节: 电力系列:介绍和聚合 | 9.13 电力系列:介绍和聚合

章节大纲

Up to this point, the infinite series that have been studied have all consisted of numbers, e.g., the series $\begin{align*}1+ \frac{1}{8}+\frac{1}{27}+\frac{1}{64}+\cdots +\frac{1}{n^3}+\cdots =\sum\limits_{n=1}^\infty \frac{1}{n^3}\end{align*}$ . There is an important type of infinite series that uses non-negative, integral powers of a variable as a way to represent a function. Whether this type of infinite series can be used for this purpose is dependent on guessed it whether the series converges for the values of the variable that are the intended domain of the function. Fortunately, you know how to determine the convergence of a series and should be able to figure out the validity of using a particular series. Alright, can you determine, before reading any farther in the concept, whether the series $\begin{align*}\sum\limits_{n=1}^\infty (-1)^{k-1}\left(\frac{x}{3}\right)^{2(k-1)}\end{align*}$ can be used to evaluate the related function at $\begin{align*}x=2\end{align*}$ ? Can you determine the function that this series represents?
::到此点为止, 所研究的无限序列全部由数字组成, 例如, 1+18+ 127+164+164%1n3\n3\n=11n3 序列。使用一个变量的非负、集成功率来代表函数的无限序列是重要的类型。这种无限序列是否可用于此目的取决于猜测该序列是否与变量的值相融合, 而变量的值是函数的指定领域。幸运的是, 您知道如何确定一个序列的趋同, 并且应该能够用特定序列来解析其有效性。好吧, 您能否在进一步阅读此概念前先确定 {n=1( 1)- k- 1 (x3)2( k- 1) 序列是否可用于评估 x=2 的相关函数 ? 您能否确定此序列所代表的函数 ? <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> As a preview of the later concepts, suppose you are given the infinite series ∞ ∑ n = 1 1.5 x n − 1 <svg focusable="false" height="6.676ex" role="img" style="vertical-align: -2.763ex;" viewbox="0 -1684.6 5211 2874.4" width="12.103ex" xmlns:xlink="http://www.w3.org/1999/xlink"> <g fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"> <g transform="translate(167,0)"> <g transform="translate(-16,0)"> <g transform="translate(0,66)"> <use x="0" xlink:href="#MJSZ2-2211" y="0"> </use> <g transform="translate(57,-1090)"> <use transform="scale(0.707)" x="0" xlink:href="#MJMATHI-6E" y="0"> </use> <use transform="scale(0.707)" x="600" xlink:href="#MJMAIN-3D" y="0"> </use> <use transform="scale(0.707)" x="1379" xlink:href="#MJMAIN-31" y="0"> </use> </g> <use transform="scale(0.707)" x="521" xlink:href="#MJMAIN-221E" y="1627"> </use> <g transform="translate(1611,0)"> <use xlink:href="#MJMAIN-31"> </use> <use x="500" xlink:href="#MJMAIN-2E" y="0"> </use> <use x="779" xlink:href="#MJMAIN-35" y="0"> </use> </g> <g transform="translate(2890,0)"> <use x="0" xlink:href="#MJMATHI-78" y="0"> </use> <g transform="translate(572,412)"> <use transform="scale(0.707)" x="0" xlink:href="#MJMATHI-6E" y="0"> </use> <use transform="scale(0.707)" x="600" xlink:href="#MJMAIN-2212" y="0"> </use> <use transform="scale(0.707)" x="1379" xlink:href="#MJMAIN-31" y="0"> </use> </g> </g> </g> </g> </g> </g> </svg> <script id="MathJax-Element-4" type="math/tex"> \begin{align*}\sum\limits_{n=1}^\infty 1.5 x^{n-1}\end{align*} , where $\begin{align*}x\end{align*}$ is a real variable, and told that it can be a positive term series, or an , and that it can converge absolutely or diverge . In addition, you are told that this infinite series can be the representation of a function $\begin{align*}f(x)\end{align*}$ . Using the concepts you have learned about infinite series, can you determine whether these statements are true or false?
::作为后来概念的预览, 假设您得到了无限序列 {n= 11. 5xn- 1 , x 是真正的变量, 并被告知它可以是正数术语序列, 或者是一个, 也可以是绝对或不同的。此外, 您被告知这个无限序列可以是函数 f( x) 的表示。使用您所了解的无限序列的概念, 您可以确定这些声明是真实的还是虚假的吗 ? <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <h3 id="x-ck12-YjNiYjQ0NzkyNWZlM2VhMDRlYTFjMGE4YWRkZTAzNTE.-ej5"> Power Series ::电力系列 </h3> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> The <span class="x-ck12-vocab-interlink" data-definition="A%20power%20series%20is%20a%20series%20of%20the%20general%20form%20%3Cmath%3E%5Csum%5Climits_%7Bn%3D0%7D%5E%5Cinfty%20a_n%20%28x-x_0%29%5En%3Da_0%2Ba_1%28x-x_0%29%2Ba_2%28x-x_0%29%5E2%2Ba_3%28x-x_0%29%5E3%2B%5Cldots%3C/math%3E%20%28called%20a%20power%20series%20centered%20at%20%3Cmath%3Ex_0%3C/math%3E%29%20where%20%3Cmath%3Ex%3C/math%3E%20is%20a%20variable%2C%20%3C/math%3Ex_0%3C/math%3E%20is%20a%20real%20constant%2C%20and%20the%20%3Cmath%3E%7Ba%7D_%7Bn%7D%3C/math%3E%27s%20are%20real%20constants%20called%20the%20coefficients%20of%20the%20series." data-id="14387" data-interlink-id="x-ck12-u2vgrldnifjw2h7n" data-json="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" data-languageid="1" data-plural="" data-term="power series" role="term" tabindex="0"> power series is a very important kind of infinite series where the terms contain powers of a variable. ::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。 <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> A power series is a series of the general form: ::电源序列是一般形式中的一系列: <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-block-math" data-mathmethod="block" data-tex="%5Csum%5Climits_%7Bn%3D0%7D%5E%5Cinfty%20a_n%20(x-x_0)%5En%3Da_0%2Ba_1(x-x_0)%2Ba_2(x-x_0)%5E2%2Ba_3(x-x_0)%5E3%2B%5Cldots"> ∞ ∑ n = 0 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 + … <div class="MathJax_SVG_Display MathJax_SVG_Processed" style="text-align: center;"> <svg focusable="false" height="6.791ex" role="img" style="vertical-align: -2.636ex; margin-bottom: -0.127ex;" viewbox="0 -1734.1 30151.9 2923.8" width="70.031ex" xmlns:xlink="http://www.w3.org/1999/xlink"> <g fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"> <g transform="translate(167,0)"> <g transform="translate(-16,0)"> <g transform="translate(0,70)"> <use x="0" xlink:href="#MJSZ2-2211" y="0"> </use> <g transform="translate(57,-1090)"> <use transform="scale(0.707)" x="0" xlink:href="#MJMATHI-6E" y="0"> </use> <use transform="scale(0.707)" x="600" xlink:href="#MJMAIN-3D" y="0"> </use> <use transform="scale(0.707)" x="1379" xlink:href="#MJMAIN-30" y="0"> </use> </g> <use transform="scale(0.707)" x="521" xlink:href="#MJMAIN-221E" y="1627"> </use> <g transform="translate(1611,0)"> <use x="0" xlink:href="#MJMATHI-61" y="0"> </use> <use transform="scale(0.707)" x="748" xlink:href="#MJMATHI-6E" y="-213"> </use> </g> <use x="2665" xlink:href="#MJMAIN-28" y="0"> </use> <use x="3054" xlink:href="#MJMATHI-78" y="0"> </use> <use x="3849" xlink:href="#MJMAIN-2212" y="0"> </use> <g transform="translate(4850,0)"> <use x="0" xlink:href="#MJMATHI-78" y="0"> </use> <use transform="scale(0.707)" x="809" xlink:href="#MJMAIN-30" y="-213"> </use> </g> <g transform="translate(5876,0)"> <use x="0" xlink:href="#MJMAIN-29" y="0"> </use> <use transform="scale(0.707)" x="550" xlink:href="#MJMATHI-6E" y="583"> </use> </g> <use x="7068" xlink:href="#MJMAIN-3D" y="0"> </use> <g transform="translate(8124,0)"> <use x="0" xlink:href="#MJMATHI-61" y="0"> </use> <use transform="scale(0.707)" x="748" xlink:href="#MJMAIN-30" y="-213"> </use> </g> <use x="9330" xlink:href="#MJMAIN-2B" y="0"> </use> <g transform="translate(10331,0)"> <use x="0" xlink:href="#MJMATHI-61" y="0"> </use> <use transform="scale(0.707)" x="748" xlink:href="#MJMAIN-31" y="-213"> </use> </g> <use x="11314" xlink:href="#MJMAIN-28" y="0"> </use> <use x="11704" xlink:href="#MJMATHI-78" y="0"> </use> <use x="12498" xlink:href="#MJMAIN-2212" y="0"> </use> <g transform="translate(13499,0)"> <use x="0" xlink:href="#MJMATHI-78" y="0"> </use> <use transform="scale(0.707)" x="809" xlink:href="#MJMAIN-30" y="-213"> </use> </g> <use x="14525" xlink:href="#MJMAIN-29" y="0"> </use> <use x="15137" xlink:href="#MJMAIN-2B" y="0"> </use> <g transform="translate(16138,0)"> <use x="0" xlink:href="#MJMATHI-61" y="0"> </use> <use transform="scale(0.707)" x="748" xlink:href="#MJMAIN-32" y="-213"> </use> </g> <use x="17121" xlink:href="#MJMAIN-28" y="0"> </use> <use x="17511" xlink:href="#MJMATHI-78" y="0"> </use> <use x="18305" xlink:href="#MJMAIN-2212" y="0"> </use> <g transform="translate(19306,0)"> <use x="0" xlink:href="#MJMATHI-78" y="0"> </use> <use transform="scale(0.707)" x="809" xlink:href="#MJMAIN-30" y="-213"> </use> </g> <g transform="translate(20333,0)"> <use x="0" xlink:href="#MJMAIN-29" y="0"> </use> <use transform="scale(0.707)" x="550" xlink:href="#MJMAIN-32" y="583"> </use> </g> <use x="21398" xlink:href="#MJMAIN-2B" y="0"> </use> <g transform="translate(22399,0)"> <use x="0" xlink:href="#MJMATHI-61" y="0"> </use> <use transform="scale(0.707)" x="748" xlink:href="#MJMAIN-33" y="-213"> </use> </g> <use x="23382" xlink:href="#MJMAIN-28" y="0"> </use> <use x="23772" xlink:href="#MJMATHI-78" y="0"> </use> <use x="24567" xlink:href="#MJMAIN-2212" y="0"> </use> <g transform="translate(25567,0)"> <use x="0" xlink:href="#MJMATHI-78" y="0"> </use> <use transform="scale(0.707)" x="809" xlink:href="#MJMAIN-30" y="-213"> </use> </g> <g transform="translate(26594,0)"> <use x="0" xlink:href="#MJMAIN-29" y="0"> </use> <use transform="scale(0.707)" x="550" xlink:href="#MJMAIN-33" y="583"> </use> </g> <use x="27659" xlink:href="#MJMAIN-2B" y="0"> </use> <use x="28660" xlink:href="#MJMAIN-2026" y="0"> </use> </g> </g> </g> </g> </svg> </div> <script id="MathJax-Element-7" type="math/tex; mode=display"> \begin{align*}\sum\limits_{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+...

