9.15 泰勒和麦克劳林系列介绍

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  泰勒和麦克劳林系列介绍

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  In the previous concept, a known function in a form resembling the sum of a convergent infinite geometric series $\begin{align*}\left(S=\frac{a}{1-r}\right)\end{align*}$ , was represented by a power series , e.g., $\begin{align*}y=\frac{1}{1-x}=\sum\limits_{k=0}^\infty x^n, |x|<1\end{align*}$ . What if the function does not resemble the form $\begin{align*}S=\frac{a}{1-r}\end{align*}$ ? To see how a function, infinitely differentiable in some interval, could be represented by a power series, try to determine the series coefficients $\begin{align*}a_n\end{align*}$  for  $\begin{align*}f(x)=e^x=\sum\limits_{n=1}^\infty a_nx^n\end{align*}$ by evaluating the series and four of its derivative at $\begin{align*}x=0\end{align*}$ . Can you determine the relationship between the values of $\begin{align*}a_n\end{align*}$  and %7D(x%3D0)"> $\begin{align*}f^{(n)}(x=0)\end{align*}$ ? 
  ::在前一个概念中,一个已知函数,其形式与集合的无限几何序列(S=a1-r)之和相类似,以一个功率序列表示,例如,y=11-xk=0xn,x1。如果函数与表S=a1-r不相像呢?看看一个在某个间隙中可以无限不同的函数如何以功率序列表示,通过在 x=0 中评估该序列及其四个衍生物,试图确定 f(x)=exn=1anxn的序列系数。您能确定一个函数和 f (x=0) 的值之间的关系吗?

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  Introduction to Taylor and Maclaurin Series
  ::泰勒和麦克劳林系列介绍

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  In the previous concept, the geometric series was used as a model to create power series useful for representing certain functions in the interval of convergence . A more general approach for creating a power series representation comes from the following theorem:
  ::在前一个概念中,几何序列被用作一种模式,用以创建在趋同间隔内用于代表某些功能的电力序列。

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  Power Series Coefficients for Function Representations
  ::职能代表的电联系列电联

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  If the function $\begin{align*}f(x)\end{align*}$  has a power series representation at $\begin{align*}x=x_0\end{align*}$  given by
  ::如果函数 f( x) 在 x=x0 时有一个功率序列表示法,则函数 f( x) 由

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  $$\begin{align*}f(x)=\sum\limits_{n=0}^\infty a_n(x-x_0)^n \ \text{for} \ |x-x_0| < R_c,\end{align*}$$
  :xx) n=0an(x-x0)n,用于 x-x0Rc,

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  then the coefficients are given by %7D(x_0)%7D%7Bn!%7D"> $\begin{align*}a_n=\frac{f^{(n)}(x_0)}{n!}\end{align*}$ , so that the power series representation $\begin{align*}x=x_0\end{align*}$ has the form:
  ::然后以 an=f(x0)n 给定系数!, 以便电源序列表示 x=x0 具有窗体:

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  %7D(x_0)%7D%7Bn!%7D%20%0A(x-x_0)%5En.">

  $$\begin{align*}f(x)=\sum\limits_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n.\end{align*}$$
  :xx)n=0fx0n!(x-x0)n。

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  Given the above, two forms of the power series are distinguished:
  ::鉴于上述情况,对权力序列有两种不同形式:

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  The Taylor series representation , $\begin{align*}T(x) \end{align*}$ , of a function $\begin{align*}f(x) \end{align*}$ at $\begin{align*}x=x_0\end{align*}$ is the power series:
  ::Taylor 序列表示, T(x), x=x0 函数 f(x) 的函数 f(x) 是电源序列 :

