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

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

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

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

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

Define the indicator function of the rational numbers on the real line by $f(x)=1$ for rational $x$ and $f(x)=0$ otherwise. Every rational shift preserves rationality and irrationality, so every nonzero rational number is a period of a function. Conversely, an irrational shift sends $x=0$ from value one to value zero, so it is not a period. The period group of a function is exactly $\mathbb{Q}$, and the positive periods $1/n$ approach zero without a smallest one. The function is nonconstant and discontinuous everywhere; this is why continuity matters in continuous periodic functions have a least positive period.

If a periodic unimodular function $a$ and its inverse have absolutely summable Fourier coefficients, replacing ${\hat{\phi }}_0$ by $\hat{\phi }=a{\hat{\phi }}_0$ preserves the orthonormal integer-translate span and integrability of an integrable scaling function. In physical space this is an absolutely summable combination of integer function translations. The new MRA low-pass filter is $m(t)=a(2t)m_0(t)/a(t)$. A rough phase can therefore preserve the multiresolution analysis while changing MRA low-pass filter regularity.

Consider the advection equation $u_t+a{u}_x=0$ with constant speed $a>0$, no sources, and periodic boundary conditions on $[0,1]$. Use $N$ equal cells of width $h=1/N$, time step $k>0$, and indices modulo $N$. The exact solution is the periodic translation $u(x,t)=u_0(x-at)$. At a face, the Riemann problem transports the left state to the right, so the two-state upwind numerical flux is $F(L,R)=aL$, with $F(U,U)=aU$.^[ref]

The finite volume method with this numerical flux and an explicit Euler method step givesHere $U_j^n$ represents a cell average; the recurrence also occurs in the upwind finite difference scheme. Impose the specific CFL condition $0<\nu \le 1$. Each updated cell average is then a convex combination of two old ones. This proves preservation of their initial minimum and maximum, including nonnegativity for nonnegative data, and contraction of the maximum norm of the difference of two solutions on this periodic grid. These statements concern this constant-speed scalar update with this time step.

For an original synthetic example, take $N=8$, $a=1$, $h=1/8$, $k=1/16$, hence Courant number $\nu =1/2$. Choose a function constant on each successive cell, with the following values, so these are exact initial cell averages. One step givesFor example, cell zero receives from cell seven across the periodic face: $U_0^1=(0+1)/2=1/2$. All eight face transfers cancel in the total by shared face flux cancellation. More generally this periodic recurrence conserves the total algebraically for any $\nu$, even though the CFL condition is needed for the convex combination argument. With the same initial data and $\nu =3/2$, one step gives $U_2^1=-1$ and $U_4^1=5/2$, despite preserving the total.

After four steps, $t=1/4$, the exact translated profile has moved two cells to the right. Repeated averaging instead gives the explicit binomial coefficient formulawith all indices modulo eight. The diagram compares these computed cell averages with the exact translated cell averages. Their integrals agree, but the numerical jumps spread across cells. The finite volume method is useful for jump discontinuities because it retains the integral balance even at a jump; conservation alone does not guarantee a sharp profile. At $\nu =1$ this particular update is an exact one-cell shift, whereas for $0<\nu <1$ it mixes neighboring states. No accuracy conclusion at a jump follows just from the conserved total.

A nonzero number $T$ is a period of a function $f$ when $f(x+T)=f(x)$ throughout its domain.

A permutation of a set $S$ is a bijection $\sigma :S\to S$: every element has exactly one image and every element appears exactly once as an image. For a finite set, a permutation can also be recorded as an ordering of all its elements, once a reference order has been chosen. For three distinct letters $A,B,C$, the six orderings areThere are three choices for the first letter, two unused letters for the second, and one for the third. Every path in the tree gives one complete ordering, and every ordering has one such path, so the rule of product gives $3\cdot 2\cdot 1=3!=6$, a factorial.

Ordering versus mapping. Label the reference positions $1,\dots ,n$. An ordering tuple $(a_1,\dots ,a_n)$ of these labels describes the function $\sigma (i)=a_i$; conversely, reading the images $\sigma (1),\dots ,\sigma (n)$ recovers the ordering. For example,means $1\mapsto 3$, $2\mapsto 1$, $3\mapsto 4$, $4\mapsto 2$. Its ordered image list is $(3,1,4,2)$, whereas its cycle decomposition of a permutation is $(1342)$: starting at $1$, follow its successive images back to $1$. Parentheses in the latter expression encode arrows around a permutation cycle, rather than an image list in reference order. ^[ref]

The two viewpoints answer different useful questions about the same object. An ordering tells us which object occupies each position; a bijection tells us where each label is sent and allows successive rearrangements through function composition. All permutations of $n$ labels form the symmetric group, whose subgroups are permutation groups. The counting viewpoint supplies $n!$ possibilities, a factorial; the mapping viewpoint reveals how rearrangements combine.

