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For an arclength-parametrized curve define the Frenet frame by $T={\alpha }^{\prime }$, $k=|{\alpha }^{\prime \prime }|>0$, $N=T^{\prime }/k$ and $B=T\times N$. Differentiating the unit lengths gives diagonal frame-derivative coefficients zero. Also $T^{\prime }=kN$ implies $N^{\prime }\cdot T=-k$, while $B^{\prime }\cdot T=-B\cdot T^{\prime }=0$. Define the torsion of a space curve by $\tau =-B^{\prime }\cdot N$. Orthogonality then givesThese are the Frenet formulae. Differentiating $B=T\times N$ also verifies the chosen torsion sign; equivalently $\tau =\det ({\alpha }^{\prime },{\alpha }^{\prime \prime },{\alpha }^{\prime \prime \prime })/|{\alpha }^{\prime }\times {\alpha }^{\prime \prime }{|}^2$.

The monodromy theorem states that a germ admitting analytic continuation along every path in a simply connected domain extends to a single-valued holomorphic function there. Equivalently continuations along homotopic paths with fixed endpoints agree. For a biholomorphism between the two tori, lift its composition with $\mathbb{C}\to \mathbb{C}/{\Lambda }_1$ through the second covering. Simple connectivity, or monodromy applied to local inverse charts, yields a holomorphic lift $F:\mathbb{C}\to \mathbb{C}$.

Lift the inverse similarly. The compositions of the lifts differ from the identity by lattice translations; adjusting the lifts shows that $F$ is bijective. By the allowed classification, $F(z)=az+b$ with $a\ne 0$. For $\lambda \in {\Lambda }_1$, its two arguments $z,z+\lambda$ represent the same torus point, so $a\lambda =F(z+\lambda )-F(z)\in {\Lambda }_2$. Thus $a{\Lambda }_1\subseteq {\Lambda }_2$. Apply the inverse lift to obtain the reverse inclusion. ThereforeThe translation $b$ affects the chosen origin, not the period lattice.

An elliptic function for a lattice $\Lambda$ is a meromorphic function on $\mathbb{C}$ satisfying $f(z+\lambda )=f(z)$ for all $\lambda \in \Lambda$. Its order is the number of poles, counted with multiplicity, in a fundamental parallelogram, equivalently the degree of its map from the torus to the Riemann sphere. Differentiation preserves the periods. If the distinct poles have orders $r_1,\dots ,r_s$, the derivative has exactly the same pole locations with orders $r_j+1$. ThereforeThe bounds use $1\le s\le m$, valid for a nonconstant elliptic function.

For the Weierstrass elliptic function $\wp$, $f=\wp {\wp }^{\prime }$ has a single pole of order five, so its derivative has order six. For the second example choose $c$ not a critical value of $\wp$ and choose $a$ away from zero and the two points where $\wp =c$. Thenhas a double pole at $a$ and three simple poles: the quotient has residue $-2$ at zero and residue 1 at each of its two other poles. No locations overlap. Hence its order is five and its derivative has order $3+2+2+2=9$, as required.

Assume $K$ is nonempty; without that necessary hypothesis the assertion is false. Put $d={\inf }_{z\in K}\Vert x-z{\Vert }_p$ and choose $y_n\in K$ with $\Vert x-y_n{\Vert }_p\to d$. Convexity gives $(y_n+y_m)/2\in K$, so its distance to $x$ is at least $d$. Apply the supplied uniform-convexity inequality to $x-y_n$ and $x-y_m$:Thus the minimizing sequence is Cauchy. Completeness of the Banach space ${\ell }^p$ gives $y_n\to y$; closedness gives $y\in K$, and continuity of the norm gives $\Vert x-y{\Vert }_p=d$. The minimum distance is attained. This is the nearest point in a uniformly convex Banach space property. The same inequality applied to two minimizers shows uniqueness, though existence alone was requested.

Apply Hölder's inequality to the finite sum with the second sequence $|x_j+y_j{|}^{p-1}$:Divide by the last power if the sum is nonzero. The resulting bound is independent of $N$, so passing to the limit proves both membership in ${\ell }^p$ and the Minkowski inequality

Let $q=p/(p-1)$. The scalar Young inequality $ab\le a^p/p+b^q/q$ follows by maximizing $ab-a^p/p$ over $a\ge 0$, obtaining $b^q/q$. If both norms are nonzero, apply it to $|x_j|/\Vert x{\Vert }_p$ and $|y_j|/\Vert y{\Vert }_q$, and sum over a finite initial segment. Passing to the increasing limit givesZero norms give the trivial case. This proves Hölder's inequality without assuming convergence of the product series in advance.