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  %7D(x_0)%7D%7Bn!%7D%20%0A(x-x_0)%5En%20%20%5C%5C%0A%26%3D%20f(x_0)%20%2B%20f%5E%7B%5Cprime%7D(x_0)(x-x_0)%20%2B%5Cfrac%7Bf%5E%7B%7B%5Cprime%7D%7B%5Cprime%7D%7D(x_0)%7D%7B2!%7D%20(x-x_0)%5E2%2B%20%5Cfrac%7Bf%5E%7B%7B%5Cprime%7D%7B%5Cprime%7D%7B%5Cprime%7D%7D(x_0)%7D%7B3!%7D(x-x_0)%5E3%2B%20%5Ccdots">

  $$\begin{align*}T(x)&= \sum\limits_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n \\ &= f(x_0) + f^{\prime}(x_0)(x-x_0) +\frac{f^{{\prime}{\prime}}(x_0)}{2!} (x-x_0)^2+ \frac{f^{{\prime}{\prime}{\prime}}(x_0)}{3!}(x-x_0)^3+ \cdots\end{align*}$$
  ::T(x)\n=0f\x0(x)\n! (x-x0)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  The Maclaurin series representation ,  $\begin{align*}M(x)\end{align*}$ , of a function  $\begin{align*}f(x)\end{align*}$  is the Taylor series for  $\begin{align*}x_0=0\end{align*}$ :
  ::函数 f( x) 的 Maclaurin 序列代表 M(x) 是 x0=0 的 Taylor 序列 :

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  %7D(0)%7D%7Bn!%7D%20x_n%20%5C%5C%0A%26%3D%20f(0)%20%2B%20f%5E%7B%5Cprime%7D(0)x%2B%20%5Cfrac%7Bf%5E%7B%7B%5Cprime%7D%7B%5Cprime%7D%7D(0)%7D%7B2!%7D%20x%5E2%2B%20%5Cfrac%7Bf%5E%7B%7B%5Cprime%7D%7B%5Cprime%7D%7B%5Cprime%7D%7D(0)%7D%7B3!%7D%20x%5E3%20%2B%20%5Ccdots">

  $$\begin{align*}M(x)&= \sum\limits_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x_n \\ &= f(0) + f^{\prime}(0)x+ \frac{f^{{\prime}{\prime}}(0)}{2!} x^2+ \frac{f^{{\prime}{\prime}{\prime}}(0)}{3!} x^3 + \cdots\end{align*}$$
  ::M(x) @n=0f(0)n!xn=f(0)+f(0)+f}(0)x+f}(0)2!x2+f}(0)3!x3}#

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  Find the power series representation of  $\begin{align*}f(x)=\cos x\end{align*}$ at:
  ::查找 f( x) =cosx 的功率序列表示值 :

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  • $\begin{align*}x_0=0\end{align*}$
    ::x0=0
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  • $\begin{align*}x_0=\frac{\pi}{3}\end{align*}$
    ::x03

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  The power series representation of  $\begin{align*}f(x)=\cos x\end{align*}$ at  $\begin{align*}x=0\end{align*}$  is the Maclaurin series given by %7D(0)%7D%7Bn!%7D%20x%5En"> $\begin{align*}M(x)= \sum\limits_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n\end{align*}$ .
  ::x=0时 f(x) =cosx 的功率序列表示为 M(x)\\n= 0f 0n!xn 给出的Maclaurin 序列。

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  Some of the required series coefficients are:
  ::要求的一些系列系数如下:

  %7D(x)"> $\begin{align*}f^{(n)}(x)\end{align*}$	%7D(x%3D0)"> $\begin{align*}f^{(n)}(x=0)\end{align*}$
  $\begin{align*}f(x)=\cos x\end{align*}$	1
  $\begin{align*}f^{\prime} (x)=-\sin x\end{align*}$	0
  $\begin{align*}f^{{\prime}{\prime}}(x)=-\cos x\end{align*}$	-1
  $\begin{align*}f^{{\prime}{\prime}{\prime}}(x)=\sin x\end{align*}$	0
  $\begin{align*}f^{(4)}(x)=\cos x\end{align*}$	1
  $\begin{align*}f^{(5)}(x)=-\sin x\end{align*}$	0

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  Notice the pattern repeats every 4 terms.
  ::注意模式每四个条件重复一次。

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  The Maclaurin series of $\begin{align*}f(x)=\cos x\end{align*}$  is:
  ::F(x)=cosx的Maclaurin系列是:

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  $$\begin{align*}M(x)=1 -\frac{1}{2!}x^2 + \frac{1}{4!}x^4 -\frac{1}{6!}x^6 + \frac{1}{8!}x^8 - \cdots =\sum\limits_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}.\end{align*}$$
  ::M(x) = 1 - 12! x2+14! x4 - 16! x6+18! x8\n= 0 (- 1)n( 2n)! x2n.

