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Aliasing is the inability to distinguish different continuous frequencies from sampled values. The identity above says that frequencies separated by an integer multiple of the sampling rate $f_s$ give identical samples of a complex Fourier mode. For real cosines, changing the sign of a frequency also gives the same waveform, explaining the $2$-versus-$8$ example at rate $10$. The Nyquist–Shannon sampling theorem prevents this ambiguity within its specified class of bandlimited functions. Before sampling, unwanted higher frequencies must be suppressed if they would violate that assumption; once two waveforms give the same samples, those samples cannot identify which was present.

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]

Apply the product-to-sum formula to the voltage model:The carrier wave component has peak voltage amplitude $A$; each AM sideband has peak voltage amplitude $Am/2$. These are temporal frequency spectrum components, unlike the spatial sidebands used in pattern stability. The separation $\Omega >0$ makes the three positive angular frequencies distinct, and $\Omega <{\omega }_c$ keeps the lower one positive. The span between sidebands is $2\Omega$, or $2f_m$ in hertz when $f_m=\Omega /(2\pi )$; this is the single-tone span, not a universal signal bandwidth for arbitrary messages.

A one-sided cosine-amplitude display assigns $A$ to the carrier line. A two-sided complex-exponential expansion instead assigns $A/2$ at each of $\pm {\omega }_c$, and $Am/4$ at each of the four sideband locations. The signal is identical; confusing the conventions creates a factor-of-two error. An ideal infinite-duration sinusoid gives a spectral line. A finite recording and finite analyser resolution broaden the displayed peak.

For a voltage applied across an ideal resistor of electrical resistance $R>0$, a component of peak voltage $B$ has root mean square voltage $B/\sqrt{2}$ and mean power $B^2/(2R)$ by Ohm's law. ThusThe proof of additivity is to average $s(t{)}^2/R$ over a long time: cross-products of distinct frequencies have zero mean. Consequently the whole signal's root mean square voltage is $A\sqrt{(1+m^2/2)/2}$, not simply its maximum wave envelope divided by $\sqrt{2}$. The carrier-component $A/\sqrt{2}$ remains unchanged as $m$ varies.

With $A=2\mathrm{V}$, $R=50\Omega$ and $m=0.5$, the carrier supplies $40\mathrm{mW}$ and each $0.5\mathrm{V}$ peak sideband supplies $2.5\mathrm{mW}$, giving $45\mathrm{mW}$ total. At $m=1.4$ each sideband is $1.4\mathrm{V}$ peak and supplies $19.6\mathrm{mW}$, giving $79.2\mathrm{mW}$ total. Neither case changes the carrier-component power.

The fraction of mean power in the message-bearing sidebands is $m^2/(2+m^2)$. It is $1/9$ at $m=0.5$ and $1/3$ at $m=1$. Conventional AM spends the remaining power on a carrier that enables an inexpensive envelope detector. Increasing $m$ beyond $1$ increases this fraction in the linear model but breaks ordinary envelope recovery. Removing the carrier gives suppressed-carrier amplitude modulation and changes the receiver requirement.

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.

Astrometry measures the positions and motions of celestial objects. Repeated position measurements can distinguish stellar parallax, caused by the observer's motion, from proper motion of the source across the sky.

An astrometry measurement needs a date and a reference frame, as well as an angle. In an equatorial coordinate system, right ascension and declination label two viewing directions. A catalogue's reference date states when its quoted position applies; comparing entries at different dates without accounting for proper motion can make the same star appear misplaced. The observational problem is to infer a consistent history from successive incoming light directions, while accounting for the observer's changing position and measurement error.

To see the separation clearly, consider just one angular coordinate $\theta$, with $t$ the numerical elapsed time in years. Assume a simplified circular orbit of the Earth, a fixed source Euclidean distance, constant proper motion in this coordinate and no other accelerations. Choose the direction and the starting date so that the annual stellar parallax projection is the cosine termHere ${\theta }_0$ is the coordinate without the annual stellar parallax offset at the reference date; the observed $\theta (0)$ is ${\theta }_0+p$. The proper motion coefficient $\mu$ is quoted in arcseconds per year, and $p$ is the annual stellar parallax in arcseconds. This is an illustrative choice of a full-amplitude projection. General sky directions have different projected amplitudes and phases in their two coordinates; their apparent paths need not be circles. Real astrometry uses the observer's calculated position rather than imposing this single cosine. ^[ref].

