big take-aways from our Class Activities.</p> <h2 style="text-align: left;">Converting between degrees and radians</h2><p><b>How many radians are in \(36^\circ\)?</b></p><blockquote> <p>There are a number of different ways you could go about this.</p> <ul> <li>You could notice that \(36^\circ\) is \(\tfrac{1}{10}\) of a circle, and a full circle is \(2\pi\) radians, so we have \(\tfrac{1}{10}(2\pi)=\tfrac{\pi}{5}\).</li> <li>You could set up a proportion, remembering that \(180^\circ\) is equivalent to \(\pi\) radians: \[ \begin{align*} \dfrac{36^\circ}{x\text{ rad}} &= \dfrac{180^\circ}{\pi\text{ rad}}\\ 36\pi &= 180x\\ x &= \dfrac{36\pi}{180}=\tfrac\pi5 \end{align*} \] You could have also had the right side as \(\dfrac{360^\circ}{2\pi}\), since both are equal to a full circle. </li> <li>You could think of it like a unit conversion or "dimensional analysis" problem like from chemistry: \[\dfrac{36^\circ}{1}\times\dfrac{\pi\text{ rad}}{180^\circ}=\tfrac\pi5\] </li> </ul> <p>All of these really come down to thinking of angles as <b>fractions of a circle</b>.</p> <div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-OfVtvRJR0V0/YCRD4-jLIgI/AAAAAAAAeN0/XNG3-RcPEsIGCedmc7bE2kB2AaGFDDyjQCLcBGAsYHQ/s1612/uc36deg.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="75%" data-original-height="1258" data-original-width="1612" src="https://1.bp.blogspot.com/-OfVtvRJR0V0/YCRD4-jLIgI/AAAAAAAAeN0/XNG3-RcPEsIGCedmc7bE2kB2AaGFDDyjQCLcBGAsYHQ/s1612/uc36deg.png"/></a></div></blockquote><p><b>How many degrees are in \(36\text{ rad}\)?</b></p><blockquote> <p>Again, we could solve this by setting up a proportion:</p> <p> \[ \begin{align*} \dfrac{36\text{ rad}}{x^\circ} &= \dfrac{\pi\text{ rad}}{180^\circ}\\ 36\cdot 180 &= \pi x\\ x &= \tfrac{6480}{\pi} \approx 2062.64062 \end{align*} \] </p> <p>Or we could do a unit conversion: \[\dfrac{36\text{ rad}}{1}\times\dfrac{180^\circ}{\pi\text{ rad}}=\tfrac{6480}{\pi}\] </p> <div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-fQ3-ClciLQ4/YCREUMil9ZI/AAAAAAAAeN8/VlpaWVcq2goBTx_-yX8iptGCH-ybQ9-1ACLcBGAsYHQ/s1488/uc36rad.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="75%" data-original-height="1488" data-original-width="1431" src="https://1.bp.blogspot.com/-fQ3-ClciLQ4/YCREUMil9ZI/AAAAAAAAeN8/VlpaWVcq2goBTx_-yX8iptGCH-ybQ9-1ACLcBGAsYHQ/s1488/uc36rad.png"/></a></div></blockquote><p>In the end, there is a "formulaic" way of thinking about these:</p><b> <p style="text-align: center;">To convert degrees to radians, multiply by \(\dfrac{\pi}{180}\).</p> <p style="text-align: center;">To convert radians to degrees, multiply by \(\dfrac{180}{\pi}\).</p></b><p>But memorizing formulas will only get you so far — what makes it stick is understanding why they work. That way you can fall back on any of the above methods even if you forget the formula (or forget which one is which).</p> <h2 style="text-align: left;">Converting between degrees and radians</h2><p><b>Are \(-1773^\circ\) and \(1467^\circ\) coterminal angles?</b></p><blockquote> <p>I was impressed by the variety of ways I saw in class!</p> <ul> <li>You could start with \(-1773^\circ\) and repeatedly add \(360^\circ\) to see if you get to \(1467^\circ\). (You do, so they <b>are</b> coterminal.)</li> <li>You could divide each by \(360^\circ\) which shows that each one is a little more than four full circles (\(1440^\circ\)) in some direction. Then, get rid of those full circles: \[ \begin{align*} 1467^\circ - 1440^\circ &= 27^\circ\\ -1773^\circ +1440^\circ &=-333^\circ \end{align*} \] Since \(27^\circ\) and \(-333^\circ\) are coterminal angles, so are our original two. (You can see this by adding another \(360^\circ\) to the negative angle.) </li> <li>You could look at the difference between the angles: \[1647^\circ-(-1773^\circ)=3240^\circ=9\cdot 360^\circ\] Since the angles are nine full circles apart — a whole number of full circles — the angles are coterminal. </li> </ul> <div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-74JSDcecLeU/YCREiSG6gJI/AAAAAAAAeOA/M_fvaitltfgMhOErkmlf_To2eqEskOtswCLcBGAsYHQ/s1454/coterm1.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="75%" data-original-height="1254" data-original-width="1454" src="https://1.bp.blogspot.com/-74JSDcecLeU/YCREiSG6gJI/AAAAAAAAeOA/M_fvaitltfgMhOErkmlf_To2eqEskOtswCLcBGAsYHQ/s1454/coterm1.png"/></a><a href="https://1.bp.blogspot.com/-gnnTgTyARBo/YCREjiLn5QI/AAAAAAAAeOI/6UzJjXTTOuEQUzO0flN9AXnhvyn5WhJngCLcBGAsYHQ/s1409/coterm2.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="75%" data-original-height="1281" data-original-width="1409" src="https://1.bp.blogspot.com/-gnnTgTyARBo/YCREjiLn5QI/AAAAAAAAeOI/6UzJjXTTOuEQUzO0flN9AXnhvyn5WhJngCLcBGAsYHQ/s1409/coterm2.png"/></a></div></blockquote><p><b>Are \(2.85\pi\) and \(3.85\pi\) coterminal angles?</b></p><blockquote> <p>Again this has to do with whether the two angles are some number of full circles apart, but this time we're in radians, so a full circle is \(2\pi\) radians.</p> <p>You can try any of the above methods, but you should find that these two angles are <b>not</b> coterminal — in fact, they're on exact opposite sides of the circle, half a circle away from each other.</p> <div class="separator" style="clear: both;"><a href="https://1.bp.blogspot.com/-yTGNyuK1yFY/YCRFE4zI0uI/AAAAAAAAeOY/bKpCRVwY37c2esRKq2FBjnyvWjjWSfsqACLcBGAsYHQ/s1534/notcoterm1.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="75%" data-original-height="1319" data-original-width="1534" src="https://1.bp.blogspot.com/-yTGNyuK1yFY/YCRFE4zI0uI/AAAAAAAAeOY/bKpCRVwY37c2esRKq2FBjnyvWjjWSfsqACLcBGAsYHQ/s1534/notcoterm1.png"/></a><a href="https://1.bp.blogspot.com/-voe-lToga2s/YCRFEwt_j5I/AAAAAAAAeOc/3CRem-y3mmcwcU54tDN1yDmGBxczSjwfACLcBGAsYHQ/s1520/notcoterm2.png" style="display: block; padding: 1em 0; text-align: center; "><img alt="" border="0" width="75%" data-original-height="1280" data-original-width="1520" src="https://1.bp.blogspot.com/-voe-lToga2s/YCRFEwt_j5I/AAAAAAAAeOc/3CRem-y3mmcwcU54tDN1yDmGBxczSjwfACLcBGAsYHQ/s1520/notcoterm2.png"/></a></div></blockquote><p>As you can see, we will repeatedly be using the fact that a full circle is \(360^\circ\) or \(2\pi\) radians. That equivalence will be so ingrained in you before long that you'll see it in your sleep!</p><hr><p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html">Return to MAT 130 main page</a></p>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com1tag:blogger.com,1999:blog-3748625510348961342.post-67078865341088668822021-02-06T19:20:00.067-08:002021-02-22T23:05:12.928-08:00The Unit Circle<p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html" >Return to MAT 130 main page</a ></p><p><a href="#previewactivity">Go to Preview Activity</a></p><hr /><p> Textbook Reference: <a href="https://openstax.org/books/precalculus/pages/5-1-angles" target="_blank" >Section 5.1 Angles</a ></p><div class="post-body"> <p> Behold, the <b><u>unit circle</u></b >: </p> <p style="text-align: center;"></p> <div class="separator" style="clear: both; text-align: center;"> <a href="https://1.bp.blogspot.com/-V3eMaHi6qfo/YCAAq7S9PuI/AAAAAAAAeKE/L28P-BEuwa8Ezf1VvJ171mUyNByqCPRWwCLcBGAsYHQ/s748/uc.png" style="margin-left: 1em; margin-right: 1em;" ><img border="0" data-original-height="748" data-original-width="744" src="https://1.bp.blogspot.com/-V3eMaHi6qfo/YCAAq7S9PuI/AAAAAAAAeKE/L28P-BEuwa8Ezf1VvJ171mUyNByqCPRWwCLcBGAsYHQ/s748/uc.png" width="320" /></a> </div> <br /> <p></p> <p> It's pretty unassuming – it's just a circle whose center is the origin \((0,0)\) and whose radius is \(1\). Simple, right? But this circle is going to be the main character in our story throughout the semester. You'll see what I mean as we go on, trust me. You and this unit circle are going to become the best of friends. </p> <h2>Angles</h2> <div> <p> Often when you think of angles, you usually think of the corners of polygons. These angles can be anywhere, which is why you usually bring a protractor with you when you need to measure them. They're also pretty <b>static</b> – they just kind of stay where they are, as if they're holding the polygon together and keeping it from collapsing. </p> <p> In this class, on the other hand, we're often going to be thinking of angles as being <b>dynamic</b> – they can open and close, like a hinge. To keep things from getting too crazy, we'll often put our angles into <b><u>standard position</u></b >, which means that one side of the angle (the <b><u>initial side</u></b >) is locked down on the x-axis, and the other side (the <b><u>terminal side</u></b >) is free to move around. We mark the initial side as \(0^\circ\), and as the angle opens <b>counterclockwise</b>, the angle gets bigger, and we can mark the number of degrees at the corresponding point on the circle. </p> <p style="text-align: center;"> <iframe frameborder="0" height="500px" src="https://www.desmos.com/calculator/mmuzjbijum?embed" style="border: 1px solid #ccc;" width="500px" ></iframe ><br /><i>Drag the blue point to change the angle.</i> </p> </div> <h2>Important Angles</h2> <div> <p> Some angles on the unit circle are so common that they're worth marking off: the multiples of \(30^\circ\) and \(45^\circ\). </p> <p style="text-align: center;"></p> <div class="separator" style="clear: both; text-align: center;"> <div class="separator" style="clear: both; text-align: center;"> <a href="https://1.bp.blogspot.com/-eFhkKYZuWsc/YCAf-Ugxo_I/AAAAAAAAeLY/qsgC0eeEKRA4GN4oXVnMtx436XwO6pRdwCLcBGAsYHQ/s1424/ucdeg.png" style="margin-left: 1em; margin-right: 1em;" ><img border="0" data-original-height="1376" data-original-width="1424" src="https://1.bp.blogspot.com/-eFhkKYZuWsc/YCAf-Ugxo_I/AAAAAAAAeLY/qsgC0eeEKRA4GN4oXVnMtx436XwO6pRdwCLcBGAsYHQ/s1424/ucdeg.png" width="100%" /></a> </div> </div> <p> Make sure you familiarize yourself with these angles. Ideally, you want to get to a point where you can visualize where they are at any time. If someone walks up to you on the street and says, "Hey, where's \(120^\circ\) on the unit circle?", you should be able to point upward and a little to the left. (You might look at them a little funny afterward, but your friend the unit circle will be pleased.) </p> </div> <h2>Coterminal Angles</h2> <div> <p>Now, you might be thinking:</p> <p style="text-align: center;"> <i>"Do angles have to be between \(0^\circ\) and \(360^\circ\)?"</i> </p> <p> Okay, well maybe you weren't thinking that, but I bet you are now. It seems pretty limiting to say that angles are only allowed to be within that certain range. Well, you know what? Let's just see what happens if we allow angles to be bigger than \(360^\circ\): </p> <!--<p style="text-align: center;"><iframe src="https://www.desmos.com/calculator/c7ryy60u8x?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe></p>--> <p style="text-align: center;"> <iframe frameborder="0" height="500px" src="https://www.desmos.com/calculator/dx6cszjwu2?embed" style="border: 1px solid #ccc;" width="500px" ></iframe> </p> <p><b>They wrap back around on themselves.</b></p> <p> You could imagine the angle revolving around and around forever, and it could get as big as you want. </p> <p>Okay, what about <b>negative</b> angles?</p> <p> If counterclockwise movement is positive, then it should be no surprise that <b>clockwise movement is negative</b>: </p> <p style="text-align: center;"> <iframe frameborder="0" height="500px" src="https://www.desmos.com/calculator/icoh2jvmvu?embed" style="border: 1px solid #ccc;" width="500px" ></iframe> </p> <p> But that's cheating, you might think! Is \(450^\circ\) really different from \(90^\circ\)? </p> <p>Well … yes and no.