电力系列:职能代表和业务

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In the previous concept, the geometric series was used as a framework to show how some functions could be represented as power series by replacing the common ratio  $\begin{align*}r\end{align*}$ by a variable expression. The applicable domain of the power series representation is the interval of convergence . If arithmetic operations like sum and product, are performed on two series, or differentiation, and integration operations are performed on a single series, what can be said about the domain (interval of convergence) of the resulting series, an the radius of convergence ? Suppose you are working with the following two series and need to find their sum: $\begin{align*}\sum\limits_{n=1}^{\infty} x^n=\frac{1}{1-x}\end{align*}$ and  $\begin{align*}\sum\limits_{n=1}^{\infty} \left(\frac{1}{3} x \right)^n=\frac{1}{1- \left(\frac{x}{3} \right)}\end{align*}$  What is the domain of the sum of the two series, and how does it compare to either one?
::在前一个概念中,几何序列被用作一个框架,以显示某些函数如何通过以变量表达方式取代通用比率 r 来作为权力序列来代表某些函数。权力序列代表的适用领域是趋同的间隔。如果算术操作,例如总和和和产品,是按两个序列进行,或分化,而集成操作是按一个单一序列进行,那么对由此产生的序列的域(趋同的相互交错),即趋同的半径,有什么看法?假设你正在与以下两个序列合作,需要找到它们的总数:n=1xn=11-x和n=1(13x)n=11-(x3)什么是两个序列的总和的域,以及它如何与两个序列的总和相比较?

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Function Representation and Operations of Power Series
::权力系列职能代表和运作

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The power series is a very important kind of infinite series where the terms contain powers of a variable. Remember that a  power series is a series of the general form:
::电源序列是一个非常重要的无限序列, 其术语包含变量的功率。 记住, 电源序列是一般形式的一系列 :

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$$\begin{align*}\sum_{n=0}^{\infty} a_n(x-x_0)^n=a_0+a_1(x-x_0)+a_2(x-x_0)^2+a_3(x-x_0)^3+ \ldots\end{align*}$$
::*n=0an(x-x0)n=a0+a1(x-x0)+a2(x-x0)2+a3(x-x0)3+...

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(called a power series centered at $\begin{align*}x_0\end{align*}$ )
:称为以x0为中心的一个电源序列)

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where $\begin{align*}x\end{align*}$  is a variable,  $\begin{align*}x_0\end{align*}$ is a real constant, and the $\begin{align*}a_n\end{align*}$ 's are real constants called the coefficients of the series.
::x 是变量, x0 是真实的常数, 而 an是实际的常数, 称为序列的系数 。

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For $\begin{align*}x_0=0\end{align*}$ , the power series representation becomes:
::对于x0=0, 电源序列表示为 :

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$$\begin{align*}\sum_{n=0}^{\infty} a_n x^n=a_0+a_1 x+a_2 x^2+a_3 x^3+ \ldots\end{align*}$$
::n=0anxn=a0+a1x+a2x2+a3x3+...

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In this concept, the following topics will be addressed:
::在这一概念中,将讨论以下专题:

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• Introduction to how a power series can be used to represent a function
  ::如何使用电源序列代表函数的导言
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• How a power series can be differentiated or integrated
  ::如何区分或整合权力序列
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• Sums (differences), products, and quotients of two power series
  ::两个电力序列的总和(差异)、产品和商数

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Function Representation by Power Series
::按电力系列分列的职能代表情况

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The power series  $\begin{align*}\sum\limits_{n=0}^{\infty} a_n x^n\end{align*}$ is an infinite series, and looks like a function of $\begin{align*}x\end{align*}$ . An easy way to illustrate the idea of a power series representing a function is to use the geometric series as an example.
::电源序列 n= 0anxn 是一个无限的序列, 看起来像 x 的函数。 用来说明代表函数的电源序列概念的一个简单方法就是以几何序列为例。

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The geometric series has the form : $\begin{align*}\sum\limits_{n=1}^{\infty}ar^{n-1}=\sum\limits_{n=0}^{\infty} ar^n=a+ar+ar^2+ar^3+ \dots\end{align*}$ .
::几何序列的窗体为: @n=1arn_1n=0arn=a+ar+ar2+ar3+...

