If f(x) = integral [cos(t^2)dt] from 0 to x,
using the known power series for cos x, find a power series for f(1).
This is f(x):
$$f(x)=\int\limits\limits_{0}^{X}\cos(t^2)dt$$

Question

If f(x) = integral [cos(t^2)dt] from 0 to x,
using the known power series for cos x, find a power series for f(1).
This is f(x):
$$f(x)=\int\limits\limits_{0}^{X}\cos(t^2)dt$$

kirbykirby · Answer

I ended up finding 
$$\sum_{n=0}^{infinity}\frac{(-1)^{n}x^{4n+1}}{(4n+1)(2n)!}=x-\frac{x^5}{5*2!}+\frac{x^9}{9*4!}-\frac{x^{13}}{13*6!}+...$$
But I'm not quite sure what it means to find the power series for f(1).. Isn't that just a number (whatever I found evaluated at 1)? There is no more x-dependence so how is it a power series :S I'm so confused

anonymous · Answer

maybe its a power series centered at x = 1

kirbykirby · Answer

Hmm maybe that's what they meant..

anonymous · Answer

You just plug in $x-1$.  Centering is not an issue here, because we're dealing with Maclaurin series which are always centered at $0$.

anonymous · Answer

I mean $x=1$