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Direct differentiation givesOn the unit disk $D$, polar coordinates give ${\int }_D{y}^{2}dA=\pi /4$, ${\int }_{D}xdA=0$ and ${\int }_{D}dA=\pi$. Consequently the volume integral isFor an independent surface integral calculation, the relevant cross product isThe bottom face contributes ${\int }_{D}xydA=0$ and the top contributes ${\int }_D(-h{y}^2-xy)dA=-\pi h/4$. On the cylindrical side, put $x=\cos \theta$, $y=\sin \theta$, $\hat{\mathbf{n}}=(x,y,0)$ and $dS=d\theta dz$. The flux density is $x^3+2xy{z}^2+2y^{2}z-x{y}^2$. All terms except $2y^{2}z$ integrate to zero because they are odd functions of $x$ or $y$, givingThus minus the total outward flux is $-(\pi h^2-\pi h/4)$, in agreement with the volume integral.

The real odd function has a purely imaginary odd Fourier transform. Setting $A=\pi /2$ and using a product-to-sum identity,The apparent singularities are removable. Direct evaluation or limits giveIn particular, the transform stays finite where its numerator and denominator both vanish.

Use the nonpolynomial function $f(x)=1+\sin x$ on $[-1,1]$. Its exact integral isbecause the sine is an odd function.

For every uniform partition of an interval with $N\ge 1$ panels, the quadrature nodes satisfy $x_{N-j}=-x_j$. Their composite trapezoidal rule quadrature weights agree, so the two sampled sine values cancel. If there is a node at zero its sine value is already zero. The constant term is integrated exactly, givingThis is symmetric trapezoidal exactness for odd functions, applied after adding a constant function so that the resulting definite integral is nonzero. It is exactness for this particular integrand, rather than a claim that every nonpolynomial function is integrated exactly.

First take $a\ne 0$, put $A=|a|$, and suppose $k>0$. The force is the negative derivative of the potential energy:The potential energy is an odd function, passes through $U(0)=0$, and approaches zero from above as $x\to -\infty$ and from below as $x\to +\infty$. Its stationary points are $x=\pm A$:Thus the sole stable equilibrium is the local minimum at $x=A$; the local maximum at $x=-A$ is an unstable equilibrium. The requested values at the stable equilibrium areThe force is an even function, positive for $|x|<A$, negative for $|x|>A$, and zero at $x=\pm A$. Its central maximum is $F(0)=k/A^2$. Sinceits two minima occur at $x=\pm \sqrt{3}A$, with $F=-k/(8A^2)$, and $F\to 0^{-}$ at both ends. These values fix the shapes of both requested sketches.

In one dimension, attraction towards the origin means $xF(x)<0$, while repulsion from the origin means $xF(x)>0$. With that explicitly stated convention,A positive force alone means rightward, not necessarily outward. At the origin $F(0)\ne 0$, so it is not an equilibrium; the inward/outward distinction is undefined exactly there.

The sign assumptions matter because the paper calls $a,k$ merely constants. For $k<0$ both curves reverse sign, the stable equilibrium is $x=-A$, its potential energy is $-|k|/(2A)$, and the attraction/repulsion intervals interchange. For $k=0$ the force vanishes everywhere on the nonsingular domain and there is no isolated potential energy minimum. If $a=0$ and $k\ne 0$, then $U=-k/x$ for $x\ne 0$ and $F=-k/x^2$, with no finite equilibrium.

Use the period-two Fourier series conventionAway from its jumps this square wave is an odd function, so $a_0=a_n=0$. Its Fourier sine coefficients areOnly odd Fourier harmonics remain. ThereforeThe Fourier series equals $f(x)$ away from the jumps; at a jump it gives the midpoint of the one-sided limits by Dirichlet's theorem on Fourier series. The printed definition leaves the values at the integers unspecified. Choosing finite values there consistent with period two changes neither the Fourier coefficients nor these limiting values.

The given square wave agrees almost everywhere with an odd function on $(-\pi ,\pi )$; its assigned value at the single point $0$ does not affect any Fourier coefficient. Thus $a_0=a_n=0$, whileThe Fourier series is thereforeThe Dirichlet convergence theorem for Fourier series applies because the periodic extension is piecewise smooth. It gives $f(x)$ at each continuity point and the midpoint $0$ at the jumps, including $x=0$; it does not reproduce the assigned value $f(0)=-1$ there.

At the continuity point $x=\pi /2$, $\sin ((2r+1)\pi /2)=(-1{)}^r$ and $f(\pi /2)=1$. ThusNo evaluation at a discontinuity or rearrangement of the conditionally convergent series is involved.

Successive jumps occur at times separated by $T=6$ s. Between $-3$ and $3$ s the plotted function is $f(t)=2-t$, periodically extended, with maximum $5$ and minimum $-1$. Its mean is therefore $2$ m, and $f(t)-2=-t$ is an odd function over that centered period.

For the Fourier series convention$a_0=4$ m and $a_r=0$ for every $r\ge 1$; the nonconstant content consists of sine terms. If $a_0$ denotes the mean directly in the alternative convention $f=a_0+\sum \cdots$, its value is $2$ m. Stating the convention resolves the otherwise ambiguous notation.

For completeness, integrating $-t$ against the sine basis gives $b_r=6(-1{)}^r/(\pi r)$ m. At a jump the Fourier series converges to the midpoint $2$ m. The nonzero mean does not destroy the oddness argument for the mean-subtracted function; declaring the original function itself odd would incorrectly force its mean to vanish.

