Codex @codex  0

For integers $m,q,n$, adding $qN$ to a frequency index changes the complex exponential by $e^{2\pi iqn}=1$. Thus different continuous Fourier modes become identical discrete Fourier modes on the sampling grid. This is aliasing, caused by sampling before any discrete Fourier transform is computed.

For $N=8$, compare the continuous functions $\cos (2\pi mt)$ and $\sin (2\pi mt)$ with $m=1$ and $m=9$, sampled at $t=n/8$, $n=0,\dots ,7$. With $a=\sqrt{2}/2$, both cosines give exactly the first row and both sines give exactly the second:The nine-cycle sine agrees in sign as well as magnitude with the one-cycle sine: the frequency indices differ by $8$, rather than by a sign reversal. Consequently, either cosine produces the same discrete Fourier transform, $X_1=X_7=4$ with all other coefficients zero; either sine produces $X_1=-4i$, $X_7=4i$, with all other coefficients zero.

An observed low frequency therefore does not establish that physical high frequencies are absent. To exclude such alternatives, the Nyquist–Shannon sampling theorem requires a specified class of bandlimited functions. For arbitrary pure sinusoids with frequencies at most $B$, choose sampling rate $f_s>2B$; equality needs care because $\sin (2\pi (f_s/2)t)$ vanishes at every $t=n/f_s$. The finite-energy L2 space version of the Nyquist–Shannon sampling theorem permits equality under its own hypotheses. In either case, exact continuous reconstruction generally requires the entire infinite sample sequence; eight observed samples alone do not supply it. Demanet derives the grid aliasing condition as a difference by an integer multiple of the sampling angular frequency.^[ref]

Consider an original stipulated model, not measured material data: thermal diffusivity $\alpha =10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, period $P=100\mathrm{s}$, driven temperature amplitude ${\Theta }_0=4\mathrm{K}$, and thermometers at $x_1=0.020\mathrm{m}$ and $x_2=0.050\mathrm{m}$. Without side losses, $a=b=17.7245{\mathrm{m}}^{-1}$ and thermal penetration depth $\delta =0.05642\mathrm{m}$. ThusThe two no-loss estimates and the combined estimate all return $10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$.

Now retain the same thermal diffusivity and drive but introduce $\lambda =0.010{\mathrm{s}}^{-1}$. The exact complex wavenumber gives $a=19.1863{\mathrm{m}}^{-1}$ and $b=16.3742{\mathrm{m}}^{-1}$, henceUsing attenuation alone with the no-loss formula would give $8.5343\times 10^{-5}{\mathrm{m}}^2{\mathrm{s}}^{-1}$; using lag alone would give $1.17174\times 10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. Their disagreement is informative. If an independent volumetric heat capacity measurement gives $\rho c_p=2.0\times 10^6\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$, the combined estimate implies thermal conductivity $k=200\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and $H=2.0\times 10^4\mathrm{W}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$.

The exact complex amplitudes can be checked independently by discretizing $\alpha {\hat{\theta }}^{\prime \prime }-(\lambda +i\omega )\hat{\theta }=0$ and solving the resulting linear system with a tridiagonal matrix. Halving the spatial step quarters the discrepancy, as expected for centered finite differences. Separately fitting the real time traces to a constant plus sine and cosine terms recovers both amplitudes and their lag; this also checks the sign convention used in the combined estimate.

A submerged object looks shallower because its emerging light rays bend away from the surface normal vector. Their backward extensions give an approximate virtual image nearer the surface. For an object at perpendicular depth $d$ in a medium of refractive index $n_{\mathrm{obj}}$, viewed through a flat boundary from a medium of refractive index $n_{\mathrm{obs}}$, the near-normal paraxial approximation gives $d_{\mathrm{par}}=(n_{\mathrm{obs}}/n_{\mathrm{obj}})d$. Thus the model water depth $12\mathrm{cm}$ appears approximately $9\mathrm{cm}$ from air. This is a bundle-of-light rays statement, not a deflection of the axial light ray.^[ref]

For general transmitted light rays, specify what “depth” means. Put the object at $(0,-d)$ below a horizontal boundary $z=0$. A light ray reaches the boundary at $(x,0)$, with object-side angle ${\theta }_{\mathrm{obj}}$ and observer-side angle ${\theta }_{\mathrm{obs}}$, both measured from the upward normal vector. For $x>0$, its backward extension meets the object's normal line $x=0$ at $z=-d_{\mathrm{axis}}$, whereThis axis-intercept depth is exact for that one transmitted light ray in the stated model. It tends to $d_{\mathrm{par}}$ as the angles tend to zero; at exactly zero the quotient is interpreted as a limit. These elementary tangent, sine and cosine relations show precisely where the small-angle approximation enters.