Set ${\theta }_0=0$, $\mu =100\mathrm{mas}\mathrm{y}{\mathrm{r}}^{-1}$ and $p=50\mathrm{mas}$, where $1\mathrm{mas}$ is one thousandth of an arcsecond. The model gives $\theta (0)=50$, $\theta (0.5)=0$ and $\theta (1)=150\mathrm{mas}$. Same-season observations one year apart differ by $100\mathrm{mas}$ because the annual stellar parallax contribution repeats. Subtracting that fitted straight trend leaves an annual signal between $-50$ and $+50\mathrm{mas}$. The existing stellar parallax distance relation then gives $20$ parsecs.

Observing only at the same annual phase would recover the proper motion trend but leave $p$ indistinguishable from a constant offset. Sampling several seasons as well as several years separates the repeating signal from the accumulating displacement. The two signatures answer different questions using the same measurements.

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.

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 first-kind Chebyshev polynomial $T_n$ converts a multiple-angle cosine into a polynomial:The cosine addition formula gives the recurrence relationThis constructs polynomials valid for every real $x$, including outside $[-1,1]$, where the real-angle description is unavailable. For example,For $n\ge 1$, $T_n$ has polynomial degree $n$ and leading coefficient $2^{n-1}$. On $[-1,1]$ its uniform norm is one, even as its oscillations become more numerous.^[ref][ref]

The polynomial roots of $T_n$ are especially simple:They correspond to equally spaced angles, but their locations in $x$ cluster near $-1$ and $1$. For $T_3$, the polynomial roots are $\sqrt{3}/2,0,-\sqrt{3}/2$. For $T_8$, the positive polynomial roots are approximately $0.980785,0.831470,0.555570,0.195090$, with their negatives completing the set. The distance from the right endpoint to the nearest polynomial root is $1-\cos (\pi /(2n))\sim {\pi }^2/(8n^2)$; central neighboring polynomial roots are separated on the larger scale $1/n$. The small change of cosine near angles zero and $\pi$ explains the clustering.

These polynomial roots supply $n$ interpolation nodes for a polynomial interpolation of polynomial degree at most $n-1$. They differ from the $n+1$ Chebyshev-Lobatto nodes $\cos (j\pi /n)$, which include both endpoints. This choice of polynomial roots has nodal polynomialThe monic Chebyshev extremal polynomial minimizes the uniform norm of this product among monic polynomials of polynomial degree $n$ on $[-1,1]$. Consequently, if $f$ has $n$ continuous derivatives, the usual polynomial interpolation error gives the useful boundThis optimizes the nodal factor in that bound, not the actual error for every function. Endpoint clustering helps control the large endpoint oscillations of equally spaced interpolation associated with Runge's phenomenon. The two node conventions and their geometric construction are discussed in Trefethen's authored text.^[ref]

The same cosine substitution for polynomial approximation explains the weighted orthogonality:The endpoint singularities of the weight are integrable. Thus the Chebyshev polynomials are orthogonal polynomials for this weight, making coefficients in a least-squares polynomial in an orthogonal-polynomial basis separate cleanly. Their polynomial roots also give Chebyshev–Gauss quadrature, connecting polynomial interpolation to Gaussian quadrature.^[ref]

Accuracy still depends on the target function. The absolute value function $|x|$ has a corner at zero, so the hypothesis requiring continuous derivatives in the displayed polynomial interpolation error bound fails. Endpoint clustering cannot remove that corner. A jump poses a stronger obstruction: polynomials are continuous functions, so they cannot converge in the uniform norm to a discontinuous function on the whole interval. Chebfun's documented examples show persistent overshoot for such jumps; these are a limitation of global polynomial interpolation, not a universal promise of rapid accuracy from Chebyshev polynomials.^[ref]

