9.18 以系列计算:二小分数功率、整体体和差异方程

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  以系列计算:二元分数功率、整体体和差分方程

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  One of the very useful features of power series is the ability to use a series to evaluate an integral that does not have a closed form anti-derivative, or to solve a differential equation . Such a power series representation can be truncated to a polynomial to evaluate a definite integral using the degree sufficient to provide a specified accuracy. If you are given the task to use a power series to evaluate the integral $\begin{align*}\int\limits^1_0 x^2 e^{-x^2} dx\end{align*}$  with accuracy to 4 decimal places, how do you do it?
  ::电源序列的一个非常有用的特征是能够使用一个序列来评价一个没有封闭式反变性的集成,或者能够解决差别方程式。这样的电源序列表示法可以被截断成一个多元体来评价一个明确的集成,使用足够精确度来提供特定准确度。如果授权您使用一个序列来评价一个精确度为小数点4位的集成 10x2e-x2dx, 那么如何进行?

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  Additional Applications of Series
  ::系列的附加应用

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  The three topics that will be presented are:
  ::将介绍的三个专题是:

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  • Taylor series representation of a power of a binomial (Binomial series)
    ::Taylor系列表示二元论(Binomial系列)的力量
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  • Evaluating non-elementary integral
    ::评价非基本组成部分
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  • Solving differential equations using power series
    ::使用电源序列解决差异方程

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  Binomial Series
  ::二进制系列丛书

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  We have learned how to generate a power (Taylor) series of a function $\begin{align*}f(x)\end{align*}$  by using the value of the function and its at a specified center . One special function form of interest is  $\begin{align*}f(x)=(1+x)^r\end{align*}$ where $\begin{align*}r\end{align*}$  is a real number and $\begin{align*}|x|<1\end{align*}$ . An important special Taylor series called the binomial series provides the Maclaurin series for this $\begin{align*}f(x)\end{align*}$ .
  ::我们学会了如何通过使用函数的值及其在指定中心的位置来生成函数 f(x) 的电源序列( Taylor) 。 一种特殊功能形式是 f( x) = ( 1+x)r , r 是真实数字 , {x\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Theorem: Binomial Series
  ::论理:二联论系列

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  Let  $\begin{align*}f(x)=(1+x)^r\end{align*}$ where $\begin{align*}r\end{align*}$  is a real number and $\begin{align*}|x|<1\end{align*}$ . Then
  ::Letf( x) =( 1+x)r , r 是真实数字, \\\ x @ 1 。 然后

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  %7D%20(0)%20%5Cfrac%7Bx%5En%7D%7Bn!%7D%3Df(0)%2Bf%5E%5Cprime%20(0)%20x%2Bf%5E%7B%5Cprime%5Cprime%7D(0)%20%5Cfrac%7Bx%5E2%7D%7B2!%7D%2Bf%5E%7B%5Cprime%5Cprime%5Cprime%7D%20(0)%20%5Cfrac%7Bx%5E3%7D%7B3!%7D%2B%5Ccdots%20f%5E%7B%7D%20(0)%20%5Cfrac%7Bx%5En%7D%7Bn!%7D%2B%5Ccdots%5C%5C%0A%26%3D%201%2Brx%2B%5Cfrac%7Br(r-1)%7D%7B2!%7Dx%5E2%2B%5Cfrac%7Br(r-1)(r-2)%7D%7B3!%7D%20x%5E3%2B%5Ccdots%20%5Cfrac%7Br(r-1)(r-2)%20%5Ccdots%20(r-n%2B1)%7D%7Bn!%7Dx%5En%2B%5Ccdots%5C%5C%0A%26%3D%20%5Csum%5Climits%5E%5Cinfty_%7Bk%3D0%7D%20%5Cbinom%7Br%7D%7Bk%7D%20x%5Ek">

  $$\begin{align*}f(x) &= \sum\limits^\infty_{n=0} f^{(n)} (0) \frac{x^n}{n!}=f(0)+f^\prime (0) x+f^{\prime\prime}(0) \frac{x^2}{2!}+f^{\prime\prime\prime} (0) \frac{x^3}{3!}+\cdots f^{(n)} (0) \frac{x^n}{n!}+\cdots\\ &= 1+rx+\frac{r(r-1)}{2!}x^2+\frac{r(r-1)(r-2)}{3!} x^3+\cdots \frac{r(r-1)(r-2) \cdots (r-n+1)}{n!}x^n+\cdots\\ &= \sum\limits^\infty_{k=0} \binom{r}{k} x^k\end{align*}$$
  ::f( x) {\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  is the binomial series for $\begin{align*}f(x)\end{align*}$ .
  ::f( x) 的二进制序列。

