# Thread: Power series for 1/cosh(z)

1. ## Power series for 1/cosh(z)

How would I go about computing the series expansion of 1/cosh(z), or at least the first few terms? I know the series of cosh(z), but trying to manipulate that, or trying to form it using the e^z definition of cosh, always gives me something complicated and looking very far away from a term-by-term definition.

Any thoughts?

2. Don't know if you're working with complex numbers, but in the real case we have $\displaystyle \frac{1}{\cosh x}=\frac{2}{e^{x}+e^{-x}}=\frac{2e^{x}}{1+e^{2x}}=2e^{x}\sum\limits_{j=0 }^{\infty }{\left( -e^{2x} \right)^{j}}.$

3. Originally Posted by Krizalid
Don't know if you're working with complex numbers, but in the real case we have $\displaystyle \frac{1}{\cosh x}=\frac{2}{e^{x}+e^{-x}}=\frac{2e^{x}}{1+e^{2x}}=2e^{x}\sum\limits_{j=0 }^{\infty }{\left( -e^{2x} \right)^{j}}.$
Sorry, I should have been clearer. I'm working over the complex numbers.

4. Originally Posted by Krizalid
Don't know if you're working with complex numbers, but in the real case we have $\displaystyle \frac{1}{\cosh x}=\frac{2}{e^{x}+e^{-x}}=\frac{2e^{x}}{1+e^{2x}}=2e^{x}\sum\limits_{j=0 }^{\infty }{\left( -e^{2x} \right)^{j}}.$
I was reprimanded before for giving a series that wasn't a power series. But, I think for all intents and purposes this is much better than some $\displaystyle \sum_{n\in\mathbb{N}}a_nx^n$ where $\displaystyle a_n$ is craaaaaaaaaaazy!

5. Let suppose that is...

$\displaystyle f(z)= \frac{1}{\cosh z}= a_{0} + a_{1}\cdot z + a_{2}\cdot z^{2} + a_{3}\cdot z^{3} + a_{4}\cdot z^{4} + \dots$ (1)

... and that we want to compute the $\displaystyle a_{n}$ till to $\displaystyle n=4$. Because is...

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

... from (1) we derive that is...

$\displaystyle (1 + \frac{z^{2}}{2!} + \frac{z^{4}}{4!} + \dots)\cdot (a_{0} + a_{1}\cdot z + a_{2}\cdot z^{2} + a_{3}\cdot z^{3} + a_{4}\cdot z^{4} + \dots)=1$ (3)

... so that is...

$\displaystyle a_{0}=1$

$\displaystyle a_{1}=0$

$\displaystyle a_{2}=-\frac{a_{0}}{2!}= -\frac{1}{2}$

$\displaystyle a_{3}=0$

$\displaystyle a_{4}=-\frac{a_{2}}{2!} - \frac{a_{0}}{4!} = -\frac{13}{24}$ (4)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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