Coherent demodulation recovers a message by multiplying the received signal by a reference carrier wave of the same frequency and known relative wave phase, then passing the result through a low-pass filter. For the ideal AM model and reference $2\cos ({\omega }_{c}t+\phi )$,A unity-gain ideal low-pass filter passing $0$ and $\Omega$ but rejecting $2{\omega }_c$ and $2{\omega }_c\pm \Omega$ leaves $Aq(t)\cos \phi$. Such a separating cutoff exists because $\Omega <2{\omega }_c-\Omega$. At $\phi =0$, subtraction of $A$ recovers $Am\cos (\Omega t)$ exactly for any finite $m\ge 0$, including overmodulation, provided the transmitter and channel preserve this linear waveform. This is the signed factor, not its absolute value.

A constant phase error scales the recovered message by $\cos \phi$: a $\pi /3$ radian error halves it, and a $\pi /2$ error removes it. A known error with nonzero cosine can be compensated in gain; a frequency mismatch makes the relative wave phase drift and no longer gives constant attenuation. The lecture illustrates synchronous multiplication, filtering and phase error.^[ref]

The extension of the cosine to the complex plane is $\cos z=(e^{iz}+e^{-iz})/2$. Splitting the complex exponential function into its real part and imaginary part gives the displayed formula in terms of hyperbolic cosine and hyperbolic sine.

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.

Use radians and take $T(x)=\cos x$ on the closed interval $[0,1]$. This is a complete metric space with distance $|x-y|$, and $T([0,1])=[\cos 1,1]\subset [0,1]$. The mean value theorem and $|T^{\prime }(x)|=\sin x\le \sin 1$ give a contraction mapping with $L=\sin 1\approx 0.841471<1$. Both the interval and its preservation matter; a small derivative by itself would not ensure that all subsequent approximations stay in the chosen space.

Starting at $x_0=0$ gives $x_1=1$, $x_2\approx 0.540302$, $x_3\approx 0.857553$, and $x_4\approx 0.654290$. The cosine is decreasing here, so successive iterates alternate around the unique fixed point $x_{\ast }\approx 0.739085133215161$. The contraction iteration error bound needs no prior knowledge of this numerical value. At step 40, $x_{40}\approx 0.739085078689123$ and the latest-step bound is below $7.20\times 10^{-7}$, certifying error below one millionth for exact iteration. The initial-step bound requires 91 iterations for the same tolerance. The observed step-40 error is only about $5.45\times 10^{-8}$; the guarantees allow less favorable behavior than this example actually exhibits.

This is the beauty of the contraction mapping theorem: a uniform local improvement controls the entire infinite process. Completeness is essential. On the incomplete interval $(0,1)$, the contraction mapping $T(x)=x/2$ keeps every iterate inside but has no fixed point there; its missing limit is zero. Banach's 1922 paper developed abstract methods for different spaces of functions with applications to integral equations, a motivation still visible whenever repeated approximation is used to construct a function.^[ref]

The discrete Fourier transform turns a finite list of samples into the weights of its discrete Fourier modes. The goal is to reveal a simple oscillatory recipe hidden in complicated-looking data, while retaining enough information to reconstruct every sample exactly. In signal processing, those weights identify sampled frequencies and wave phases; in numerical analysis, they simplify circular convolution and the discrete sine transform Poisson solver.

Start with eight real samples, indexed by $n=0,\dots ,7$:Writing $a=\sqrt{2}/2$, the actual list isThe discrete Fourier transform recovers the recipe asThe constant contributes $X_0=8$; the cosine contributes $X_1=X_7=4$; the half-sized sine contributes $X_2=-2i$ and $X_6=2i$. All other coefficients vanish exactly. The imaginary coefficients carry the sine's wave phase, which a plot of absolute values alone would hide.

