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

Show that $$\sec{z}=1+\sum_{k=1}^{\infty}{\frac{E_{2k}}{(2k)!}\,z^{2k}}$$ for some constants $E_2, E_4,\ldots$ known as the Euler numbers.

Is there a clever way to do this that doesn't involve taking several derivatives? I thought at first of taking advantage of $\sec{z}=\frac{1}{\cos{z}}$ and using the expansion of cosine, but I'm not sure where to go with it

Thanks

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marked as duplicate by J. M., Sasha, Mike Spivey, t.b., Srivatsan Dec 11 '11 at 0:44

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There should be a $(-1)^k$ in the summation. The series you wrote is the power series for $\operatorname{sech}z$. –  SL2 Dec 10 '11 at 17:40
    
The way it is worded ... "for some constants" ... I would guess it is asking you to show that there is a series expansion with odd terms all equal to zero and constant term $1$. –  GEdgar Dec 10 '11 at 19:03
    
@SL2: I think the sign issue is resolved by the Euler numbers. –  Bey Dec 10 '11 at 20:22
    
@JM: Thank you for that. Somehow that did not come up when I searched for similar questions –  Bey Dec 10 '11 at 20:23

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up vote 3 down vote accepted

You already noted that $$\sec(z)\cos(z) = 1$$ Expand $\cos(z)$ into a (known) powerseries and $\sec(z)$ into an (unknown) powerseries $\sum b_n z^n$ (why can you do that)? Now use the Cauchy product and compare coefficients (inductively).

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We know $\sec{z}$ admits a power series representation within a disk of radius $\pi/2$ about zero. Thanks for your help! –  Bey Dec 10 '11 at 20:25

If the problem is only to prove that there exist such constants, and not to find them or prove anything about them, then what you need is that the secant function is an even function ($\sec(-z) = \sec z$) and that it's analytic. The first fact you heard when you learned trigonometry. How to prove it rigorously might depend on the context in which you're working. If you know that $\cos z = \text{a power series in }z$, then you've got that cosine is analytic. That the reciprocal of an analytic function (i.e. capable of expansion as a convergent power series) is analytic is the Lagrange inversion theorem.

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