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Recall that there is a unique value for $\begin{align*}\sum\limits_{n=1}^{\infty} ar^{n-1}\end{align*}$ , of $\begin{align*}\frac{a}{1-r}\end{align*}$ , only if $\begin{align*}|r| < 1\end{align*}$ ; otherwise for  $\begin{align*}|r| > 1\end{align*}$ there is no value for the infinite summation, it diverges . These are the convergence and divergence characteristics of a geometric series.
::回顾 a1-r 的 {n= 1arn- 1 有独特的价值, 只有 {r} 1 才能有; 否则, r} 1 的无限相加没有价值, 但它有差异。 这些是几何序列的趋同和差异特性 。

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Now, by replacing $\begin{align*}r\end{align*}$  in $\begin{align*}\sum\limits_{n=0}^{\infty} ar^n\end{align*}$  with the variable $\begin{align*}x\end{align*}$ , the power series $\begin{align*}\sum\limits_{n=0}^{\infty} ax^n\end{align*}$  is created, with every $\begin{align*}a_n=a\end{align*}$ . However, a finite and unique value exists for $\begin{align*}\sum\limits_{n=0}^{\infty} ax^n\end{align*}$ only if $\begin{align*}|x| < 1\end{align*}$ . The unique value is $\begin{align*}\frac{a}{1-x}\end{align*}$ , and we can write:  $\begin{align*}f(x)=\frac{a}{1-x}=\sum\limits_{n=0}^{\infty} ax^n\end{align*}$ for the domain $\begin{align*}|x| < 1\end{align*}$ .
::现在,通过以变量 x 替换 {n= 0arn 中的 r, 将 {n= 0axn 创建为 {n= 0a= a> 。 但是, {n= 0ax= 0axn 只有 {x\\\\\\\ = = a1- x= a1-x= 0axn 的特性值才存在。 我们可以写下: f(x)= a1- xn= 0axn 用于域 {x= *1 。

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Now, let's find the function represented by the following infinite series and state the domain:
::现在,让我们找到由以下无限序列代表的函数, 并声明域名 :

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• $\begin{align*}\sum\limits_{n=0}^{\infty} 3x^{2n}\end{align*}$
  ::*n=03x2n
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• $\begin{align*}\sum\limits_{n=0}^{\infty} 2 \left[\frac{1}{3}(x-1) \right]^n \end{align*}$
  ::n=02[13(x-1)]n

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$\begin{align*}\sum\limits_{n=0}^{\infty} 3x^{2n}=\sum\limits_{n=0}^{\infty} 3(x^2)^n\end{align*}$  is a geometric series with $\begin{align*}a=3\end{align*}$  and $\begin{align*}r=x^2\end{align*}$ .
::n=03x2nn=03(x2n)是一个几何序列,有 a=3 和 r=x2 。

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The series converges if $\begin{align*}|r|=x^2 < 1\end{align*}$ , i.e., the interval of convergence is $\begin{align*}(-1, 1)\end{align*}$ .
::如果 \ rx2 < 1, 即聚合间隔为 (-1, 1) , 则序列会趋同 。

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Within $\begin{align*}(-1, 1)\end{align*}$ , the series has the sum $\begin{align*}\sum\limits_{n=0}^{\infty} 3(x^2)^n=\frac{3}{1-x^2}\end{align*}$ .
::在(-1,1)之内,该系列的数值为n=03(x2,n)=31-x2。

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Therefore, $\begin{align*}f(x)=\frac{3}{1-x^2}\end{align*}$  in the domain $\begin{align*}(-1, 1)\end{align*}$ . The series does not represent this function outside of the domain because the series diverges.
::因此,在域中 f(x) = 31- x2 (-1, 1) 。 该序列不代表域外的此函数, 因为序列不同 。