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  This is the same power series representation introduced in the previous concept. It is called the Maclaurin power series representation of $\begin{align*}f(x)=\cos x\end{align*}$ .
  ::这是上一个概念中引入的相同的权力序列表示法。 它被称为 Maclaurin 权力序列表示法 f( x) = cosx 。

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  The power series representation of $\begin{align*}f(x)=\cos x\end{align*}$  at $\begin{align*}x=\frac{\pi}{3}\end{align*}$  is the Taylor series given by %7D(x_0)%7D%7Bn!%7D(x-x_%7B0%7D)%5En"> $\begin{align*}T(x)=\sum\limits_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x-x_{0})^n\end{align*}$ .
  ::f( x) =cosx 的功率序列表示 x *% 3 是 T( x) n= 0f( n)( x0) n! (x- x0) n 给出的 Taylor 序列 。

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  Some of the required series coefficients are:
  ::要求的一些系列系数如下:

  %7D(x)"> $\begin{align*}f^{(n)}(x)\end{align*}$	%7D%20%5Cleft(x%3D%5Cfrac%7B%5Cpi%7D%7B3%7D%20%5Cright)"> $\begin{align*}f^{(n)} \left(x=\frac{\pi}{3} \right)\end{align*}$
  $\begin{align*}f(x)=\cos x\end{align*}$	$\begin{align*}\frac{1}{2}\end{align*}$
  $\begin{align*}f^{\prime}(x)=-\sin x\end{align*}$	$\begin{align*}- \frac{\sqrt{3}}{2}\end{align*}$
  $\begin{align*}f^{\prime \prime}(x)=-\cos x\end{align*}$	$\begin{align*}- \frac{1}{2}\end{align*}$
  $\begin{align*}f^{\prime \prime \prime}(x)=\sin x\end{align*}$	$\begin{align*}\frac{\sqrt{3}}{2}\end{align*}$
  $\begin{align*}f^{(4)}(x)=\cos x\end{align*}$	$\begin{align*}\frac{1}{2}\end{align*}$
  $\begin{align*}f^{(5)}(x)=- \sin x\end{align*}$	$\begin{align*}- \frac{\sqrt{3}}{2}\end{align*}$

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  The Taylor series of $\begin{align*}f(x)=\cos x\end{align*}$  centered at $\begin{align*}x=\frac{\pi}{3}\end{align*}$  is:
  ::F(x) =cosx 的泰勒系列居于 x% 3 的中心是 :

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  $$\begin{align*}T(x)=\frac{1}{2}- \frac{\sqrt{3}}{2} \left(x- \frac{\pi}{3} \right) - \frac{1}{2} \frac{1}{2!} \left(x- \frac{\pi}{3} \right)^2 + \frac{\sqrt{3}}{2} \frac{1}{3!} \left(x - \frac{\pi}{3} \right)^3+\frac{1}{2} \frac{1}{4!} \left(x- \frac{\pi}{3} \right)^4 - \cdots\end{align*}$$
  ::T(x) = 1232(x3)- 1212! (x3) 23213! (x3) 3+1214! (x3) 44

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  $$\begin{align*}T(x)=-1+(x-2)-2 \frac{1}{2 !}(x-2)^2+6 \frac{1}{3 !}(x-2)^3-24 \frac{1}{4 !}(x-2)^4+ \cdots =\sum_{n=0}^{\infty}(-1)^{n+1} (x-2)^n\end{align*}$$
  ::T(x)1+(x-2)-2-2-212!(x-2)2+613!(x-2)3-2414!(x-2)4n=0(-1)n+1(x-2)n

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  Common Maclaurin Series
  ::共同的麦克劳林系列

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  There are many functions which can be represented by power series. Listed below are some common functions, their Maclaurin series forms, and the domain of applicability (convergence interval).
  ::有许多功能可以用电源序列来代表,下面列出一些共同功能、其Maclaurin序列形式和适用性领域(趋同间隔)。

