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For strictly decreasing integer tuples $\kappa ,\eta$, let $A_{\kappa }(z)=\det (z_i^{{\kappa }_j})$ on the diagonal maximal torus of the unitary group. Expanding both determinants, Fourier orthogonality makes the integral zero unless the exponent tuples have the same permutation orbit. Strict decreasing order makes that equivalent to $\kappa =\eta$; the $n!$ matching permutation terms then each contribute one. An alternating Laurent polynomial is a finite linear combination of these alternants. For an irreducible character ${\chi }_{\lambda }$, its highest monomial gives coefficient one for $A_{\lambda +\delta }$ in ${\chi }_{\lambda }{A}_{\delta }$, with $\delta =(n-1,\dots ,0)$. The Weyl integration formula and Schur orthogonality relations force the sum of squared coefficients to be one, so all other coefficients vanish. This proves the unitary Weyl character formula directly.

Choose $N$ equal-amplitude, monochromatic odd components with orders $q_j=q_0+2j$, $j=0,\dots ,N-1$, and odd $q_0$. Fix the complex-field conventionHere $I$ is a normalized light intensity after averaging the fast carrier oscillations. Setting every ${\varphi }_j=0$ and summing a geometric series givesAt the removable singularities take the continuous limit. Maxima repeat every $T/2=\pi /\omega$; the first zeros on either side of a maximum lie at $t=\pm T/(2N)$. The zero-to-zero width is $T/N$, which is not the full width at half maximum.

Choose orders $13,15,17,19,21$ and driving wavelength $\lambda =800\mathrm{nm}$. Using $T=\lambda /c$ with the speed of light $c$ gives $T=2.66851\mathrm{fs}$ and burst spacing $1.33426\mathrm{fs}$. Solving the exact half-maximum equation gives a full width at half maximum of approximately $240.6$ attoseconds. These are values of the specified five-component model, not measurements from the 2001 experiment.

A deliberately chosen spectral phase ${\varphi }_q=q\omega t_0$ translates the whole attosecond pulse train by $t_0$ without changing its shape. Adding $\kappa (q-17{)}^2/2$ makes neighboring phase increments unequal: the same component intensities then produce a different temporal profile. For the choice $t_0=0.20\mathrm{fs}$ and $\kappa =0.08$ radians per squared harmonic order, the adjacent-component delay estimates arenamely $0.0981$, $0.1660$, $0.2340$ and $0.3019\mathrm{fs}$ for $q=13,15,17,19$. This finite difference estimates the group delay at the pair's midpoint. It is not a single common arrival time.

The infinite monochromatic sum is periodic forever and has infinite total energy. A real attosecond pulse train has a finite envelope and finite component linewidths. This calculation isolates coherent addition; it cannot describe an isolated burst or establish its contrast against satellites. As a further check, averaging $I(t)$ over one half-period gives $N|a{|}^2$ for any constant component spectral phases, because the cross terms integrate to zero by Fourier orthogonality. Spectral phase redistributes the light intensity in time while leaving that average fixed.

A lattice-translation-invariant interaction conserves total crystal momentum modulo a reciprocal lattice vector $G$. On a periodic chain of $M$ sites with unit spacing, the discrete Fourier transform supplies ${\sum }_{j=1}^M{e}^{i(k_1+k_2-k_3-k_4)j}=M{\delta }_{k_1+k_2,k_3+k_4(\mathrm{mod}2\pi )}$. This Fourier orthogonality proves the selection rule directly; it permits transfer by a reciprocal lattice vector rather than requiring equality of unrestricted momenta.

If symmetric partial sums ${\sum }_{r=-n}^n{a}_r{e}^{irt}$ converge almost everywhere on the circle, then $a_n,a_{-n}\to 0$. Their successive differences $T_n=a_n{e}^{int}+a_{-n}{e}^{-int}$ tend to zero almost everywhere. If $M_n=|a_n|+|a_{-n}|$ failed to tend to zero, choose a subsequence with $M_n\ge \epsilon$. The normalized $T_n/M_n$ are bounded by one and tend to zero almost everywhere, so dominated convergence makes their squared integrals tend to zero. Fourier orthogonality instead makes those integrals $(|a_n{|}^2+|a_{-n}{|}^2)/M_n^2\ge 1/2$, a contradiction.