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$\begin{align*}\sum\limits_{n=0}^{\infty} 2 \left[\frac{1}{3}(x-1) \right]^n\end{align*}$ is a geometric series with $\begin{align*}a=2\end{align*}$  and $\begin{align*}r= \frac{1}{3}(x-1)\end{align*}$ .
::n=02[13(x-1)]n是一个几何序列,有 a=2 和 r=13(x-1) 。

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The series converges if $\begin{align*}|r|=\Big|\frac{1}{3}(x-1) \Big| < 1\end{align*}$ , i.e., the interval of convergence is $\begin{align*}(-2, 4)\end{align*}$ .
::如果 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\I\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\融合的融合的融合。

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Within $\begin{align*}(-2, 4)\end{align*}$ , the series has the sum $\begin{align*}\sum\limits_{n=0}^{\infty} 2 \left[\frac{1}{3}(x-1) \right]^n=\frac{2}{1- \frac{1}{3}(x-1)}=\frac{6}{4-x}\end{align*}$ .
::在(-2,4)之内,该系列的数值为+n=02[13(x-1)]n=21-13(x-1)=64-x。

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Therefore, $\begin{align*}f(x)=\frac{6}{4-x}\end{align*}$  in the domain $\begin{align*}(-2, 4)\end{align*}$ . The series does not represent this function outside of the domain because the series diverges.
::因此,F(x)=64-x在域(-2,4)中,该序列不代表域外的该函数,因为序列不同。

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There are many functions which can be represented by power series, and how to generate them will be covered in the next concept. Listed below are some common functions, their power series forms, and the domain of applicability (convergence interval).
::有许多功能可以用电源序列来代表,下一个概念将涵盖如何产生这些功能。 下面列出一些共同功能、其电源序列形式和适用性领域(趋同间隔)。

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Differentiation and Integration of Power Series
::电力系列的差别和统一

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Having generated a power series representation of a function, an important interest is how to handle the differentiation and integration of the series. We expect to achieve the same results as direct differentiation or integration of the explicit function . 
::产生一个功能的权力序列代表,一个重要的利益是如何处理该序列的区别和整合,我们期望取得与直接区别或整合明确功能相同的结果。

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The Differentiation and Integration of a Power Series Theorem states:
::《权力系列理论的差别和整合》指出:

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Suppose the power series $\begin{align*}\sum\limits_{n=0}^{\infty} a_n(x-x_0)^n\end{align*}$  has radius of convergence $\begin{align*}R_c\end{align*}$ .
::假设电源序列n=0an(x-x0)n具有半径趋同 Rc。

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Then, the function defined by $\begin{align*}f(x)=\sum\limits_{n=0}^{\infty} a_n(x-x_0)^n\end{align*}$  is:
::然后,f(x)\\\\\n=0an(x-x0)n定义的函数是:

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• differentiable for $\begin{align*}|x-x_0| < R_c\end{align*}$ , i.e. on $\begin{align*}(x_0-R_c, x_0+R_c)\end{align*}$ , and $\begin{align*}f^\prime (x)=\sum\limits_{n=1}^{\infty} na_n(x-x_0)^{n-1}\end{align*}$ .
  ::可用于 x-x0Rc, 即 (x0- Rc,x0+Rc) 和 f(x) n= 1nan(x-x0)n- 1。

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i.e., $\begin{align*}\frac{d}{dx} \sum\limits_{n=0}^{\infty} a_n(x-x_0)^n=\sum\limits_{n=1}^{\infty} \frac{d}{dx} [a_n(x-x_0)^n]\end{align*}$
::i. , ddxn=0an(x-x0)nn=1ddx[an(x-x0)n]

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• integrable for $\begin{align*}|x-x_0| < R_c\end{align*}$ , and $\begin{align*}\int f(x) dx=\sum\limits_{n=0}^{\infty} \frac{a_n(x-x_0)^{n+1}}{n+1}+C\end{align*}$ .
  ::x-x0Rc 和f(x)dxn=0an(x-x0)n+1n+1+C 的不可调试性。

