Finding $n$th derivative in an unusual way

Question

If $f(z) = \frac{e^{iz}}{z^2-1}$, then $f^{(4)}(z)$ can be found by differentiating $f(z)$ four times. I tried to use Cauchy's integral formula, but the integrand is not holomorphic at $z=0$, so we can't use it in this case. What can possibly be another clever method of finding the 4th (or $n$th derivative) of this function using an indirect approach?

Answer 1

$$(z^2-1) f(z) = e^{iz}$$ Differentiate this four times: $$ \eqalign{2 z f(z) + (z^2-1) f'(z) &= i e^{iz}\cr 2 f(z) + 4 z f'(z) + (z^2-1) f''(z) &= -e^{iz}\cr 6 f'(z) + 8 z f''(z) + (z^2-1) f'''(z) &= -i e^{iz}\cr 14 f''(z) + 16 z f'''(z) + (z^2-1) f''''(z) &= e^{iz}\cr}$$ Now substitute $z=0$: $$ \eqalign{-f(0) &= 1\cr -f'(0) &= i\cr 2 f(0) - f''(0) &= -1\cr 6 f'(0) - f'''(0) &= -i\cr 14 f''(0) - f''''(0) &= 1\cr}$$ so that $$\eqalign{f(0) &= -1\cr f'(0) &= -i\cr f''(0) &= -1\cr f'''(0) &= -5i\cr f''''(0) &= -13\cr}$$

Answer 2

May be, another way : since you want the derivatives for $z=0$, you could use Taylor expansion of $e^{x}$ and replace $x$ by $iz$; this would give $$e^{iz}=1+i z-\frac{z^2}{2}-\frac{i z^3}{6}+\frac{z^4}{24}+\frac{i z^5}{120}+O\left(z^6\right)$$ Now, use long division to get $$f(z) = \frac{e^{iz}}{z^2-1}=-1-i z-\frac{1}{2}z^2-\frac{5 i }{6}z^3-\frac{13 }{24}z^4-\frac{101 i }{120}z^5+O\left(z^6\right)$$ and use $$f(z)=\sum_{i=0}^\infty\frac{z^i}{i!}f^{(i)}(0)$$ Then, identification becomes simple and leads to the results already given by Robert Israel in his answer.

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets $\ds{\verts{z} < 1}$:

\begin{align} \color{#f00}{\mathrm{f}\pars{z}} & = {\expo{\ic z} \over z^{2}-1} = \overbrace{% \bracks{-\sum_{k = 0}^{\infty}\pars{z^{2}}^{k}}}^{\ds{1 \over z^{2} - 1}}\ \overbrace{\bracks{\sum_{n = 0}^{\infty}{\pars{\ic z}^{n} \over n!}}} ^{\ds{\expo{\ic z}}}\ =\ -\sum_{k = 0}^{\infty}\sum_{n = 0}^{\infty}{\ic^{n} \over n!}z^{n + 2k} \\[3mm] & = -\sum_{k = 0}^{\infty}\sum_{n = 2k}^{\infty} {\ic^{n - 2k} \over \pars{n - 2k}!}z^{n} = -\sum_{n = 0}^{\infty}z^{n}\,\ic^{n}\sum_{k = 0}^{k\ \leq\ n/2} {\pars{-1}^{k} \over \pars{n - 2k}!} = \sum_{n = 0}^{\infty}\bracks{-\ic^{n}n!\sum_{k = 0}^{\left\lfloor n/2\right\rfloor} {\pars{-1}^{k} \over \pars{n - 2k}!}}{z^{n} \over n!} \end{align}

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$$ \begin{array}{|c|}\hline\mbox{}\\ \ds{\quad\color{#f00}{\mathrm{f}^{\mathrm{\pars{n}}}\pars{0}} = \color{#f00}{-\ic^{n}n!\sum_{k = 0}^{\left\lfloor n/2\right\rfloor} {\pars{-1}^{k} \over \pars{n - 2k}!}}\quad} \\ \mbox{}\\ \hline \end{array} $$

By the way, $$ \color{#f00}{\mathrm{f}^{\mathrm{\pars{4}}}\pars{0}} = -\ic^{4}\,4!\sum_{k = 0}^{2} {\pars{-1}^{k} \over \pars{4 - 2k}!} = -24\pars{{1 \over 4!} - {1 \over 2!} + {1 \over 0!}} = \color{#f00}{-13} $$