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  The Binomial coefficients are denoted by  $\begin{align*}\binom{r}{k}\end{align*}$ where :
  ::Binomial 系数用(rk)表示,其中:

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  $$\begin{align*}\binom{r}{0} &= 1 \ \text{for} \ k=0\\ \binom{r}{k} &= \frac{r(r-1) \ldots (r-k+1)}{k!} \ \text{for} \ k \ge 1.\end{align*}$$
  ::k=0( rk) =r( r- 1) = 1 。 (r- k+1) k_ 1 ! k_ 1 = 1 。 (r- k+1) k_ 1 !

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  Note:
  ::注:

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  1. If $\begin{align*}r\end{align*}$  is a non-negative integer $\begin{align*}n\end{align*}$ , then the series terminates and the binomial series reduces to a polynomial of degree  $\begin{align*}n\end{align*}$ which converges for all $\begin{align*}x\end{align*}$ . 
     ::如果 r 是非负整 n,则序列终止,二进制序列减为全 x 的一进制 n。
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  2. Otherwise, the series is infinite. The series diverges if $\begin{align*}|x| < 1\end{align*}$ ; the behavior of the series for $\begin{align*}|x|=1\end{align*}$  depends on the value of $\begin{align*}r\end{align*}$ .
     ::否则,该序列是无限的。 如果 \\\ x\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  The above series, which has an infinite number of terms, looks similar to the binomial equation, which has a finite number of terms:
  ::上述系列有无限数目的术语,与二元式公式类似,二元式公式有一定数目的术语:

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  $$\begin{align*}(a+b)^n &= a^n+na^{n-1} b+\frac{n(n-1)}{2!} a^{n-2} b^2+\ldots+nab^{n-1}+b^n\\ & && \ldots \text{the Binomial coefficients are denoted by:}\\ & =\sum\limits^n_{k=0} \binom{n}{k} a^{n-k}b^k && \binom{n}{k} = \frac{n!}{k!(n-k)!} \ \text{for} \ k \ge 0.\end{align*}$$
  :a+b)n=an+nan-1b+n(n-1)2!an-2b2+...+Nabn-1+bn...bn_Ben_BAR_BAR__BAR_BAR__BAR_BAR__BAR_BAR_BAR__BAR__BAR_BAR__BAR_BAR__BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_B_BAR_BAR_BAR_B_B_B_B_B_B_B_B_B_BAR_B_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_B_B_B_B_B_B_________________________________________________________________________________________________________________________

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  Let's apply the theorem from above and find a power series representation of $\begin{align*}\sqrt{1+x}\end{align*}$ .
  ::让我们应用上方的定理 找到一个=% 1+x 的电源序列表示 。

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  For $\begin{align*}\sqrt{1+x}, r=\frac{1}{2}\end{align*}$ . For $\begin{align*}|x|<1\end{align*}$ , the function can be written as the binomial series
  ::%% 1+x, r=12。 对于\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  $\begin{align*}\sqrt{1+x}=\sum\limits^\infty_{k=0} \dbinom{\frac{1}{2}}{k} x^k\end{align*}$ , with binomial coefficients given by:
  ::+1+xk=0(12k)xk,二元系数为:

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  $$\begin{align*}\binom{\frac{1}{2}}{k} &= \frac{\left(\frac{1}{2}\right) \left(-\frac{1}{2}\right) \left(-\frac{3}{2}\right) \cdots \left(-\frac{2k-3}{2}\right)}{k!}\\ &= \frac{(-1)^{k-1} (2k-2)!}{[2 \times 4 \times 6 \cdots \times (2k-2)]2^k k!}\\ &= \frac{(-1)^{k-1}(2k-2)!}{2^{k-1}k!(k-1)2^k k!}\\ &= \frac{(-1)^{k-1} (2k-2)!}{2^{2k-1}k!(k-1)!}\end{align*}$$
  :12) = (12) (12) (-12) (-32) (- 2k) (- 2k) 32k! (-1) 1) 1 (2k) 2! [2) 4x6 (2k) 2) 2kk! = (-1) 1x1 (2k) 2k- 1k! (k) 2k- 1k! (- 1) = (1) 1x1 (2k) 2x2! 22k- 1k! (k) ! (1) !

