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August 21

Deriving the Taylor series of the secant function

Basically, how does one show that the explicit formula(s) of the Euler numbers (given in the article) define the coefficients of the Taylor series of the secant function (up to a change in sign)?

One guess I have is to take advantage of the integral definition of the inverse secant, $\operatorname {arcsec}(x)=\int _{1}^{x}{\frac {dt}{t{\sqrt {t^{2}-1}}}}$ , expand the square root with a binomial series, antidifferentiate, and then apply the Lagrange inversion theorem, but I do not have experience applying that.

Once this series is found, the Cauchy product of it with the series for sine produces the tangent function's series.--Jasper Deng (talk) 01:40, 21 August 2014 (UTC)[reply]

Is secant(t) =

{\frac {1}{\cosh(it)}}=\sum _{n=0}^{\infty }{\frac {i^{n}E_{n}}{n!}}\cdot t^{n}\!

helpful?--Wikimedes (talk) 18:45, 27 August 2014 (UTC)[reply]