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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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    @JM: Thank you for that. Somehow that did not come up when I searched for similar questions2011-12-10

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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!2011-12-10
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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.