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The Arago spot is the bright axial point in the shadow of a circular obstruction under suitably coherent illumination. In ideal scalar diffraction theory, secondary fields from the circular edge arrive in phase on axis. Fresnel diffraction by an opaque disk gives an axial intensity equal to the unobstructed intensity in its paraxial model; this is not a claim that the whole geometric shadow is bright.

At the geometric shadow boundary of a straight opaque half-plane, the transmitted Fresnel integral is exactly half the unobstructed complex amplitude for ideal point illumination. The Cornu spiral chord is from its origin to one limiting endpoint, rather than between both limiting endpoints. Therefore $U/U_0=1/2$ and $I/I_0=1/4$. Half the unobstructed intensity would confuse a field fraction with its square. Edge thickness, source extent and alignment can modify the ideal result.

A spectrogram displays a signal's changing local frequency spectrum, usually by color proportional to squared short-time Fourier transform magnitude. With window function $w$, one convention isThe window selects a neighborhood of time $t$ before taking its Fourier transform. A sinusoid gives a persistent frequency ridge, while a brief pulse spreads across frequencies and occupies a short time range.

A longer window separates nearby persistent frequencies more clearly but mixes events across a longer interval. A shorter window localizes transients but broadens their spectral features. Changing the window function changes this tradeoff and the spectral leakage. Zero padding can sample the displayed frequency curve more finely; it does not add observed data or restore the resolution lost through a short measurement interval. The phase is discarded in the plotted squared magnitude, so the spectrogram should not be confused with a complete set of complex transform coefficients.

For four identical long slits of width $a$ centred at $x=\pm D/2,\pm 3D/2$, the aperture Fourier transform givesHere $\mathrm{sinc}x=\sin x/x$. Normalized to the on-axis light intensity,Principal array directions have $qD=2\pi m$; a single-slit diffraction envelope zero cancels such an order when $ma/D$ is a nonzero integer. The closest central array zeros satisfy $qD=\pm \pi /2$. Finite slit width generally shifts the actual maxima of the complete product slightly away from the nominal principal array directions.

A uniformly illuminated rectangle of width $w$ and height $h$ has field proportional to $\mathrm{sinc}(q_{x}w/2)\mathrm{sinc}(q_{y}h/2)$, with angular spatial wavenumbers $q_x,q_y$ measured in inverse length. The first envelope zeros give $w=2\pi /|q_{x,0}|$ and $h=2\pi /|q_{y,0}|$. Two identical rectangles with horizontal centre separation $s$ introduce an additional $\cos (q_{x}s/2)$ factor, so the interference period is $\Delta q_x=2\pi /s$.

The factors $2\pi$ depend on using angular spatial frequency; axes in cycles per length would use the reciprocal widths directly. Fourier light intensity alone cannot determine an overall aperture translation or distinguish a mask from its inverted copy. Extra assumptions about shape and coherent illumination are necessary for a unique geometrical reconstruction.

Extend $F_L(x)=x(L-|x|)$ from $[-L,L]$ to a periodic function of period $2L$, with $L>0$. It is odd and continuous, so its Fourier series contains only sine terms. Its Fourier coefficients areTwo applications of integration by parts prove the coefficient formula. The first has no endpoint contribution because $x(L-x)$ vanishes at both endpoints; the second leaves a multiple of $1-\cos (n\pi )$. Thus only odd frequencies survive, giving the displayed Fourier sine series. Its absolutely summable coefficients give uniform convergence, and Fourier orthogonality identifies its sum with $F_L$.

Taking $L=1$, ${\int }_{-1}^1{F}_1(x{)}^{2}dx=2{\int }_0^1{x}^2(1-x{)}^{2}dx=1/15$. Parseval's identity therefore gives

On $0\le x\le L$, the Fourier sine series of ${\sin }^2(\pi x/L)$ has zero even-index coefficients and odd-index coefficients $b_n=8/[\pi n(4-n^2)]$. Substitute $u=\pi x/L$ in the coefficient integral and use ${\sin }^{2}u\sin (nu)=\frac{1}{2}\sin (nu)-\frac{1}{4}\sin ((n+2)u)-\frac{1}{4}\sin ((n-2)u)$. For odd $n$, integrating gives $b_n=(2/\pi )[1/n-\frac{1}{2}(1/(n+2)+1/(n-2))]$; reflection about the midpoint makes the even coefficients zero. The $O(n^{-3})$ decay gives absolute and uniform convergence, and Fourier orthogonality identifies the sum. In the heat equation with zero endpoint temperatures, multiply each coefficient by $\exp [-\kappa (n\pi /L{)}^{2}t]$. With insulated endpoints, the same initial profile is instead just the constant and second cosine modes of a Fourier cosine series.