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  Therefore,
  ::因此,

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  $\begin{align*}\sqrt{1+x} = \sum\limits^\infty_{k=0} \dbinom{\frac{1}{2}}{k} x^k=1+\sum\limits^\infty_{k=0} \frac{(-1)^{k-1} (2k-2)!}{2^{2k-1} k! (k-1)!} x^k \ \text{for} \ |x|<1\end{align*}$ .
  ::+1+xk=0( 12k) xk=1k=0( - 1k)- 1( 2k-2)! 22k- 1k! (k- 1)! xk for *x# 1)! xk for *x# 1

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  Evaluating Non-Elementary Integrals
  ::评价非组成部分综合评估

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  There are many simple-looking functions that have no explicit formula for their integral in the form of elementary functions. However, in some cases their integral can be written as a Taylor Series in their interval of convergence .
  ::有许多简单看得见的职能,没有以基本职能的形式对其整体性作出明确的公式,但在某些情况下,其整体性可以写成泰勒系列,在它们交汇的间隔内写成泰勒系列。

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  For example, let's find a power series representation of $\begin{align*}\int \frac{e^x}{x}dx\end{align*}$ .
  ::例如,让我们来找一个权力序列代表 exxdx。

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  Since $\begin{align*}\frac{e^x}{x}\end{align*}$  is not defined at $\begin{align*}x=0\end{align*}$ , we apply the Taylor Series of $\begin{align*}e^x\end{align*}$  at, say, $\begin{align*}x=1\end{align*}$  by writing $\begin{align*}e^x=e \cdot e^{x-1}\end{align*}$  with a change of variable $\begin{align*}u=x-1\end{align*}$ .
  ::由于exx=0没有定义,因此我们适用Taylor Ex系列,例如,x=1, 写出 ex=eex-1, 变数u=x-1。

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  $$\begin{align*}\int \frac{e^x}{x} dx &= \int \frac{e \cdot e^u}{1+u}du\\ &= \int \left[e \cdot \frac{1}{1+u} \sum\limits^\infty_{n=0} \frac{(u)^n}{n!}\right] du\\ &= \int e(1-u+u^2-u^3+\ldots) \left(1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\frac{u^4}{4!}+\ldots \right) du && \ldots \text{where} \ \frac{1}{1+u}=\sum\limits^\infty_{n=0} (-1)^n u^n\\ & && \text{for} \ |u|<1.\\ &= \int e \left(1+\frac{1}{2}u^2-\frac{1}{3}u^3+\frac{3}{8}u^4-\frac{11}{30} u^5+\ldots \right)du && \text{for} \ |u|<1.\\ & = e \left(u+\frac{1}{6}u^3-\frac{1}{12}u^4+\frac{3}{40} u^5-\frac{11}{180}u^6+\ldots \right)+C\end{align*}$$
  ::=======================================================================================================================================================+=====================================================================================================================================================================================================================================================================================================================================================================

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  Therefore,
  ::因此,

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  $\begin{align*}\int \frac{e^x}{x} dx=e \left[(x-1)+\frac{1}{6} (x-1)^3-\frac{1}{12}(x-1)^4+\frac{3}{40} (x-1)^5-\frac{11}{180} (x-1)^6+\ldots \right)+C\end{align*}$ .
  ::exxdx=e[(x-1)+16(x-1)3-112(x-1)4+340(x-1)5-11180(x-1)6+...])+C。

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  Solving differential equations using power series
  ::使用电源序列解决差异方程

