求n阶导数

hbghlyj

证明$$\frac{d^{n}}{dx^n}(\frac{x^n}{1+x^2})=n!\sin(y)\left[\sin(y)-\binom{n}{1}\cos(y)\sin(2y)+\binom{n}{2}\cos^2(y)\sin(3y)-\dots\right], $$
其中$x=\cot(y)$

hbghlyj

AOPS
$E={{\left( {{x}^{n}}\frac{1}{1+{{x}^{2}}} \right)}^{(n)}}=\frac{i}{2}{{\left[ {{x}^{n}}\left( \frac{1}{x+i}-\frac{1}{x-i} \right) \right]}^{(n)}}=\frac{i}{2}\sum\limits_{k=0}^{n}{\left( \begin{matrix} n \\ k \\ \end{matrix} \right){{\left( {{x}^{n}} \right)}^{(n-k)}}}{{\left( \frac{1}{x+i}-\frac{1}{x-i} \right)}^{(k)}}$
So $E=\frac{i}{2}\sum\limits_{k=0}^{n}{k{{(-1)}^{k}}\left( \begin{matrix} n \\ k \\ \end{matrix} \right){{\left( {{x}^{n}} \right)}^{(n-k)}}}\left( \frac{1}{{{\left( x+i \right)}^{k}}}-\frac{1}{{{\left( x-i \right)}^{k}}} \right)$
But using the polar coordinates and the Moivre formula, we can write: ${{\left( x\pm i \right)}^{k}}=\frac{\cos ky\pm i\sin ky}{{{\sin }^{k}}y}\Rightarrow \frac{1}{{{(x+i)}^{k}}}-\frac{1}{{{(x-i)}^{k}}}=-2i{{\sin }^{k}}y\cdot \sin (ky)$ and just replace this in the expression of the $E$. Also you must use ${{\left( {{x}^{n}} \right)}^{(n-k)}}=\frac{n!}{k!}{{x}^{k}}=\frac{n!}{k!}\frac{{{\cos }^{k}}y}{{{\sin }^{k}}y}$.