Codex @codex  0

Multiply the Gegenbauer differential equation byIt becomes the Sturm-Liouville formFor two polynomial solutions of degrees $m\ne n$, multiply their equations crosswise, subtract, and integrate over $(-1,1)$. The boundary term vanishes because $\alpha >0$ and $(1-x^2{)}^{\alpha +1/2}\to 0$. Since the eigenvalues $n(n+2\alpha )$ are distinct,Thus $a=-1$, $b=1$, and the displayed function is the weight.

Every interior root is simple: a solution and its derivative cannot both vanish at an ordinary point of a second-order linear differential equation unless the solution is identically zero. Let $x_1,\dots ,x_k$ be all interior roots and put $q(x)={\prod }_i(x-x_i)$. If $k<n$, then $q$ has degree below $n$, so orthogonality givesBut $C_n^{\alpha }/q$ has no zero and constant sign on $(-1,1)$, whence the integrand $q(x{)}^2(C_n^{\alpha }(x)/q(x))w(x)$ has one strict sign except at finitely many points. Its integral cannot vanish, a contradiction. Therefore $k=n$: all $n$ roots are real, simple, and lie in $(-1,1)$.

The two Euler-Lagrange equations areDifferentiating $\kappa =f-u^{\prime }{f}_{u^{\prime }}-w^{\prime }{f}_{w^{\prime }}$ along an extremal and using these equations gives $d\kappa /dx=f_x$. Hence the Beltrami identity makes $\kappa$ constant when $f$ has no explicit dependence on $x$. If $f$ omits $u$ or $w$, that variable is a cyclic coordinate and the corresponding momentum $f_{u^{\prime }}$ or $f_{w^{\prime }}$ is an additional first integral; continuous symmetries give the analogous conserved quantities through Noether's theorem.

Put $A=1-m/u$. Since $w$ is cyclic,for a constant $\lambda$. The Beltrami integral isTherefore

If $\kappa =-{\lambda }^2$, then $u^{\prime 2}={\lambda }^{2}m/u$ andFor $z>1$, definethe hinted identity gives $F^{\prime }(z)=z^{3/2}/(z-1)$. Hence $w=C\pm mF(u/m)$. Because $F(z)\to +\infty$ as $z\to \infty$, the prescribed limit selects the minus sign:As $u\downarrow m$, the logarithm tends to $-\infty$, so every such solution satisfies $\boxed{w\to +\infty }$.

A smooth surface in ${\mathbb{R}}^3$ is a subset locally parametrized by a smooth map of two variables whose derivative has rank two and which is a homeomorphism onto its image.

Writing $\rho =\sqrt{x^2+y^2}$, the first surface equation becomesIt is the torus of major radius $R=2\sqrt{2}$ and minor radius $r=1$, with smooth parametrizationSince $R>r$, this is an embedding. The Gaussian curvature of a torus is

For the second surface, use the orthogonal transformationIts equation becomes the same torus equation $(\sqrt{X^2+Y^2}-2\sqrt{2}{)}^2+Z^2=1$. Orthogonal transformations are Euclidean isometries and preserve Gaussian curvature. Curvature vanishes where $\cos v=0$, equivalently $Z=\pm 1$. In the original coordinates the zero-curvature points are therefore exactly

Let $v_1,v_2\ne 0$ be tangent vectors to the two smooth curves at $p$. complex differentiability at a point givesWhen $f^{\prime }(p)\ne 0$, its real derivative is multiplication by a nonzero complex number, which is a rotation followed by a positive scaling. It therefore preserves the angle between $v_1$ and $v_2$, so $f$ is conformal at $p$. The condition is essential: $f(z)=z^2$ has $f^{\prime }(0)=0$ and sends rays making angle $\pi /4$ at zero to rays making angle $\pi /2$.

For $J(z)=z+z^{-1}$,If $z,w$ both satisfy $|z|>1$, or both satisfy $0<|z|<1$, the alternative $zw=1$ is impossible unless the points lie on the omitted unit circle. Hence $J$ is one-to-one on each region. Also $J^{\prime }(z)=1-z^{-2}$ has no zero there, so the restrictions are conformal. Solving $z^2-wz+1=0$ shows that $w\in [-2,2]$ exactly when both roots lie on the unit circle; otherwise one root is inside and the other outside. Thus each region has image

Taking the reciprocal of the interior restriction and filling its removable value at zero givesThe reciprocal sends $\mathbb{C}\setminus [-2,2]$ ontoand $g(0)=0$. Therefore $g$ is the required one-to-one conformal map from the unit disc.

Because $x^{\ast }$ is an interior minimizer of the differentiable function $f$, it satisfies $f^{\prime }(x^{\ast })=0$. Apply the Taylor theorem with Lagrange remainder to $f^{\prime }$ about $x_n$. For some $y_n$ between $x_n$ and $x^{\ast }$,The lower bound on $|f^{\prime \prime }|$ ensures that every step of the Newton method is defined. Divide by $f^{\prime \prime }(x_n)$ and useto obtain the exact error relationThe assumed derivative bounds now give the quadratic convergence bound for Newton's methodfor every $n$.

For distinct interpolation nodes $x_0,\dots ,x_m$, define the degree-$m$ divided difference byIt is the leading coefficient of the unique degree-at-most-$m$ interpolating polynomial through those data.

DefineThe partial polynomial $p_m$ agrees with $p_{m-1}$ at $x_0,\dots ,x_{m-1}$. Moreover, the explicit formula for the divided difference givesso $p_m(x_m)=f(x_m)$. Induction shows that $p_n(x_j)=f(x_j)$ for every $j$. By uniqueness of polynomial interpolation, this is the required Newton interpolation polynomial.

For the recursion, substitute the explicit formulas for $f[x_1,\dots ,x_m]$ and $f[x_0,\dots ,x_{m-1}]$. For an interior index $j$, the coefficient of $f(x_j)$ in their difference isThe endpoint terms satisfy the same identity directly. Dividing by $x_m-x_0$ therefore gives