Suppose $f$ is a real valued function with $f(0) = 0$ and $f'(x) = 1/\sqrt{1+x^2}$. I want to calculate the coefficient of the $x^7$ term around $0$.
My approch: Although I know the answer if given $$f^{(7)}(0)/7!,$$ I am unable to calculate that far. I was hoping if there is shorter way to get the answer given the fact that this question was previously asked in an exam.
It is not difficult to find the Taylor series of $(1-t)^{\alpha}$ around $0$: $$(1+t)^{\alpha}=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3+\cdots$$
Put $t=x^2$, $\alpha=-1/2$, the Taylor series of $f'(x)=(1+x^2)^{-\frac12}$ around $0$ is given by $$f'(x)=1-\frac12x^2+\cdots+\frac{-\frac12(-\frac12-1)(-\frac12-2)}{3!}x^6+\cdots$$ Take the sixth derivative and put $x=0$, $$f^{(7)}(0)=\left.\frac{d^6}{dx^6}f'(x)\right|_{x=0}=0+\frac{-\frac12(-\frac12-1)(-\frac12-2)}{3!}(6!)+0.$$