Taylor Series Section 4. Examples

<body bgcolor="#ffffff"> <hr> <center> <table border="2"> <tr valign="top"> <td><a href="node3.shtml"><img height="44" width="44" src="/UCESIcons/tools/previous.gif"></a></td> <td><a href="taylor_series.html"><img height="44" width="44" src="/UCESIcons/tools/up.gif"></a></td> <td><a href="node5.shtml"><img height="44" width="44" src="/UCESIcons/tools/next.gif"></a></td> <td><a href="contents-node4.html"><img height="44" width="44" src="/UCESIcons/tools/toc.gif"></a></td> <td><a href="copyright.html"><img height="44" width="44" src="/UCESIcons/tools/copyright.gif"></a></td> <td><a href="mailto:uces@krellinst.org"><img height="44" width="44" src="/UCESIcons/tools/feedback.gif"></a></td> </tr> </table> </center> <hr> <!DOCTYPE HTML PUBLIC "-//IETF//DTD HTML 2.2//EN">  <HTML> <HEAD> <TITLE>Section 4. Examples</TITLE> <BASE TARGET="_parent"> <META NAME="description" CONTENT="Section 4. Examples"> <META NAME="keywords" CONTENT="taylor_series"> <META NAME="resource-type" CONTENT="document"> <META NAME="distribution" CONTENT="global"> <LINK REL=STYLESHEET HREF="taylor_series.css"> </HEAD> <BODY LANG="EN" BGCOLOR="#FFFFFF"> <P> <H1>Section 4.   Examples</H1> <P> <B>Example 1:</B> Let's begin with an example motivated by our first point in Section 1. Namely, let's determine the Taylor series for the function <IMG WIDTH=83 HEIGHT=23 ALIGN=MIDDLE ALT="f(x)=sin(x)" SRC="img23.gif"> . In order to do this, we must choose a value of <I>c</I>. To start, let's find the Taylor series for <IMG WIDTH=29 HEIGHT=12 ALIGN=BOTTOM ALT="sin(x)" SRC="img1.gif"> about <nobr><I>c</I>=0.</nobr> <P> (A bit of terminology is in order here. A Taylor series about <nobr><I>c</I>=0</nobr> is also known as a <B>Maclaurin series</B>.) <P> The first task we must accomplish is to find the derivatives of <IMG WIDTH=29 HEIGHT=12 ALIGN=BOTTOM ALT="sin(x)" SRC="img1.gif"> and then evaluate these at <nobr><I>c</I>=0.</nobr> (This information is needed as part of the coefficient of each term in the Taylor series.) <P> We compile this information in the following table: <P> <center> <TABLE COLS=3 cellpadding=3 BORDER WIDTH="200" FRAME=BOX RULES=GROUPS> <COLGROUP><COL ALIGN=RIGHT> <COLGROUP><COL ALIGN=RIGHT> <COLGROUP><COL ALIGN=RIGHT> <TBODY> <TR> <TD VALIGN=Bottom ALIGN=center><I>n</I></TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=47 HEIGHT=19 ALIGN=MIDDLE ALT="f^(n) (x)" SRC="img9.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=45 HEIGHT=19 ALIGN=MIDDLE ALT="f^(n) (c)" SRC="img24a.gif"></TD> </TR> </TBODY> <TBODY> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>0</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=29 HEIGHT=12 ALIGN=BOTTOM ALT="sin(x)" SRC="img1.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>0</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>1</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=31 HEIGHT=7 ALIGN=BOTTOM ALT="cos(x)" SRC="img25.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>1</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>2</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=45 HEIGHT=22 ALIGN=MIDDLE ALT="-sin(x)" SRC="img26.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>0</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>3</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=46 HEIGHT=13 ALIGN=MIDDLE ALT="-cos(x)" SRC="img27.