  播放段落 Maclaurin Power Series ::Maclaurin电力系列	播放段落 Interval of Convergence ::趋同的间隔
  播放段落 $\begin{align*}\sin x=\sum\limits_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{(2n+1)!}=x- \frac{x^3}{3 !}+ \frac{x^5}{5 !}- \frac{x^7}{7 !}+ \cdots\end{align*}$ ::sinxn=0(- 1)nx2n+1( 2n+1)!=x- x33!+x55!- x77!) @ @ @	$\begin{align*}(- \infty, \infty)\end{align*}$
  播放段落 $\begin{align*}\cos x=\sum\limits_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}=1- \frac{x^2}{2 !}+ \frac{x^4}{4 !}- \frac{x^6}{6 !}+ \cdots\end{align*}$ ::comsxn=0(- 1)nx2n( 2n)!=1- x22!+x44!- x66!) @ {\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ </span> </p> </td> <td> <p id="x-ck12-xku"> <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="(-%20%5Cinfty%2C%20%5Cinfty)"> <span class="MathJax_Preview" style="color: inherit;"> <span class="MJXp-math" id="MJXp-Span-1373"> <span class="MJXp-mtable" id="MJXp-Span-1374"> <span> <span class="MJXp-mtr" id="MJXp-Span-1375" style="vertical-align: baseline;"> <span class="MJXp-mtd" id="MJXp-Span-1376" style="text-align: right;"> <span class="MJXp-mo" id="MJXp-Span-1377" style="margin-left: 0em; margin-right: 0em;"> ( </span> <span class="MJXp-mo" id="MJXp-Span-1378" style="margin-left: 0.267em; margin-right: 0.267em;"> − </span> <span class="MJXp-mi" id="MJXp-Span-1379"> ∞ </span> <span class="MJXp-mo" id="MJXp-Span-1380" style="margin-left: 0em; margin-right: 0.222em;"> , </span> <span class="MJXp-mi" id="MJXp-Span-1381"> ∞ </span> <span class="MJXp-mo" id="MJXp-Span-1382" style="margin-left: 0em; margin-right: 0em;"> ) </span> </span> </span> </span> </span> </span> </span> <span class="MathJax_SVG MathJax_SVG_Processing" id="MathJax-Element-64-Frame" style="font-size: 100%; display: inline-block;" tabindex="-1"> </span> <script id="MathJax-Element-64" type="math/tex"> \begin{align*}(- \infty, \infty)\end{align*}
  播放段落 $\begin{align*}e^x=\sum\limits_{n=0}^{\infty} \frac{x^n}{n !}=1+x+ \frac{x^2}{2 !}+ \frac{x^3}{3 !}+ \cdots\end{align*}$ ::exn=0xnn!=1+x+x22!+x33!\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ x22\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\	$\begin{align*}(- \infty, \infty)\end{align*}$
  播放段落 $\begin{align*}\ln(1+x)=\sum\limits_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}=x- \frac{x^2}{2 !}+ \frac{x^3}{3 !}- \frac{x^4}{4 !}+ \cdots\end{align*}$ ::In( 1+x)\\\ n=0 (- 1nxn+1n+1n+1=x- x22! +x33!- x44!)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\	$\begin{align*}(-1,1]\end{align*}$
  播放段落 $\begin{align*}\ln(1-x)=\sum\limits_{n=0}^{\infty} (-1) \frac{x^{n+1}}{n+1}=-x- \frac{x^2}{2 !}- \frac{x^3}{3 !}- \frac{x^4}{4 !}+ \cdots\end{align*}$ ::In(1-x)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\	$\begin{align*}[-1,1)\end{align*}$
  播放段落 $\begin{align*}\arcsin x= \sin^{-1}x=\sum\limits_{n=0}^{\infty} \frac{(2n)! x^{2n+1}}{2^{2n} (n !)^2(2n+1)}\end{align*}$ ::arcsinx=sin- 1xn= 0( 2n)! x2n+122n( n!) 2( 2n+1)	$\begin{align*}[-1,1]\end{align*}$
  播放段落 $\begin{align*}\arctan x= \tan^{-1}x=\sum\limits_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}\end{align*}$ ::arctanx=tan- 1xn=0(- 1)nx2n+12n+1)	$\begin{align*}(-1, 1)\end{align*}$
  播放段落 $\begin{align*}\sinh x=\sum\limits_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}\end{align*}$ ::sinhxn= 0x2n+1 (2n+1)!	$\begin{align*}(- \infty, \infty)\end{align*}$
  播放段落 $\begin{align*}\cosh x=\sum\limits_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}\end{align*}$ ::COshxn= 0x2n( 2n) !	$\begin{align*}(- \infty, \infty)\end{align*}$