(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)

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。

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

$\begin{align*}\sum\limits_{n=0}^\infty a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+\ldots\end{align*}$
::n=0anxn=a0+a1x+a2x2+a3x3+...

Power series are a generalization of polynomials, with infinitely many terms. As noted, the indices , $\begin{align*}n\end{align*}$ , are non-negative, so there are no negative integral exponents of $\begin{align*}x\end{align*}$ , e.g. $\begin{align*}\frac{1}{x}\end{align*}$ does not appear in a power series.
::电源序列是多元星系的概括,用无限多的术语。如前所述, n 指数是非负值的, 所以 x 没有负整体指数, 例如 1x 没有出现在电源序列中。

Power series are very useful, as we will explore in a later concept, for approximating functions in a form more easily useful for calculations.
::电源系列非常有用,我们将在以后的概念中探讨这一点,以便以更便于计算的形式,使功能接近一致。

For example, the function $\begin{align*}f(x)=e^x\end{align*}$ can be represented by the series $\begin{align*}f(x)=e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots = \sum\limits_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}=\sum\limits_{n=0}^\infty \frac{x^n}{n!}\end{align*}$ for any $\begin{align*}x\end{align*}$ .
::例如,函数 f( x) =ex 可以用序列 f( x) =ex= 1+x1!+x22!\\n= 1xn- 1( n- 1)!\\n= 0xnn! 对于任何 x 来说, 函数 f( x) = ex 可以用序列 f( x) = 1+x1!+x22!\n= 1xn- 1( n- 1)!\\n= 0xnn!

This power series representation opens up the possibility of approximating $\begin{align*}f(x)=e^x\end{align*}$ by a finite number of terms of the series, or using the series to evaluate difficult integrals, e.g. $\begin{align*}\int x^4e^xdx\end{align*}$ by replacing the function by a partial series when only an approximation is required.
::这种权力序列表示法有可能使 f(x) =ex 接近于该序列的限定数目,或利用该序列来评价困难的积分,例如 x4exdx ,在只需要近似值时将函数替换为部分序列。

For a given power series to be useful, one of the things we need to know is for what value(s) of $\begin{align*}x\end{align*}$ does the series converge and diverge. Fortunately you know how to evaluate the convergence or divergence of a series.
::要使给定的权力序列有用,我们需要知道的一件事是 x 的值是该序列的趋同和异差。幸运的是,你知道如何评价一个序列的趋同或异差。

Using what you already know, does the series $\begin{align*}\sum\limits_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}=\sum\limits_{n=0}^\infty \frac{x^n}{n!}\end{align*}$ converge?
::使用您已经知道的, 序列 *n= 1xn - 1 (n- 1)!\\\ n= 0xn! 集合吗 ?

