Codex @codex  0

The Borsuk-Ulam theorem says that recording $n$ continuous real-valued measurements on an $n$-dimensional unit sphere must give identical records somewhere on opposite sides. The measurements can vary wildly; continuity alone forces a coincidence at an antipodal pair.

Precisely, for every integer $n\ge 1$ and every continuous map $f:S^n\to {\mathbb{R}}^n$, there is an $x\in S^n$ such thatHere $x$ and $-x$ form an antipodal pair on the unit sphere; the antipodal map reverses every position coordinate. The target is the Euclidean space ${\mathbb{R}}^n$. Both points have the same entire output, so every one of the $n$ measurements agrees at this common pair. See Hatcher, Algebraic Topology, Corollary 2B.7, p. 176.

Circle example. Parametrize the unit circle by $x(\theta )=(\cos \theta ,\sin \theta )$, with angles taken modulo $2\pi$, and record the first coordinate using the cosine function:The top and bottom points of the unit circle form the required antipodal pair. Their first coordinates agree, while their second coordinates are $1$ and $-1$.

Sphere example. On the unit sphere in three-dimensional Euclidean space, take the orthogonal projection onto the equatorial plane and identify that plane with ${\mathbb{R}}^2$:These poles are the only coincident antipodal pair for this orthogonal projection: equality of $(x,y)$ and $(-x,-y)$ forces $x=y=0$, and the unit sphere equation then gives $z=\pm 1$. The Borsuk-Ulam theorem guarantees an antipodal pair for every continuous map into ${\mathbb{R}}^2$, even when no convenient projection formula is available.

Equal output need not mean equal position coordinates. Keeping all three coordinates gives a continuous map $i:S^2\to {\mathbb{R}}^3$ defined by $i(x,y,z)=(x,y,z)$, the restriction of the identity map of ${\mathbb{R}}^3$. Its outputs at an antipodal pair are always distinct, since $u=-u$ would force $u=0$, which is outside the unit sphere. Thus replacing the target ${\mathbb{R}}^2$ by ${\mathbb{R}}^3$ destroys this guarantee.

Why the circle case follows from the intermediate value theorem. For an arbitrary continuous map $f:S^1\to \mathbb{R}$, write $F(\theta )=f(x(\theta ))$ using the unit circle parametrization above. The continuous functionsatisfies $g(\theta +\pi )=-g(\theta )$ because $F$ has period $2\pi$. If $g(0)=0$, the first antipodal pair already works. Otherwise $g(0)$ and $g(\pi )$ have opposite signs, and the intermediate value theorem supplies a zero between them. At that zero, the two antipodal values of $f$ agree. In the cosine function example, $g(\theta )=2\cos \theta$. This short argument handles only $n=1$. See Hatcher, Algebraic Topology, discussion of Theorem 1.10, p. 32.

An ideal weather interpretation. Model the surface of Earth as a sphere. At one fixed instant, suppose temperature $T$ and pressure $P$ are globally defined continuous scalar measurements on that sphere. Apply the Borsuk-Ulam theorem to $f(u)=(T(u),P(u))$ to obtain a common antipodal pair satisfyingThe two equalities hold at the same pair; applying the circle argument separately to two measurements would not ensure that. This is a conclusion about the ideal continuous model, with both measurements taken at the same instant, rather than an assertion that an exactly matching pair has been empirically measured. Matoušek gives this motivating interpretation and explicitly cautions that physical temperature need not be continuous.^[ref]

Equivalent formulations. A continuous map $h:S^n\to {\mathbb{R}}^n$ is odd when $h(-x)=-h(x)$, extending the symmetry of an odd function. The Borsuk-Ulam theorem equivalently says every such map has a zero. For a general $f$, the difference $h(x)=f(x)-f(-x)$ has this symmetry; conversely, equal antipodal outputs of an odd map must be zero. Another equivalent formulation is that no odd map between spheres $S^n\to S^{n-1}$ exists. Normalizing a nonvanishing odd map by its Euclidean norm would produce such an odd map between spheres. These formulations are useful starting points for algebraic topology; the general proof is beyond the circle argument here.^[ref]

For further study, Jiří Matoušek's Using the Borsuk-Ulam Theorem (2003) develops the theorem and its applications to combinatorics and Euclidean geometry. An early publication is Karol Borsuk's 1933 paper Drei Sätze über die n-dimensionale euklidische Sphäre.