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>-1</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>4</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=29 HEIGHT=12 ALIGN=BOTTOM ALT="sin(x)" SRC="img1.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>0</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>5</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=31 HEIGHT=7 ALIGN=BOTTOM ALT="cos(x)" SRC="img25.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>1</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>6</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=45 HEIGHT=22 ALIGN=MIDDLE ALT="-sin(x)" SRC="img26.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>0</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>7</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=46 HEIGHT=13 ALIGN=MIDDLE ALT="-cos(x)" SRC="img27.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>-1</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>8</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=29 HEIGHT=12 ALIGN=BOTTOM ALT="sin(x)" SRC="img1.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>0</TD> </TR> </TBODY> </TABLE> </center> <P> Thanks to the cyclical nature of the derivatives of the sine function, we see a pattern emerging in the table. We really need not build any other derivatives because of this pattern. <P> From our definition of Taylor series above, we now see that <P> <IMG WIDTH=576 HEIGHT=124 ALIGN=BOTTOM ALT="[Taylor series for sin(x)about x=0]" SRC="img28.gif"> <P> If we now wanted to approximate <IMG WIDTH=38 HEIGHT=23 ALIGN=MIDDLE ALT="sin(x)" SRC="img29.gif"> for some value of <I>x</I> close to 0 (the <nobr><I>c</I>-value),</nobr> we could do so by simply using a Taylor polynomial derived by truncating the above Taylor series. For example, from the work above, we see that <P> <IMG WIDTH=325 HEIGHT=36 ALIGN=BOTTOM ALT="P_5(x) = x - x^3/3! + x^5/5!" SRC="img30.gif"> <P> in this case. We could approximate <IMG WIDTH=50 HEIGHT=23 ALIGN=MIDDLE ALT="sin(0.2)" SRC="img31.gif"> by simply evaluating <IMG WIDTH=48 HEIGHT=23 ALIGN=MIDDLE ALT="P_5(0.2)" SRC="img32.gif"> . Here <P> <IMG WIDTH=415 HEIGHT=36 ALIGN=BOTTOM ALT="P_5(0.2) = 0.198669333..." SRC="img33.gif"> <P> while my hand-held calculator yields <P> <IMG WIDTH=337 HEIGHT=15 ALIGN=BOTTOM ALT="sin(0.2) = 0.19866933..." SRC="img34.gif"> <P> That's not a bad approximation!!! <P> One last comment to close this out. Notice that this <IMG WIDTH=36 HEIGHT=17 ALIGN="middle" ALT="P_5(x)" SRC="img18.gif"> is NOT a good approximator of <IMG WIDTH=29 HEIGHT=12 ALIGN=BOTTOM ALT="sin(x)" SRC="img1.gif"> if the value of <I>x</I> chosen is NOT close to <nobr><I>c</I>=0.</nobr> As an example, notice that <IMG WIDTH=144 HEIGHT=18 ALIGN="middle" ALT="P_5(10) = 676.666..." SRC="img35.gif"> while <IMG WIDTH=183 HEIGHT=18 ALIGN="absmiddle" ALT="sin(10)= -0.54402111..." SRC="img36.gif"> . <P> <B>Example 2:</B> Determine the Taylor series for the function <IMG WIDTH=77 HEIGHT=25 ALIGN=MIDDLE ALT="f(x) = ln(x)" SRC="img37a.gif"> about <nobr><I>c</I>=1.</nobr> (Question: Why can't <nobr><I>c</I>=0?)