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  Knowing the above series can often simplify the determination of a power series representation for another function.
  ::了解上述系列往往可以简化确定另一职能的权力序列代表。

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  Find a power series representation of the function $\begin{align*}f(x)=x \sin x^3\end{align*}$  centered at $\begin{align*}x_0=0\end{align*}$ .
  ::查找函数 f( x) =xsinx3 的功率序列表示值, 以 x0=0 为中心 。

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  The series $\begin{align*}f(x)=x \sin x^3\end{align*}$  can be written as the product of $\begin{align*}x\end{align*}$  and the power series for the sine function, using the argument $\begin{align*}x^3\end{align*}$  instead of $\begin{align*}x\end{align*}$ .
  ::f(x) =xsinx3 序列可以用参数 x3 而不是 x 来写成 x 的产物和正弦函数的功率序列。

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  $$\begin{align*}x \sin x^3 &=x \sum\limits_{n=0}^{\infty}(-1)^n \frac{(x^3)^{2n+1}}{(2n+1)!} \\ &=x \sum\limits_{n=0}^{\infty}(-1)^n \frac{x^{6n+3}}{(2n+1)!} \\ &=\sum\limits_{n=0}^{\infty}(-1)^n \frac{x^{6n+4}}{(2n+1)!}\end{align*}$$
  ::xinx3=xn=0(- 1)n(x3)2n+1( 2n+1)!=x_n=0(- 1)nx6n+3( 2n+1)!\n=0(- 1)nx6n+4( 2n+1)!

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to determine the series coefficients $\begin{align*}a_n\end{align*}$  for  $\begin{align*}f(x)=e^x=\sum\limits_{n=1}^{\infty}a_n x^n\end{align*}$ by evaluating the series and four of its derivative at $\begin{align*}x=0\end{align*}$ . You were also asked about the relationship between the values of $\begin{align*}a_n\end{align*}$  and %7D(x%3D0)"> $\begin{align*}f^{(n)}(x=0)\end{align*}$ .
  ::早些时候,有人要求您通过在 x=0 时对 f(x) = exn= 1anxn 序列及其四个衍生物进行估评来确定 f(x) = exn = 1anxn 的序列系数。 也有人要求您了解 a 和 f (x=0) 的值之间的关系。

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  Were you able to figure this out? Just list the results:
  ::你能弄清楚吗?

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  $$\begin{align*}f(0) = e^0=1 &=\sum\limits_{n=0}^{\infty} a_n x^n=a_0 \Rightarrow a_0=f(0)=1,\\ f^\prime (0) = e^0=1 &=\sum\limits_{n=1}^{\infty} na_n x^{n-1}=a_1 \Rightarrow a_1=\frac{f^\prime (0)}{1}=1,\\ f^{\prime \prime} (0)=e^0=1 &=\sum\limits_{n=2}^{\infty} n(n-1)a_n x^{n-2}=2a_2 \Rightarrow a_2=\frac{f^{\prime \prime} (0)}{2}=\frac{1}{2},\\ f^{\prime \prime \prime} (0)=e^0=1 &= \sum\limits_{n=3}^{\infty} n(n-1)(n-2)a_n x^{n-3}=6a_3 \Rightarrow a_3=\frac{f^{\prime \prime \prime} (0)}{6}=\frac{1}{3 !},\\ f^{(4)} (0)=e^0=1 &= \sum\limits_{n=4}^{\infty} n(n-1)(n-2)(n-3)a_n x^{n-4}=24a_4 \Rightarrow a_4=\frac{f^{(4)} (0)}{24}=\frac{1}{4 !}.\end{align*}$$
  ::f( 0) = e0= 1\\\\ n0= 0xn= 10\\\\\\\\\\\\0\\\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0=0\0\0\0\0\1\1\1\1\1\1\1\\\1\\\\0\2\2\2\2\0\2\0\0\0\0\0\0\2\0\0\2\0\0\0\0\0\0\0\2\2\0\0\0\0\4\4\4\4\\1\1\\\\\\\1\1\1\0\0\\\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\\\\\0\0\0\0\\\\0\0\\\\0\0\0\0\\\\0\0\0\0\0\0\0\0\\\\\\\0\0\0\0\\\\0\0\0\\\\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\1\1\1\1\1\0\1\1\1\1\1\1\1\1\1\1\1\1\1\1\1\1\1\1\1\1\0\\1\