Because the series $\begin{align*}\sum\limits_{n=0}^\infty \frac{x^n}{n!}\end{align*}$ can be a positive term series or an alternating series depending on the value of $\begin{align*}x\end{align*}$ , use a positive and negative term series test for convergence. Try the for absolute convergence.
::因为 {n= 0xnn!} 序列可以是正数术语序列或依 x 值而定的交替序列, 使用正数和负数术语序列测试来求同。尝试绝对趋同。

$\begin{align*}\lim\limits_{n\rightarrow \infty}\left|\frac{a^{n+1}x^{n+1}}{a^nx^n}\right| &=\lim\limits_{n\rightarrow \infty}\left|\frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}}\right|\\ &=\lim\limits_{n\rightarrow \infty}\left|\frac{x^{n+1}}{x^n}\frac{n!}{(n+1)!}\right|\\ &=\lim\limits_{n\rightarrow \infty}\left|\frac{x}{n+1}\right|\\ &=x\lim\limits_{n\rightarrow \infty}\left|\frac{1}{n+1}\right|\\ &=x\cdot 0=0<1 \end{align*}$
::limnan+1xn+1xn+1anxn+1xn+1(n+1)! xnn! @ limn}xn+1xn+1xnn! (n+1) @xn+1@xlimn+1}1n+1xlimn+1}1n+1}xxxxn+1}1n+1}0=0<1

The series $\begin{align*}\sum\limits_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}=\sum\limits_{n=0}^\infty\frac{x^n}{n!}\end{align*}$ converges absolutely for all values of $\begin{align*}x\end{align*}$ , i.e., $\begin{align*}-\infty < x< \infty\end{align*}$ .
::==1xn-1(n- 1)!\\\n=0xnn!对于 x 的所有值, 即\\\ xx\\\\\\\\\ 绝对一致。

Let’s look at an example of a series with a more restricted range of $\begin{align*}x\end{align*}$ for convergence.
::让我们看看一个例子,

Does the series $\begin{align*}\sum\limits_{n=0}^\infty \frac{x^n}{n^2}\end{align*}$ converge or diverge?
::序列 *n=0xnn2 相交或相左吗 ?

Because the series $\begin{align*}\sum\limits_{n=0}^\infty \frac{x^n}{n^2}\end{align*}$ can be a positive term series or an alternating series depending on the value of $\begin{align*}x\end{align*}$ , use a positive and negative term series test for convergence. Try the Ratio Test for absolute convergence.
::因为 n= 0xnn2 序列可以是正数术语序列或依 x 值而定的交替序列, 使用正数和负数术语序列测试以求同。尝试比率测试以求绝对趋同。

Let $\begin{align*}b_n=\frac{x^n}{n^2}\end{align*}$ .
::bn=xnn2 允许 bn=xnn2 。

Then,
::然后,

$\begin{align*}\lim\limits_{n\rightarrow \infty}\left|\frac{b_{n+1}}{b_n}\right| &=\lim\limits_{n\rightarrow \infty}\left|\frac{x^{n+1}}{(n+1)^2}\cdot \frac{n^2}{x^n}\right|\\ &=\lim\limits_{n\rightarrow \infty}\left|\left(\frac{1}{1+\frac{1}{n}}\right)^2 x\right|\\ &=|x|\end{align*}$
:

11+1n)2xxx

By the Ratio Test, the series is:
::根据比率测试,该系列为:

absolutely convergent for $\begin{align*}|x|<1\end{align*}$ , i.e., in the open interval $\begin{align*}-1 < x < 1\end{align*}$ ; and
::*%x%1 绝对趋同, 即开放间隔 - 1 < x < 1; 和

divergent for $\begin{align*}|x|>1\end{align*}$ .
::%x% 1 的差异。

But what happens at the endpoints?
::但是在终点会发生什么事呢?

If $\begin{align*}x=\pm 1\end{align*}$ , then the series $\begin{align*}\sum\limits_{n=0}^\infty \frac{x^n}{n^2}\end{align*}$ becomes the series $\begin{align*}\sum\limits_{n=0}^\infty \frac{1}{n^2}\end{align*}$ or $\begin{align*}\sum\limits_{n=0}^\infty \frac{(-1)^n}{n^2}\end{align*}$ , both of which are absolutely convergent $\begin{align*}p\end{align*}$ -series with $\begin{align*}p=2\end{align*}$ .
::如果 x1, 那么序列= 0xnn2 变成序列\\ n= 010n2 或\ n= 0(- 1) nn2, 两者都是与 p=2 绝对一致的 p- 系列 p- 系列。

Hence the series $\begin{align*}\sum\limits_{n=0}^\infty \frac{x^n}{n^2}\end{align*}$ is absolutely convergent for $\begin{align*}|x|\le 1\end{align*}$ , i.e., the closed interval $\begin{align*}-1 \le x\le 1\end{align*}$ .
::因此,%n= 0xnn2 序列绝对合为 {x}1, 也就是说, 闭合间隔 - 1x}1。