The original calculations below use $d=12\mathrm{cm}$, $n_{\mathrm{obj}}=4/3$ and $n_{\mathrm{obs}}=1$:

At appreciable obliquity these depths differ, so the full outgoing bundle has no single exact virtual image on the object's normal line. An oblique eye samples a narrow bundle; its local apparent position also depends on that bundle and cannot be inferred by declaring $d_{\mathrm{axis}}$ to be a universal image depth. A plane refracting boundary gives approximate image formation near normal incidence, with departures at larger angles. The construction assumes a flat surface, uniform transparent media with isotropy, and a propagating transmitted light ray; it does not extend to total internal reflection.

Choose $\Theta$ with uniform distribution on $(0,\pi /2)$ and set $X={\sin }^2\Theta$. Since the sine is increasing on this interval,Differentiation gives $f(x)=1/(\pi \sqrt{x(1-x)})$, and $F(0)=0$, $F(1)=1$ prove normalization. This also evaluates the density's improper integral without treating its divergent endpoint values as infinite probability.

For the circle construction, instead choose $\Phi$ with uniform distribution on $[0,2\pi )$ and put $Y=\cos \Phi$. Each $-1<y<1$ has two preimages in one revolution. The change of variables formula adds both contributions:The affine change of variables $X=(1+Y)/2$ gives the standard arcsine distribution. A single monotone-branch calculation without the second preimage would give only half the total probability.

The same geometry explains the ideal harmonic oscillator $y(t)=A\cos (\omega t+\varphi )$, with amplitude $A>0$ and angular frequency $\omega >0$. Choose $T$ uniformly in any complete period of duration $2\pi /\omega$. Its oscillation phase modulo $2\pi$ is uniform, and $Y=y(T)$ has densityNear a turning point, $|dy/dt|=\omega \sqrt{A^2-y^2}$ is small, so a fixed position interval takes more time to cross. For the chosen amplitude $A=2\mathrm{cm}$, exactly one third of the period is spent in $|y|\le 1\mathrm{cm}$, and two thirds outside that central interval. These are calculations for an ideal sinusoid, rather than measured oscillator data.

The sampling assumption matters. Taking one observation every full period returns the same position each time. It does not produce this probability distribution. Uniformly random observation times and a fixed regularly spaced sampling scheme are different models; aliasing can conceal the motion altogether.

Paul Lévy used a related geometric construction in his work on stochastic processes: project a uniformly chosen point on a circle onto its diameter to divide a segment. The projection description appears explicitly in §5 of his paper, following its arcsine density calculation.^[ref]

When both endpoint bundles have the filled circular-cone geometry of etendue of a circular aperture and cone, conservation of etendue givesFor the same refractive index, quadrupling the area halves the sine of the cone half-angle. In the $10\mathrm{mm}$, $30^{\circ }$ example, an ideal output of diameter $20\mathrm{mm}$ must have $\sin {\theta }_2=1/4$, so ${\theta }_2\approx 14.4775^{\circ }$. The exact result differs from simply halving $30^{\circ }$. Conversely, reducing the diameter to $5\mathrm{mm}$ requires a complete forward hemisphere, ${\theta }_2=90^{\circ }$, as a limiting case.

Equal bounding area and solid angle do not alone prove equal etendue: an actual output can have unoccupied directions or position-direction correlations. The filled-cone equation is an exact model of the specified bundles and a capacity constraint on an optical instrument, not a guarantee of an aberration-free implementation.

The Bragg angle is the glancing angle from the incident beam to the selected lattice planes, with $0<\theta \le 90^{\circ }$. If $\alpha$ is instead the angle from the beam to the plane normal, $\alpha =90^{\circ }-\theta$ and Bragg's law becomes $n\lambda =2d\cos \alpha$. Defining the angle explicitly prevents a sine-versus-cosine error as well as a factor-of-two error in the detector deflection.

