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<h2 class="hd hd-2 unit-title">1. Motivation</h2>
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<h3 class="hd hd-2">Power series and Taylor series</h3>
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<h2 class="hd hd-2 unit-title">2. Power series and Taylor series</h2>
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<p><b class="bfseries">Objectives</b></p><ul class="itemize"><li><p>
Represent functions by <span style="color:#27408C"><b class="bf">power series</b></span> over the <span style="color:#27408C"><b class="bf">interval of convergence</b></span>. </p></li><li><p>
Find the <span style="color:#27408C"><b class="bf">radius of convergence</b></span> for power series. </p></li><li><p>
Use <span style="color:#27408C"><b class="bf">Taylor's formula</b></span> to define power series. </p></li><li><p>
Apply rules for addition, multiplication, division, differentiation and integration of polynomials to power series. </p></li><li><p><span style="color:#27408C"><b class="bf">Estimate the error</b></span> in approximations using power series. </p></li></ul><p><b class="bfseries">Contents: 21 pages</b></p><p>
16 videos (123 minutes 1x speed) 34 questions </p>
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<h2 class="hd hd-2 unit-title">3. First example</h2>
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<h3 class="hd hd-2">First example of a power series</h3>
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Convergence
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Recall that </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001330" style="table-layout:auto" width="100%">
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } x^ n[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle \lim _{N\rightarrow \infty }\sum _{n=0}^{N} x^ n.[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.342)</td>
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For what values of [mathjaxinline]x[/mathjaxinline] does the power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } x^ n[/mathjaxinline] converge?<br/></p>
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<text> All real numbers.</text>
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<text> This series is not well defined for any [mathjaxinline]x[/mathjaxinline].</text>
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Domains of definition
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Let us define the function [mathjaxinline]\ f[/mathjaxinline] by </p>
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[mathjaxinline]\displaystyle \displaystyle f(x)= \sum _{n=0}^{\infty } x^ n\,[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.343)</td>
</tr>
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<p>
where the power series converges. <br/></p>
<p>
Consider also the function [mathjaxinline]\ g(x) = \displaystyle \frac{1}{1-x}[/mathjaxinline]. </p>
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What is the domain of definition </td>
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What is the domain of definition</td>
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of the function [mathjaxinline]\, \displaystyle f(x)[/mathjaxinline]? </td>
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<p>
For simplicity, we will start with power series centered at the point [mathjaxinline]x=0.\,[/mathjaxinline] In the following text, the term “power series" is short for such a power series.<br/></p><p>
A <span style="color:#27408C"><b class="bf">power series </b></span> is a series involving <b class="bf">only non-negative</b> powers of a variable [mathjaxinline]x[/mathjaxinline]: </p><table id="a0000001334" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001335"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle a_0+a_1x + a_2 x^2 + a_3 x^3 + \dotsb[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.344)</td></tr></table><p>
Note that the coefficients [mathjaxinline]a_ i[/mathjaxinline] can be zero, and therefore a power series may have finitely many non-zero terms.<br/></p><p>
The <span style="color:#27408C"><b class="bf">radius of convergence</b></span> [mathjaxinline]R[/mathjaxinline] of the power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n[/mathjaxinline], is a real number [mathjaxinline]0 \leq R < \infty[/mathjaxinline] such that </p><ul class="itemize"><li><p>
for [mathjaxinline]|x|<R[/mathjaxinline], the power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n[/mathjaxinline] converges (to a finite number); </p></li><li><p>
for [mathjaxinline]|x|>R[/mathjaxinline], the power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n[/mathjaxinline] diverges; </p></li><li><p>
for [mathjaxinline]|x| = R[/mathjaxinline], the power series may converge or diverge. But we will mostly ignore what happens at the end points of the interval of convergence. </p></li></ul><p><b class="bfseries">Examples:</b></p><ul class="itemize"><li><p>
Geometric series: [mathjaxinline]\displaystyle 1+ x + x^2 + x^3 + \dotsb = \sum _{n=0}^{\infty } x^ n[/mathjaxinline], radius of convergence is [mathjaxinline]1[/mathjaxinline]. </p></li><li><p>
Polynomials [mathjaxinline]\displaystyle a_0 + a_1x + a_2x^2 + \dotsb a_ N x^ N = \sum _{n=0}^{N} a_ n x^ n[/mathjaxinline] are power series, because polynomials can be written as infinite series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n[/mathjaxinline] where [mathjaxinline]a_ m=0[/mathjaxinline] for all [mathjaxinline]m>N[/mathjaxinline]. The radius of convergence for any polynomial is [mathjaxinline]\infty[/mathjaxinline]. In other words, the sum converges for all [mathjaxinline]x[/mathjaxinline]. </p></li></ul>
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Identify the power series
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Which of the following expressions are power series? (Check all that apply.) </p>
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<text>[mathjaxinline]1[/mathjaxinline]</text>
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<text>[mathjaxinline]x[/mathjaxinline]</text>
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<text>[mathjaxinline]1/x[/mathjaxinline]</text>
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<text>[mathjaxinline]\displaystyle 3-x+\frac{x^3}{2}[/mathjaxinline]</text>
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<text>[mathjaxinline]\displaystyle \sum _{n=-5}^{5} n\, x^ n[/mathjaxinline]</text>
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<text>[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } n\, x^ n[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">5. Radius of Convergence</h2>
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<p><b class="bfseries">Note about video</b></p><p>
In the first minute of video that follows, Christine used the word “Taylor series." What she means is “power series".<br/></p><p>
When you hear the words "Taylor series," replace them by "power series."<br/></p><p>
There is also an error in Example 4. Christine says the example is the series for [mathjaxinline]\cos x[/mathjaxinline], but that is an error. You'll see the correct series for [mathjaxinline]\cos x[/mathjaxinline] shortly, and discover what series this represents later in this lecture. </p>
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<h3 class="hd hd-2">Ratio test and radius of convergence</h3>
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<p><b class="bfseries">Finding the radius of convergence</b></p><p>
Given a power series [mathjaxinline]\displaystyle \, \sum _{n=0}^{\infty } a_ n x^ n[/mathjaxinline], the ratio test implies that the power series converges if </p><table id="a0000001336" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001337"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \lim _{n \rightarrow \infty } \left| \frac{a_{n+1} x^{n+1}}{a_ n x^ n}\right|[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle \lim _{n \rightarrow \infty } \left| \frac{a_{n+1} }{a_ n}\right||x|[/mathjaxinline]
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[mathjaxinline]\displaystyle <[/mathjaxinline]
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[mathjaxinline]\displaystyle 1.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.345)</td></tr></table><p>
There are 3 possibilities: </p><ol class="enumerate"><li value="1"><p>
There is a finite number [mathjaxinline]R[/mathjaxinline] such that </p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle |x|<R \Longrightarrow \lim _{n \rightarrow \infty } \left| \frac{a_{n+1} }{a_ n}\right| |x|< 1[/mathjaxinline], <b class="bf">and</b> </p></li><li><p>
[mathjaxinline]\displaystyle |x|>R \Longrightarrow \lim _{n \rightarrow \infty } \left| \frac{a_{n+1} }{a_ n}\right| |x|>1[/mathjaxinline]. </p></li></ul><p>
We say the radius of convergence is [mathjaxinline]R[/mathjaxinline]. </p></li><li value="2"><p>
For all [mathjaxinline]x[/mathjaxinline] [mathjaxinline]\displaystyle \lim _{n \rightarrow \infty } \left| \frac{a_{n+1} }{a_ n}\right||x| < 1[/mathjaxinline]. We say the radius of convergence is [mathjaxinline]\infty[/mathjaxinline]. (All [mathjaxinline]x[/mathjaxinline] satisfy [mathjaxinline]|x| < \infty .[/mathjaxinline]) </p></li><li value="3"><p>
[mathjaxinline]\displaystyle \lim _{n \rightarrow \infty } \left| \frac{a_{n+1} }{a_ n}\right| |x| > 1[/mathjaxinline] for all [mathjaxinline]x \neq 0[/mathjaxinline]. We say the radius of convergence is [mathjaxinline]0[/mathjaxinline]. </p></li></ol>
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Radius of convergence practice 1
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Find the radius of convergence of </p>
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[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } \frac{x^ n}{n}[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.346)</td>
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( If [mathjaxinline]R=\infty[/mathjaxinline], type <b class="bf">infty</b>.)<br/></p>
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<p style="display:inline">[mathjaxinline]R=[/mathjaxinline]</p>
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Radius of convergence practice 2
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Find the radius of convergence of the power series [mathjaxinline]1-x[/mathjaxinline]. </p>
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(Enter [mathjaxinline]R[/mathjaxinline] such that for [mathjaxinline]|x|&lt;R[/mathjaxinline] the power series converges. If [mathjaxinline]R=\infty[/mathjaxinline], type <b class="bf">infty</b>.) </p>
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<p style="display:inline">[mathjaxinline]R=[/mathjaxinline]</p>
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Enter <font color="#0078b0">lambda </font> for \(\lambda \)
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<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
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<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
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<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
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<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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Radius of convergence practice 3
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Find the radius of convergence of </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001343" style="table-layout:auto" width="100%">
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } \frac{n! x^ n}{n^ n}[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.349)</td>
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<p>
(Enter [mathjaxinline]R[/mathjaxinline] such that for [mathjaxinline]|x|&lt;R[/mathjaxinline] the power series converges. If [mathjaxinline]R=\infty[/mathjaxinline], type <b class="bf">infty</b>.) </p>
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<p style="display:inline">[mathjaxinline]R=[/mathjaxinline]</p>
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<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
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<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
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<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" rowspan="3" scope="row">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Differentials</th>
<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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Radius of convergence practice 4
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Find the radius of convergence of </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001356" style="table-layout:auto" width="100%">
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[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } n^ n x^ n[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.358)</td>
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(Enter [mathjaxinline]R[/mathjaxinline] such that for [mathjaxinline]|x|&lt;R[/mathjaxinline] the power series converges. If [mathjaxinline]R=\infty[/mathjaxinline], type <b class="bf">infty</b>.) </p>
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<p style="display:inline">[mathjaxinline]R=[/mathjaxinline]</p>
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<p><b class="bfseries">Remark: Alternative method using ratio test</b></p><p>
(Note that in the method that follows, the [mathjaxinline]n+1[/mathjaxinline] term is in the denominator and the [mathjaxinline]n[/mathjaxinline] term is in the numerator, which is the opposite of the ratio test.) </p><p>
Given a power series [mathjaxinline]\displaystyle \, \sum _{n=0}^{\infty } a_ n x^ n,[/mathjaxinline] </p><p>
if [mathjaxinline]\displaystyle \lim _{n \rightarrow \infty } \left|\frac{a_ n}{a_{n+1}} \right| = R[/mathjaxinline] (where [mathjaxinline]R[/mathjaxinline] exists or is [mathjaxinline]\infty[/mathjaxinline]), </p><p>
then the radius of convergence for the power series is [mathjaxinline]R[/mathjaxinline]. </p><p><b class="bfseries">Example</b></p><p>
Consider [mathjaxinline]\displaystyle \, \sum _{n=0}^{\infty } 2^ n x^ n,[/mathjaxinline] then [mathjaxinline]\displaystyle \lim _{n \rightarrow \infty } \left| \frac{2^ n}{2^{n+1}}\right| = \frac12 \quad \Longrightarrow \quad R = \frac12.[/mathjaxinline] </p>
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<p><b class="bfseries">Root test for radius of convergence</b></p><p>
Given a power series [mathjaxinline]\displaystyle \, \sum _{n=0}^{\infty } a_ n x^ n[/mathjaxinline], the root test implies that the power series converges if </p><table id="a0000001363" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001364"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \lim _{n \rightarrow \infty } \sqrt [n]{\left|a_ n x^ n\right|}[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \lim _{n \rightarrow \infty }\sqrt [n]{\left|a_ n\right|}|x|[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle <[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle 1.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.362)</td></tr></table><p>
There are 3 possibilities: </p><ol class="enumerate"><li value="1"><p>
There is a finite number [mathjaxinline]R[/mathjaxinline] such that </p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle |x|<R \Longrightarrow \lim _{n \rightarrow \infty } \sqrt [n]{\left| a_ n\right|} |x|< 1[/mathjaxinline], </p></li><li><p>
[mathjaxinline]\displaystyle |x|>R \Longrightarrow \lim _{n \rightarrow \infty } \sqrt [n]{\left| a_ n\right|} |x|>1[/mathjaxinline]. </p></li></ul><p>
We say the radius of convergence is [mathjaxinline]R[/mathjaxinline]. </p></li><li value="2"><p>
For all [mathjaxinline]x[/mathjaxinline] [mathjaxinline]\displaystyle \lim _{n \rightarrow \infty }\sqrt [n]{\left|a_ n\right|}|x| < 1[/mathjaxinline]. We say the radius of convergence is [mathjaxinline]\infty[/mathjaxinline]. (All [mathjaxinline]x[/mathjaxinline] satisfy [mathjaxinline]|x| < \infty .[/mathjaxinline]) </p></li><li value="3"><p>
