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At a non-planar point of a minimal surface, the shape operator is not zero. Minimality gives nonzero opposite principal curvatures, so $K<0$ nearby and $dN$ is invertible. The inverse function theorem makes the Gauss map a local diffeomorphism onto a patch of the sphere; part (ii) makes it conformal.

Choose stereographic coordinates $(u,v)$ on that sphere patch. Their metric is $4(d{u}^2+d{v}^2)/(1+u^2+v^2{)}^2$. Pulling them back by $N$, the conformal identity $N^{\ast }{g}_{S^2}=\lambda g_S$ givesThe positive coefficient shows that these are local isothermal coordinates. No general isothermal-coordinate existence theorem is needed: the nondegenerate conformal Gauss map supplies them explicitly.

Let $A=-dN$ be the self-adjoint shape operator. In an orthonormal principal basis, $⟨dN(v),dN(w)⟩=⟨Av,Aw⟩$, with diagonal coefficients $k_1^2,k_2^2$. The Gauss map is conformal with nonzero factor exactly when $k_1^2=k_2^2=\lambda >0$.

If the surface is minimal and has no umbilics, $k_2=-k_1$ and $k_1\ne 0$, so this equality holds. Conversely equality of the squares gives either $k_1=k_2$ or $k_1=-k_2$. The first possibility is excluded by the no-umbilic hypothesis. Hence $H=0$, proving minimality is equivalent to conformality of the Gauss map under the stated hypothesis. The factor is automatically positive because it equates positive definite inner products. The hypothesis is essential: a sphere has conformal Gauss map and nonzero mean curvature.

A minimal surface has zero mean curvature $H=(k_1+k_2)/2$; equivalently its area is stationary under compactly supported normal variations. Its principal curvatures satisfy $k_2=-k_1$, so its Gaussian curvature isEquality occurs precisely where its second fundamental form vanishes.

The element $ϵ=3+\sqrt{10}>1$ has norm $-1$. To prove that it is a fundamental unit, let $u=x+y\sqrt{10}$ be a positive real unit with $1<u<ϵ$. Its conjugate is $\pm 1/u$, so $y=(u-\overline{u})/(2\sqrt{10})>0$. For norm $-1$, the increasing function $u+1/u$ gives $y<(ϵ+1/ϵ)/(2\sqrt{10})=1$; for norm $+1$, $y<(ϵ-1/ϵ)/(2\sqrt{10})=3/\sqrt{10}<1$. Either contradicts integer $y$. There is no positive unit between one and $ϵ$. For any positive unit, multiply by a suitable integer power of $ϵ$ to place it in $[1,ϵ)$; it must then equal one. HenceTo classify the norm-six elements, a principal ideal generated by such an element has norm 6. Unique prime-ideal factorization and part (b) force it to be either ${\mathfrak{p}}_2{\mathfrak{p}}_3=(ϵ+1)$ or ${\mathfrak{p}}_2{\overline{\mathfrak{p}}}_3=(ϵ-1)$. Therefore its generator is an associate of one of these two elements. Since their norms are respectively 6 and $-6$, the unit multiplying the first must have even exponent and the unit multiplying the second must have odd exponent. Combining the two cases gives all integer solutions:Conversely every displayed element is integral, since $ϵ$ is a unit, and has norm 6. This proves completeness of the norm-six equation in the quadratic field of square root ten.

The PDF's restriction to $n=0,1,2,\dots$ misses conjugate solutions. In particular $(x,y)=(4,-1)$ satisfies the equation but is not in that restricted family; it is the $n=-1$ member. A nonnegative-index version becomes correct by adding the conjugates of every displayed element, with both signs. The restriction as printed cannot describe all signed integer pairs.

Write $r=\sqrt{10}$. The ideals are ${\mathfrak{p}}_2=(2,r)$, ${\mathfrak{p}}_3=(3,r+1)$ and ${\overline{\mathfrak{p}}}_3=(3,r-1)$, with residue fields of orders 2, 3 and 3 respectively. Thus their ideal norms are 2, 3 and 3.

The generators of ${\mathfrak{p}}_2^2$ are $4,2r,10$. They are multiples of 2, and the integer combination of 4 and 10 giving 2 proves ${\mathfrak{p}}_2^2=(2)$. Similarly ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$ is generated by $9,3r+3,3r-3,r^2-1=9$. The difference of the middle generators is 6; together with 9 this generates 3. Therefore the product is $(3)$.