Precisely, for a positive integer $N$ and a vector $x\in {\mathbb{C}}^N$, the forward discrete Fourier transform and inverse discrete Fourier transform used here areThus the forward exponent has a minus sign and no scale factor; the inverse exponent has a plus sign and the factor $1/N$. The coefficients in the reconstruction are $X_k/N$, rather than $X_k$. In the example, $X_0/8=1$, the pair $4/8$ reconstructs the cosine, and the pair $-2i/8,2i/8$ reconstructs the half-sized sine. These conventions agree with the scaled Fourier coefficients in MIT's authored notes.^[ref]

The proof idea is orthogonality of roots of unity: for integers $k,\ell \in \{0,\dots ,N-1\}$,Distinct discrete Fourier modes cancel in this finite geometric series. Substituting the forward formula into the inverse discrete Fourier transform leaves only the desired sample. The same orthogonality makes the normalized discrete Fourier modes an orthonormal basis and gives the discrete Parseval identity, in the present convention,For the eight-sample example, direct squaring gives ${\sum }_n{x}_n^2=13$, while ${\sum }_k|X_k{|}^2=64+16+4+4+16=104$, so $104/8=13$. This identity preserves the squared L2 norm after accounting for normalization. The existing discrete Parseval identity instead scales the forward discrete Fourier transform by $N^{-1/2}$, producing a unitary matrix and equal squared L2 norms without the factor $1/N$.

For real input, discrete Fourier conjugate symmetry pairs positive and negative frequencies. The label $k$ is a residue modulo $N$: $k=7$ when $N=8$ represents the same discrete Fourier mode as $k=-1$, since $e^{2\pi i7n/8}=e^{-2\pi in/8}$. It therefore represents negative frequency $-1/8$ cycles per sample. If samples are $\Delta t$ seconds apart, the corresponding signed frequency is $-1/(8\Delta t)$ cycles per second. Similarly, index $6$ represents $-2$, while index $4$ is its own negative modulo $8$.

The discrete Fourier transform is an invertible map between two finite vectors. The continuous Fourier transform instead describes a function on the real line by an integral over a continuous frequency variable. A finite list of samples does not determine the full continuous frequency spectrum: aliasing of discrete Fourier modes already supplies distinct waveforms with identical observations. Applying the inverse formula at arbitrary integer indices gives the periodic extension of a finite sequence, repeating the recorded block every $N$ samples. This is a property of that representation, rather than evidence that the physical signal repeats beyond the recording. Laurent Demanet's MIT notes place the discrete Fourier transform alongside the continuous Fourier transform and Fourier series to explain the role of sampling and repetition.^[ref]

The beauty is that a change of basis exposes cancellation and structure without losing data. Direct evaluation needs $N$ sums of $N$ terms; a Fast Fourier transform computes the same discrete Fourier transform with far less work. Computational speed and the mathematical meaning of the coefficients are separate questions.

A dot product measures how much two Euclidean vectors point along each other. This article uses real Euclidean vectors in ${\mathbb{R}}^n$, with the usual Cartesian basis: multiply matching coordinates and add,The result is a scalar, while $\Vert \mathbf{a}\Vert$ is the Euclidean norm. This real convention has no complex conjugation; for vectors over the complex numbers, use the convention specified in inner product.

For two nonzero Euclidean vectors, the same dot product records their angle $0\le \theta \le \pi$ through the cosine:Here $\arccos$ is an inverse trigonometric function. The Cauchy-Schwarz inequality guarantees that the normalized ratio lies in $[-1,1]$. The zero vector has dot product zero with every vector, but its angle is undefined: the ratio would divide by zero. To outline the geometric identity, expand $\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=\Vert \mathbf{a}{\Vert }^2+\Vert \mathbf{b}{\Vert }^2-2\mathbf{a}\cdot \mathbf{b}$ and compare with the law of cosines.^[ref]

Take $\mathbf{a}=(3,4)$ and $\mathbf{c}=(2,0)$. Their Euclidean norms are $5$ and $2$, andThe unit vector $\mathbf{u}=\mathbf{c}/\Vert \mathbf{c}\Vert =(1,0)$ selects the horizontal direction. The scalar projection $\mathbf{a}\cdot \mathbf{u}=3$ says how far $\mathbf{a}$ reaches along that direction; its vector projection is $3\mathbf{u}=(3,0)$. Thus $\mathbf{a}=(3,0)+(0,4)$ splits into an aligned part and an orthogonal projection residual.