</p> <p> On one hand, if this were a circular track, and you ran from \(0^\circ\) to \(450^\circ\), you'd feel a lot more tired than if you just ran from \(0^\circ\) to \(90^\circ\). But, you have to admit that the angles <b>end in the same place</b>. We have a name for that – we say that \(90^\circ\) and \(450^\circ\) are <b><u>coterminal angles</u></b >. </p> <p> Starting at any angle, you can always go around a full circle (\(360^\circ\)) in either direction, and when you stop, you'll be at an angle that's <b>coterminal</b> to the one you were at before. </p> <p> In other words, <b><i>an angle can be ANY real number</i></b >. </p> <p>This will be really useful.</p> <h2>Turns</h2> <p> In mathematics, you <b>always </b>have to be asking <b>why </b>something is true. Sometimes doing that leads you to some pretty amazing insights, especially when you realize that some things you always took for granted don't always have to be set in stone. </p> <p> You probably didn't even blink when I said that there were \(360^\circ\) in a full circle. But did you ever think about why? </p> <p> Well, one reason why is that it's actually what the <b>Babylonians </b>used. But when it really comes down to it, \(360^\circ\) is just a convenient-yet-arbitrary number for the degrees in a circle. </p> <p> We could have divided a circle into, say, \(400\) pieces instead. Then a right angle would have been \(100\) pieces, and so on. (France actually tried to do this during the French Revolution, when they turned everything into the decimal system. They called these divisions of the circle "<a href="https://en.wikipedia.org/wiki/Gradian" >grads</a >." Unlike the meter and the kilogram, the grad didn't catch on.) </p> <p> To really understand circles, we need to come up with a measure that doesn't depend on some arbitrary number. We need a more <b>natural </b>measure. </p> <p> And you might have thought of one already, something that brings back memories of drawing little pizzas in elementary school: </p> <p><b>Fractions.</b></p> <p> Whether we think of a right angle as being \(90\) degrees or \(100\) "grads", it's still \(1/4\) of a circle. Thinking of our angles actively again, we can imagine that the angle <b>turned </b>\(1/4\) of the way around the circle. We can use the Greek letter \(\tau\) (<b><u>tau</u></b >) to stand for a "turn" and write it as follows:\[90^\circ=\tfrac14\tau\] </p> <p> This arguably makes it even <b>easier </b>to work with angles than when we had degrees! If you want to go \(5/12\) of the way around the circle, then your angle is just \(\tfrac{5}{12}\tau\): five-twelfths of a turn. Simple and clean. </p> </div> <h2>Radians</h2> <div> <p> We're definitely closer to having a natural unit of angle measurement, but we haven't really linked it to the <b>circle </b>yet. We could just as easily apply our fraction concept to a square, or a hexagon: </p> <p style="text-align: center;"></p> <div class="separator" style="clear: both; text-align: center;"> <div class="separator" style="clear: both; text-align: center;"> <a href="https://1.bp.blogspot.com/-dHfm_bpJ_5Q/YCAe2Y0vFxI/AAAAAAAAeLM/Vca_7F7OihIoK_-O_zVyMLm2QbvFuQsawCLcBGAsYHQ/s1714/ucalts.png" style="margin-left: 1em; margin-right: 1em;" ><img border="0" data-original-height="836" data-original-width="1714" src="https://1.bp.blogspot.com/-dHfm_bpJ_5Q/YCAe2Y0vFxI/AAAAAAAAeLM/Vca_7F7OihIoK_-O_zVyMLm2QbvFuQsawCLcBGAsYHQ/s1714/ucalts.png" width="100%" /></a> </div> </div> <p></p> <p>So what makes the circle special?</p> <p> <b>As the angle opens, the arc of the circle grows proportionally with it.</b> With any other shape, the amount marked off on the perimeter will grow faster or slower depending on how close the shape is to the origin, but with a circle, double the angle always means double the arc. </p> <p> You could imagine wrapping a string around the circle, starting at the initial side and ending at the terminal side, and then measuring how long it is. The problem is that would depend on how big the circle's radius is. But we can fix that too, by always basing our measurements on the simplest possible circle: the unit circle, with a radius of \(1\). </p> <p> If we wrapped our string all the way around the unit circle, you'd notice it would be about \(6.28\) units long. That number is actually a really important number in mathematics – and, in fact, it's actually called tau!\[\tau=6.2831853071795864769228\ldots\] </p> <p> Tau is the circumference of a unit circle, and it's one of those deep and fundamental constants that "makes math tick." Wherever there's a circle, you can be sure that \(\tau\) is lurking somewhere behind the scenes. </p> <p> This means that we can still use our measurements we said before, though we do it in a funny way. We say that a string 1 unit long would be one <b><u>radian</u></b >, since it's exactly as long as the radius (which we also said was 1 unit long on the unit circle). That means that \(1\) radian is about \(57.3^\circ\), which is a little inconvenient. </p> <div class="separator" style="clear: both; text-align: center;"> <a href="https://upload.wikimedia.org/wikipedia/commons/2/28/Circle_radians_tau.gif" style="margin-left: 1em; margin-right: 1em;" ><img border="0" data-original-height="450" data-original-width="80%" src="https://upload.wikimedia.org/wikipedia/commons/2/28/Circle_radians_tau.gif" /></a> </div> <i ><div style="text-align: center;"> <i >By Lucas Vieira - Own work, Public Domain,<br /><a href="https://commons.wikimedia.org/w/index.php?curid=25139980" >https://commons.wikimedia.org/w/index.php?curid=25139980</a ></i > </div></i > <p> But we almost never use whole-number measurements for radians – instead we use fractions of \(\tau\). So, the whole circle is \(\tau\) radians, halfway around the circle is \(\tfrac12\tau\) radians, a right angle is \(\tfrac14\tau\) radians, an eighth of a circle is \(\tfrac18\tau\), and so on. If you keep things in terms of \(\tau\), you get to keep those intuitive fractions. </p> <p style="text-align: center;"></p> <div class="separator" style="clear: both; text-align: center;"> <a href="https://1.bp.blogspot.com/-irc7J1EcSdY/YCAOnRxou4I/AAAAAAAAeKg/FafbkSM8Bj4T_Yxlivmlt_WP2CBAL2koACLcBGAsYHQ/s1557/uctau.png" style="margin-left: 1em; margin-right: 1em;" ><img border="0" data-original-height="1557" data-original-width="1521" src="https://1.bp.blogspot.com/-irc7J1EcSdY/YCAOnRxou4I/AAAAAAAAeKg/FafbkSM8Bj4T_Yxlivmlt_WP2CBAL2koACLcBGAsYHQ/s1557/uctau.png" width="100%" /></a> </div> <i ><div style="text-align: center;"> <i>(Click the figure to enlarge it.)</i> </div></i > <p></p> <p>There's just one catch.</p> <p> The unfortunate thing about \(\tau\) is that, due to how history has gone, it has a much more famous little brother: \(\pi\) (pi), which is half of \(\tau\). Or, put another way, \(\tau=2\pi\). </p> <p> So, the way mathematicians usually measure radians is in terms of \(\pi\) instead of \(\tau\). A full circle has a total of \(2\pi\) radians, since (using the formula \(C=2\pi r\) that you learned back in elementary school) the circumference of the unit circle is \(2\pi\). That means that halfway around the circle is \(\pi\), a right angle is \(\tfrac12\pi\) (or often written \(\tfrac\pi2\)), an eighth of a circle is \(\tfrac14\pi\) (or \(\tfrac\pi4\)), and so on. There's a lot of multiplying by \(2\) and dividing by \(2\), and it can get very confusing if you're not careful. </p> <div class="separator" style="clear: both; text-align: center;"> <a href="https://1.bp.blogspot.com/-R893Jo0Su-M/YCEZEcnz7-I/AAAAAAAAeMI/agFcK0EvmSkT4NK6RZrtEBKStGjIZnjzgCLcBGAsYHQ/s1437/ucpi.png" style="margin-left: 1em; margin-right: 1em;" ><img border="0" data-original-height="1548" data-original-width="1543" src="https://1.bp.blogspot.com/-R893Jo0Su-M/YCEZEcnz7-I/AAAAAAAAeMI/agFcK0EvmSkT4NK6RZrtEBKStGjIZnjzgCLcBGAsYHQ/s1437/ucpi.png" width="100%" /></a> </div> <p style="text-align: center;"><i>(Click the figure to enlarge it.)</i></p> <p> It would be much nicer if we could just always do things in terms of \(\tau\). <a href="https://tauday.com/" >And there are some people who are proposing that we do exactly that.</a > However, until that day comes, the rest of the world still uses \(\pi\). But you'll be okay if you just remember one thing: </p> <p style="text-align: center;"> <b>Radians are angles measured as fractions of \(2\pi\).</b> </p> <p> If you're looking at \(\tfrac18\) of a circle, that's \(\tfrac18\) of \(2\pi\):\[\dfrac18\cdot 2\pi=\dfrac{2\pi}{8}=\dfrac\pi4\] </p> <p> In other words, as long as you remember the <b>intuition</b> that made radians work in the first place, you'll always have something to fall back on. </p> </div></div><hr /><a id="previewactivity"></a><h1 style="text-align: left;">Preview Activity 1</h1><p> Answer these questions and submit your answers as a document <b>on Moodle</b>. (Please submit as .docx or .pdf if possible.) </p><ol style="text-align: left;"> <li> List out as many divisors (that is, factors) of \(360\) as you can think of. Why would this have made it convenient for the Babylonians to split up the circle into \(360^\circ\)? </li> <li>Find at least three angles that are coterminal to \(20^\circ\).</li> <li> Remember that the four quadrants are laid out as follows.<br /> <div style="text-align: center;"> <ol style="display: inline;"> <li style="display: inline;"> <div class="separator" style="clear: both; text-align: center;"> <a href="https://1.bp.blogspot.com/-Jlb-dvK5h8U/YCEa7yx_YFI/AAAAAAAAeMU/4SA9dckuAHg-0RJJr2pwqTXKSd4_SGqUACLcBGAsYHQ/s922/quadrants.png" style="margin-left: 1em; margin-right: 1em;" ><img border="0" data-original-height="890" data-original-width="922" src="https://1.bp.blogspot.com/-Jlb-dvK5h8U/YCEa7yx_YFI/AAAAAAAAeMU/4SA9dckuAHg-0RJJr2pwqTXKSd4_SGqUACLcBGAsYHQ/s922/quadrants.png" width="320" /></a> </div> </li> </ol> </div> In which quadrants would each of the following angles be found? (It might help you to sketch these out.) <ol style="list-style-type: lower-alpha;"> <li style="padding-bottom: 0.25em;">\(110^\circ\)</li> <li style="padding-bottom: 0.25em;">\(215^\circ\)</li> <li style="padding-bottom: 0.25em;">\(-250^\circ\)</li> <li style="padding-bottom: 0.25em;">\(950^\circ\)</li> </ol> </li> <li> Each of the following is the radian measure of an angle. Multiply each one out, writing your answers in lowest terms. Then try to figure out the corresponding angle in degrees based on <b>visualizing</b> it on the unit circle. <ol style="list-style-type: lower-alpha;"> <li style="padding-bottom: 0.45em;">\(\tfrac{1}{2}(2\pi)\)</li> <li style="padding-bottom: 0.45em;">\(\tfrac{2}{3}(2\pi)\)</li> <li style="padding-bottom: 0.45em;">\(\tfrac{3}{4}(2\pi)\)</li> <li style="padding-bottom: 0.45em;">\(\tfrac{5}{6}(2\pi)\)</li> <li style="padding-bottom: 0.45em;">\(\tfrac{3}{8}(2\pi)\)</li> <li style="padding-bottom: 0.45em;">\(\tfrac{5}{12}(2\pi)\)</li> </ol> </li> <li> Answer <b>AT LEAST</b> one of the following questions: <ol style="list-style-type: lower-alpha;"> <li>What was something you found interesting about this reading?</li> <li>What was an "a-ha" moment you had while doing this reading?</li> <li>What was the muddiest point of this reading for you?</li> <li>What question(s) do you have about anything you've read?</li> </ol> </li></ol><p> <a href="https://drive.google.com/file/d/1B3HbBYQDin-4zqsLKUJ4yn9lmIk3lTHh/view?usp=sharing" target="_blank" ><b>Answers</b></a ></p><hr /><p> <a href="https://www.solidangl.es/p/spring-2021-mat-130-advanced-functions.html" >Return to MAT 130 main page</a ></p> <script>const headers = document.querySelectorAll(".post-body h2"); const rightArr = " [Expand]"; const downArr = " [Collapse]"; headers.forEach(node => { const sectionDiv = node.nextElementSibling; const expandBtn = document.createElement("span"); const btnText = document.createTextNode(downArr); expandBtn.append(btnText); node.append(expandBtn); node.addEventListener("click", e => { const curr = btnText.textContent; btnText.textContent = curr == rightArr ? downArr : rightArr; const currDisplay = sectionDiv.style.display; sectionDiv.style.display = currDisplay == "none" ? "block" : "none"; }); console.log(sectionDiv.tagName) }) </script><!