# Euler numbers generating sequence radius of convergence

Printable View

• Nov 21st 2010, 07:33 PM
MSUMathStdnt
Euler numbers generating sequence radius of convergence
I'm not sure which section a post on finding the Euler numbers using series belongs, so I put it here.

Find (a) the radius of convergence, and (b) the first six Euler numbers $E_n$ from:

$\frac1{\cosh(z)}=\sum_{n=0}^\infty \frac{E_n}{n!}z^n$

I haven't any idea how to start the part (a).

I can't put the finishing touches on part (b):

$\cosh(z) = \frac{e^z+e^{-z}}{2}$

$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$
$e^{-z} = \sum_{n=0}^\infty \frac{(-1)^n z^n}{n!}$

$e^z+e^{-z} = \sum_{n=0}^\infty \frac{z^n}{n!} + \sum_{n=0}^\infty \frac{(-1)^n z^n}{n!} = \sum_{n=0}^\infty \frac{(2z)^n}{(2n)!}$

So,

$\frac{1}{\cosh(z)} = \frac{2}{2\sum_{n=0}^\infty \frac{(2z)^n}{(2n)!}} = \frac{1}{\sum_{n=0}^\infty \frac{(2z)^n}{(2n)!}}$

However, I can't quite get this to a format where I can get the Euler numbers.

I know I can't flip the bottom fraction since it's an infinite sum. But I only need the first six, yet I still can't do it. The first one I can do with the first term of the series (right?):

$\frac{1}{\cosh(z)} = \frac{1}{\sum_{n=0}^\infty \frac{z^n}{(2n)!}} = \frac{1}{1}=1$ and $E_0=1$ (so the back of my book says).

The next term, and the one after that, is:
$\frac{1}{\cosh(z)} = \frac{1}{1+z^2/2}=\frac{2}{2+z^2}$
$\frac{1}{\cosh(z)} = \frac{1}{1+z^2/2+z^4/4!}=\frac{24}{24+12z^2+z^4}$

Now what? How do I find the coefficient's of z, z^2, etc. from this?

I'm still working on part (a), so I don't have a specific question yet.

Thanks.

P.S. How do you make a new line in this forum? I tried \\, \\*, and \newline with no success and had to make separate 'math' tags for each line.
• Nov 21st 2010, 07:58 PM
chisigma
The function $\cosh (z)$ is an entire function so that it has 'only zeroes' and its McLaurin expansion converges in the whole complex plane. Its inverse $\frac{1}{\cosh z}$ has 'only poles' and they are located on the imaginary axis where $z= i\ \omega$ and is...

$\cosh (i \omega)= \cos \omega = 0 \implies \omega = (2n+1)\ \frac{\pi}{2}$ (1)

... so that the radious of convergence of the McLaurin expansion of $\frac{1}{\cosh (z)}$ is $r= \frac{\pi}{2}$...

Kind regards

$\chi$ $\sigma$
• Nov 21st 2010, 08:56 PM
MSUMathStdnt
Quote:

Originally Posted by chisigma
The function $\cosh (z)$ is an entire function so that it has 'only zeroes' and its McLaurin expansion converges in the whole complex plane. Its inverse $\frac{1}{\cosh z}$ has 'only poles' and they are located on the imaginary axis where $z= i\ \omega$ and is...

$\cosh (i \omega)= \cos \omega = 0 \implies \omega = (2n+1)\ \frac{\pi}{2}$ (1)

... so that the radious of convergence of the McLaurin expansion of $\frac{1}{\cosh (z)}$ is $r= \frac{\pi}{2}$...

Kind regards

$\chi$ $\sigma$

Hi chi, could you give me some more details.

Can you explain "cosh(z) has only zeros", "1/cosh(z) has only poles".

I understand that cosh(z) is analytic (that is, it has no singularities and is differentiable on the entire z-plane). Is that what you mean by "only zeros"? I don't understand the "only poles" comment at all. Can you also explain how that leads to your following statement, too?

