9.17 泰勒系列计算:选择中心

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  泰勒系列计算:选择中心

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  Two researchers discussing a plot of some data from an experiment, realized that the data of interest followed the function $\begin{align*}f(x)=\frac{1}{x}\end{align*}$  in the vicinity of $\begin{align*}x=200\end{align*}$ , but that further analysis would be easier if they used a 2 ^nd degree power series to represent the data. They agreed that the function did not have a Maclaurin series representation, but could be represented by a 2 ^nd degree Taylor polynomial with center at 200. When they looked at the resulting Taylor polynomial, they realized that it was equivalent to the Maclaurin series of a different, but related function. Do you know why $\begin{align*}f(x)=\frac{1}{x}\end{align*}$  cannot be represented by a Maclaurin series? What is the 2 ^nd degree Taylor polynomial with center at 200? What could be the related function that has an equivalent Maclaurin series?
  ::两位研究人员讨论从实验中得出的一些数据图,意识到引人注意的数据跟随了x=200附近的函数 f(x)=1x,但如果他们使用二级功率序列来代表数据,则进一步的分析会更容易些。他们一致认为,该函数没有Maclaurin系列的表示,但可以用第二级泰勒多式和中点的200来代表。当他们查看由此得出的泰勒多式多式手式时,他们意识到它相当于一个不同但相关功能的Maclaurin系列。你知道为什么F(x)=1x不能由Maclaurin系列来代表吗?第二级泰勒多式系列与中点200来代表什么?什么相关功能可以具有等效的Maclaurin系列?

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  Choosing the Center of a Taylor Series
  ::选择泰勒系列中的中心

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  Recall the definition of a Taylor series and a Maclaurin series:
  ::回顾泰勒系列和麦克劳林系列的定义:

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  For a function $\begin{align*}f(x)\end{align*}$  that is differentiable at $\begin{align*}x=x_0\end{align*}$ , the Taylor series representation of $\begin{align*}f(x)\end{align*}$  centered at $\begin{align*}x=x_0\end{align*}$  is the power series:
  ::对于在 x=x0 可区分的 f(x) 函数, 以 x=x0 为中心的 f(x) 泰勒序列表示法是功率序列 :

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

  $$\begin{align*}T(x) & =\sum\limits_{n=0}^\infty\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\\ & = f(x_0)+f^\prime(x_0)(x-x_0)+\frac{f^{\prime\prime}(x_0)}{2!}(x-x_0)^2+\frac{f^{\prime\prime\prime}(x_0)}{3!}(x-x_0)^3+\ldots\end{align*}$$
  ::T(x)\n=0fx0(x)n! (x-x0)n=f(x0)+f(x0)+f(x0)(x-x0)+f*(x0)2!(x-x0)2+f*(x0)2+f(x0)3!(x-x0)3!(x-x0)3+...

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  The Maclaurin series representation , of $\begin{align*}f(x)\end{align*}$  is the Taylor series centered at $\begin{align*}x_0=0\end{align*}$ :
  ::Maclaurin系列代表,f(x)是泰勒系列,以x0=0为中心:

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

  $$\begin{align*}M(x)& =\sum\limits_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n\\ & = f(0)+f^\prime(0)x+\frac{f^{\prime\prime}(0)}{2!}x^2+\frac{f^{\prime\prime\prime}(0)}{3!}x^3+\ldots\end{align*}$$
  ::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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  The Maclaurin series for some commonly used functions is shown again below:
  ::以下再次列出一些常用功能的Maclaurin系列:

  Maclaurin Power Series	播放段落 Interval of Convergence ::相互交汇
  $\begin{align*}\sin x=\sum\limits_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\end{align*}$	$\begin{align*}(-\infty,\infty)\end{align*}$
  $\begin{align*}\cos x=\sum\limits_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\end{align*}$	$\begin{align*}(-\infty,\infty)\end{align*}$
  $\begin{align*}e^x=\sum\limits_{n=0}^\infty\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\end{align*}$	$\begin{align*}(-\infty,\infty)\end{align*}$
  $\begin{align*}\ln(1+x)=\sum\limits_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\end{align*}$	$\begin{align*}(-1,1]\end{align*}$
  $\begin{align*}\ln(1-x)=\sum\limits_{n=0}^\infty(-1)\frac{x^{n+1}}{n+1}=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+\cdots\end{align*}$	$\begin{align*}[-1,1)\end{align*}$
  $\begin{align*}\arcsin x=\sin^{-1}x=\sum\limits_{n=0}^\infty\frac{(2n)!x^{2n+1}}{2^{2n}(n!)^2(2n+1)}\end{align*}$	$\begin{align*}[-1,1]\end{align*}$
  $\begin{align*}\arctan x=\tan^{-1}x=\sum\limits_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}\end{align*}$	$\begin{align*}(-1,1)\end{align*}$
  $\begin{align*}\sinh x=\sum\limits_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}\end{align*}$	$\begin{align*}(-\infty,\infty)\end{align*}$
  $\begin{align*}\cosh x=\sum\limits_{n=0}^\infty\frac{x^{2n}}{(2n)!}\end{align*}$	$\begin{align*}(-\infty,\infty)\end{align*}$

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  A differentiable function can be represented by Taylor (Maclaurin) series, which has an infinite number of terms. Truncating the Taylor series to generate an $\begin{align*}n\end{align*}$ th degree polynomial allows us to represent a function with a finite number of terms that achieves some specified accuracy requirement.
  ::一个不同的功能可以用泰勒(Maclaurin)系列来表示,该系列有无限的术语。 将泰勒系列划线以产生 nnth ylumial , 使我们可以代表一个能达到某些特定精确要求的有限数量术语的函数。

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  In choosing a center, $\begin{align*}x_0\end{align*}$ , of a Taylor series (polynomial) that represents a function, the following considerations apply:
  ::在选择代表功能的泰勒系列(Polynomial)的x0中心时,适用以下考虑:

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  1. Choose an  $\begin{align*}x_0\end{align*}$ such that the function is defined and differentiable at $\begin{align*}x_0\end{align*}$ .
     ::选择 x0 , 以便函数在 x0 时被定义且可区分 。
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  2. Choose an  $\begin{align*}x_0\end{align*}$ that allows all values of  $\begin{align*}x\end{align*}$ that will be evaluated to lie within the interval of convergence of the series, i.e. must $\begin{align*}|x-x_0|< R_c\end{align*}$ , or $\begin{align*}-R_c+x_0< x< R_c+x_0\end{align*}$ .
     ::选择一个 x0 , 允许被评估的 x 的所有值都位于序列的趋同间隔内, 即 {x_ x0\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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  3. Choose an  $\begin{align*}x_0\end{align*}$ such that accurate values of  $\begin{align*}f(x_0),f^{\prime}(x_0),f^{\prime\prime}(x_0),\ldots\end{align*}$ can be computed.
     ::选择一个 x0, 这样可以计算 f( x0) 的准确值 。
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  4. Choose an  $\begin{align*}x_0\end{align*}$ such that the number of series terms needed to achieve a specified degree of accuracy holds for all values of $\begin{align*}x\end{align*}$ .
     ::选择 x0 , 使 x 所有值的精确度达到一定的精确度所需的序列条件数保持不变 。

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  The first and easiest choice for a series center is $\begin{align*}x_0=0\end{align*}$ , i.e. use a Maclaurin series. This may not always be a valid choice for the function.
  ::序列中心的首选和最简单的选择是 x0=0, 即使用 Maclaurin 序列。 这不一定是函数的有效选择 。

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  Take number  $\begin{align*}\ln 0.99\end{align*}$  and try to approximate it.
  ::取编号0.99 并尝试接近它。

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  The problem requires evaluation of the natural logarithm $\begin{align*}\ln x\end{align*}$ . However, a Maclaurin series representation of the function does not exist for a center at $\begin{align*}x_0=0\end{align*}$  because  $\begin{align*}f(x)=\ln x\end{align*}$ is undefined there, and has undefined .
  ::这个问题需要评估自然对数 Inx 。 但是, 在 x0=0 时, 函数的 Maclaurin 序列表示并不存在, 因为 f( x) = lnx 在那里没有定义, 而且没有定义 。

