Divide the differential equation by $\sqrt{1-x^2}$. Its Sturm-Liouville form isHere $p(x)=\sqrt{1-x^2}$ and weight $w(x)=(1-x^2{)}^{-1/2}$. Multiply the equations for $T_n,T_m$ by the other polynomial and subtract. Integration by parts gives a boundary term $p(T_m{T}_n^{\prime }-T_n{T}_m^{\prime })$, which vanishes at both endpoints because polynomial derivatives are finite and $p\to 0$. HenceWith the standard nonnegative degree indices $n,m$, this proves orthogonality for $n\ne m$. If signed integer indices are used literally, the condition is $|n|\ne |m|$: $T_{-n}=T_n$, so merely $n\ne m$ would not suffice.

Set $x=\cos \theta$ and $Y(\theta )=y(\cos \theta )$. The chain rule gives $(1-x^2)y^{\prime \prime }-x{y}^{\prime }=Y^{\prime \prime }$, so $Y^{\prime \prime }+n^{2}Y=0$. For $n>0$, $Y=A\cos (n\theta )+B\sin (n\theta )$. A polynomial $y$ has $Y^{\prime }(0)=0$, forcing $B=0$. At $n=0$, $Y=A+B\theta$ and the same endpoint condition gives $B=0$. Therefore every polynomial solution is proportional toThe cosine recurrence yields $T_{n+1}=2x{T}_n-T_{n-1}$, $T_0=1,T_1=x$, confirming that these functions are polynomials. In this normalization the Chebyshev polynomial weight identity isIt is simply cosine orthogonality after $x=\cos \theta$.

Use the Fourier transform conventionThis convention also gives the sign of the inverse transform in 15A. The Parseval identity, in its bilinear form, isIn particular $\int |f{|}^2=(2\pi {)}^{-1}\int |\hat{f}{|}^2$. Initially take $f,g$ to be Schwartz functions; the formula then extends to $L^2$ with Fourier transforms interpreted by completion.

Here is a proof which also justifies the needed limit operations. For $\epsilon >0$, formAbsolute integrability allows Fubini's theorem to replace the transformed factors by their defining integrals. The elementary Gaussian integral givesso $I_{\epsilon }=\int f(x)\overline{(K_{\epsilon }\ast g)(x)}dx$, with convolution $(h\ast g)(x)=\int h(x-y)g(y)dy$. The kernel is nonnegative, has integral one, and its mass outside any fixed neighborhood of zero tends to zero. Since a Schwartz function is bounded and uniformly continuous, splitting the convolution difference into small and large $|x-y|$ proves $K_{\epsilon }\ast g\to g$ uniformly. As $f\in L^1$, the last integral consequently tends to $\int f\overline{g}$.

On the spectral side, repeated integration by parts makes the transforms rapidly decreasing; their product is integrable. The dominated convergence theorem therefore gives $I_{\epsilon }\to (2\pi {)}^{-1}\int \hat{f}\overline{\hat{g}}$. Equating the two limits proves the displayed identity. For general $L^2$ functions, approximate by Schwartz functions. The norm identity makes their transforms Cauchy in $L^2$ with the prescribed factor; the limit defines the $L^2$ Fourier transform independently of the approximation. Taking limits in the bilinear identity proves it for all $L^2$ pairs. Thus the proof includes the hypotheses and avoids unjustified exchanges of nonabsolutely convergent integrals.

The map is differentiable at $u$ when there exists a linear map $A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe increments are taken sufficiently small to stay in the open domain; $A=Df(u)$ is the derivative. This is a joint approximation in all directions, not just the existence of coordinate partial derivatives.

At $(a,b)$ with $ab\ne 0$, choose $\delta <\min (|a|,|b|)$. If $\Vert (h,k)\Vert <\delta$, neither coordinate changes sign, andThus the remainder in the differentiability definition is identically zero throughout that neighborhood. Consequentlywhich proves the requested differentiability directly.

Each coordinate lies in $[0,1]$. Since $0<c<1$, $x_i^c\ge x_i$, with equality exactly when $x_i=0$ or $1$. Summing gives $f(\mathbf{x})\ge 1$. Equality under the simplex constraint forces exactly one coordinate to be one. Thus the minimum is $1$, attained exactly at the vertices of the probability simplex.

For the maximum, $t\mapsto t^c$ is strictly concave on $[0,\infty )$. The Jensen inequality givesStrict concavity makes equality possible exactly when all coordinates agree. Hence the extrema of a concave power sum on a simplex areFor $n=1$ the simplex is a single point, and both descriptions coincide.

For the given orthonormal basis, the orthogonal projection isIt belongs to $W$, and $⟨v-\pi v,e_j⟩=0$ for every basis vector. Thus $v-\pi v\in W^{\perp }$. For any $w\in W$, decompose $v-w=(v-\pi v)+(\pi v-w)$. The summands are orthogonal, so the Pythagorean theorem in an inner-product space for the inner product givesEquality holds exactly when $w=\pi v$. Hence $\pi v$ is the unique closest point of $W$ to $v$. This proves the minimizing property as well as the formula, including $W=\{0\}$ and $W=V$.