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  Do you see the pattern? The function can be written as:
  ::您看到这个模式吗? 函数可以写为:

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  $\begin{align*}f(x)=e^x=\sum\limits_{n=1}^{\infty} a_n x^n=1+x+ \frac{1}{2 !}x^2+ \frac{1}{3 !}x^3+ \frac{1}{4}x^4+ \cdots =\sum\limits_{n=0}^{\infty} \frac{x^n}{n !}a_n\end{align*}$ . This is how Maclaurin series are generated.
  ::f(x) =exn= 1anxn= 1+x+12! x2+13! x3+14x4@n=0xnn! an。 这是 Maclaurin 序列是如何生成的 。

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  Example 2
  ::例2

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  Find the power series representation of $\begin{align*}f(x)=\frac{1}{1-x}\end{align*}$  centered on:
  ::查找 f( x) = 11 - x 的电源序列表示法, 以 :

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  • $\begin{align*}x=0\end{align*}$
    ::x=0x=0
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  • $\begin{align*}x=2\end{align*}$
    ::x=2x=2

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  The power series representation of $\begin{align*}f(x)=\frac{1}{1-x}\end{align*}$  at $\begin{align*}x=0\end{align*}$  is the Maclaurin series given by %7D(0)%7D%7Bn%20!%7D%20x%5En"> $\begin{align*}M(x)=\sum\limits_{n=0}^{\infty} \frac{f^{(n)}(0)}{n !} x^n\end{align*}$ .
  ::x=0时的 f(x)=11-x 的功率序列表示为 M(x)\\n= 0fn!xn 给出的 Maclaurin 序列。

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  Some of the required series coefficients are:
  ::要求的一些系列系数如下:

  %7D(x)%0A"> $\begin{align*}f^{(n)}(x) \end{align*}$	%7D(x%3D0)%0A"> $\begin{align*}f^{(n)}(x=0) \end{align*}$
  $\begin{align*}f(x)=\frac{1}{1-x} \end{align*}$	1
  $\begin{align*}f^\prime (x)=\frac{1}{(1-x)^2} \end{align*}$	1
  $\begin{align*}f^{\prime \prime}(x)=\frac{2}{(1-x)^3} \end{align*}$	2
  $\begin{align*}f^{\prime \prime \prime}(x)=\frac{6}{(1-x)^4} \end{align*}$	6
  $\begin{align*}f^{(4)}(x)=\frac{24}{(1-x)^5} \end{align*}$	24
  $\begin{align*}f^{(5)}(x)=- \sin x \end{align*}$	0

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  The Maclaurin series of $\begin{align*}f(x)=\frac{1}{1-x}\end{align*}$ is:
  ::F(x)=11-x的Maclaurin系列是:

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  $$\begin{align*}M(x)=1+x+2 \frac{x^2}{2 !}+6 \frac{x^3}{3 !}+24 \frac{x^4}{4 !}+ \cdots =\sum_{n=0}^{\infty} x^n.\end{align*}$$
  ::M(x) = 1+x+2x22!+6x33!+24x44! @n=0xn。

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  This is the same power series representation introduced in the previous concept. It is a geometric series.
  ::这是前一个概念中引入的相同的权力序列表示。 这是一个几何序列 。

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  The power series representation of $\begin{align*}f(x)=\frac{1}{1-x}\end{align*}$  at $\begin{align*}x=2\end{align*}$  is the Taylor series given by %7D(x_0)%7D%7Bn%20!%7D%20(x-x_0)%5En"> $\begin{align*}T(x)=\sum\limits_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n !} (x-x_0)^n\end{align*}$ .
  ::F(x)=11-x在 x=2 时的功率序列表示为 T(x)\\n= 0f (x0)n!(x-x0)n 给出的泰勒序列。