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i.e., $\begin{align*}\int \sum\limits_{n=0}^{\infty} (a_n x^n)dx= \sum\limits_{n=0}^{\infty} \int (a_n x^n)dx\end{align*}$
:nxn) dxn=0 (anxn) dxx

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Both the derivative and the integral above have the same radius of convergence $\begin{align*}R_c\end{align*}$ .
::衍生物和上述整体体的趋同半径相同。

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Note that even though the power series, its derivative, and its integral have the same radius of convergence, the endpoints of the interval of convergence may be different.
::请注意,尽管动力序列、其衍生物及其整体体具有相同的趋同半径,但趋同间隔的终点可能不同。

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Now, using the theorem above, let's find a power series for each of the functions and its radius of convergence:
::现在,使用上面的定理, 让我们为每个函数及其趋同半径 找到一个电源序列:

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• $\begin{align*}g(x)=\frac{1}{(1-x)^2}\end{align*}$
  ::g(x)=1(1-x)2
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• $\begin{align*}h(x)=\tan^{-1} x\end{align*}$
  ::h(x) =tan-1x

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You should recognize that $\begin{align*}g(x)=\frac{1}{(1-x)^2}=\frac{d}{dx} \left(\frac{1}{1-x} \right)=\frac{d}{dx} \left(\sum\limits_{n=0}^{\infty} x^n \right)\end{align*}$ . The radius of convergence of the power series $\begin{align*}\sum\limits_{n=0}^{\infty} x^n\end{align*}$  is $\begin{align*}R_c=1\end{align*}$ , with interval of convergence $\begin{align*}-1 < x < 1\end{align*}$ .
::您应该认识到 g( x) =1( 1- x) 2=ddx( 11- x) =ddx( n= 0xn) 。 电源序列的趋同半径 { n= 0xn 是 Rc=1, 与趋同的间隔- 1 < x < 1 > 。

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By the differentiation theorem above,
::以上面的区别定理,

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$\begin{align*}g(x)=\frac{d}{dx} \sum\limits_{n=0}^{\infty} x^n=\sum\limits_{n=1}^{\infty} nx^{n-1}\end{align*}$ , and has radius of convergence $\begin{align*}R_c=1\end{align*}$ . The interval of convergence is also $\begin{align*}-1 < x < 1\end{align*}$ .
::g(x) =ddxn=0xnn=1nxn- 1, 并具有趋同Rc=1的半径。 趋同的间隔也是- 1 < x<1 。

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We recognize $\begin{align*}h(x)=\tan^{-1} x=\int \frac{1}{1+x^2}dx=\int \left[\sum\limits_{n=0}^{\infty} (-1)^n x^{2n} \right]dx\end{align*}$ , i.e.,  $\begin{align*}h(x)\end{align*}$ is the of $\begin{align*}\frac{1}{1+x^2}\end{align*}$ which can be represented by a power series with $\begin{align*}R_c=1\end{align*}$ , and interval of convergence $\begin{align*}-1 < x < 1\end{align*}$ .
::我们确认 h(x) = tan- 1x11+x2dx[n= 0(- 1)nx2n] dx, 即 h(x) 是 11+x2 的电源序列, 可以用 Rc=1, 和 趋同的间隔- 1 < x < 1 > 来表示 。

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By the integration theorem above,
::根据上面的融合理论,

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$\begin{align*}h(x)=\tan^{-1} x=\int \left[\sum\limits_{n=0}^{\infty} (-1)^n x^{2n} \right]dx=\sum\limits_{n=0}^{\infty} (-1)^n x^{2n} dx=\sum\limits_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}+C\end{align*}$ and has radius of convergence $\begin{align*}R_c=1\end{align*}$ .
::h(x) = tan- 1x[n= 0(- 1,nx2n) dxn= 0(- 1,nx2ndxn=0(- 1,nx2n+12n+1+C) , 并具有Rc=1的趋同半径。