Composition convention. Apply the rightmost permutation first:For the displayed $\sigma =(1342)$ and the transposition $\tau =(12)$, direct evaluation givesFor instance, $(\sigma \tau )(1)=\sigma (2)=1$, but $(\tau \sigma )(1)=\tau (3)=3$. Thus function composition of permutations need not be commutative; the omitted labels in these permutation cycles are fixed points.

Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

The Picard theorem has two forms concerning the image of a holomorphic function. The Little Picard theorem permits at most one omitted value for a nonconstant entire function. The Great Picard theorem is local: near an isolated essential singularity, every complex value except possibly one occurs infinitely often. The complex exponential function $e^z$ and the function $e^{1/z}$ show that the possible exception is necessary.

Émile Picard presented the research in 1879 and published proofs in his 1880 memoir. Its opening footnote lists communications to the Académie des Sciences on 28 April, 19 May, and 20 October 1879; sections 3 and 15 contain the global and local results respectively. ^[ref]

A function is piecewise continuous on a finite interval if it is continuous between finitely many breakpoints and has finite one-sided limits at those points and at the endpoints. On $[0,\infty )$, require this on every finite interval. A Heaviside step function is an example; a function with an infinite blow-up at an interior point is not. This condition guarantees local integrability while allowing jumps in an input.

Place $N$ objects into $m\ge 1$ boxes, assigning each object to exactly one box. Here $N$ is a nonnegative integer and $m$ is a positive integer. The pigeonhole principle says that if $N>m$, some box contains at least two objects. Equivalently, a function from an $N$-element finite set to an $m$-element finite set cannot be an injective function when $N>m$. The boxes may stand for categories rather than physical containers. ^[ref]

For thirteen objects and twelve boxes, a box containing at least two is unavoidable. Thinking of thirteen people and their birth months gives the same conclusion: at least two share a birth month. Twelve people could occupy the twelve months one apiece, so thirteen is the smallest number that forces a shared month. The generalized pigeonhole principle quantifies how many objects must share a box when there are still more objects.

The useful surprise is that the pigeonhole principle proves existence without needing to locate the objects involved. Choosing a good assignment rule is often the creative step: equal labels can mean equal remainders, nearby points, or repeated states of an algorithm. Once the rule is chosen, a short count can force a relationship among a much larger collection of objects. The pigeonhole principle itself does not name the crowded box. When the objects and their assignments are explicitly available, a simple algorithm finds one: scan the objects, keep the first object seen in each box, and stop when a later object enters an occupied box. At most $m+1$ scanned objects are needed when $N>m$.

A geometric example. Any five points in a square of side one contain two whose Euclidean distance is at most $1/\sqrt{2}$. Give each point Cartesian coordinates $(x,y)\in [0,1{]}^2$. Divide both ranges into the intervals $[0,1/2)$ and $[1/2,1]$, making four cells. Thus $x<1/2$ belongs to the left half and $x\ge 1/2$ to the right half; $y<1/2$ belongs to the lower half and $y\ge 1/2$ to the upper half. This convention assigns every boundary point to exactly one cell.

The pigeonhole principle puts two of the five points in the same cell. Their differences in Cartesian coordinates have magnitude at most $1/2$, so the Pythagorean theorem gives their Euclidean distance $d$ the boundFor the plotted points $(0.1,0.1)$, $(0.4,0.4)$, $(0.8,0.2)$, $(0.2,0.8)$ and $(0.8,0.8)$, the first two occupy the lower-left cell and have Euclidean distance $\sqrt{0.3^2+0.3^2}=\sqrt{0.18}\approx 0.424264$. These Cartesian coordinates exhibit the forced pair; the argument applies to every placement of five points.

An arithmetic example. Assign eight integers to their seven residue classes modulo the integer modulus $7$. The pigeonhole principle gives two with the same residue class, so $7$ is an integer divisor of their difference. In $\{2,5,9,12,16,20,24,27\}$, for example, $2$ and $9$ satisfy the integer congruence $9\equiv 2(\mathrm{mod}7)$, and $9-2=7$. Likewise, assigning fractional parts to short intervals gives the Dirichlet approximation theorem; its existing article explains the approximation bound.