[mathjaxinline]\displaystyle \lim _{n \rightarrow \infty }\sqrt [n]{\left|a_ n\right|}|x| > 1[/mathjaxinline] for all [mathjaxinline]x \neq 0[/mathjaxinline]. We say the radius of convergence is [mathjaxinline]0[/mathjaxinline]. </p></li></ol><p><b class="bfseries">Example</b></p><p>
Consider [mathjaxinline]\displaystyle \, \sum _{n=0}^{\infty } 2^ n x^ n,[/mathjaxinline] then [mathjaxinline]\displaystyle \lim _{n \rightarrow \infty } \sqrt [n]{\left| 2^ nx^ n\right|} = 2|x| < 1[/mathjaxinline] when [mathjaxinline]|x|<\frac12[/mathjaxinline]. This implies that the radius of convergence is [mathjaxinline]R = \frac12.[/mathjaxinline] </p>
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<p>
Add, subtract, multiply, divide, differentiate, and integrate convergent power series as one does for polynomials. We will discuss multiplication and division at a later time. </p><p>
Consider the power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n,[/mathjaxinline] which converges for [mathjaxinline]|x|< A[/mathjaxinline]. </p><ul class="itemize"><li><p>
The derivative [mathjaxinline]\displaystyle \frac{d}{dx} \sum _{n=0}^{\infty } a_ n x^ n = \sum _{n=1}^{\infty } n \, a_ n x^{n-1}[/mathjaxinline] also converges for [mathjaxinline]|x| < A[/mathjaxinline]. </p></li><li><p>
The integral [mathjaxinline]\displaystyle \int \left( \sum _{n=0}^{\infty } a_ n x^ n \right) \, dx = c + \sum _{n=0}^{\infty } a_ n \frac{x^{n+1}}{n+1}[/mathjaxinline] also converges for [mathjaxinline]|x| < A[/mathjaxinline]. Note that [mathjaxinline]c[/mathjaxinline] is the constant of integration. </p></li></ul><p>
Consider another power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } b_ n x^ n[/mathjaxinline], which converges for [mathjaxinline]|x|<B[/mathjaxinline]. </p><ul class="itemize"><li><p>
If [mathjaxinline]A \neq B[/mathjaxinline], then [mathjaxinline]\displaystyle \left( \sum _{n=0}^{\infty } a_ n x^ n\right) \pm \left( \sum _{n=0}^{\infty } b_ n x^ n\right) = \sum _{n=0}^{\infty } (a_ n \pm b_ n) x^ n[/mathjaxinline] converges for [mathjaxinline]|x|< \textrm{min}(A,B)[/mathjaxinline]. </p></li><li><p>
If [mathjaxinline]A = B[/mathjaxinline] , then [mathjaxinline]\displaystyle \left( \sum _{n=0}^{\infty } a_ n x^ n\right) \pm \left( \sum _{n=0}^{\infty } b_ n x^ n\right) = \sum _{n=0}^{\infty } (a_ n \pm b_ n) x^ n[/mathjaxinline] has a radius of convergence which is at least [mathjaxinline]A[/mathjaxinline], but it <em>could</em> have a larger radius of convergence. </p></li></ul>
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Adding power series
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Consider the sum of the two power series [mathjaxinline]\displaystyle \sum _{n=1}^{\infty } \frac{x^ n}{7^ n}[/mathjaxinline] and [mathjaxinline]\displaystyle \sum _{n=1}^{\infty } n^ n x^ n[/mathjaxinline]. </p>
<p>
Find the power series, and find its radius of convergence [mathjaxinline]R[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\displaystyle \Large \sum _{n=1}^{\infty }[/mathjaxinline]</p>
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(If the power series does not converge, type [mathjaxinline]0[/mathjaxinline]. If it converges for all [mathjaxinline]x[/mathjaxinline] type <b class="bf">infty</b> for [mathjaxinline]\infty[/mathjaxinline].) </p>
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<p style="display:inline">[mathjaxinline]R=[/mathjaxinline]</p>
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<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
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<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
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<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
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Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
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<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
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<th class="formulainput" rowspan="3" scope="row">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
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<td class="formulainput">arcsin, arccos, arctan, etc.</td>
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<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
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<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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Differentiating power series
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Differentiate the power series for [mathjaxinline]\displaystyle \frac{1}{1-x}[/mathjaxinline] to find the power series for [mathjaxinline]\displaystyle \frac{1}{(1-x)^2}[/mathjaxinline]. (Note the summation index starts at [mathjaxinline]n=1[/mathjaxinline].)<br/></p>
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<p style="display:inline">[mathjaxinline]\displaystyle \Large \sum _{n=1}^{\infty }[/mathjaxinline]</p>
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What is the radius of convergence [mathjaxinline]R[/mathjaxinline] for this series? </p>
<p>
(If the power series does not converge, type [mathjaxinline]0[/mathjaxinline]. If it converges for all [mathjaxinline]x[/mathjaxinline] type <b class="bf">infty</b> for [mathjaxinline]\infty[/mathjaxinline].) </p>
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<p style="display:inline">[mathjaxinline]R=[/mathjaxinline]</p>
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<font color="#0078b0">2520</font>
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<font color="#0078b0">2/3</font>
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<td class="formulainput">Decimals </td>
<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
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<th class="formulainput" rowspan="4" scope="row">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
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<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
</td>
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<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" rowspan="3" scope="row">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
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<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
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<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Differentials</th>
<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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Integrating power series
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Use power series integration on the power series for [mathjaxinline]\displaystyle \frac{1}{1-x}[/mathjaxinline] to find the power series for [mathjaxinline]\ -\ln (1-x)[/mathjaxinline] of the form </p>
<table cellpadding="7" cellspacing="0" class="equation" id="a0000001368" style="table-layout:auto" width="100%">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle c + \sum _{n=1}^{\infty }a_ n x^ n.[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
(Note the summation index starts at [mathjaxinline]n=1[/mathjaxinline].)<br/></p>
<p>
<p style="display:inline">[mathjaxinline]c=[/mathjaxinline]</p>
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\(\)
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<p style="display:inline">[mathjaxinline]a_ n =[/mathjaxinline]</p>
<div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="formulaequationinput_taylor-tab6-problem3_3_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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\(\)
<img src="/static/images/spinner.bc34f953403f.gif" class="loading" alt="Loading"/>
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<div class="script_placeholder" data-src="/static/js/capa/src/formula_equation_preview.b1967ab28c31.js"/>
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<p>
What is the radius of convergence [mathjaxinline]R[/mathjaxinline] for this series? </p>
<p>
(If the power series does not converge, type [mathjaxinline]0[/mathjaxinline]. If it converges for all [mathjaxinline]x[/mathjaxinline] type <b class="bf">infty</b> for [mathjaxinline]\infty[/mathjaxinline].) </p>
<p>
<p style="display:inline">[mathjaxinline]R=[/mathjaxinline]</p>
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\(\)
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<th class="formulainput" scope="col">Allowable Entries</th>
<th class="formulainput" scope="col">Descriptions</th>
<th class="formulainput" scope="col">Example Entries</th>
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<th class="formulainput" rowspan="3" scope="row">Numbers</th>
<td class="formulainput">Integers</td>
<td class="formulainput">
<font color="#0078b0">2520</font>
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</tr>
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<td class="formulainput">Fractions</td>
<td class="formulainput">
<font color="#0078b0">2/3</font>
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<td class="formulainput">Decimals </td>
<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
</tr>
<tr class="formulainput">
<th class="formulainput" rowspan="4" scope="row">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" rowspan="3" scope="row">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Differentials</th>
<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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<h2 class="hd hd-2 unit-title">7. Review of approximations</h2>
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Review linear and quadratic approximations
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<p>
Compare the function [mathjaxinline]\ f(x)[/mathjaxinline] and its linear approximation [mathjaxinline]\ L(x)[/mathjaxinline] through the point [mathjaxinline]x=0[/mathjaxinline]: </p>
<table cellpadding="7" cellspacing="0" class="equation" id="a0000001372" style="table-layout:auto" width="100%">
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<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle L(x) = f(0) + f'(0)x[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
Which of the following are necessarily true? </p>
<p>
(Check all that apply.) </p>
<p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_taylor-tab7-problem1_2_1">
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<text>[mathjaxinline]f(0) = L(0)[/mathjaxinline]</text>
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<text>[mathjaxinline]f(x) = L(x)[/mathjaxinline]</text>
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<text>[mathjaxinline]f'(0) = L'(0)[/mathjaxinline]</text>
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<input type="checkbox" name="input_taylor-tab7-problem1_2_1[]" id="input_taylor-tab7-problem1_2_1_choice_3" class="field-input input-checkbox" value="choice_3"/><label id="taylor-tab7-problem1_2_1-choice_3-label" for="input_taylor-tab7-problem1_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_taylor-tab7-problem1_2_1">
<text>[mathjaxinline]f'(x) = L'(x)[/mathjaxinline]</text>
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<text>[mathjaxinline]f^{\prime \prime }(0) = L^{\prime \prime }(0)[/mathjaxinline]</text>
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Compare the function [mathjaxinline]\ f(x)[/mathjaxinline] and its quadratic approximation [mathjaxinline]\, q(x)[/mathjaxinline] through the point [mathjaxinline]x=0[/mathjaxinline]: </p>
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<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle q(x) = f(0) + f'(0)x + \frac{f^{\prime \prime }(0)}{2} x^2[/mathjax]</td>
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<p><b class="bfseries">Food for thought</b></p><p>
Now we can imagine coming up with a best fit cubic function to our function [mathjaxinline]\ f(x)[/mathjaxinline] at [mathjaxinline]x=0[/mathjaxinline], the best fit 4th degree polynomial, 5th degree polynomial etc. If we continue this process to infinitely many terms, we end up with a <b class="bf"><em>power series</em></b> whose derivatives all agree with the derivatives of our original function [mathjaxinline]\ f[/mathjaxinline] at the point [mathjaxinline]x=0[/mathjaxinline]. </p>
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Cubic approximations
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Find the best cubic approximation to a function [mathjaxinline]\ f(x)[/mathjaxinline] at the point [mathjaxinline]x=0[/mathjaxinline]. </p>
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<text> [mathjaxinline]\displaystyle f(0) + f'(0)x + f^{\prime \prime }(0) x^2 + f^{\prime \prime \prime }(0) x^3[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle f(0) + f'(0)x + \frac{f^{\prime \prime }(0)}{2} x^2 + \frac{f^{\prime \prime \prime }(0)}{3} x^3[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle f(0) + f'(0)x + \frac{f^{\prime \prime }(0)}{2} x^2 + \frac{f^{\prime \prime \prime }(0)}{6} x^3[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle f(0) + f'(0)x + \frac{f^{\prime \prime }(0)}{2} x^2 + \frac{f^{\prime \prime \prime }(0)}{12} x^3[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle f(0) + \frac{f'(0)}{2}x + \frac{f^{\prime \prime }(0)}{6} x^2 + \frac{f^{\prime \prime \prime }(0)}{24} x^3[/mathjaxinline]</text>
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<p>
Recall that [mathjaxinline]\, \displaystyle n!=n (n-1) (n-2)\cdots (3)(2)(1)\,[/mathjaxinline] for all integers [mathjaxinline]\, n \geq 1\,[/mathjaxinline]. </p><p>
We define [mathjaxinline]\, 0!=1\,[/mathjaxinline]. This is a very valuable convention that simplifies many formulas. </p><p>
The <b class="bf"><span style="color:#27408C">Taylor series</span></b> of a function [mathjaxinline]f(x)[/mathjaxinline] is the following power series: </p><table id="a0000001384" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001385"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \frac{f(0)}{0!}x^0 + \frac{f'(0)}{1!}x + \frac{f^{\prime \prime }(0)}{2!} x^2 + \frac{f^{\prime \prime \prime }(0)}{3!} x^3 + \dotsb[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } \frac{f^{(n)}(0)}{n!}x^ n,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.374)</td></tr></table><p>
Just like any other power series, the Taylor series converges when [mathjaxinline]\, |x|<R\,[/mathjaxinline] where [mathjaxinline]\, R\,[/mathjaxinline] is the radius of convergence. <br/></p><p>
In this course, we only consider functions [mathjaxinline]f(x)[/mathjaxinline] that equal their Taylor series within the radius of convergence. This is a large class of functions that include many of the functions we regularly use. For all such functions: </p><table id="a0000001386" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001387"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } \frac{f^{(n)}(0)}{n!}x^ n \qquad (\text {for } \, |x|<R).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.375)</td></tr></table><p><b class="bfseries">Important examples</b></p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle e^ x = \sum _{n=0}^{\infty } \frac{x^ n}{n!}\, =\, 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots[/mathjaxinline]<br/></p></li><li><p>
[mathjaxinline]\displaystyle \sin x = \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n+1}}{(2n+1)!}\, =\, x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots[/mathjaxinline] </p></li><li><p>
[mathjaxinline]\displaystyle \cos x = \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n}}{(2n)!}\, =\, 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots[/mathjaxinline] </p></li></ul><p>
We will use Taylor's formula to derive the series for [mathjaxinline]\, \sin (x)\,[/mathjaxinline] and [mathjaxinline]\, \cos (x)\,[/mathjaxinline] on the next page. </p><p>
Notice that the factorial appears in the denominator of all terms in all three power series above.<br/></p><p>
