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The period-one extension of $x(1-x)$ on $[0,1]$ has Fourier series $1/6-{\pi }^{-2}{\sum }_{n\ge 1}{n}^{-2}\cos (2\pi nx)$. Endpoint agreement makes the periodic function continuous. Its derivative series is $(2/\pi ){\sum }_{n\ge 1}{n}^{-1}\sin (2\pi nx)$, which equals $1-2x$ in the open unit interval and converges to zero at integers, the midpoint of the derivative's jump. Termwise differentiation of a Fourier series applies locally away from the corners.

The period-$L$ extension of $f(x)=Kx/L$ on $-L/2<x<L/2$ has Fourier coefficients $a_0=a_n=0$ and $b_n=K(-1{)}^{n+1}/(\pi n)$. ConsequentlyThe Fourier series converges to the ramp away from its jumps and to zero at every jump, the average of the one-sided limits. For $K\ne 0$, Termwise differentiation of a Fourier series fails here as an ordinary convergent series: the differentiated summands have nondecaying amplitude $2K/L$. The classical derivative equals $K/L$ between jumps, whereas the periodic distributional derivative includes negative point masses of weight $K$ at the jumps.

Let $S_N(x)={\sum }_{n=1}^N[a_n\cos (nx)+b_n\sin (nx)]$. Under the first coefficient bounds,This proves the uniform Cauchy condition on the whole real line and hence uniform convergence to $f$. It is the Weierstrass M-test applied with the summable majorant $2c{n}^{-1-\epsilon }$. Every partial sum is continuous. To see continuity of the limit explicitly, at a chosen $x_0$ first choose $N$ with small uniform error; then make $|S_N(x)-S_N(x_0)|$ small by continuity. The inequalityproves continuity. Also $S_N(x+2\pi )=S_N(x)$ for every $N$, so taking limits gives $f(x+2\pi )=f(x)$.

Under the stronger coefficient bounds, the derivative series has summable uniform majorant $2c{n}^{-1-\epsilon }$. Thus $S_N^{\prime }$ converges uniformly to a continuous function $g$ given by that series. The original series also converges uniformly. On any finite interval, the fundamental theorem of calculus gives $S_N(x)-S_N(0)={\int }_0^x{S}_N^{\prime }(t)dt$. Uniform convergence permits passage to the limit inside this finite integral, since the error is at most $|x|\Vert S_N^{\prime }-g{\Vert }_{\infty }$. Therefore $f(x)-f(0)={\int }_0^{x}g(t)dt$, and continuity of $g$ gives $f^{\prime }=g$. This justifies Termwise differentiation of a Fourier series rather than assuming it:

The endpoint agreement $f(1)=f(0)=0$ makes the boundary terms vanish when integrating the Fourier coefficients of $f^{\prime }$ by parts. If $a_n^{\prime },b_n^{\prime }$ denote these coefficients, thenThese are exactly the coefficients obtained by differentiating the preceding Fourier series:On every compact subinterval of $(0,1)$ the differentiated series converges uniformly by the Dirichlet test, since partial sums of the sine factors are uniformly bounded there. The Termwise differentiation of a Fourier series is therefore valid away from integers and gives $1-2x$ on $(0,1)$. At an integer the periodic original has a corner, so the differentiated series is not asserting the existence of an ordinary derivative there.

Expand the unknown in Fourier modes:By Termwise differentiation of a Fourier series,Also,Therefore the coefficient of $e^{i\pi mx}$ on the left-hand side isThe right-hand side has coefficientsbecause the coefficient of each of $e^{\pm i\pi nx}$ is $1/(n^2+1)$ and the constant coefficient is one. Hence the Fourier spectral system for a cosine-modulated second derivative isIts coefficient array is an infinite tridiagonal matrix.

For a spectral truncation, retain the modes $-M\le n\le M$, setand impose the displayed coefficient equation for every $-M\le m\le M$. This gives an explicit $(2M+1)$-dimensional tridiagonal system. The resulting finite Fourier series automatically satisfies the periodic boundary conditions.

Termwise differentiation of a Fourier series multiplies a mode by its wavenumber, soThe convolution theorem turns pointwise multiplication into discrete convolution:soThe assumed real analyticity justifies these termwise operations through rapid decay of the Fourier coefficients.

Not as ordinary convergent Fourier series, for a nontrivial moving ramp. Formal spatial differentiation would giveand formal time differentiation multiplies that expression by $-v$. The summands have nondecaying amplitude and do not tend to zero, so the ordinary series cannot converge. The periodic extension has jumps, violating the regularity needed for Termwise differentiation of a Fourier series.

Away from the moving jumps, the classical derivatives exist and are $g_x=K/L$ and $g_t=-vK/L$. A distributional derivative also includes the Dirac delta distribution contributions at the jumps, so distributional differentiation is possible, but it is a different interpretation. If $v=0$, the time derivative is trivially zero; if $K=0$, both derivatives are zero.

Termwise differentiation of a Fourier series, applied to the continuous periodic extension in part (c), gives for $-1<x<1$Hence the required Fourier sine series isThe endpoint deserves care: at $x=1$ every term is zero, so the Fourier series sums to $0$, the average of the periodic one-sided limits $1$ and $-1$. It therefore does not reproduce the specified endpoint value $g(1)=1$. Changing a value at one point does not change any Fourier coefficient, so this is still the Fourier series of the stated function on $(-1,1]$.