On a rectangle $(0,L_x)\times (0,L_y)$, products of the one-dimensional Fourier sine basis give an orthogonal basis of L2 space. Fourier orthogonality in each variable givesFor a constant function one, these coefficients are $4[1-(-1{)}^m][1-(-1{)}^n]/({\pi }^{2}mn)$: only odd indices occur. The product modes vanish on the rectangle's boundary, making them natural Dirichlet Laplacian eigenfunctions. In the heat equation, their amplitudes decay as $e^{-\kappa {\pi }^2(m^2/L_x^2+n^2/L_y^2)t}$. Initial data not vanishing at the boundary are still represented in the interior and in the square-integrable sense; the positive-time boundary values need not agree continuously with the initially assigned boundary values.

Fourier analysis describes a function by the oscillations that compose it. A graph records values at successive positions or times; a frequency spectrum records how much of each frequency is present, together with its wave phase. The same complicated shape can become a short list of simple components. For periodic functions, the components form a Fourier series; on the whole real line, the Fourier transform uses a continuous range of frequencies, and the Fourier inversion theorem puts them back together.

For a concrete example, consider the trigonometric polynomialWith $t$ measured in seconds, its three Fourier harmonics have frequencies $1$, $3$, and $5$ cycles per second and amplitudes $1$, $1/2$, and $1/4$. Each cosine has zero wave phase. Adding them creates the sharper peaks in the figure. The frequency spectrum makes the recipe immediately visible; changing an amplitude changes the shape without changing the component's frequency. Fourier orthogonality lets us recover each Fourier coefficient by averaging against the corresponding sine or cosine, because the other Fourier harmonics cancel over a full period of a function.

The deeper reason these components are so useful is that they simplify operations. By Euler's formula, the exponential function $e^{i\omega x}$ packages a cosine and a sine into one Fourier mode. A translation or a derivative changes only its multiplier:Thus Fourier modes are simultaneous eigenfunctions of these operations. In Fourier series, this gives a literal diagonalization in the Fourier basis; on the real line, the Fourier transform gives the continuous counterpart, with each frequency handled separately. The Translation property of the Fourier transform and the Fourier transform of a derivative turn shifting and differentiating a whole function into multiplication of its Fourier transform. A convolution likewise becomes multiplication by the convolution theorem. This is beautiful because one description simplifies several apparently different operations at once.

For example, the heat equation $u_t=\kappa u_{xx}$, with thermal diffusivity $\kappa >0$, evolves a spatial Fourier mode asFine ripples have larger angular frequencies and decay faster: the formula explains why heating smooths a jagged temperature profile. For a general initial profile with sufficient regularity, the Fourier transform gives $\hat{u}(\omega ,t)=e^{-\kappa {\omega }^{2}t}\hat{u}(\omega ,0)$, and Fourier inversion recovers the temperature. Similarly, multiplying selected parts of a frequency spectrum suppresses unwanted components. Claude E. Shannon's Communication in the Presence of Noise (1949), Section III, describes ideal filtering as a projection that retains some frequency coordinates and removes others.^[ref]

The physical motivation is already visible in Joseph Fourier's Théorie analytique de la chaleur (1822). Chapter III constructs heat profiles using sums of sines and cosines; Chapter IX, articles 355–359, passes from Fourier series to integrals as the spatial interval becomes unbounded. These heat problems explain the motivation for the discrete and continuous descriptions. Their modern formulations specify which functions admit reconstruction and in what sense the expansions converge.^[ref][ref]

A Fourier series describes a periodic function through sines and cosines whose frequencies are integer multiples of one base frequency. These Fourier harmonics give a useful alternative to describing the value at every input: each Fourier coefficient measures how much of one oscillating pattern is present. For a real, integrable function of period $2\pi$, the real form isThe cosine formula includes $n=0$; the sine formula uses $n\ge 1$. Thus $a_0/2$ is the mean value over a period, not $a_0$ itself. The symbol $\sim$ associates a function with its Fourier coefficients; an equality requires a specified kind of convergence, such as pointwise convergence. For general period $T>0$, replace $nx$ by $2\pi nt/T$ and the coefficient prefactor by $2/T$. Fourier orthogonality explains why integrating against one mode removes the others. ^[ref].

