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A Fourier series separates a periodic physical signal into harmonics. For example, a periodic voltage applied to a linear circuit can be treated one frequency at a time; the harmonic responses then superpose to give the output waveform.

Write $w=\Delta T$ for the pulse width and $\delta =w/T$ for its duty cycle, with $0<w<T$. The waveform is a train of positive rectangular pulses centred on $t=jT$, each with area one. It is even about $t=0$, so every sine coefficient vanishes. Cosine orthogonality givesand, for $n\ge 1$,This is the Fourier series of a periodic rectangular pulse:The actual constant term in the series is $a_0/2=1/T$, not $a_0$. As usual for a jump discontinuity, the series converges to the average of the one-sided values at each pulse edge, namely $1/(2w)$, irrespective of the value assigned at that isolated endpoint. Elsewhere it converges to the rectangular waveform.

Counting the constant as the first term, the first three terms areso their sum isIf “first three” instead means the first three nonconstant harmonics, add $a_3\cos (6\pi t/T)$; the original plot shows that alternative partial sum as well. The three-term approximation rounds the narrow peaks and ripples between them; many harmonics are needed to represent a sharp pulse.

The original sketches use the illustrative duty cycle $\delta =0.15$ and cover $-T\le t\le T$. The coefficient panel also displays the first spectral zero.

For $n\delta \ll 1$, expand the sinc function:Thus the low-index coefficients, including $a_0$, have nearly equal values, although their constant-series contribution still has the factor $1/2$. The first sinc zero occurs at $n\delta =1$, or $n=T/w$. If this is not an integer, the closest integer coefficient is near zero rather than exactly zero: writing $n\delta =1+\epsilon$ gives $a_n\simeq -(2/T)\epsilon$. This pulse-width scale of Fourier coefficients means that a narrower pulse requires significant harmonics extending to proportionally larger indices, the reciprocal time-width/frequency-width relation.

For $T>0$, values of a function $f$ on a half-open interval $[a,a+T)$ determine its periodic extension to the real line: choose the unique integer $k$ with $x-kT\in [a,a+T)$ and set $F(x)=f(x-kT)$. The resulting periodic function has period $T$. If the endpoint limits of $f$ differ, the extension has a jump discontinuity at every period boundary. Under the usual piecewise-smooth convergence hypotheses, its Fourier series takes the average of the two one-sided limits there; an assigned value at an isolated endpoint does not change the Fourier coefficients.

Consider the advection equation $u_t+a{u}_x=0$ with constant speed $a>0$, no sources, and periodic boundary conditions on $[0,1]$. Use $N$ equal cells of width $h=1/N$, time step $k>0$, and indices modulo $N$. The exact solution is the periodic translation $u(x,t)=u_0(x-at)$. At a face, the Riemann problem transports the left state to the right, so the two-state upwind numerical flux is $F(L,R)=aL$, with $F(U,U)=aU$.^[ref]

The finite volume method with this numerical flux and an explicit Euler method step givesHere $U_j^n$ represents a cell average; the recurrence also occurs in the upwind finite difference scheme. Impose the specific CFL condition $0<\nu \le 1$. Each updated cell average is then a convex combination of two old ones. This proves preservation of their initial minimum and maximum, including nonnegativity for nonnegative data, and contraction of the maximum norm of the difference of two solutions on this periodic grid. These statements concern this constant-speed scalar update with this time step.

For an original synthetic example, take $N=8$, $a=1$, $h=1/8$, $k=1/16$, hence Courant number $\nu =1/2$. Choose a function constant on each successive cell, with the following values, so these are exact initial cell averages. One step givesFor example, cell zero receives from cell seven across the periodic face: $U_0^1=(0+1)/2=1/2$. All eight face transfers cancel in the total by shared face flux cancellation. More generally this periodic recurrence conserves the total algebraically for any $\nu$, even though the CFL condition is needed for the convex combination argument. With the same initial data and $\nu =3/2$, one step gives $U_2^1=-1$ and $U_4^1=5/2$, despite preserving the total.

After four steps, $t=1/4$, the exact translated profile has moved two cells to the right. Repeated averaging instead gives the explicit binomial coefficient formulawith all indices modulo eight. The diagram compares these computed cell averages with the exact translated cell averages. Their integrals agree, but the numerical jumps spread across cells. The finite volume method is useful for jump discontinuities because it retains the integral balance even at a jump; conservation alone does not guarantee a sharp profile. At $\nu =1$ this particular update is an exact one-cell shift, whereas for $0<\nu <1$ it mixes neighboring states. No accuracy conclusion at a jump follows just from the conserved total.

A real-valued function $f:D\to \mathbb{R}$ is a strictly increasing function if every pair $x<y$ in its domain of a function satisfies $f(x)<f(y)$. It is therefore injective, and each output has at most one preimage. Its inverse function on its image of a function is also strictly increasing. Every strictly increasing function is a nondecreasing function; the converse fails when outputs repeat.

The polynomial $f(x)=x^3$ is a strictly increasing function on the real line, although its derivative $f^{\prime }(x)=3x^2$ is zero at zero. Indeed, for $x<y$,because $x^2+xy+y^2=(x+y/2{)}^2+3y^2/4$ can vanish only when $x=y=0$. Thus a positive derivative everywhere is sufficient on a real interval, but is not necessary for a strictly increasing function.

Strict monotonicity does not ensure continuity: $g(x)=x$ for $x<0$ and $g(x)=x+1$ for $x\ge 0$ is a strictly increasing function with a jump discontinuity. Its image of a function omits $[0,1)$, so the inverse function exists only on the actual image of a function, rather than on all the real numbers.

A vanishing moment is a cancellation condition: positive and negative parts of an analyzing function balance when weighted by a monomial. Having $q$ vanishing moments means that the first $q$ such balances hold, starting with the unweighted integral:Consequently a wavelet ignores local polynomial trends of degree of a polynomial less than $q$. A smooth feature is close to its Taylor polynomial, so only the small Taylor remainder contributes to its detail coefficient. A jump discontinuity cannot be approximated this way across its edge; its jump response of a localized analyzing function can remain large relative to details from a smooth region.^[ref]

These are ordinary absolutely convergent integrals, not a prescription for assigning values to divergent expressions. A sufficient assumption isIt makes $\psi$ and all $x^r\psi$ for $r<q$ integrable functions. Being square-integrable alone does not ensure these weighted integrals exist. For a square-integrable function with compact support, the Cauchy-Schwarz inequality guarantees every such weighted integral is finite. Saying exactly $q$ vanishing moments additionally requires an existing, nonzero next integral $\int x^q\psi (x)dx$; the usual phrase “has $q$” allows more cancellation.