Suppose $y^2+21=x^3$. Its positive left side gives $x>0$. Reduction modulo four shows that $x$ is odd and $y$ even. Moreover $3\mid y$ would make the valuation of $y^2+21$ at 3 equal to one, impossible for a cube. The same argument excludes $7\mid y$.

In $\mathbb{Z}[s]$, factor $(y+s)(y-s)=(x{)}^3$. Any common prime ideal divisor divides both $2y$ and $2s$, so must lie above 2, 3 or 7. The odd norm excludes 2, and the previous divisibility arguments exclude 3 and 7. The two ideals are therefore coprime. Unique factorization of ideals gives $(y+s)=I^3$. The class group has exponent two, so $[I{]}^3=1$ implies $[I]=1$. Thus $I=(a+bs)$, and the only units, $\pm 1$, can be absorbed into a cube. It follows thatComparing the coefficient of $s$ gives $1=3b(a^2-7b^2)$, an impossibility over the integers. Consequently there are no integer solutions: this is the no integer solutions of the Mordell equation with constant minus twenty-one result. The class-group argument is essential: element factorization was not assumed to be unique.

For the ideal class group of the imaginary quadratic field of discriminant minus eighty-four, write $s=\sqrt{-21}$. Here ${\mathcal{O}}_K=\mathbb{Z}[s]$, $D_K=-84$, and the Minkowski bound for ideal classes is $(2/\pi )\sqrt{84}<6$. Thus every class has a representative of norm at most five. The relevant prime ideals areThe first two are ramified: ${\mathfrak{p}}_2^2=(2)$ and ${\mathfrak{p}}_3^2=(3)$. Also ${\mathfrak{p}}_5^2=(2-s)$, as both sides have norm 25 and $s\equiv 2(\mathrm{mod}{\mathfrak{p}}_5)$. The element $3+s$ has norm 30 andThere are no elements of norms 2, 3, 5 or 6, since the element norm is $a^2+21b^2$. Hence the classes of ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$ and their product are all nontrivial and distinct. The relation makes the last product equal to the class of ${\mathfrak{p}}_5$. Ideals of norm four are principal, and the two primes above five have the same class of order two. The norm bound therefore gives no further classes. Thus

Choose a nonzero integral ideal $I$ in the inverse of any desired ideal class. For the nonzero element $x\in I$ supplied by the norm bound, $J=(x)I^{-1}$ is integral, belongs to the desired class and satisfiesThere are only finitely many integral ideals of bounded norm: identify ${\mathcal{O}}_K$ with ${\mathbb{Z}}^n$ as an additive group, so ideals of norm $m$ are among its subgroups of index $m$, of which there are finitely many. Since every ideal class has one of these bounded-norm representatives, the ideal class group is finite.

Use the given square-class assignments and ${\omega }^3=1$, $1+\omega +{\omega }^2=0$. The character values, in the specified class order, areFor example the fourth entry of ${\chi }_S$ is $((-{\omega }^2{)}^2-\omega )/2=0$, and that of ${\chi }_A$ is $\omega$. A class of centralizer order $c$ has size $24/c$, so a character norm is the sum of its squared moduli divided by these centralizer orders. ThereforeBoth are genuine characters by their symmetric/exterior-square construction. A character norm is the sum of squares of its irreducible multiplicities, so both characters are irreducible.

For a representation space $V$, let ${\chi }_S$ and ${\chi }_A$ be the characters on ${\mathrm{Sym}}^{2}V$ and ${\bigwedge }^{2}V$. If the eigenvalues of $\rho (g)$ are ${\lambda }_1,\dots ,{\lambda }_d$, the respective traces are ${\sum }_{i\le j}{\lambda }_i{\lambda }_j$ and ${\sum }_{i<j}{\lambda }_i{\lambda }_j$. Since $\chi (g)=\sum {\lambda }_i$ and $\chi (g^2)=\sum {\lambda }_i^2$, this givesFinite-order matrices over $\mathbb{C}$ are diagonalizable, so this eigenvalue argument applies to every group element. Equivalently the symmetric and antisymmetric projections on $V\otimes V$ are $(I\pm \text{swap})/2$.