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.

On $(0,1)$, integration by parts gives the displayed Fourier sine series. For normalized Dirichlet eigenfunctions $y_i=\sqrt{2}\sin [(i+1)\pi x]$, indexed from $i=0$, it has the form $A[y_0+{\sum }_{i\ge 1}{c}_i{y}_i]$, with $A=4\sqrt{2}/{\pi }^3$ and $c_i=[1-(-1{)}^{i+1}]/[2(i+1{)}^3]$. Its Rayleigh quotient for $-d^2/d{x}^2$ is $10$, since ${\int }_0^1(u^{\prime }{)}^{2}dx=1/3$ and ${\int }_0^1{u}^{2}dx=1/30$. Reflection symmetry forces $c_1=0$, because that first excited eigenfunction is antisymmetric about $1/2$. Thus the first excited state's contribution to the error is exactly zero; the first nonzero correction is $c_2=1/27$.

An unwanted ellipse is an experimental warning: a finite-amplitude pendulum can develop additional orbit-axis precession even without Earth's rotation. A symmetrical, firm support and careful launch reduce this effect.^[ref] Distinguish that nonlinear effect from the tiny ellipse already present in the Exact linear Foucault pendulum trajectory. The linear calculation establishes a controlled approximation, not the complete motion of an arbitrary large swing.

Support anisotropy creates a second competing mechanism. As an illustrative modification, let the two horizontal restoring angular frequencies be ${\omega }_x$ and ${\omega }_y$. Even with $\alpha =0$, a diagonal release from rest givesWhen ${\omega }_x\ne {\omega }_y$, the relative phase changes on the scale $1/|{\omega }_x-{\omega }_y|$: the trace changes shape without any terrestrial Coriolis force. It need not be uniform precession of a Foucault pendulum. To cleanly resolve the terrestrial effect over times of order $1/|\alpha |$, a useful design target is $|{\omega }_x-{\omega }_y|\ll |\alpha |$. Near the equator that target becomes particularly demanding. Compare fast periods along perpendicular launch directions to look for this frequency splitting; repeat at several launch orientations to detect preferred suspension axes.

Other controls address distinct alternatives: shelter the apparatus from air currents, track the support against the building, measure both orbit axes, and compare early and late fitted directions as the amplitude decays. A sustaining drive, if used in a different apparatus, must supply mechanical energy without imposing an azimuthal bias; an undriven run provides a useful comparison. A change in swing direction alone cannot exclude these effects. Agreement of the signed rate with independently known geographical latitude and Earth's rotation, together with stable repeat launches and small competing effects, is the inference the experiment is designed to support.

For release from rest relative to the building, take $x(0)=A>0$, $y(0)=\dot{x}(0)=\dot{y}(0)=0$. The exact solution of the stated coupled linear differential equations isIndeed $u(0)=A$ and $\dot{u}(0)=i\alpha A$, so $\dot{z}(0)=0$. In the slowly rotating coordinates, the ellipse has semiaxes $A$ and $A|\alpha |/\beta$. A bob released from rest therefore does not describe a perfectly straight line even in this ideal linear model. Replacing the bracket by $\cos {\omega }_{0}t$ gives the familiar leading approximation but introduces a spurious initial transverse velocity $-i\alpha A$. The exact expression keeps the initial condition and separates that tiny orbit width from the slow rotation of its long axis.

For the illustrative $\ell =20\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the fast period $2\pi /\beta$ is approximately $8.9714\mathrm{s}$. At $45^{\circ }$ north, $|\alpha |/\beta \simeq 7.3624\times 10^{-5}$: the transverse semiaxis is too small to distinguish at the figure's scale. Nevertheless the long-axis orientation changes by about $-63.8138^{\circ }$ in six hours. Changing the sign of geographical latitude conjugates the entire solution, reflecting the trajectory across the eastward axis; setting $\alpha =0$ gives $z=A\cos {\omega }_{0}t$ exactly. These are stronger checks than merely drawing a rotating line.

