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For $\displaystyle\sum\limits_{n \geq 0}a_n z^n$ the Maclaurin series of $f(z)=\frac{e^z}{\cos{\frac{\pi}{2}z}}$, what is an approximation of $a_{100}$ with error $0.0001$ at most?

This is a past qualifying exam question, so no computer. I am looking for a solution that could be given in a timed exam setting, by a student with one semester of graduate Complex Analysis, familiar with first half of Stein's book, with Churchill's book, and somewhat with first half of Ahlfor's book.

I write $c(z)\,=\,\cos{\frac{\pi}{2}z}\,=\,\displaystyle\sum\limits_{\;n\geq0\\\text{(even)}}c_n z^n,$ and also $\theta=\frac{2}{\pi}<1.$ We have $c_n\,=\,\frac{(-1)^{n/2}}{n!\theta^n}.$

$$f(z)\,=\,\frac{1\,+\,z\,+\,\frac{z^2}{2!}\,+\,\frac{z^3}{3!}\,+\,\frac{z^4}{4!}\,+\,\cdots}{1\;\quad-\quad\frac{z^2}{2!\theta^2}\quad+\quad\frac{z^4}{4!\theta^4}\,-+\,\cdots}\qquad\qquad(z\in\mathbb{D}).$$

Using the usual method, I found $a_0,a_1,a_2,a_3,a_4$ by hand, so we have $\,f(z)\,=\,$

$$\displaystyle\sum\limits_{n \geq 0}a_n z^n\,=\,1+z+\left(\frac{1}{2}+\frac{1}{2\theta^2}\right)z^2+\left(\frac{1}{6}+\frac{1}{2\theta^2}\right)z^3+\left(\frac{1}{24}+\frac{1}{4\theta^2}+\frac{5}{24\theta^4}\right)z^4+\cdots.$$

I tried to discern a pattern by keeping the expressions more abstract, i.e. keeping the $a_n$ and $c_n$. Doing so we can write e.g.,

$$a_2\,=\,\frac{1}{2!}\,-\,c_2\,=\,\frac{1}{2!}\,-\,\displaystyle\sum\limits^{2}_{\,\ell\,=\,2\\\text{(even)}}c_\ell a_{n-\ell},$$

$$a_7\,=\,\frac{1}{7!}\,-\,(c_2 a_5+c_4 a_3+c_6)\,=\,\frac{1}{7!}\,-\,\displaystyle\sum\limits^{7}_{\,\ell\,=\,2\\\text{(even)}}c_\ell a_{n-\ell},$$

$$a_{100}\,=\,\frac{1}{100!}\,-\,(c_2 a_{98}+c_4 a_{96}+\cdots+c_{98}a_2+c_{100}).$$

I am not sure what to do next. I did consider Cauchy's Integral Inequalities, but could not see how to get a useful estimate from them. If $0<r<1,$

$$a_{100} = \frac{f^{(100)}(0)}{100!\,} = \frac{1}{2\pi i } \oint_{|z| = r} \frac{e^z}{c(z)\,z^{101}} dz,$$

$$|a_{100}| \leq \frac{1}{r^{100}}\max_{|z| = r}\left|\frac{e^z}{\cos(\frac{\pi}{2}z)} \right|,$$

but for either $r$ very near zero, or very near one, or say for $r=$ one-half, the estimate seems to not be useful.

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  • $\begingroup$ What would you get if you substituted functions.wolfram.com/01.11.06.0017.01 into the Cauchy integral formula and integrated term-by-term? $\endgroup$
    – Gary
    Jul 27, 2021 at 5:30
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    $\begingroup$ Please avoid using \displaystyle when using MathJax on question titles. For more information, see Guidelines for good use of MathJax on question titles. $\endgroup$
    – soupless
    Jul 27, 2021 at 14:03
  • $\begingroup$ avoid displaystyle in titles, I will remember. I will change it right away. @soupless $\endgroup$
    – 311411
    Jul 27, 2021 at 14:18

1 Answer 1

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I think the idea is to subtract the principal parts $P$ at the poles at $z = \pm 1$, after which the resulting function $g=f-P$ is analytic in the disk $|z| < 3$, so that Cauchy inequalities should give you that the coefficient $b_{100}$ of $g$ is very small, certainly $|b_{100}| < 0.0001$. This means that within this small error, $a_{100}$ is just the corresponding coefficient of the sum of the principal parts $P$, which can be easily evaluated by geometric series.

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  • $\begingroup$ There is a Laurent Series for $f(z)$ in $0<|z-1|<2,$ the pole is simple, and the Principal Part is $\frac{-e\theta}{z-1}.$ (I believe.) This is the way to do it? Then $f(z)+\frac{e\theta}{z-1}$ is analytic in $|z-1|<2.$ But the Laurent Series for $f(z)$ in $0<|z+1|<2$ is different. So I am not sure how to get to the big disc $|z|<3$. Will you advise me please? (For $z=-1$, I think the P.P. is $\frac{\theta}{e(z+1)}.$) $\endgroup$
    – 311411
    Aug 4, 2021 at 14:49
  • $\begingroup$ @311411: You want all the series centered at 0, i.e., the Taylor series centered at 0 of the principal parts at both poles. $\endgroup$ Aug 6, 2021 at 21:31

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