How to find asymptotic expansion of $\frac{1}{\cos(z)}$

Question

Find the asymptotic expantion of coefficients of the exponential generating function $f(z)=\frac{1}{\cos(z)}$ using all of its poles.

My work:

I know that: $$ \frac{1}{\cos \left( z \right)}=\sum_{n\ge 0}{E_{2n}}\frac{z^{2n}}{\left( 2n \right) !} $$

To find the poles we solve $\cos(z)=0$ where z is a complex number. Getting, $\chi_k=\pi/2+\pi k$ where $k \in Z$. Now wrie $\frac{1}{\cos(z)} \sim \frac{1}{z-\chi_k}= \frac{-1}{\chi_k} \frac{1}{1-z/\chi_k}$.(I'm not sure if this what I am supposed to do). Then $$ E_{2n}=-\left( 2n \right) !\sum_{k\in Z}{\frac{1}{\chi _k}}\frac{1}{\chi _{k}^{n}} \\ E_{2n}=-\left( 2n \right) !\left( \frac{1}{\left( \frac{\pi}{2} \right) ^{n+1}}+\sum_{k\ge 1}{\left( \frac{1}{\chi _{k}^{n+1}}+\frac{1}{\chi _{k_-}^{n+1}} \right)} \right) $$

Then I can look at the internal sum in the last equation and try to simplify it.

Edited:

$f(z)=\frac{1}{cos(z)} \sim \frac{Res(f(z),\chi_k)}{z-\chi_k}$ where $\chi_k=\frac{\pi}{2}+2\pi k$ the poles of $f(z)$ and $Res(f(z),\chi_k)= lim_{z\to \chi_k} (z-\chi_k) f(z)$.

After calculations we get that:

$\chi_0= \pi /2$ with Res= -1, $\chi_{-1} = -\pi / 2$ with Res = 1 , $\chi_1=3\pi /2$ with Res = 1, $\chi_{-2}= -3\pi /2$ with Res = -1 etc.

So, $f(z)=\frac{1}{cos(z)} \sim \frac{Res(f(z),\chi_k)}{z-\chi_k} \sim \frac{-1}{z-1/2 \pi} +\frac{1}{z+1/2 \pi} +\frac{1}{z-3/2 \pi} + \frac{-1}{z+3/2 \pi}+....$

Then if we expand these geometric coulomns and look at their coefficients we get:

$\frac{E_{2n}}{n!} \sim (2/ \pi)^{n+1} + (-2/ \pi)^{n+1} - (2/ 3\pi)^{n+1} +(2/ 3\pi)^{n+1} - (2/ 5\pi)^{n+1} -(-2/ 5\pi)^{n+1}-(2/ 7\pi)^{n+1} +(-2/7\pi)^{n+1}...)$

Thus, $E_{2n} \sim (2n)! ( (2/ \pi)^{n+1} + (-2/ \pi)^{n+1} - (2/ 5\pi)^{n+1} -(-2/ 5\pi)^{n+1}-(2/ 7\pi)^{n+1} +(-2/7\pi)^{n+1}...)$

Thus $E_{2n} \sim (2n)! ( (2/ \pi)^{n+1} + (-2/ \pi)^{n+1} +O(2/ 5\pi)^{n})$

What do you think. Any guide will be so appreciated.