The Radius and Interval of Convergence of a Power Series
::电力系列集集集的半径和半径间隙

In the above examples, the value of $\begin{align*}x\end{align*}$ determines whether the power series is convergent or divergent. Convergence can occur in an interval of $\begin{align*}x\end{align*}$ that may be open or closed.
::在上述例子中, x 的值决定电源序列是集合的还是不同的。聚合可以在x的间隔内发生,可以是开放的,也可以是封闭的。

The theorem for Convergence of a Power Series is as follows:
::电力系列集集的理论理论论如下:

Let $\begin{align*}\sum\limits_{n=0}^\infty a_n (x-x_0)^n\end{align*}$ be a power series centered at $\begin{align*}x_0\end{align*}$ .
::@n=0an(x-x0)n 是一个以 x0 为中心的电源序列。

Then exactly one of the following three cases describes all the values of $\begin{align*}x\end{align*}$ where the series is convergent:
::然后精确地描述以下三个案例之一, 描述序列集合的 x 的所有值 :

Case 1. The series converges only at $\begin{align*}x=x_0\end{align*}$ .
::案例1. 本序列只在x=x0时趋同。

Case 2. The series absolutely converges for all $\begin{align*}x\end{align*}$ .
::案例2. 所有x的系列绝对一致。

Case 3. There is a real number $\begin{align*}R_c>0\end{align*}$ such that the series absolutely converges if $\begin{align*}|x-x_0| < R_c\end{align*}$ and diverges if $\begin{align*}|x-x_0|>R_c\end{align*}$ .
::案例3. 有真实的 Rc>0 号,因此如果 {x_x0}\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Note that $\begin{align*}|x-x_0|< R_c\end{align*}$ means the open interval $\begin{align*}(x_0-R_c)< x< (x_0+R_c)\end{align*}$ . The theorem cannot confirm what happens at the endpoints $\begin{align*}x=x_0\pm R_c\end{align*}$ ; that must be determined separately.
::请注意, x- x0Rc 是指开放间隔( x0- Rc) < x < (x0+Rc) 。定理无法确认端点sx=x0Rc 发生的情况; 这必须单独确定。

In the above theorem, the number $\begin{align*}R_c\end{align*}$ is called the radius of convergence of the power series. The complete $\begin{align*}x\end{align*}$ -interval where the power series converges is called the interval of convergence .
::在上述定理中, Rc 的数值被称为电源序列的趋同半径。功率序列相交的完整 X 间距,称为汇合间隔。

The theorem for the Radius and Interval of Convergence of a Power Series is as follows:
::电力系列的拉迪乌斯和汇合时间间隔的理论理论如下:

Let $\begin{align*}\sum\limits_{n=0}^\infty a_n (x-x_0)^n\end{align*}$ be a power series centered at $\begin{align*}x_0\end{align*}$ .
::@n=0an(x-x0)n 是一个以 x0 为中心的电源序列。

Then exactly one of the following three cases describes the relationship between the radius of convergence and the interval of convergence:
::然后,以下三个例子中正好有一个说明了趋同半径与趋同间隔之间的关系:

Case	Radius of Convergence	Interval of Convergence
1. Series converges only at $\begin{align}x = x_0\end{align}$ .	$\begin{align}R_c=0\end{align}$	One value: $\begin{align}x_0\end{align}$
2. Series absolutely converges for all $\begin{align}x\end{align}$ .	$\begin{align}R_c=\infty\end{align}$	$\begin{align}(-\infty ,\infty )\end{align}$
3. Series absolutely converges if $\begin{align}\|x-x_0\| < R_c\end{align}$ , and diverges if $\begin{align}\|x-x_0\|>R_c\end{align}$ . ::3. 如果 x-x0Rc,则绝对一致,如果 x-x0Rc,则绝对一致;如果 x-x0Rc,则绝对一致。	$\begin{align}R_c>0\end{align}$	$\begin{align}(x_0-R_c, x_0+R_c)\end{align}$ , or $\begin{align}(x_0-R_c, x_0+R_c] ^\end{align}$ , or $\begin{align}[x_0 - R_c, x_0+R_c)^\end{align}$ , or $\begin{align}[x_0-R_c, x_0+R_c]^\end{align*}$

* The convergence at the endpoints $\begin{align*}x=x_0\pm R_c\end{align*}$ must be determined separately.
::* 端点x=x0-Rc的趋同必须分开确定。

To determine the interval or radius of convergence for a power series, you must apply one of the convergence tests learned in previous concepts, and find the conditions on $\begin{align*}x\end{align*}$ that make convergence true.