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Then $\begin{align*}C=\tan^{-1} 0=0\end{align*}$  and $\begin{align*}h(x)=\sum\limits_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}\end{align*}$ .
::然后C=tan- 10=0和h(x)\\\\n=0(- 1)nx2n+12n+1。

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The interval of convergence of $\begin{align*}h(x)=\sum\limits_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}\end{align*}$  can be shown to be $\begin{align*}-1 \le x \le 1\end{align*}$ , i.e., the series converges at the endpoints.
::h(x)\\\n=0(- 1)nx2n+12n+1的趋同间隔可以显示为-1x+1,也就是说,该序列在端点会合。

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Sums (differences), Products, and Quotients of Power Series
::电力系列总和(差异)、产品和引号

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The following theorems provide guidance on how to handle arithmetic operations on power series.
::以下定理为如何处理电源序列的算术操作提供指导。

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Theorems on Operations on Power Series
::电力系列操作理论集

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Given two power series $\begin{align*}f(x)=\sum\limits_{n=0}^{\infty} a_n x^n\end{align*}$  and $\begin{align*}g(x)=\sum\limits_{n=0}^{\infty} b_n x^n\end{align*}$  with radii of convergence $\begin{align*}R_f\end{align*}$  and $\begin{align*}R_g\end{align*}$  respectively, then the following properties hold:
::根据两个电源序列 f(x)n=0anxn 和 g(x)n=0bnxn, 并分别使用Rf和Rg的半径趋同, 然后以下属性持有:

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1. the sum of the two series is given by: $\begin{align*}f(x)+g(x)=\sum\limits_{n=0}^{\infty}(a_n+b_n)x^n\end{align*}$ , with $\begin{align*}|x| < R\end{align*}$ , where $\begin{align*}R=\min(R_f, R_g)\end{align*}$ .
   ::两个序列的总和由 f( x)+g( x)\\\n=0( an+bn) xn, 加上 {x}R, 其中R=min( Rf, Rg) 。

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Note that the difference of two series can be considered their sum with a change in sign for $\begin{align*}b_n\end{align*}$ .
::请注意,两个系列的差额可视为两者之和,但须更改bn的标志。

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2. the product of the two series is given by: $\begin{align*}f(x) \cdot g(x)=\sum\limits_{n=0}^{\infty} c_n x^n=\sum\limits_{n=0}^{\infty} \left(\sum\limits_{n=0}^{n} a_k b_{n-k} \right)x^n\end{align*}$ , with $\begin{align*}|x| < R\end{align*}$ , where $\begin{align*}R=\min(R_f, R_g)\end{align*}$ .
   ::以 f(x)\\g(x)\\n=0cnxn\n=0(nn)=0akbn-k)xn 表示两个序列的产物。
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3. the quotient of the two series is given by: $\begin{align*}\frac{f(x)}{g(x)}=\sum\limits_{n=0}^{\infty} c_n x^n\end{align*}$ , with $\begin{align*}|x| < R^\prime\end{align*}$ , where  $\begin{align*}R^\prime=\min(R_f, R_g, x_{zero})\end{align*}$ and $\begin{align*}x_{zero}\end{align*}$  is the zero of $\begin{align*}g(x)\end{align*}$  closest to $\begin{align*}x=0\end{align*}$ .
   ::两个序列的商数由下列单位给出: f(x)g(x)n=0cnxn,其中xR},Rmin(Rf,Rg,x0)和x0为零,最接近x=0的g(x)为零。

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$\begin{align*}c_n\end{align*}$  is determined recursively $\begin{align*}(b_0 \ne 0)\end{align*}$ : $\begin{align*}c_0=\frac{a_0}{b_0}\end{align*}$ ,  $\begin{align*}c_n=\frac{1}{b_0} \left[a_n- \sum\limits_{k=1}^n b_k c_{n-k} \right] n=1, 2, 3 \ldots\end{align*}$
::cn 被递归确定 (b00): c0=a0b0, cn=1b0 [an-nk=1bkcn-kn] n=1,2,3...

