Review Note

设\(y\left( x \right) =\mathrm{arc}\tan x,\text{求}y^{\left( n \right)}\left( 0 \right) \)

泰勒展开求高阶导
抽象展开

\(f\left( x \right) =f\left( x_0 \right) +f\prime\left( x_0 \right) \left( x-x_0 \right) +\frac{f''\left( x_0 \right)}{2!}\left( x-x_0 \right) ^2+\cdots +\frac{f^{\left( n \right)}\left( x_0 \right)}{n!}\left( x-x_0 \right) ^n\)
具体展开

\(\mathrm{arc}\tan x=x-\frac{1}{3}x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+\cdots +\frac{\left( -1 \right) ^k}{2k+1}x^{2k+1}+o\left( x^{2k+1} \right) \)
\(//\text{奇函数泰勒展开只有奇数次项}\)
利用唯一性

\(n=2k,\frac{f^{\left( n \right)}\left( 0 \right)}{n!}=0\Rightarrow f^{\left( n \right)}\left( 0 \right) =0\)
\(n=2k+1,\frac{f^{\left( n \right)}\left( 0 \right)}{n!}=\frac{\left( -1 \right) ^k}{2k+1}\Rightarrow f^{\left( n \right)}\left( 0 \right) =\frac{\left( -1 \right) ^k}{2k+1}\cdot \left( 2k+1 \right) !=\left( -1 \right) ^k\left( 2k \right) !\)
\(k=\frac{n-1}{2},f^{\left( n \right)}\left( 0 \right) =\left( -1 \right) ^k\left( 2k \right) !=\left( -1 \right) ^{\frac{n-1}{2}}\left( n-1 \right) !\)
综上\(y^{(n)}(0)=\begin{cases}
    0,&        n\,\,\text{为偶数}\\
    (-1)^{\frac{n-1}{2}}(n-1)!,&        n\,\,\text{为奇数}\\
\end{cases}\)
这个还有一种思路就是求导以后再用莱布尼兹公式，也是大部分习题集给出的答案."