Using the Taylor series of [mathjaxinline]\, e^ x,\,[/mathjaxinline] we find a formula for the number [mathjaxinline]e[/mathjaxinline] as the rapidly converging series: </p><table id="a0000001388" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001389"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle e= \sum _{n=0}^{\infty } \frac{1}{n!}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.376)</td></tr></table>
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Radius of convergence
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<p>
Find the radii of convergence [mathjaxinline]\, R\,[/mathjaxinline] of the 3 following Taylor series. </p>
<ul class="itemize">
<li>
<p>
[mathjaxinline]\displaystyle e^ x = \sum _{n=0}^{\infty } \frac{x^ n}{n!}[/mathjaxinline] </p>
</li>
<li>
<p>
[mathjaxinline]\displaystyle \sin x = \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n+1}}{(2n+1)!}[/mathjaxinline] </p>
</li>
<li>
<p>
[mathjaxinline]\displaystyle \cos x = \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n}}{(2n)!}[/mathjaxinline] </p>
</li>
</ul>
<p>
(If [mathjaxinline]\, R=\infty \,[/mathjaxinline] enter <b class="bf">infty</b>.)<br/></p>
<p>
<p style="display:inline">For [mathjaxinline]\, e^ x,\,[/mathjaxinline] [mathjaxinline]R=\,[/mathjaxinline]</p>
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<p style="display:inline">For [mathjaxinline]\, \sin (x),\,[/mathjaxinline] [mathjaxinline]R=\,[/mathjaxinline]</p>
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<p style="display:inline">For [mathjaxinline]\, \cos (x),\,[/mathjaxinline] [mathjaxinline]R=\,[/mathjaxinline]</p>
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Evenness and oddness
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Find the Taylor series for [mathjaxinline]\sin (-x)[/mathjaxinline] up to degree [mathjaxinline]5[/mathjaxinline] by substituting [mathjaxinline]-x[/mathjaxinline] into the Taylor series of [mathjaxinline]\, \sin (x)\,[/mathjaxinline] recalled below: </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001399" style="table-layout:auto" width="100%">
<tr id="a0000001400">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sin (x) \, =\, \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n+1}}{(2n+1)!}[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\cdots \, .[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.383)</td>
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(Your answer should be of the form [mathjaxinline]a_0+a_1x+a_2 x^2+a_3x^3+a_4 x^4+a_5 x^5[/mathjaxinline]. Do not add more terms. The grader recognizes factorials; for example, type <b class="bf">factorial(8)</b> for [mathjaxinline]8![/mathjaxinline].)<br/></p>
<p>
Up to degree [mathjaxinline]\, 5\,[/mathjaxinline]: <p style="display:inline">[mathjaxinline]\sin (-x)\, =\,[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_taylor-tab8-problem2_2_1" class="text-input-dynamath capa_inputtype inline textline">
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Find the Taylor series for [mathjaxinline]\cos (-x)[/mathjaxinline] up to degree [mathjaxinline]5[/mathjaxinline] by substituting [mathjaxinline]-x[/mathjaxinline] into the Taylor series of [mathjaxinline]\, \cos (x)\,[/mathjaxinline] recalled below: </p>
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[mathjaxinline]\displaystyle \displaystyle \cos (x) \, =\, \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n}}{(2n)!}[/mathjaxinline]
</td>
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[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
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[mathjaxinline]\displaystyle 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\cdots \, .[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.384)</td>
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(Your answer should be of the form [mathjaxinline]a_0+a_1x+a_2 x^2+a_3x^3+a_4 x^4[/mathjaxinline]. Do not add more terms. The grader recognizes factorials; for example, type <b class="bf">factorial(8)</b> for [mathjaxinline]8![/mathjaxinline].)<br/></p>
<p>
Up to degree [mathjaxinline]\, 4\,[/mathjaxinline]: <p style="display:inline">[mathjaxinline]\cos (-x)\, =\,[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="inputtype_taylor-tab8-problem2_3_1" class="text-input-dynamath capa_inputtype inline textline">
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The geometric series
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<p>
Use Taylor's formula to verify that the Taylor's series of [mathjaxinline]\, \displaystyle \frac1{1+x}\,[/mathjaxinline] is indeed the geometric series.<br/>(Recall that we obtained the formula for the geometric series using the limit of the partial sums. ) <br/></p>
<p>
To use Taylor's formula, first find the [mathjaxinline]\, \displaystyle n^{\text {th}}\,[/mathjaxinline] derivative of [mathjaxinline]\displaystyle \, \frac1{1+x}[/mathjaxinline].<br/>(Enter your answer in terms of [mathjaxinline]n[/mathjaxinline]. Type <b class="bf">factorial(n)</b> for [mathjaxinline]n![/mathjaxinline].)<br/></p>
<p>
<p style="display:inline">[mathjaxinline]\displaystyle \frac{d^ n}{(dx)^ n}\frac1{1+x} \, =\quad[/mathjaxinline]</p>
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Suppose the Taylor series of [mathjaxinline]\, \displaystyle \frac1{1+x}\,[/mathjaxinline] is </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001410" style="table-layout:auto" width="100%">
<tr id="a0000001411">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \frac1{1+x}[/mathjaxinline]
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<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n \qquad (\text {for }\, |x|&lt;1)[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.390)</td>
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<p>
Use the derivative you computed above to get the coefficient [mathjaxinline]a_ n[/mathjaxinline] of the [mathjaxinline]x^ n[/mathjaxinline] term in the Taylor's series of [mathjaxinline]\, \displaystyle \frac1{1+x}\,[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\displaystyle \, a_ n \, =\quad[/mathjaxinline]</p>
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<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
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<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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What is the radius of convergence [mathjaxinline]R[/mathjaxinline] of the Taylor's series of a polynomial [mathjaxinline]p(x)[/mathjaxinline] of degree [mathjaxinline]200[/mathjaxinline]?<br/>(Enter <b class="bf">infty</b> for infinity.)<br/></p>
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<p><b class="bfseries">Video note</b></p><p>
In the video below, when Prof Miller discussed the radius of convergence of the Taylor series of [mathjaxinline]\sin (x)[/mathjaxinline], what he showed is that it is believable that the radius of convergence is [mathjaxinline]\infty[/mathjaxinline]. The complete proof is in text after video.<br/></p>
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<h3 class="hd hd-2">Taylor series of sine</h3>
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<p><b class="bfseries">Radius of convergence of Taylor series of sine</b></p><p>
In the video, we find the Taylor series of [mathjaxinline]\sin (x)\,[/mathjaxinline] to be </p><table id="a0000001427" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001428"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sin x[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.400)</td></tr><tr id="a0000001429"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n+1}}{(2n+1)!}\qquad (|x|<R).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.401)</td></tr></table><p>
We now use the ratio test to find the radius of convergence [mathjaxinline]R[/mathjaxinline]: </p><table id="a0000001430" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001431"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } \left|\frac{(-1)^{n+1} x^{2(n+1)+1}}{\left(2(n+1)+1\right)!}\right| \, \left| \frac{(2n+1)!}{(-1)^ n x^{2n+1}}\right|[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle |x|^2 \, \lim _{n\rightarrow \infty } \frac{1}{(2n+2)(2n+3)}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.402)</td></tr><tr id="a0000001432"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle |x|^2\, (0)\, =\, 0 \qquad \text {for all }\, x.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.403)</td></tr></table><p>
Hence, the Taylor series is absolutely convergent for all [mathjaxinline]\, x\,[/mathjaxinline] and the radius of convergence is [mathjaxinline]\, R=\infty[/mathjaxinline]. </p><p><b class="bf">Remark:</b> If we are concerned with only convergence but not absolute convergence, we can use the alternating series test at each [mathjaxinline]\, x\,[/mathjaxinline] to show that the Taylor series for [mathjaxinline]\, \sin (x)\,[/mathjaxinline] converges for all [mathjaxinline]\, x.\, \,[/mathjaxinline] Since for any [mathjaxinline]x[/mathjaxinline], </p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } \left| \frac{(-1)^ n x^{2n+1}}{(2n+1)!}\right|=0\,[/mathjaxinline] </p></li><li><p>
[mathjaxinline]\left| \frac{(-1)^ n x^{2n+1}}{(2n+1)!}\right|[/mathjaxinline] decreases as [mathjaxinline]\, n\,[/mathjaxinline] increases, </p></li></ul><p>
the alternating series [mathjaxinline]\, \displaystyle \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n+1}}{(2n+1)!}\,[/mathjaxinline] converges for all [mathjaxinline]\, x[/mathjaxinline].<br/></p>
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Taylor series of cosine
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We will use Taylor's formula to obtain the coefficients in the Taylor series of [mathjaxinline]\displaystyle \, \cos (x).[/mathjaxinline]<br/></p>
<p>
First, compute [mathjaxinline]\, \cos ^{(n)} (x)[/mathjaxinline] and [mathjaxinline]\, \displaystyle \frac{\cos ^{(n)} (x)}{n!}\,[/mathjaxinline] for [mathjaxinline]\, n=0\,[/mathjaxinline] to [mathjaxinline]\, 4[/mathjaxinline]. <br/>(Note that [mathjaxinline]\cos ^{(n)}(x)\,[/mathjaxinline] is the [mathjaxinline]\, n^{\text {th}}\,[/mathjaxinline] derivative, not the [mathjaxinline]\, \displaystyle n^{\text {th}}\,[/mathjaxinline] power, of [mathjaxinline]\cos (x)[/mathjaxinline]. Type <b class="bf">factorial(n)</b> for [mathjaxinline]n![/mathjaxinline].)<br/></p>
<p>
<p style="display:inline">[mathjaxinline]\displaystyle \cos '(x) \, =\,[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\qquad \displaystyle \frac{\cos '(0)}{1!} \, =\,[/mathjaxinline]</p>
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<p>
<p style="display:inline">[mathjaxinline]\displaystyle \cos ^{\prime \prime }(x) \, =\,[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\qquad \displaystyle \frac{\cos ^{\prime \prime }(0) }{2!} \, =\,[/mathjaxinline]</p>
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<p>
<p style="display:inline">[mathjaxinline]\displaystyle \cos ^{(3)} (x) \, =\,[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\qquad \displaystyle \frac{\cos ^{(3)}(0) }{3!} \, =\,[/mathjaxinline]</p>
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<p>
<p style="display:inline">[mathjaxinline]\displaystyle \cos ^{(4)} (x) \, =\,[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\qquad \displaystyle \frac{\cos ^{(4)}(0) }{4!} \, =\,[/mathjaxinline]</p>
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Find the coefficient of [mathjaxinline]x^{18}[/mathjaxinline] in the Taylor series of [mathjaxinline]\, \cos (x)\,[/mathjaxinline].<br/>(Type <b class="bf">factorial(n)</b> for [mathjaxinline]n![/mathjaxinline].)<br/></p>
<p>
<p style="display:inline">[mathjaxinline]\displaystyle \text {Coefficient of }\, x^{18} \, =\,[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">11. New series from old: multiplying series</h2>
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So far, we have used Taylor's formula to obtain the following Taylor series: </p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle \frac1{1-x} = \sum _{n=0}^{\infty } x^ n\,[/mathjaxinline] for [mathjaxinline]\, |x|<1[/mathjaxinline]. </p></li><li><p>
[mathjaxinline]\displaystyle e^ x = \sum _{n=0}^{\infty } \frac{x^ n}{n!}[/mathjaxinline] </p></li><li><p>
[mathjaxinline]\displaystyle \sin x = \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n+1}}{(2n+1)!}[/mathjaxinline] </p></li><li><p>
[mathjaxinline]\displaystyle \cos x = \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n}}{(2n)!}[/mathjaxinline]. </p></li></ul><p>
We have also integrated the geometric series to obtain a power series for [mathjaxinline]\, \displaystyle \ln (1-x)[/mathjaxinline]: </p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle -\ln (1-x)\, =\, \sum _{n=1}^{\infty } \frac{x^ n}{n}[/mathjaxinline]. </p></li></ul><p>
We will now obtain more Taylor series from the ones above by performing addition, multiplication, differentiation, and integration on them. </p>
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Review: multiply two polynomials
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Find the coefficient of [mathjaxinline]x^3[/mathjaxinline] of [mathjaxinline]\, p(x)q(x)[/mathjaxinline], which is the product of the two polynomials [mathjaxinline]\, p(x)\,[/mathjaxinline] and [mathjaxinline]\, q(x)\,[/mathjaxinline] below: </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001442" style="table-layout:auto" width="100%">
<tr id="a0000001443">
<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \displaystyle p(x)[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle 1+x-2x^2+\frac{2}{3}x^3-3x^4+9x^6[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.410)</td>
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[mathjaxinline]\displaystyle q(x)[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle 3-2x+x^2+x^3+\frac{4}{3}x^5.[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.411)</td>
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<p style="display:inline">Coefficient of [mathjaxinline]\, x^3[/mathjaxinline] in the [mathjaxinline]\, p(x)q(x)\,[/mathjaxinline] is </p>
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Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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<p><b class="bfseries">Multiplying two power series</b></p><p>
We multiply two powers series using the same rule as when we multiply two polynomials.<br/></p><p>
Consider the power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n,[/mathjaxinline] which converges for [mathjaxinline]|x|< A[/mathjaxinline], and the power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } b_ n x^ n[/mathjaxinline], which converges for [mathjaxinline]|x|<B[/mathjaxinline]. </p><ul class="itemize"><li><p>
The product of the two power series converges for [mathjaxinline]|x|< \textrm{min}(A,B)[/mathjaxinline], but it <em>could</em> have a larger radius of convergence. </p></li></ul>
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Multiply two power series
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Find the coefficient of [mathjaxinline]x^3[/mathjaxinline] of [mathjaxinline]P(x)Q(x)[/mathjaxinline], which is the product of the two power series [mathjaxinline]\, P(x)\,[/mathjaxinline] and [mathjaxinline]\, Q(x)\,[/mathjaxinline] below: </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001450" style="table-layout:auto" width="100%">
<tr id="a0000001451">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle P(x)[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle 1+x-2x^2+\frac{2}{3}x^3-3x^4+9x^6+\cdots[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.415)</td>
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[mathjaxinline]\displaystyle Q(x)[/mathjaxinline]
</td>
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[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
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[mathjaxinline]\displaystyle 3-2x+x^2+x^3+\frac{4}{3}x^5+\cdots .[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.416)</td>