The ideals ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ are coprime, so their product equals their intersection. The element $4+r$ belongs to both, hence $(4+r)\subseteq {\mathfrak{p}}_2{\mathfrak{p}}_3$. Both ideals have norm 6, so equality holds. Likewise $2+r$ belongs to ${\mathfrak{p}}_2$ and ${\overline{\mathfrak{p}}}_3$, and its norm is $-6$. ThusThe discriminant is 40. The Minkowski bound for ideal classes in a real quadratic field is $\sqrt{40}/2=\sqrt{10}<4$, so every ideal class has an integral representative of norm 1, 2 or 3. The only possible prime ideals with those norms are the ones just listed. Their classes satisfy $[{\mathfrak{p}}_2{]}^2=1$ and $[{\mathfrak{p}}_3]=[{\overline{\mathfrak{p}}}_3]=[{\mathfrak{p}}_2]$. Thus there are at most two classes. The ideal ${\mathfrak{p}}_2$ is not principal, since a generator would have norm $\pm 2$, already ruled out. Therefore the class number is exactly two.

Since $10\equiv 2(\mathrm{mod}4)$, the ring of integers is ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{10}]$, with field norm $N(x+y\sqrt{10})=x^2-10y^2$. Norms $\pm 2$ and $\pm 3$ are impossible: modulo five they would require a square congruent to 2 or 3. An algebraic integer is a unit exactly when its norm is $\pm 1$.

If 2 factored into two nonunits, their absolute norms would both be 2, since $N(2)=4$; this is impossible. The same argument for $N(3)=9$ excludes a factorization of 3. Finally $ϵ+1=4+\sqrt{10}$ has norm 6, so a nonunit factorization would require absolute norms 2 and 3, also impossible. Hence 2, 3 and $ϵ+1$ are irreducible. The conjugate $\overline{ϵ}+1$ is irreducible by the same argument.

The equality $6=2\cdot 3=(4+\sqrt{10})(4-\sqrt{10})$ gives inequivalent factorizations into irreducibles: no norm-six factor is associated to either 2 or 3, whose norms are 4 and 9. A principal ideal domain is a unique factorization domain. Consequently ${\mathcal{O}}_K$ is not a principal ideal domain.

A canonical transformation preserves the symplectic form and therefore the form of Hamilton's equations for arbitrary Hamiltonians. In one degree of freedom,Thus the transformation is canonical exactly when $[Q,P]=1$. Equivalently the chain rule gives $[F,G{]}_{q,p}=[Q,P{]}_{q,p}[F,G{]}_{Q,P}$ for arbitrary transformed functions, so the bracket and Hamiltonian dynamics are preserved precisely with this factor one. The condition also makes the Jacobian nonzero and hence supplies local invertibility; a global canonical change of variables additionally requires the stipulated global bijectivity on its domain.

For a time-independent function $F$, Hamilton's equations giveThus a constant of motion has $[F,H]=0$. For $F={\sum }_k{p}_k$ and the given Hamiltonian, $[F,H]=-{\sum }_k{V}_{q_k}$. Simultaneous translation $q_k\mapsto q_k+s$ leaves every difference $q_i-q_j$, and hence $V$, unchanged. Differentiation with respect to $s$ gives ${\sum }_k{V}_{q_k}=0$, so $[F,H]=0$ also in this example. This is conservation of total momentum from translational symmetry.

Hamilton's equations generate the phase-space vector field $(H_{p_1},\dots ,H_{p_n},-H_{q_1},\dots ,-H_{q_n})$. Its divergence vanishes:by equality of mixed derivatives. The Jacobian $J(t)$ of the flow obeys $\dot{J}=(\mathrm{div}V)J=0$ with $J(0)=1$. Consequently $d\tau =d{q}_1\cdots d{q}_{n}d{p}_1\cdots d{p}_n$ is preserved for every time on which the smooth flow exists. This proves Liouville theorem in Hamiltonian mechanics directly; explicit time dependence of the Hamiltonian would not change the divergence argument.

Choose a polynomial $P$ with $\Vert f-P{\Vert }_{\infty }<ϵ$, using part (i). For all sufficiently large $n$, its degree is at most $2n-1$, so $I_n(P)={\int }_0^{1}P$. Positive weights summing to one give $|I_n(f-P)|\le ϵ$, and $|{\int }_0^1(f-P)|\le ϵ$. HenceSince $ϵ$ is arbitrary, the quadratures converge to the integral of every continuous $f$. The positivity established in part (ii), not just polynomial exactness on its own, controls the approximation error uniformly in $n$.