A pointed set is a set $X$ with a distinguished element $x_0\in X$. A morphism of pointed sets is a function carrying the distinguished element to the distinguished element. The resulting category of pointed sets has every singleton as a zero object.

For a sequence of functions $f_n:X\to \mathbb{R}$, pointwise convergence to a function $f:X\to \mathbb{R}$ meansIt is the ordinary limit of a sequence at each fixed input. The threshold $N=N(x,\epsilon )$ may depend on the input. Uniform convergence instead requires one threshold for the entire domain of a function.^[ref]

Consider the polynomial functions $f_n(x)=x^n$, for positive integers $n$, on the interval $[0,1]$. For $0\le x<1$ the geometric sequence $(x^n)$ tends to zero, while $f_n(1)=1$. ThusEach $f_n$ is a continuous function, but its pointwise limit is discontinuous at one. Holding an input still hides what happens near that endpoint. The analogous even-power example on $[-1,1]$ appears in the source.^[ref]

For $0<x<1$ and $0<\epsilon <1$, taking the logarithm of $x^n<\epsilon$ and dividing by the negative real number $\log x$ reverses the inequality:The floor function followed by adding one handles the strict inequality even when the ratio is an integer. For tolerance $\epsilon =0.01$, the first successful indices at $x=0.5$, $0.8$, and $0.99$ are respectively $7$, $21$, and $459$. At $x=0.5$, $2^{-6}=0.015625$ fails and $2^{-7}=0.0078125$ succeeds. At $x=0$ and $x=1$, the error from this pointwise limit is already zero, so $N=1$ works. Nevertheless, $N(x,\epsilon )$ grows without bound as $x\to 1$ from below. A finite list of accurate test inputs cannot establish uniform convergence.

An even sharper warning concerns integration. On $[0,1]$, define the continuous functionsTheir supports are $[0,1/n]$, with height $n$ at $1/(2n)$ and value zero at both ends. Every fixed $x>0$ eventually lies beyond the support, and $g_n(0)=0$ always, so pointwise convergence gives the zero function. Nevertheless, each integral is the area of a triangle: ${\int }_0^1{g}_n(x)dx=\frac{1}{2}(1/n)n=\frac{1}{2}$. Shrinking width compensates increasing height. Hence the limit of a sequence of these integrals differs from the integral of the pointwise limit; interchanging them requires additional conditions, such as uniform convergence preserves integrals on a compact interval.

For a periodic function that is piecewise continuously differentiable with finitely many pieces, its Fourier series converges at each point to the average of the two one-sided limits. At a point of continuity this is the function value. This includes the identified interval endpoints, whose two limits belong to the periodic extension. Absolute summability of the Fourier coefficients additionally gives uniform convergence.

A pointwise limit assigns to each input the limit of a sequence of function values there, on the specified domain of a function. It is unique wherever it exists, by uniqueness of the numerical limit of a sequence. Its existence supplies no common approximation rate over the domain of a function. For $f_n(x)=x^n$ on $[0,1]$, it equals zero below one and one at one; restricting to $[0,a]$ with $a<1$ gives the zero function.

Two Poisson structures are compatible when every linear combination ${\Pi }_1+\lambda {\Pi }_2$ is a Poisson structure. This family is a Poisson pencil. In terms of the Schouten-Nijenhuis bracket, compatibility means $[{\Pi }_1,{\Pi }_2]=0$, in addition to the two individual Jacobi identities. Constant Poisson bivectors are compatible. Over a real manifold, complex parameter values refer to the complexification of its function algebra.

The Poisson summation formula equates sample sums of a function and its Fourier transform. Exchanging broad and narrow sums can save work. Use the conventionFor every Schwartz function $f\in \mathcal{S}(\mathbb{R})$,Both series have absolute convergence. A sufficient hypothesis is that $f$ is smooth and ${\sup }_x(1+|x|{)}^M|f^{(r)}(x)|<\infty$ for all nonnegative integers $M,r$; its Fourier transform is then also a Schwartz function.^[ref] This assumption controls both decay and regularity; mere membership of an L1 space does not control values at individual sampling points, as periodization of an integrable function illustrates.

Here $\xi$ counts cycles per unit length. In the angular convention used elsewhere in Fourier analysis, $\tilde{f}(\omega )=\int f(x)e^{-i\omega x}dx$, so $\hat{f}(\xi )=\tilde{f}(2\pi \xi )$ and the right-hand side becomes ${\sum }_k\tilde{f}(2\pi k)$. Keeping the Fourier transform convention explicit prevents stray $2\pi$ factors. Changing the sample spacing leads to the Poisson summation formula for a Euclidean lattice.