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<p>
(Hint: Does this look familiar? )</p>
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<p style="display:inline">Coefficient of [mathjaxinline]\, x^3[/mathjaxinline] is </p>
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Multiply two series 2
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Consider the power series of [mathjaxinline]\, e^ x \cos (x)\,[/mathjaxinline] obtained by multiplying the Taylor series for [mathjaxinline]\, e^ x\,[/mathjaxinline] and [mathjaxinline]\, \cos (x)[/mathjaxinline]. </p>
<p>
Find its radius of convergence [mathjaxinline]\, R\,[/mathjaxinline] and the coefficient of [mathjaxinline]x^4[/mathjaxinline].<br/>(Enter <b class="bf">infty</b> for infinity.)<br/></p>
<p>
<p style="display:inline">[mathjaxinline]R=\quad[/mathjaxinline]</p>
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The coefficient of [mathjaxinline]\, x^4[/mathjaxinline] in the power series of [mathjaxinline]\, e^ x\cos (x)\,[/mathjaxinline] is: <div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="formulaequationinput_taylor-tab11-problem3_3_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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Consider the product of two exponential series: </p><table id="a0000001463" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001464"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle e^ ae^{b}[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \left(1+a+\frac{a^2}{2!}+\frac{a^3}{3!}+\frac{a^4}{4!}+\cdots \right)\left(1+b+\frac{b^2}{2!}+\frac{b^3}{3!}+\frac{b^4}{4!}+\cdots \right)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.424)</td></tr></table><p>
Is the product of these two series the same as the Taylor series of [mathjaxinline]e^{a+b}[/mathjaxinline]?<br/></p><p>
We say that a term in this product has <span style="color:#27408C"><b class="bf">total degree 2</b></span> if the power of [mathjaxinline]a[/mathjaxinline] plus the power of [mathjaxinline]b[/mathjaxinline] is 2. </p><p>
If we look at the all total degree 2 term in [mathjaxinline]a[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline] in this product we get: </p><table id="a0000001465" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001466"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \text {Terms of total degree }\, 2 \,[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{a^2}{2!}+ab+\frac{b^2}{2!}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.425)</td></tr><tr id="a0000001467"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{1}{2}\left(a^2+2ab+b^2\right)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.426)</td></tr><tr id="a0000001468"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{(a+b)^2}{2!}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.427)</td></tr></table><p>
Can the terms of total degree [mathjaxinline]\, 3\,[/mathjaxinline] be expressed in terms of [mathjaxinline](a+b)^3[/mathjaxinline]? </p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">See answer<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><table id="a0000001469" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001470"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \text {Terms of total degree }\, 3 \,[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{a^3}{3!}+ \frac{a^2}{2!} (b)+(a) \frac{b^2}{2!} +\frac{b^3}{3!}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.428)</td></tr><tr id="a0000001471"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{1}{3!}\left(a^3+3a^2 b+3a b^2+b^3 \right)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.429)</td></tr><tr id="a0000001472"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
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[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{(a+b)^3}{3!}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.430)</td></tr></table><p><b class="bf">Remark:</b> The pattern above continues for all [mathjaxinline]\, n[/mathjaxinline]. This is consistent with [mathjaxinline]e^ ae^ b = e^{a+b}[/mathjaxinline]. To verify that this holds from the series formulas requires use of the Binomial Theorem: </p><table id="a0000001473" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001474"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle (a+b)^ n[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{p=0}^{n} \frac{n!}{p!(n-p)!} a^ p \, b^{n-p}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.431)</td></tr></table><p>
Here's the computation: </p><table id="a0000001475" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001476"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \text {Terms of total degree }\, n \,[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{p=0}^{n} \frac{a^ p}{p!}\, \frac{b^{n-p}}{(n-p)!}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.432)</td></tr><tr id="a0000001477"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{1}{n!} \sum _{p=0}^{n} \frac{n!}{p!(n-p)!} \, a^ p b^{n-p}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.433)</td></tr><tr id="a0000001478"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{(a+b)^ n}{n!}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.434)</td></tr></table><p>
where in the last step we have used the Binomial Theorem. </p></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p><SCRIPT src="/assets/courseware/v1/9567be3f4e91361f26a8beaadf749715/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/latex2edx.js" type="text/javascript"/><LINK href="/assets/courseware/v1/daf81af0af57b85a105e0ed27b7873a0/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/latex2edx.css" rel="stylesheet" type="text/css"/>
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If </p><table id="a0000001479" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001480"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle F(x)[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle \frac{G(x)}{H(x)}[/mathjaxinline]
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where [mathjaxinline]\, F(x), \, G(x),\, H(x)\,[/mathjaxinline] are all power series, then we can find [mathjaxinline]\, F(x)\,[/mathjaxinline] by solving the following equation of power series degree by degree: </p><table id="a0000001481" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001482"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle F(x)H(x)[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle G(x).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.436)</td></tr></table><p>
The radius of convergence of [mathjaxinline]\, F(x)\,[/mathjaxinline] is more difficult to track, since [mathjaxinline]\, F(x)\,[/mathjaxinline] diverges whenever [mathjaxinline]\, H(x)\, =\, 0\,[/mathjaxinline] and [mathjaxinline]\, G(x)\neq 0[/mathjaxinline]. </p>
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Taylor series of tangent
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Compute the Taylor series for [mathjaxinline]\displaystyle \, \tan (x)\,[/mathjaxinline] up to degree [mathjaxinline]5[/mathjaxinline] by using </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001483" style="table-layout:auto" width="100%">
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \, \displaystyle \tan (x)\cos (x)\, =\, \sin (x).[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.437)</td>
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(Your answer should be of the form [mathjaxinline]a_0+a_1x+a_2 x^2+a_3x^3+a_4 x^4+a_5 x^5[/mathjaxinline].)<br/></p>
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Up to degree [mathjaxinline]\, 5\, :[/mathjaxinline] <p style="display:inline"> [mathjaxinline]\, \tan (x)\, =\quad[/mathjaxinline] </p><div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="formulaequationinput_taylor-tab12-problem1_2_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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Preparation: exponential of negative x
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Substitute [mathjaxinline]-x[/mathjaxinline] into the Taylor series for [mathjaxinline]e^ x[/mathjaxinline] to get a power series for [mathjaxinline]e^{-x}[/mathjaxinline]. Also find the radius of convergence of the resulting power series. </p>
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(Enter the series up to degree [mathjaxinline]4[/mathjaxinline]. Type <b class="bf">factorial(4)</b> for [mathjaxinline]4![/mathjaxinline].)<br/></p>
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Up to degree [mathjaxinline]4[/mathjaxinline]: <p style="display:inline"> [mathjaxinline]e^{-x}\, =[/mathjaxinline] </p><div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_taylor-tab13-problem1_2_1" class="text-input-dynamath capa_inputtype inline textline">
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Find the radius of convergence [mathjaxinline]\, R\,[/mathjaxinline] of the series you obtained.<br/>( If [mathjaxinline]R=\infty[/mathjaxinline], type <b class="bf">infty</b>.)<br/></p>
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<p style="display:inline">[mathjaxinline]R=[/mathjaxinline]</p>
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Taylor series of Hyperbolic sine
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<p>
Recall that the hyperbolic sine function is defined as </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001502" style="table-layout:auto" width="100%">
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<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sinh (x)[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
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<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{e^ x-e^{-x}}{2}.[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.448)</td>
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<p>
Find the Taylor series of [mathjaxinline]\, \sinh (x)\,[/mathjaxinline] up to degree [mathjaxinline]5[/mathjaxinline]. </p>
<p>
(Your answer should be of the form [mathjaxinline]a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5.\, \,[/mathjaxinline] Enter <b class="bf">factorial(n)</b> for [mathjaxinline]n![/mathjaxinline].) </p>
<p>
Up to degree [mathjaxinline]5[/mathjaxinline]: <p style="display:inline">[mathjaxinline]\sinh (x)\, =[/mathjaxinline] </p><div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_taylor-tab13-problem2_2_1" class="text-input-dynamath capa_inputtype inline textline">
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Find the radius of convergence [mathjaxinline]\, R\,[/mathjaxinline] of the Taylor series of [mathjaxinline]\sinh (x)[/mathjaxinline].<br/>( If [mathjaxinline]R=\infty[/mathjaxinline], type <b class="bf">infty</b>.)<br/></p>
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<p style="display:inline">[mathjaxinline]R=[/mathjaxinline]</p>
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<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
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<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
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<td class="formulainput">_ (add a subscript)</td>
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<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
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<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
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<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
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<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
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<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
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<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
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<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
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<th class="formulainput" scope="row">Differentials</th>
<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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Differentiate and integrate the Taylor series of hyperbolic sine
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Let [mathjaxinline]\, S(x)\,[/mathjaxinline] be the Taylor series for [mathjaxinline]\, \sinh (x)\,[/mathjaxinline] that you obtained in the previous problem.<br/></p>
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Compute [mathjaxinline]\, \displaystyle \frac{d}{dx}\, S(x)\,[/mathjaxinline] up to degree [mathjaxinline]4[/mathjaxinline].<br/>(Enter <b class="bf">factorial(n)</b> for [mathjaxinline]n![/mathjaxinline].)<br/></p>
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Up to degree [mathjaxinline]4,\,[/mathjaxinline] <p style="display:inline"> [mathjaxinline]\, \displaystyle \frac{d}{dx}\, S(x)\, =\quad[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_taylor-tab13-problem3_2_1" class="text-input-dynamath capa_inputtype inline textline">
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Compute [mathjaxinline]\, \displaystyle \int S(x) \, dx\,[/mathjaxinline] up to degree [mathjaxinline]4[/mathjaxinline].<br/>(Enter <b class="bf">C</b> as the constant of integration.)<br/></p>
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Up to degree [mathjaxinline]4,\,[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\, \displaystyle \int \, S(x)\, dx\, =\quad[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">14. Substitution</h2>
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<p>
We can find the composition of two power series [mathjaxinline]\, f(g(x))\,[/mathjaxinline] by using similar rules as composition of polynomials. In this course, we will only substitute a polynomial [mathjaxinline]p(x)[/mathjaxinline] into a power series [mathjaxinline]f(x)[/mathjaxinline].<br/></p><p>
If the radius of convergence of the power series [mathjaxinline]\, f(x)\,[/mathjaxinline] is [mathjaxinline]\, R,\,[/mathjaxinline] then the power series [mathjaxinline]\, f(p(x))\,[/mathjaxinline] converges whenever [mathjaxinline]\, |p(x)|<R\,[/mathjaxinline].<br/></p><p>
In particular, we can find a Taylor series for a function [mathjaxinline]\ f(x)\,[/mathjaxinline] at [mathjaxinline]x=0[/mathjaxinline] and then substitute in a polynomial of the form [mathjaxinline]p(x) = ax^ n[/mathjaxinline] for [mathjaxinline]x[/mathjaxinline] where [mathjaxinline]n\geq 1[/mathjaxinline].</p>
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Practice substitution
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Use substitution and the Taylor series of [mathjaxinline]\, \sin (x)\,[/mathjaxinline] and [mathjaxinline]\, \sinh (x)\,[/mathjaxinline] to find the first three non-zero terms of the power series representing [mathjaxinline]\, \displaystyle \sin \left(2x^5\right)+ \sinh \left(2x^5\right)[/mathjaxinline].<br/></p>
<p>
(Your answer should be the sum of the three non-zero terms with the lowest power of [mathjaxinline]x[/mathjaxinline]. Remember that the constant term counts as the degree [mathjaxinline]0[/mathjaxinline] term. Enter <b class="bf">factorial(n)</b> for [mathjaxinline]n![/mathjaxinline].)<br/></p>
<p>
The 3 lowest non-zero terms:<br/><p style="display:inline"> [mathjaxinline]\, \displaystyle \sin \left(2x^5\right)+ \sinh \left(2x^5\right)\, =\quad[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_taylor-tab14-problem1_2_1" class="text-input-dynamath capa_inputtype inline textline">
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<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
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<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
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<td class="formulainput">dx, dy</td>
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<h2 class="hd hd-2 unit-title">15. Recognizing power series</h2>
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<h3 class="hd hd-2">Recognizing power series </h3>
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Review: radius of convergence