-- -->Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-20609431438906592652019-10-15T13:40:00.010-07:002021-02-10T21:28:36.444-08:00Orbit-Stabilizer Theorem<div><i>My song on the dance game Pump it Up XX is named after a famous theorem from group theory. Here's an explanation of what it's all about.</i></div><div><br /></div><span><a name='more'></a></span><div><br /></div><div><br /></div>This is going to be a different sort of post from the ranting-about-math-education kinds that usually I make.<br /><br />As many of you know, when I'm not doing math (yes, I do take a break sometimes), I write electronic music, some of which makes its way into rhythm games. Most recently, I wrote a track called <b>Orbit Stabilizer </b>for the Korean dance game Pump It Up XX.<br /><br /><div class="separator" style="clear: both; text-align: center;"><p style="text-align:center"><iframe allowfullscreen="" class="YOUTUBE-iframe-video" data-thumbnail-src="https://i.ytimg.com/vi/uYRJwzH043U/0.jpg" frameborder="0" src="https://www.youtube.com/embed/uYRJwzH043U?feature=player_embedded" width="480" height="360"></iframe></p></div>If you didn't figure it out from watching the background animation, there's a mathematical connection to the song. This isn't the first time I've made references to mathematics in my music:<br /><ul><li>One of my songs on Dance Dance Revolution, called <b><a href="https://www.youtube.com/watch?v=x2PQuUe2KG0">ΔMAX</a></b>, changes by one BPM every beat. It goes up to 573 BPM, which is significant because the number 573 can be <a href="https://en.wikipedia.org/wiki/Japanese_wordplay">read out loud as</a> "Konami" , who makes DDR.</li><li>My track on J-Rave Nation is called <b><a href="https://www.youtube.com/watch?v=WK7lVan2duw">π・ρ・maniac</a></b>, which is really just a play on the word "pyromaniac" but uses the well-known constant \(\pi\). (Also, there's a <a href="https://www.youtube.com/watch?v=c5v0as3KWT8">mission</a> in Pump It Up Infinity that makes you do math problems in the middle of the song.)</li><li>Another song from Pump It Up Prime, called <b><a href="https://www.youtube.com/watch?v=ivIITIMYWXs">Annihilator Method</a></b>, is a reference to <a href="https://en.wikipedia.org/wiki/Annihilator_method">a technique for solving differential equations</a>. (You have to admit that's a pretty hardcore-sounding name for something from a math class!)</li></ul><div>This time, the reference is to something from a branch of mathematics called <u style="font-weight: bold;"><a href="https://en.wikipedia.org/wiki/Group_theory">group theory</a></u>, which is part of the deep abstract foundations for algebra. It was pioneered simultaneously by French mathematician <a href="https://en.wikipedia.org/wiki/%C3%89variste_Galois">Évariste Galois</a> (who shows up in the background of the Orbit Stabilizer video!) and Norwegian mathematician <a href="https://en.wikipedia.org/wiki/Niels_Henrik_Abel">Niels Henrik Abel</a>. Both are worth reading up on ― heartbreaking stories of geniuses who died tragically young.</div><div><br /></div><div>Group theory is, in essence, the study of <i>symmetry</i>. It can be used to describe everything from the shuffling of cards and the positions of a Rubik's cube to the fundamental laws of physics that govern our universe. I'm not going to go into an in-depth explanation of group theory ― there are <a href="https://www.youtube.com/playlist?list=PLL0ATV5XYF8AQZuEYPnVwpiFy0jEipqN-">plenty of YouTube videos</a> that do that (<a href="https://www.youtube.com/watch?v=WwndchnEDS4">including my own</a>). Rather I'm going to go just far enough to explain what the <b><u>orbit-stabilizer theorem</u></b> is.</div><div><br /></div><div>Imagine you have a square sheet of paper in front of you.<br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" data-original-height="182" data-original-width="182" src="https://1.bp.blogspot.com/-mSVajUr0alU/XaX73fovSGI/AAAAAAAAaps/DCT99yLIETcjUSeQm0hZDgdwWxJ6fm0RgCLcBGAsYHQ/s1600/smallgraysquare.png" /></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><br /></div><div>You look away, and then I sneak in and potentially move the paper in some way. When you look back, the square looks the same as it did before. What could I have done to the paper?<br /><br />The set of possible transformations I could have done are called <u style="font-weight: bold;">symmetries</u> of the square. In this case, there are three overall types of such symmetries:<br /><ul><li>I could have left the paper alone entirely, doing nothing to it. This is called the <u style="font-weight: bold;">identity transformation</u>.</li><li>I could have rotated the paper by \(90^\circ\), \(180^\circ\), or \(270^\circ\).</li><li>I could have flipped the paper over across one of its four axes of symmetry ― the horizontal axis, the vertical axis, or one of the two diagonal axes.</li></ul><div>All in all, I have <i>eight</i> possible transformations that would take the square back to itself. Notice that translations (shifts) aren't included ― if I'd moved the paper a bit to the left, you would have noticed (we assume)! We're only considering transformations that would leave you completely unable to tell what I did. To visualize the effects of these transformations better, we can label the square and watch what happens:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-2V0PdayTch4/XaYBJxE_6pI/AAAAAAAAap4/dN07hZ45WZodnBYDAyixavTl7gD88fVBwCLcBGAsYHQ/s1600/squares.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="723" data-original-width="917" src="https://1.bp.blogspot.com/-2V0PdayTch4/XaYBJxE_6pI/AAAAAAAAap4/dN07hZ45WZodnBYDAyixavTl7gD88fVBwCLcBGAsYHQ/s1600/squares.png" width="100%" /></a></div><div><br /></div><div><br /></div><div>These transformations form what mathematicians call a <u style="font-weight: bold;">group</u>, which means that they obey four fundamental laws (axioms):</div><div><ol><li>Combining any two of these transformations ― such as, say, rotating by \(90^\circ\) counterclockwise and then flipping across the horizontal axis ― also takes the square back to itself, and is equivalent to one of the original eight transformations ― in this case, flipping across the "backslash" diagonal axis. We call this the <u style="font-weight: bold;">closure</u> law.</li><li>Combining transformations is <b><u>associative</u></b>, which means that if I have any transformations \(a\), \(b\), and \(c\), then \((ab)c\) ― that is, doing \(c\) first and then doing whatever \((ab)\) is equivalent to ― gives the same result as \(a(bc)\) ― that is, doing whatever \((bc)\) is equivalent to first and then doing \(a\) afterward. This is just like how you can move parentheses around when doing addition and multiplication.</li><li>The set of transformations includes an <u style="font-weight: bold;">identity</u> transformation, which essentially does nothing. (I say "essentially" because you could also think of a \(360^\circ\) rotation as an identity transformation since it has no overall effect.)</li><li>Every transformation has an <u style="font-weight: bold;">inverse</u> transformation which "undoes" it. For example, a rotation of \(90^\circ\) can be undone by a rotation of \(270^\circ\), or a reflection can be undone by doing that same reflection again.</li></ol></div><div>We usually use the letter \(G\) to talk about a group in general. We'll also use the notation \(|G|\) to talk about the "size" of a group ― so in this case, we have \(|G|=8\).<br /><br />So we know that these transformations take the square back to itself, but what if we ask what happens to just a single point inside of the square?<br /><br />Suppose we let \(x\) mark the midpoint of segment \(AB\). Then we can see where \(x\) lands after each of the transformations:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-oMziX-M3ag0/XaYTZGXIivI/AAAAAAAAaqE/Hxk8A64egloiyBBHnCCcPelRQqL0ckWUgCLcBGAsYHQ/s1600/midpoint.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="723" data-original-width="917" src="https://1.bp.blogspot.com/-oMziX-M3ag0/XaYTZGXIivI/AAAAAAAAaqE/Hxk8A64egloiyBBHnCCcPelRQqL0ckWUgCLcBGAsYHQ/s1600/midpoint.png" width="100%" /></a></div><br />The set of possible places where \(x\) can land using our transformations is called the <u style="font-weight: bold;">orbit</u> of \(x\) under our group, and we denote it as \(\text{orb}(x)\). Since there are four different places that \(x\) can land ― it can end up on any one of the four midpoints of one of the sides ― we say that \(|\text{orb}(x)|=4\).<br /><br />Looking a little closer, we can find that there are two particular transformations that do something special. Focus on the identity transformation and the vertical axis flip. What happens to \(x\) under these two transformations? Well, it stays where it is! The other six transformations, on the other hand, all move \(x\) to a different point of the square. These two special transformations form what's called the <u style="font-weight: bold;">stabilizer</u> of \(x\) under our group, and we denote it as \(\text{stab}(x)\). The stabilizer forms what's called a <u style="font-weight: bold;">subgroup</u> of our original group, because if you limit yourself to only doing those things, you still satisfy the same basic group laws. Again, we can measure the "size" of the stabilizer, which in this case is \(|\text{stab}(x)|=2\).<br /><br />Now, look at the three "sizes" we've computed.<br />\[<br />\begin{align*}<br />|G|&=8\\<br />|\text{orb}(x)|&=4\\<br />|\text{stab}(x)|&=2<br />\end{align*}<br />\]Those three numbers (\(8\), \(4\), and \(2\)) are just begging to be related to each other!<br /><br />\[|\text{orb}(x)|\cdot|\text{stab}(x)|=|G|\]<br /><br /></div><div>Is this a coincidence? Well, you can find out for yourself by picking different places for \(x\) (like, say, a corner point, or the very center, or just some random point in the square), then calculating the sizes of its orbit and its stabilizer. Or you can even try it for a different shape, like a triangle or a rhombus or a pentagon, which will have a different size group. You'll notice that you always get that same relationship!<br /><br />That relationship is called the <u style="font-weight: bold;">Orbit-Stabilizer Theorem</u>.<br /><br />...well, almost.<br /><br />We actually run into a bit of a problem if our group is <i>infinite</i>. If that's the case, then multiplying and dividing using infinite numbers can get a bit hairy and make it a bit difficult to show that the relationship is meaningful. To deal with that case, we usually write the Orbit-Stabilizer Theorem in a slightly different way:<br /><br />\[|\text{orb}(x)|=[G : \text{stab}(x)]\]</div><div><br /></div><div>The quantity \([G : \text{stab}(x)]\) is called the <u style="font-weight: bold;">index</u> of the stabilizer as a subgroup of \(G\). Essentially this stands for how many copies of \(\text{stab}(x)\) can fit inside \(G\). (Technically it's how many "cosets" the stabilizer has, if you want to look that up.) This is the version of the equation that shows up in the video.</div><div><br /></div><div>Okay, so that explains what the Orbit-Stabilizer Theorem <i>is</i>. But, you might be wondering, is it also <i>useful</i>?<br /><br />Well, one famous result in combinatorics (the study of counting) that makes great use of the Orbit-Stabilizer Theorem is <u style="font-weight: bold;"><a href="https://en.wikipedia.org/wiki/Burnside%27s_lemma">Burnside's Lemma</a></u>. Burnside's Lemma can be used to solve problems like counting the number of possible ways that the sides of a cube can be painted with three different colors. (Painting the top side blue and all the others white is considered the same as, say, painting the front side blue and all the others white, since you can just rotate one to get the other.) Looking for something more "real-world"? Burnside's Lemma has applications in <a href="https://www.springer.com/us/book/9781461291053">chemistry</a> ― like if you need to find out how many different ways certain groups can be placed around a central carbon atom ― as well as other areas such as electronic circuits and even (to bring things full circle) music theory!