For a finite example, takeIts mean is $1/2$, its real Fourier coefficients are $a_0=1$, $a_1=1$, $b_2=1/2$, and all other $a_n,b_n$ vanish. At $x=\pi /4$, the three contributions add to $1/2+1/\sqrt{2}+1/2=1+1/\sqrt{2}\approx 1.70710678$. The function is already a trigonometric polynomial, so retaining these modes reconstructs it exactly. Here $x$ is an angular coordinate in radians; if it instead represents time, a base angular frequency must supply the corresponding units.

The change of viewpoint is useful because many linear problems act separately on these modes. The second derivative maps $\sin (nx)$ to $-n^2\sin (nx)$ and maps $\cos (nx)$ in the same way. Consequently a differential equation that is complicated for a whole profile can become a simpler calculation for each Fourier harmonic. A linear time-invariant system similarly has a separate response to each complex exponential. Superposition combines the responses when the relevant sums and operations are justified. ^[ref].

The infinite case contains a subtlety absent from the finite example. Under pointwise convergence of a piecewise smooth Fourier series, the value at a jump discontinuity is the average of the two one-sided limits. For a square-integrable function, the complex Fourier series instead converges in the $L^2$ norm: the integrated squared error tends to zero. Neither statement says that the error becomes small everywhere at once. A square wave shows the difference vividly through the Gibbs phenomenon.

A historical source for this connection between profiles and modes is Joseph Fourier's 1822 book Théorie analytique de la chaleur. Chapter III develops expansions in sines and cosines while studying the propagation of heat. The heat equation gives a continuing application of this approach. ^[ref].

For an integrable function on $[-L,L]$, with $L>0$ and integer $n\ge 1$, the displayed Fourier coefficient multiplies the mode $\sin (n\pi x/L)$ in its real Fourier series. Fourier orthogonality fixes the normalization because the squared mode has integral $L$. If $f$ is odd, the product with the sine is even, so $b_n=(2/L){\int }_0^{L}f(x)\sin (n\pi x/L)dx$. These are also the coefficients of the Fourier sine series of the restriction to $(0,L)$.

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.

Let $\alpha ,\beta$ be nonzero real constants and let $f$ be a smooth periodic function of period $2\pi$, with real Fourier coefficients $c_n,d_n$. Write the unknown periodic solution with Fourier coefficients $a_n,b_n$. Fourier orthogonality applied to the equation givesSolving these two-by-two linear systems givesThe smooth forcing has rapidly decreasing Fourier coefficients, so the resulting series and its derivatives converge and satisfy the equation. Uniqueness follows because the difference of two solutions is $C{e}^{-\beta x/\alpha }$, which is periodic only when $C=0$. A single sinusoidal forcing therefore gives a single sinusoidal response with changed amplitude and phase.

For $0<\theta <\gamma$ and $0<r<R$, a bounded harmonic function that vanishes on the straight edges and has outer boundary data $\phi$ has coefficientsThis follows by separation of variables and Fourier orthogonality. Negative radial powers are excluded by boundedness at the vertex. Reflection-symmetric data $\phi (\theta )=\phi (\gamma -\theta )$ have only odd Fourier modes. Knowing the resulting solution on the bisector determines all these odd coefficients by its radial power series. Without reflection symmetry, the even angular modes vanish on the bisector and cannot be recovered from that restriction alone.

For a periodic box of volume $V$, expanding $\psi (\mathbf{r})=V^{-1/2}{\sum }_{\mathbf{k}}{e}^{i\mathbf{k}\cdot \mathbf{r}}{a}_{\mathbf{k}}$ in $H_{\mathrm{int}}=\frac{u}{2}\int d^{3}r{\psi }^{†}{\psi }^{†}\psi \psi$ gives the displayed normal ordering. The spatial integral is $V{\delta }_{{\mathbf{k}}_1+{\mathbf{k}}_2,{\mathbf{k}}_3+{\mathbf{k}}_4}$ by Fourier orthogonality, enforcing momentum conservation and leaving the factor $V^{-1}$. The coupling $u$ in three dimensions has units of energy times volume.