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Let’s look at a series product problem:
::让我们看看一系列产品问题:

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Find a power series for the function $\begin{align*}\frac{1}{(1-x)(1-2x)}\end{align*}$ .
::查找函数 1( 1) - x( 1) - 2x 的电源序列 。

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We see that $\begin{align*}\frac{1}{(1-x)(1-2x)}=\frac{1}{1-x} \cdot \frac{1}{1-2x}\end{align*}$ , and each factor has the following power series representations:
::我们看到1(1-1-x(1-2x)=11-x11-2x,每个系数的功率序列表示如下:

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$\begin{align*}\frac{1}{1-x}=\sum\limits_{n=0}^{\infty} x^n\end{align*}$  with radius of convergence $\begin{align*}R_c=1\end{align*}$ , and interval of convergence $\begin{align*}-1 < x < 1\end{align*}$ ; and  $\begin{align*}\frac{1}{1-2x}=\sum\limits_{n=0}^{\infty} (2x)^n\end{align*}$ with radius of convergence $\begin{align*}R_c=\frac{1}{2}\end{align*}$ , and interval of convergence $\begin{align*}- \frac{1}{2} < x < \frac{1}{2}\end{align*}$ .
::11-xn=0xn,半径为Rc=1,趋同间隔为-1 <x <1;和11-2xn=0(2x)n,半径为Rc=12,趋同间隔为-12 <x <12。

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By the Product Theorem above, the radius of convergence of the product is $\begin{align*}R=\min \left(1, \frac{1}{2} \right)=\frac{1}{2}\end{align*}$ .
::根据以上产品定理,该产品的合并半径为R=min(1,12)=12。

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So for $\begin{align*}|x| < \frac{1}{2}\end{align*}$ ,
::所以对于Zx12,

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$$\begin{align*}\frac{1}{(1-x)(1-2x)} &=\sum\limits_{n=0}^{\infty}x^n \cdot \sum\limits_{n=0}^{\infty}(2x)^n \\ &=(1+x+x^2+ \ldots)(1+2x+4x^2+ \ldots) \\ &=1+3x+7x^2+15x^3+ \ldots && \ldots \text{By the above product Theorem algorithm}. \\ &=\sum\limits_{n=0}^{\infty} (2^{n+1}-1)x^n.\end{align*}$$
::1(1)-(1)-(1)-2(2)n=0xn*n=0(2)n=(1)+x+x+x2+...(1)+2x+4x2+...)=1+3x+7x2+15x3+...(根据上述产品,)Theorem算法=0(2n+1)xn。

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Examples
::实例

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Example 1
::例1

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Earlier, you were asked to find the domain of the sum of the following two series: $\begin{align*}\sum\limits_{n=1}^{\infty} x^n=\frac{1}{1-x}\end{align*}$  and  $\begin{align*}\sum\limits_{n=1}^{\infty} \left(\frac{1}{3} x \right)^n=\frac{1}{1-\left(\frac{x}{3} \right)}\end{align*}$ . You were also asked to compare how the domain of the sum relates to each series.
::早些时候,有人要求您找到以下两个序列的总和的域 : @n=1xn=11-x 和\n=1(13x)n=11-(x3) 。 还要求您比较总和域与每个序列的关系。

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Were you able to work through to find an answer?
::你能够找到答案吗?