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<p>
Find the radius of convergence [mathjaxinline]\, R\,[/mathjaxinline] of the 4 power series presented in the video: (If [mathjaxinline]\, R=\infty ,\,[/mathjaxinline] enter <b class="bf">infty</b>.)<br/></p>
<p>
<p style="display:inline">For [mathjaxinline]\, \displaystyle \sum _{n=0}^{\infty }\, \frac{x^{n+2}}{n!},\,[/mathjaxinline] &#8195;[mathjaxinline]R=\,[/mathjaxinline]</p>
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<p style="display:inline">For [mathjaxinline]\, \displaystyle \sum _{n=2}^{\infty }\, x^ n,\,[/mathjaxinline] &#8195;&#8195;[mathjaxinline]R=\,[/mathjaxinline]</p>
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<p style="display:inline">For [mathjaxinline]\, \displaystyle \sum _{n=0}^{\infty }\left(\frac{x^ n}{n!}+x^ n\right),\,[/mathjaxinline] &#8195;[mathjaxinline]R=\,[/mathjaxinline]</p>
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<p>
<p style="display:inline">For [mathjaxinline]\, \displaystyle \sum _{n=-1}^{\infty } x^{n+1},\,[/mathjaxinline] &#8195;&#8195;[mathjaxinline]R=\,[/mathjaxinline]</p>
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<th class="formulainput" scope="col">Descriptions</th>
<th class="formulainput" scope="col">Example Entries</th>
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<td class="formulainput">Integers</td>
<td class="formulainput">
<font color="#0078b0">2520</font>
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<font color="#0078b0">2/3</font>
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<td class="formulainput">Decimals </td>
<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
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<th class="formulainput" rowspan="4" scope="row">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
</tr>
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<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
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<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
</tr>
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<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
</tr>
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<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
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<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
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<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
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<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
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<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
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<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
</tr>
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<th class="formulainput" scope="row">Differentials</th>
<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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Recognizing power series
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Find the functions whose Taylor series are given below. (Recall that [mathjaxinline]\, 0!=1[/mathjaxinline].)<br/>(<i class="itshape">Hint</i>: You will need to recall the power series of the trigonometric and hyperbolic trigonometric functions, as well as how power series behave under differentiation/integration.)<br/></p>
<p>
For [mathjaxinline]\, |x|&lt;\infty[/mathjaxinline]:<br/><p style="display:inline">[mathjaxinline]\displaystyle x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+\cdots \, =\quad[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="formulaequationinput_taylor-tab15-problem2_2_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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\(\)
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For [mathjaxinline]\, |x|&lt;1[/mathjaxinline]:<br/><p style="display:inline">[mathjaxinline]\, \displaystyle \sum _{n=0}^{\infty } (n+2)(n+1)x^{n}\, =\quad[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="formulaequationinput_taylor-tab15-problem2_3_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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\(\)
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For [mathjaxinline]\, |x|&lt;\infty[/mathjaxinline]:<br/><p style="display:inline">[mathjaxinline]\, \displaystyle \sum _{n=0}^{\infty }\, \frac{(-x)^{6n+1}}{(2n)!}\, =\quad[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 3" role="group"><div id="formulaequationinput_taylor-tab15-problem2_4_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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\(\)
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<th class="formulainput" scope="col">Descriptions</th>
<th class="formulainput" scope="col">Example Entries</th>
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<th class="formulainput" rowspan="3" scope="row">Numbers</th>
<td class="formulainput">Integers</td>
<td class="formulainput">
<font color="#0078b0">2520</font>
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<td class="formulainput">Fractions</td>
<td class="formulainput">
<font color="#0078b0">2/3</font>
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<td class="formulainput">Decimals </td>
<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
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<th class="formulainput" rowspan="4" scope="row">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
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<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
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<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
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<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
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<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
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<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
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<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
</td>
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<th class="formulainput" rowspan="3" scope="row">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
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<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
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<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
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<th class="formulainput" scope="row">Differentials</th>
<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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<h2 class="hd hd-2 unit-title">16. Taylor series of the Error function</h2>
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<p><b class="bf">Error function</b><br/></p><p>
Recall the <b class="bf"><span style="color:#27408C">error function</span></b> is defined by the following integral formula: </p><table id="a0000001552" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001553"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \text {erf}(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{2}{\sqrt {\pi }}\int _0^ x e^{-t^2}\, dt,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.483)</td></tr></table><p>
and cannot be expressed in terms of functions that we already know with algebraic operations such as addition and multiplication. <br/></p><p>
To obtain the Taylor series of the error function, we replace the integrand [mathjaxinline]\, e^{-t^2}\,[/mathjaxinline] with its Taylor series: </p><table id="a0000001554" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001555"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \text {erf}(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle \frac{2}{\sqrt {\pi }}\int _0^ x\, e^{-t^2}\, dt[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.484)</td></tr><tr id="a0000001556"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle \frac{2}{\sqrt {\pi }}\int _0^ x\, \left(\sum _{n=0}^{\infty } \frac{(-1)^ n t^{2n}}{n!}\right)\, dt[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.485)</td></tr><tr id="a0000001557"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle \frac{2}{\sqrt {\pi }} \left(\sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n+1}}{(2n+1)n!}\right).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.486)</td></tr></table>
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Taylor series of Fresnel integral
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Find the first three non-zero terms of Taylor series of the Fresnel integral [mathjaxinline]\, \displaystyle \int _0^ x \sin \left(t^2\right)\, dt\,[/mathjaxinline]. Recall that there is no formula for this integral. </p>
<p>
(Type <b class="bf">factoria(n)</b> for [mathjaxinline]n![/mathjaxinline].) </p>
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The first three non-zero terms of the Taylor series of <p style="display:inline">[mathjaxinline]\, \displaystyle \int _0^ x \sin \left(t^2\right)\, dt\, \, =\quad[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_taylor-tab16-problem1_2_1" class="text-input-dynamath capa_inputtype inline textline">
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Also find the radius of convergence [mathjaxinline]\, R[/mathjaxinline] of the Taylor series.<br/>(If [mathjaxinline]R=\infty ,\,[/mathjaxinline] enter <b class="bf">infty</b>.) <p style="display:inline">[mathjaxinline]R=\,[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="inputtype_taylor-tab16-problem1_3_1" class="text-input-dynamath capa_inputtype inline textline">
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<h2 class="hd hd-2 unit-title">17. Taylor polynomials and remainder theorem</h2>
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<p><b class="bf">Taylor polynomials</b><br/></p><p>
If the Taylor series of [mathjaxinline]\, f(x)\,[/mathjaxinline] is </p><table id="a0000001562" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001563"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(x)=a_0+a_1x +a_2 x^2 +a_3 x^3 + \cdots \qquad (\text {for }|x|<R),[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.490)</td></tr></table><p>
then the polynomial </p><table id="a0000001564" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001565"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle P_ n(x)=a_0+a_1x +a_2 x^2 +a_3 x^3 + \cdots a_ n x^ n[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.491)</td></tr></table><p>
is called the <b class="bf"><span style="color:#27408C">Taylor polynomial</span></b> of degree [mathjaxinline]\, n\,[/mathjaxinline] of [mathjaxinline]\, f(x)[/mathjaxinline]. In other words, the degree n Taylor polynomial is the polynomial obtained by truncating the Taylor series to degree n. </p><p>
The degree 1 and degree 2 Taylor polynomials for [mathjaxinline]\, f(x)\,[/mathjaxinline] are respectively the linear and quadratic approximations of [mathjaxinline]\, f(x)\,[/mathjaxinline] near [mathjaxinline]x=0[/mathjaxinline]. The degree n Taylor polynomial [mathjaxinline]P_ n(x)\,[/mathjaxinline] is the best fit degree n polynomial of [mathjaxinline]\, f(x)\,[/mathjaxinline] near [mathjaxinline]\, x=0,\, \,[/mathjaxinline] in the sense that </p><table id="a0000001566" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001567"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \, \displaystyle P_ n^{(k)}(0) = f^{(k)}(0)\qquad \text {for } 0<k\leq n.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.492)</td></tr></table><p>
In other words, at the point [mathjaxinline]x=0[/mathjaxinline], all the derivatives of the polynomial [mathjaxinline]P_ n[/mathjaxinline] matches the respective derivatives of [mathjaxinline]f[/mathjaxinline].<br/></p><p>
Just as we use linear and quadratic approximations, we can use Taylor polynomials of any degree to approximate a function near a point. But how do we know how accurate the approximation is? And what degree of Taylor polynomial is needed to achieve a certain accuracy? The answer is in the <b class="bf"><span style="color:#27408C">Taylor Remainder Theorem</span></b> below. </p><p><b class="bf">Taylor remainder theorem</b><br/></p><p>
Let [mathjaxinline]\, P_ n(x)\, =a_0+a_1x +a_2 x^2 + \cdots +a_ n x^ n\,[/mathjaxinline] be the degree [mathjaxinline]n[/mathjaxinline] Taylor polynomial of [mathjaxinline]\, f(x)[/mathjaxinline].<br/></p><p>
Then, if</p><ul class="itemize"><li><p>
[mathjaxinline]\, f\,[/mathjaxinline] is [mathjaxinline]n+1[/mathjaxinline] times differentiable on the open interval [mathjaxinline]\, (0,x),\,[/mathjaxinline] i.e. [mathjaxinline]\, f',\, f^{\prime \prime },\, \ldots ,f^{(n)},\, f^{(n+1)}[/mathjaxinline] exist on the open interval [mathjaxinline]\, (0,x),\,[/mathjaxinline] and </p></li><li><p>
[mathjaxinline]\, f',\, f^{\prime \prime },\, \ldots ,\, f^{(n)}[/mathjaxinline] are all continuous on the <b class="bf">closed</b> interval [mathjaxinline]\, [0,x],\,[/mathjaxinline] </p></li></ul><p>
then there is a number [mathjaxinline]\, c[/mathjaxinline] in [mathjaxinline]\, (0,x)\,[/mathjaxinline] such that </p><table id="a0000001568" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001569"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle R_{n}(x)\, =\, f(x)-P_ n(x)=\frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} \qquad (0<c<x).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.493)</td></tr></table><p>
This is the <b class="bf"><span style="color:#27408C">Taylor remainder theorem</span></b>, and [mathjaxinline]\, \displaystyle R_ n(x)\,[/mathjaxinline] is called the <b class="bf"><span style="color:#27408C">remainder</span></b> or <b class="bf"><span style="color:#27408C">error of approximation</span></b>. </p><p>
The [mathjaxinline]\, n=0\,[/mathjaxinline] case is the Mean Value Theorem (MVT). As in the MVT, we do not know exactly where [mathjaxinline]\, c\,[/mathjaxinline] is. Nonetheless, the Taylor remainder theorem is useful in giving upper bounds on errors of approximation by Taylor polynomials. </p><p><b class="bf">Remark:</b> The formula for [mathjaxinline]\, R_ n[/mathjaxinline] is true even if [mathjaxinline]x[/mathjaxinline] is outside the radius of convergence [mathjaxinline]\, R\,[/mathjaxinline] of the Taylor series. If [mathjaxinline]|x|<R,\,[/mathjaxinline] then in addition, </p><table id="a0000001570" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001571"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \, \displaystyle \lim _{n\rightarrow \infty } R_ n(x)=0\qquad (|x|<R).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.494)</td></tr></table><p>
This means that within the radius of convergence, we can obtain an approximation of arbitrarily high accuracy by using a Taylor polynomial of high enough degree.<br/></p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Proof of the Taylor remainder theorem<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
Given the hypothesis of the Taylor remainder theroem, we need to show there is a [mathjaxinline]c[/mathjaxinline] in the open interval [mathjaxinline](0,x)\,[/mathjaxinline] such that </p><table id="a0000001572" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001573"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle R_{n}(x)\, =\, f(x)-P_ n(x)=\frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} \qquad (0<c<x).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.495)</td></tr></table><p>
In the proof below, we will use Rolle's theorem (a special case of the MVT in which the function takes the same value at the two end points), so let us first recall the statement of <span style="color:#27408C"><b class="bf">Rolle's Theorem</b></span>: </p><p>
If [mathjaxinline]\, F\,[/mathjaxinline] is differentiable on [mathjaxinline]\, (0,x)\,[/mathjaxinline] and continuous on [mathjaxinline]\, [0,x]\,[/mathjaxinline] (as in the hypothesis of the MVT), and if [mathjaxinline]\, F(0)=F(x),\,[/mathjaxinline] then there is [mathjaxinline]c[/mathjaxinline] in [mathjaxinline](0,x)\,[/mathjaxinline] such that [mathjaxinline]F'(c)=0[/mathjaxinline].<br/></p><p>