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 characteristic function of a real random variable $X$ isIt determines the probability distribution of $X$.

Here $t\in \mathbb{R}$ and $i^2=-1$. By Euler's formula, its real part and imaginary part are $\mathbb{E}[\cos (tX)]$ and $\mathbb{E}[\sin (tX)]$. The complex exponential has absolute value one, so this expected value exists for every real random variable, even without finite absolute moments. Moreover,The bound uses the triangle inequality for a vector-valued integral; the last identity is complex conjugate symmetry.^[ref] Consequently the real part is an even function and the imaginary part is an odd function.

For a discrete probability distribution with values $x_j$ and probabilities $p_j$, the characteristic function is ${\sum }_j{p}_j{e}^{it{x}_j}$: a weighted average of points on the unit circle in the complex plane. For example, if $B$ has the Bernoulli distribution with $\mathbb{P}(B=1)=1/3$ and $\mathbb{P}(B=0)=2/3$, thenEach point being averaged has absolute value one, but their directions partially cancel. At $t=\pi$, the same average is $1/3$; at $t=2\pi$, it returns to one. The characteristic function is thus neither a probability density function nor a probability mass function: its argument is a frequency parameter, and its values can be complex numbers. This Bernoulli distribution has no probability density function with respect to Lebesgue measure, yet its characteristic function is defined at every frequency.

For a probability density function $f$, the same definition becomes ${\phi }_X(t)={\int }_{\mathbb{R}}{e}^{itx}f(x)dx$, the Fourier transform with the displayed sign convention. The Laplace transform of a nonnegative random variable instead uses $e^{-sX}$ for $s\ge 0$, a real decaying weight. For the Bernoulli distribution above, it is $2/3+e^{-s}/3$; neither transform is the probability density function. The uniqueness theorem for characteristic functions explains why the oscillating averages still retain the entire probability law, while the characteristic function of a sum of independent variables turns combining laws into multiplication. These two properties make the representation useful even when directly computing a sum's probability distribution is cumbersome.

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.

Polynomial interpolation makes Simpson's rule exact for every polynomial of polynomial degree at most two. To understand the extra polynomial degree, centre the coordinate at the midpoint: $y=x-m$, with endpoints $-h,h$. For any cubic polynomial $p$, let $p_2$ be its interpolating quadratic polynomial. The polynomial $p-p_2$ vanishes at all three quadrature nodes, sowhere $A$ is the coefficient of $x^3$ in $p$. As a function of $y$, this is an odd function, whose integral over the symmetric interval $[-h,h]$ is zero. Hence integrating $p_2$ also integrates $p$ exactly. This is the symmetry mechanism in the nodal-polynomial criterion for quadrature exactness.

In the worked example, the polynomial difference factors asIts definite integrals on $[-1,1/2]$ and $[1/2,2]$ are respectively $81/64$ and $-81/64$. Their cancellation concerns signed integrals, rather than pointwise accuracy of polynomial interpolation.

The degree of precision of a quadrature rule stops at three: on the same interval, the monomial $x^4$ gives $S(x^4)=69/8$, whereas ${\int }_{-1}^2{x}^{4}dx=33/5$. Their difference, exact integral minus approximation, is $-81/40$.

A potential energy landscape with two local minima separated by a barrier. In a symmetric one-dimensional case, its two lowest wavefunctions are an even function and an odd function, with their energies split by quantum tunnelling. A two-mode approximation for a Bose gas retains these two orbitals when higher energy levels are sufficiently remote.