</div><div><br /></div><div>Hopefully this gives you an idea of why the Orbit-Stabilizer Theorem is interesting, and why I would name a song after it (besides just being a cool-sounding name). Feel free to let me know if you have any questions. And of course, if you play Pump It Up, give the track a shot!<br /><br /><i>P.S. If you want to know more about group theory, I would highly recommend <a href="https://www.youtube.com/playlist?list=PLL0ATV5XYF8AQZuEYPnVwpiFy0jEipqN-">this YouTube series</a> by Dr. Matt Salomone at Bridgewater State University. He has a knack for making these sorts of abstract algebraic topics very accessible with excellent examples and intuitive explanations!</i></div></div>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com0tag:blogger.com,1999:blog-3748625510348961342.post-6415839226314388482019-08-01T10:27:00.004-07:002021-02-06T19:22:11.083-08:00When Does Nothing Mean Something?<div><i>Those deliberately ambiguous order-of-operations problems that plague social media may actually be evidence of a mathematical "language shift."</i></div><div><br /></div><span><a name='more'></a></span><div><br /></div><br />I thought I was done writing about this topic, but it just keeps coming back. The internet just cannot seem to leave this sort of problem alone:<br /><blockquote class="twitter-tweet"><div dir="ltr" lang="en">oomfies solve this <a href="https://t.co/0RO5zTJjKk">pic.twitter.com/0RO5zTJjKk</a></div>— em ♥︎ (@pjmdolI) <a href="https://twitter.com/pjmdolI/status/1155598050959745026?ref_src=twsrc%5Etfw">July 28, 2019</a></blockquote><script async="" charset="utf-8" src="https://platform.twitter.com/widgets.js"></script> I don't know what it is about expressions of the form \(a\div b(c+d)\) that fascinates us as a species, but fascinate it does. I've <a href="http://www.solidangl.es/2014/07/the-implications-of-being-implicit.html">written about this before</a> (as well as <a href="http://www.solidangl.es/2014/07/poorly-executed-mnemonics-definitely.html">why "PEMDAS" is terrible</a>), but the more I've thought about it, the more sympathy I've found with those in the minority of the debate, and as a result my position has evolved somewhat.<br /><br />So I'm going to go out on a limb, and claim that the answer <i style="font-weight: bold;">should</i> be \(1\).<br /><br />Before you walk away shaking your head and saying "he's lost it, he doesn't know what he's talking about", let me assure you that I'm obviouly not denying the left-to-right convention for how to do explicit multiplication and division. Nobody's arguing that.* Rather, there's something much more subtle going on here.<br /><br />What we may be seeing here is evidence of a mathematical "language shift".<br /><br />It's easy to forget that mathematics did not always look as it does today, but has arrived at its current form through very human processes of invention and revision. There's an <a href="http://jeff560.tripod.com/mathsym.html">excellent page</a> by Jeff Miller that catalogues the earliest recorded uses of symbols like the operations and the equals sign -- symbols that seem timeless, symbols we take for granted every day.<br /><br />People also often don't realize that this process of invention and revision still happens to this day. The <a href="http://www.solidangl.es/2014/09/throw-outdated-notation-to-floor.html">modern notation for the floor function</a> is a great example that was only developed within the last century. Even today on the internet, you occasionally see discussions in which people debate on how mathematical notation can be improved. (I'm still holding out hope that my <a href="http://www.solidangl.es/2015/04/a-radical-new-look-for-logarithms.html">alternative notation for logarithms</a> will one day catch on.)<br /><br />Of particular note is the evolution of <a href="http://jeff560.tripod.com/grouping.html">grouping symbols</a>. We usually think only of parentheses (as well as their variations like square brackets and curly braces) as denoting grouping, but an even earlier symbol used to group expressions was the <u style="font-weight: bold;">vinculum</u> -- a horizontal bar found over or under an expression. Consider the following expression: \[3-(1+2)\] If we wrote the same expression with a vinculum, it would look like this: \[3-\overline{1+2}\] Vincula can even be stacked: \[13-\overline{\overline{1+2}\cdot 3}=4\] This may seem like a quaint way of grouping, but it does in fact survive in our notation for fractions and radicals! You can even see both uses in the quadratic formula: \[x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\]<br /><br />Getting back to the original problem, what I think we're seeing is evidence that <u style="font-weight: bold;">concatenation</u> -- placing symbols next to each other with no sort of explicit symbol -- has become another way to represent grouping.<br /><br />"But wait", you might say, "concatenation is used to represent <i>multiplication</i>, not <i>grouping</i>!" That's certainly true in many cases, for example in how we write polynomials. However, there are a few places in mathematics that provide evidence that there's more to it than that.<br /><br />First of all, as <a href="https://twitter.com/EoN_Tweets/status/1156757840880656384">a beautifully-written Twitter thread by EnchantressOfNumbers (@EoN_tweets)</a> points out, we use concatenation to show a special importance of grouping when we write out certain trigonometric expressions without putting their arguments in parentheses. Consider the following identity:<br />\[\sin 4u=2\sin 2u\cos 2u\] When we write such an equation, we're saying that not only do \(4u\) and \(2u\) represent multiplications, but that this grouping is so tight that they constitute the entire arguments of the sine and cosine functions. In fact, the space between \(\sin 2x\) and \(\cos 2x\) can also be seen as a somewhat looser form of concatention. Then again, so does the space between \(\sin\) and \(x\), which represents a different thing -- the connection of a function to its argument. Perhaps this is why the popular (and amazing) online graphing calculator <a href="https://www.desmos.com/">Desmos</a> is only so permissive when it comes to parsing concatenation:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-Ne57iJO2yGY/XUMUP2-EkWI/AAAAAAAAaic/unI8YtWxEr8C6GhrpOxuL5d-GO0VFabeQCLcBGAs/s1600/concatsin.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="192" data-original-width="457" src="https://1.bp.blogspot.com/-Ne57iJO2yGY/XUMUP2-EkWI/AAAAAAAAaic/unI8YtWxEr8C6GhrpOxuL5d-GO0VFabeQCLcBGAs/s1600/concatsin.png" /></a></div>In contrast, where we do draw the line is with an expression like the following:\[\sin x+y\] We always interpret this as \((\sin(x))+y\), never \(\sin(x+y)\). To drive home just how much stronger implicit multiplication feels to us than explicit multiplication, just take a look at the following expression: \[\sin x\cdot y\] Does this mean \((\sin(x))\cdot y\) or \(\sin(x\cdot y)\)? If that expression makes you writhe uncomfortably, while if it had been written as \(\sin xy\) it would be fine, then you might see what I'm getting at.<br /><br />An even more curious case is <i>mixed numbers</i>. When writing mixed numbers, concatenation actually stands for addition, not multiplication. \[3\tfrac{1}{2}=3+\tfrac{1}{2}\] In fact, concatenation actually makes addition come <i>before</i> multiplication when we multiply mixed numbers! \[3\tfrac{1}{2}\cdot 5\tfrac{5}{6}=(3+\tfrac{1}{2})\cdot(5+\tfrac{5}{6})=20\tfrac{5}{12}\]<br /><br />Now, you may feel that this example shows how mixed numbers are an inelegance in mathematical notation (and I would agree with you). Even so, I argue that this is evidence that we <i>fundamentally</i> view concatenation as a way to represent <i>grouping</i>. It just so happens that, since multiplication takes precedence over addition anyway in the absence of other grouping symbols, we use concatenation when we write it. This all stems from a sort of "laziness" in how we write things -- - laying out precedence rules allows us to avoid writing parentheses, and once we've established those precedence rules, we don't even need to write out the multiplication at all.<br /><br />So how does the internet's favorite math problem fit into all this?<br /><br />The most striking feature of the expression \(8\div 2(2+2)\) is that <i>it's written all in one line</i>.<br /><br />Mathematical typesetting is difficult. <a href="https://en.wikibooks.org/wiki/LaTeX/Mathematics">LaTeX</a> is powerful, but has a steep learning curve, though various other editors have made it a bit easier, such as <a href="https://www.youtube.com/watch?v=oOtZW95ZcGg">Microsoft Word's Equation Editor</a> (which has much improved since when I first used it!). Calculators have also recognized this difficulty, which is why TI calculators now have <a href="https://www.youtube.com/watch?v=Qqc3zPNfJGw">MathPrint</a> templates (though its entry is quite clunky compared to Desmos's "as-you-type" formatting via <a href="http://mathquill.com/">MathQuill</a>).<br /><br />Even so, all of these input methods exist in very specific applications. What about when you're writing an email? Or sending a text? Or a Facebook message? (If you're wondering "who the heck writes about math in a Facebook message", the answer at least includes "students who are trying to study for a test".) The evolution of these sorts of media has led to the importance of one-line representations of mathematics with easily-accessible symbols. When you don't have the ability (or the time) to neatly typeset a fraction, you're going to find a way to use the tools you've got. And that's even more important as we realize that <i>everybody</i> can (and should!) engage with mathematics, not just mathematicians or educators.<br /><br />So that might explain why a physics student might type "hbar = h / 2pi", and others would know that this clearly means \(\hbar=\dfrac{h}{2\pi}\) rather than \(\hbar=\dfrac{h}{2}\pi\). Remember, mathematics is not about just answer-getting. It's about communication of those ideas. And when the medium of communication limits how those ideas can be represented, the method of communication often changes to accomodate it.<br /><br />What the infamous problem points out is that while almost nobody has laid out any explicit rules for how to deal with concatenation, we seem to have developed some implicit ones, which we use without thinking about them. We just never had to deal with them until recently, as more "everyday" people communicate mathematics on more "everyday" media.<br /><br />Perhaps it's time that we address this convention explicitly and admit that <i style="font-weight: bold;">concatenation really has become a way to represent grouping</i>, just like parentheses or the vinculum. This is akin to taking a more descriptivist, rather than prescriptivist, approach to language: all we would be doing is recognizing that this is <i>already </i>how we do things everywhere else.<br /><br />Of course, this would throw a wrench in PEMDAS, but that just means we'd need to <a href="http://www.solidangl.es/2014/07/poorly-executed-mnemonics-definitely.html">actually talk about the mathematics behind it</a> rather than memorizing a silly mnemonic. After all, as inane as these internet math problems can be, they've shown that (whether they admit it or not) people really <i>do</i> want to get to the bottom of mathematics, to truly understand it.<br /><br />I'd say that's a good thing.