</nobr> <P> We start by finding the derivatives in question. Again, we consolidate them in the table below: <P> <center> <TABLE COLS=3 cellpadding=3 BORDER WIDTH="200" FRAME=BOX RULES=GROUPS> <COLGROUP><COL ALIGN=RIGHT> <COLGROUP><COL ALIGN=RIGHT> <COLGROUP><COL ALIGN=RIGHT> <TBODY> <TR> <TD VALIGN=bottom ALIGN=center><I>n</I></TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=47 HEIGHT=19 ALIGN=MIDDLE ALT="f^(n) (x)" SRC="img9.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=45 HEIGHT=19 ALIGN=MIDDLE ALT="f^(n) (c)" SRC="img24.gif"></TD> </TR> </TBODY> <TBODY> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>0</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=23 HEIGHT=12 ALIGN=BOTTOM ALT="ln(x)" SRC="img38.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>0</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>1</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=24 HEIGHT=14 ALIGN=BOTTOM ALT="x^(-1)" SRC="img39.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>1</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>2</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=36 HEIGHT=14 ALIGN=BOTTOM ALT="-x^(-2)" SRC="img40.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>-1</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>3</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=32 HEIGHT=14 ALIGN=BOTTOM ALT="2*x^(-3)" SRC="img41.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>2</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>4</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=45 HEIGHT=14 ALIGN=BOTTOM ALT="-6*x^(-4)" SRC="img42.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>-6</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>5</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=39 HEIGHT=14 ALIGN=BOTTOM ALT="24*x^(-5)" SRC="img43.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>24</TD> </TR> <TR> <TD VALIGN=BASELINE ALIGN=RIGHT>6</TD> <TD VALIGN=BASELINE ALIGN=RIGHT> <IMG WIDTH=60 HEIGHT=14 ALIGN=BOTTOM ALT="120*x^(-6)" SRC="img44.gif"></TD> <TD VALIGN=BASELINE ALIGN=RIGHT>-120</TD> </TR> </TBODY> </TABLE> </center> <P> A pattern emerges. We note that <IMG WIDTH=214 HEIGHT=30 ALIGN=MIDDLE ALT="[formula for f^(n) (x)]" SRC="img45.gif"> , which allows us to write the Taylor series quickly. We have <P ALIGN=center> <IMG WIDTH=516 HEIGHT=36 ALIGN=BOTTOM ALT="[unsimplified formula for series for ln(x) about x=1]" SRC="img46.gif"> </P> <P> which can be simplified to <P> <IMG WIDTH=443 HEIGHT=36 ALIGN=BOTTOM ALT="[simplified form of previous formula]" SRC="img47.gif"> <P> Again, if we now want to approximate <IMG WIDTH=23 HEIGHT=12 ALIGN=BOTTOM ALT="ln(x)" SRC="img38.gif"> for some <I>x</I> near 1, we may quickly do so. Using <P> <IMG WIDTH=429 HEIGHT=36 ALIGN=BOTTOM ALT="[P_4(x)]" SRC="img48.gif"> <P> as our approximating function, we see that <IMG WIDTH=180 HEIGHT=23 ALIGN=MIDDLE ALT="P_4(1.5) = 0.401041666..." SRC="img49.gif"> while <IMG WIDTH=177 HEIGHT=23 ALIGN=MIDDLE ALT="ln(1.5) = 0.405465108..." SRC="img50.gif"> (according to my calculator). <P> At this point, you might rightly say, "That's not a very good approximation of <IMG WIDTH=45 HEIGHT=23 ALIGN=MIDDLE ALT="ln(1.5)" SRC="img51.gif"> ." To a certain extent, I would agree. What do we do then? In general, using a Taylor polynomial of higher degree should yield a better approximation. Indeed, if we