Odd & Even Function Power Series

Question

Let $g$ be some power series $g(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$, $\forall x$.

(a) If $g$ is an odd function, prove that it must be the case that $$ a_{0} = a_{2} = a_{4} = \cdots $$

(b) If $g$ is an even function, prove that it must be the case that $$ a_{1} = a_{3} = a_{5} = \cdots $$

Thanks!

Answer 1

• $f$ is odd if and only if $f(x)+f(-x)=0$ for all $x$. This yields the identity $$2\sum_{k=0}^\infty a_{2k}x^{2k}=0$$ which should need no further explanation.

• $f$ is even if and only if $f(x)-f(-x)=0$ for all $x$. This yields the identity $$2\sum_{k=0}^\infty a_{2k+1}x^{2k+1}=0$$ which should need no further explanation.

Answer 2

Suppose $g$ is an odd function.
Then, $g(x)+g(-x)=0$ for any $x$ whose absolute value is sufficiently small.
$\sum_{n=0}a_nx^n+\sum_{n=0}a_n(-x)^n=0$ for any $x$ whose absolute value is sufficiently small.
$\sum_{n=0}\left(a_n+(-1)^na_n\right)x^n=0$ for any $x$ whose absolute value is sufficiently small.
If $n$ is odd, then $a_n+(-1)^na_n=0$.
If $n$ is even, then $a_n+(-1)^na_n=2a_n$.
So, $2a_0+0x+2a_2x^2+0x^3+2a_4x^4+\dots=0$ for any $x$ whose absolute value is sufficiently small.
$0+0x+0x^2+\dots$ is the power series expansion of the zero function at $0$.
And $2a_0+0x+2a_2x^2+0x^3+2a_4x^4+\dots=0+0x+0x^2+\dots$ for any $x$ whose absolute value is sufficiently small.
By the uniqueness of power series expansion, $2a_0=2a_2=\dots=0$.
So, $a_0=a_2=\dots=0$.