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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 $E_0(z)={\int }_0^z{e}^{-t^2}{\int }_0^t{e}^{u^2}dudt$ 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.

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

Write $g_2=60G_4$ and $g_3=140G_6$, with the lattice sums taken over nonzero lattice elements. Reindexing the normally convergent series shows that the Weierstrass elliptic function $\wp$ is an even function, and consequently ${\wp }^{\prime }$ is an odd function. Expanding $(z-w{)}^{-2}$ near zero and cancelling the odd lattice sums gives the Laurent coefficients of the Weierstrass elliptic functionThus the elliptic function ${\wp }^{\prime 2}-4{\wp }^3+60G_4\wp$ has no pole at zero and its constant term is $-140G_6$. By periodicity it has no poles anywhere. An entire elliptic function is bounded on a fundamental parallelogram and hence everywhere, so Liouville theorem makes it constant. This proves the Weierstrass elliptic differential equation

We first verify that the cubic is smooth, rather than presupposing it when using its group law. There are three nonzero two-torsion points of a complex torus $h_j\in (\frac{1}{2}\Lambda )/\Lambda$. Oddness and periodicity imply ${\wp }^{\prime }(h_j)=0$. Since ${\wp }^{\prime }$ has a triple pole and no other poles on the torus, value multiplicity of an elliptic function says that these three distinct zeros exhaust its zeros and are simple. Put $e_j=\wp (h_j)$. Each $\wp -e_j$ has a double zero at $h_j$, since ${\wp }^{\prime }(h_j)=0$ and the zero of ${\wp }^{\prime }$ is simple. If $e_i=e_j$ with $i\ne j$, this same function would have at least four zeros counted with multiplicity, although its only pole is double. Therefore the half-period values of the Weierstrass elliptic function are distinct. They are roots of $4x^3-g_{2}x-g_3$, so all its roots are distinct and the affine curve is a smooth algebraic curve. Its projective closure isIt has the single point $O=(0:1:0)$ at infinity. The derivative with respect to $Z$ of the defining homogeneous polynomial is nonzero at $O$, proving smoothness there too.

For any finite value $a$, value multiplicity of an elliptic function gives exactly two solutions of $\wp (z)=a$, counting multiplicities. Evenness supplies the pair $z,-z$. At a nonzero half-period these coincide and the zero is double; elsewhere they are distinct. An even elliptic function $f$ therefore descends to a meromorphic function of $x=\wp (z)$ on the Riemann sphere. To justify descent at a half-period, put $z=h+u$: both $f(h+u)$ and $\wp (h+u)$ are even in $u$, and $\wp (h+u)-\wp (h)$ is $u^2$ times a nonvanishing even analytic function. The local Laurent series of $f$ is consequently meromorphic in the coordinate $x-e$. At zero the same argument uses $1/x=z^2+O(z^6)$. Every meromorphic function on the Riemann sphere is a rational function, proving that even elliptic functions are rational in the Weierstrass function.

For an arbitrary elliptic function $f$, its even part $(f(z)+f(-z))/2$ is $A(\wp )$. Its odd part divided by ${\wp }^{\prime }$ is an even meromorphic function, hence is $B(\wp )$ with $B$ rational. We have therefore proved the elliptic function-field decompositionUniqueness follows by taking even and odd parts. In particular the field of elliptic functions is $\mathbb{C}(\wp ,{\wp }^{\prime })$.

Define $\Phi (0)=O$ and $\Phi (z)=(\wp (z),{\wp }^{\prime }(z))$ elsewhere on $\mathbb{C}/\Lambda$. Periodicity makes the map well defined. Near zero, the projective chart $Y\ne 0$ has coordinates $X/Y=\wp /{\wp }^{\prime }=-z/2+O(z^5)$ and $Z/Y=1/{\wp }^{\prime }=-z^3/2+O(z^7)$, so the extension is holomorphic. The fiber description above proves injectivity: the two candidates $z,-z$ are distinguished by ${\wp }^{\prime }$ unless they coincide at a half-period. It also proves surjectivity, since every $x$ occurs and the two derivatives are the two allowed $y$ values, with $y=0$ at a branch value.

Finally, a nonvertical line $y=ax+b$ pulls back to the elliptic function ${\wp }^{\prime }-a\wp -b$, whose only pole is triple at zero. Its three zeros $z_1,z_2,z_3$, counted with intersection multiplicity, satisfy $z_1+z_2+z_3=0$ modulo $\Lambda$ by the zero-pole sum of an elliptic function. On the cubic, reflection $y\mapsto -y$ corresponds to $z\mapsto -z$. The chord-and-tangent group law therefore gives $\Phi (z_1)+\Phi (z_2)=\Phi (-z_3)=\Phi (z_1+z_2)$. Vertical lines give inverse pairs; tangencies are included through repeated zeros, and the identity cases follow from the extension at zero. This completes the Weierstrass uniformization by half-period values: $\Phi$ is a group isomorphism.