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  Alternative approaches are to find a Taylor series representation of $\begin{align*}f(x)=\ln x\end{align*}$  at a differentiable center; or a Maclaurin series for a different, but related function. Since $\begin{align*}0.99\end{align*}$  equals $\begin{align*}[1-(+0.01)]\end{align*}$ or $\begin{align*}[1+(-0.01)]\end{align*}$ , the following are options:
  ::替代办法是在不同的中心找到泰勒系列表示 f(x) = lnx;或Maclaurin系列表示不同但相关的函数。0.99等于[1-(+0.01)]或[1+(- 0.01)],以下是备选办法:

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  Taylor series for  $\begin{align*}f(x)=\ln x\end{align*}$ with $\begin{align*}x_0=1\end{align*}$ :  %7D(1)"> $\begin{align*}f(1),f^\prime(1),\ldots,f^{(n)}(1)\end{align*}$ are defined.
  ::f(x) 的泰勒序列= xxxxxxxxxx0=1: f(1), f-(1),...f-n) (1) 定义。

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  The series looks like:
  ::这个系列看起来像:

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  $\begin{align*}f(x)=\ln x=0+(x-1)-\frac{(x-1)^2}{2!}+2\frac{(x-1)^3}{3!}-6\frac{(x-1)^4}{4!}\cdots=\sum\limits_{n=0}^\infty(-1)^n\frac{(x-1)^{n+1}}{n+1}\end{align*}$ with interval of convergence $\begin{align*}(0,2]\end{align*}$ .
  ::f(x) = lnx=0+0+(x-1)-(x-1)-(x-1)22! +2(x-1) 333!- 6(x-1)- 44!\\ n=0(x-1)n(x-1)n+1n+1, 并有交汇间隔( 0, 2) 。

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  In this case, $\begin{align*}x=0.99\end{align*}$ .
  ::在这种情况下,x=0.99。

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  or
  ::或

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  Maclaurin series for  $\begin{align*}\ln(1-x)\end{align*}$ with $\begin{align*}x=0.1\end{align*}$ , or  $\begin{align*}\ln(1+x)\end{align*}$ with $\begin{align*}x=-0.1\end{align*}$ :  %7D(1)"> $\begin{align*}f(1),f^\prime(1),\ldots,f^{(n)}(1)\end{align*}$ are defined.
  ::IN( 1- x) 的 Maclaurin 序列, 使用 x=0. 1 , 或 IN( 1+x) , 使用 x0. 1 : f(1), f(1),..., f( n) (1) 的定义。

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  The series looks like:
  ::这个系列看起来像:

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  $\begin{align*}\ln(1+x)=\sum\limits_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\end{align*}$ with interval of convergence $\begin{align*}(-1,1]\end{align*}$ .
  ::In( 1+x) n=0 (- 1nxn+1n+1=x- x22+x33- x44) 并有趋同(-1,1) 的间隔 。

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  In this case, $\begin{align*}x=-0.1\end{align*}$ .
  ::在这种情况下, x0.1。

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  Both series look identical when their respective values of $\begin{align*}x\end{align*}$  are evaluated.
  ::在评价各自的 x 值时,这两个序列看起来都是一样的。

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  Using the 2 ^nd degree Maclaurin polynomial for  $\begin{align*}\ln(1+x)\end{align*}$ with $\begin{align*}x=-0.1\end{align*}$ :
  ::使用 2 度的 Maclaurin 多元性, 用于 In( 1+x) 和 x0. 1 :

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  $$\begin{align*}\ln 0.99 & = \ln(1-0.01)\\ & \thickapprox -0.01-\frac{(0.01)^2}{2}\\ & \thickapprox -0.01005 \qquad\qquad \ldots \text{Calculator value:} \ \ln 0.99=-0.010050336\end{align*}$$
  ::In. 99=ln( 1- 0.01) 0.01 - ( 0.01) 22 0.005... 计算器值: IN. 99 0.0100336