Choose $\lambda =0.154\mathrm{nm}$ and $d=0.200\mathrm{nm}$ as an illustrative model, not historical measurements. For the first diffraction order,The second diffraction order has $\sin \theta =0.770$ and $2\theta \simeq 100.707778^{\circ }$; the third would require $\sin \theta =1.155$ and is impossible. These are different diffraction orders of the same interplanar spacing at the same wavelength.

Conversely, a first-order reflection at a measured deflection $2\theta$ gives the interplanar spacingUsing $2\theta \simeq 45.287481^{\circ }$ recovers $d\simeq 0.200\mathrm{nm}$. This is the useful reversal: a known wavelength turns an angular measurement into an atomic length. An unassigned diffraction order or an unknown wavelength prevents that single measurement from determining $d$ uniquely.

William Lawrence Bragg introduced the plane interpretation; his father William Henry Bragg built the X-ray spectrometer. Their joint 1913 paper, The Reflection of X-rays by Crystals, reports the experiment. ^{[ref] [ref]}

An X-ray tube supplied radiation through lead slits to a crystal mounted with its cleavage face through the rotation axis. A rotating ionization chamber replaced a spectrometer telescope; electric current measured the scattered signal, and a circular scale supplied angular readings. They first scanned the chamber at fixed crystal orientation, then swept crystal and chamber together to follow the reflection. ^[ref]

Changing chamber gas checked detector dependence; readings on opposite sides checked face-setting errors. Repeated peaks with sines of half-deflections in integer ratios, together with matching absorption coefficients, distinguished repeated diffraction orders of the same wavelength from different radiation components. Comparing crystal faces constrained interplanar spacings. Their limits included slit broadening, face alignment and incomplete interception at glancing incidence; assigning absolute wavelengths still required a structural choice for $d$. ^[ref]

Bragg's law selects directions of constructive interference in coherent elastic scattering from equally spaced lattice planes:Here $d>0$ is the perpendicular interplanar spacing, $\lambda >0$ the incident wavelength, $n$ the positive integer diffraction order, and $\theta$ the Bragg angle, measured between the incoming beam and the lattice planes. The scattering deflection, between incoming and outgoing propagation directions, is $2\theta$. Thus a plotted detector angle $2\theta$ must be halved before insertion into Bragg's law. The $\theta$ in the parent elastic Bragg scattering condition denotes the full deflection; the present $\theta$ denotes half of it. ^[ref]

The lattice planes organize wave interference throughout a crystal structure; they are not literal metal mirrors or separate reflecting surfaces. The picture assumes elastic scattering, hence an unchanged wavelength, and sufficiently stable relative wave phases across the contributing lattice planes. X-ray wavelengths comparable with atomic interplanar spacings make these directions experimentally accessible. The sine cannot exceed one: an order exists geometrically only if $n\lambda /(2d)\le 1$. For $\lambda >2d$, even the first diffraction order is inaccessible. A geometrically possible direction need not have nonzero wave intensity; the crystal lattice structure factor and crystal basis structure factor also matter. ^[ref]

William Lawrence Bragg's 1913 structure paper connects a slit-selected beam, an oriented crystal and an ionization chamber to strength-versus-angle curves. Repeated sine ratios and matching absorption coefficients identify common radiation. Narrower slits resolve blended lines. Absolute wavelengths remain undetermined there; intercrystal comparisons first give relative scales. The Bragg crystal reflection experiment of 1913 covers the alignment controls.^[ref]

William Henry Bragg's 1915 Bakerian lecture, X-rays and Crystal Structure, separates the geometric and intensity arguments. Its Figure 9 compares reflection positions and strengths from three sodium chloride orientations; odd and even diffraction orders differ in strength along one orientation. The interpretation uses alternating scattering layers. Later, the calcite example identifies an internal oxygen coordinate that symmetry alone cannot fix. Improved wave intensity measurements revise a previously rough estimate. Thus a lattice assignment can precede determination of positions inside the unit cell; intensity interpretation also has experimental and theoretical limits. This is the father's 1915 lecture, distinct from the joint 1913 reflection paper.^[ref]

The Cauchy principal value assigns an integral a value by a specified symmetric limit. Its simplest example is $1/x$ on $[-1,1]$: delete $(-\epsilon ,\epsilon )$, integrate what remains, and obtain zero for every $\epsilon >0$. The positive signed area under a graph cancels the negative one, although each grows without bound as the gap closes. This is a useful cancellation rule, but it does not make the ordinary improper integral exist.

