calculating a higher order derivative

Question

My task is to find the values $f^{(2017)}(0)$ and $f^{(2018)}(0)$ for $f(x)=\frac{\arccos(x)}{\sqrt{1-x^2}}$.

Basically, it's about finding the $n^{th}$ derivative of $f$. So I noticed if I let $g(x)=arccos(x)$, then $f(x)=-g(x)\cdot g'(x)$. I was able to prove by induction that for all $n\geq 2$ and $k\in \Bbb{N}$ the $n^{th}$ derivative of $g'$ is $$[(g')^{k}]^{(n)}(0)=k\cdot(n-1)\cdot[(g')^{k+2}]^{(n-2)}(0)$$ But even with applying the Leibniz rule to $f=g\cdot g'$ I don't understand how to get the final result. Did I make the wrong approach to the problem or is the general formula above useful? If so, how should I apply it?

Answer 1

Partial answer: $(g\cdot g')^{(n)}=\sum_{i+j=n}g^{(i)}\cdot g^{(1+j)}$ so even with a formula for $g^{(n)}$ we still have that sum in our way. Perhaps it is tractable but it may not be.

For the $2018$ case, in fact, this small miracle does seem to happen: for $|x|<1$ we have that

\begin{align*} \arccos x&=\frac{\pi}{2}-\sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1} \\ \frac{1}{\sqrt{1-x^{2}}}&=\sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}x^{2n} \end{align*}

Now, we wish to find $f^{(2018)}(0)$, which means finding the coefficient of $x^{2018}$ in the product of the above two series. The detail is that the second series only has even powers of $x$, and the first series only has odd powers of $x$, save for the constant. This mean that the coefficient $c_{n}$ of $x^{2n}$ in the product of the series must be $=a_{0}b_{2n}$, where $a_{i}$ is the coefficient of the term of degree $i$ in the first series, and $b_{j}$ is the coefficient of the term of degree $j$ in the second series:

$$ f^{(2018)}(0) = \frac{\pi}{2}\frac{4035!!}{4036!!} $$

Answer 2

Let $ n\in\mathbb{N} $, if $ f $ is a $ \mathcal{C}^{n+1} $ function on $ \left(-1,1\right) $, then we can use Leibniz formula for the $ n $-th derivative of the product :

$$ \left(\forall x\in\left(-1,1\right)\right),\ \left(ff'\right)^{\left(n\right)}\left(x\right)=\sum_{k=0}^{n}{\binom{n}{k}f^{\left(k\right)}\left(x\right)f^{\left(n+1-k\right)}\left(x\right)} $$

Setting $ f:x\mapsto\arccos{x} $, and $ g:x\mapsto\frac{1}{\sqrt{1-x^{2}}} $, we have : \begin{aligned} \left(\forall p\geq 1\right)\left(\forall x\in\left(-1,1\right)\right),\ f^{\left(p\right)}\left(x\right)&=-\frac{\mathrm{d}^{p-1}}{\mathrm{d}x^{p-1}}\left(\frac{1}{\sqrt{1-x^{2}}}\right)\\ &=-\sum_{k=0}^{p-1}{\binom{p-1}{k}\frac{\mathrm{d}^{k}}{\mathrm{d}x^{k}}\left(\left(1-x\right)^{-\frac{1}{2}}\right)\frac{\mathrm{d}^{p-1-k}}{\mathrm{d}x^{p-1-k}}\left(\left(1+x\right)^{-\frac{1}{2}}\right)}\\ &=-\sum_{k=0}^{p-1}{\binom{p-1}{k}\left(-1\right)^{k}\prod_{j=0}^{k-1}{\left(-\frac{1}{2}-j\right)}\left(1-x\right)^{-\frac{1}{2}-k}\prod_{j=0}^{p-2-k}{\left(-\frac{1}{2}-j\right)}\left(1+x\right)^{\frac{1}{2}+k-p}}\\ f^{\left(p\right)}\left(x\right)&=-\left(p-1\right)!\sum_{k=0}^{p-1}{\left(-1\right)^{k}\binom{-\frac{1}{2}}{k}\binom{-\frac{1}{2}}{p-1-k}\left(1-x\right)^{-\frac{1}{2}-k}\left(1+x\right)^{\frac{1}{2}+k-p}}\end{aligned}

Thus : $$ f^{\left(p\right)}\left(0\right)=-\left(p-1\right)!\sum_{k=0}^{p-1}{\left(-1\right)^{k}\binom{-\frac{1}{2}}{k}\binom{-\frac{1}{2}}{p-1-k}}=-\frac{1-\left(-1\right)^{p}}{2}\times\frac{\Gamma\left(\frac{p}{2}\right)\Gamma\left(p\right)}{\sqrt{\pi}\Gamma\left(\frac{p+1}{2}\right)} $$

Hence : \begin{aligned} \left(ff'\right)^{\left(n\right)}\left(0\right)&=f\left(0\right)f^{\left(n+1\right)}\left(0\right)-\sum_{k=1}^{n}{\binom{n}{k}\frac{1-\left(-1\right)^{k}}{2}\times\frac{\Gamma\left(\frac{k}{2}\right)\Gamma\left(k\right)}{\sqrt{\pi}\Gamma\left(\frac{k+1}{2}\right)}\frac{1-\left(-1\right)^{n+1-k}}{2}\times\frac{\Gamma\left(\frac{n+1-k}{2}\right)\Gamma\left(n+1-k\right)}{\sqrt{\pi}\Gamma\left(\frac{n+2-k}{2}\right)}}\\ &=\fbox{$\begin{array}{rcl}-\displaystyle\frac{\pi\left(1+\left(-1\right)^{n}\right)}{4}-\frac{n!\left(1-\left(-1\right)^{n}\right)}{2\pi}\sum_{k=1}^{n}{\left(\frac{1-\left(-1\right)^{k}}{2k}\right)\frac{\Gamma\left(\frac{k}{2}\right)\Gamma\left(\frac{n+1-k}{2}\right)}{\Gamma\left(\frac{k+1}{2}\right)\Gamma\left(\frac{n+2-k}{2}\right)}}\end{array}$} \end{aligned}