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  The above approximation is good to 6 decimal places.
  ::以上近似值为小数点后6位。

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  Given a Taylor series (polynomial) representation of a function, always make sure that evaluation of the series (polynomial) is made for values that are within the interval of convergence.
  ::鉴于泰勒系列(球体)代表一种功能,必须始终确保对系列(球体)的评价是针对在趋同的间隔内的价值。

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  Now, approximate $\begin{align*}\ln 2.1\end{align*}$  using a 2 ^nd degree Maclaurin polynomial.
  ::现在,约IN2.1 使用二度Maclaurin多面体学。

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  In the previous problem , the Maclaurin series (polynomial)  $\begin{align*}\ln(1+x)=\sum\limits_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}\thickapprox x-\frac{x^2}{2}\end{align*}$ is used to evaluate $\begin{align*}\ln 0.99\end{align*}$ . Using the same polynomial yields
  ::在前一个问题中,使用Maclaurin系列(Polynomial) In( 1+x) @n=0( 1)nxn+1n+1n+1x-x22)来评估IN0.99。

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  $$\begin{align*}\ln (2.1) & = \ln (1+1.1)\\ & \thickapprox 1.1- \frac{1.1^2}{2}\\ & \thickapprox 0.495 \qquad\qquad \ldots \text{Calculator value:} \ \ln 2.1=0.741937345 \end{align*}$$
  ::In( 2.1) = ln( 1+1.1)\\\\\\ 1.1- 1.122\\\\\\\\ 0. 495... 计算器值: In2.1= 0. 74193745

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  The above approximation is not very good!
  ::上面的近似性不是很好!

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  Using the 3 ^rd degree polynomial yields: $\begin{align*}\ln 2.1 \thickapprox 0.9387\end{align*}$ .
  ::使用第三度多元产量:IN2.10.9387。

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  Using the 4 ^th degree polynomial yields: $\begin{align*}\ln 2.1 \thickapprox 0.5727\end{align*}$ .
  ::使用四度多元产量:IN2.10.5727。

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  What’s wrong?
  ::有什么问题吗?

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  The problem is that $\begin{align*}x=2.1\end{align*}$  is outside the interval of convergence $\begin{align*}(-1,1]\end{align*}$ , and the Maclaurin series does not converge . This means the series (polynomial) used to represent (approximate) $\begin{align*}\ln(1+x)\end{align*}$  is not valid.
  ::问题在于 x=2.1 是在趋同( -1, 1) 间隔之外, 而Maclaurin 序列并不趋同。 这意味着用于表示( 近似) In( 1+x) 的序列( polyynomial) 无效 。

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  To remedy this problem choose a center for a Taylor series of $\begin{align*}\ln x\end{align*}$ , or a related function for a Maclaurin series, that keeps $\begin{align*}x=2.1\end{align*}$  within the interval of convergence.
  ::为解决这一问题,选择泰勒系列 Inx 的中心,或Maclaurin系列的相关功能,将x=2.1维持在汇合间隔内。

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  Let's fix the approximation from above and approximate $\begin{align*}\ln 2.1\end{align*}$  using a 2 ^nd degree Taylor polynomial.
  ::让我们从上到下修正近似值 大约是IN2.1 使用第二度的泰勒多面体