Integrate a polynomial times the conjugate of another over the product of unit circles, with normalized product angular measure. Fourier orthogonality makes the monomials $z_1^{{\alpha }_1}\cdots z_d^{{\alpha }_d}$ an orthonormal basis of the polynomial vector space. The Hermitian form is positive definite because the squared norm is the sum of squared absolute values of the finitely many coefficients. No completion or infinite series is needed to obtain this algebraic basis.

For a square-integrable function of period $T$, use Fourier coefficients $u_r=T^{-1}{\int }_0^{T}f(t)e^{-2\pi irt/T}dt$. Fourier orthogonality proves the displayed identity for a finite trigonometric polynomial. The mean-square convergence of the Fourier series then passes it to the limit. It includes the zero-frequency contribution $|u_0{|}^2$; excluding that contribution gives the variance rather than the full mean-square signal power.

Let $V=L^3$ be the box volume. The number operator is the sum of the occupation numbers of all momentum eigenstates, including the zero mode:In the Fourier series for the contact interaction, the two creation operators supply phase $e^{-i({\mathbf{k}}_1+{\mathbf{k}}_2)\cdot \mathbf{r}}$ and the two annihilation operators supply $e^{i({\mathbf{k}}_3+{\mathbf{k}}_4)\cdot \mathbf{r}}$. Fourier orthogonality with periodic boundary conditions givesThe four normalization factors together contribute $V^{-2}$, leaving the momentum-space contact-interaction HamiltonianThe Kronecker delta enforces momentum conservation. An equivalent expression that resolves the momentum transfer isBoth are already in normal ordering; no extra commutator constant is introduced.

Write $\mathbf{r}={\sum }_i{s}_i{\mathbf{a}}_i$. The electron density is periodic with period one in each $s_i$, so its three-dimensional Fourier series has integer harmonics:If $\Omega$ is a primitive unit cell of volume ${\Omega }_c$, Fourier orthogonality givesConversely, translation by ${\mathbf{a}}_i$ multiplies a general plane wave by $e^{i\mathbf{q}\cdot {\mathbf{a}}_i}$. Periodicity permits only $\mathbf{q}\cdot {\mathbf{a}}_i\in 2\pi \mathbb{Z}$, precisely the reciprocal lattice vectors. For a physical bounded periodic electron density, the expansion exists in the usual square-integrable sense.

The product-to-sum formula givesWhen both indices are positive the two integer frequencies integrate to zero. If either index is zero the integrand already vanishes. This is Fourier orthogonality.

Taking the inner product with $f_m$ and using Fourier orthogonality isolates $a_m$. Since $x\sin (mx)$ is even, integration by parts givesConsequently the Fourier sine series isThe series converges to $x$ at each interior point and in the L2 norm. Its periodic extension has jumps at $\pm \pi$, where the Fourier series instead converges to the midpoint value zero. The overshoot near those jumps is the Gibbs phenomenon.

For the real even function, let $s_N=a_0/2+{\sum }_{n=1}^N{a}_n\cos nx$. Fourier orthogonality gives ${\int }_{-\pi }^{\pi }\cos nxdx=0$ for $n\ge 1$ and ${\int }_{-\pi }^{\pi }\cos nx\cos mxdx=\pi {\delta }_{nm}$. ConsequentlyFor an even square-integrable function, the complete Fourier cosine series converges in the L2 norm; hence its norms converge. Taking $N\to \infty$ proves Parseval's identity in the required normalization:This limit argument avoids multiplying an infinite series without a justification of L2 norm convergence.

For the quadratic function, the Fourier coefficients areThe last equality follows from two integrations by parts, using $\sin n\pi =0$. ThusThis Fourier series is absolutely convergent; the periodically extended quadratic is continuous, so it represents its values also at the endpoints. Substituting its coefficients into Parseval's identity gives