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  Some of the required series coefficients are:
  ::要求的一些系列系数如下:

  %7D(x)%0A"> $\begin{align*}f^{(n)}(x) \end{align*}$	%7D(x%3D2)%0A"> $\begin{align*}f^{(n)}(x=2) \end{align*}$
  $\begin{align*}f(x)=\frac{1}{1-x} \end{align*}$	-1
  $\begin{align*}f^\prime (x)=\frac{1}{(1-x)^2} \end{align*}$	1
  $\begin{align*}f^{\prime \prime}(x)=\frac{2}{(1-x)^3} \end{align*}$	-2
  $\begin{align*}f^{\prime \prime \prime}(x)=\frac{6}{(1-x)^4} \end{align*}$	6
  $\begin{align*}f^{(4)}(x)=\frac{24}{(1-x)^5} \end{align*}$	-24

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  The Taylor series of $\begin{align*}f(x)=\cos x\end{align*}$  centered at $\begin{align*}x=\frac{\pi}{3}\end{align*}$  is:
  ::F(x) =cosx 的泰勒系列居于 x% 3 的中心是 :

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  Example 3
  ::例3

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  Find the following representations:
  ::发现以下陈述:

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  • Taylor series for $\begin{align*}f(x)=e^{2x}\end{align*}$  centered on $\begin{align*}x=3\end{align*}$ .
    ::f( x) =e2x 的 Taylor 序列, 以 x= 3 为中心 。
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  • Maclaurin series for $\begin{align*}\cosh x=\frac{e^x+e^{-x}}{2}\end{align*}$  (by using known series); compare it to the series for  $\begin{align*}\cos x\end{align*}$ .
    ::cosx=ex+e-x2的Maclaurin系列(使用已知序列);将其与cosx的序列进行比较。

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  The Taylor series representation for $\begin{align*}f(x)=e^{2x}\end{align*}$  centered on $\begin{align*}x=3\end{align*}$  is %7D(x_0)%7D%7Bn%20!%7D(x-x_0)%5En"> $\begin{align*}T(x)=\sum\limits_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n !}(x-x_0)^n\end{align*}$ .
  ::f( x) =e2x 以 onx = 3 为中心的 Taylor 序列表示为 T( x) = 0f (x0)n! (x- x0)n 。

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  Here $\begin{align*}x_0=3\end{align*}$ , and have the form:
  ::这里 x0=3, 表格如下:

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  %7D(x)%20%26%3D2%5En%20e%5E%7B2x%7D">

  $$\begin{align*}f^\prime (x) &=2e^{2x} \\ f^{\prime \prime}(x) &=2^2 e^{2x} \\ f^{\prime \prime \prime}(x) &=2^3 e^{2x} \\ &\vdots \\ f^{(n)}(x) &=2^n e^{2x}\end{align*}$$
  ::f\( x) = 2e2xf}( x) = 222e2xf}( x) = 23e2x}* f( n( x) = 2ne2x

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  Therefore,
  ::因此,

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  $$\begin{align*}T(x)=\sum\limits_{n=0}^{\infty} \frac{2^n e^6}{n !}(x-3)^n=\sum\limits_{n=0}^{\infty} \frac{e^6}{n !} [2(x-3)]^n.\end{align*}$$
  :xx)n=02ne6n! (x-3)n\n=0e6n! [2(x-3)]n.

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  Using the known Maclaurin series for $\begin{align*}e^x\end{align*}$ , for  $\begin{align*}\cosh x=\frac{e^x+e^{-x}}{2}\end{align*}$  we get:
  ::使用已知的 Maclaurin 序列前, 对于 coshx=ex+e- x2, 我们得到 :

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  $$\begin{align*}\cosh x &=\frac{e^x+ e^{-x}}{2} \\ &=\frac{1}{2} \left[\sum_{n=0}^{\infty} \frac{x^n}{n !}+ \sum_{n=0}^{\infty} \frac{(-x)^n}{n !} \right] \\ &=\sum_{n=0}^{\infty} \frac{1+(-1)^n}{2(n !)} x^n && =\frac{2}{2(0 !)} x^0+0 \cdot x^1+ \frac{2}{2(2 !)} x^2+0 \cdot x^3+ \frac{2}{2(4 !)}x^4+ \cdots \\ &=\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}\end{align*}$$
  ::comshx=ex+e- x2=12[n=0xnn!]\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Therefore, $\begin{align*}\cosh x=\frac{e^x+e^{-x}}{2}=M(x)=\sum\limits_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \end{align*}$ .
  ::因此, cosx=ex+e-x2=M(x)\\n=0x2n( 2n)!!