Now, we proceed to prove the Remainder Theorem for any fixed degree [mathjaxinline]n[/mathjaxinline].<br/></p><p>
To begin, fix [mathjaxinline]\, x\,[/mathjaxinline] ([mathjaxinline]x\neq 0\,[/mathjaxinline]), and define </p><table id="a0000001574" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001575"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle a[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{R_ n(x)}{x^{n+1}} \qquad (\text {such that }\, R_ n(x)=ax^{n+1}).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.496)</td></tr></table><p>
With this [mathjaxinline]a[/mathjaxinline], define a new function [mathjaxinline]F_ n(t)[/mathjaxinline] with [mathjaxinline]\, 0\leq t\leq x[/mathjaxinline]: </p><table id="a0000001576" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001577"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle F_ n(t)=R_ n(t)-a t^{n+1}[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \left(f(t)-P_ n(t)\right)-a t^{n+1}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.497)</td></tr><tr id="a0000001578"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle f(t)-\left(f(0)+f'(0)\, t+\frac{f^{\prime \prime }(0)}{2} \, t^2+\cdots + \frac{f^{(n)}(0)}{n!}t^ n\right) - a t^{n+1}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.498)</td></tr></table><p>
Then, this new function has the following properties: </p><table id="a0000001579" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001580"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle F_ n(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle 0[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.499)</td></tr><tr id="a0000001581"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle F_ n(0)\, =\, F_ n^{\prime }(0)\, =\, F_ n^{\prime \prime }(0)\, =\, \cdots \, =F_ n^{(n)}(0)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle 0\qquad (\text {check this}).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.500)</td></tr></table><p>
Moreover, [mathjaxinline]\, F_ n\,[/mathjaxinline] is [mathjaxinline]n+1[/mathjaxinline] times differentiable on the interval [mathjaxinline]\, (0,x)\,[/mathjaxinline] with the first [mathjaxinline]n[/mathjaxinline] derivatives continuous on [mathjaxinline]\, [0,x],\,[/mathjaxinline] because the hypothesis of the theorem says that [mathjaxinline]\, f\,[/mathjaxinline] is [mathjaxinline]\, n+1\,[/mathjaxinline] times differentiable on [mathjaxinline]\, (0,x)\,[/mathjaxinline] with the first [mathjaxinline]n[/mathjaxinline] derivatives continuous on [mathjaxinline]\, [0,x],\,[/mathjaxinline] and [mathjaxinline]\, P_ n+at^{n+1}\,[/mathjaxinline] is a polynomial.<br/></p><p>
Now, we find [mathjaxinline]a[/mathjaxinline] by iterated applications of Rolle's Theorem. Since [mathjaxinline]\, F_ n\,[/mathjaxinline] is continuous on [mathjaxinline]\, [0,x],\,[/mathjaxinline] differentiable on [mathjaxinline]\, (0,x),\,[/mathjaxinline] and [mathjaxinline]\, F_ n(x)=F_ n(0)=0,\,[/mathjaxinline] by Rolle's Theorem, there is [mathjaxinline]\, c_0,\,[/mathjaxinline] with [mathjaxinline]\, 0<c_0<x\,[/mathjaxinline] such that </p><table id="a0000001582" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001583"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \, \displaystyle F_ n^{\prime }\left(c_0\right)=0\qquad (0<c_0<x) .[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.501)</td></tr></table><p>
But now since [mathjaxinline]\, F_ n^{\prime }(0)=0\,[/mathjaxinline] as well, and [mathjaxinline]\, F_ n'\,[/mathjaxinline] is continuous on [mathjaxinline]\, [0,c_0]\,[/mathjaxinline] and differentiable on [mathjaxinline]\, (0,c_0),\,[/mathjaxinline] we can apply Rolle's Theorem to [mathjaxinline]\, F_ n^{\prime }[/mathjaxinline] on [mathjaxinline]\, [0,c_0]\,[/mathjaxinline] and obtain a number [mathjaxinline]\, c_1\,[/mathjaxinline] with [mathjaxinline]0<c_1<c_0<x\,[/mathjaxinline] such that </p><table id="a0000001584" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001585"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \, \displaystyle F_ n^{\prime \prime }\left(c_1\right)=0 \qquad (0<c_1<c_0).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.502)</td></tr></table><p>
Again, [mathjaxinline]F_ n^{\prime \prime }(0)=0,\,[/mathjaxinline] and [mathjaxinline]F_ n^{\prime \prime }[/mathjaxinline] satisfies the hypothesis of Rolle's Theorem, so we can apply Rolle's Theorem to [mathjaxinline]\, F_ n^{\prime \prime }\,[/mathjaxinline] and obtain [mathjaxinline]c_2\,[/mathjaxinline] with [mathjaxinline]\, 0<c_2<c_1<c_0<x\,[/mathjaxinline] such that </p><table id="a0000001586" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001587"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \, \displaystyle F_ n^{(3)}\left(c_2\right)=0\qquad (0<c_2<c_1).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.503)</td></tr></table><p>
We can continue to apply Rolle's Theorem to the higher derivatives of [mathjaxinline]\, F_ n\,[/mathjaxinline] on shrinking intervals, with the last application to [mathjaxinline]F_ n^{(n)}[/mathjaxinline]. The end result is a number [mathjaxinline]c_ n[/mathjaxinline] satisfying </p><table id="a0000001588" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001589"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \, 0<c_ n<c_{n-1}<\cdots <c_2<c_1<c_0<x,\,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.504)</td></tr></table><p>
such that </p><table id="a0000001590" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001591"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \, \displaystyle F_ n^{(n+1)}\left(c_ n\right)=0\qquad (0<c_ n<x) .[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.505)</td></tr></table><p>
This final equation gives an expression for [mathjaxinline]a[/mathjaxinline]. Let [mathjaxinline]\, c=c_ n.\, \,[/mathjaxinline] We compute the derivative: </p><table id="a0000001592" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001593"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \, \displaystyle F_ n^{(n+1)}\left(c\right)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \left.\frac{d^{n+1}}{(dt)^{n+1}}\left( f(t)-\left(f(0)+f'(0)t+\frac{f^{\prime \prime }(0)}{2} \, t^2+\cdots + \frac{f^{(n)}(0)}{n!}t^ n \right) - a t^{n+1}\right)\right|_{t=c}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.506)</td></tr><tr id="a0000001594"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle f^{(n+1)}(c)- (n+1)! \, a\, =\, 0.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.507)</td></tr></table><p>
This implies [mathjaxinline]\, \displaystyle a=\frac{f^{(n+1)}(c)}{(n+1)!}\,[/mathjaxinline] for [mathjaxinline]0<c<x[/mathjaxinline]. <br/></p><p>
We have shown that given any fixed degree [mathjaxinline]n[/mathjaxinline] and fixed [mathjaxinline]x \, (x\neq 0)[/mathjaxinline], there is a number [mathjaxinline]\, c[/mathjaxinline] in [mathjaxinline](0,x)\,[/mathjaxinline] such that </p><table id="a0000001595" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001596"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle R_ n(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.508)</td></tr></table></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p><SCRIPT src="/assets/courseware/v1/9567be3f4e91361f26a8beaadf749715/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/latex2edx.js" type="text/javascript"/><LINK href="/assets/courseware/v1/daf81af0af57b85a105e0ed27b7873a0/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/latex2edx.css" rel="stylesheet" type="text/css"/>
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<h2 class="hd hd-2 unit-title">18. Worked example: Taylor approximation</h2>
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<p>
We approximate [mathjaxinline]\, \text {erf}(1)\,[/mathjaxinline] by a Taylor polynomial of degree up to 2 and find an upper bound on the error of approximation. </p><p>
Recall the Taylor series of [mathjaxinline]\, \displaystyle \text {erf}(x)\,[/mathjaxinline] is </p><table id="a0000001597" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001598"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \text {erf}(x)\, =\, \frac{2}{\sqrt {\pi }} \left(\sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n+1}}{(2n+1)n!}\right).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.509)</td></tr></table><p>
Since this series contains only odd power terms, the degree 1 and degree 2 Taylor polynomials are the same: </p><table id="a0000001599" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001600"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle P_2(x)\, =\, P_1(x) \, =\, \frac{2}{\sqrt {\pi }}x.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.510)</td></tr></table><p>
Hence, the approximation of [mathjaxinline]\, \text {erf}(1)\,[/mathjaxinline] by the Taylor polynomial of degree 2 is </p><table id="a0000001601" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001602"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \text {erf}(1)\, \approx \, P_2(1)\, =\, \frac{2}{\sqrt {\pi }}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.511)</td></tr></table><p>
Now, we find an upper bound on the error of approximation [mathjaxinline]\, \displaystyle \text {erf}(1)-P_2(1).\, \,[/mathjaxinline] By the Taylor remainder theorem, </p><table id="a0000001603" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001604"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \text {erf}(x)-P_2(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{\text {erf}^{(3)}(c)}{3!}\left(x^3\right)\qquad \text {for some }\, 0<c<x[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.512)</td></tr><tr id="a0000001605"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \Longrightarrow[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \text {erf}(1)-P_2(1)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{\text {erf}^{(3)}(c)}{3!}\left(1^3\right)\qquad \text {for some } \, 0<c<1.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.513)</td></tr></table><p>
Now, we compute [mathjaxinline]\displaystyle \, \text {erf}^{(3)}(x) \,[/mathjaxinline]: </p><table id="a0000001606" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001607"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \text {erf}'(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{2}{\sqrt {\pi }} e^{-x^2}\qquad (\text {by }\, \text {FTC2})[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.514)</td></tr><tr id="a0000001608"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \text {erf}^{\prime \prime }(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{2}{\sqrt {\pi }} (-2x) e^{-x^2}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.515)</td></tr><tr id="a0000001609"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \text {erf}^{(3)}(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{2}{\sqrt {\pi }} \left(-2e^{-x^2}+4x^2 e^{-x^2}\right)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.516)</td></tr><tr id="a0000001610"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{4}{\sqrt {\pi }} e^{-x^2}\left(2x^2-1\right).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.517)</td></tr></table><p>
Since both [mathjaxinline]\, \left|e^{-x^2}\right|< 1\,[/mathjaxinline] and [mathjaxinline]\, \displaystyle \left|2x^2-1\right|<1\,[/mathjaxinline] for [mathjaxinline]\, 0< x<1,\,[/mathjaxinline] we conclude that </p><table id="a0000001611" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001612"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \left|\text {erf}^{(3)}(c)\right|[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle <[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{4}{\sqrt {\pi }} \qquad \text {for all }\, 0<c<1.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.518)</td></tr></table><p>
In other words, [mathjaxinline]\, 4/\sqrt {\pi }\,[/mathjaxinline] is an upper bound of [mathjaxinline]\, \displaystyle \left|\text {erf}^{(3)}\right|\,[/mathjaxinline] on the interval [mathjaxinline]\, (0,1)[/mathjaxinline]. This gives an upper bound on the error of approximation:<br/></p><table id="a0000001613" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001614"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \left|\text {erf}(1)-P_2(1)\right|[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle <[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \left|\frac{4/\sqrt {\pi }}{3!}\right| \, =\, \frac13\left(\frac{2}{\sqrt {\pi }}\right).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.519)</td></tr></table><p>
In other words, the error of the approximation of [mathjaxinline]\, \text {erf}(1)\,[/mathjaxinline] by [mathjaxinline]\, \displaystyle P_2(1) = \frac{2}{\sqrt {\pi }}\,[/mathjaxinline] is less than [mathjaxinline]\, \displaystyle \frac13\left(\frac{2}{\sqrt {\pi }}\right)[/mathjaxinline]. </p>
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Approximating square root of two
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Often, numerical tools such as graphing tools and calculators use Taylor polynomials to approximate numbers.<br/></p>
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Approximate [mathjaxinline]\, \sqrt {2}\,[/mathjaxinline] by [mathjaxinline]P_3(1),\,[/mathjaxinline] where [mathjaxinline]P_3(x)\,[/mathjaxinline] is the degree [mathjaxinline]3[/mathjaxinline] Taylor polynomial of [mathjaxinline]\displaystyle \, \sqrt {1+x}[/mathjaxinline] centered at [mathjaxinline]x=0[/mathjaxinline]. <p style="display:inline">[mathjaxinline]\, \displaystyle \sqrt {2}\approx P_3(1)\, =\, \quad[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="formulaequationinput_taylor-tab18-problem1_2_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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Use the Taylor remainder theorem to give an expression of [mathjaxinline]\, \sqrt {2}-P_3(1)\,[/mathjaxinline] in terms of [mathjaxinline]c,\,[/mathjaxinline] where [mathjaxinline]c[/mathjaxinline] is some number between [mathjaxinline]0[/mathjaxinline] and [mathjaxinline]1[/mathjaxinline].<br/>(Your answer should be a function of [mathjaxinline]c[/mathjaxinline]. Enter <b class="bf">factorial(n)</b> for [mathjaxinline]n![/mathjaxinline]. )<br/><p style="display:inline">[mathjaxinline]\displaystyle \sqrt {2}-P_3(1) =\quad[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="formulaequationinput_taylor-tab18-problem1_3_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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Find the maximum over the interval [mathjaxinline]\, [0,1][/mathjaxinline] of the absolute value of the expression you entered above. This gives an upper bound on [mathjaxinline]\, \left|\sqrt {2}-P_3(1)\right|[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\displaystyle \left|\sqrt {2}-P_3(1)\right|\leq \quad[/mathjaxinline]</p>
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Approximating the number e
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Let [mathjaxinline]P_ n(x)[/mathjaxinline] be the degree [mathjaxinline]n[/mathjaxinline] Taylor polynomial of [mathjaxinline]\, e^ x\,[/mathjaxinline]. Then the approximation of the number [mathjaxinline]\, e\,[/mathjaxinline] by [mathjaxinline]\, P_ n(1)\,[/mathjaxinline] is </p>
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[mathjaxinline]\displaystyle e\, \approx \, P_ n(1)\, =\, 1+1+\frac1{2!}+\frac{1}{3!}+\cdots + \frac{1}{n!}.[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.529)</td>
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Follow the two steps below to find [mathjaxinline]n[/mathjaxinline] such that [mathjaxinline]P_ n(1)\,[/mathjaxinline] is accurate to 4 decimal places, that is, to within [mathjaxinline]0.00005=5\times 10^{-5}[/mathjaxinline]. <br/></p>
