# Even and odd functions for Taylor serie

I have asked this question before, sorry, but I'm still confused about how I can show it. Hope anybody can help me?

We let $$f:\mathbb{R}\to\mathbb{R}$$ be infinitely often differentiable function and we let the Taylor series be: $$\displaystyle\sum_{n=0}^{\infty}\left(\left(\frac{f^{n}(0)}{n!}\right)x^n\right)$$Let $$\{a_n\}_{n\in \mathbb N }$$ be $$a_n=\frac{f^{n}(0)}{n!}$$. We have to assume that the Taylor series converges toward $$f$$ in an open interval $$(-r,r)$$ around zero. Then I have to show that if $$a_{2n-1}=0$$ for all $$n\in \mathbb N$$ so is $$f(-x)=f(x)$$ for all $$x\in(-r,r)$$. How can I do it? I think I maybe can see on $$kx^{2n}$$ while all odd joints are zero while $$a_{2n-1}=0$$? But how can I prove it?

• Isn't it enough to say that all terms are even functions ? – Yves Daoust Jun 2 at 8:04

We have $$f(x)= \sum_{n=0}^{\infty}a_{2n}x^{2n}$$ for $$x\in (-r,r).$$

Firthermore: $$(-x)^{2n}=x^{2n}.$$

Can you proceed ?

For all $$x\in(-r,r)$$ and all $$m$$,

$$\sum_{n=0}^m\frac{f^{(2n)}(0)}{(2n)!}(-x)^{2n}=\sum_{n=0}^m\frac{f^{(2n)}(0)}{(2n)!}x^{2n}$$

so that $$\lim_{m\to\infty}\sum_{n=0}^m\frac{f^{(2n)}(0)}{(2n)!}(-x)^{2n}=\lim_{m\to\infty}\sum_{n=0}^m\frac{f^{(2n)}(0)}{(2n)!}x^{2n}.$$

If all coefficients $$a_{2n-1}$$ of Taylor expansion are zero, then the only remaining non-zero coefficients are the even ones, that is the coefficients of the form $$a_{2n}x^{2n}$$. Then notice that $$x^2=(-x)^2$$, therefore $$a_{2n}x^{2n}=a_{2n}(-x)^{2n}$$ and then \begin{align*} f(x)&=\sum_{n=0}^{\infty}a_{2n}x^{2n}\nonumber\\ &=\sum_{n=0}^{\infty}a_{2n}(-x)^{2n}\nonumber\\ &=f(-x)\nonumber \end{align*}