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  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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    Ok, I can do that. Thank you. Do you have an idea how to go about item 3?2011-11-29

1 Answers 1

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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.

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    yes, excellent!2011-11-29