# Showing that $\sec z = \frac1{\cos z} = 1+ \sum\limits_{k=1}^{\infty} \frac{E_{2k}}{(2k)!}z^{2k}$

1. Show that there are complex numbers $E_2,E_4,E_6,\dotsc$ such that $\sec z = \frac{1}{\cos z} = 1+ \sum\limits_{k=1}^{\infty} \frac{E_{2k}}{(2k)!}z^{2k}$ in a neighborhood of $0$.

2. What is the radius of convergence?

3. Show that: $E_{2n}-{2n\choose 2n-2} E_{2n-2} + {2n\choose 2n-4}E_{2n-4}+ \dotsb -(-1)^{n}{2n \choose 2}E_{2} + (-1)^{n} = 0$.

4. Compute $E_2, E_4$ and $E_6$.

Can you show me how to solve this problem? I am not able to do any of them. Thanks.

Edit: Ideas have been given for 1,2,4. Thank you. What about item 3 ?

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1. and 4. Note that $\sec\,z$ is even; its Maclaurin expansion ought to consist solely of even powers. You should then be able to figure out your $E_{2k}$ (Euler numbers, they're called) from your series expansion. – J. M. is back. Nov 29 '11 at 12:38
2. Where's the nearest pole(s) of $\sec\,z$ from the origin? – J. M. is back. Nov 29 '11 at 12:38
I do not understand your suggestion for task 1) and 4). For 2) The nearest pole is at 90 degrees. Why is the question only about one radius of convergence if there are more than one poles? Thanks for trying to explain. – user20318 Nov 29 '11 at 12:59
I was presuming you knew how to do a series expansion for a function like $\sec\,z$; that should help in doing the first and fourth items... for item 2, you do know that the secant is even, yes? Thus, the two (yes, that's a hint) nearby poles are... – J. M. is back. Nov 29 '11 at 13:02
If you cannot do 1, then you need to find a calculus textbook and read the part on Taylor series. Also, $z$ should be in radians, so don't say 90 degrees. – GEdgar Nov 29 '11 at 13:27

About 3, note that $$1=\frac{1}{\cos z}\cdot\cos(z) =\left( 1+ \sum\limits_{k=1}^{\infty} \frac{E_{2k}}{(2k)!}z^{2k}\right)\cdot\left(\sum\limits_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}z^{2k}\right).$$ For each $n\geqslant1$, the coefficient of $z^{2n}$ in the product of these two series is $$0=\frac{(-1)^n}{(2n)!}+\sum\limits_{k=1}^n\frac{E_{2k}}{(2k)!}\frac{(-1)^{n-k}}{(2n-2k)!}=\frac1{(2n)!}\left((-1)^n+\sum\limits_{k=1}^n(-1)^{n-k}{2n\choose 2k}E_{2k}\right).$$ The last parenthesis is the alternated sum you are interested in.