We let {$a_n$}$_{n\in N}$ be $a_n$=$\frac{f^{n}(0)}{n!}$. I have to show that if $f$ is an even function so is $a_{2n-1}$$=0$ for all n$\in$N. How can I show it? By induction maybe? Can anyone give a hint? If we try induction:
Induction start: n=1: $$f(x)=f(-x)$$ $$f'(x)=-f'(x)$$ $$x=0 ->2f'(0)=0->f'(0)=0$$
Induction assumption is then: $$f^{2n-1}(0)=0$$
But now I can't move on my own. What to do then?
On
Hint you may look this answer
Here is a simple proof:
We know that the derivative of an odd function is even and the derivative of an even function is odd. Obviously, an odd function evaluated at $0$ must vanish, otherwise we would get a contradiction. Consequently, if $f(x)$ is odd, then $f^{(n)}(x)$ is either even or odd. In particular, if $n$ is even, then $f^{(n)}(x)$ is odd and as a result $f^{(n)}(0) = 0$. Hence, the Maclaurin series expansion of $f(x)$ can consists only of the odd degree terms.
Hints/Outline:
If $f$ is even, then $f(-x) = f(x)$. Equivalently, $f(x) - f(-x) = 0$.
Try differentiating this equation with respect to $x$ an odd number of times, call it $2n-1$ times. Then you get $f^{(2n-1)}(x) + f^{(2n-1)}(-x) = 0$. Be careful with the signs for the $f(-x)$ term's derivative as you verify this yourself.
Let $x=0$. This will imply that $f^{(2n-1)}(0) = 0$. This in turn implies what you seek.