Prove the derivative

Question

Let $f(x) = (x^2-1)^{\frac{1}{2}}, x>1$. How do I prove that the $n$th derivative of $f(x) > 0$ for odd $n$, and the $n$th derivative of $f(x) < 0$ for even $n$?

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $x > 1$: \begin{align} \pars{x^{2} - 1}^{1/2}&=x\pars{1 - {1 \over x^{2}}}^{1/2} =x\sum_{\ell = 0}^{\infty}{\Gamma\pars{3/2} \over \ell!\,\Gamma\pars{3/2 - \ell}} \pars{-1}^{\ell}x^{-2\ell} \\[3mm]&=x\pars{1 - {1 \over 2x^{2}}} +x\sum_{\ell = 2}^{\infty}{\Gamma\pars{3/2} \over \ell!\,\Gamma\pars{3/2 - \ell}} \pars{-1}^{\ell}x^{-2\ell} \end{align}

However, $$ \Gamma\pars{{3 \over 2} - \ell}= {\pi \over \Gamma\pars{\ell - 1/2}\sin\pars{\pi\bracks{\ell - 1/2}}} =-\pars{-1}^{\ell}\,{\pi \over \Gamma\pars{\ell - 1/2}} $$

$$ \pars{x^{2} - 1}^{1/2} =x\pars{1 - {1 \over 2x^{2}}} -{1 \over 2\root{\pi}}\sum_{\ell = 2}^{\infty} {\Gamma\pars{\ell - 1/2} \over \ell!}\,x^{-2\ell + 1} $$

\begin{align} \totald{\pars{x^{2} - 1}^{1/2}}{x} &=1 + {1 \over 2x^{2}} + {1 \over 2\root{\pi}}\sum_{\ell = 2}^{\infty} {\Gamma\pars{\ell - 1/2} \over \ell!}\,\pars{2\ell - 1}x^{-2\ell} \color{#00f}{\large > 0} \\[3mm] \totald[2]{\pars{x^{2} - 1}^{1/2}}{x} &=-\,{1 \over x^{3}} - {1 \over 2\root{\pi}}\sum_{\ell = 2}^{\infty} {\Gamma\pars{\ell - 1/2} \over \ell!}\pars{2\ell}\pars{2\ell - 1}x^{-2\ell - 1} \color{#00f}{\large < 0} \\[3mm] \totald[3]{\pars{x^{2} - 1}^{1/2}}{x} &={3 \over x^{4}} + {1 \over 2\root{\pi}}\sum_{\ell = 2}^{\infty} {\Gamma\pars{\ell - 1/2} \over \ell!}\, \pars{2\ell + 1}\pars{2\ell}\pars{2\ell - 1}x^{-2\ell - 2} \color{#00f}{\large > 0} \\[3mm] \totald[4]{\pars{x^{2} - 1}^{1/2}}{x} &=-\,{4 \times 3 \over x^{5}} - {1 \over 2\root{\pi}}\sum_{\ell = 2}^{\infty} {\Gamma\pars{\ell - 1/2} \over \ell!}\, {\pars{2\ell + 2}! \over \pars{2\ell - 2}!}x^{-2\ell - 3} \color{#00f}{\large < 0} \end{align} and so on.

For $n \geq 2$ $\pars{~\mbox{for}\ n = 1\ \mbox{it's evident}~}$: $$ \totald[n]{\pars{x^{2} - 1}^{1/2}}{x} =\pars{-1}^{n + 1}\ \overbrace{\bracks{{n! \over 2x^{n + 1}} + {1 \over 2\root{\pi}}\sum_{\ell = 2}^{\infty} {\Gamma\pars{\ell - 1/2} \over \ell!}\, {\pars{2\ell + n - 2}! \over \pars{2\ell - 2}!}x^{-2\ell - n + 1}}}^{\ds{\large > 0}} $$