The coupled equations give an additional independent constraint. Multiplying them by $\dot{x}$ and $\dot{y}$ and adding cancels the Coriolis force's contributions:This is mechanical energy per unit mass in the reduced model; precession does not require a loss of mechanical energy. Also, direct differentiation givesFor this release the constant is $\alpha A^2$. The exact trajectory satisfies both identities. They test the coupled dynamics separately from the visible slow rotation and express the fact that the Coriolis force does no work.

Use local east, north, and up unit vectors, with $x$ eastward, $y$ northward, and positive orientation $\psi$ anticlockwise as viewed from above. Let $\phi$ be signed geographical latitude, positive north, and $\Omega >0$ the magnitude of Earth's angular velocity. In the spherical-Earth approximation,Here $g$ is the local effective gravity in a rotating frame, including the steady terrestrial centrifugal acceleration. For small horizontal displacements compared with $\ell$, the horizontal restoring acceleration from tension is $-{\omega }_0^2(x,y)$. Vertical displacement is second order in the swing amplitude, so vertical velocity can be omitted at this order. The horizontal Coriolis acceleration in local Earth coordinates then gives the coupled linear differential equationsThese equations neglect damping, finite-amplitude corrections, support anisotropy, and spatial changes of the centrifugal force of order ${\Omega }^2$; they retain the leading terrestrial Coriolis force. A related derivation uses south/east axes instead.^[ref]

Introduce the complex number $z=x+iy$. Multiplying the second equation by $i$ and adding givesSet $z=e^{-i\alpha t}u$. Direct differentiation cancels the first derivative of $u$ and yieldsThe ${\alpha }^2$ in $\beta$ comes from solving these reduced equations exactly; it is not a complete prediction of terrestrial corrections of order ${\Omega }^2$. Thus the horizontal trajectory is an ellipse in coordinates rotating at angular velocity $-\alpha$ relative to the building. When $|\alpha |\ll {\omega }_0$, each rapid swing samples an almost fixed long axis; its slow precession isThe sidereal day $T_{\mathrm{sid}}=2\pi /\Omega$ is a rotation relative to distant directions. $T_{\mathrm{p}}$ denotes a full $2\pi$ turn of a continuously followed orientation; an unlabelled swing line repeats after $T_{\mathrm{p}}/2$. At the equator there is no finite precession period in this model. The Coriolis parameter is $f=2\alpha$, so the plane's rate is $-f/2$, not $-f$.

The limiting cases check the sign and scale. At the north pole $\alpha =\Omega$: the swing direction turns clockwise once per sidereal day. At the equator $\alpha =0$: both horizontal equations reduce to independent simple harmonic motions and a line launch stays on its original line. At the south pole $\alpha =-\Omega$: the direction turns anticlockwise once per sidereal day. Vanishing horizontal precession at the equator does not mean that every component of Coriolis acceleration vanishes there.

For a stipulated $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, the sidereal day is approximately $23.9345\mathrm{h}$. At $45^{\circ }$ north the predicted rate is $-10.6356^{\circ }/\mathrm{h}$ and $T_{\mathrm{p}}\simeq 33.8485\mathrm{h}$; at $45^{\circ }$ south the rate changes sign. At the illustrative geographical latitude $34^{\circ }$ south, the prediction is $+8.4109^{\circ }/\mathrm{h}$. These calculations are model predictions, not new observations of the UNSW apparatus.

A swing plane fixed among the stars is generally only a polar picture. A north–south swing at the equator turns with Earth in an inertial frame, while having no precession against the building.^[ref]

The University of New South Wales documents an undriven, visitor-started Foucault pendulum. A fixed vertical wire and wall line define its initial north–south swing plane. The documented comparison one hour later shows reduced amplitude and about nine degrees of anticlockwise precession. The apparatus swings for several hours without an electromagnetic drive.^[ref]

To turn that demonstration into a quantitative experiment, the following is a proposed measurement protocol. Establish a rigid support with no preferred horizontal swing direction, measure the wire length to the bob's centre, and mark two perpendicular horizontal axes fixed to the building. Start with a small displacement along the sighting plane, then release with as little transverse velocity as possible. Record the actual launch method; burning a fine restraining thread is one possible proposed release, not a claim about the documented UNSW launch. Keep people clear of the bob and avoid touching the suspension during the observation.