::要确定一个电源序列趋同的间隔或半径,您必须应用在先前概念中所学到的趋同测试之一,并找到使趋同成为事实的x的条件。

Using what you now know, determine if the series $\begin{align*}\sum\limits_{n=0}^\infty (x-3)^n\end{align*}$ converges or diverges; if it does, find the radius and interval of convergence.
::使用您现在所知道的, 确定序列 *n=0(x- 3)n 是否趋同或相异; 如果相异, 则找到趋同的半径和间隔。

Try the $\begin{align*}n\end{align*}$ th for absolute convergence.
::尝试在 nth 中绝对趋同。

$\begin{align*}\lim\limits_{n \rightarrow \infty}{\sqrt[n]{|(x-3)^n|}} &=\lim\limits_{n \rightarrow \infty}|(x-3)|\\ &=|(x-3)|\end{align*}$
:

x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3) (x-3)

The series will be absolutely convergent if $\begin{align*}|(x-3)|<1\end{align*}$ , and divergent if $\begin{align*}|(x-3)|>1\end{align*}$ .
::如果 {(x-3)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(x\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\和分裂的差异和不同的差异和和和和和不同的差异的差异。。和不同。\\\和差异\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> | ( x − 3 ) | < 1 <script id="MathJax-Element-85" type="math/tex"> \begin{align*}|(x-3)|<1\end{align*} means a radius of convergence $\begin{align*}R_c=1\end{align*}$ over an interval of convergence of $\begin{align*}2 < x < 4\end{align*}$ .
:x-3-3)1 是指在2 < x < 4的交汇间隔内Rc=1的趋同半径。

To determine if the endpoints need to be included, look at the series at the endpoints:
::为确定是否需要包括终点,请在终点处查看该系列:

At $\begin{align*}x=2\end{align*}$ : $\begin{align*}\sum\limits_{n=0}^\infty (2-3)^n = \sum\limits_{n=0}^\infty (-1)^n\end{align*}$ which diverges absolutely, does not satisfy the conditions, or settle on a single limit of the partial sum. This endpoint does not produce convergence.
::x=2: @n=0(2-3)nn=0(-1)n 绝对分歧,不符合条件,或只解决部分总和的单一限额。这个终点没有产生趋同。

At $\begin{align*}x=3\end{align*}$ : Same condition as above: This endpoint does not produce convergence.
::x=3:与上述条件相同:这一终点不产生趋同。

Therefore, the series $\begin{align*}\sum\limits_{n=0}^\infty (x-3)^n\end{align*}$ has a radius of convergence $\begin{align*}R_c=1\end{align*}$ over an interval of convergence of $\begin{align*}2 < x < 4\end{align*}$ .
::因此,Xn=0(x-3)n系列在2 <x < 4 " 的交汇间隔内,有Rc=1的趋同半径。

Examples
::实例

Example 1
::例1

Earlier, you were asked to determine whether the series $\begin{align*}\sum\limits_{n=1}^\infty (-1)^{k-1}\left(\frac{x}{3}\right)^{2(k-1)}\end{align*}$ can be used to evaluate the related function at $\begin{align*}x=2\end{align*}$ and what function this series represents.
::之前曾要求您确定该序列 *n=1 (- 1) -1 (x3)(k-1) 是否可用于评估 x=2 的相关函数以及该序列代表什么函数。

The series can be rewritten as follows: $\begin{align*}\sum\limits_{n=1}^\infty (-1)^{k-1}\left(\frac{x}{3}\right)^{2(k-1)} = \sum\limits_{n=1}^\infty\left(-\frac{x^2}{9}\right)^{k-1}\end{align*}$ . This is a geometric series with $\begin{align*}a=1\end{align*}$ and $\begin{align*}r=-\frac{x^2}{9}\end{align*}$ . The series will absolutely converge (yield meaningful results) as long as $\begin{align*}|r|=\left|-\frac{x^2}{9}\right|=\frac{x^2}{9}<1\end{align*}$ , which means $\begin{align*}-3 < x < 3\end{align*}$ . Since $\begin{align*}x=2\end{align*}$ is within the interval of convergence, the series can be evaluated at $\begin{align*}x=2\end{align*}$ .
::该序列可重写如下:n=1(-1)(-1)-1(x3)(k-1)(2--1)n=1(-x29)-1.这是一个几何序列,有=1和rx29。只要rx29(x29)x29 <1),该序列将绝对趋同(产生有意义的结果),这意味着-3 < x < 3. 因为 x=2 在交汇的间隔内,该序列可以在x=2时评估。

But what function is to be evaluated? Well $\begin{align*}\sum\limits_{n=1}^\infty \left(-\frac{x^2}{9}\right)^{k-1}=\frac{a}{1-r}=\frac{1}{1-\left(-\frac{x^2}{9}\right)}=\frac{9}{9+x^2}\end{align*}$ .
::但是要评估的功能是什么? @ n=1(-x29k- 1)=a1-r=11- (-x29)=99+x2。

Therefore, $\begin{align*}f(x)=\frac{9}{9+x^2}\end{align*}$ , and $\begin{align*}f(2)=\frac{9}{9+2^2}=\frac{9}{13}\end{align*}$ .
::因此,f(x)=99+x2,f(2)=99+22=913。