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  Use of a power series is a basic method for solving differential equations. This method yields an approximate solution to a differential equation near a single point $\begin{align*}x=x_0\end{align*}$ . The method uses the differential equation to find the coefficients of the Taylor series about the point $\begin{align*}x=x_0\end{align*}$ .
  ::使用电源序列是解决差别方程式的基本方法。 这个方法可以大致解决接近一个点 x=x0 的差异方程式。 这个方法使用差方程式来找到关于点 x=x0 的泰勒序列系数。

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  The problem below shows application of the power series to solving a linear differential equation:
  ::以下问题显示了电源序列用于解决线性差方程的情况:

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  Solve the linear first order differential  $\begin{align*}y^\prime -y=0\end{align*}$ using a power series.
  ::使用一个电源序列解决线性第一顺序差值 yy=0 。

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  Assume the solution has the form $\begin{align*}y=\sum\limits^\infty_{n=0} a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+\ldots\end{align*}$ .
  ::假设溶液为yn=0anxn=a0+a1x+a2x2+a3x3+...

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  Then $\begin{align*}y^\prime =\sum\limits^\infty_{n=1} na_nx^{n-1}=a_1+2a_2x+3a_3x^2+\ldots\end{align*}$ .
  ::然后您= 1nanxn - 1 = a1+2a2x+3a3x2+...

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  Substituting these in the differential equation
  ::以差异方程替代这些公式

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  $$\begin{align*}y^\prime -y &= 0\\ (a_1+2a_2x+3a_3x^2+\ldots) - (a_0+a_1x+a_2x^2+a_3x^3+\ldots) &= 0\\ (a_1-a_0)+(2a_2-a_1)x+(3a_3-a_2) x^2+\ldots &= 0\end{align*}$$
  ::yy=0(a1+2a2x+3a3x2+...)-(a0+a1x+a2x2+a3x3+...)=0(a1-a0)+(2a2-a1)x+(3a3-a2)xx2+...=0

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  Equating the coefficient of each power of $\begin{align*}x\end{align*}$  to 0 gives:
  ::将x至0的每个功率系数均分给:

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  $$\begin{align*}(a_1-a_0) = 0 & \Rightarrow a_1=a_0\\ (2a_2-a_1)=0 & \Rightarrow a_2 =\frac{a_1}{2}=\frac{a_0}{2}\\ (3a_3-a_2)=0 & \Rightarrow a_3=\frac{a_2}{3}=\frac{a_0}{6}\\ \vdots \qquad \qquad & \qquad \qquad \vdots\end{align*}$$
  :a1-a0)=0a1=a0(2a2-a1)=0a2=a12=a02(3a3-a2)=0a3=a23=a06*

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  Using the coefficients in the power series for $\begin{align*}y\end{align*}$  yields:
  ::使用功率序列中的系数 y 产量 :

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  $$\begin{align*}y &= a_0+a_1x+a_2x^2+a_3x^3+\ldots\\ &= a_0+a_0x+\frac{a_0}{2}x^2+\frac{a_0}{6}x^3+\ldots\\ &= a_0(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots)\\ &= a_0 e^x\end{align*}$$
  ::y=a0+a1x+a2x2+a3x3+...=a0+a0x+a02x2+a06x3+...=a0+1+x+x22!+x33!+...=a0ex

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  The solution to the differential equation $\begin{align*}y^\prime - y=0\end{align*}$  is $\begin{align*}y=a_0 e^x\end{align*}$ .
  ::yy=0的公式是y=a0ex。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked how to use a power series to evaluate the integral $\begin{align*}\int\limits^1_0 x^2 e^{-x^2} dx\end{align*}$  with accuracy to 4 decimal places.
  ::早些时候,您被问及如何使用一个电源序列来评估 =0x2e-x2dx 的积分,精确到小数点后4位。

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  Using the power (Maclaurin) series representation for $\begin{align*}e^{-x}\end{align*}$ , the integral can be written as:
  ::e-x的电源(Maclaurin)序列表示法,集成可以写为:

