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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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    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.2011-11-29
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    2. Where's the nearest pole(s) of $\sec\,z$ from the origin?2011-11-29
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    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.2011-11-29
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    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...2011-11-29
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    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.2011-11-29
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    J.M, the two nearest poles are at $-\pi / 2 $ and $\pi / 2$. so that means the radius of convergence is $\pi / 2$. GEdgar, I know how to compute the first few terms of the expansion by derivation. I am having trouble to find the closed form of the series of the form $\sum a_{n}(z-z_{0})^{n}$.2011-11-29
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    The question was not asking you to derive any closed forms. Just as long as you know how to take successive derivatives of the secant, you can get $E_{2k}$ values.2011-11-29
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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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    A convolution... nice.2011-11-29
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    yes, excellent!2011-11-29