now use <IMG WIDTH=36 HEIGHT=23 ALIGN=MIDDLE ALT="P_8(x)" SRC="img52.gif"> instead of <IMG WIDTH=36 HEIGHT=23 ALIGN=MIDDLE ALT="P_4(x)" SRC="img53.gif"> , we find that <IMG WIDTH=172 HEIGHT=23 ALIGN=MIDDLE ALT="P_8(1.5) = 0.40531529..." SRC="img54.gif">   This is a better approximation to the actual answer, but still leaves something to be desired. <P> Why aren't we "homing in" on the "exact" value of <IMG WIDTH=45 HEIGHT=23 ALIGN=MIDDLE ALT="ln(1.5)" SRC="img51.gif"> in a quicker fashion? In the jargon of infinite series, notice that the Taylor series for <IMG WIDTH=23 HEIGHT=12 ALIGN=BOTTOM ALT="ln(x)" SRC="img38.gif"> about <nobr><i>c</i>=1</nobr> is an <i>alternating series</i> which does converge (by the Alternating Series Test). However, it is converging quite slowly. Because of this, in order to get a good approximation of the full infinite series, one must use many, many terms. This is equivalent to needing a large degree for the Taylor polynomial. <P> <B>Example 3:</B> Consider an example similar to one of the motivating examples at the beginning of this document. Namely, let's approximate <P> <IMG WIDTH=305 HEIGHT=41 ALIGN=BOTTOM ALT="[I_1 = integral of e^(x^2) from 0 to 1]" SRC="img55.gif"> <P> As noted above, we cannot find this integral in closed form. However, we can approximate it with a tool such as Simpson's rule. Note that <IMG WIDTH=119 HEIGHT=22 ALIGN=MIDDLE ALT="I_1 ~ 1.462653625" SRC="img56.gif"> if we use Simpson's Rule with 20 subintervals. (This will give us something to compare to our approximation below using a Taylor polynomial.) <P> As a second approach, we want to approximate the integrand <IMG WIDTH=20 HEIGHT=17 ALIGN=BOTTOM ALT="e^(x^2)" SRC="img6.gif"> using a Taylor polynomial <IMG WIDTH=38 HEIGHT=23 ALIGN=MIDDLE ALT="P_n(x)" SRC="img7.gif"> about <nobr><I>c</I>=0</nobr> and then consider a new integral <P> <IMG WIDTH=315 HEIGHT=41 ALIGN=BOTTOM ALT="I_2 = integral of P_n(x) from 0 to 1" SRC="img57.gif"> <P> Since <IMG WIDTH=82 HEIGHT=32 ALIGN=MIDDLE ALT="e^(x^2) ~ P_n(x)" SRC="img58.gif"> , we would assume <IMG WIDTH=48 HEIGHT=22 ALIGN=MIDDLE ALT="I_1 ~ I_2" SRC="img59.gif"> . (Keep your fingers crossed!!) <P> The "difficult" part of this process is to find this Taylor polynomial. We begin by finding the Taylor series for the function <IMG WIDTH=13 HEIGHT=12 ALIGN=BOTTOM ALT="e^x" SRC="img2.gif"> and then transition to the Taylor series for <IMG WIDTH=20 HEIGHT=17 ALIGN=BOTTOM ALT="e^(x^2)" SRC="img6.gif"> , the function in question. Given <IMG WIDTH=67 HEIGHT=23 ALIGN=MIDDLE ALT="f(x) = e^x" SRC="img60.gif"> , we know that <IMG WIDTH=87 HEIGHT=30 ALIGN=MIDDLE ALT="f^(n) (x) = e^x" SRC="img61.gif"> as well for all positive integers <I>n</I>. Thus, the contribution of <IMG WIDTH=46 HEIGHT=30 ALIGN=MIDDLE ALT="f^(n) (0)" SRC="img62.gif"> to each of the terms in the Taylor series is just <IMG WIDTH=13 HEIGHT=14 ALIGN=BOTTOM ALT="e^0" SRC="img63.gif"> or 1. Therefore, we know that the Taylor series expansion of <IMG WIDTH=13 HEIGHT=12 ALIGN=BOTTOM ALT="e^x" SRC="img2.gif"> about <nobr><I>c</I>=0</nobr> is given by <P> <IMG WIDTH=389 HEIGHT=36 ALIGN=BOTTOM ALT="[series for e^x]" SRC="img64.gif"> <P> Now use this in finding the Taylor series for <IMG WIDTH=20 HEIGHT=17 ALIGN=BOTTOM ALT="e^(x^2)" SRC="img6.gif"> . If