<br /><br /><br /><i>* If your argument for why the answer is \(16\) starts with "Well, \(2(2+2)\) means \(2\cdot(2+2)\), so...", then you have missed the point entirely.</i>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com1tag:blogger.com,1999:blog-3748625510348961342.post-18947808448249202942019-01-12T19:19:00.005-08:002021-02-06T16:51:01.020-08:00Who Says You Can't Do That? --- Trig Identities<i>Everyone knows you can't do the same thing to both sides of a trig identity to prove it... right?</i><div><br /></div><span><a name='more'></a></span><div><br /></div><div><br />Ahh, trig identities... a rite of passage for any precalculus student.<br /><br />This is a huge stumbling block for many students, because up until this point, many have been perfectly successful (or at least have gotten by) in their classes by learning canned formulas and procedures and then doing a bunch of exercises that just change a \(2\) to a \(3\) here and a plus to a minus there. Now, all of a sudden, there's no set way of going about things. No "step 1 do this, step 2 do that". Now they have to rely on their intuition and "play" with an identity until they prove that it's correct.<br /><br />And to make matters worse, many textbooks --- and, as a result, many teachers --- make this subject arbitrarily and artificially harder for the students.<br /><br />They insist that students are not allowed to work on both sides of the equation, but instead must specifically start at one end and work their way to the other. I myself once subscribed to this "rule", because it's how I'd always been taught, and I always fed students the old line of "you can't assume the thing you're trying to prove because that's a logical fallacy".<br /><br />Then one of my Honors Precalculus students called me on it.<br /><br />He asked me to come up with an example of a trig non-identity where adding the same thing on both sides would lead to a false proof that the identity was correct. After some thought, I realized that not only couldn't I think of one, but that mathematically, there's no reason that one should exist.<br /><br />To begin with, one valid way to prove an identity is to work with each side of the equation <i>separately</i> and show that they are both equal to the same thing. For example, suppose you want to verify the following identity:<br /><br />\[\dfrac{\cot^2{\theta}}{1+\csc{\theta}}=\dfrac{1-\sin{\theta}}{\sin{\theta}}\]<br />Trying to work from one side to the other would be a nightmare, but it's much simpler to show that each side is equal to \(\csc{\theta}-1\). This in fact demonstrates one of the oldest axioms in mathematics, as written by Euclid: "things which are equal to the same thing are equal to each other."<br /><br />But what about doing the same thing to <i>both</i> sides of an equation?<br /><br />There are two important points to realize about what's going on behind the scenes here.<br /><br />The first is that if your "thing you do to both sides" is a <i>reversible step</i> --- that is, if you're applying a <i>one-to-one function</i> to both sides of an equation --- then it's perfectly valid to use that as part of your proof because it establishes an <i>if-and-only-if</i> relationship. If that function is not one-to-one, all bets are off. You can't prove that \(2=-2\) by squaring both sides to get \(4=4\), because the function \(x\mapsto x^2\) maps multiple inputs to the same output.<br /><br />It baffles me that most Precalculus textbooks mention one-to-one functions in the first chapter or two, yet completely fail to understand how this applies to solving equations.* A notable exception is UCSMP's Precalculus and Discrete Mathematics book, which establishes the following on p. 169:<br /><b><br /></b><br /><blockquote class="tr_bq"><u style="font-weight: bold;">Reversible Steps Theorem</u></blockquote><blockquote class="tr_bq">Let \(f\), \(g\), and \(h\) be functions. Then, for all \(x\) in the intersection of the domains of functions \(f\), \(g\), and \(h\),</blockquote><blockquote class="tr_bq"><ol><li>\(f(x)=g(x) \Leftrightarrow f(x)+h(x)=g(x)+h(x)\)</li><li>\(f(x)=g(x) \Leftrightarrow f(x)\cdot h(x)=g(x)\cdot h(x)\) <i>[We'll actually come back to this one in a bit -- there's a slight issue with it.]</i></li><li>If \(h\) is 1-1, then for all \(x\) in the domains of \(f\) and \(g\) for which \(f(x)\) and \(g(x)\) are in the domain of \(h\), \[f(x)=g(x) \Leftrightarrow h(f(x))=h(g(x)).\]</li></ol></blockquote><br />Later on p. 318, the book says:<br /><br /><blockquote class="tr_bq"><b><i>"...there is no new or special logic for proving identities. Identities are equations and all the logic that was discussed with equation-solving applies to them."</i></b></blockquote><br />Yes, that whole "math isn't just a bunch of arbitrary rules" thing applies here too.<br /><br />The second important point, which you may have noticed while looking at the statement of the Reversible Steps Theorem, is that the implied <i>domain</i> of an identity matters a great deal. When you're proving a trig identity, you are trying to establish that it is true for all inputs that are in the <i>domain</i> of both sides. Most textbooks at least pay lip service to this fact, even though they don't follow it to its logical conclusion.<br /><br />To illustrate why domain is so important, consider this example:<br /><br />\[\dfrac{\cos{x}}{1-\sin{x}} = \dfrac{1+\sin{x}}{\cos{x}}\]<br />To verify this identity, I'm going to do something that may give you a visceral reaction: I'm going to "cross-multiply". Or, more properly, I'm going to multiply both sides by the expression \((1 - \sin x)\cos x\). I claim that this is a perfectly valid step to take, and what's more, it makes the rest of the proof downright easy by reducing to everyone's favorite Pythagorean identity:<br /><br />\[<br />\begin{align*}<br />(\cos{x})(\cos{x}) &= (1+\sin{x})(1-\sin{x})\\<br />\cos^2{x} &= 1-\sin^2{x}\\<br />\sin^2{x} + \cos^2{x} &= 1 \quad\blacksquare<br />\end{align*}<br />\]<br />"But wait," you ask, "what if \(x=\pi/2\)? Then you're multiplying both sides by zero, and that's certainly not reversible!"<br /><br />True. But if \(x=\pi/2\), then the denominators of both sides of the equation are zero, so the identity isn't even true in the first place. For any value of \(x\) that does <i>not</i> yield a zero in either denominator, though, multiplying both sides of an equation by that value is a reversible operation and therefore completely valid.<br /><br />Now, this isn't to say that multiplying both sides of an equation by a function can't lead to problems --- for example, if \(h(x)=0\) (as in the zero <i>function</i>), then \(f(x)\cdot h(x)=g(x)\cdot h(x)\) no matter what. This can even lead to problems in more subtle cases: suppose \(f\) and \(g\) are equal everywhere but a single point \(a\); for example, perhaps \(f(a)=1\) and \(g(a)=2\). If it just so happens that \(h(a)=0\), then \(f\cdot h\) and \(g\cdot h\) will be equal <i>as functions</i>, even though \(f\) and \(g\) are not themselves equal.<br /><br />The real issue here can be explained via a quick foray into higher mathematics. Functions form what's called a <u style="font-weight: bold;">ring</u> -- basically meaning you can add, subtract, and multiply them, and these operations have all the nice properties we'd expect. But being able to preserve that if-and-only-if relationship when multiplying a function by both sides of an equation requires a special kind of ring called an <u style="font-weight: bold;">integral domain</u>, which means that it's impossible to multiply two nonzero functions together and get a zero function.<br /><br />Unfortunately, functions in general don't form an integral domain --- not even continuous functions, or differentiable functions, or even <i>infinitely</i> differentiable functions do! But if we move up to the <i>complex numbers</i> (where everything works better!), then the set of <u style="font-weight: bold;">analytic</u> functions --- functions that can be written as power series (infinite polynomials) on an open domain --- <i>is </i>an integral domain. And most of the functions that precalculus students encounter generally turn out to be analytic**: polynomial, rational, exponential, logarithmic, trigonometric, and even inverse trigonometric. This means that when proving trigonometric identities, multiplying both sides by the same function is a "safe" operation.<br /><br />So in sum, when proving trigonometric identities, as long as you're careful to only use reversible steps (what a great time to spiral back to one-to-one functions, by the way!), you are welcome to apply all the same algebraic operations that you would when solving equations, and the chain of equalities you establish will prove the identity. Even "cross-multiplying" is fair game, because any input that would make the denominator zero would invalidate the identity anyway.*** Since trigonometric functions are generally "safe" (analytic), we're guaranteed to never run into any issues.<br /><br />Now, none of this is to say that there isn't intrinsic merit to learning how to prove an identity by working from one side to the other. Algebraic "tricks" --- like multiplying by an expression over itself (\(1\) in disguise!) to conveniently simplify certain expressions --- are important tools for students to have under their belts, especially when they encounter limits and integrals next year in calculus.<br /><br />What we need to do, then, is encourage our students to come up with multiple solution methods, and perhaps present working from one side to the other as an added challenge to build their mathematical muscles. And if students are going to work on both sides of an equation at once, then we need to hold them to high standards and make them <i>explicitly</i> state in their proofs that all the steps they have taken are reversible! If they're unsure on whether or not a step is valid, have them investigate it until they're convinced one way or the other.<br /><br />If we're artificially limiting our students by claiming that only one solution method is correct, we're sending the wrong message about what mathematics really is. Instead, celebrating and cultivating our students' creativity is the best way to prepare them for problem-solving in the real world.<br /><br />--<br /><br /><i>* Rather, I <b>would</b> say it baffles me, but actually I'm quite used to seeing textbooks treat mathematical topics as disparate and unconnected, like how a number of Precalculus books teach vectors in one chapter and matrices in the next, yet never once mentione how they are so beautifully tied together via transformations.</i><br /><i><br /></i><i>** Except perhaps at a few points. The more correct term for rational functions and certain trigonometric functions is actually <u style="font-weight: bold;">meromorphic</u>, which describes functions that are analytic everywhere except a discrete set of points, called the <u style="font-weight: bold;">poles</u> of the function, where the function blows up to infinity because of division by zero.</i><br /><br /><i>*** If you extend the domains of the trig functions to allow for division by zero, you do need to be more careful. <a href="https://1dividedby0.com/">Not because there's anything intrinsically wrong with dividing by zero</a>, but because \(0\cdot\infty\) is an indeterminate expression and causes problems that algebra simply can't handle.</i></div>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com1tag:blogger.com,1999:blog-3748625510348961342.post-53575738391182138112019-01-02T20:23:00.005-08:002021-02-06T16:51:01.821-08:00When Math is in Jeopardy!<div><i>What kind of message does it send that game shows so often get mathematics wrong?</i></div><div><i><br /></i></div><div><span><a name='more'></a></span></div><div style="font-style: italic;"><i><br /></i></div>(Forgive the somewhat dramatic title, but it was too good to pass u<i>p.)