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I am a bit confused about this: I want to calculate the Fourier series $S^f$ of $f(x)$, where $f$ is periodic with period $k\in \mathbb{R}$. I know that the equations for my terms are: $$a_n=\frac{\color{red}{2}}{L} \int_{-L}^L f(x)\cos {\frac{n\pi x}{L}} dx$$ $$b_n=\frac{\color{red}{2}}{L} \int_{-L}^L f(x)\sin {\frac{n\pi x}{L}} dx$$ How do I know what is $L$? Example:$$ f(x)=\left\{\begin{matrix} L+x, & -L\leq x <0\\ L-x, & 0 \leq x<L \end{matrix}\right.$$ Assume that the given function is periodically extended outside the original interval. I am getting confused between $L=2 \pi$ and $L=\pi$ in this example. Another one: $$f(x)=\left\{\begin{matrix} 0, & -1\leq x <0\\ x^2/4, & 0 \leq x<1 \end{matrix}\right.$$

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  • $\begingroup$ For the first map, $L$ can't be define, in other word, you can take any $L\in\mathbb R$. For the other map, $L=2$. $\endgroup$
    – idm
    Commented Sep 8, 2014 at 13:00

2 Answers 2

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For the formula for Fourier coefficients, the function is originally defined on $[-L,L]$, and then the function is extended to a periodic function on ${\mathbb R}$.

So in your first example the $L$ in the formula for Fourier coefficients is the same as the $L$ in the definition of the function, while in the second case the $L$ in the formula for Fourier coefficients is equal to $1$.

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In math terms $[-L,L]$ is the interval where the function is defined on, which is going to be extended on $\mathbb{R}$.

Physically speaking, for example, in heat conduction, $L$ is considered to be the length of a bar with transversal section of area S. See figure.

$\hskip1.2in$enter image description here

Imagine the bar is placed over the x-axis. Then the temperature of a point x in the x-axis is represented by $u(x,t)$ at the time $t$. Notice that the temperature independs of the y and z coordinates. As shown in the next figure.

$\hskip0.5in$enter image description here

In other words we take the region $R$ determined by $0<x<L$ and $t>0$, and $\overline{R}$ to be $\lbrace 0 \leq x \leq L, t\geq 0 \rbrace $, where the function is determined.

To understand more the intuition of it, see here

In future approaches, as in disc, spheres, among others, I believe this the example above might help you see things through.

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    $\begingroup$ Very helpful. I am sure going to watch some of those videos. $\endgroup$
    – E Be
    Commented Sep 8, 2014 at 15:26
  • $\begingroup$ Glad I could help. $\endgroup$ Commented Sep 8, 2014 at 16:38

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