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  $$\begin{align*}\int\limits^1_0x^2 e^{-x^2}dx &= \int\limits^1_0 x^2 \left[1+(-x)^2+\frac{(-x^2)^2}{2!}+\frac{(-x^2)^3}{3!}+\ldots\right] dx=\int\limits^1_0 \left[x^2-x^4+\frac{x^6}{2!}-\frac{x^8}{3!}+\ldots \right]dx\\ &= \left[\frac{x^3}{3}-\frac{x^5}{5}+\frac{x^7}{7 \cdot 2!}-\frac{x^9}{9 \cdot 3!}+\ldots \right]^1_0=\left[\sum\limits^\infty_{n=0} \frac{(-1)^n x^{2n+3}}{(2n+3)n!}\right]^1_0=\sum\limits^\infty_{n=0} \frac{(-1)^n}{(2n+3)n!}.\end{align*}$$
  ::0x2e- x2dx=10x2[1+(-x)2+(-x2)22!+(-x2)33!+...]dx=10[x2-x4+x62!-x83!+...]dx=[x33-x55+x772]!-x993]!+...]10=[n=0(-1)x2n+3(2n+3n!]]10\n=0(-1)n(2n+3n!n!]

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  Because this is an , $\begin{align*}\Big| \frac{(-1)^n}{(2n+3)n!} \Big| \le 5 \times 10^{-5}\end{align*}$  for 4 decimal place accuracy occurs at $\begin{align*}n=6\end{align*}$ .
  ::因为这是 {( 1) n( 2n+3) n!}\\ 5x10- 5 = 4 位小数精确度为 n= 6 。

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  Therefore, $\begin{align*}\int\limits^1_0 x^2 e^{-x^2} dx \approx \sum\limits^5_{n=0} \frac{(-1)^n}{(2n+3)n!}=0.1894\end{align*}$ .
  ::因此, 10x2e-x2dx=5n=0(- 1)n( 2n+3n! =0. 1894) 。

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  Example 2
  ::例2

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  Find a power series representation of   $\begin{align*}\frac{1}{(1-x)^m}\end{align*}$  where  $\begin{align*}m\end{align*}$ is a positive integer.
  ::在 m 是正整数的地方, 找到一个电源序列代表 1( 1) - x( m) 。

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  In $\begin{align*}(1+x)^r\end{align*}$ , replace $\begin{align*}x\end{align*}$  by $\begin{align*}-x\end{align*}$ , and let $\begin{align*}r=-m\end{align*}$ . The Binomial coefficients for $\begin{align*}r=-m\end{align*}$  are given by:
  ::在 (1+x)r 中,将x 替换为-x,并让 rm 。

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  $$\begin{align*}\binom{-m}{k} &= \frac{(-m)(-m-1)(-m-2) \ldots (-m-k+1)}{k!}\\ &= \frac{(-1)^k m(m+1)(m+2)\ldots (m+k-1)}{k!}\\ &= (-1)^k \binom{m+k-1}{k}\end{align*}$$
  :-mk) = (-m) = (-m)(-m-1)(-m-2) 。 (-m-k+1-1) ! (-m) = (-1) 。 (-m) = (-m+k+1)(m+1)(m+2) 。 (m+k-1) ! (m+k) = (-1) k(m+k-1k)

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  Therefore,
  ::因此,

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  $\begin{align*}(1+x)^r=\sum\limits^\infty_{k=0} (-1)^k \binom{m+k-1}{k}x^k\end{align*}$ .
  :1+x)rk=0(-1)k(m+k-1k)xk。

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  Example 3
  ::例3

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  Find a power representation of  $\begin{align*}\int \frac{\sin x^2}{x}dx\end{align*}$ .
  ::查找 sinx2xdx 的功率表示法 。

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  Direct substitution of $\begin{align*}x^2\end{align*}$  in the Maclaurin Series of $\begin{align*}\sin x\end{align*}$  gives
  ::直接替换 Maclaurin 罪过系列中的x2

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  $$\begin{align*}\sin x^2 &= \sum\limits^\infty_{n=0} \frac{(-1)^n}{(2n+1)!} (x^2)^{2n+1}\\ &= \sum\limits^\infty_{n=0} \frac{(-1)^n}{(2n+1)!}x^{4n+2}\\ \frac{\sin x^2}{x} &= \sum\limits^\infty_{n=0} \frac{(-1)^n}{(2n+1)!}x^{4n+1}\end{align*}$$
  :x2)2n+1)! (x2)2n+1n+1n=0(- 1)n( 2n+1)!x4n+2sinx2xxx_n=0(- 1)n( 2n+1)!x4n+1