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  Because $\begin{align*}2.1=2+0.1\end{align*}$ , and the derivatives of $\begin{align*}f(x)=\ln x\end{align*}$  are $\begin{align*}\thicksim \frac{1}{x^n}\end{align*}$ , choose  $\begin{align*}x_0=2\end{align*}$ as a center. Taylor series for $\begin{align*}f(x)=\ln x\end{align*}$  with $\begin{align*}x_0=2\end{align*}$ : %7D(2)"> $\begin{align*}f(2),f^{\prime}(2),\ldots,f^{(n)}(2)\end{align*}$  are defined. The series looks like: $\begin{align*}f(x)=\ln x=\ln 2+\frac{1}{2}(x-2)-\frac{1}{2^2}\frac{(x-2)^2}{2!}+2\frac{1}{2^3}\frac{(x-2)^3}{3!}-6\frac{1}{2^4}\frac{(x-2)^4}{4!}\cdots=\sum\limits_{n=0}^\infty(-1)^n\frac{(x-2)^{n+1}}{2^{n+1}(n+1)}\end{align*}$    
  ::因为 2. 1= 2+0. 1, F( x) = lnx 的衍生物为 1xn, 请选择 x0= 2 作为中心 。 f( x) = xx = lnxxxxxxxxxxxxxxxxxxxxxxx0=x0=2: f(2), f=2,..., f( n) (2) 定义了 f( 2) : f( x) =x=ln2+12( x-2) 122( x-2)22! +2123( x-2) 33! - 6124( x-2)44!\n= 0( 1)n( x- 2)n+12n( n+1) 。

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  with interval of convergence $\begin{align*}(0,4]\end{align*}$ .
  ::与趋同间隔(0.4)。

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  Using the polynomial
  ::使用多面体

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  $$\begin{align*}\ln(2.1) & \thickapprox \ln 2+\frac{1}{2}(x-2)-\frac{1}{2^2}\frac{(x-2)^2}{2!}\\ & \thickapprox 0.69315+\frac{1}{2}(2.1-2)-\frac{1}{2^2}\frac{(2.1-2)^2}{2!}\\ & \thickapprox 0.69315+0.05-0.00125\\ & \thickapprox 0.7419 \qquad\qquad\qquad\qquad \ldots \text{Calculator value:} \ \ln 2.1=0.741937345\end{align*}$$
  ::In( 2. 1) @ ln2+12(x-2)- 122(x-2)22! @ 0. 69315+12(2.1-2)- 122(2.1-2)- 122(2.1-2)-222! @ 0. 69315+0.05-0.005-0.00125_ 0.7419...计算器值: In2.1=0. 7419373445

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  This approximation is considerably better, good to 4 decimal places. Choosing the right center made the difference!
  ::这个近似值好得多,好到小数点后4位。选择正确的中心可以改变!

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked why $\begin{align*}f(x)=\frac{1}{x}\end{align*}$  cannot be represented by a Maclaurin series. You were also asked what the 2 ^nd degree Taylor polynomial approximation of  $\begin{align*}f(x)\end{align*}$ with center at 200 is and what could be the related function that has an equivalent Maclaurin series.
  ::早些时候,有人问你为什么f(x)=1x不能用Maclaurin系列来表示。有人还问你,以200为中点的f(x)为第2级Taylor多元近似值是多少,还有哪些相关函数具有等效的Maclaurin系列。

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  $\begin{align*}f(x)=\frac{1}{x}\end{align*}$  is not defined at $\begin{align*}x=0\end{align*}$ , and neither are any of its derivatives. It therefore, does not have a Maclaurin series (polynomial) representation.
  ::f(x)=1x没有在 x=0 上定义,也没有任何衍生物。 因此,它没有Maclaurin系列(Polynomial)代表。

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  The 2 ^nd degree Taylor polynomial centered at  $\begin{align*}x=200\end{align*}$ is:
  ::以x=200为轴心的泰勒多面体第二度为:

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  $$\begin{align*}T_2(x)=\frac{1}{200}-\frac{1}{200^2}(x-200)+\frac{1}{200^3}(x-200)^2.\end{align*}$$
  ::T2(x) = 1200-12002(x-200) +12003(x-200) 2。

