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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?

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

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

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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*}

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