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First, use the Taylor remainder theorem to give an expression of the error of approximation, [mathjaxinline]\, e-P_ n(1),\,[/mathjaxinline] in terms of [mathjaxinline]c[/mathjaxinline] and [mathjaxinline]n,\,[/mathjaxinline] where [mathjaxinline]c[/mathjaxinline] is some number between [mathjaxinline]0[/mathjaxinline] and [mathjaxinline]1[/mathjaxinline].<br/>(Enter <b class="bf">factorial(n)</b> for [mathjaxinline]n![/mathjaxinline].)<br/><p style="display:inline">[mathjaxinline]\displaystyle e-P_ n(1) =\quad[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_taylor-tab18-problem2_2_1" class="text-input-dynamath capa_inputtype inline textline">
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Use the fact that [mathjaxinline]\, e^ c&lt;3[/mathjaxinline] for [mathjaxinline]0&lt;c&lt;1\,[/mathjaxinline] to find the smallest [mathjaxinline]n[/mathjaxinline] such that the expression for [mathjaxinline]\, e-P_ n(1)\,[/mathjaxinline] above is [mathjaxinline]\, &lt;5\times 10^{-5}[/mathjaxinline]. <p style="display:inline">[mathjaxinline]\displaystyle n=[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="formulaequationinput_taylor-tab18-problem2_3_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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Using the [mathjaxinline]n[/mathjaxinline] you found above, compute [mathjaxinline]\, \displaystyle P_ n(1)\,[/mathjaxinline] and check whether it is correct to 4 decimal places. <p style="display:inline">[mathjaxinline]\displaystyle e\approx[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 3" role="group"><div id="formulaequationinput_taylor-tab18-problem2_4_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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<h2 class="hd hd-2 unit-title">19. Taylor polynomials centered about other points</h2>
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What if we want to find a Taylor polynomial approximation to a function at a point that is not [mathjaxinline]x=0[/mathjaxinline]? </p><p>
There is nothing special about the point [mathjaxinline]x=0[/mathjaxinline]. In fact, we can translate any function to the right or to the left by some amount. So it makes sense that we should be able to find power series and Taylor approximations centered about other points. In fact, you have already seen linear and quadratic approximations centered about other points. </p><p>
The mathlet below allows you to see the Taylor approximation up to order [mathjaxinline]9[/mathjaxinline] at any point [mathjaxinline]x=a[/mathjaxinline]. </p><iframe src="https://mathlets1801.surge.sh/taylorPolynomials.html" width="1100 px" height="630 px" style="border:0px; display: block;"/>
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Find the linear approximation [mathjaxinline]\, T_1[/mathjaxinline] of [mathjaxinline]f(x)[/mathjaxinline] at [mathjaxinline]x=a[/mathjaxinline]. This is the Taylor polynomial of degree [mathjaxinline]1[/mathjaxinline] at [mathjaxinline]x=a[/mathjaxinline]. </p>
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Find the quadratic approximation [mathjaxinline]T_2(x)\,[/mathjaxinline] of [mathjaxinline]f(x)[/mathjaxinline] at [mathjaxinline]x=a[/mathjaxinline], by adding a quadratic term to [mathjaxinline]\, T_1(x)[/mathjaxinline]. This is the degree 2 Taylor polynomial of [mathjaxinline]f(x)[/mathjaxinline] at [mathjaxinline]x=a[/mathjaxinline]. </p>
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<h2 class="hd hd-2 unit-title">20. Taylor series centered about other points</h2>
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Let [mathjaxinline]\, g(t)=f(t+a).\, \,[/mathjaxinline] That is, [mathjaxinline]\, g\,[/mathjaxinline] is the translation of [mathjaxinline]f[/mathjaxinline] to the left by [mathjaxinline]a[/mathjaxinline]. </p><p>
Recall the Taylor series of [mathjaxinline]\, g(t)\,[/mathjaxinline] at [mathjaxinline]t=0[/mathjaxinline] is </p><table id="a0000001639" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001640"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle g(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } \frac{g^{(n)}(0)}{n!}t^ n \qquad (|t|<R).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.533)</td></tr></table><p>
Then </p><table id="a0000001641" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001642"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(t+a)=g(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } \frac{g^{(n)}(0)}{n!}t^ n \qquad (|t|<R)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.534)</td></tr><tr id="a0000001643"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } \frac{f^{(n)}(a)}{n!}t^ n \qquad (|t|<R).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.535)</td></tr></table><p>
Now let [mathjaxinline]\, x=t+a.\, \,[/mathjaxinline] In terms of [mathjaxinline]x,[/mathjaxinline] the above formula becomes </p><table id="a0000001644" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001645"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } \frac{f^{(n)}(a)}{n!}(x-a)^ n \qquad (|x-a|<R).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.536)</td></tr></table><p>
This is the Taylor series of [mathjaxinline]f(x)[/mathjaxinline] at [mathjaxinline]x=a[/mathjaxinline].<br/></p><p>
Note that the radius of convergence of the Taylor series of [mathjaxinline]f(x)[/mathjaxinline] at [mathjaxinline]x=a\,[/mathjaxinline] is the number [mathjaxinline]\, R\,[/mathjaxinline] such that [mathjaxinline]\, f(x)\,[/mathjaxinline] converges when [mathjaxinline]\, |x-a|<R,\,[/mathjaxinline] and diverges when [mathjaxinline]\, |x-a|>R[/mathjaxinline]. </p><p>
If [mathjaxinline]a=0[/mathjaxinline], we get the formula for the Taylor series that we started with in this section. This special case of Taylor's formula gives us a power series often referred to as the <span style="color:#27408C"><b class="bf">Maclaurin series</b></span>. </p>
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Taylor series of a polynomial at a
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The Taylor series for the polynomial [mathjaxinline]\displaystyle \ f(x) = 3x^3 + 4x^2 - 2x -1[/mathjaxinline] at [mathjaxinline]x=-1[/mathjaxinline] takes the form: </p>
<p>
[mathjaxinline]\displaystyle f(x) = a_0 + a_1 (x+1) + a_2 (x+1)^2 + a_3 (x+1)^3.[/mathjaxinline] Find the coefficients of this Taylor series. </p>
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<p style="display:inline">[mathjaxinline]a_0=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]a_1=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]a_2=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]a_3=[/mathjaxinline]</p>
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Taylor series of the exponential at a
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Find the term of order [mathjaxinline]11[/mathjaxinline] in the Taylor series of [mathjaxinline]e^ x[/mathjaxinline] at [mathjaxinline]x=a[/mathjaxinline].<br/>(Be sure to include the entire term including the power of [mathjaxinline]x[/mathjaxinline]. Enter <b class="bf">factorial(n)</b> for [mathjaxinline]n![/mathjaxinline].)<br/></p>
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Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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Find the first 4 nonzero terms of the Taylor series for the function [mathjaxinline]\displaystyle \frac{1}{1-x}[/mathjaxinline] at [mathjaxinline]x=2[/mathjaxinline]. <br/>(Do not include extra terms.)<br/></p>
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\(\)
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This series converges when <p style="display:inline">[mathjaxinline]\, |x-2|&lt;\quad[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="formulaequationinput_taylor-tab20-problem3_3_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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<h2 class="hd hd-2 unit-title">21. Summary</h2>
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<p><b class="bfseries">General power series</b></p><p>
A <span style="color:#27408C"><b class="bf">power series </b></span> is an infinite series involving positive powers of a variable [mathjaxinline]x[/mathjaxinline]: </p><table id="a0000001665" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001666"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle a_0+a_1x + a_2 x^2 + a_3 x^3 + \dotsb[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.550)</td></tr></table><p>
The <span style="color:#27408C"><b class="bf">radius of convergence</b></span> [mathjaxinline]R[/mathjaxinline] of the power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n[/mathjaxinline], is a real number [mathjaxinline]0 \leq R < \infty[/mathjaxinline] such that </p><ul class="itemize"><li><p>
for [mathjaxinline]|x|<R[/mathjaxinline], the power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n[/mathjaxinline] converges (to a finite number); </p></li><li><p>
for [mathjaxinline]|x|>R[/mathjaxinline], the power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n[/mathjaxinline] diverges; </p></li><li><p>
for [mathjaxinline]|x| = R[/mathjaxinline], the power series may converge or diverge. But we will mostly ignore what happens at the end points of the interval of convergence. </p></li></ul><p><b class="bfseries">Examples:</b></p><ul class="itemize"><li><p>
Geometric series: [mathjaxinline]\displaystyle 1+ x + x^2 + x^3 + \dotsb = \sum _{n=0}^{\infty } x^ n[/mathjaxinline], radius of convergence is [mathjaxinline]1[/mathjaxinline]. </p></li><li><p>
Polynomials: [mathjaxinline]\displaystyle a_0 + a_1x + a_2x + \dotsb a_ N x^ N = \sum _{n=0}^{N} a_ n x^ n[/mathjaxinline], radius of convergence [mathjaxinline]\infty[/mathjaxinline]. In other words, the sum converges for all [mathjaxinline]x[/mathjaxinline]. </p></li></ul><p><b class="bfseries">Finding the radius of convergence</b></p><p>
Given a power series [mathjaxinline]\displaystyle \, \sum _{n=0}^{\infty } a_ n x^ n[/mathjaxinline], the ratio test implies that the power series converges if </p><table id="a0000001667" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001668"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \lim _{n \rightarrow \infty } \left| \frac{a_{n+1} x^{n+1}}{a_ n x^ n}\right|[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \lim _{n \rightarrow \infty } \left| \frac{a_{n+1} }{a_ n}\right||x|[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle <[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle 1.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.551)</td></tr></table><p>
There are 3 possibilities: </p><ol class="enumerate"><li value="1"><p>
There is a finite number [mathjaxinline]R[/mathjaxinline] such that </p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle |x|<R \Longrightarrow \lim _{n \rightarrow \infty } \left| \frac{a_{n+1} }{a_ n}\right| |x|< 1[/mathjaxinline], </p></li><li><p>
[mathjaxinline]\displaystyle |x|>R \Longrightarrow \lim _{n \rightarrow \infty } \left| \frac{a_{n+1} }{a_ n}\right| |x|>1[/mathjaxinline]. </p></li></ul><p>
We say the radius of convergence is [mathjaxinline]R[/mathjaxinline]. </p></li><li value="2"><p>
For all [mathjaxinline]x[/mathjaxinline] [mathjaxinline]\displaystyle \lim _{n \rightarrow \infty } \left| \frac{a_{n+1} }{a_ n}\right||x| < 1[/mathjaxinline]. We say the radius of convergence is [mathjaxinline]\infty[/mathjaxinline]. (All [mathjaxinline]x[/mathjaxinline] satisfy [mathjaxinline]|x| < \infty .[/mathjaxinline]) </p></li><li value="3"><p>
[mathjaxinline]\displaystyle \lim _{n \rightarrow \infty } \left| \frac{a_{n+1} }{a_ n}\right| |x| > 1[/mathjaxinline] for all [mathjaxinline]x \neq 0[/mathjaxinline]. We say the radius of convergence is [mathjaxinline]0[/mathjaxinline]. </p></li></ol><p><b class="bfseries">Remark: Alternative method using ratio test</b></p><p>
(Note that in the method that follows, the [mathjaxinline]n+1[/mathjaxinline] term is in the denominator and the [mathjaxinline]n[/mathjaxinline] term is in the numerator, which is the opposite of the ratio test.) </p><p>
Given a power series [mathjaxinline]\displaystyle \, \sum _{n=0}^{\infty } a_ n x^ n,[/mathjaxinline] </p><p>
if [mathjaxinline]\displaystyle \lim _{n \rightarrow \infty } \left|\frac{a_ n}{a_{n+1}} \right| = R[/mathjaxinline] (where [mathjaxinline]R[/mathjaxinline] exists or is [mathjaxinline]\infty[/mathjaxinline]), </p><p>
then the radius of convergence for the power series is [mathjaxinline]R[/mathjaxinline]. </p><p><b class="bfseries">Example</b></p><p>
Consider [mathjaxinline]\displaystyle \, \sum _{n=0}^{\infty } 2^ n x^ n,[/mathjaxinline] then [mathjaxinline]\displaystyle \lim _{n \rightarrow \infty } \left| \frac{2^ n}{2^{n+1}}\right| = \frac12 \quad \Longrightarrow \quad R = \frac12.[/mathjaxinline] </p><p><b class="bfseries">Root test for radius of convergence</b></p><p>
Given a power series [mathjaxinline]\displaystyle \, \sum _{n=0}^{\infty } a_ n x^ n[/mathjaxinline], the root test implies that the power series converges if </p><table id="a0000001669" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001670"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \lim _{n \rightarrow \infty } \sqrt [n]{\left|a_ n x^ n\right|}[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \lim _{n \rightarrow \infty }\sqrt [n]{\left|a_ n\right|}|x|[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle <[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle 1.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.552)</td></tr></table><p>
There are 3 possibilities: </p><ol class="enumerate"><li value="1"><p>
There is a finite number [mathjaxinline]R[/mathjaxinline] such that </p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle |x|<R \Longrightarrow \lim _{n \rightarrow \infty } \sqrt [n]{\left| a_ n\right|} |x|< 1[/mathjaxinline], </p></li><li><p>
[mathjaxinline]\displaystyle |x|>R \Longrightarrow \lim _{n \rightarrow \infty } \sqrt [n]{\left| a_ n\right|} |x|>1[/mathjaxinline]. </p></li></ul><p>
We say the radius of convergence is [mathjaxinline]R[/mathjaxinline]. </p></li><li value="2"><p>
For all [mathjaxinline]x[/mathjaxinline] [mathjaxinline]\displaystyle \lim _{n \rightarrow \infty }\sqrt [n]{\left|a_ n\right|}|x| < 1[/mathjaxinline]. We say the radius of convergence is [mathjaxinline]\infty[/mathjaxinline]. (All [mathjaxinline]x[/mathjaxinline] satisfy [mathjaxinline]|x| < \infty .[/mathjaxinline]) </p></li><li value="3"><p>
[mathjaxinline]\displaystyle \lim _{n \rightarrow \infty }\sqrt [n]{\left|a_ n\right|}|x| > 1[/mathjaxinline] for all [mathjaxinline]x \neq 0[/mathjaxinline]. We say the radius of convergence is [mathjaxinline]0[/mathjaxinline]. </p></li></ol><p><b class="bfseries">Example</b></p><p>
Consider [mathjaxinline]\displaystyle \, \sum _{n=0}^{\infty } 2^ n x^ n,[/mathjaxinline] then [mathjaxinline]\displaystyle \lim _{n \rightarrow \infty } \sqrt [n]{\left| 2^ nx^ n\right|} = 2|x| < 1[/mathjaxinline] when [mathjaxinline]|x|<\frac12[/mathjaxinline]. This implies that the radius of convergence is [mathjaxinline]R = \frac12.[/mathjaxinline] </p><p><b class="bfseries">Properties of power series</b></p><p>
Add, subtract, multiply, divide, differentiate, and integrate convergent power series as one does for polynomials. We will discuss multiplication and division at a later time. </p><p>
Consider the power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n,[/mathjaxinline] which converges for [mathjaxinline]|x|< A[/mathjaxinline]. </p><ul class="itemize"><li><p>
The derivative [mathjaxinline]\displaystyle \frac{d}{dx} \sum _{n=0}^{\infty } a_ n x^ n = \sum _{n=1}^{\infty } n \, a_ n x^{n-1}[/mathjaxinline] also converges for [mathjaxinline]|x| < A[/mathjaxinline]. </p></li><li><p>
The integral [mathjaxinline]\displaystyle \int \left( \sum _{n=0}^{\infty } a_ n x^ n \right) \, dx = c + \sum _{n=0}^{\infty } a_ n \frac{x^{n+1}}{n+1}[/mathjaxinline] also converges for [mathjaxinline]|x| < A[/mathjaxinline]. Note that [mathjaxinline]c[/mathjaxinline] is the constant of integration. </p></li></ul><p>
Consider another power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } b_ n x^ n[/mathjaxinline], which converges for [mathjaxinline]|x|<B[/mathjaxinline]. </p><ul class="itemize"><li><p>