we replace every instance of <I>x</I> above by <nobr>x<sup><small>2</small></sup> ,</nobr>  we have <P> <IMG WIDTH=427 HEIGHT=36 ALIGN=BOTTOM ALT="[unsimplified series for e^(x^2)]" SRC="img66.gif"> <P> or <P> <IMG WIDTH=400 HEIGHT=36 ALIGN=BOTTOM ALT="[unsimplified series for e^(x^2)]" SRC="img67.gif"> <P> This is <B>much</B> easier than attempting to find the Taylor series of <IMG WIDTH=20 HEIGHT=17 ALIGN=BOTTOM ALT="e^(x^2)" SRC="img6.gif"> in a direct fashion. This is easily seen as one begins to compute the derivatives of <IMG WIDTH=20 HEIGHT=17 ALIGN=BOTTOM ALT="e^(x^2)" SRC="img6.gif"> . This becomes less than enjoyable rather quickly! <P> Now to the task at hand. Using <IMG WIDTH=43 HEIGHT=23 ALIGN=MIDDLE ALT="P_10(x)" SRC="img68.gif"> , the truncation of the Taylor series at the <IMG WIDTH=22 HEIGHT=14 ALIGN=BOTTOM ALT="x^10" SRC="img69.gif"> term, we see that <P> <IMG WIDTH=403 HEIGHT=41 ALIGN=BOTTOM ALT="[I_2 using P_10(x)]" SRC="img70.gif"> <P> This integral is easily computed, because we are now simply integrating polynomial terms! Going through the calculations, we see that <P> <IMG WIDTH=388 HEIGHT=34 ALIGN=BOTTOM ALT="[result of term-by-term integration for I_2]" SRC="img71.gif"> <P> or <IMG WIDTH=124 HEIGHT=22 ALIGN=MIDDLE ALT="I_2 = 1.462530063 ." SRC="img72.gif"> <P> As we close out this example, let's compare the approximations of <IMG WIDTH=11 HEIGHT=22 ALIGN=MIDDLE ALT="I_1" SRC="img73.gif"> . Recall that Simpson's Rule yielded an approximation of 1.462653625. Our Taylor series approximation gives 1.462530063. These agree at least at the thousandths digit and differ by only one unit in the ten-thousandths digit. <P> <B>Example 4:</B> As a final example, consider the function <P> <IMG WIDTH=299 HEIGHT=34 ALIGN=BOTTOM ALT="f(x) = 1/(1-x)" SRC="img74.gif"> <P> This function occurs quite frequently in infinite series discussions. Its most obvious connection to this area of mathematics is the geometric series. Recall that, for <span class="nobr">|<I>x</I>| < 1 ,</span> we know <P> <IMG WIDTH=359 HEIGHT=34 ALIGN=BOTTOM ALT="1 + x + x^2 + x^3 + ... = 1/(1-x)" SRC="img75.gif"> <P> Thus, thanks to work in geometric series, we already know the Maclaurin series representation of <P> <IMG WIDTH=271 HEIGHT=34 ALIGN=BOTTOM ALT="1/(1-x)" SRC="img76.gif"> <P> As a matter of practice, I leave it to the reader to actually “compute” this series using the derivatives and the definition of Maclaurin series. <P> <HR> <ADDRESS> <I>James A. Sellers<br> <a href="mailto:sellersj@math.psu.edu">sellersj@math.psu.edu</a></I> </ADDRESS> <P> </BODY> </HTML> <hr> <center> <table border="2"> <tr valign="top"> <td><a href="node3.shtml"><img height="44" width="44" src="/UCESIcons/tools/previous.gif"></a></td> <td><a href="taylor_series.html"><img height="44" width="44" src="/UCESIcons/tools/up.gif"></a></td> <td><a href="node5.shtml"><img height="44" width="44" src="/UCESIcons/tools/next.gif"></a></td> <td><a href="contents-node4.html"><img height="44" width="44" src="/UCESIcons/tools/toc.gif"></a></td> <td><a href="copyright.html"><img height="44" width="44" src="/UCESIcons/tools/copyright.gif"></a></td> <td><a href="mailto:uces@krellinst.org"><img height="44" width="44" src="/UCESIcons/tools/feedback.gif"></a></td> </tr> </table> </center> <hr> </body>