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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.

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  • $\begingroup$ 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 $f(x) =\sin^{-1}x,g(x)=(1-x^2)^{-1/2}$. $\endgroup$
    – Paramanand Singh
    May 6, 2020 at 16:03
  • $\begingroup$ @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. $\endgroup$
    – s1mple
    Jun 20, 2020 at 12:53
  • $\begingroup$ This guy Taylor tells us that coefficient of $x^n$ in $f(x) $ is given by $f^{(n)} (0)/n!$ $\endgroup$
    – Paramanand Singh
    Jun 20, 2020 at 13:19
  • $\begingroup$ 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 $x^6$. $\endgroup$
    – s1mple
    Jun 20, 2020 at 13:21
  • $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    Jun 20, 2020 at 13:23

1 Answer 1

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HINT: $$\frac{d \mathrm{sin}^{-1}(x)}{dx} = \frac{1}{\sqrt{1-x^2}}$$

$$\frac{dx^7}{dx} = 7x^6$$

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    $\begingroup$ The second equation should be $7x^6$. $\endgroup$
    – Paramanand Singh
    May 6, 2020 at 15:59
  • $\begingroup$ @ParamanandSingh Thanks for the correction. $\endgroup$
    – SagarM
    May 6, 2020 at 21:09

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