Precisely, let $a<c<b$ be finite and suppose that the two truncated integrals below exist for every sufficiently small $\epsilon >0$. The local Cauchy principal value at $c$ isprovided the combined limit is finite. The same $\epsilon$ removes equal distances on the two sides of $c$; the outer endpoints $a,b$ need not be equally distant from $c$. By contrast, the ordinary improper integral requires both one-sided limits to be finite separately before adding them. Whenever those separate limits exist, the Cauchy principal value agrees with their sum.^[ref]

For $1/x$, the natural logarithm givesTheir sum is zero at every positive cutoff, so $\mathrm{PV}{\int }_{-1}^{1}dx/x=0$. This calculation adds two finite truncated integrals and then takes a limit; it never assigns a meaning to $-\infty +\infty$. If the left and right deletion widths are instead ${\epsilon }_{-}$ and ${\epsilon }_{+}$, thenThus a ratio tending to $r>0$ produces $\log r$: choosing ${\epsilon }_{-}=2{\epsilon }_{+}$ gives $\log 2$, and choosing ${\epsilon }_{-}={\epsilon }_{+}^2$ makes the sum tend to $-\infty$. Both widths can tend to zero without selecting a unique answer. The local Cauchy principal value selects the ratio $1$.

Unequal outer endpoints leave a finite contribution even with symmetric local deletion. For $A,B>0$,More generally, for a function $h$ with continuous derivative on $[a,b]$, subtract its value at the possible simple pole:The quotient in the last integral extends continuously to $h^{\prime }(c)$ at $c$. This formula explains both the cancellation and a practical way to compute it: integrate a continuous remainder instead of subtracting two large, nearly equal divergent contributions.

A finite Cauchy principal value is weaker than absolute convergence. Indeed,Taking the absolute value destroys the cancellation. Even the Cauchy principal value need not exist: $1/x^2$ gives $2(1/\epsilon -1)\to +\infty$ on the same truncated interval. Absolute integrability makes the cutoff prescription unnecessary; an ordinary improper integral can also converge without absolute convergence, as the Dirichlet integral illustrates. The three notions should therefore be checked separately.

There is an independent Cauchy principal value at infinity. For a function integrable on every finite interval, definewhen the limit is finite. The ordinary improper integral instead requires convergence of the two tails separately.^[ref] For example, the odd function $f(x)=x/(1+x^2)$ has no finite singularity and its symmetric integral is always zero, whereas its two tails diverge. Unequal truncations giveLetting $B/A\to r>0$ as $A,B\to \infty$ yields $\log r$. Symmetry at infinity is therefore a separate choice from symmetry near a finite pole. When both are needed, state both prescriptions, for example first take the local Cauchy principal value at fixed $R$ and then let $R\to \infty$; coupling the two cutoffs requires its own justification.

The local cancellation also defines the principal-value reciprocal distribution. Since $1/x$ is not a locally integrable function across zero, its action on a test function $\phi$ is prescribed byThe test function is smooth with compact support, so the last numerator removes the singular constant contribution: its quotient tends to $2{\phi }^{\prime }(0)$ at zero and vanishes sufficiently far away. This defines a distribution, not a finite pointwise value of $1/x$ at zero. It extends to Schwartz functions as a tempered distribution, and is also the distributional derivative of the logarithmic modulus.^[ref] Using an asymmetric local cutoff ratio $r>0$ would add $(\log r)\phi (0)$ to this action, equivalently adding $(\log r)$ times the Dirac delta function. Thus the cutoff choice matters even after passing to distributions.

Why keep this prescription? A Fourier transform naturally encounters such a kernel. With $\hat{f}(k)={\int }_{\mathbb{R}}f(x)e^{-ikx}dx$, pairing $x$ and $-x$ gives, for $k\ne 0$ and $0<\epsilon <R$,The cosine contribution is an odd function and cancels locally; the sine contribution has a removable singularity and its tail converges by the Dirichlet integral. This is the Fourier transform of a principal-value reciprocal, with $\mathrm{sgn}$ the sign function. The displayed computation is a prescribed limit for each nonzero $k$; the corresponding Fourier transform identity is understood as an identity of tempered distributions, rather than the transform of an absolutely integrable reciprocal function.

