Consider the power series $\sum_{n=0}^\infty{\frac{x^{2n}}{(2n)!}}$
From this, it follows that its sum defines an infinitely differentiable function $f$, given by $f(x):=\sum_{n=0}^\infty{\frac{x^{2n}}{(2n)!}}$
- Determine whether $f$ is even or odd.
To do this, do I simply show that $f(x)=f(-x)$ and hence the function is even or is there more to the justification than that?
- Is there a simple relationship between $f''$ and $f$?
I found that the series can be expressed as the function $\cosh x$. The power series of the first derivative is or $\sinh x$ or $\sum_{n=1}^\infty{\frac{x^{2n+1}}{(2n+1)!}}$ and the second derivative is $\sum_{n=0}^\infty{\frac{x^{2n}}{(2n)!}}$, but I'm not sure what is meant by 'a simple relationship between $f''$ and $f$.