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Let $f(x)$ be the sum function of the power series
$$\sum_{n=0}^{\infty} a_nx^n$$
on $(-R,R)$ for $R>0$. If $f(x)+f(-x)=0$
for all $x∈(-R,R)$, prove that $a_n=0$ for all even $n$

Now What I have done is as follows:-

As we know, $1/(1+x) =$ $$\sum_{n=0}^{\infty} (-1)^nx^n$$

Similarly I write,

$$\sum_{n=0}^{\infty} a_nx^n$$ $=1/(1-a_nx^n)=f(x)$, $x∈(-R,R)$

then $f(-x)=$ $$\sum_{n=0}^{\infty} (-1)^na_nx^n$$
$=1/(1+a_nx^n)$, $x∈(-R,R)$

On the basis of given information we have
$f(x)+f(-x)=0$

$→$ $$\sum_{n=0}^{\infty} a_nx^n$$ $+$ $$\sum_{n=0}^{\infty} (-1)^na_nx^n$$ $=0$
$→$ $1/(1-a_nx^n)+1/(1+a_nx^n)=0$
$→$ $2/(a_nx^n)=0$

$→$ $2*1/a_n*x^-n=0$

My question is from here can we conclude that the $a_n=0$ for all even $n$ ?
And if it is not the correct way please someone suggest me how to proceed. Thank you in advance.

Here is my second approach

If I break the general terms in this way then the following stuffs come:-

If $f(x)= $$\sum_{n=0}^{\infty} a_nx^n$
$→$ $=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...$

On the other hand I write

$f(-x)= $$\sum_{n=0}^{\infty}(-1)^n a_nx^n$
$→$ $=a_0-a_1x+a_2x^2-a_3x^3+a_4x^4-a_5x^5+...$

Now from $f(x)+f(-x)=0$ we have
$2*[a_0+a_2x^2+a_4x^4+a_6x^6+...]=0$
$→$ $2*$$\sum_{n=0}^{\infty} a_2nx^{2n}=0$

Now $x^{2n}$ are positive terms so that $a_n=0$ for even $n$. Is it correct?

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  • $\begingroup$ Hint: $f$ is odd. There's a quick kill to this $\endgroup$ – Sean Roberson Dec 19 '17 at 21:15
  • $\begingroup$ Yes $f$ is odd. but I'm still confused to proceed. $\endgroup$ – vbm Dec 19 '17 at 21:17
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Your first approach fails since you have used the following step in your working: $$\sum_{n=0}^{\infty} a_nx^n=\frac1{1-a_n x^n}$$ This is false, since the right hand side has some dependence on $n$ whereas the left hand side is a sum over $n$, so the final result doesn't depend on any particular value of $n$. You can't really deduce anything of this form from the sum without knowing something further about the coefficients.

Your second method is good, and is the way to go. You have $$a_0+a_2x^2+a_4 x^4+\cdots=0$$ which holds for all $x$. Straight away this implies that $a_0=0$, since this must hold for $x=0$. So you have $x^2(a_2+a_4 x^2+\cdots)=0$ which holds for all $x$ (in particular, in the region close to $0$), and so this implies that $a_2+a_4x^2+\cdots=0$ also. So from this you can conclude (by setting $x=0$) that $a_2=0$. You can continue this for all further even coefficients.

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  • $\begingroup$ Thank you so much for help $\endgroup$ – vbm Dec 19 '17 at 21:27
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The second approach is correct. For $x\rightarrow 0$ and using continuity, we conclude that $a_{0}=0$. So we have $\displaystyle\sum_{n=1}a_{2n}x^{2n}=0$. For all $x$ with $0<|x|<R$, then $\displaystyle\sum_{n=1}a_{2n}x^{2n-2}=0$, taking $x\rightarrow 0$ again, we have $a_{2}=0$, proceed in this eay we get all $a_{2n}=0$, $n=0,1,...$

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  • $\begingroup$ Thank you so much for help $\endgroup$ – vbm Dec 19 '17 at 21:23

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