I'm trying to compute the Taylor series for $\sec(x)$ but the derivatives are getting way too complicated. So how do I go about this without having to calculate all the derivatives? I tried to build some kind of relationship with the series for $\cos(x)$ but I didn't get anything meaningful.
Calculus – Taylor Series for sec(x)
calculuspower series
Best Answer
Look at the Boustrophedon table: $$\matrix{1\\0&1\\1&1&0\\0&1&2&2\\5&5&4&2&0\\0&5&10&14&16&16\\61&61&56&46&32&16&0\\0&61&122&178&224&256&272&272\\1385&1385&1324&1202&1024&800&544&272&0}$$ etc. Each row is the series of partial sums of the previous row, but at each stage one reverses the order we add up and enter the partial sums. Any, from the first column we read off $$\sec x=1+\frac{x^2}{2!}+\frac{5x^4}{4!}+\frac{61x^6}{6!}+\frac{1385x^8}{8!}+\cdots.$$ The right-most elements also give $$\tan x=x+\frac{2x^3}{3!}+\frac{16x^5}{5!}+\frac{272x^7}{7!}+\cdots.$$
There's a good discussion on this in Concrete Mathematics by Graham, Knuth and Patashnik.