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:
$$f(x)=\left\{\begin{matrix}
0, & -1\leq x <0\\
x^2/4, & 0 \leq x<1
\end{matrix}\right.$$
[Math] the $L$ in the Fourier series term
calculusfourier series
Related Solutions
Since you define the Fourier series via Lebesgue integral, you should be able to observe the following general fact:
Suppose $f$ and $g$ are periodic of period $T$. Also suppose that $f$ is Lebesgue integrable on its period. If the set $\{x:f(x)\ne g(x)\}$ has Lebesgue measure zero, then the Fourier series of $g$ is the same as the Fourier series of $f$.
Indeed, modifying a function on a null set does not change any of the integrals that determine the Fourier coefficients.
In your example $f$ is identically $1$, $g$ is the Dirichlet function, and $T$ can be any rational number.
Yes, the Fourier series of a discontinuous function need not converge to that function pointwise. (Even for some continuous functions the pointwise convergence fails, though examples are harder to come by.)
"Periodically extended" in the context of Fourier series, in an intuitive sense, basically is like copying and pasting the graph over an interval to all of $\mathbb{R}$, with a potential slight alteration.
A periodic extension that isn't really said to be anything special is generally just assumed to be straight up copying and pasting. In other words, the function appears the same on $[0,L]$ as on $[L,2L]$ or $[-L,0]$, aside from just being at different x-values.
An "even" periodic extension first extends the function by flipping it horizontally onto an interval of length $2L$ (assuming we start at $[0,L]$ - the idea is similar on other intervals, just the math becomes a little nastier). So in other words, if we have $a \in [0,L]$, then $f(\alpha) = f(- \alpha)$ - because we extend it so it becomes an even function. And then we just "copy and paste" this length $2L$ interval all over the real line.
An "odd" periodic extension has it flip horizontally and vertically (across the y-axis) first and sticks it to either end of the function, onto an interval of length $2L$. So basically it's like the even extension, but you flip it upside down too, to give you an odd function. So if we have $\alpha \in [0,L]$, then $f(\alpha) = -f(- \alpha)$.
A more graphic example, using $f(x) = |x|$ over $[0,2]$:
Insofar as uniqueness is concerned, I believe when this came up in my class, it was insofar as the period of the function was concerned. Having a full copy of the question might be more helpful since this was a few months ago for me, though, just for clarity's sake.
But if so, if I remember correctly, was mostly a matter of considering two functions, $f,g$, which were the same function but you chose a different period for each (say $p$ for one and $kp, k \in \mathbb{Z}$ for another) and then worked through the formulas and definitions to show that the choice of period was irrelevant by both functions ultimately having the same Fourier expansion.
Best Answer
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 $1$.