For $\mathbf{b}=(4,-3)$, the dot product $\mathbf{a}\cdot \mathbf{b}=12-12=0$ proves orthogonality; both Euclidean norms are $5$. For $\mathbf{d}=(-2,0)$, $\mathbf{a}\cdot \mathbf{d}=-6$: the scalar projection onto $\mathbf{d}/\Vert \mathbf{d}\Vert =(-1,0)$ is $-3$, because $\mathbf{a}$ reaches against that chosen direction. Its angle with $\mathbf{d}$ is $126.8699^{\circ }$.

Keeping $\mathbf{a}=(3,4)$ fixed and choosing $\mathbf{b}$ of Euclidean norm $2$ gives $\mathbf{a}\cdot \mathbf{b}=10\cos \theta$. At angles $0^{\circ },60^{\circ },90^{\circ },120^{\circ },180^{\circ }$, the dot products are $10,5,0,-5,-10$. Positive values mean partial alignment, zero means orthogonality, and negative values mean a component against the other direction.

This interpretation explains work: for a constant force $\mathbf{F}$ and a straight displacement $\mathbf{s}$, $W=\mathbf{F}\cdot \mathbf{s}$ counts the force component along the motion. A sideways force contributes zero work, and an opposing force contributes negative work.^[ref]

Electric flux measures how much electric field crosses a chosen oriented surface, with a sign distinguishing the two crossing directions. A strong electric field running along a surface contributes nothing to its electric flux: only the normal component counts. This lets Gauss's law turn a potentially complicated electric field into one number recording the net enclosed electric charge.

For a surface $S$ with chosen unit normal $\mathbf{n}$, the electric flux is the surface integralHere $dA$ is positive surface area, while $d\mathbf{S}$ is the vector area element. The dot product makes electric flux a signed scalar, rather than a vector like the electric field. Its SI unit is $\mathrm{N}{\mathrm{m}}^2/C$: electric field measured using the newton per coulomb, multiplied by surface area in square metres. Reversing the orientation of a surface replaces $\mathbf{n}$ by $-\mathbf{n}$ and reverses the electric flux. Calling it electric flux does not imply that electric charge is moving through the surface. ^[ref].

For an open patch in a plane, with surface area $A$ and a constant electric field, the unit normal is constant and the surface integral simplifies toThe angle $\theta$ is between the electric field and the chosen unit normal, not between the electric field and the plane. The cosine describes the signed projected surface area seen by the electric field. Thus electric flux is positive for $0\le \theta <90^{\circ }$, zero at $90^{\circ }$, and negative for $90^{\circ }<\theta \le 180^{\circ }$. ^[ref].

For example, take $E=100N/C$ and $A=0.02{\mathrm{m}}^2$, so $EA=2\mathrm{N}{\mathrm{m}}^2/C$. Three choices of angle giveKeeping the $60^{\circ }$ patch fixed and reversing its unit normal gives $120^{\circ }$ and changes $+1$ to $-1\mathrm{N}{\mathrm{m}}^2/C$. The electric field and physical patch have not changed. If the electric field is instead tangent to the patch, its dot product with the unit normal vanishes: electric flux is zero even though $E=100N/C$.

For a closed surface enclosing a volume $V$, choose the outward unit normal. Gauss's law givesHere ${\epsilon }_0$ is the vacuum permittivity, and $Q_{\mathrm{enclosed}}$ is the signed total electric charge inside. The electric field in the surface integral includes all sources, including external electric charges. Assume no ideal point charge lies on the integration surface. Zero enclosed electric charge implies zero total electric flux; it does not imply zero electric field everywhere on the closed surface or inside it. Positive and negative contributions to the surface integral can cancel. ^[ref].

