to find coefficient of $x^{7}$ in Taylor series expansion of arcsin(x)

Question

Question is to find coefficient of $x^{7}$ in Taylor series expansion of arcsin(x) around x=0. It will be very lengthy to go on computing seven derivatives. Am I missing some trick here. What I did is I compute first derivative which is $ 1/\sqrt{1-x^2}$. Then second derivative is $2x/(1-x^{2})$. While going for third derivative I applied product rule and consider only first term which is $2/(1-x^{2})$ as we are interested to find derivative at 0. Doing this way my only required term at seventh derivative came as $4.2.2/(1-x^{2})^{3}$. But my final answer is not matching with answer key. Any help. Thanks.

Answer 1

$\arcsin''(x)=\frac{x}{(1-x^2)^{3/2}}$ and not what you've written.

The Taylor series of $\frac1{\sqrt{1-x^2}}$ is easy to compute because it is $$(1-x^2)^{-1/2}=\sum_{k=0}^\infty \binom{-1/2}{k} (-x^2)^k=\sum_{k=0}^\infty(-1)^k\binom{-1/2}k x^{2k}$$

Therefore $\arcsin x=\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\binom{-1/2}k x^{2k+1}$.

Answer 2

$$[x^7]\arcsin(x)=\frac{1}{7}[x^6]\frac{1}{\sqrt{1-x^2}}=\frac{1}{7}[x^3]\frac{1}{\sqrt{1-x}}=\frac{\binom{6}{3}}{7\cdot 4^3}=\frac{5}{112} .$$

Answer 3

As

$$y'(x)=\frac1{\sqrt{1-x^2}}$$

you can consider the development of $$(1-t)^{-1/2}=1+\frac12t+\frac12\frac32\frac{t^2}2+\frac12\frac32\frac52\frac{t^3}{2\cdot3}+\cdots$$ and substitute $x^2$ for $t$.

Then by integration, the coefficient of $x^7$ is

$$\frac{1\cdot3\cdot5}{2\cdot2\cdot2}\frac{1}{2\cdot3}\frac1{7}.$$

More generally,

$$\frac{(2k-1)!!}{2^k}\frac1{k!}\frac1{2k+1}x^{2k+1}.$$