# What is the coefficient of $x^7$ in the Taylor series expansion, around $x=0$ of the function $f(x)=\sin^{-1}(x)$?

Question: What is the coefficient of $$x^7$$ in the Taylor series expansion, around $$x=0$$ of the function $$f(x)=\sin^{-1}(x)$$?

I know how to solve using the Taylor expansion, but in that case I will have to take derivative so many times and the calculation becomes cumbersome. I want to see the reasoning behind the solution given below:

Solution: $$\;\;$$ $$\frac{f^7(0)}{7!}= \;\text{coefficient of }x^7\;\;\text{ in power series expansion of } \sin^{-1}(x)$$ $$=\frac{1}{7}\cdot(\;\text{coefficient of }x^6\;\text{in power series expansion of }(1-x^2)^{-\frac{1}{2}})\;\;\cdots(*)$$ $$=\frac{1}{7}\cdot\frac{\frac{1}{2}(\frac{1}{2}+1)(\frac{1}{2}+2)}{3!}$$

Can someone please explain me how the step labelled $$(*)$$ follows? Thank you for the help.

• Just note that if $g(x) =f'(x)$ then $g^{(6)}(x)=f^{(7)}(x)$ and hence $f^{(7)}(0)/7!=(1/7)(g^{(6)}(0)/6!)$. Now use $f(x) =\sin^{-1}x,g(x)=(1-x^2)^{-1/2}$. May 6, 2020 at 16:03
• @ParamanandSingh I completely understand your hint/solution. But my doubt is different. Here it says $\frac{1}{7}\cdot(\;\text{coefficient of }x^6\;\text{in power series expansion of }(1-x^2)^{-\frac{1}{2}})$ and what you have written in the above comment, are different right? Finding coeff of $x^6$ in power series expansion and finding $g^{(6)}(0)$ are different and the latter one is more difficult to compute. In the solution we have just calculated the coeff of $x^6$ using binomial series.I am sorry for the late reply. Jun 20, 2020 at 12:53
• This guy Taylor tells us that coefficient of $x^n$ in $f(x)$ is given by $f^{(n)} (0)/n!$ Jun 20, 2020 at 13:19
• Yes that is perfectly alright. But finding coeff of $x^6$ using Taylor series and using binomial expansion are different isn't it? Here in the solution, the solution uses the binomial expansion for finding coeff of $x^6$. Jun 20, 2020 at 13:21
• Find coefficients by any means, the relation between coefficients of $f$ and $f'$ is as given in my comment and is a consequence of Taylor theorem. Jun 20, 2020 at 13:23

HINT: $$\frac{d \mathrm{sin}^{-1}(x)}{dx} = \frac{1}{\sqrt{1-x^2}}$$
$$\frac{dx^7}{dx} = 7x^6$$
• The second equation should be $7x^6$. May 6, 2020 at 15:59