For each node define the Lagrange interpolation polynomialThe nodes are distinct, so this is well defined, has degree $n-1$, and satisfies $L_j(x_{k,n})={\delta }_{jk}$. Its square has degree $2n-2$, within the assumed exactness range. ThereforeThe strict inequality follows because $L_j$ is not the zero polynomial. Exactness for the constant polynomial also gives ${\sum }_{j=1}^n{a}_{j,n}=1$. Thus the quadrature is a positive weighted average; the requested nonnegativity follows in stronger form.

For fixed $t$, let $B$ have the binomial distribution with parameters $m,t$. The Bernstein polynomial is $f_m(t)=\mathbb{E}[f(B/m)]$. Continuity on the compact interval gives uniform continuity and a finite bound $M=\Vert f{\Vert }_{\infty }$.

Given $ϵ>0$, choose $\delta >0$ so $|f(s)-f(t)|<ϵ$ whenever $|s-t|<\delta$. Since $\mathbb{E}[B/m]=t$ and $\mathrm{Var}(B/m)=t(1-t)/m\le 1/(4m)$, Chebyshev's inequality givesThe bound is independent of $t$, including both endpoints. Taking the supremum and then letting $m$ tend to infinity proves $f_m\to f$ uniformly on $[0,1]$. This provides the required polynomial-approximation theorem with its proof.

The Runge-Kutta method has $A=\left(\begin{array}{cc}1/4& 1/4-a\\ 1/4+a& 1/4\end{array}\right)$ and $b=(1/2,1/2{)}^T$. Solving the stage equations for $y^{\prime }=\lambda y$ gives the stability functionFor $a\ne 0$, the denominator roots have positive real parts: real roots are positive from their positive sum and product, while a complex conjugate pair has real part $1/(4a^2)>0$. For $a=0$ its sole root is $2$. Thus there are no poles or singular stage systems in the left half-plane.

Writing $z=x+iy$, direct expansion givesThis proves the A-stability of a symmetric two-stage implicit Runge-Kutta family:

Apply the numerical method to the test equation $y^{\prime }=\lambda y$ and put $z=h\lambda$. Its linear stability domain is the set of $z$ for which the computed amplification is stable, for a one-step method $|R(z)|\le 1$ with all stage solves well-defined. The method is A-stable if this includes the whole closed left half-plane (or its interior with boundary limits).

Balancing the time derivative, cubic nonlinearity and third spatial derivative gives $x\sim t^{1/3}$ and $u\sim t^{-1/3}$. Choose $\zeta =x/(3t{)}^{1/3}$ and $u=(3t{)}^{-1/3}w(\zeta )$. ThenThus the similarity solution isThe extra spherical term changes the coefficient of $w$. This third-order equation cannot be integrated directly into the standard second-order Painlevé II equation of the ordinary modified KdV reduction.

The initial data require $\Phi (\xi )=\exp [-(2\nu {)}^{-1}{\int }_{x_{\ast }}^{\xi }U(s)ds]$, up to a positive constant that cancels from $u$. Substituting this into the heat kernel formula gives the exponent $-G/(2\nu )$ withDifferentiate the kernel integral in $x$ and apply the Cole-Hopf transformation. Its constant prefactor cancels, leavingThe choice of $x_{\ast }$ adds an irrelevant constant to $G$; the usual convergence assumptions on the initial data justify the integrals and differentiation.

Set $u=-2\nu {\partial }_x\log \varphi$. Direct differentiation givesThus the Cole-Hopf transformation maps every nonvanishing solution of the heat equation to a solution of the viscous Burgers equation. Conversely the Burgers equation makes the ratio inside the derivative a function of time alone; a time-dependent rescaling of $\varphi$, which leaves $u$ unchanged, removes that function and gives the heat equation exactly.

False. A function of $\cos t$ is even in $t$, whereas $\sin (nt)$ is odd. If the identity held, replacing $t$ by $-t$ would imply $\sin (nt)=0$ for every $t$, contradicted by $t=\pi /(2n)$. Thus no polynomial of any degree has the required property.

True, with $R_n=T_n$. The hyperbolic cosine identity $\cosh ((n+1)t)=2\cosh t\cosh (nt)-\cosh ((n-1)t)$ and the initial values $1,\cosh t$ give $T_n(\cosh t)=\cosh (nt)$ by mathematical induction, using the same Chebyshev polynomial recurrence.

True. Define the Chebyshev polynomials by $T_0(x)=1$, $T_1(x)=x$, and $T_{n+1}(x)=2x{T}_n(x)-T_{n-1}(x)$. The cosine addition identities give the same recurrence for $\cos (nt)$, so mathematical induction proves $T_n(\cos t)=\cos (nt)$. For $n\ge 1$, the leading coefficient is $2^{n-1}$, proving that its polynomial degree is exactly $n$.