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$$ f(t) =\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos(n\pi tL)+\sum_{n=1}^\infty b_n \sin(n\pi tL) $$

In something like this, the coefficients are found for $\cos(1\pi tL)$, $\cos(2\pi tL)$ $\cos(3\pi tL)$ ...

Is there some reason $n$ needs to be a whole number, why can't n be a fraction $\cos(0.5\pi tL)$, $\cos(1\pi tL)$ $\cos(1.5\pi tL)$ ...

If this works, does it lead to faster convergence?

Does this also hold for the complex Fourier series?

vonbrand
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calcUser98
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    A Fourier series should be a periodic function, something such that $f(t + L) = f(t)$ for all $t \in \mathbb{R}$. $\cos(1.5 \pi t L)$ is not periodic in $L$. – Joppy Mar 21 '20 at 23:58
  • Why is it required to be periodic? Is it because the function you are try to approximate is periodic? isn't it very rarely true that you're trying to approximate a periodic function? – calcUser98 Mar 22 '20 at 00:04
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    If your function is not periodic then what you will want is the Fourier Transform. It is formulated as an integral, so is not confined to whole numbers. The series is a special case. https://en.wikipedia.org/wiki/Fourier_transform – Ali Mar 22 '20 at 00:06
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    I thought with Fourier series you were always trying to approximate something that is periodic. It may be that the original function is only defined over an interval like $[0, L]$, but the assumption is that it is extended outside this interval to the whole of $\mathbb{R}$ by periodicity. – Joppy Mar 22 '20 at 00:21
  • Part of the question is: convergence to what? What problem are you solving? If, for example, you are solving the Poisson equation, then the entire point of the method is to write down the solution as a combination of eigenfunctions of the Laplacian satisfying the homogeneous boundary condition that you have imposed. This determines the appropriate form of the eigenfunctions in the expansion. With periodic boundary conditions in 1D, the ordinary Fourier series is suitable. – Ian Mar 22 '20 at 00:37
  • The frequencies in Fourier series are whole numbers for the same reason that the exponents in polynomials (and power series) are whole numbers, namely, it's the simplest choice for getting the job done. Anyway, your suggestion of using half-integers reduces to the original on expressing everything in terms of $L^$, defined by letting $L=2L^$. – Gerry Myerson Mar 22 '20 at 00:38
  • @Joppy as it is written the period of the function is not $L$ but rather its inverse – lcv Mar 22 '20 at 00:38
  • (Cont.) Different forms of the Fourier series are obtained for homogeneous Dirichlet boundary conditions (sine series) or homogeneous Neumann boundary conditions (cosine series). Still other homogeneous boundary conditions will give less familiar Fourier series. – Ian Mar 22 '20 at 00:38

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