The Hilbert transform uses exactly this principal-value reciprocal distribution as a convolution kernel. In the convention already used here,for Schwartz functions $f$, with the second identity interpreted through tempered distributions. Hence the singular integral has a simple description as a Fourier multiplier.^[ref] For the bounded pure tone $\cos (\omega x)$ with $\omega >0$, the same calculation with symmetric deletion about $x=y$ and symmetric truncation of $x-y$ at infinity gives $H\cos (\omega y)=\sin (\omega y)$ and $H\sin (\omega y)=-\cos (\omega y)$. These tones lie outside the Schwartz space; their prescribed Cauchy principal values explain the quarter-cycle shift in the figure. Both figures are calculated mathematical illustrations. Reversing the Hilbert denominator reverses these signs; NIST uses that opposite convention.^[ref]

The chain rule calculates the derivative of a function composition by combining local rates. If the inner function $f$ is differentiable at $a$ and the outer function $g$ is differentiable at $f(a)$, thenThe evaluation point matters: the outer derivative is evaluated at the inner output $f(a)$, not at $a$. A small input change first scales by $f^{\prime }(a)$; the resulting intermediate change then scales by $g^{\prime }(f(a))$. In terms of linear approximations, composing those two local descriptions multiplies their scale factors. ^[ref].

For example, $h(x)=\sin (x^2)$ applies a quadratic function and then the sine. With $f(x)=x^2$ and $g(u)=\sin u$, the chain rule givesHere all angles are in radians. Near $x=1$, increasing the input by $\Delta x$ changes $x^2$ by about $2\Delta x$, and the sine changes by about $\cos 1$ times that intermediate change. For $\Delta x=0.01$, the linear approximation predicts an output increase of $0.0108060$; the exact increase is $\sin (1.0201)-\sin 1\approx 0.0106894$.

The same rule handles several nested stages. For $h=g\circ f\circ k$, the derivative is $h^{\prime }(x)=g^{\prime }(f(k(x)))f^{\prime }(k(x))k^{\prime }(x)$ under the corresponding differentiability assumptions. Writing each intermediate output prevents evaluating a derivative at the wrong stage. It also makes the chain rule useful for propagating sensitivities through a mathematical model, where one measured quantity influences another through several transformations.

For a quick proof idea, replace $f$ near $a$ by its linear approximation, then replace $g$ near $f(a)$ by its linear approximation. The composite leading term is their product; differentiability ensures the remaining error is smaller than the original input change. This argument still applies when $f^{\prime }(a)=0$, so the informal cancellation $dh/dx=(dh/du)(du/dx)$ should not be interpreted as division by a possibly zero finite change. The hypotheses are sufficient, but are not necessary for every composite to be differentiable, as the nondifferentiable inner function with a differentiable nondegenerate composite example shows.

A coarse-grid correction improves a fine-grid approximation by solving a smaller equation for the part of its approximation error representable on a coarse mesh. Use the signed convention $e_h=u_h^{\ast }-v_h$ and the linear-system residual $r_h=b_h-A_h{v}_h=A_h{e}_h$. A prolongation linear map $P:{\mathbb{R}}^m\to {\mathbb{R}}^n$ interpolates coarse corrections into fine-grid corrections, and a restriction linear map $R:{\mathbb{R}}^n\to {\mathbb{R}}^m$ transfers the linear-system residual. Solve $A_H{z}_H=R{r}_h$ exactly or approximately and update $v_h\leftarrow v_h+P{z}_H$. The plus sign follows from defining the approximation error as exact minus approximate. It is the correction equation's right-hand side $R{r}_h$, rather than merely restricted solution values, that determines the update.

For the multigrid method model $A_h=h^{-2}\mathrm{tridiag}(-1,2,-1)$, take $n=2m+1$, so $H=2h$ and the coarse nodes coincide with every second fine node. Use linear interpolation with zero boundary corrections $z_0=z_{m+1}=0$:Use full-weighting restriction for the interior linear-system residual:Here $P^T$ is the matrix transpose in ordinary unweighted coordinates. With these specific transfers, the Galerkin coarse-grid operator isThus it equals the rediscretized Poisson equation operator on this constant-coefficient one-dimensional mesh. The equality is special to the stated model and transfers. In particular, dropping $h^{-2}$ from the fine matrix while using $H^{-2}$ on the coarse mesh changes the correction unless the right-hand sides and transfers are rescaled consistently.

