9 Fourier series flow chart

(added Oct. 20, 2016)

Figure 2: Fourier series

9.1 Theorem on when we can do term by term differentiation

If \(f\left ( x\right ) \) on \(-L\leq x\leq L\) is continuous (notice, NOT piecewise continuous), this means \(f\left ( x\right ) \) has no jumps in it, and that \(f^{\prime }\left ( x\right ) \) exists on \(-L<x<L\) and \(f^{\prime }\left ( x\right ) \) is either continuous or piecewise continuous (notice, that \(f^{\prime }\left ( x\right ) \) can be piecewise continuous (P.W.C.), i.e. have finite number of jump discontinuities), and also and this is very important, that \(f\left ( -L\right ) =f\left ( L\right ) \) then we can do term by term differentiation of the Fourier series of \(f\left ( x\right ) \) and use \(=\) instead of \(\sim \). Not only that, but the term by term differentiation of the Fourier series of \(f\left ( x\right ) \) will give the Fourier series of \(f^{\prime }\left ( x\right ) \) itself.

So that main restriction here is that \(f\left ( x\right ) \) on \(-L\leq x\leq L\) is continuous (no jump discontinuities) and that \(f\left ( -L\right ) =f\left ( L\right ) \). So look at \(f\left ( x\right ) \) first and see if it is continuous or not (remember, the whole \(f\left ( x\right ) \) has to be continuous, not piecewise, so no jump discontinuities). If this condition is met, look at see if \(f\left ( -L\right ) =f\left ( L\right ) \).

For example \(f\left ( x\right ) =x\) on \(-1\leq x\leq 1\) is continuous, but \(f\left ( -1\right ) \neq f\left ( 1\right ) \) so the F.S. of \(f\left ( x\right ) \) can’t be term be term differentiated (well, it can, but the result will not be the Fourier series of \(f^{\prime }\left ( x\right ) \)). So we should not do term by term differentiation in this case.

But the Fourier series for \(f\left ( x\right ) =x^{2}\) can be term by term differentiated. This has its \(f^{\prime }\left ( x\right ) \) being continuous, since it meets all the conditions. Also Fourier series for \(f\left ( x\right ) =\left \vert x\right \vert \) can be term by term differentiated. This has its \(f^{\prime }\left ( x\right ) \) being P.W.C. due to a jump at \(x=0\) but that is OK, as \(f^{\prime }\left ( x\right ) \) is allowed to be P.W.C., but it is \(f\left ( x\right ) \) which is not allowed to be P.W.C.

There is a useful corollary that comes from the above. If \(f\left ( x\right ) \) meets all the conditions above, then its Fourier series is absolutely convergent and also uniformly convergent. The M-test can be used to verify that the Fourier series is uniformly convergent.

9.2 Relation between coefficients of Fourier series of \(f\left ( x\right ) \) Fourier series of \(f^{\prime }\left ( x\right ) \)

If term by term differentiation allowed, then let

\begin{align*} f\left ( x\right ) & =\frac {a_{0}}{2}+\sum _{n=1}^{\infty }a_{n}\cos \left ( n\frac {\pi }{L}x\right ) +b_{n}\sin \left ( n\frac {\pi }{L}x\right ) \\ f^{\prime }\left ( x\right ) & =\frac {\alpha _{0}}{2}+\sum _{n=1}^{\infty }\alpha _{n}\cos \left ( n\frac {\pi }{L}x\right ) +\beta _{n}\sin \left ( n\frac {\pi }{L}x\right ) \end{align*}

Then

\begin{align*} \alpha _{n} & =nb_{n}\\ \beta _{n} & =-na_{n}\end{align*}

And Bessel’s inequality instead of \(\frac {a_{0}^{2}}{2}+\sum _{n=1}^{\infty }\left ( a_{n}^{2}+b_{n}^{2}\right ) <\infty \) now becomes \(\sum _{n=1}^{\infty }n^{2}\left ( a_{n}^{2}+b_{n}^{2}\right ) <\infty \). So it is stronger.

9.3 Theorem on convergence of Fourier series

If \(f\left ( x\right ) \) is piecewise continuous on \(-L<x<L\) and if it is periodic with period \(2L\) and if on any point \(x\) on the entire domain \(-\infty <x<\infty \) both the left sided derivative and the right sided derivative exist (but these do not have to be the same !) then we say that the Fourier series of \(f\left ( x\right ) \) converges and it converges to the average of \(f\left ( x\right ) \) at each point including points that have jump discontinuities.