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The series  $\begin{align*}\sum\limits_{n=1}^{\infty} x^n=\frac{1}{1-x}\end{align*}$ is like a geometric series with $\begin{align*}r=x\end{align*}$  with $\begin{align*}|r|=|x| < 1\end{align*}$ ; $\begin{align*}R_c=1\end{align*}$ .
::序列 *n=1xn=11-x就像一个有 r=x 和 r=x 的几何序列, Rc=1; Rc=1; Rc=1 。

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The series  $\begin{align*}\sum\limits_{n=1}^{\infty} \left(\frac{1}{3} x \right)^n=\frac{1}{1- (\frac{x}{3})}\end{align*}$ is like a geometric series with $\begin{align*}r=\frac{x}{3}\end{align*}$  with $\begin{align*}|r|=|\frac{x}{3}| < 1\end{align*}$ ; $\begin{align*}R_c=3\end{align*}$ .
::序列 *n=1(13x)n=11-(x3)是像 r=x3 和 r=x3=1; Rc=3 的几何序列。

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The sum of the two series is $\begin{align*}\sum\limits_{n=1}^{\infty} x^n+ \sum\limits_{n=1}^{\infty} \left(\frac{1}{3} x \right)^n=\sum\limits_{n=1}^{\infty}\left[1+ \left(\frac{1}{3} \right)^n \right] x^n\end{align*}$ . The can be used to show the condition for convergence:
::这两个序列的总和是n=1xnn=1(13x)nn=1[1+(13)n]xn。可用于显示趋同条件:

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$\begin{align*}\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n}=\lim\limits_{n \to \infty} \Bigg|\frac{\left[1+ \left(\frac{1}{3} \right)^{n+1} \right]x^{n+1}}{\left[1+ \left(\frac{1}{3} \right)^{n} \right]x^{n}} \Bigg|=\lim\limits_{n \to \infty}|x|\end{align*}$ . This means that convergence occurs for $\begin{align*}|x| < 1\end{align*}$ , and $\begin{align*}R_c=1\end{align*}$ . The convergence interval for the sum is the smaller of the two series. This is consistent whit the theorem in this concept!
::limnan+1an=limn[1+(13)n+1xn+1[1+(13)xn]xnx]。这意味着x1和Rc=1会出现趋同。总和的趋同间隔是两个序列中较小的。在这个概念中,这是一致的定理!

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Example 2
::例2

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Find the power series representation for the following:
::找到下列权力序列表示法:

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1. $\begin{align*}f(x)=\frac{20 x^2}{25+9 x^2}\end{align*}$
   :x) = 20x225+9x2
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2. $\begin{align*}f^\prime (x)\end{align*}$
   ::f_( x)

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$\begin{align*}f(x)=\frac{20 x^2}{25+9 x^2}=\frac{20 x^2}{25} \cdot \frac{1}{1-(- \frac{9 x^2}{25})}\end{align*}$ , and
::f(x) = 20x225+9x2 = 20x225}11 - (- 9x225) 和

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$$\begin{align*}\frac{1}{1+ \frac{9 x^2}{25}} &=\frac{a}{1-r} && \ldots \frac{a}{1-r}=\sum\limits_{n=1}^{\infty} ar^{n-1} \ \text{for a geometric series with} \ |r| < 1. \\ &=\sum\limits_{n=1}^{\infty} \left(- \frac{9 x^2}{25} \right)^{n-1} && \ldots \bigg|- \frac{9 x^2}{25} \bigg| < 1 \Rightarrow |x| < \frac{5}{3}, \ \text{and} \ R_c=\frac{5}{3}. \\ &=\sum\limits_{n=0}^{\infty} \left(- \frac{9 x^2}{25} \right)^n \\ &=\sum\limits_{n=0}^{\infty} (-1)^n \left(\frac{9}{25} \right)^n x^{2n} \end{align*}$$
::11+9x225=a1-r...a1-rn=1arn-1几何序列,其几何序列数为 @r1. @n=1(- 9x225n- 1...)\\\\9x225\\1_x53和Rc=53.n=0(- 9x225n)n=0(- 1n)n(- 925nx2n)n

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Therefore $\begin{align*}f(x)=\frac{20 x^2}{25} \cdot \sum\limits_{n=0}^{\infty} (-1)^n \left(\frac{9}{25} \right)^n x^{2n}\end{align*}$ , or $\begin{align*}f(x)=\frac{20}{9} \cdot \sum\limits_{n=0}^{\infty} (-1)^n \left(\frac{9}{25} \right)^{n+1} x^{2(n+1)}\end{align*}$ .
::因此 f( x) = 20x225 @n= 0( - 1) n( 925nx2n) 或 f( x) = 209\n= 0( - 1) n( 925n+2n+1)。

