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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!

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    $\begingroup$ What do you mean by "proper"? It sure is a Fourier series (the general form is $$a_0/2+\sum_{n=1}^{\infty}\Big(a_n\sin\frac{n\pi x}{L}+b_n\cos\frac{n\pi x}{L}\Big) ~).$$ $\endgroup$ – Nikolaos Skout Mar 23 '16 at 16:00
  • $\begingroup$ Wait, so we can count ${\frac9 \pi} {\frac 1 {2n - 1}} {\sin\left(\frac{2n - 1} 2 \pi \right)}$ as $a_n$, and $-{\frac9 \pi} {\frac 1 {2n - 1}} {\cos\left(\frac{2n - 1} 2 \pi \right)}$ as $b_n$? Huh...did not know that... Always thought the coefficients weren't supposed to contain functions... $\endgroup$ – Arjuna Mar 23 '16 at 16:09
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    $\begingroup$ basically, note that the expressions $\sin(\ldots),~ \cos(\ldots)$ you noted above are not functions but ... numbers (there is no $x$)! $\endgroup$ – Nikolaos Skout Mar 23 '16 at 16:10
  • $\begingroup$ Right. That just kicked me in the head just now. Thanks a lot for your help! $\endgroup$ – Arjuna Mar 23 '16 at 16:13
  • $\begingroup$ you're welcome ;-) ! $\endgroup$ – Nikolaos Skout Mar 23 '16 at 16:14
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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.

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Note that the two trigonometric factors that do not contain the variable $t$ can be simplified as follows

$$ \cos\left(\frac{2n-1}2 \pi\right) = 0 \space, \forall \space n \in \mathbb{Z^+} $$

$$ \sin\left(\frac{2n-1}2 \pi\right) = (-1)^{n-1} \space, \forall \space n \in \mathbb{Z^+} $$

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This allows us to simplify $a_n$ and $b_n$ as follows

$$ a_n = \frac{9(-1)^{n-1}}{\pi\left(2n-1\right)} $$

$$ b_n = 0 $$

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