Answer 1

You found that $$ \frac{1}{{\cos z}} = \sum\limits_{k = 0}^\infty {\left( {\frac{{( - 1)^k }}{{z + \left( {k + \frac{1}{2}} \right)\pi }} - \frac{{( - 1)^k }}{{z - \left( {k + \frac{1}{2}} \right)\pi }}} \right)} = \sum\limits_{k = 0}^\infty {( - 1)^k \frac{{(2k + 1)\pi }}{{\left( {k + \frac{1}{2}} \right)^2 \pi ^2 - z^2 }}} . $$ This expansion is actually convergent for all $z \neq (k+1/2)\pi$, $k \in \mathbb{Z}$. If $|z|<\frac{\pi}{2}$, then, by the geometric series, $$ \frac{1}{{\left( {k + \frac{1}{2}} \right)^2 \pi ^2 - z^2 }} = \frac{1}{{\left( {k + \frac{1}{2}} \right)^2 \pi ^2 }}\frac{1}{{1 - \left( {z/\left( {k + \frac{1}{2}} \right)\pi } \right)^2 }} = \sum\limits_{n = 0}^\infty {\frac{{z^{2n} }}{{\left( {k + \frac{1}{2}} \right)^{2n + 2} \pi ^{2n + 2} }}} . $$ Thus \begin{align*} \frac{1}{{\cos z}} & = \sum\limits_{k = 0}^\infty {( - 1)^k (2k + 1)\pi \sum\limits_{n = 0}^\infty {\frac{{z^{2n} }}{{\left( {k + \frac{1}{2}} \right)^{2n + 2} \pi ^{2n + 2} }}} } \\ & = \sum\limits_{n = 0}^\infty {( - 1)^n \left[ {( - 1)^n \left( {\frac{2}{\pi }} \right)^{2n + 1} 2(2n)!\sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }}{{(2k + 1)^{2n + 1} }}} } \right]\frac{{z^{2n} }}{{(2n)!}}} , \end{align*} provided $|z|<\frac{\pi}{2}$. Consequently, $$ E_{2n} = ( - 1)^n \left( {\frac{2}{\pi }} \right)^{2n + 1} 2(2n)!\sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }}{{(2k + 1)^{2n + 1} }}} $$ for all $n\geq 0$.

Answer 2

Here are the two facts you need to solve the problem:

1. If $f(z)=\sum_{n\ge 0} a_nz^n$ is analytic at the origin with radius of convergence $R$, then $a_n\in O\big((1/R+\epsilon )^{n}\big)$ for all $\epsilon>0$. When $f$ is meromorphic, recall that $R$ is the infimum of the magnitudes of the poles of $f$.

2. If $f(z)$ has a pole of order $1$ at $a$, then $f(z)-\frac{\text{Res}(f,a)}{z-a}$ has a removable singularity at $a$, and the same poles as $f$ elsewhere. Here, $\text{Res}(f,a)=\lim_{z\to a}(z-a)f(z)$.

This means that if $f$ only has simple poles, then by subtracting out $\frac{\text{Res}(f,a)}{z-a}$ as $a$ ranges over the poles of $f$, then coefficients of what remains will go to zero at an arbitrarily fast rate.

For example, $\frac1{\cos z}$ has a pole at $\chi_0=\pi/2$, and the residue at that pole is $\text{Res}(f,\pi/2)=\lim_{z\to\pi/2}\frac{z-\pi/2}{\cos z}=-1$. Therefore, $f-\frac{-1}{z-\pi/2}$ has one fewer pole. This is akin to what you attempted, but you forgot to compute the residue.

Similarly, you can subtract the pole at $\chi_{-1}=-\pi/2$, whose residue is $1$. When you do this, the current state of affairs is $$ \frac{1}{\cos z}+\frac1{z-\pi/2}-\frac1{z-(-\pi/2)} $$ and the poles of what remains are only at $3\pi/2,5\pi/2,\dots$ and $-3\pi/2,-5\pi/2,\dots$. The infimum the magnitude of these values is $3\pi/2$, which proves that $$ {E_{n}}/{n!}-(2/\pi)^{n+1}+(-2/\pi)^{n+1}\in O\big((2/3\pi+\epsilon)^n) $$ You can then keep subtracting out poles, accumulating more terms on the left hand side, and decreasing the error on the right hand side.

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In general, this method does not give an equality for the coefficients of $f(z)$ in terms of an infinite series of the subtracted poles. It happens to work for $f(z)=1/\cos z$. However, in general, even after subtracting out all the poles of $f$, there may still be an entire function left over which contributes to the coefficents of $f$.