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  Therefore,
  ::因此,

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  $$\begin{align*}\int \frac{\sin x^2}{x}dx &= \sum\limits^\infty_{n=0} \frac{(-1)^n}{(2n+1)!} \int x^{4n+1}dx\\ &= \sum\limits^\infty_{n=0} \frac{(-1)^n}{(2n+1)!} \frac{x^{4n+2}}{4n+2}\\ &= \sum\limits^\infty_{n=0} \frac{(-1)^n}{(4n+2)} \frac{x^{4n+2}}{(2n+1)!}\end{align*}$$
  ::@sinx2xxn=0(- 1)n( 2n+1)! @x4n+1dxn=0( 1)n( 2n+1)!x4n+24n+2n=0( 1)n( 4n+2)x4n+2( 2n+1)!

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  Review
  ::回顾

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  For #1-4, find a power series representation the function.
  ::#1-4, 找到一个电源序列代表函数 。

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  1. $\begin{align*}(9-x)^{-\frac{1}{2}}\end{align*}$ at $\begin{align*}x=0\end{align*}$ .
     :9-x)-12 at x=0.
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  2. $\begin{align*}\frac{1}{\sqrt{1+x}}\end{align*}$ at $\begin{align*}x=0\end{align*}$ .
     ::11+xxxx=0。
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  3. $\begin{align*}\frac{1}{(2-x)^2}\end{align*}$ at $\begin{align*}x=0\end{align*}$ .
     ::1(2-x)2 时x=0。
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  4. $\begin{align*}\left(1-\frac{1}{x}\right)^{\frac{1}{2}}\end{align*}$ at $\begin{align*}x=0\end{align*}$ .
     :1-1x)12x=0时。
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  5. $\begin{align*}\sqrt{(1+x+x^2)}\end{align*}$ at $\begin{align*}x=-\frac{1}{2}\end{align*}$ . In what interval is the equality true? Hint: Notice that $\begin{align*}1+x+x^2=\frac{3}{4}+\left(x+\frac{1}{2}\right)^2\end{align*}$ .
     ::{( 1+x+xx2) at x\\\\\12。 平等在什么时间段是真实的? 提示: 注意 1+x+x2=34+( x+12) 2。

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  For #6-8, evaluate the integral using the power series representation (Maclaurin Series) of the integrand.
  ::对于#6-8,使用原格朗的电力序列(Maclaurin Series)表示法评估整体。

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  6. $\begin{align*}\int e^{x^2} dx\end{align*}$ , and approximate  $\begin{align*}\int\limits^1_0 e^{x^2} dx\end{align*}$ to 6 decimal places.
     ::ex2dx, 和大约 10ex2dx 到小数点后 6 位数 。
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  7. $\begin{align*}\int \sin x^2 dx\end{align*}$ .
     ::辛克斯2dx。
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  8. $\begin{align*}\int \frac{\arctan x}{x}dx\end{align*}$
     ::

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  For #9-13, find the power series solution to each of the differential equations:
  ::在9-13中, 找到每个差异方程式的电源序列解决方案 :

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  9. $\begin{align*}y^\prime - y=0\end{align*}$ .
     ::y'y=0。 y"y"y"y"y"y"y"y"y"y"y"y"y"yy"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"y"s"。
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  10. $\begin{align*}y^\prime + 3y=0\end{align*}$
      ::y3y=0 y3y=0
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  11. $\begin{align*}y^\prime = 2xy\end{align*}$
      ::y2xy y2xy
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  12. $\begin{align*}y^\prime +3xy=0\end{align*}$
      ::y3xy=0 y3xy=0
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  13. $\begin{align*}y^{\prime\prime}+y=0\end{align*}$ Hint: There will be two power series coefficients that are arbitrary for the general solution.
      ::yy=0 提示 : 将有两个权力序列系数, 对于一般解决方案是任意的 。

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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 。