To see the cancellation with sources inside, place point charges $+q$ at $(-a,0,0)$ and $-q$ at $(a,0,0)$, where $q>0$, $a=0.30\mathrm{m}$, and both lie inside the sphere of radius $R=1\mathrm{m}$ centered at the origin. This separated pair has the electric dipole geometry. Its enclosed electric charge is $q-q=0$, so its total outward electric flux is zero. At the point $P=(0,0,R)$ on the sphere, Coulomb's law and the superposition principle give the individual electric fieldsThe $z$ components cancel and the positive $x$ components add:For the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $q=1\mathrm{nC}$, this is about $4.740N/C$ along positive $x$. At $P$, the outward unit normal is $\hat{\mathbf{z}}$, so this nonzero electric field is tangent to the sphere. At other points of the sphere, its normal component can be positive or negative. The source-field formula applies away from the two point charges, where the electric field has zero divergence; it is singular at the sources. ^[ref].

The diagram shows streamlines of the three-dimensional electric field in the section $y=0$. The circle is a cross-section of the sphere, not the integration surface: Gauss's law uses the entire three-dimensional surface integral. A two-dimensional line integral of outward field around that circle would be a different quantity, with units $\mathrm{N}m/C$ rather than the electric flux units $\mathrm{N}{\mathrm{m}}^2/C$.

Electromagnetic radiation is a propagating disturbance of the electric field and magnetic field that carries energy and momentum away from its source. It can be a sustained electromagnetic wave or a finite wave packet: periodic oscillation is not required. In vacuum the disturbance travels at the speed of light. A receiver responds to fields arriving now from an earlier source event, as expressed by retarded time. This is how electrical activity at one place can deliver energy and information elsewhere without a connecting wire.

Stored electromagnetic energy alone does not establish outgoing electromagnetic radiation. A stationary electric charge in electrostatics, or a steady localized electric current in magnetostatics, can store electromagnetic energy with zero Poynting vector. Static combinations can have circulating energy flux without net outward power. Near a changing source, reactive energy can return later. The near and far field distinction separates these contributions from outgoing radiation zone fields.^[ref]

A single traveling vacuum plane electromagnetic wave obeying the Maxwell equations has $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$. Its Poynting vector gives the energy density and flux of a plane electromagnetic wave:Here $I$ is wave intensity; ${ϵ}_0$ and ${\mu }_0$ are vacuum permittivity and vacuum permeability. For a monochromatic wave with linear polarization and peak amplitude $E_0$, squared cosine averages to $1/2$. Pulses require the instantaneous Poynting vector; arbitrary incident-and-reflected superpositions need not obey this relation.^[ref]

The electromagnetic momentum density is $\mathbf{g}=\mathbf{S}/c^2$. Completely absorbing a unidirectional vacuum pulse of energy $U$ transfers momentum $U/c$; see radiation pressure on a perfectly absorbing surface. A calculated $1.0\mathrm{mJ}$ pulse transfers $3.34\times 10^{-12}\mathrm{kg}m/s$. Symmetric emission can lose energy without net recoil because opposite momenta cancel.^[ref]

The direction of emission matters as much as its total power. For an oscillating electric dipole moment along $z$, the angular distribution of electric dipole radiation is strongest across the axis and vanishes along it. Its origin is geometric: the electric dipole radiation field is the transverse projection of the second time derivative of the electric dipole moment, so its magnitude contains $\sin \theta$. Squaring this electric field to obtain wave intensity gives ${\sin }^2\theta$. The electric-dipole approximation for radiation requires a source much smaller than its wavelength and observation in the radiation zone. ^[ref]

For example, the power per solid angle at $\theta =30^{\circ }$ is one quarter of its equatorial value, while the electric field amplitude is one half. To normalize the angular distribution to total power, use spherical coordinates with $d\Omega =\sin \theta d\theta d\varphi$:The integral follows by substituting $x=\cos \theta$, giving ${\int }_{-1}^1(1-x^2)dx=4/3$. This is a probability-like normalization over direction, with units of inverse steradian; it does not make the dipole an isotropic radiator. The same angular integral underlies the Larmor formula for nonrelativistic accelerated electric charge.

Detection requires the incident electromagnetic field to do something observable. An antenna develops an electrical response; a thermopile converts absorbed energy into a thermal and then electrical signal; a photodetector can respond through photon absorption. Hertz spark transmission and detection experiment makes this conversion visible as a tiny spark. Detector orientation, sensitivity, and surrounding reflections all matter: failure to see a signal does not prove that no electromagnetic radiation reached the receiver.