</i><br /><div><br /></div><div>I absolutely love game shows. It's always fun to imagine you're there on stage ... or at least to yell at the people on TV when they don't know a question that was just <i>so easy</i>!</div><div><br /></div><div>One of my favorite game shows, of course, is Jeopardy --- it has an air of intellectualism about it, with such an eclectic collection of topics. I don't even mind that I'm admittedly pretty terrible at it... I always find myself thinking, "Man, if they just had some <i style="font-weight: bold;">math</i> questions, I could knock those out of the park!"</div><div><br /></div><div>Well, a couple of days ago, as we were getting ready to celebrate the advent of 2019, I was elated to see that there was in fact a math question! (Or math answer, rather. You know what, I'm just going to call them "clue" and "response" to avoid confusion.)</div><div><br /></div><div>The category was "Ends in 'ITE'", and the <strike>question</strike> clue was as follows:</div><blockquote class="tr_bq"><b>"In math, when the number of elements in a set is countable, it's this type of set."</b></blockquote>As I was trying to think of what <strike>answer</strike> response would end in those three letters, a contestant buzzed in and said:<br /><div><blockquote class="tr_bq"><b>"What is 'finite'?"</b></blockquote>Alex Trebek notified the contestant that they were correct.<br /><br />Meanwhile, I was speechless because I knew they weren't.<br /><br />Well, at least not completely correct. See, in math, words have very specific definitions in order to describe very specific phenomena. In this case, the word "countable" is used in set theory to describe a set --- a collection of things --- that can be put into one-to-one correspondence with a subset of the natural numbers, \(\mathbb{N}=\{0,1,2,3,\ldots\}\).* So, while finite sets are indeed countable, there are also infinite countable sets --- the integers, the even numbers, and the rational numbers are all well-known examples. (That last one still amazes me --- in some sense, there are exactly as many fractions as there are whole numbers, even though it seems like the former should outnumber the latter!)<br /><br />That means that the statement "When the number of elements in a set is countable, it's a finite set" is actually incorrect.<br /><br />Naturally, I took to the internet to voice my displeasure with how my favorite subject was represented on national television. When talking with a friend of mine (who knows more about game shows than I ever will), I learned that there had been two other cases recently where math-centric Jeopardy questions had issues with them in the past couple of months.<br /><ul><li>In one case, the clue was, <b>"If \(x^2+2=18\), then \(x\) equals this."</b> The response that was judged to be correct was <b>"What is </b>'<b>\(4\)'?"</b>. Any algebra teacher reading this right now is shaking their head, because that answer won't get you full credit on any test. There are two possible values of \(x\): \(4\) and \(-4\).</li><li>In another case, the response to the clue was supposed to be <b>"What is the commutative property?"</b>. Nobody got it correct (which makes me sad in itself), but when Alex Trebek read the correct response out loud, he said it as "COM-myoo-TAY-tive" instead of "com-MYOO-tuh-TIVE".</li></ul>Now, in isolation, any one of these would be only a minor annoyance. After all, there are plenty of clues on Jeopardy that have multiple possible correct responses, and contestants aren't expected to give all correct responses, but rather just one.<br /><br />But the fact that questions about mathematics appear so infrequently on the show compared to topics such as history, combined with the fact that these kinds of details are not attended to, seems to send a message that mathematics is considered to be not as important, not worth researching fully in the spirit of the subject.<br /><br />We already live in a culture in which any time I tell somebody I teach math, the inevitable response is "Oh, haha, I was never any good at math." Somehow people seem proud to admit and even proclaim this. I'm willing to wager (maybe even make this a true Daily Double) that those people would be much more reluctant to say something like "Oh, haha, I was never any good at reading." There's a pervasive attitude that mathematics is a torturous and frivolous subject, devoid of the <a href="https://medium.com/q-e-d/stop-selling-math-for-its-usefulness-d9143e80d78d">awe-inspiring beauty and sheer fun</a> that those who embrace the subject know it to have.<br /><br />With that said, I'd like to challenge the writers of Jeopardy --- and perhaps other game shows as well --- to make a conscious effort not only to ask more questions about mathematics, but to take care to do them well, perhaps even consulting one or more mathematicians to make sure the precision and nuance of the subject are properly represented. (I know math teachers who have come up with versions of game shows for their classes with only mathematically-oriented questions... the students love it!)<br /><br />It doesn't have to be something like "\(1\times 2\times 3\times 4\times 5\)"** either. There's such a rich amount of material to pull from --- why not ask questions about, say, fractals? Or famous mathematicians? Or even unsolved problems (and those who eventually solved them)? I would be giddy to see something like,<br /><blockquote class="tr_bq"><b>"Shot and killed in a duel when he was only twenty, this mathematician spent the last night of his life writing down what he'd discovered about quintic equations."</b></blockquote>Doesn't that make you want to find out what the story was, why he felt fifth-degree equations were so important that he just had to share it, knowing he would soon die? Mathematics is full of stories like this, and perhaps letting people know those stories exist, that there's more to math than doing arithmetic problems, might change how people view the subject.<br /><br />Of course, if winning prize money is what you like about game shows, there's always the million-dollar <a href="http://www.claymath.org/millennium-problems/millennium-prize-problems">Millennium Prize Problems</a>...<br /><br />--<br /><br /><i>* You might be thinking, "But zero isn't a natural number!" As it turns out, there's no real consensus on whether zero is considered a natural number. Some mathematicians choose to include it, while others don't. To some extent it depends on your field of study --- for example, number theorists may be more likely to disinclude zero because it doesn't play very nicely with things like prime factorizations, while computer scientists are used to counting from zero instead of one. Peano actually started off his axiomatization of the natural numbers with one being the starting point, but then changed his mind later and started with zero!</i><br /><i><br /></i><i>** This was actually a clue on a Kid's Jeopardy episode, under the category "Non-Common Core Math". That will eventually be the impetus for another blog post in the future on the way we currently view mathematics teaching.</i></div>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com1tag:blogger.com,1999:blog-3748625510348961342.post-21107092009774900532017-10-21T17:23:00.004-07:002021-02-06T16:51:02.504-08:00Discreet Discrete Calculus<div><i>A presentation I gave at the Georgia Mathematics Conference, about slyly introducing precalculus students to a discrete version of calculus.</i></div><div><br /></div><span><a name='more'></a></span><div><br /></div><div><br /></div>Over the past week, I went to the Georgia Mathematics Conference (GMC) at Rock Eagle, held by the Georgia Council of Teachers of Mathematics (GCTM). The GMC is one of the events I look forward to most every year --- tons of math educators and advocates sharing lessons, techniques, and ideas about how to best teach math to students from kindergarten through college. I always enjoy sharing my own perspectives as well (even when they do get a bit bizarre!)<br /><br />This time, I got to share the results of a lesson that I guinea-pigged on my Honors Precalculus class last year, where they explored the relationships between polynomial sequences, common differences, and partial sums. The presentation from the GMC uses the techniques we looked at to develop the formula for the sum of the first \(n\) perfect squares:<br /><br />\[1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}\]<br /><br />At the GMC, we did go a bit further than my class did --- they didn't do the full development of discrete calculus --- but some knowledge of where the ideas lead is never a bad thing, and a different class at a different school may even be able to go further!<br /><br /><b><a href="https://drive.google.com/open?id=0B-uVGGkZosoPM0kwMllTVzJPQmc">Here is the PowerPoint from the presentation.</a></b> If you find it useful, or have any questions, please don't hesitate to leave a comment!<br /><br /><i>(I recommend downloading the PPTX file and viewing the slide show in PowerPoint, rather than using Google's online viewer.)</i>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com1tag:blogger.com,1999:blog-3748625510348961342.post-14626193589859870922017-09-15T14:12:00.005-07:002021-02-06T19:22:11.995-08:00I'm Done with Two Column Proofs.<div><i>Statement: They're terrible.</i></div><div><i>Reason: They don't help students reason — if anything, they just get in the way.</i></div><div><br /></div><span><a name='more'></a></span><div><br /></div><div>Wow.</div><br />It really has been a while since I've posted here, hasn't it?<br /><br />But I suppose having a mini-crisis in my geometry class that has forced me to reject our textbook's philosophy on what "proofs" should be is a good enough reason to resurrect this blog. I finally have the time this year, and I feel the need to share my probably overly opinionated beliefs about math education.<br /><br />This is the first year I have tried to integrate proofs into our school's geometry curriculum across the board. (In the past, proofs were only discussed in honors classes, but I felt somehow that reasoning through how you knew something was true was important for everyone.) I was trying to justify the two-column format, as much as I hate it, as a way to "scaffold" student thinking --- yay educational buzzwords! But when I actually did it, it got exactly the reaction that I knew it would --- it just served to overly obfuscate the material and utterly drain the life out of it. I realized I should have stuck to my guns and listened to the likes of <a href="https://www.maa.org/external_archive/devlin/LockhartsLament.pdf">Paul Lockhart</a> and <a href="https://mathwithbaddrawings.com/2013/10/16/two-column-proofs-that-two-column-proofs-are-terrible/">Ben Orlin</a>.<br /><br />So, after some reflection and course correction, here's the email I just sent my students.<br /><br />---<br /><blockquote class="tr_bq">Hello mathematicians. I have a rather bizarre request. </blockquote><blockquote class="tr_bq">Don't do your geometry homework this weekend. </blockquote><blockquote class="tr_bq">Yes, you read that right. Don't. </blockquote><blockquote class="tr_bq">Let me explain.