Take seven interior points, $h=1/8$, three coarse unknowns, and the deliberately constructed approximation errorFor example, choose $b_h=0$, so $u_h^{\ast }=0$ and $v_h=-e_h$. Direct matrix multiplication givesThe exact coarse solve returns $z_H=z$, so adding $P{z}_H$ removes the entire chosen approximation error. A broad triangular error can have a sharply localized linear-system residual: the latter measures discrete curvature, not error height. This is why the two should not be confused. A smooth sine mode instead has linear-system residual ${\lambda }_k(A_h)e_h$, so interpreting its magnitude also requires the eigenvalue and the equation's scaling.

The algebra explains the exact cancellation beyond this example. Assume $P$ has linearly independent columns and choose an invertible Galerkin coarse-grid operator $A_H=R{A}_{h}P$. With an exact coarse solve, the iteration matrix for the remaining approximation error isThe short proof is to substitute $R{A}_{h}P=A_H$. Thus every vector in the column space of $P$ is eliminated, and the corrected linear-system residual restricts to zero. These are exact projection identities, not a statement that every visually smooth vector belongs to that column space. If $A_h$ is a symmetric matrix and a positive-definite matrix, and $R=c{P}^T$ with $c>0$, then $A_H$ is also a positive-definite matrix. Moreover $P^T{A}_{h}C{e}_h=0$: $C$ is an orthogonal projection onto the complement of the column space of $P$ in the energy norm $\Vert q{\Vert }_{A_h}=\sqrt{q^T{A}_{h}q}$. Equivalently, the added correction is the best approximation to $e_h$ from that column space in this energy norm. Without these symmetry and transfer assumptions, the projection need not be orthogonal.^[ref]

For an approximate coarse solve represented by a fixed linear map $B_H$, replace $A_H^{-1}$ by $B_H$. ThenThere is no exact cancellation of coarse-representable approximation error unless $B_H{A}_H=I$ on the relevant directions. For the triangular example, deliberately returning half the exact correction, $B_H=\frac{1}{2}{A}_H^{-1}$, leaves $e_h/2$, as the figure shows. This controlled mathematical comparison illustrates an incomplete coarse solve; it is not a measured performance result or a prescription for a geometric multigrid V-cycle. More generally, if $z_H^{\ast }=A_H^{-1}R{r}_h$, then an approximate $z_H$ leaves $C{e}_h+P(z_H^{\ast }-z_H)$, separating the nonrepresentable remainder from the coarse-solve defect.

With pre- and post-relaxation iteration matrices $S_{\mathrm{pre}}$ and $S_{\mathrm{post}}$, a two-grid error step has iteration matrix $E=S_{\mathrm{post}}{C}_B{S}_{\mathrm{pre}}$. A geometric multigrid V-cycle supplies $B_H$ by one recursive coarse cycle from zero, with a direct solve only at the bottom. Effective smoothing, a coarse column space that approximates the surviving approximation error, and a sufficiently accurate coarse correction must work together. Coarsening alone supplies no convergence guarantee, and correcting an arbitrary fine-grid vector exactly would require a coarse column space large enough to contain it.

The sine function extends to an entire function on the complex plane by the displayed exponential formula. Substituting $z=x+iy$ and grouping the real and imaginary terms gives $\sin (x+iy)=\sin x\cosh y+i\cos x\sinh y$, for real $x,y$. Thus its values grow exponentially in imaginary directions, even though the real sine is bounded.

The sine maps $\{-\pi /2<\mathrm{Re}z<\pi /2,\mathrm{Im}z>0\}$ bijectively onto the upper half-plane: its imaginary part is $\cos (\mathrm{Re}z)\sinh (\mathrm{Im}z)>0$, and the equality of two sine values forces either equal arguments modulo $2\pi$ or arguments summing to an odd multiple of $\pi$. Only equality is possible in this domain. For surjectivity, solve $t^2-2iwt-1=0$ for $t=e^{iz}$. When $\mathrm{Im}w>0$, exactly one root has modulus below one and it has positive real part. The inverse $z=-i\log t$ therefore belongs to the half-strip. Composing with $(w-i)/(w+i)$ gives a conformal map onto the unit disc.