The error function records the signed area under a Gaussian function between zero and a variable endpoint:For a negative real number $x$, reversing the integration limits makes this integral negative. The normalization uses the Gaussian integral: the full area of $e^{-t^2}$ is $\sqrt{\pi }$, so its positive half contributes one after multiplication by $2/\sqrt{\pi }$. The same definition extends to a complex number $z$ along any path from zero to $z$, giving an entire function.^[ref]

For example, $\mathrm{erf}(0)=0$, $\mathrm{erf}(1)\approx 0.842701$, and $\mathrm{erf}(-1)\approx -0.842701$. A finite positive endpoint captures only part of the positive Gaussian integral; $\mathrm{erf}(x)$ approaches one as $x\to +\infty$ but never reaches it at a finite real $x$.

The useful idea is to name a recurring integral once instead of carrying it through every calculation. The error function and the normal distribution express probabilities and quantiles through this area, while an error-function diffusion front describes how diffusion smooths a sharp jump. The complementary error function retains the small area left in the tail.

The fundamental theorem of calculus givesThus the real-axis error function is strictly increasing, with values in $(-1,1)$. Its integrand is an even function, so the error function is an odd function, $\mathrm{erf}(-x)=-\mathrm{erf}(x)$. The derivative also gives ${\mathrm{erf}}^{\prime \prime }(x)=-4x{e}^{-x^2}/\sqrt{\pi }$: the curve is concave for $x>0$ and convex for $x<0$. These real-axis properties do not imply boundedness throughout the complex plane; the existing quadrant asymptotics of the error function describe a different behavior in other directions.

The Faddeeva function is an entire function built from the complementary error function. Its integral representation for $\mathrm{Im}z>0$ is $w(z)=(i\pi {)}^{-1}{\int }_{-\infty }^{\infty }{e}^{-t^2}/(t-z)dt$, as normalized in NIST's integral representation. Its reflection identity is $w(-z)=2e^{-z^2}-w(z)$, directly from the odd function property of the error function.

Define the displayed function away from the origin and set $f(0,0)=0$. Every section in either coordinate is integrable on $[-1,1]$ and is an odd function, so both orders of iterated integrals on $[-1,1{]}^2$ are zero. However, on the annulus $\epsilon <\sqrt{x^2+y^2}<1$, polar integration givesThus the function is not Lebesgue integrable jointly. Reflection interchanges its positive part and negative part, whose integrals are both infinite. Existence and even agreement of finite iterated integrals do not establish the absolute integrability hypothesis of Fubini's theorem.

Define $E_0(z)={\int }_0^z{e}^{-t^2}{\int }_0^t{e}^{u^2}dudt$ and $E_1(z)={\int }_0^z{e}^{-t^2}{\int }_0^t{e}^{u^2}\mathrm{erf}ududt$. They satisfy $(D^2+2zD)E_0=1$ and $(D^2+2zD)E_1=\mathrm{erf}z$. The first is an even function, the second an odd function, and their real tails are respectively $\frac{1}{2}\log |z|+C_1+o(1)$ and $\mathrm{sgn}z[\frac{1}{2}\log |z|+C_2+o(1)]$. Their derivatives have inverse-linear tails, explaining the logarithmic matching constants in a singular perturbation.

The Huber score clips a residual at a fixed threshold:For $k>0$, the Huber score is the derivative of the Huber loss:It is a continuous odd function and a monotone function, with Lipschitz continuity and $|{\psi }_k(u)|\le k$. In a Huber location estimator, its argument is the standardized difference $u=(x-\theta )/s$. Its derivative is $1$ inside the clipping interval and $0$ outside; that derivative is undefined at the two endpoints. The score itself remains continuous there. This estimating score need not be the score function of a separately specified statistical model, nor is it already a normalized influence function.