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  In $\begin{align*}T_2(x)\end{align*}$ , the factor $\begin{align*}(x-200)\end{align*}$  is the excursion from 200. The function $\begin{align*}f(x)=\frac{1}{x}(x\ne 0)\end{align*}$ can be written to apply in the vicinity of  $\begin{align*}x=200\end{align*}$ as follows:  $\begin{align*}g(u)=\frac{1}{200+u}\end{align*}$ , where  $\begin{align*}u\end{align*}$ takes the place of $\begin{align*}(x-200)\end{align*}$ . The function  $\begin{align*}g(u)\end{align*}$ and all of its derivates is defined at $\begin{align*}u=0\end{align*}$ , and so has a Maclaurin series (polynomial): $\begin{align*}M_2(u)=\frac{1}{200}-\frac{1}{200^2}u+\frac{1}{200^3}u^2\end{align*}$ .
  ::在 T2( x) 中, 系数( x- 200) 是200 的外移值。 函数 f( x) = 1x( x0) 可以写成在 x=200 附近应用的函数 f( x) = 1x( x0) 如下: g( u) = 1200+u, 其中 u 取代 ( x- 200) 。 函数 g( u) 及其所有衍生物的定义为 u=0, 因此, Maclaurin 系列( 球体) m2( u) = 1200 - 12002u+12003u 2 。

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  Example 2
  ::例2

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  Approximate $\begin{align*}\frac{1}{1.9^2}\end{align*}$  to  $\begin{align*}4\end{align*}$ decimal places using a Taylor polynomial.
  ::大约11.92至4小数点使用泰勒多面体。

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  Note that $\begin{align*}\frac{1}{1.9^2}=\frac{1}{(2-0.1)^2}=\frac{1}{4\left(1-\frac{0.1}{2}\right)^2}\end{align*}$ .
  ::注意11.92=1(2-0.1)2=14(1-0.12)2。

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  This suggests that a series representation of the function  $\begin{align*}\frac{1}{(1-x)^2}\end{align*}$ might help.
  ::这表明,职能1(1-x)2的系列表示可能有所帮助。

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  Recall that $\begin{align*}\frac{1}{(1-x)^2}=\frac{d}{dx}\left[\frac{1}{1-x}\right]\end{align*}$ , and  $\begin{align*}\frac{1}{1-x}=\sum\limits_{n=0}^\infty x^n\end{align*}$ is the Maclaurin series for $\begin{align*}|x|<1\end{align*}$ . The center is at $\begin{align*}x_0=0\end{align*}$ .
  ::回顾 1( 1 - x) 2=ddx[ 11 - x] 和 11 -xn=0xn 是 x1 的Maclaurin序列。 中心在 x0=0 。

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  Through term-by- term differentiation of the series for $\begin{align*}\frac{1}{1-x}\end{align*}$ we have $\begin{align*}\frac{1}{(1-x)^2}=\sum\limits_{n=1}^\infty nx^{n-1}\end{align*}$ .
  ::通过对11-x系列的逐期区分,我们有1-1-x)2n=1nxn-1。

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  We can also write $\begin{align*}\frac{1}{\left(1-\frac{x}{2}\right)^2}=\sum\limits_{n=1}^\infty n\left(\frac{x}{2}\right)^{n-1}\end{align*}$  for $\begin{align*}|x|<2\end{align*}$ .
  ::我们也可以写 1,1 -x2,2,2n=1n(x2,2n-1)

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  Then
  ::然后

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  $$\begin{align*}\frac{1}{1.9^2} & = \frac{1}{4\left(1-\frac{0.1}{2}\right)^2}\\ & = \sum\limits_{n=1}^\infty\frac{n}{4}\left(\frac{0.1}{2}\right)^{n-1}\\ & = \frac{1}{4}[1+2(0.05)+3(0.05)^2+4(0.05)^3+\cdots]\end{align*}$$
  ::11.92=14(1-0.12.2)2n=1n4(0.12)n-1=14[1+2(0.005)+3(0.05)2+4(0.05)]

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  Using the first, second, and third degree polynomials gives the following estimates:
  ::使用第一、第二和第三多度多米亚数提供了以下估计数字:

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  1 ^st degree: $\begin{align*}\frac{1}{1.9^2}\thickapprox \frac{1.1}{4}=0.275\end{align*}$ , with an underestimate of $\begin{align*}0.0020\end{align*}$ ;
  ::第一级:11.921.14=0.275,低估为0.0020;

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  2 ^nd degree: $\begin{align*}\frac{1}{1.9^2}\thickapprox \frac{1.1075}{4}=0.276875\end{align*}$ , with an underestimate of $\begin{align*}0.000133\end{align*}$ ;
  ::第二级:11.921.10754=0.276875,低估为0.00033;

