Expansions in Calc 2

[erai0208]

Come up with a formula for the function f[x] whose expansion in powers of x is
1 - [(x^2)/(2^2*2!)] + [(x^4)/(2^4*4!)] - [(x^6)/(2^6*6!)] + ...
I know that the Cos[x] in powers of x:
1 - (x^2)/2! + (x^4)/4! + ...+
So I need to do something to Cos[x] so that in the denominator there is a 2^2, a 2^4, and a 2^6...

How do I do that to the denominator?

 

[galactus]

Note, the general summation of your series looks like this:

\(\displaystyle \sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{2^{2k}(2k)!}\)

\(\displaystyle =\sum_{k=0}^{\infty}\frac{(-1)^{k}(\frac{x}{2})^{2k}}{(2k)!}\)

See it now?. It is the series for cos, but exactly what cos?.