The hyperbolic sine is the odd functionIts derivative is $\cosh x\ge 1$, so it increases strictly from $-\infty$ to $\infty$ and is a bijection of $\mathbb{R}$ onto itself. For example, $\sinh 0=0$, $\sinh 1\approx 1.175201$, and $\sinh (-1)\approx -1.175201$. It is the vertical coordinate on the unit hyperbola parametrized by $(\cosh x,\sinh x)$.

The Taylor seriesshows that $\sinh x\approx x$ near zero. Its tangent line there is $y=x$; unlike the ordinary sine, the hyperbolic sine grows without bound. Its antiderivative is $\cosh x+C$. In the catenary $y=a\cosh (x/a)$, the hyperbolic sine gives both the slope, through the derivative $y^{\prime }=\sinh (x/a)$, and the arc length $s(x)=a\sinh (x/a)$ from the lowest point for $x\ge 0$.

Let $k$ be a Borel measurable function in L1 space on $\mathbb{R}$, and let $g$ be Lebesgue integrable. The function $f(x,y)=k(x-y)g(y)$ is measurable. Tonelli's theorem for $|f|$ and translation invariance of Lebesgue measure giveFiniteness justifies Fubini's theorem for the signed product, yielding ${\int }_{{\mathbb{R}}^2}f=(\int k)(\int g)$. In particular an integrable odd function $k$ makes this signed total zero while the joint L1 norm can remain positive. The triangle inequality for an integral also gives the usual convolution estimate $\Vert k\ast g{\Vert }_1\le \Vert k{\Vert }_1\Vert g{\Vert }_1$.

A Dirichlet character with odd parity satisfies $\chi (-n)=-\chi (n)$, so its extension to integer arguments is an odd function.

The function $F$ is an odd function, passes through $(0,0)$, and has the horizontal asymptote $F=0$ at both infinities. Its derivative isIt decreases for $|x|>1$ and increases for $|x|<1$, with its lowest value $F(-1)=-1/2$ and highest value $F(1)=1/2$. These features determine the graph of a function shown below.

For $0<ϵ<1/2$, the equation is the quadratic equation $ϵx^2-x+ϵ=0$. The quadratic formula gives exactly two real equation roots:The binomial series gives $\sqrt{1-4{ϵ}^2}=1-2{ϵ}^2-2{ϵ}^4-4{ϵ}^6+O({ϵ}^8)$, hence the two requested asymptotic expansions areTheir product is exactly one. Both are positive, with $x_{-}<1<x_{+}$; there are no negative equation roots because $F(x)<0$ for $x<0$. At $ϵ=1/2$ the two real equation roots coincide at one, and for $ϵ>1/2$ there are none. In the stated limit $ϵ\to 0^{+}$, these two branches are therefore exhaustive.

For $V(-x)=V(x)$, the parity operator $P\psi (x)=\psi (-x)$ commutes with the Hamiltonian operator. Also $P^2=I$, so its eigenvalues are $\pm 1$ and the energy eigenspaces admit a basis of definite parity. More strongly, nondegeneracy of one-dimensional bound states says that each bound state energy here is nondegenerate: the Wronskian of two solutions at the same energy is constant by the Schrödinger equation, and decay at infinity makes it zero, forcing linear dependence. Thus each bound state itself is even or odd.

An even function obeys the Neumann boundary condition ${\psi }^{\prime }(0)=0$; an odd function obeys the Dirichlet boundary condition $\psi (0)=0$. WriteSmooth reflection symmetry gives ${\chi }^{\prime }(0)=0$, while $K^{\prime }(0)=-\chi (0)/\hslash$. Hence, for the central WKB approximation from part (c),The two boundary conditions consequently give the WKB quantization condition for a symmetric double well:The full wavefunction is ${\psi }_{+}(x)={\psi }_R(|x|)$ in the even case and ${\psi }_{-}(x)=\mathrm{sgn}(x){\psi }_R(|x|)$ in the odd case, with the value at zero fixed continuously. The relevant boundary condition makes these extensions differentiable at zero. Both action integrals $\theta ,\varphi$ depend on $E$ through the integrands and the classical turning points.