Thanks for your help.
• Nov 22nd 2010, 12:54 AM
chisigma
As all the entire functions $\cosh (z)$ is expandable as 'infinite product'...

$\displaystyle \cosh (z)= \prod_{n=0}^{\infty} \{1+\frac{4 z^{2}}{(2n+1)^{2}\ \pi^{2}}\}$ (1)

... and that means that the function vanishes for $z= \pm i (2n+1)\ \frac{\pi}{2}$ and it has 'only zeroes'. Its inverse $\frac{1}{\cosh z}$ has infinite poles at $z= \pm i (2n+1)\ \frac{\pi}{2}$ and no zeroes and that means that it has 'only poles'...

Kind regards

$\chi$ $\sigma$
• Nov 22nd 2010, 05:24 AM
MSUMathStdnt
Thanks again, chi sig. But we haven't learned this yet, so I don't think I can use this answer.
• Nov 22nd 2010, 08:38 AM
chisigma
First we suppose that Euler's number $E_{n}$ are defined by the 'identity'...

$\displaystyle \frac{1}{\cosh z} = \sum_{n=0}^{\infty} \frac{E_{n}}{n!}\ z^{n}$ (1)

... then we found them with the following procedure. The function $\frac{1}{\cosh z}$ is analytic in $z=0$, so that we can write...

$\displaystyle \frac{1}{\cosh z} = 1 + a_{1}\ z + a_{2}\ z^{2} + a_{3}\ z^{3} + a_{4}\ z^{4} + a_{5}\ z^{5} + a_{6}\ z^{6} + ...$ (2)

... and it is well known that...

$\displaystyle \cosh z = 1 + \frac{z^{2}}{2} + \frac{z^{4}}{4!} + \frac{z^{6}}{6!} + ...$ (3)

Now we multiply (2) and (3) and obtain...

$\displaystyle (1 + a_{1}\ z + a_{2}\ z^{2} + a_{3}\ z^{3} + a_{4}\ z^{4} + a_{5}\ z^{5} + a_{6}\ z^{6} + ...)\ (1 + \frac{z^{2}}{2} + \frac{z^{4}}{4!} + \frac{z^{6}}{6!} + ...)= 1$ (4)

... and from (4)...

$a_{0}=1 \implies E_{0}=1$

$a_{0}\ 0 + a_{1} =0 \implies a_{1}=0 \implies E_{1}=0$

$\frac{a_{0}}{2} + a_{1}\ 0 + a_{2} =0 \implies a_{2}= -\frac{1}{2} \implies E_{2}=-1$

$a_{0}\ 0 + \frac{a_{1}}{2} + a_{2}\ 0 + a_{3} =0 \implies a_{3}=0 \implies E_{3}=0$

$\frac{a_{0}}{4!} + a_{1}\ 0 + \frac{a_{2}}{2} + a_{3}\ 0 + a_{4} =0 \implies a_{4}= \frac{5}{4!} \implies E_{4}= 5$

$a_{0}\ 0 + \frac{a_{1}}{4!} + a_{2}\ 0 + \frac{a_{3}}{2} + a_{4}\ 0 + a_{5} =0 \implies a_{5}=0 \implies E_{5}=0$

$\frac{a_{0}}{6!} + a_{1}\ 0 + \frac{a_{2}}{4!} + a_{3}\ 0 + \frac{a_{4}}{2} + a_{5}\ 0 + a_{6} =0 \implies a_{6}= - \frac{61}{6!} \implies E_{6}= -61$

Kind regards

$\chi$ $\sigma$
• Nov 22nd 2010, 10:03 AM
MSUMathStdnt
Hi chi. Good answer. Thanks. This is the answer me and my classmates were working towards. But I was stuck on the expansion of 1/cosh(x). That one is not in my book. I could generate it using Maple, but I could not create that expansion analytically. If it's well-known, maybe I will just use it as is.