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  By comparison, the series for $\begin{align*}\cos x\end{align*}$  is $\begin{align*}\cos x=\sum\limits_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}\end{align*}$ , which is an version of $\begin{align*}\cosh x\end{align*}$ .
  ::相比之下, cosx 的序列为 cosx=0(- 1)nx2n( 2n)!, 这是 Coshx 的版本 。

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  Review
  ::回顾

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  For #1-11, find a power series representation of the function centered at the given $\begin{align*}x_0\end{align*}$ .
  ::对于 # 1- 11, 找到以给定 x0 为中心的函数的电源序列表示 。

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  1. $\begin{align*}f(x)=\sin x\end{align*}$ , $\begin{align*}x_0=0\end{align*}$ .
     ::f(x) =sinx, x0=0。
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  2. $\begin{align*}f(x)=\sin 2x\end{align*}$ , $\begin{align*}x_0=0\end{align*}$ .
     ::f(x) =sin2x, x0=0。
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  3. $\begin{align*}f(x)=\tan x\end{align*}$ , $\begin{align*}x_0=0\end{align*}$ (first 4 terms only)
     ::f(x) = tanx, x0=0(仅前4个术语)
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  4. $\begin{align*}f(x)=\tan x\end{align*}$ , $\begin{align*}x_0=\frac{\pi}{4}\end{align*}$ (first 5 terms only)
     ::f(x) = tanx, x04 (仅前5个术语)
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  5. $\begin{align*}f(x)=\sqrt{1+x}\end{align*}$
     :xx)%1+x
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  6. $\begin{align*}\sqrt{1+x+x^2}\end{align*}$ at $\begin{align*}x_0=- \frac{1}{2}\end{align*}$  (first 4 terms)
     ::++x+x2 时为 x0 @% 12 (前4个条件)
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  7. $\begin{align*}f(x)=\cos^2 x\end{align*}$ , $\begin{align*}x_0=0\end{align*}$ . Hint: Use $\begin{align*}\cos^2 x=\frac{1}{2}+ \frac{1}{2} \cos 2x\end{align*}$
     ::f(x) = cos2x, x0=0. 提示: 使用 cos2x= 12+12cos2x
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  8. $\begin{align*}\frac{x}{e^x}\end{align*}$ , $\begin{align*}x_0=0\end{align*}$
     ::xex, x0=0xxxxex, x0=0
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  9. $\begin{align*}(1+x)^{\alpha}\end{align*}$ , $\begin{align*}\alpha\end{align*}$ a real constant; $\begin{align*}x_0=0\end{align*}$
     : 1+x) α, α a 实值常数; x0=0
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  10. $\begin{align*}\ln(3+9x)\end{align*}$ , $\begin{align*}x_0=0\end{align*}$
      ::In( 3+9x), x0=0
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  11. Use series expansions to evaluate $\begin{align*}\lim\limits_{x \to 0} \frac{\sin kx}{x}\end{align*}$
      ::使用序列扩展来评价 limx0sinkxxxx
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  12. Expand  $\begin{align*}e^{-x} \cos x\end{align*}$ about $\begin{align*}x=\frac{\pi}{2}\end{align*}$ .
      ::展开 e - xcosx 大约 x% 2 。
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  13. If $\begin{align*}f(x)=\sum\limits_{n=1}^{\infty} \frac{(-1)^n(x-7)^{n+1}}{n !}\end{align*}$ , find $\begin{align*}f^{(6)}(7)\end{align*}$ .
      ::如果 f( x) n=1 (- 1) n( x-7) n+1n!,请查找 f(6)(7)。
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  14. Evaluate the indefinite integral $\begin{align*}\int \frac{e^x -1}{x}dx\end{align*}$  as an infinite series.
      ::将无限的 ex- 1xdx 作为无限序列来评估 。

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  Review (Answers)
  ::回顾(答复)

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  Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。