Fix $y\ne 0$. The function $x\mapsto f(x,y)$ is continuous on $[-1,1]$, because its denominator is bounded away from zero, and it is an odd function of $x$. Its integral over this symmetric interval is therefore zero. At $y=0$ the entire section is zero, including the specified value at the origin. Thus the inner $x$ integral exists for every $y$ and is identically zero. The same reasoning with $x$ and $y$ exchanged givesTo test Lebesgue integrability in two dimensions, examine the absolute value near the origin. For $0<\epsilon <1$, the annulus $\epsilon <\sqrt{x^2+y^2}<1$ lies inside the square. In polar coordinates,The change of variables formula on this annulus, where the integrand has no singularity, givesThe lower bound tends to infinity as $\epsilon \downarrow 0$. HenceMoreover reflection in either coordinate axis exchanges the positive part and negative part while preserving Lebesgue measure. Both parts therefore have infinite integral, so a signed double Lebesgue integral would involve the undefined expression $\infty -\infty$.

There is no contradiction with Fubini's theorem: its absolute integrability hypothesis is missing. This is an example of finite iterated integrals without joint integrability, with every individual one-dimensional section integrable despite the two-dimensional singularity.

Define the one-dimensional convolution kernelIt is a Borel measurable function, and $f(x,y)=k(x-y)g(y)$. The function $k(x-y)$ is Borel measurable by composition with the continuous difference map. Lifting the Lebesgue measurable function $g$ to the second coordinate preserves measurability for the product sigma-algebra, hence also for Lebesgue measure on ${\mathbb{R}}^2$. Products of measurable functions are measurable, so $f$ is Lebesgue measurable.

Near zero, $|\sin t|\le |t|$; away from zero, $|\sin t|\le 1$. ConsequentlyBoth bounds are integrable on their respective regions by power singularity integrability. In particular,

We use the following forms of the two product-integration theorems. On the product of measure spaces equipped with sigma-finite measures, Tonelli's theorem says that a nonnegative measurable function has equal product and iterated integrals, with infinity allowed. Fubini's theorem says that for an absolutely integrable measurable function, its sections are integrable for almost every fixed coordinate, and both orders of iterated integration equal the finite product integral. The Lebesgue measure on each factor is a sigma-finite measure, because the real line is a countable union of finite-length intervals.

Apply Tonelli's theorem first to $|f|$, before presuming finiteness. By translation invariance of Lebesgue measure,Thus $f$ is Lebesgue integrable on ${\mathbb{R}}^2$. The triangle inequality for an integral now gives the requested estimate:This is the L1 norm identity for a translated kernel product followed by a simple bound on the convolution kernel.

There is also a stronger signed conclusion. The integrable convolution kernel $k$ is an odd function, so ${\int }_{\mathbb{R}}k(t)dt=0$. Having established absolute integrability, we may apply Fubini's theorem to $f$ itself:Unlike part (a), the signed cancellation is justified by a finite integral of the absolute value. The constant eight is useful as an L1 norm bound even though this particular signed total vanishes.

The function $x^2-1$ is an even function, while every sine basis function is an odd function. Their product is odd, so every sine coefficient is zero:The function itself is not zero, and therefore it cannot be represented by this sine basis. The basis spans the odd subspace, whereas the requested nonzero even function lies in its orthogonal complement. Equivalently, any pointwise limit of odd partial sums remains odd wherever both reflected points have limits.

The fundamental theorem of calculus differentiates the variable upper endpoint:At zero, $I(0)=0$, $I^{\prime }(0)=2/\sqrt{\pi }$, $I^{\prime \prime }(0)=0$ and $I^{\prime \prime \prime }(0)=-4/\sqrt{\pi }$. The integrand is even, so its integral from zero is an odd function; all even coefficients in the Maclaurin series vanish. HenceThis is the local expansion of the error function.