Under mild conditions regarding a piecewise smooth periodic function #f#, the Fourier series of its derivative #f'# can be determined from the Fourier series of #f# by differentiating term by term.
Let #L# be a positive number and #f# a real continuous #2L#-periodic function that is piecewise smooth on #\ivcc{-L}{L}#. If #a_0#, #a_1#, #b_1#, #a_2#, #b_2,\ldots# are the Fourier coefficients of #f#, then the Euler coefficients #a'_0#, #a'_1#, #b'_1#, #a'_2#, #b'_2,\ldots# of the derivative #f'# of #f# satisfy
\[\begin{array}{rcl} \displaystyle a_0' = 0,\quad a_k=-\frac{b_k '}{k}, \quad b_k=\frac{a_k '}{k} \end{array}\]
If, in addition, #f'# is piecewise smooth on #\ivcc{-L}{L}#, then the Fourier series #s_{f'}(x)# of #f'# is
\[s_{f'}(x) = \sum_{n=1}^{\infty} \frac{n\pi}{L}\left(b_n \cos\left(\frac{n\pi x}{L}\right) -a_n \sin\left(\frac{n\pi x}{L}\right)\right) \]
This means that the Fourier series of #f'# can be obtained by differentiating the Fourier series of #f# term by term.
It suffices to prove the theorem in the case where #L=\pi#, since, #g(x) = f\left (\frac{L}{\pi }x\right)# has period #2\pi# and, from the result for #g#, the result for #f# immediately follows since #f'(x) = \frac{\pi}{L} g'\left(\frac{\pi x}{L} \right)# by the chain rule for differentiation. Therefore, we assume #L = \pi#.
By the Euler formulas and integration by parts for piecewise smooth functions, we have
\[\begin{array}{rcl}\displaystyle a'_0 &=&\displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f'(x) \dd x \\&&\phantom{xx}\color{blue}{\text{Euler formula}}\\&=&\displaystyle\frac{1}{\pi}\left( f(\pi)-f(-\pi)\right)\\&&\phantom{xx}\color{blue}{\text{Fundamental theorem of Calculus and continuity of }f}\\&=&0\\ &&\phantom{xx}\color{blue}{f(-\pi) = f(\pi)\text{ because }f\text{ is } 2\pi\text{-periodic}}\\ \\ \displaystyle a'_n &=& \displaystyle\frac{1}{\pi} \int_{-\pi}^{\pi} f'(x)\cos(nx)\dd x \\&&\phantom{xx}\color{blue}{\text{Euler formula}}\\&=& \displaystyle\frac{1}{\pi} f(x)\cos(nx) \Bigr|_{-\pi}^{\pi} + \frac{n}{\pi} \int_{-\pi}^{\pi} f(x)\sin(nx) \dd x \\&&\phantom{xx}\color{blue}{\text{integration by parts applied; }f, \cos(nx) \text{ are continuous and piecewise smooth}}\\&=& nb_n \\&&\phantom{xx}\color{blue}{\text{Euler formula}}\\ \\ b'_n &=& \displaystyle\frac{1}{\pi} \int_{-\pi}^{\pi} f'(x)\sin(nx) \dd x \\&&\phantom{xx}\color{blue}{\text{Euler formula}}\\&=& \displaystyle\frac{1}{\pi} f(x)\sin(nx) \Bigr|_{-\pi}^{\pi} - \frac{n}{\pi} \int_{-\pi}^{\pi} f(x)\cos(nx) \dd x\\&&\phantom{xx}\color{blue}{\text{integration by parts applied; }f, \sin(nx)\text{ are continuous and piecewise smooth}}\\ &=& -na_n\\&&\phantom{xx}\color{blue}{\text{Euler formula}}\end{array}\]
Under the additional assumption that #f'# is piecewise smooth on #\ivcc{-L}{L}#, Dirichlet's Theorem applies to #f'#, so #f'# has a Fourier series that converges to #f'(x)# at all points where #f'# is continuous. Hence the Fourier series #s_{f'}(x)# for #f'# is
\[ \begin{array}{rcl}\displaystyle s_{f'}(x)&=&\displaystyle \frac{a'_0}{2} + \sum_{n=1}^{\infty}(a'_n\cos(nx) +b'_n\sin(nx))\\& =&\displaystyle\sum_{n=1}^{\infty} n(b_n \cos(nx) -a_n \sin(nx)) \end{array}\]
This is exactly what we obtain by differentiating the Fourier series of #f# term by term.
