Time-Shifted Trigonometric Fourier Series Coefficients I'm trying to find the Trigonometric Fourier series coefficients for a particular periodic function.
Given
$$f(t) = 2 - \frac9\pi \sum_{n=1}^\infty\frac1{2n-1}\sin\left(\frac{2n-1}2 \pi t\right)$$
The $a_0$, $a_n$, and $b_n$ coefficients would be
$$a_0=2$$
$$a_n=0 \ for \ all \ positive \ n$$
$$b_n=-\frac9\pi\frac1{2n-1} \ for \ all \ positive \ n$$
But in the case that I need to find the coefficients for $f(t - 1)$, how would I go about doing it?
My first thought was to substitute the value of $t - 1$ into the equation, followed by trigonometric identity to expand the function, giving:
$$f(t-1) = 2 - \frac9\pi \sum_{n=1}^\infty\frac1{2n-1}\sin\left(\frac{2n-1}2 \pi (t-1)\right)$$
$$= 2 - \frac9\pi \sum_{n=1}^\infty\frac1{2n-1}\sin\left(\frac{2n-1}2 \pi t - \frac{2n-1}2 \pi \right)$$
$$= 2 - \frac9\pi \sum_{n=1}^\infty\frac1{2n-1}\left[\sin\left(\frac{2n-1}2 \pi t\right)\cos\left(\frac{2n-1}2 \pi\right) - \cos\left(\frac{2n-1}2 \pi t\right)\sin\left(\frac{2n-1}2 \pi\right)\right]$$
At which point I get stuck. I can't find a way to factorize the cosine and sine terms out to form a "proper" Fourier Series representation in order to determine the coefficients.
Is there something that I'm doing very wrong here? Any help you can provide would be great! Thanks in advance!
 A: You are actually very close to solving the problem
$$f(t-1) = {2 - \frac9\pi \sum_{n=1}^\infty\frac1{2n-1}\left[\sin\left(\frac{2n-1}2 \pi t\right)\cos\left(\frac{2n-1}2 \pi\right) - \cos\left(\frac{2n-1}2 \pi t\right)\sin\left(\frac{2n-1}2 \pi\right)\right] }$$
$$\space$$
The new coefficients are as follows
$$ a_0 = 2 $$
$$ a_n = \frac{9}{\pi\left(2n-1\right)}\sin\left(\frac{2n-1}2 \pi\right) $$
$$ b_n = \frac{-9}{\pi\left(2n-1\right)}\cos\left(\frac{2n-1}2 \pi\right) $$
You might think that it is wrong to have $cosine$ and $sine$ factors in $a_n$ and $b_n$, but it is perfectly acceptable as these are not functions of the variable $t$, so relative to $t$ they are constants.
$$\space$$
Note that the two trigonometric factors that do not contain the variable $t$ can be simplified as follows
$$ \cos\left(\frac{2n-1}2 \pi\right) = 0 \space, \forall \space n \in \mathbb{Z^+} $$
$$ \sin\left(\frac{2n-1}2 \pi\right) = (-1)^{n-1} \space, \forall \space n \in \mathbb{Z^+} $$
$$\space$$
This allows us to simplify $a_n$ and $b_n$ as follows
$$ a_n = \frac{9(-1)^{n-1}}{\pi\left(2n-1\right)} $$
$$ b_n = 0 $$