</blockquote><blockquote class="tr_bq">We've spent the past couple of days looking at "proofs" in geometry. The reason I say "proofs" in quotations is that, in all honesty, I don't believe the two-column proofs that our book does are are all that useful. I have actually been long opposed to them, but against my better judgment, decided to give them a shot anyway and make them sound reasonable. But you know what they say ... if you put lipstick on a chazir*, it's still a chazir. (They do say that, right?) The thing is, that style of proof just ends up sounding like an overly repetitive magical incantation rather than an actual logical argument --- as some of you pointed out in class today. I truly do value that honesty, by the way, and I hope you continue to be that honest with me. </blockquote><blockquote class="tr_bq">Here is what I actually will expect of you going forward. It's quite simple: </blockquote><blockquote class="tr_bq">I expect you to be able to tell me how you know something is true, and back it up with evidence. </blockquote><blockquote class="tr_bq">That's it. </blockquote><blockquote class="tr_bq">It may be a big-picture kind of question, or it may be telling me how we get from step A to step B, but when it really comes down to it, it's all just "here's why I know this is true, based on this evidence". It doesn't have to be some stilted-sounding name like the "Congruent Supplements Theorem" either --- just explain it in your own words. That doesn't mean that any explanation is correct --- it still has to be valid mathematical reasoning. You can't tell me that two segments on a page are congruent because they're drawn in the same color, or something silly like that. But it doesn't have to be in some prescribed way --- just as long as you show me you really do understand it. </blockquote><blockquote class="tr_bq">With that in mind, by the way, I'm also not going to be giving you a quiz on Monday, either. Instead, we're going to focus on how to make arguments that are a lot more convincing than just saying the same thing in different words. I think you'll find that Monday's class will make a lot more sense than the past few classes combined. </blockquote><blockquote class="tr_bq">So, relax, take a much-deserved Shabbat, and when we come back, I hope to invite you to see geometry the way I see it --- not as a set of arbitrary rules, but as something both logical and beautiful. </blockquote><blockquote class="tr_bq">Shabbat Shalom.</blockquote><br /><i>* I teach at a Jewish private school. "Chazir" is Hebrew for "pig", which has the added bonus of being non-Kosher. Two-column proofs are treif... at least in the context of introductory geometry.</i><br /><i><br /></i><i>P.S. I am not saying that two-column proofs NEVER have a place in mathematics. I am merely saying that introductory geometry, when kids are still getting used to much of geometry as a subject, is not the proper place to introduce the building of an axiomatic system. Save that for later courses for the students who choose to become STEM majors.</i>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com2tag:blogger.com,1999:blog-3748625510348961342.post-69831562498495885932015-08-14T20:22:00.004-07:002021-02-06T19:22:12.825-08:00A Real-Life Paradox: The Banach-Tarski Burrito<i>Who knew the Axiom of Choice could help me decide whether to get guacamole for an extra $1.95?</i><br /><div><br /></div><span><a name='more'></a></span><div><br /></div><div><div>A couple of weeks ago, the popular YouTube channel <a href="https://www.youtube.com/user/Vsauce">Vsauce</a> released a video that tackles what it details as “one of the strangest theorems in modern mathematics”: the <a href="https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox">Banach-Tarski Paradox</a>. In the video, Michael Stevens explains how a single sphere can be decomposed into peculiar-looking sets, after which those sets can be recombined to form two spheres, each perfectly identical to the original in every way. If you haven’t had a chance to watch the video, go ahead and do so here:</div><div><br /><div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen="" class="YOUTUBE-iframe-video" data-thumbnail-src="https://i.ytimg.com/vi/s86-Z-CbaHA/0.jpg" frameborder="0" height="266" src="https://www.youtube.com/embed/s86-Z-CbaHA?feature=player_embedded" width="320"></iframe></div><br />Although this seems like a purely theoretical abstraction of mathematics, the video leaves us wondering if perhaps there could be a real-world application of such a bizarre phenomenon. Stevens asks, “is [the Banach-Tarski paradox] a place where math and physics separate? We still don’t know … The Banach-Tarski Paradox could actually happen in our real world … some scientists think it may be physically valid.”<br /><br />Well, my friends, I would like to make the bold claim that I have indeed discovered a physical manifestation of this paradox.<br /><br />And it happened a few years ago at my local <a href="http://www.chipotle.com/">Chipotle</a>.<br /><br />Let me start off by saying this: I <b><i>love</i> </b>Chipotle. It’s a particularly good day for me when I walk in and get my burrito with brown rice, fajita veggies, steak, hot salsa, cheese, pico de gallo, corn, sour cream, guacamole (yes I know it’s extra, just put it on my burrito already!), and a bit of lettuce. No chips, Coke, and about a half hour later I’m one happily stuffed math teacher.<br /><br />The only thing that I don’t like about Chipotle is that the construction of said burritos often ends up failing at the most crucial step – the rolling into one coherent, tasty package. Given the sheer amount of food that gets crammed into a Chipotle burrito, it’s unsurprising that they eventually lose their structural integrity and burst, somewhat defeating the purpose of ordering a burrito in the first place.<br /><br />If you have ever felt the pain of seeing your glorious Mexican monstrosity explode with toppings like something out of an Alien movie because of an unlucky burrito-roller, you have probably been offered the opportunity to “double-wrap” your burrito for no extra charge, giving it an extra layer of tortilla to ensure the safe deliverance of guacamole-and-assorted-other-ingredients into your hungry maw.<br /><br />Now, being a mathematically-minded kind of guy, I asked the employee who made me this generous offer:<br /><br />“Well, could I just get my ingredients split between two tortillas instead?”<br /><br />The destroyer-of-burritos gave that look that you always get from anybody who works at a business that bandies about words like “company policy” when they realize they have to deny a customer’s request even in the face of logic, and said:<br /><br />“If you do that, we’ll have to charge you for two burritos.”<br /><br />I was dumbfounded.<br /><br />“Wait … so you’re saying that if you put a second tortilla around my burrito, you’ll charge me for one burrito, but if you rearrange the <i>exact same ingredients</i>, you’ll charge me for two?”<br /><br />“Yes sir – company policy.”<br /><br />Utterly defeated, I begrudgingly accepted the offer to give my burrito its extra layer of protection, doing my best to smile at the girl who probably knew as well as I did the sheer absurdity of the words that had come out of her mouth. I paid the cashier, let out an audible “oof” as I lifted the noticeably heavy paper bag covered with trendy lettering, and exited the store.<br /><br />When I arrived home, I took what looked like an aluminum foil-wrapped football out of the bag (which was a great source of amusement for my housemates), laid it out on the kitchen table, and decided to dismantle the burrito myself and arrange it into two much more manageable Mexican morsels. I wondered whether I should have done this juggling of ingredients right there at Chipotle, just to see whether the staff’s heads would explode. <br /><br />It was in that moment, with my head still throbbing from the madness of the entire experience, that I began to realize what had just happened. How <i>was</i> it possible that a given mass of food could cost one amount one moment and another amount the next? I immediately began to deconstruct my burrito, laying out the extra tortilla onto a plate and carefully making sure that precisely one-half of the ingredients – especially the guacamole – found their way into their new home. As I carefully re-wrapped both tortillas, my suspicions were confirmed. Sitting right in front of me were two delicious burritos, each identical in price to my original.<br /><br />I had discovered the Banach-Tarski Burrito.<span><!--more--></span><span><!--more--></span></div><span><!--more--></span></div>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com2tag:blogger.com,1999:blog-3748625510348961342.post-2272902034172907032015-04-24T17:47:00.004-07:002021-02-06T19:22:16.575-08:00A Radical New Look for Logarithms<div><i>What if we could invent a new notation for logarithms that was consistent with how we write exponents and radicals?</i></div><div><i><br /></i></div><div><span><a name='more'></a></span></div><i><div><i><br /></i></div>"A good notation has a subtlety and suggestiveness which at times make it almost seem like a live teacher."</i> — Bertrand Russell<br /><br /><i>"We could, of course, use any notation we want; do not laugh at notations; invent them, they are powerful. In fact, mathematics is, to a large extent, invention of better notations."</i> — Richard Feynman<br /><br />"<i>By relieving the brain of all unnecessary work, a good notation sets it free to concentrate on more advanced problems, and in effect increases the mental power of the race.</i>" — Alfred North Whitehead<br /><br /><br />Notation is perhaps one of the most important aspects of mathematics. The right choice of notation can make a concept clear as day; the wrong choice can make extracting its meaning hopeless. Of course, one great thing about notation is that even if there's a poor choice of notation out there (such as \(\left[x\right]\) or \(\pi\)), often someone comes along and creates a better one (such as \(\lfloor x\rfloor\) for the <a href="http://www.solidangl.es/2014/09/throw-outdated-notation-to-floor.html">floor function</a> or multiples of <a href="http://tauday.com/tau-manifesto">tau</a>, \(\tau\approx 6.28318\), for radian measure of angles).<br /><br />Which brings me to one such poor choice of notation, one that I believe needs fixing: the rather asymmetrical notation of powers, roots, and logarithms.<br /><br />Here we have three very closely related concepts — both roots and logarithms are ways to invert exponentiation, the former returning the base and the latter returning the exponent. And yet their notation couldn't be more different:<br />\[2^3=8\\<br />\sqrt[3]{8}=2\\<br />\log_2{8}=3\]This always struck me as annoyingly inelegant. Wouldn't it be nice if these notations bore at least some resemblance to each other?<br /><br />After giving it some thought, I believe I have found a possible solution. As an alternative to writing \(\log_2 {8}\), I propose the following notation:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-vyIGboQ3DkY/VTrVNLnYilI/AAAAAAAACmA/eAkvCZAFl3Y/s1600/new%2Blog%2Bnotation.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-vyIGboQ3DkY/VTrVNLnYilI/AAAAAAAACmA/eAkvCZAFl3Y/s1600/new%2Blog%2Bnotation.png" /></a></div><br />This notation makes use of a reflected radical symbol, such that the base of the logarithm is written in a similar manner to the index of a radical but below the "point" (the pointy part of the radical symbol), and the argument of the logarithm is written "inside". The use of this notation has a number of advantages:<br /><br /><ol><li>The symmetry between the normal radical for roots and the reflected radical for logarithms highlights both their similarities and their differences — each one "undoes" an exponential expression, but each one gives a different part of the expression (the base and the exponent, respectively.)</li><li>The radical symbol can be looked at as a modified lowercase letter "r". (This may actually be the origin of the symbol, where the "r" stands for <i>radix</i>, the Latin word for "root".) In a similar way, the new symbol for logarithms resembles a capital "L".</li><li>The placement of the "small number" and the "point" can take on a secondary spatial meaning: </li><ul><li>The "small number" represents a piece of information we <u>know</u> about an exponential expression, and its placement indicates <u>which part</u> we know.