User:Egm6341.s11.team4.yang/hw1

=HW 1.2 problem=

Find $${{p}_{n}}(x)$$ and $${{R}_{n+1}}(x)$$ of

1).$${{e}^{x}}$$

2).$${{f}_{2}}\left( x \right)=\frac{{{e}^{x}}-1}{x}$$

at point $${{x}_{0}}=0$$.

=Answer:=

According to the Taylor series from the lecture [[media:nm1.s11.mtg3.djvu|Mtg 3 (d)]], the Taylor polynomials is as following:

$${{p}_{n}}(x)=f({{x}_{0}})+\frac{(x-{{x}_{0}})}{1!}{{f}^{(1)}}({{x}_{0}})+\frac{2!}{{f}^{(2)}}({{x}_{0}})....+\frac{n!}{{f}^{(n)}}({{x}_{0}})$$

$${{R}_{n+1}}(x)=\frac{1}{n!}\int\limits_^{x}{{{(x-t)}^{n}}{{f}^{(n+1)}}(t)}dt=\frac{(n+1)!}{{f}^{(n+1)}}(\xi )\begin{matrix} , & \xi \in \left[ {{x}_{0}},x \right] \\ \end{matrix}$$

1). For $$f(x)={{e}^{x}}$$, take the i-th order derivative:

$$ {{f}^{(1)}}(x)={{f}^{(2)}}(x)=...={{f}^{(n)}}(x)={{e}^{x}}$$.

At point $${{x}_{0}}=0$$,

$$ {{f}^{(1)}}(0)={{f}^{(2)}}(0)=...={{f}^{(n)}}(0)=1$$.

Thus, the n-th order polynomial and residual is:
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$$\displaystyle {{p}_{n}}(x)=1+\frac{x}{1!}+\frac{2!}+...+\frac{n!} $$ $$
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$$\displaystyle {{R}_{n+1}}(x)=\frac{1}{n!}\int\limits_{0}^{x}dt=\frac{(n+1)!}{{e}^{\xi }}\begin{matrix} , & \xi \in \left[ 0,x \right] \\ \end{matrix} $$ $$
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2). For $${{f}_{2}}\left( x \right)=\frac{{{e}^{x}}-1}{x}$$, based on the result of 1): $${{f}_{1}}(x)={{p}_{n}}(x)+{{R}_{n+1}}(x)$$ , we obtain:

$$\begin{align} & {{f}_{2}}^ – (x)=\frac{{{f}_{1}}^ – (x)-1}{x} \\ & \begin{matrix} {} & \begin{matrix} {} & =\frac{{{p}_{n}}(x)+{{R}_{n+1}}(x)-1}{x} \\ \end{matrix} \\ \end{matrix} \\ & \begin{matrix} {} & \begin{matrix} {} & = \\ \end{matrix}  \\ \end{matrix}\frac{1+\frac{x}{1!}+\frac{2!}+...+\frac{n!}+\frac{(n+1)!}{{e}^{\xi }}-1}{x} \\ & \begin{matrix} {} & \begin{matrix} {} & = \\ \end{matrix}  \\ \end{matrix}\frac{1}{1!}+\frac{x}{2!}+...+\frac{n!}+\frac{(n+1)!}{{e}^{\xi }} \\ \end{align}$$

Thus, At point $${{x}_{0}}=0$$ ,the n-th order polynomial and residual for $${{f}_{2}}\left( x \right)=\frac{{{e}^{x}}-1}{x}$$ is:


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$$\displaystyle {{p}^{'}}_{n}(x)=\frac{1}{1!}+\frac{x}{2!}+...+\frac{n!} $$ $$
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$$\displaystyle {{R}^{'}}_{n+1}(x)=\frac{(n+1)!}{{e}^{\xi }}\begin{matrix} , & \xi \in \left[ 0,x \right] \\ \end{matrix} $$ $$
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