The Lusternik-Schnirelmann-Borsuk theorem has the following two equivalent covering formulations. For every integer $d\ge 0$, a cover of the sphere $S^d$ by $d+1$ closed sets has a member containing an antipodal pair. The same assertion holds with open sets in place of closed sets. Thus $d+1$ antipodal-pair-free open sets, or $d+1$ antipodal-pair-free closed sets, cannot cover $S^d$. The dimension-zero case simply says that one set covering the two-point sphere contains both points.

Here is why the two versions agree. A finite open cover of a compact metric space admits a closed shrinking that still covers: sufficiently small closed balls subordinate to the open cover can be grouped according to their containing open member. Applying the closed version to that shrinking proves the open version. Conversely, if a nonempty closed subset $C$ of $S^d$ avoids antipodal pairs, compactness gives positive distance between $C$ and $-C$. A sufficiently small open neighbourhood of $C$ still avoids antipodal pairs. Enlarge each member of a hypothetical closed counterexample in this way; the open version rules it out. Empty members cause no difficulty.

Another common equivalent formulation is the Borsuk-Ulam theorem: every continuous $f:S^d\to {\mathbb{R}}^d$ has $f(x)=f(-x)$ for some $x$, or, equivalently, every continuous odd function $g:S^d\to {\mathbb{R}}^d$ has a zero. For example, if $d+1$ antipodal-pair-free open sets covered $S^d$, a subordinate partition of unity $({\varphi }_1,\dots ,{\varphi }_{d+1})$ would give the odd functionIt cannot vanish, since some ${\varphi }_i(x)>0$, whereas antipodal-pair-freeness forces ${\varphi }_i(-x)=0$. In the other direction, if an odd function $g:S^d\to {\mathbb{R}}^d$ never vanishes, choose $d+1$ vectors $u_i$ forming a regular simplex centred at the origin in ${\mathbb{R}}^d$. For $d\ge 1$, the open sets $\{x:u_i\cdot g(x)>0\}$ cover $S^d$ and none contains an antipodal pair, contradicting the covering theorem.

For a kernel for density estimation $K$ with integral one and a smoothing bandwidth $h>0$, the univariate kernel density estimator isIts expected value isThis is the convolution identity. It holds wherever the integral is defined, in particular everywhere for the bounded kernel used below. For a nonnegative kernel for density estimation, Tonelli theorem also gives the identity as an extended nonnegative integral. A signed kernel for density estimation instead requires absolute integrability for this interchange.

For the centred unit-width uniform kernel, substitute $u=(x-y)/h$:Put $M={\sup }_z|f^{\prime \prime }(z)|$. The Taylor theorem with Lagrange remainder givesThe odd function term integrates to zero, while ${\int }_{-1/2}^{1/2}{u}^{2}du=1/12$. Hence the bias of a kernel density estimator obeysThis is a second-order pointwise bias bound for kernel density estimation. Existence and boundedness of the second derivative suffice for the stated remainder; continuity of that derivative is not an extra assumption.

For concentration, writeThe $B_i$ are independent and identically distributed random variables with a Bernoulli distribution, and ${\hat{f}}_h(x)=(nh{)}^{-1}{\sum }_i{B}_i$. The required concentration bound follows from the Hoeffding lemma. We prove its Bernoulli case directly. Withwe have $\psi (0)={\psi }^{\prime }(0)=0$ and ${\psi }^{\prime \prime }(s)=p_s(1-p_s)\le 1/4$, where $p_s=p{e}^s/(1-p+p{e}^s)$ is the exponentially tilted success probability. Therefore $\psi (s)\le s^2/8$ for every real $s$. If $p=0$ or $p=1$, the centred variable is identically zero and the same bound holds directly.

Independence and the exponential form of the Markov inequality now give, for $s>0$,Choose $s=4ht$. Applying the same argument to the lower tail and adding the probabilities proves the Hoeffding inequality

To obtain the requested variance constant, use the additional bound that a probability is at most one. Set $Z={\hat{f}}_h(x)-\mathbb{E}{\hat{f}}_h(x)$ and $a=2n{h}^2$. The tail integral formula for moments, justified by Tonelli theorem, yieldsThusThe clipping at one is essential for this integration argument to give that constant. It is an instance of a second moment bound from a clipped Gaussian tail. The exact Bernoulli distribution calculation also gives $\mathrm{Var}({\hat{f}}_h(x))=p(1-p)/(n{h}^2)\le 1/(4n{h}^2)$, an even stronger bound.