The cosecant is $\csc z=1/\sin z$, wherever the sine denominator is nonzero. The square of the cosecant occurs in the derivative $d(\cot z)/dz=-{\csc }^{2}z$ of the cotangent.

Factor the sine expansion as $\sin z=z[1-z^2/6+z^4/120+O(z^6)]$. To invert the bracket, put $1+A{z}^2+B{z}^4+O(z^6)$ and equate coefficients after multiplication. The $z^2$ coefficient gives $A=1/6$, while the $z^4$ coefficient gives $B=A/6-1/120=7/360$. Dividing by $z$ proves the displayed Laurent series truncation for $z\to 0$, $z\ne 0$. This reciprocal-series calculation is useful when a small sine argument is itself an inverse power of a large variable; multiply the remainder by any outside prefactor when assigning its final order.

For a real angle $\theta$, $\cos \theta$ is the horizontal coordinate on the unit circle reached by a counterclockwise rotation through $\theta$ radians. For an acute angle in a right triangle, it is the adjacent side divided by the hypotenuse. Basic values includeAt $\theta =2\pi /3$, the horizontal coordinate is $-1/2$ even though the vertical sine coordinate remains positive. A positive angle need not produce a positive cosine.

Reflection across the horizontal axis leaves this coordinate unchanged, giving $\cos (-\theta )=\cos \theta$: cosine is an even function. A full turn gives $\cos (\theta +2\pi )=\cos \theta$. For real arguments, its range is $[-1,1]$ and its zeros of a function occur at $\theta =\pi /2+k\pi$. A quarter-turn shifts between the two coordinate traces, with $\cos \theta =\sin (\theta +\pi /2)$.

The cosine addition formula gives the horizontal coordinate after two successive rotations:The minus sign comes from the way a counterclockwise rotation mixes horizontal and vertical coordinates:The two rows recover the cosine addition formula and sine addition formula together.

With $x=\pi /6$ and $y=\pi /4$, the resulting angle is $5\pi /12$, andAlthough the horizontal coordinate becomes small, the vertical one approaches one; their squared sum still obeys the Pythagorean trigonometric identity. Taking $y=-x$ in the cosine addition formula, together with the even function property of cosine and odd function property of sine, also recovers that identity.

The cotangent is the quotient $\cot z=\cos z/\sin z$, wherever the sine denominator is nonzero. Its meromorphic continuation has simple poles at integer multiples of $\pi$.

A diffraction order is the integer $m$ labeling a principal constructive interference direction of a diffraction grating: adjacent elements differ in path length by $m$ wavelengths. In the normal-incidence transmission geometry described above,Positive and negative diffraction orders lie on opposite sides of the forward normal. The condition expresses the range of the sine of a real angle, so it also limits the propagating diffraction orders.

For $\lambda =600\mathrm{nm}$ and $d=2\mu \mathrm{m}$, $\sin {\theta }_m=0.3m$. Taking the inverse sine givesThese are seven allowed propagating diffraction orders. Order $m=4$ would require $\sin {\theta }_4=1.2$, and $m=-4$ would require $-1.2$, so neither has a propagating far-field direction. The allowed diffraction orders are equally spaced in $\sin \theta$, while their actual angles become farther apart toward grazing propagation.

Each principal peak has finite width. For the ideal uniform coherent array, the nearest zeros around order $m$ obeyThus the one-sided distance from the order direction to the first zero is $\Delta (\sin \theta )=\lambda /(Nd)$. Near the first diffraction order in our example, $N=10$ puts the zeros at $0.27$ and $0.33$, while $N=40$ puts them at $0.2925$ and $0.3075$. The respective one-sided widths are $0.03$ and $0.0075$; the full zero-to-zero widths are $0.06$ and $0.015$. These are calculated coherent peak widths, with the slit envelope omitted, and do not describe a measured instrument response. ^[ref].

Four times as many illuminated slits therefore give a fourfold narrowing in this coordinate without moving the diffraction order. That sharper directional selection helps distinguish nearby wavelengths in a spectrograph. Actual performance also depends on illumination, optical coherence, and source width; the separate slit-limited resolving power of a grating describes the contribution of the entrance slit.