Example 2
::例2

Determine if the series $\begin{align*}\sum\limits_{n=0}^\infty (n+1)^2(x+4.5)^n\end{align*}$ converges or diverges; if it does, find the radius and interval of convergence.
::确定 {n=0(n+1)2(x+4.5)n 序列是否趋同或相异;如果相异,则找到趋同的半径和间隔。 <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> Try the n <script id="MathJax-Element-107" type="math/tex"> \begin{align*}n\end{align*} th Root Test for absolute convergence.
::尝试 nth 根测试, 以绝对趋同。

$\begin{align*}\lim\limits_{n\rightarrow \infty}\sqrt[n]{|(n+1)^2(x+4.5)^n|} &= \lim\limits_{n\rightarrow \infty}\left|(n+1)^{\frac{2}{n}}(x+4.5)\right|\\ &=|(x+4.5)| \lim\limits_{n\rightarrow \infty}(n+1)^{\frac{2}{n}} && \ldots \text{Let } y=\lim\limits_{n\rightarrow \infty}(n+1)^{\frac{2}{n}}, \text{then}\\ & && \lim\limits_{n\rightarrow \infty}\ln y= \lim\limits_{n\rightarrow \infty}\left[\frac{2\ln (n+1)}{n}\right]\\ & && \qquad \quad \ \ = \lim\limits_{n\rightarrow \infty}\left[\frac{\frac{2}{ (n+1)}}{1}\right] = 0.\\ & && \text{Therefore, } y=e^0=1= \lim\limits_{n\rightarrow \infty} (n+1)^{\frac{2}{n}}.\\ &=|(x+4.5)|\end{align*}$
::limn@ (n+1) 2(x+4.5) n(n+1) 2n(x+4.5) (x+4.5) _(n+1) 2n... (n+1) (n+1) 2n, thenlimnlny=limn[2ln(n+1) =limn[2(n+1) 1] =0. 。因此, y=e0=1=limn(n+1) 2n. (x+4.5) (y=1==1=limn}(n+1) (x+4.5) *(x+4.5) *(y+4.5) *

The series will be absolutely convergent if $\begin{align*}|(x+4.5)|<1\end{align*}$ , and divergent if $\begin{align*}|(x+4.5)|>1\end{align*}$ .
::如果 {(x+4.5)\}\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(x+4.5\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\和不同的差异和不同差异和不同的差异和不同。

$\begin{align*}|(x+4.5)|<1\end{align*}$ means a radius of convergence $\begin{align*}R_c=1\end{align*}$ over an interval of convergence of $\begin{align*}-5.5 < x <-3.5\end{align*}$ .
:x+4.5)1 是指在-5.5<x3.5的交汇间隔内,Rc=1的趋同半径。

To determine if the endpoints need to be included in the interval of convergence, look at the series at the endpoints:
::为确定是否需要将终点包括在趋同间隔内,请在终点处查看该系列:

At $\begin{align*}x=-5.5\end{align*}$ : $\begin{align*}\sum\limits_{n=0}^\infty (n+1)^2(-5.5+4.5)^n = \sum\limits_{n=0}^\infty (n+1)^2(-1)^n\end{align*}$ which diverges absolutely, does not satisfy the Alternating Series Test conditions, but diverges by the $\begin{align*}n\end{align*}$ th term divergence test. The endpoint does not produce convergence.
::在 x5.5 : n= 0(n+1)(2--5.5+4.5)nn= 0(n+1)(-2- 1)n, 绝对有差异, 不符合交替序列试验条件, 但因 n 值差异试验而有差异。终点不产生趋同。

At $\begin{align*}x=-3.5\end{align*}$ : $\begin{align*}\sum\limits_{n=0}^\infty (n+1)^2(-3.5+4.5)^n = \sum\limits_{n=0}^\infty (n+1)^2\end{align*}$ diverges. This endpoint does not produce convergence.
::x3.5: n=0(n+1)(2-3.5+4.5)nn=0(n+1)2 差异。此端点不会产生趋同。

Therefore, the series $\begin{align*}\sum\limits_{n=0}^\infty (n+1)^2(x+4.5)^n\end{align*}$ has a radius of convergence $\begin{align*}R_c=1\end{align*}$ over an interval of convergence of $\begin{align*}-5.5 < x < -3.5\end{align*}$ .
::因此,n=0(n+1)2(x+4.5)n系列在-5.5<x3.5的交汇间隔内,有Rc=1的半径趋同。

Review
::回顾

For #1-5, determine where the series converges and diverges. If it does converge, find the radius and the interval of convergence.
::对于 # 1 5 , 确定序列的相交点和差异点。如果相交点, 则会找到半径和趋同间隔。

$\begin{align*}\sum\limits_{n=1}^\infty nx^n\end{align*}$
::*n=1nxn

$\begin{align*}\sum\limits_{n=1}^\infty n!x^n\end{align*}$
::#Nn=1n! xn! # # Nn=1n! # xn! #

$\begin{align*}\sum\limits_{n=1}^\infty \frac{(-1)^n x^n}{n}\end{align*}$
::*n=1(- 1)nxnn

$\begin{align*}\sum\limits_{n=1}^\infty \frac{x^{\frac{n}{3}}}{n!}\end{align*}$
::1xn3n! 一xn3n! 一xn3n! 一xn3n!