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  3 ^rd degree: $\begin{align*}\frac{1}{1.9^2}\thickapprox \frac{1.108}{4}=0.2770\end{align*}$ , with an underestimate of $\begin{align*}0.0000083\end{align*}$ .
  ::第三级:11.921.1084=0.2770,低估为0.000083。

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  Review
  ::回顾

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  For #1-4, choose a center and evaluate a Taylor (Maclaurin) polynomial approximation that meets the stated conditions:
  ::在 #1-4中, 选择一个中心, 并评估泰勒(Maclaurin)符合规定条件的多元近似值:

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  1. $\begin{align*}\sqrt{4.1}\end{align*}$  with a 1 ^st degree polynomial. Can the same polynomial be used to accurately evaluate $\begin{align*}\sqrt{0.1}\end{align*}$ ?
     ::4.1 具有第1度多元性。能否用同一多元性来准确评估 °0.1 ?
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  2. $\begin{align*}\ln 0.9\end{align*}$  to  $\begin{align*}4\end{align*}$ decimal places.
     ::零点九到四小数点
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  3. $\begin{align*}\sin(0.8)\end{align*}$  to  $\begin{align*}6\end{align*}$ decimal places.
     ::sin(0.8) 至 小数点后 6 位数。
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  4. $\begin{align*}\sin(6)\end{align*}$  to an accuracy of $\begin{align*}0.001\end{align*}$ .
     :六) 精确度为0.001。
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  5. $\begin{align*}\frac{1}{9^{3}}\end{align*}$  to  $\begin{align*}6\end{align*}$ decimal places.
     ::193至6十进制位数。
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  6. Compute  $\begin{align*}\sqrt{10}\end{align*}$ using a Taylor polynomial of degree 1. Give an estimate on the size of the error. Compare the estimate to the calculator value.
     ::使用泰勒1度的多元度计算 10。请对错误的大小作出估计。将估计值与计算值相比较。
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  7. Can a Maclaurin polynomial for  $\begin{align*}\sqrt{x}\end{align*}$ be used to estimate the value of $\begin{align*}\sqrt{0.9}\end{align*}$ ? Justify your answer. How should the value be estimated?
     ::是否可使用 x 的 Maclaurin 多元数值来估计 0. 9 的值? 请说明答案的理由。 如何估计该值 ?
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  8. Estimate  $\begin{align*}\sqrt{99}\end{align*}$ to within $\begin{align*}\pm 0.001\end{align*}$ .
     ::99年估计数在0.001内。
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  9. What happens when the Maclaurin series for  $\begin{align*}\ln(1-x)\end{align*}$ is used to estimate  $\begin{align*}\ln 5\end{align*}$ to 7 decimal places by setting $\begin{align*}x=-4\end{align*}$ ? Why?
     ::当使用内( 1- x) 的 Maclaurin 序列设置 x% 4 来估计 5 至 7 位小数时会怎样? 为什么?
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  10. Calculate  $\begin{align*}e^{-0.2}\end{align*}$ correct to five decimal places.
      ::计算 e-0.2 正确到小数点后五位数。
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  11. Approximate  $\begin{align*}\cos (0.01)\end{align*}$ with an error less than $\begin{align*}10^{-20}\end{align*}$ .
      ::误差小于 10-20 的近似cos( 0.0 1) 。
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  12. To evaluate $\begin{align*}\tan(0.8)\end{align*}$ , predict whether a two-term Maclaurin polynomial or a two-term Taylor polynomial centered at  $\begin{align*}x=\frac{\pi}{4}\end{align*}$ would give the more accurate estimate.
      ::为了评估Tan(0.8),预测双期Maclaurin多年度或双期Taylor多年度(以x4为中心)是否会提供更准确的估计。
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  13. Approximate  $\begin{align*}\sin(1.1)\end{align*}$ to 4 decimal places.
      ::近似罪(1.1) 至小数点后位数 4。

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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 。