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$\begin{align*}f^\prime (x)\end{align*}$ also has $\begin{align*}R_c=\frac{5}{3}\end{align*}$ .
::f§(x) 也有 Rc=53。

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$$\begin{align*}\frac{d}{dx}f(x) &=\frac{d}{dx} \left[\frac{20}{9} \cdot \sum\limits_{n=0}^{\infty} (-1)^n \left(\frac{9}{25} \right)^{n+1} x^{2(n+1)} \right] \\ &=\left[\frac{20}{9} \cdot \sum\limits_{n=0}^{\infty} (-1)^n \left(\frac{9}{25} \right)^{n+1} \frac{d}{dx} x^{2(n+1)} \right] \\ &=\left[\frac{40}{9} \cdot \sum\limits_{n=0}^{\infty} (-1)^n \left(\frac{9}{25} \right)^{n+1} (n+1)x^{2n+1} \right]\end{align*}$$
::ddxf( x) =ddx[209\\n=0( 1)n( 925n+12( n+1)] =[209\n=0( 1)n( 925n+925n+1)] =[409\n=0( 1)n( 925n+1n+1)x2n+1]

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Therefore $\begin{align*}f^\prime x=\left[\frac{40}{9} \cdot \sum\limits_{n=0}^{\infty} (-1)^n \left(\frac{9}{25} \right)^{n+1} (n+1)x^{2n+1} \right]\end{align*}$ , and a check of endpoints shows $\begin{align*}|x| < \frac{5}{3}\end{align*}$ .
::因此 f'x = [409\\n=0(- 1)n( 925)n+1( n+1)x2n+1), 端点检查显示\\ x53 。

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Review
::回顾

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1. Find a function, if possible, represented by the series $\begin{align*}\sum\limits_{n=0}^{\infty} 2^n x^{2^n}\end{align*}$ .
   ::如果可能,查找由序列 'n=02nx2n' 表示的函数。

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For #2-9, find a power series and the radius of convergence for the  function.
::2- 9, 找到函数的电源序列和趋同半径 。

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2. $\begin{align*}\frac{x}{2-x}\end{align*}$
   ::x2 - xx
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3. $\begin{align*}\frac{1}{(1-x)^2}\end{align*}$
   ::1(1-1x)2
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4. $\begin{align*}\frac{x^2}{(1-x)^3}\end{align*}$
   ::x2( 1) 3
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5. $\begin{align*}\frac{12 x}{7-2 x^2}\end{align*}$
   ::12x7-2x2
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6. $\begin{align*}\ln (1+x^2)\end{align*}$
   ::In In( 1+x2)
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7. $\begin{align*}\int \tan^{-1} xdx\end{align*}$
   ::- 1xdx
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8. $\begin{align*}\ln (1+x+x^2)\end{align*}$ in $\begin{align*}x+ \frac{1}{2}\end{align*}$
   ::在 x+12 中插入( 1+x+xx2)
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9. $\begin{align*}\int x^2 e^x dx\end{align*}$ [Hint: use the representation for $\begin{align*}e^x\end{align*}$ ]
   ::x2exdx [提示: 使用此代表符作为前
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10. Find a power series for  $\begin{align*}\frac{1}{(1-rx)(1-sx)}\end{align*}$ where $\begin{align*}r\end{align*}$ ,  $\begin{align*}s > 0\end{align*}$ are real numbers and
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    1. $\begin{align*}r \ne s\end{align*}$
       ::瑞 兹
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    2. $\begin{align*}r=s\end{align*}$
       ::r=s
    
    ::查找 1,1-rx (1-sx) 的电源序列, 其中 r, s>0 是真实数字, rs r=s

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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 。