If $f(x)=\sum\limits_{n=0}^\infty a_n x^n$ is even and continous, then $a_n=0$ for odd n, and if f is an odd function, then $a_n=0$ for even n.

Question

I need help with this problem:

Show that If $f(x)=\sum\limits_{n=0}^\infty a_n x^n$ is an even and continous function, then $a_n=0$ for odd $n$, and if f is an odd function, then $a_n=0$ for even $n$.

I tried solving it by using the fact that if $f(x)$ is even, the $f(x)=f(-x)$, and ended up with this:$$f(x)=\sum_{n=0}^\infty a_n x^n = a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...$$ and $$f(-x)=\sum_{n=0}^\infty a_n (-x)^n = a_0-a_1x+a_2x^2-a_3x^3+a_4x^4-...$$ where for odd $n$, the terms in $f(-x)$ are negative. I did the same for odd functions $f(-x)=-f(x)$: $$f(-x)=\sum_{n=0}^\infty a_n (-x)^n = a_0-a_1x+a_2x^2-a_3x^3+a_4x^4-...$$and$$-f(x)=\sum_{n=0}^\infty -(a_n (x)^n) = -a_0-a_1x-a_2x^2-a_3x^3-a_4x^4-...$$After that I add both,like this:

if $f$ is even $2f(x)=f(x)+f(-x)$ thus $$f(x)=\frac{f(x)+f(-x)}{2}=a_0+a_2x^2+a_4x^4+a_6x^6+...$$

and if $f$ is odd $2f(x)=f(-x)+(-f(x))$ thus $$f(x)=\frac{f(-x)+(-f(x))}{2} = -a_1x-a_3x^3-a_5x^5-a_7x^7-...$$ I'm stuck here, I don't know what else should I do. Help me please.

It looks like your work is correct. Im not sure what else you are looking for. — D.B., Feb 13 '19 at 05:19
I guess you don't know coefficients of a power series are unique? — YuiTo Cheng, Feb 13 '19 at 05:19
In your llast couple lines, it should be $$-2f(x) = f(-x)-f(x).$$ — D.B., Feb 13 '19 at 05:20
I'm supposed to show that if $f$ is even then $a_n=0$ for odd n and that if $f$ is odd then $a_n=0$ for even n. — davidllerenav, Feb 13 '19 at 05:21
This is exactly what you showed. In your series expansion, you only have even powers for your even function. — D.B., Feb 13 '19 at 05:22
@D.B. Why? I'm adding both of them, and both of them are equal, so wouln't it be $2f(x)$? — davidllerenav, Feb 13 '19 at 05:23
@davidllerenav See https://math.stackexchange.com/questions/856932/can-there-be-more-than-one-power-series-expansion-for-a-function — YuiTo Cheng, Feb 13 '19 at 05:25
If $f$ is odd, then,$$-2f(x) = f(-x)-f(x).$$ If $f$ is even, $$2f(x) = f(-x)+f(x).$$ — D.B., Feb 13 '19 at 05:25
@D.B. Oh, I see. So there isn't any odd powers on the even function, so for odd $n$, $a_n=0$? — davidllerenav, Feb 13 '19 at 05:26
@D.B. So all the terms would be positive instead of negative? — davidllerenav, Feb 13 '19 at 05:27
@D.B. Ok! So all I have to do know is wrtie down in words that for even functios, if n is odd then $a_n=0$ and the same for the other case, right? — davidllerenav, Feb 13 '19 at 05:28

If $f(x)=\sum\limits_{n=0}^\infty a_n x^n$ is even and continous, then $a_n=0$ for odd n, and if f is an odd function, then $a_n=0$ for even n.

0 Answers0