$\begin{align*}\sum\limits_{n=1}^\infty \sqrt{n}(x-x_0)^n\end{align*}$
::=1n(x-x0)n

Given $\begin{align*}\sum\limits_{n=0}^\infty a_n x^n\end{align*}$ converges at $\begin{align*}x=5\end{align*}$ and diverges at $\begin{align*}x=-7\end{align*}$ , deduce where possible, the convergence or divergence of these series:
1. $\begin{align*}\sum\limits_{n=0}^\infty a_n\end{align*}$
  ::n=0an
2. $\begin{align*}\sum\limits_{n=0}^\infty a_n 3^n\end{align*}$
  ::=0an3n =0an3n =0an3n =0an3n =0an3n =0an3n
3. $\begin{align*}\sum\limits_{n=0}^\infty a_n(-8)^n\end{align*}$
  ::n=0an(-8)n
4. $\begin{align*}\sum\limits_{n=0}^\infty a_n 9^n\end{align*}$
  ::*n=0an9n* *n=0an9n*
5. $\begin{align*}\sum\limits_{n=0}^\infty a_n 6^n\end{align*}$
  ::= 0an6n = 0an6n = 0an6n = 0an6n = 0an6n = 0an6n
::鉴于n=0anxn在x=5时会趋同,在x=5时会相异,在x=7时会有所不同,尽可能推断出这些序列的趋同或分歧:n=0an3n=0an3n=0an(-8)n=0an9n=n=0an6n

For #7-15, determine where the series converges and diverges. If it does converge, find the radius and the interval of convergence.
::对于 # 7-15 , 确定序列的相交点和不同点。如果相交点, 则找到半径和趋同的间隔。

$\begin{align*}\sum\limits_{n=0}^\infty \frac{n(x+4)^n}{5^{n+1}}\end{align*}$
::*n=0n(x+4)n5n+1

$\begin{align*}\sum\limits_{n=1}^\infty 3^n (x-2)^n\end{align*}$
::n=13n(x-2)n

$\begin{align*}\sum\limits_{n=1}^\infty (-1)^n(x+4)^n\end{align*}$
::*n=1(- 1)n( x+4)n

$\begin{align*}\sum\limits_{n=0}^\infty \frac{n!x^n}{2^n}\end{align*}$
::#Nnn#xn2n##nn#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#n#xn#n#n#n#n#n#n#n#n#n#n#xn#n#n#n#n#n#xxn#n#n#n#n#n#n#n#n#n#n#n##n#n#n#n#n######################################################################################################################################################################################################################################################################################################################################################

$\begin{align*}\sum\limits_{n=0}^\infty \frac{(2x)^n}{n^2}\end{align*}$
::*n=0(2x)nn2

$\begin{align*}\sum\limits_{n=0}^\infty \frac{x^n}{4n^2+3}\end{align*}$
::*n=0xn4n2+3

$\begin{align*}\sum\limits_{n=0}^\infty \frac{4^n x^{2n}}{n+3}\end{align*}$
::*n=04nx2nn+3

$\begin{align*}\sum\limits_{n=1}^\infty \frac{5^n(x-3)^n}{n4^n}\end{align*}$
::=15n(x-3)nn4n

$\begin{align*}\sum\limits_{n=0}^\infty \frac{n(x+1)^{2n}}{7^n}\end{align*}$
::*n=0n(x+1)2n7n

Review (Answers)
::回顾(答复)

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。

Case	Radius of Convergence	Interval of Convergence
1. Series converges only at $\begin{align}x = x_0\end{align}$ .	$\begin{align}R_c=0\end{align}$	One value: $\begin{align}x_0\end{align}$
2. Series absolutely converges for all $\begin{align}x\end{align}$ .	$\begin{align}R_c=\infty\end{align}$	$\begin{align}(-\infty ,\infty )\end{align}$
3. Series absolutely converges if $\begin{align}\|x-x_0\| < R_c\end{align}$ , and diverges if $\begin{align}\|x-x_0\|>R_c\end{align}$ . ::3. 如果 x-x0Rc,则绝对一致,如果 x-x0Rc,则绝对一致;如果 x-x0Rc,则绝对一致。	$\begin{align}R_c>0\end{align}$	$\begin{align}(x_0-R_c, x_0+R_c)\end{align}$ , or $\begin{align}(x_0-R_c, x_0+R_c] ^\end{align}$ , or $\begin{align}[x_0 - R_c, x_0+R_c)^\end{align}$ , or $\begin{align}[x_0-R_c, x_0+R_c]^\end{align*}$

章节大纲

The Radius and Interval of Convergence of a Power Series ::电力系列集集集的半径和半径间隙

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Review ::回顾