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Mathematics


将下列函数在指定点处展开成幂级数

Posted by haifeng on 2015-08-24 22:56:47 last update 2015-08-24 22:56:47 | Edit | Answers (2)

(1) $f(x)=\frac{3}{x^2+x-2}$ 在 $x=2$ 处.

(2) 将函数 $f(x)=\arctan\frac{1-2x}{1+2x}$ 展开成为关于 $x$ 的幂级数, 并求 $\sum\limits_{n=0}^{\infty}\frac{(-1)^n}{2n+1}$ 的和.

1

Posted by haifeng on 2015-08-27 10:02:25

\[
\begin{split}
f(x)&=\frac{3}{(x+2)(x-1)}=\frac{1}{x-1}-\frac{1}{x+2}\\
&=\frac{1}{1+(x-2)}-\frac{1}{4}\cdot\frac{1}{1+\frac{x-2}{4}}\\
&=\sum_{n=0}^{\infty}(-1)^n (x-2)^n-\frac{1}{4}\cdot\sum_{n=0}^{\infty}(-1)^n (\frac{x-2}{4})^n\\
&=\sum_{n=0}^{\infty}(-1)^n \Bigl[1-\frac{1}{4^{n+1}}\Bigr](x-2)^n .
\end{split}
\]

2

Posted by haifeng on 2015-08-27 10:26:12

\[
f'(x)=\frac{1}{1+(\frac{1-2x}{1+2x})^2}\cdot\frac{-2(1+2x)-(1-2x)\cdot 2}{(1+2x)^2}=\frac{-2-4x-2+4x}{(1+2x)^2+(1-2x)^2}=\frac{-2}{1+4x^2}.
\]

$f(0)=\arctan\frac{1-0}{1+0}=\frac{\pi}{4}$.

\[
f'(x)=\frac{-2}{1+4x^2}=(-2)\cdot\sum_{n=0}^{\infty}(-1)^n(4x^2)^n=(-2)\cdot\sum_{n=0}^{\infty}(-1)^n 4^n x^{2n}.
\]

因此

\[
\begin{split}
f(x)-f(0)&=\int_0^x f'(t)dt\\
&=\int_0^x(-2)\cdot\sum_{n=0}^{\infty}(-1)^n 4^n t^{2n}dt\\
&=(-2)\cdot\sum_{n=0}^{\infty}\int_0^x (-1)^n 4^n t^{2n}dt\\
&=(-2)\cdot\sum_{n=0}^{\infty}(-1)^n 4^n \frac{x^{2n+1}}{2n+1}.
\end{split}
\]

因此, $f(x)=\frac{\pi}{4}-2\cdot\sum\limits_{n=0}^{\infty}(-1)^n 4^n \frac{1}{2n+1}x^{2n+1}$. 收敛区间是 $(-\frac{1}{2},\frac{1}{2})$. 事实上,

\[
\lim_{n\rightarrow\infty}\Bigl|\frac{u_{n+1}}{u_n}\Bigr|=\lim_{n\rightarrow\infty}\Bigl|\frac{(-1)^{n+1}4^{n+1}x^{2n+2}}{(-1)^n4^nx^{2n}}\Bigr|=4x^2.
\]

当 $x=\pm\frac{1}{2}$ 时, 级数也收敛. 因此收敛域是 $[-\frac{1}{2},\frac{1}{2}]$.

特别的, 当 $x=\frac{1}{2}$ 时, 

\[
f(\frac{1}{2})=\frac{\pi}{4}-2\cdot\sum\limits_{n=0}^{\infty}(-1)^n 4^n \frac{1}{2n+1}(\frac{1}{2})^{2n+1}=\frac{\pi}{4}-\sum_{n=0}^{\infty}(-1)^n\frac{1}{2n+1}.
\]

\[
\sum_{n=0}^{\infty}(-1)^n\frac{1}{2n+1}=\frac{\pi}{4}-f(\frac{1}{2})=\frac{\pi}{4}-\arctan\frac{1-2\cdot\frac{1}{2}}{1+2\cdot\frac{1}{2}}=\frac{\pi}{4}-0=\frac{\pi}{4}.
\]


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