For an elliptically polarized wave, the tip of its transverse vector traces an ellipse at a fixed point during each wave period. For polarization of an electromagnetic wave, this is the electric field vector. With the real components in polarization of waves, assume $A_x,A_y>0$ and $\sin \delta \ne 0$. Eliminating time givesA short derivation uses $\varphi =\omega t$ and the cosine addition formula to obtain $E_y/A_y=(E_x/A_x)\cos \delta -\sin \varphi \sin \delta$; squaring and using the sine and cosine identity ${\sin }^2\varphi =1-{\cos }^2\varphi$ gives the displayed equation. Outside phase quadrature, the mixed term gives a tilted ellipse; $A_x,A_y$ are component wave amplitudes and need not be its semi-major and semi-minor axes.

In phase quadrature, $\delta =\pm \pi /2$, the mixed term vanishes. For $(A_x,A_y)=(2,1)V/m$, the ellipse is $E_x^2/4+E_y^2=1$, its semi-major and semi-minor axes have lengths $2$ and $1V/m$, and its electric field magnitude varies between $1$ and $2V/m$. At $\varphi =\pi /4$ with $\delta =\pi /2$, the electric field is $(\sqrt{2},-1/\sqrt{2})V/m$ and has magnitude $\sqrt{5/2}V/m$. Equal wave amplitudes instead give circular polarization, a special case of elliptical polarization with constant magnitude. As $\delta$ approaches $0$ or $\pi$, the ellipse collapses to a line segment and gives linear polarization; this limit is degenerate rather than a nondegenerate ellipse.

Changing $\delta$ from $\pi /2$ to $-\pi /2$ reverses traversal of the same ellipse. In the displayed $x,y$ plot, $\delta =\pi /2$ moves from $(A_x,0)$ toward negative $y$ as time increases. A label such as right- or left-handed circular polarization additionally requires a stated viewing direction and naming convention: viewing along propagation and viewing against propagation reverse the apparent rotation. ^[ref].

The ellipse records both the relative wave amplitudes and the phase difference of two transverse oscillations. This separates a change of polarization of waves from a change of overall wave amplitude: multiplying both components by the same positive factor enlarges the ellipse while preserving its shape and orientation.

A rotating ideal linear polarizer can probe this geometry. For the $2:1$ ellipse in phase quadrature, orthogonal projection onto a pass axis at angle $\theta$ gives $E_{\mathrm{out}}(t)=2\cos \theta \cos \omega t-\sin \theta \sin \omega t$, in $V/m$. Averaging its square over a wave period gives $(4{\cos }^2\theta +{\sin }^2\theta )/2$, while the incoming squared electric field magnitude averages to $5/2$. Thus the transmitted fraction of light intensity is $(4{\cos }^2\theta +{\sin }^2\theta )/5$: it is $4/5$ at $0^{\circ }$, $1/2$ at $45^{\circ }$ and $1/5$ at $90^{\circ }$. There is no single incident line of linear polarization to insert into Malus's law; projecting the electric field and averaging its square is the appropriate generalization. This measurement cannot distinguish $\delta =\pi /2$ from $\delta =-\pi /2$, because reversing the traversal leaves all these squared averages unchanged.

For a single traveling vacuum plane electromagnetic wave, $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$ gives equal electric and magnetic contributions to electromagnetic energy:For the fields with linear polarization along $x$ and magnetic orientation along $y$ specified above,Angle brackets mean an average over one period at fixed position. The factor $1/2$ comes from averaging squared cosine; $E_0$ is the peak amplitude, whereas the root mean square field is $E_{\mathrm{rms}}=E_0/\sqrt{2}$. The equal energy contributions oscillate together, rather than trading energy as in an RLC circuit. ^[ref].

For $E_0=100V/m$, using $c=299792458m/s$ and ${ϵ}_0\simeq 8.854\times 10^{-12}F/m$,A $0.010{\mathrm{m}}^2$ surface normal to $+z$ therefore has average crossing power $0.133\mathrm{W}$. This is incident power; the amount absorbed depends on the surface. Doubling the field amplitude quadruples the average flux.