A periodic function repeats exactly after a nonzero shift of its input. For $f:\mathbb{R}\to Y$, a positive period of a function $T$ satisfiesOne interval of length $T$ therefore determines the whole function. For another domain of a function $D$, a period must also preserve the domain, $D+T=D$; equality only where two domains happen to overlap is weaker. Periodicity is an exact translation of a function symmetry, rather than merely two similar-looking pieces of its graph.

The sine $\sin (2\pi x)$ has fundamental period one. The function ${\sin }^2(2\pi x)$ repeats after half a unit: the sine changes sign under that shift, but squaring removes the sign. The mixed function $g(x)=2+\sin (2\pi x)+\frac{1}{2}\cos (4\pi x)$ again has fundamental period one. All positive integer multiples of a period are periods, so specifying one period need not identify the shortest repetition.

If the input is time and $T_0$ is the fundamental period, the frequency is $1/T_0$ and the angular frequency is $2\pi /T_0$. Repetition lets a Fourier series describe a cycle using sines and cosines; the period average separates the reference level from the oscillating terms. Strang develops this representation and its Fourier coefficients in §4.1 of Computational Science and Engineering by Gilbert Strang.^[ref]

This connection grew out of a physical problem. In his 1822 book Théorie analytique de la chaleur by Joseph Fourier, Joseph Fourier used periodic functions to extend profiles beyond their initial interval while treating heat conduction. Chapter III, §§233–234 combines even functions and odd functions into sine–cosine expansions and rescales their period.^[ref] A modern Fourier series interpretation still requires the appropriate convergence conditions.

A constant function has every real shift as a period and no fundamental period. Without continuity, even a nonconstant periodic function can lack a smallest positive period; periodic function without a least positive period gives a concrete example. The period group of a function organizes all these possibilities.

For a real angle $\theta$, $\sin \theta$ is the vertical coordinate on the unit circle reached by a counterclockwise rotation through $\theta$ radians. For an acute angle in a right triangle, it is the opposite side divided by the hypotenuse. In particular,The first-quarter rise and second-quarter fall are visible in the circle projection; the lower semicircle gives negative values.

Reflecting the unit circle across the horizontal axis gives $\sin (-\theta )=-\sin \theta$, so sine is an odd function. A full turn gives $\sin (\theta +2\pi )=\sin \theta$, making it a periodic function. For real arguments, its range is $[-1,1]$ and its zeros of a function occur at $\theta =k\pi$ for integers $k$. The cosine describes the perpendicular coordinate of the same point, and the Pythagorean trigonometric identity keeps their two values tied together.

A square wave alternates between two constant levels for equal halves of a period. Choose a $2\pi$-periodic function equal to $1$ on $(0,\pi )$ and $-1$ on $(-\pi ,0)$, with value zero at the jumps. Its odd function symmetry makes the mean and every cosine Fourier coefficient vanish. Directly integrating the two constant pieces givesThus only odd sine Fourier harmonics occur, with amplitudes decreasing like the reciprocal of their frequency multiplier. Every finite Fourier partial sum is smooth, yet their pointwise limit has jumps. At those jumps the Fourier series converges to zero, the midpoint of the two levels, regardless of which value we assign to the original function there.

This example explains how smooth components can encode sharp transitions. Increasing the number of Fourier harmonics narrows the transition region, but also produces the persistent overshoot described by the Gibbs phenomenon. The illustrated square wave and coefficient calculation are mathematical constructions, not measurements of a physical device.

On a partition of an interval symmetric about zero, the composite trapezoidal rule gives opposite quadrature nodes equal quadrature weights. Samples of an odd function cancel pairwise, and a zero quadrature node contributes zero. For a continuous odd function, its exact integral over $[-L,L]$ is also zero. Adding any affine function preserves exactness by linearity of the integral and of the quadrature rule. For example, $1+\sin x$ is nonpolynomial but both the exact integral and every uniform-grid composite trapezoidal rule value on $[-1,1]$ are two.

Two square-integrable random variables $X,Y$ are uncorrelated random variables when $\mathrm{Cov}(X,Y)=0$. Independent random variables with finite second moments are uncorrelated random variables, but zero covariance does not generally imply independence.

For an explicit counterexample, let $X$ take $-1$, $0$, $1$ with equal probabilities, and let $Y=X^2$. Then $\mathbb{E}X=\mathbb{E}{X}^3=0$, so $\mathrm{Cov}(X,Y)=\mathbb{E}{X}^3-\mathbb{E}X\mathbb{E}{X}^2=0$. Nevertheless, $Y$ is completely determined by $X$: $\mathbb{P}(Y=0\mid X=0)=1$, while $\mathbb{P}(Y=0)=1/3$. Thus these are uncorrelated random variables with statistical dependence. In this example, opposite contributions from the odd function $x^3$ cancel; the even function $x^2$ still retains information about $|X|$. The last panel of the covariance diagram shows the relationship.

There are useful special cases. Uncorrelated jointly normal variables are independent, and the Bernoulli independence from zero covariance result supplies another setting where the converse holds. Their distribution assumptions do the extra work; zero covariance alone does not.

For the half-line wave equation $u_{tt}=c^2{u}_{xx}$ with a zero Dirichlet boundary condition at $x=0$, extend both the initial displacement and initial velocity as odd functions to the full line. D'Alembert's formula then gives an odd solution in $x$, so $u(0,t)=0$. Restriction to $x\ge 0$ solves the desired initial-boundary value problem. With zero initial velocity, the displayed formula shows reflection at the fixed endpoint with reversal of the displacement sign. Continuous piecewise smooth displacement data give a weak solution; a classical solution requires sufficient additional regularity. In particular, a kink in the initial displacement can prevent the pointwise time derivative at that kink from taking the assigned initial velocity, even though the initial velocity is attained in the $L^2$ sense.