</li><ul><li>For a root, the "small number" is on top, so <i>we know the <u>exponent</u></i>.</li><li>For a logarithm, the "small number" is on bottom, so <i>we know the <u>base</u></i>.</li></ul><li>The symbol seems to "point" to the piece of information that we are <u>looking for</u>.</li><ul><li>For a root, the "point" is pointing downward, so <i>we are looking for the <u>base</u></i>.</li><li>For a logarithm, the "point" is pointing upward, so <i>we are looking for the <u>exponent</u></i>.</li></ul><li>Looking at the image above, the new notation seems to say "We know the <i>base</i> is 2, so <i>what's the exponent</i> that will get us to 8?"</li><li>Similarly, the expression \(\sqrt[3]{8}\) now can be interpreted as saying "We know the <i>exponent</i> is 3, so <i>what's the base</i> that will get us to 8?"</li></ul></ol><div><div><br /></div><div>This notation would obviously not make much of a difference for seasoned mathematicians who are perfectly comfortable with the \(\log\) and \(\ln\) functions. But from a pedagogical standpoint, the reflected radical, with its multi-layered meaning and auto-mnemonic properties, could help students become more comfortable with a concept that many look at as just meaningless manipulation of symbols.<br /><br />When I first came up with this reflected-radical notation, I had originally imagined that it should <i>replace</i> the current notation. However, after some feedback from various people and some further consideration, I think a better course of action would be to have this notation be used <i>alongside</i> the current notation, much in the way that we have multiple notations for other concepts in math (such as the many ways to write derivatives). However, I would suggest that, if it were to become commonplace*, this notation would be best to use when first introducing the concept in schools. The current notation isn't <i>wrong</i> per se — it's just not very evocative of the underlying concept. Anything that can better elucidate that concept can't be a bad thing when it comes to students learning mathematics!</div><div><br /></div><div>It may seem like a <b>radical </b>idea.</div><div>But it's a <b>logical</b> one.<br /><br /><br /><i>* Of course, for this notation to become commonplace, somebody would need to figure out how to replicate it in LaTeX. Any takers?</i></div></div>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com12tag:blogger.com,1999:blog-3748625510348961342.post-68697028658403006552014-09-15T20:31:00.004-07:002021-02-06T19:22:17.660-08:00Infinity is my favorite number.<i>Yes, you read that right.</i><div><br /><div><span><a name='more'></a></span></div><br />I've recently been embroiled in a lovely debate on <a href="http://numberphile.com/">Numberphile</a>'s video, "Infinity is bigger than you think", in which Dr. James Grime starts off: "We're going to break a rule. We're breaking one of the rules of Numberphile. We're talking about something that isn't a number. We're going to talk about infinity."<br /><br /><div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/elvOZm0d4H0?feature=player_embedded' frameborder='0' /></div><br /><br />I, too, was a longtime believer of what high school students all over are told: "Infinity is not a number; infinity is a concept." As my studies of mathematics progressed, however, I began to see that perhaps the things I had always taken for granted were not as black-and-white as they had seemed. There was a lot more nuance to mathematics than I had ever realized, and learning those nuances opened up an entire new level of understanding, unlocking all sorts of links between concepts that had previously seemed worlds apart. So it's no wonder that "Infinity Is Not A Number" (which I will occasionally abbreviate as "IINAN") was one of the first claims to which I took a fine-tooth comb. What I learned changed my stance on infinity and firmly cemented it as my favorite number - not just concept, but honest-to-god <i>number</i>.*<br /><br />The most common argument made IINAN proponents involves the curious property that \(\infty +1=\infty\). This, they say, leads to all sorts of contradictions, because all one has to do is simply subtract \(\infty\) from both sides:<br />\[<br />\infty+1=\infty\\<br />\underline{-\infty\ \ \ \ \ \ \ \ -\infty}\ \ \\<br />\ \ \ \ \ \ 1=0\]Oh no! We know that the statement \(1=0\) is obviously false, so there must be a false assumption somewhere. Many IINAN defenders claim that the false assumption was that we tried to treat \(\infty\) as a number. But that's not actually where the problem with infinity lies.<br /><br />The problem is that we tried to do algebra with it.<br /><br />For mathematicians, the most convenient place to do algebra is in a structure called a <b>field</b>. If you're already familiar with what a field is, great, but if not, you can think of a field as a number system in which the age-old operations of addition, subtraction, multiplication, and division <span face="arial, sans-serif" style="color: #545454; font-size: x-small;"><span style="line-height: 18.2px;">—</span></span> the four operations that my father often notes are the only ones he ever needs when I talk about the kinds of math I teach <span face="arial, sans-serif" style="color: #545454; font-size: x-small;"><span style="line-height: 18.2px;">—</span></span> work exactly as we'd like them to. The fields with which we are most familiar are the rational numbers (\(\mathbb{Q}\)), the badly-named so-called "real" numbers (\(\mathbb{R}\)), and often the complex numbers (\(\mathbb{C}\)). One basic thing about a field is that the <b>subtraction property of equality</b> holds: For any numbers \(a\), \(b\), and \(c\) in our field, if \(a=b\), then \(a-c=b-c\).<br /><br />What about \(\infty\) though? When we attempted to use \(\infty\) in an algebra problem, we got back complete garbage. And we know that the subtraction property of equality should hold for <span style="font-style: italic;">any</span> numbers in a field. What this means, then, is that \(\infty\) <i>is not part of that field</i> (or any field as far as I'm aware). So, when someone says "Infinity is not a number", what they <i>really</i> mean is <b>"Infinity is not a </b><i style="font-weight: bold;">real </i><b>number."</b><i style="font-weight: bold;"> </i>(It's not a complex number, either, for that matter.) It doesn't follow the same rules that the real numbers do.<br /><br />But that doesn't mean it's not a number at all.<br /><br />We've seen this sort of thing happen before. The Greek mathematician Diophantus, when faced the equation \(4x+20=0\), called its solution of \(-5\) "absurd" — yet now students learn about negative numbers as early as elementary school, and we barely blink an eye at their use in everyday life. Square roots of negative numbers seemed equally preposterous to the Italian mathematician Gerolamo Cardano, and the French mathematician René Descartes called them "imaginary", a term that we're unfortunately stuck with today. But imaginary numbers — and the complex numbers we build from them — are a vital part of physics, from alternating currents to quantum mechanics.<br /><br />So what makes infinity any different from \(i\)?<br /><br />Sure, it seems bizarre that a number plus one could equal itself. But it's equally bizarre that the square of a number could be negative. And sure, we can get a contradiction if we do certain things to infinity. But that happens with \(i\) as well! If we attempt to use the identity \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\), we can arrive at a similar contradiction:<br />\[\sqrt{-1}\cdot\sqrt{-1}=\sqrt{-1\cdot-1}\\<br />\ \ \ i\cdot i=\sqrt{1}\\<br />-1=1\ \]When this equation fails, you don't see mathematicians clamoring that "\(i\) isn't a number"! Instead, the response is that the original equation doesn't work like we thought it did when we extend our real number system to include the complex numbers — instead, the square root function takes on a new life as a multi-valued function. There's that nuance again! For the same reason, infinity makes us look closer at something as simple as subtraction, at which point we find that \(\infty-\infty\) is an <b>indeterminate form</b>, something that we need the tools of calculus to properly deal with.<br /><br />The truth is, mathematicians have been treating \(\infty\) as a number** for quite some time now.<br /><br />In <b>real analysis</b>, which was developed to give the techniques of calculus a rigorous footing, points labelled \(+\infty\) and \(-\infty\) can be added to either end of the real number line to give what we call the <b>extended real number line</b>, often denoted \(\overline{\mathbb{R}}\) or \(\left[-\infty,+\infty\right]\). The extended real number line is useful in describing concepts in measure theory and integration, and it has algebraic rules of its own, though analysts are still careful to mention that these two extra points are not <i>real</i> numbers. What's more, the extended real line is not a <i>field</i>, because it doesn't satisfy all the nice properties that a field does. (But that just makes us appreciate working in a field that much more!)<br /><br /><b>Projective geometry</b> gives us a different sort of infinity, what I like to call an "unsigned infinity", one that is obtained by letting \(-\infty\) and \(+\infty\) overlap and creating what is known as the <b>real projective line</b>. And <b>complex analysis</b>, which extends calculus to the complex plane,<b> </b>takes it even further, letting <i>all</i> the different infinities in all directions overlap to create a sort of "complex infinity", sometimes written \(\tilde{\infty}\), sitting atop the <b>Riemann sphere</b>. What I particularly like about these projective infinities is that, using them, <i>you can actually divide by zero!</i> ***<br /><br />So, since there are actually a number of different kinds of infinity that can be referred to, I would say that, more specifically, <i>complex infinity is my favorite number</i>.<br /><br />The tough thing about this situation is that the concept of "number" is a very difficult one to precisely and universally define — similar to how linguists still struggle to come up with a universal definition of "word". By trying to come up with such a description, you end up either including things that you don't want to be numbers (such as matrices) or excluding things that you do want to be numbers (such as complex numbers). The best we can really do is keep an open mind about what a "number" is.<br /><br />After all, there's infinitely many of them already — so there's bound to be new ones we haven't seen yet sooner or later.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-GCL0xZc_dCI/VBeqZtjo_nI/AAAAAAAAATM/HSzchObC2l4/s1600/Infinity.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="179" src="http://3.bp.blogspot.com/-GCL0xZc_dCI/VBeqZtjo_nI/AAAAAAAAATM/HSzchObC2l4/s1600/Infinity.jpg" width="320" /></a></div><br /><br />∎<br /><br />*<i> I'm not saying that infinity isn't a concept. When it really comes down to it, every number is a concept. That's the beauty of having abstracted the number "two" as an adjective, as in "two sheep", to "two" as a noun.</i><br /><i><br /></i>**<i> There's an argument to be made that <b>treating</b> something like a number doesn't mean it <b>is</b> a number. But at some point, the semantic distinction between these two becomes somewhat blurred.</i><br /><i><br /></i>***<i> Don't worry, I'll make a post about how to legitimately divide by zero in the near future!</i><span><!--more--></span><span><!--more--></span><span><!--more--></span></div>Bill Shillitohttp://www.blogger.com/profile/17774101901445053590noreply@blogger.com4