Record the horizontal position from above with a stationary, calibrated camera, or sight the bob against the wire and wall line at recorded times. Fit the long axis of each short group of swings to estimate its orientation $\psi$; use the extrema on both sides rather than following the bob's rapidly reversing velocity. A swing line has orientation modulo $\pi$, so unwrap successive small direction changes continuously. Rotate no camera or reference between measurements. Record the fast oscillation period separately, and compare the slope of $\psi (t)$ with the Latitude law of Foucault precession. Repeat launches along different directions and monitor ellipses, pivot motion, and air currents before attributing the slow change to Earth's rotation.

The condition required of a final program state when a program finishes under its precondition. The summation loop verification requires returned value $s$ to satisfy $2s=n(n+1)$ and preserves the input $n$. Relating the answer to the original input matters: allowing the program to replace $n$ by zero would change the intended contract.

A postcondition alone gives no promise of algorithm termination. It also need not specify a unique output: $s\ge 0$ allows many answers, while the exact triangular number equation determines one integer answer for each permitted $n$. Stronger postconditions express more obligations; proving only a weaker consequence cannot justify the stronger requirement.

The condition required of a starting program state for a stated correctness guarantee. In the summation loop verification, $n\in \mathbb{Z}$ and $n\ge 0$ admit $n=0$ and $n=4$, but exclude $n=-1$. The accompanying model assumes exact integer arithmetic and terminating elementary operations. The initial values of $i,s$ need no restriction because initialization overwrites them before use.

A precondition is a logical assumption, not automatically an executable check. Writing require in a procedure states its contract; an implementation that checks or rejects invalid inputs needs that behavior specified too. Weakening a precondition admits more starts and therefore demands more proof: dropping $n\ge 0$ here makes $n=-2$ a terminating counterexample to $2s=n(n+1)$, since initialization returns zero while $n(n+1)=2$.

A forest containing one minimum spanning tree of each connected component of a graph of a finite undirected weighted graph, including isolated vertices. For disconnected input, Kruskal's algorithm returns a minimum spanning forest.

For a lightly damped forced harmonic oscillator with known force amplitude $F_0$, measure the static displacement $x_s$, the resonance angular frequency ${\omega }_r$ and the peak displacement $x_p$. The response is $|x_0|=F_0/\sqrt{(k-m{\omega }^2{)}^2+b^2{\omega }^2}$. To leading order at large quality factor,The exact displacement resonance is ${\omega }_r^2=k/m-b^2/(2m^2)$, whereas the undamped natural angular frequency is ${\omega }_0=\sqrt{k/m}$. A lightly damped half-power bandwidth obeys $\Delta \omega \simeq b/m$ and $Q\simeq {\omega }_0/\Delta \omega$. Half power corresponds to displacement $x_p/\sqrt{2}$, not $x_p/2$. Graph readings should be reported as estimates rather than extra significant digits from the model inversion.

For a supported mass $m$, total spring constant $k$, and viscous coefficient $b$, a harmonic applied force has response $|x|=F_0/\sqrt{(k-m{\omega }^2{)}^2+b^2{\omega }^2}$. Far above the natural angular frequency, light damping gives $|x|\simeq F_0/(m{\omega }^2)$. For a prescribed displacement bound $x_{\max }$, this estimates $m\gtrsim F_0/({\omega }^2{x}_{\max })$, but the finite spring term can make the estimated equality slightly violate the bound. If $F_0/x_{\max }>b\omega$, the high-mass exact threshold isThis model describes a specified external force. Motion transmitted through a moving support instead has a base-displacement forcing term and a different dependence on damping.

For overdamped Brownian motion at uniform temperature in a potential energy $U$, constant diffusion coefficient $D$ gives the displayed Fokker-Planck equation, with $\beta =(k_{B}T{)}^{-1}$. The probability current is $\mathbf{J}=-D(\nabla p+\beta p\nabla U)$. Zero current recovers the Boltzmann distribution; absorbing boundaries instead permit transport and mean first-passage times. This diffusion equation should be distinguished from the similarly named coagulation equation.