Choose the square root with $-\pi /2<\arg z<3\pi /2$, so the branch cut is the negative imaginary axis. The change of variable $u=e^{-i\pi /4}\sqrt{z}$ maps this sheet onto $\mathrm{Re}u>0$ and gives $z=i{u}^2$. Writing the exponential function as $e^{k\varphi (z)}$, its phase becomes exactly quadratic:The saddle point is therefore $u=1$, or $z_s=i$. On the descending line $u=1+iv$, $v\in \mathbb{R}$, the phase is $-1-v^2$. Its image is the parabolaThe contour runs from left to right, corresponding to $v$ decreasing from $+\infty$ to $-\infty$. A contour deformation in the right half of the $u$-plane moves the original indented contour to this line. The connecting tails vanish in the descending sectors, and no branch point or pole is crossed. Since $dz=-2(1+iv)dv$, the Gaussian integral and the odd function $v{e}^{-k{v}^2}$ giveHere the method of steepest descent actually gives an exact answer for $k>0$, because the quadratic phase and linear transformed amplitude have no further even correction.

The first of the Gaussian drift primitives, $E_0$, is an even function; the second, $E_1$, is an odd function. Their derivatives follow by the fundamental theorem of calculus. In particular $E_0^{\prime }$ is the Dawson function, which behaves as $1/(2z)+O(z^{-3})$. Since the error function tends to $\pm 1$, the derivative of $E_1$ behaves as $1/(2|z|)+O(|z{|}^{-3})$ at either end. Integration therefore givesAlso $F\sim z+\sqrt{\pi }/4$ on the left and $F\sim \sqrt{\pi }/4$ on the right. The two inner overlaps are consequentlyExpressing the outer expansions in $r=\epsilon z$ requires the left constant $1+A\log \epsilon$ and the right constant $B\log \epsilon$. Equating those constants gives the stated $\alpha ,\beta$. This is exactly how a logarithmic overlap creates a switchback term: $\log |r|=\log \epsilon +\log |z|$. Adding the outer solutions and removing these overlaps leaves the displayed additive composite expansion.

Jordan lemma states that if $a>0$ and $g$ is analytic in the upper half-plane outside a fixed circle, withthen on the upper semicircle $C_R$,Indeed, $|e^{iaR{e}^{i\theta }}|=e^{-aR\sin \theta }$. Using symmetry and $\sin \theta \ge 2\theta /\pi$ on $0\le \theta \le \pi /2$ gives

The residue of $f$ at an isolated singularity $a$ is the coefficient of $(z-a{)}^{-1}$ in its Laurent series, equivalentlyIf $f(z)=g(z)/(z-a{)}^k$, the Taylor expansion of the analytic function $g$ shows that

For the real integral, apply contour integration in the upper half-plane toJordan's lemma removes the semicircular arc. The only enclosed singularity is the double pole at $z=i$, whose residue isThe residue theorem therefore givesIts real part vanishes because $x^3\cos x/(1+x^2{)}^2$ is an odd function, while its imaginary part is the requested integral. Hence

For $p=0$ and $\lambda =1$, the linear ordinary differential equation is $y^{\prime \prime \prime \prime }=y$, whose general solution isBy the stated parity reduction, an even eigenfunction has the form $y=A\cos x+C\cosh x$. The conditions $y(c)=y^{\prime }(c)=0$ have a nonzero solution exactly whenthat is,An odd eigenfunction has the form $y=B\sin x+D\sinh x$, and the corresponding determinant givesThus $\lambda =1$ is an eigenvalue precisely under one of the two stated conditions.

For the minus sign, division by $\cos c\cosh c$ away from $c=\pi /2$ gives $\tan c=\tanh c$. On $(0,\pi /2)$, $\tan c>\tanh c$ for $c>0$, while on $(\pi /2,\pi )$ their signs differ, so there is no root. For the plus sign the equation isThere is no root in $(0,\pi /2)$. On $(\pi /2,\pi )$, the function $\tan c+\tanh c$ increases strictly from $-\infty$ to $\tanh \pi >0$, so it has exactly one root by the intermediate value theorem. Numerically,Hence the combined $\pm$ condition has exactly one solution in $0<c<\pi$.

The momentum operator in the position representation is $\hat{p}=-i\hslash d/dx$. Differentiating the given Gaussian wave packet givesThe probability density $|\Psi {|}^2$ is an even function, so the term proportional to the odd function $x$ has integral zero. Normalization supplies ${\int }_{-\infty }^{\infty }|\Psi {|}^{2}dx=1$, and hence