# Thread: Even/Odd proof involving Taylor Series

1. ## Even/Odd proof involving Taylor Series

I do not understand how to go about the following proof, any suggestions?

Suppose f(x) = a0 + a1x + a2x2 + a3x3 + a4x4 + ... for x in an interval
(-R, R).

If f is even, prove that:
0 = a1 = a3 = a5 = a7 = ...

If f is odd, prove that:
0 = a0 = a2 = a4 = a6 = ...

Hint. The coefficients of a power series converging on an interval have to
be the Taylor coefficients of the function that the series converges to. Thus
it is not possible for different power series to converge to the same function
on an interval (-R, R).

2. Perhaps this is the wrong method but I'll post it up...

For the case when $f$ is even, $f(x) = f(-x)$.

$f(x) = a_0 + a_1 x + a_2x^2 + a_3 x^3 + \dots$

$f(-x) = a_0 - a_1 x + a_2 x^2 - a_3 x^3 + \dots$

So equating the series you get...

$a_0 + a_1 x + a_2x^2 + a_3 x^3 + \dots = a_0 - a_1 x + a_2x^2 - a_3 x^3 + \dots$

Canceling gives you...
$a_1 x + a_3 x^3 + \dots = - a_1 x - a_3 x^3 - \dots$

Hence equating coefficients gives...

$a_1 = -a_1$, $a_1 = 0$,

$a_3 = -a_3$, $a_3 = 0$,

etc...

Similar for the $f$ is odd case (in which $f(x) = -f(-x)$)

3. Ok, that makes sense. Thanks for the quick reply .

4. Originally Posted by crymorenoobs
Ok, that makes sense. Thanks for the quick reply .
Like I said it might not be the right method but was the first thing I thought of when I looked at it. Seems right enough though...

Perhaps this is the wrong method but I'll post it up...

For the case when $f$ is even, $f(x) = f(-x)$.

$f(x) = a_0 + a_1 x + a_2x^2 + a_3 x^3 + \dots$

$f(-x) = a_0 - a_1 x + a_2 x^2 - a_3 x^3 + \dots$

So equating the series you get...

$a_0 + a_1 x + a_2x^2 + a_3 x^3 + \dots = a_0 - a_1 x + a_2x^2 - a_3 x^3 + \dots$

Canceling gives you...
$a_1 x + a_3 x^3 + \dots = - a_1 x - a_3 x^3 - \dots$

Hence equating coefficients gives...

$a_1 = -a_1$, $a_1 = 0$,

$a_3 = -a_3$, $a_3 = 0$,

etc...

Similar for the $f$ is odd case (in which $f(x) = -f(-x)$)
Correct method although maybe a cleaner way to say it:

For the even case $2f(x)=f(x)+f(-x)=2\sum_{j=0}^{\infty}a_{2j}\implies a_{2j-1}=0$