If [mathjaxinline]A \neq B[/mathjaxinline], then [mathjaxinline]\displaystyle \left( \sum _{n=0}^{\infty } a_ n x^ n\right) \pm \left( \sum _{n=0}^{\infty } b_ n x^ n\right) = \sum _{n=0}^{\infty } (a_ n \pm b_ n) x^ n[/mathjaxinline] converges for [mathjaxinline]|x|< \textrm{min}(A,B)[/mathjaxinline]. </p></li><li><p>
If [mathjaxinline]A = B[/mathjaxinline] , then [mathjaxinline]\displaystyle \left( \sum _{n=0}^{\infty } a_ n x^ n\right) \pm \left( \sum _{n=0}^{\infty } b_ n x^ n\right) = \sum _{n=0}^{\infty } (a_ n \pm b_ n) x^ n[/mathjaxinline] has a radius of convergence which is at least [mathjaxinline]A[/mathjaxinline], but it <em>could</em> have a larger radius of convergence. </p></li></ul><p><b class="bfseries">Taylor's formula</b></p><p>
Recall that [mathjaxinline]\, \displaystyle n!=n (n-1) (n-2)\cdots (3)(2)(1)\,[/mathjaxinline] for all integers [mathjaxinline]\, n \geq 1\,[/mathjaxinline]. </p><p>
We define [mathjaxinline]\, 0!=1\,[/mathjaxinline]. This is a very valuable convention that simplifies many formulas. </p><p><b class="bf"><span style="color:#27408C">Taylor's formula</span></b> says that </p><table id="a0000001671" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001672"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle f(0) + f'(0)x + f^{\prime \prime }(0) \frac{x^2}{2} + f^{\prime \prime \prime }(0) \frac{x^3}{6} + \dotsb[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.553)</td></tr><tr id="a0000001673"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{f(0)}{0!}x^0 + \frac{f'(0)}{1!}x + \frac{f^{\prime \prime }(0)}{2!} x^2 + \frac{f^{\prime \prime \prime }(0)}{3!} x^3 + \dotsb[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.554)</td></tr><tr id="a0000001674"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } \frac{f^{(n)}(0)}{n!}x^ n[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.555)</td></tr></table><p>
when [mathjaxinline]\, |x|<R\,[/mathjaxinline] where [mathjaxinline]\, R\,[/mathjaxinline] is the radius of convergence of the power series above.<br/></p><p>
The power series in Taylor's formula is called the <b class="bf"><span style="color:#27408C">Taylor series</span></b> of [mathjaxinline]\, f(x)\,[/mathjaxinline]. </p><p><b class="bfseries">Important examples</b></p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle e^ x = \sum _{n=0}^{\infty } \frac{x^ n}{n!}\, =\, 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots[/mathjaxinline]<br/></p></li><li><p>
[mathjaxinline]\displaystyle \sin x = \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n+1}}{(2n+1)!}\, =\, x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots[/mathjaxinline] </p></li><li><p>
[mathjaxinline]\displaystyle \cos x = \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n}}{(2n)!}\, =\, 1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots[/mathjaxinline] </p></li></ul><p>
We will use Taylor's formula to derive the series for [mathjaxinline]\, \sin (x)\,[/mathjaxinline] and [mathjaxinline]\, \cos (x)\,[/mathjaxinline] on the next page. </p><p>
Notice that the factorial appears in the denominator of all terms in all three power series above.<br/></p><p>
Using the Taylor series of [mathjaxinline]\, e^ x,\,[/mathjaxinline] we find a formula for the number [mathjaxinline]e[/mathjaxinline] as the rapidly converging series: </p><table id="a0000001675" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001676"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle e= \sum _{n=0}^{\infty } \frac{1}{n!}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.556)</td></tr></table><p><b class="bfseries">Known Maclaurin series</b></p><p>
So far, we have used Taylor's formula to obtain the following Taylor series: </p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle \frac1{1-x} = \sum _{n=0}^{\infty } x^ n\,[/mathjaxinline] for [mathjaxinline]\, |x|<1[/mathjaxinline]. </p></li><li><p>
[mathjaxinline]\displaystyle e^ x = \sum _{n=0}^{\infty } \frac{x^ n}{n!}[/mathjaxinline] </p></li><li><p>
[mathjaxinline]\displaystyle \sin x = \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n+1}}{(2n+1)!}[/mathjaxinline] </p></li><li><p>
[mathjaxinline]\displaystyle \cos x = \sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n}}{(2n)!}[/mathjaxinline]. </p></li></ul><p>
We have also integrated the geometric series to obtain a power series for [mathjaxinline]\, \displaystyle \ln (1-x)[/mathjaxinline]: </p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle -\ln (1-x)\, =\, \sum _{n=1}^{\infty } \frac{x^ n}{n}[/mathjaxinline]. </p></li></ul><p><b class="bfseries">Multiplying two power series</b></p><p>
We multiply two powers series using the same rule as when we multiply two polynomials.<br/></p><p>
Consider the power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n x^ n,[/mathjaxinline] which converges for [mathjaxinline]|x|< A[/mathjaxinline], and the power series [mathjaxinline]\displaystyle \sum _{n=0}^{\infty } b_ n x^ n[/mathjaxinline], which converges for [mathjaxinline]|x|<B[/mathjaxinline]. </p><ul class="itemize"><li><p>
The product of the two power series converges for [mathjaxinline]|x|< \textrm{min}(A,B)[/mathjaxinline], but it <em>could</em> have a larger radius of convergence. </p></li></ul><p><b class="bfseries">Dividing two power series</b></p><p>
If </p><table id="a0000001677" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001678"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle F(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{G(x)}{H(x)}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.557)</td></tr></table><p>
where [mathjaxinline]\, F(x), \, G(x),\, H(x)\,[/mathjaxinline] are all power series, then we can find [mathjaxinline]\, F(x)\,[/mathjaxinline] by solving the following equation of power series degree by degree: </p><table id="a0000001679" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001680"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle F(x)H(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle G(x).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.558)</td></tr></table><p>
The radius of convergence of [mathjaxinline]\, F(x)\,[/mathjaxinline] is more difficult to track, since [mathjaxinline]\, F(x)\,[/mathjaxinline] diverges whenever [mathjaxinline]\, H(x)\, =\, 0\,[/mathjaxinline] and [mathjaxinline]\, G(x)\neq 0[/mathjaxinline]. </p><p><b class="bfseries">Substitution and Taylor series</b></p><p>
We can find the composition of two power series [mathjaxinline]\, f(g(x))\,[/mathjaxinline] by using similar rules as composition of polynomials. In this course, we will only substitute a polynomial [mathjaxinline]p(x)[/mathjaxinline] into a power series [mathjaxinline]f(x)[/mathjaxinline].<br/></p><p>
If the radius of convergence of the power series [mathjaxinline]\, f(x)\,[/mathjaxinline] is [mathjaxinline]\, R,\,[/mathjaxinline] then the power series [mathjaxinline]\, f(p(x))\,[/mathjaxinline] converges whenever [mathjaxinline]\, |p(x)|<R\,[/mathjaxinline].<br/></p><p>
In particular, we can find a Taylor series for a function [mathjaxinline]\ f(x)\,[/mathjaxinline] at [mathjaxinline]x=0[/mathjaxinline] and then substitute in polynomial of the form [mathjaxinline]p(x) = ax^ n[/mathjaxinline] for [mathjaxinline]x[/mathjaxinline] since [mathjaxinline]p(0) = 0[/mathjaxinline]. </p><p><b class="bfseries">Error function</b></p><p>
Recall the <b class="bf"><span style="color:#27408C">error function</span></b> is defined by the following integral formula: </p><table id="a0000001681" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001682"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \text {erf}(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{2}{\sqrt {\pi }}\int _0^ x e^{-t^2}\, dt,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.559)</td></tr></table><p>
and cannot be expressed in terms of functions that we already know with algebraic operations such that addition and multiplication. <br/></p><p>
To obtain the Taylor series of the error function, we replace the integrand [mathjaxinline]\, e^{-t^2}\,[/mathjaxinline] with its Taylor series: </p><table id="a0000001683" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001684"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \text {erf}(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{2}{\sqrt {\pi }}\int _0^ x\, e^{-t^2}\, dt[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.560)</td></tr><tr id="a0000001685"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{2}{\sqrt {\pi }}\int _0^ x\, \left(\sum _{n=0}^{\infty } \frac{(-1)^ n t^{2n}}{n!}\right)\, dt[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.561)</td></tr><tr id="a0000001686"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{2}{\sqrt {\pi }} \left(\sum _{n=0}^{\infty } \frac{(-1)^ n x^{2n+1}}{(2n+1)n!}\right).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.562)</td></tr></table><p><b class="bfseries">Taylor polynomials</b></p><p>
If the Taylor series of [mathjaxinline]\, f(x)\,[/mathjaxinline] is </p><table id="a0000001687" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001688"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(x)=a_0+a_1x +a_2 x^2 +a_3 x^3 + \cdots \qquad (\text {for }|x|<R),[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.563)</td></tr></table><p>
then the polynomial </p><table id="a0000001689" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001690"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle P_ n(x)=a_0+a_1x +a_2 x^2 +a_3 x^3 + \cdots a_ n x^ n[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.564)</td></tr></table><p>
is called the <b class="bf"><span style="color:#27408C">Taylor polynomial</span></b> of degree [mathjaxinline]\, n\,[/mathjaxinline] of [mathjaxinline]\, f(x)[/mathjaxinline]. In other words, a Taylor polynomial is the polynomial obtained by truncating the Taylor series to degree [mathjaxinline]\, n[/mathjaxinline]. </p><p>
The degree [mathjaxinline]\, n\,[/mathjaxinline] Taylor polynomial [mathjaxinline]P_ n(x)\,[/mathjaxinline] is the best fit degree [mathjaxinline]\, n\,[/mathjaxinline] polynomial of [mathjaxinline]\, f(x)\,[/mathjaxinline] at [mathjaxinline]\, x=0,\, \,[/mathjaxinline] in the sense that </p><table id="a0000001691" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001692"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \, \displaystyle \frac{d^ k}{dx^ k} P_ n(x) = \left. \frac{d^ k}{dx^ k} f(x)\right|_{x=0}\qquad \text {for } 0<k\leq n.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.565)</td></tr></table><p>
Hence, the degree [mathjaxinline]1[/mathjaxinline] and degree [mathjaxinline]2[/mathjaxinline] Taylor polynomials for [mathjaxinline]\, f(x),[/mathjaxinline] are the linear and quadratic approximations of [mathjaxinline]\, f(x)\,[/mathjaxinline] respectively.<br/></p><p>
Taylor polynomials are especially useful as for approximating functions, functions that cannot be expressed algebraically in terms of the elementary functions that we know, such as the error function. Often, numerical tools, such as the graphing tool on your calculator or computer, use Taylor polynomials to approximate these functions. <br/></p><p>
We can use Taylor polynomials to approximate a function with arbitrarily high accuracy inside the radius of convergence of the Taylor series. But how do we know the degree of the Taylor polynomial needed to achieve a certain accuracy? The answer is in the <b class="bf"><span style="color:#27408C">Taylor remainder theorem</span></b> below. </p><p><b class="bfseries">Taylor remainder theorem</b></p><p>
Suppose the Taylor series of [mathjaxinline]\, f(x)\,[/mathjaxinline] is </p><table id="a0000001693" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001694"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(x)=a_0+a_1x +a_2 x^2 +a_3 x^3 + \cdots \qquad \text {for } |x|<R,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.566)</td></tr></table><p>
and let [mathjaxinline]\, P_ n(x)\,[/mathjaxinline] be the degree [mathjaxinline]n[/mathjaxinline] Taylor polynomial of [mathjaxinline]\, f(x)[/mathjaxinline]: </p><table id="a0000001695" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001696"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle P_ n(x)=a_0+a_1x +a_2 x^2 + \cdots +a_ n x^ n.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.567)</td></tr></table><p>
Then, for any [mathjaxinline]\, |x|<R,\,[/mathjaxinline] </p><ul class="itemize"><li><p>
if [mathjaxinline]\, f(x),\,[/mathjaxinline] is [mathjaxinline]n+1[/mathjaxinline] times differentiable on the open interval [mathjaxinline]\, (0,x),\,[/mathjaxinline] that is, [mathjaxinline]\, f^{(n+1)}(x)\,[/mathjaxinline] and all lower derivatives [mathjaxinline]\, f\,[/mathjaxinline] exist on [mathjaxinline]\, (0,x)[/mathjaxinline], and </p></li><li><p>
if [mathjaxinline]\, f^{(n)}\,[/mathjaxinline] is continuous on the closed interval [mathjaxinline]\, [0,x],\,[/mathjaxinline] </p></li></ul><p>
then there is a number [mathjaxinline]\, c[/mathjaxinline] in [mathjaxinline]\, (0,x)\,[/mathjaxinline] such that </p><table id="a0000001697" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001698"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(x)-P_ n(x)=\frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.568)</td></tr></table><p>
This is the <b class="bf"><span style="color:#27408C">Taylor remainder theorem</span></b>. Note that the [mathjaxinline]\, n=0\,[/mathjaxinline] case is the Mean Value Theorem (MVT). As in the MVT, we do not know exactly where [mathjaxinline]\, c\,[/mathjaxinline] is. <br/></p><p>
Nevertheless, we can use the Taylor remainder theorem to find upper bounds on the error caused by approximating a function by a Taylor polynomial. </p><p><b class="bfseries">Taylor series centered at [mathjaxinline]x=a[/mathjaxinline]</b></p><p>
Let [mathjaxinline]\, g(t)=f(t+a).\, \,[/mathjaxinline] That is, [mathjaxinline]\, g\,[/mathjaxinline] is the translation of [mathjaxinline]f[/mathjaxinline] to the left by [mathjaxinline]a[/mathjaxinline]. </p><p>
Recall the Taylor series of [mathjaxinline]\, g(t)\,[/mathjaxinline] at [mathjaxinline]t=0[/mathjaxinline] is </p><table id="a0000001699" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001700"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle g(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } \frac{g^{(n)}(0)}{n!}t^ n \qquad (|t|<R).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.569)</td></tr></table><p>
Then </p><table id="a0000001701" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001702"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(t+a)=g(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } \frac{g^{(n)}(0)}{n!}t^ n \qquad (|t|<R)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.570)</td></tr><tr id="a0000001703"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } \frac{f^{(n)}(a)}{n!}t^ n \qquad (|t|<R).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.571)</td></tr></table><p>
Now let [mathjaxinline]\, x=t+a.\, \,[/mathjaxinline] In terms of [mathjaxinline]x,[/mathjaxinline] the above formula becomes </p><table id="a0000001704" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001705"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } \frac{f^{(n)}(a)}{n!}(x-a)^ n \qquad (|x-a|<R).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.572)</td></tr></table><p>
This is the Taylor series of [mathjaxinline]f(x)[/mathjaxinline] at [mathjaxinline]x=a[/mathjaxinline].<br/></p><p>
Note that the radius of convergence of the Taylor series of [mathjaxinline]f(x)[/mathjaxinline] at [mathjaxinline]x=a\,[/mathjaxinline] is the number [mathjaxinline]\, R\,[/mathjaxinline] such that [mathjaxinline]\, f(x)\,[/mathjaxinline] converges when [mathjaxinline]\, |x-a|<R,\,[/mathjaxinline] and diverges when [mathjaxinline]\, |x-a|>R[/mathjaxinline]. </p><p>
If [mathjaxinline]a=0[/mathjaxinline], we get the formula for the Taylor series that we started with in this section. This special case of Taylor's formula gives us a power series often referred to as the <span style="color:#27408C"><b class="bf">Maclaurin series</b></span>. </p>
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