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Multiplication by $(1-x^2{)}^{-1/2}$ puts the equation in Sturm-Liouville theory form:Multiply the equations for $T_n,T_m$ by $T_m,T_n$, subtract and integrate. The boundary term vanishes because $\sqrt{1-x^2}=0$ at both endpoints and the polynomial derivatives are bounded. Therefore, for $n^2\ne m^2$,This is the usual Chebyshev polynomial orthogonality.

Differentiating the original equation and writing $U_n=T_n^{\prime }$ givesIts self-adjoint form isThe same subtraction argument now yields, for $n^2\ne m^2$,These two relations form the Chebyshev derivative Sturm-Liouville pair.

A map $h:X\to X$ is a contraction if there is a constant $q<1$ such thatfor every $x,y\in X$.

The contraction mapping theorem states that a contraction of a nonempty complete metric space has a unique fixed point, and that the iterates from every starting point converge to it. To prove this, choose $x_0\in X$ and put $x_{n+1}=h(x_n)$. Thenso, for $m>n$,Thus $(x_n)$ is Cauchy and converges, by completeness, to some $x_{\ast }$. A contraction is continuous, soIf $y_{\ast }$ is another fixed point, thenforcing $x_{\ast }=y_{\ast }$.

For the Newton mapone has $g(r)=r$ andOn the given neighbourhood,Since $f(r)=0$, choose a closed interval $U^{\prime }$ centred at $r$ and contained in $U$ so small that $M|f(x)|/{\delta }^2\le 1/2$ there. Thenso $g(U^{\prime })\subseteq U^{\prime }$ and $g$ is a contraction on the complete interval $U^{\prime }$. The local contraction proof for Newton iteration therefore shows that $r$ is the unique fixed point of $g$ on $U^{\prime }$.

The Eisenstein criterion says that a primitive polynomialis irreducible if there is a prime $p$ such thatBy the Gauss lemma for polynomials, irreducibility over $\mathbb{Q}$ and over $\mathbb{Z}$ agree for primitive polynomials.

If the urn contains $g$ green balls, it contains $g+2$ red balls: both update rules preserve the difference $R-G=2$. Until absorption at $g=0$, the green count is therefore a birth-death chain withThe functionis harmonic, since

Let ${\tau }_0$ and ${\tau }_N$ be the hitting times of $0$ and $N$. The optional sampling theorem for a supermartingale, applied to the bounded stopped martingale $h(G_{n\wedge {\tau }_0\wedge {\tau }_N})$, givesSolving,Letting $N\to \infty$, the events on the left increase to eventual termination. The harmonic hitting probability for the balanced-difference urn is therefore

A gauge transformation isBecause partial derivatives commute, the added contribution to $F_{\mu \nu }$ isso the electromagnetic field tensor is gauge invariant.

DefineSince ${\partial }_0=c^{-1}{\partial }_t$ and $A_0=-\Phi /c$,

The identityfollows directly from $F=dA$. Its spatial and mixed components areFor the other two equations define the four-currentThen ${\partial }_{\nu }{F}^{0\nu }={\mu }_0{j}^0$ givesand ${\partial }_{\nu }{F}^{i\nu }={\mu }_0{j}^i$ givesThis is the Covariant Maxwell equation with the minus-plus-plus-plus metric.

Usingone findsThis is the electromagnetic energy density.

For a null vector, the trace term in $T^{\mu \nu }{k}_{\mu }{k}_{\nu }$ vanishes. PutAntisymmetry gives $q\cdot k=0$. A vector orthogonal to a null vector has nonnegative Minkowski norm, soExplicitly, in a frame with $k^{\mu }=\kappa (1,1,0,0)$,Hence the null energy condition for the electromagnetic field is strict whenever the contraction is nonzero:

The convolution isWe claim that the $n$-fold convolution isThis is true for $n=1$. If it holds for $n-1$ and $x\ge 0$, thenand the convolution vanishes for $x<0$. This is the gamma density from repeated exponential convolution.

The Fourier transform is

The convolution theorem statesIndeed, Fubini and $u=x-y$ giveSince $\hat{F}(k)=(1+ik{)}^{-1}$, induction immediately verifies

For Parseval identity, take $g(x)=\overline{f(-x)}$. ThenUsing Fourier inversion at zero, which follows from the supplied delta identity,and the convolution theorem gives

Apply this to $F_{n+1}$. Sinceone obtains the Even rational Parseval integral

For a variation $y+\epsilon \eta$,Integration by parts givesFor fixed endpoints, $\eta (0)=\eta (x_0)=0$. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationA solution makes the first variation vanish for every admissible variation, so it is a stationary candidate; whether it is a minimum or maximum is decided by higher variations. If endpoint values are free, $\eta$ is arbitrary there, and the boundary term instead vanishes under the natural conditionsThus the same Euler-Lagrange solution is stationary for all free-endpoint variations. These are the natural boundary conditions for a free endpoint.

Forthe two equations areSet $u=y+z$ and $v=y-z$. ThenThe conditions $y(0)=z(0)=0$ giveand hence the most general solution is

Free conditions at $x_0$ are $y^{\prime }(x_0)=z^{\prime }(x_0)=0$. Adding and subtracting them givesThus $A=0$. The zero solution exists for every $x_0$, while nonzero solutions exist precisely whenFor those values they form the one-parameter familywhere $C$ is arbitrary. This is the free-endpoint normal mode of a coupled variational functional.

For $0<a<2$, the beta-integral evaluation givesDifferentiation under the integral sign near $a=1$ is justified by domination at zero and infinity. ThereforeWriting $a=1+t$ givesSincethe logarithmic moments of the Cauchy kernel yieldThe same expansion gives $I^{\prime \prime }(1)={\pi }^3/8$, agreeing with the supplied identity.