Under the conditions of this theorem, the Fourier coefficients #a'_n#, #b'_n# for the derivative #f'# of #f# satisfy
\[a'_0= 0, \quad a'_n=\frac{n\pi b_n}{L} \quad \text{ and } \quad b'_n=-\frac{n\pi a_n}{L}\quad \text{ for } \quad n=0,1,2,\ldots\]
This follows immediately from equating the expression for #s_{f'}(x)# of the statement to \[\frac{a'_0}{2} + \sum_{n=1}^{\infty}\left(a'_n\cos\left(\frac{n\pi x}{L}\right) +b'_n\sin\left(\frac{n\pi x}{L}\right)\right)\]
Convergence of coefficients as discussed in Dirichlet's theorem for Fourier series gives that the Fourier coefficients of a continuous piecewise smooth function with a piecewise smooth derivative converge to #0# as #n \rightarrow \infty#:
\[\lim_{n \rightarrow \infty} n a_n =\lim_{n \rightarrow \infty} n b_n = 0\]
Consider the function
\[f(x) = \begin{cases} 0 & -\pi \leq x \lt 0 \\ 1 & 0 \leq x \lt \pi \end{cases}\]
The Fourier series for this function, extended to the whole line by #2\pi#-periodicity, is
\[ \frac{1}{2}+ \sum_{n = 1}^{\infty} \dfrac{1-(-1)^n}{\pi n} \sin(nx) \]
Differentiating this series term by term, as outlined in the theorem, we obtain
\[ \sum_{n = 1}^{\infty} \dfrac{1-(-1)^{n}\cos(nx)}{\pi} \]
But this series does not even converge at #x=0#. Also the cosine coefficients do not converge to #0# for #n\to\infty#. So this series is not the Fourier series of the derivative of #f#. The theorem cannot be applied because the condition that #f# be continuous is not satisfied.
Let #m\ge1#. The theorem can be applied #m# times to derive that, for a real #2\pi#-periodic #m+1# times differentiable function #f# such that its #(m+1)#-st derivative is piecewise continuous, the Fourier series of the first #m# derivatives can be computed by differentiating the Fourier series of #f# term by term. All of these #m# series converge to the corresponding derivatives. Moreover, the Fourier coefficients of #f# satisfy the relations
\[\lim_{n \rightarrow \infty} n^m a_n = \lim_{n \rightarrow \infty} n^m b_n = 0 \]
The convergence of all the Fourier series follows again from application of Dirichlet's Theorem. The relations describing the behaviour at infinity of the Fourier coefficients can be deduced by iteratively applying the relation found in the theorem. Thus,
\[a_n= -\frac{b'_n}{n} = -\frac{a''_n}{n^2} = \frac{b'''_n}{n^3} = \cdots = \frac{\alpha^{(m)}_n}{n^m}\]
\[b_n= \frac{a'_n}{n} = -\frac{b''_n}{n^2} = -\frac{a'''_n}{n^3} = \cdots = \frac{\beta^{(m)}_n}{n^m}\]
where #a'_n#, #a''_n, \ldots, b'_n, b''_n# are the Fourier coefficients of the functions #f'#, #f'', \ldots#, and #\alpha^{(m)}_n, \beta^{(m)}_n, \ldots# are the Fourier coefficients of #f^{(m)}# taken with the appropriate sign.
By uniform convergence of the Fourier series, which we will prove later, the first #m-1# series converge uniformly.
****MOVE THIS****
Consider the #2\pi#-periodic function #f# determined by \[ f(x)=\text{sign}(x) = \begin{cases} -1 & -\pi \leq x \leq 0 \\ 1 & 0 \lt x \lt \pi \end{cases} \]
It is the derivative of the #2\pi#-periodic function #F# determined by \[ F(x)=|x| \text{ for } x \in \ivcc{-\pi}{\pi} \] whose Fourier series is
\[ s_F(x) = \frac{\pi}{2} - \frac{4}{\pi} \sum_{n = 0}^{\infty} \frac{\cos((2n+1)x)}{(2n+1)^2} \]
Calculate the Fourier series of #f#.
#s_f(x) =# # \frac{4}{\pi}\sum_{n=1}^{\infty} \frac{\sin((2n+1)x)}{2n+1} #
The function #F# satisfies the conditions of the theorem
Derivatives of Fourier series. Hence, knowing that its Fourier series is \( \frac{\pi}{2} - \frac{4}{\pi} \sum_{n = 0}^{\infty} \frac{\cos((2n+1)x)}{(2n+1)^2} \) and using term by term differentiation, we find that the Fourier series of #f# is \[\begin{array}{rcl} s_{f}(x)&=&\displaystyle \frac{\dd }{\dd x}\left(\frac{\pi}{2} - \frac{4}{\pi} \sum_{n = 0}^{\infty} \frac{\cos((2n+1)x)}{(2n+1)^2}\right) \\ &=&\displaystyle - \frac{4}{\pi} \sum_{n = 0}^{\infty} \frac{\dd }{\dd x}\left(\ \frac{\cos((2n+1)x)}{(2n+1)^2}\right) \\ &=&\displaystyle \frac{4}{\pi}\sum_{n=1}^{\infty} \frac{\sin((2n+1)x)}{2n+1} \\ \end{array} \]
Differentiation of Fourier series
