Question regarding Fourier Series

Question

Things I understand (scroll down to see question in bold):

Let $T$ be the function's period
Let $w_0 = \frac{2π}{T}$

• A function $x(t)$ can be written as the sum of its even and odd parts, that is to say

  $x(t) = x_e(t) + x_o(t)$

  where

  $x_e(t) = \frac{1}{2} (x(t) + x(-t))$

  and

  $x_o(t) = \frac{1}{2} (x(t) - x(-t))$

(Also important to note that a function's even and odd parts will have the same value $T$ and therefore the same value $w_0$)

• An even periodic function can be written as

  $x_e(t) = a_0 + \sum ^{\infty}_{n=1} a_n\cos(nw_0t)$

  where

  $a_0 = \frac{1}{T} \int_T x_e(t) dt$

  and

  $a_n = \frac{2}{T} \int_T x_e(t)\cos(nw_0t) dt$

• An odd periodic function can be written as

  $x_o(t) = \sum^{\infty}_{n=1} b_n\sin(nw_0t)$

  where

  $b_n = \frac{2}{T} \int_T x_o(t)\sin(nw_0t) dt$

Now here is what I don't get. Why can you just use $x(t)$ as opposed to $x_e(t)$ and $x_o(t)$ when finding $a_n$ and $b_n$? If you split the function into its even and odd parts:

$a_n= \frac{2}{T} \int_T \frac{1}{2} (x(t)+x(-t))\cos(nw_0t) dt$

$b_n= \frac{2}{T} \int_T \frac{1}{2} (x(t)-x(-t))\sin(nw_0t) dt$

Because we can use $x(t)$ (for reasons unknown to me), and say

$a_n= \frac{2}{T} \int_T x(t)\cos(nw_0t) dt$

$b_n= \frac{2}{T} \int_T x(t)\sin(nw_0t) dt$

it is implied that

1) $\int_T x(-t) \cos(nw_0t) dt = \int_T x(t) \cos(nw_0t) dt$

(true for even functions, but x(t) is not necessarily even)

2) $\int_T -x(-t) \sin(nw_0t) dt = \int_T x(t) \sin(nw0t) dt$

(true for odd functions, but x(t) is not necessarily odd)

I do not see how either of these implications holds true.

Answer 1

So I posted this on the r/math subreddit and someone (u/skatanic28182) figured it out for me. Here is the link: http://www.reddit.com/r/math/comments/3519zq/question_regarding_fourier_series

Here is the proof in case you are curious (Note $\int_T$ and $\int_a^{a+T}$ express the same idea): $\int_a^{a+T} x(-t)\cos(nw_0t) dt $

Using $U$ substitution where $u=-t$

= $\int_{-a}^{-a-T}-x(u)\cos(nw_0(-u)) du$

Using the properties

• negating an integral reverses the limits

• $\cos(x) = \cos(-x)$

= $\int_{-a-T}^{-a}x(u)\cos(nw_0u) du$

Because $a$ and $u$ are essentially dummy variables

= $\int_a^{a+T}x(t)\cos(nw_0t) dt $

$\int_a^{a+T}-x(-t)\sin(nw_0t) dt =\int_a^{a+T}x(t)\sin(nw_0t) dt $ can be proven similarly