Find first five terms of the power series representation for the function f(x) = ${e^x cos(x^2)}$
So I have the answer which is ${1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+(\frac{1}{4!}-\frac{1}{2!})x^4+...}$
So I know that ${e^x = \sum\frac{x^n}{n!}}$ and that ${cos(x)=\sum(-1)^n\frac{x^{2n}}{(2n)!}}$
However I don't see/understand how this is the answer. The first 4 terms look like ${\sum e^x}$ then it changes which is due to the cosine I am guessing.
How can I represent ${cos(x^2)}$ as a series? I think it is ${\sum(-1)^n\frac{x^{4n}}{(2n)!}}$.
Furthermore how can I represent f(x) as a power series? 
Thanks for any help in advance!
 A: Your series for $e^x$ and $\cos(x^2)$ are correct. If you want to multiply two series, think about $(1+a_1x+a_2x^2+a_3x^3+\dots )(1+b_1x+b_2x^2+b_3x^3+\dots )=1+(a_1+b_1)x+(a_2+b_2+a_1b_1)x^2+\dots$
Can you see where each term in $x, x^2$ comes from?  Can you do the $x^3$ term?
A: Or you could use the power series. Notice that $f(x) = e^x \cos(x^2)$ is infinitely differentiable. So if a power series expansion exists and is convergent in an interval then it is the same as the Taylor series for that function on that particular interval. 
Hence $$f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + ... + \frac{f^{(n)}(0)}{n!} x^n + ...$$
where $f'(0), f''(0), f^{(n)(0)}$ are the first, second and $n$th derivative of $f(x)$ evaluated at $x = 0$. 
Sop all you have to do is to diferentiate $f(x)$ 5 times and plug in $0$. Do this if you feel the computation is a little easier than multiplying the two power series for $e^x$ and $\cos (x^2)$. This method does not require the knowledge of those two either. 