Edit: OK, I had more time to look at this. So here's my last question - I promise. I just want to be sure I understand the this right, and create a record for when I return in a few months. This:
$\displaystyle (1 + a_{1}\ z + a_{2}\ z^{2} + a_{3}\ z^{3} + a_{4}\ z^{4} + a_{5}\ z^{5} + a_{6}\ z^{6} + ...)\ (1 + \frac{z^{2}}{2} + \frac{z^{4}}{4!} + \frac{z^{6}}{6!} + ...)= 1$

is actually:
$\displaystyle (a_{0} + a_{1}\ z + a_{2}\ z^{2} + a_{3}\ z^{3} + a_{4}\ z^{4} + a_{5}\ z^{5} + a_{6}\ z^{6} + ...)\ (1 + 0 \ z + \frac{z^{2}}{2} + 0\ z^3 + \frac{z^{4}}{4!} + 0\ z^5 + \frac{z^{6}}{6!} + ...)= 1$

Then you multiply and compare coefficients with the rhs (which are all zero except for the constant term). So you compared the constant terms:
$a_0 \cdot 1 = 1 \implies a_0=1$

Then compared the z term:
$a_0 \cdot (0 \cdot z) + (a_1 \cdot z) \cdot 1 = 0 \implies a_1=0$

And one more (z^2 term):
$a_0 \cdot \frac{z^2}{2} + (a_1 \cdot z) \cdot (0 \cdot z) + (a_2 \cdot z^2) \cdot (1) = 0\implies (\frac12 + a_2) \cdot z^2 = 0 * z^2 \implies a_2 = \frac{-1}{2}$

And so on...
-Jeff
• Nov 26th 2010, 12:55 AM
MSUMathStdnt
Alternative Solution
I just learned an alternative solution for finding the Eulers which might be a little more straightforward and be easier to work if you need to find more than just the first 6 numbers (and have a CAS handy), so here goes.

First, we need a few simple (well-known) definitions:
$cosh(z)=cos(iz)=\frac{e^z+e^{-z}}{2}$
$\frac1{cosh(z)}=\frac2{e^z+e^{-z}}$

$\frac1{cosh(z)}=\sum_{n=0}^\infty \frac{E_n}{n!} \cdot z^n = E_0 + E_1 \cdot z + \frac{E_2}{2!} \cdot z^2 + ...$

Evaluate at $z=0$:
$\frac1{cosh(0)}=1=E_0+E_1 \cdot (0) + E_n \cdot (0)^n=E_0 \implies E_0=1$

In the complex plane, cosh(z) is never zero, so 1/cosh(z) is analytic and can be derived ad nauseum (that's how many times an undergrad has to evaluate it for his homework). The first term of the expansion drops out. Keep deriving, evaluating at zero, and solving for the next Euler number:

$\frac{d(\frac{1}{cosh(0)})}{dz}=0=E_1+E_2 \cdot (0) \implies E_1=0$
$\frac{d^{(2)}(\frac{1}{cosh(0)})}{dz^2}=-1=E_2+E_3 \cdot (0) \implies E_2=-1$

etc.

In other words, $E_n$ is the $n^{th}$ derivative of $\frac1{cosh(z)}, evaluated at 0$.
• Nov 26th 2010, 01:26 AM
chisigma
Quote:

Originally Posted by MSUMathStdnt
... in the complex plane cosh(z) is never zero...

That isn't true!... $\cosh z$ vanishes for $z= \pm i (2n+1) \frac{\pi}{2}$ and that's why the radius of convergence of the Taylor expansion of $\displaystyle \frac{1}{ \cosh z}$ around $z=0$ is $\displaystyle \frac{\pi}{2}$ ...

Kind regards

$\chi$ $\sigma$
• Nov 26th 2010, 01:50 AM
MSUMathStdnt
You're right. Thanks.

I think the correct explanation is that 1/cosh(z) is analytic near z=0, and for that reason we can take as many derivatives as we want. Apparently, I need to do some more studying before my final in two weeks. :D