For steady incompressible flow of constant density without a body force, the Euler equation and mass conservation giveApply the divergence theorem componentwise to a fluid volume. With $\mathbf{n}$ its outward normal, the integral momentum flux balance isIf a solid bounds part of the volume, retain its pressure contribution; that contribution supplies the force on the fluid. This derivation includes pressure as well as transported momentum.

Choose the axis in the incoming jet direction. The printed diagram measures $\alpha$ between that axis and the downstream sheet: it is the cone's half-angle. Along a streamline, the steady Euler equation integrates to the Bernoulli equation $p/\rho +|\mathbf{u}{|}^2/2=\text{constant}$. At the incoming free jet and the outgoing free sheet the pressure is atmospheric. With gravity neglected, the outgoing speed is therefore $u$. This argument uses far free surfaces; it does not assert atmospheric pressure on the wetted sphere.

At radial distance $r$, the sheet's circumference is $2\pi r\sin \alpha$. Its cross-section normal to its velocity is this circumference times its small thickness $d$. Conservation of mass gives $Au=(2\pi r\sin \alpha )du$, henceFor the axial momentum balance, transverse contributions cancel by axial symmetry. The outgoing axial momentum rate is $\rho A{u}^2\cos \alpha$ and the incoming rate is $\rho A{u}^2$. Uniform atmospheric pressure cancels on the closed control surface when expressed as gauge pressure. The sphere's force on the fluid is therefore $\rho A{u}^2(\cos \alpha -1)$ in the axis direction. By action and reaction, the fluid force on the sphere is downstream, with magnitudeThe downstream direction in the original diagram fixes this sign and the factor $1-\cos \alpha$. If one calls the full cone opening its vertex angle, that full opening is $2\alpha$; using $\alpha$ as the full opening would conflict with the printed thickness formula.

With metric signature $(+,-,-,-)$, four-momentum is $p^{\mu }=(E/c,\mathbf{p})$; for a particle of rest mass $m$ it satisfies $p^{\mu }{p}_{\mu }=m^2{c}^2$. For a massive particle $p^{\mu }=m{u}^{\mu }$, where $u^{\mu }$ is its four-velocity. The total four-momentum is the sum of the individual four-momenta. If that sum is timelike, its center-of-momentum frame exists and the total energy in that frame isThis follows because the spatial momentum is zero in that frame and the Minkowski norm is invariant under a Lorentz boost. A null total momentum, such as parallel photons, would not have such a rest frame.

For the photon striking an electron initially at rest, the initial invariant gives $W^2=m_e^2{c}^4+2m_e{c}^2{E}_{\gamma }$. The three final particles have total rest energy $3m_e{c}^2$. Their center-of-momentum energy is at least this value, with equality when all have zero relative momentum. Energy-momentum conservation therefore requiresThe photon-induced triplet pair-production threshold is consequently four electron rest-energy units. This is attainable as a kinematic threshold: at equality the center-of-momentum speed in the laboratory is $4c/5$, and boosting the three resting final particles gives each energy $5m_e{c}^2/3$ and momentum $4m_{e}c/3$. Their totals are the initial $5m_e{c}^2$ and $4m_{e}c$. Recoil of the original electron is why twice the rest energy is insufficient.

For the radial Helmholtz Dirichlet Green function, set $H(x,\xi )=xG(x,\xi )$. Multiplication of the radial equation by $x$ givesThe first endpoint follows from the required boundedness of $G$ at zero. For $k\ne 0$ and $\sin k\ne 0$, use the left solution $\sin kx$ and right solution $\sin k(1-x)$. Their Wronskian is $-k\sin k$. Matching continuity and the derivative jump therefore gives, with $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$,This is finite at zero, vanishes at one, is continuous at $x=\xi$, and has $G_x(\xi +,\xi )-G_x(\xi -,\xi )=1$. Thus it has exactly the prescribed delta source, not its negative. Its weighted reciprocity is $x^{2}G(x,\xi )={\xi }^{2}G(\xi ,x)$, as expected for the radial weight.

Integrating the Green function against the constant forcing givesThe integrals are respectively $-x\cos (kx)/k+\sin (kx)/k^2$ and $(1-x\cos (k(1-x)))/k-\sin (k(1-x))/k^2$. After cancellation their weighted sum is $(\sin (kx)-x\sin k)/k$. HenceThe values at zero are interpreted by continuity. At $k=0$ the limit exists: $G(x,\xi )=-\xi x_{<}(1-x_{>})/x$ and $y(x)=(x^2-1)/6$.

The exceptional nonzero values $k\in \pi \mathbb{Z}$ need a genuine qualification. There is then a nonzero bounded homogeneous solution $\sin (kx)/x$ satisfying both endpoint conditions, so an inverse Green function does not exist. Moreover the stated forcing is incompatible, not merely nonunique. If $h=xy$ solved the problem, multiply $h^{\prime \prime }+k^{2}h=x$ by $\sin (kx)$ and integrate by parts using $h(0)=h(1)=0$. The left side is zero, whereasThus there is no solution for nonzero $k\in \pi \mathbb{Z}$. This explains exactly when the generic formula is applicable.