If $\rho$ affords an irreducible character $\chi$, entrywise complex conjugation gives a representation with character $\overline{\chi }$. An invariant complex subspace for the conjugate representation conjugates back to one for $\rho$, so irreducibility is preserved. If representations afford $\chi ,\psi$, their tensor product has trace $\chi (g)\psi (g)$, proving that their product is a character.

With the usual character inner product,The character orthogonality and the first assertion therefore make this equal to one if $\chi =\overline{\psi }$ and zero otherwise. This also counts the invariant vectors in the tensor-product representation.

For $r\ge 3$, join a clique on $r-2$ vertices to an independent set on $n-r+2$ vertices. The resulting complete $(r-1)$-partite graph hasIts largest clique has size $r-1$. Every missing edge joins two vertices of the independent set; adding it produces a $K_r$ with the original $(r-2)$-clique. It is therefore maximal $K_r$-free. For $n>r$ its part sizes are unbalanced, so its edge count is strictly less than that of $T_{r-1}(n)$.

For $r=3$ the construction is a star with $n-1$ edges. Every maximal triangle-free graph on $n>3$ vertices is connected: an edge between distinct components would create no triangle. A connected graph has at least $n-1$ edges, so the minimum is $n-1$, attained by the star. Literally the requested construction is impossible for $r=2$, since $T_1(n)$ has zero edges. The nontrivial assertion thus requires the customary $r\ge 3$ hypothesis.

The Turán graph $T_r(n)$ is the complete $r$-partite graph with part sizes differing by at most one. Turan theorem states that a graph on $n$ vertices containing no $K_r$ has at most $e(T_{r-1}(n))$ edges, with equality precisely for that balanced complete multipartite graph.

Here is a symmetrization proof, including equality. Choose a $K_r$-free graph with the maximum number of edges. Replacing a vertex $u$ by a nonadjacent twin of a nonadjacent vertex $v$ preserves $K_r$-freeness: a clique using the replacement would give the same clique using $v$. Thus nonadjacent vertices must have equal degrees, or cloning the higher-degree vertex increases the edge count. In fact their neighborhoods are identical. Otherwise choose $w$ adjacent to $u$ but not $v$. Cloning $u$ to $v$ preserves the total number of edges but lowers $w$'s degree by one, leaving $v$'s degree unchanged. The still nonadjacent pair $w,v$ now has unequal degrees, allowing an edge-increasing clone and contradicting maximality.

Nonadjacency is consequently an equivalence relation, and the extremal graph is complete multipartite. It has at most $r-1$ parts because choosing one vertex from each part forms a clique. For $n\ge r-1$, using fewer parts cannot be extremal if a part can be split. Finally its number of edges is $\frac{1}{2}(n^2-\sum n_i^2)$, maximized uniquely by balanced sizes: moving a vertex from a part at least two larger than another decreases the sum of squares. This proves the theorem and its equality characterization. The cases with fewer vertices than the forbidden clique are immediate.

Add to the group axioms the sentences $\forall x(x^n=e\Rightarrow x=e)$ for every $n\ge 2$. Their models are precisely the torsion-free groups. This theory is not finitely axiomatizable. If a finite sentence or finite theory were equivalent to it, the compactness theorem would imply that finitely many of these power axioms already entailed the proposed axiomatization. A cyclic group of prime order greater than all the finitely listed exponents satisfies those power axioms but is not torsion-free, a contradiction.

The class of all torsion groups is not first-order axiomatizable at all. If a theory $T$ axiomatized it, add a new constant $c$ and the sentences $c^n\ne e$ for all $n\ge 1$. Every finite subset has a model: use a cyclic group of sufficiently large prime order, with $c$ a generator. Compactness gives a model of all of $T$ in which $c$ has infinite order, contradicting the claimed torsion property. These conclusions concern the ordinary first-order language of groups.

One Hilbert axiomatization of classical propositional logic uses the schemas $A\to (B\to A)$, $[A\to (B\to C)]\to [(A\to B)\to (A\to C)]$ and $(\neg B\to \neg A)\to (A\to B)$, with modus ponens as its inference rule. Other connectives can be defined from implication and negation. Syntactic entailment $\Gamma ⊢A$ means that a finite formal proof derives $A$ from the schemas and members of $\Gamma$. Consistency means that no formula and its negation are both derivable, equivalently no contradiction is derivable.

Use the standard deduction theorem, closure of derivability under modus ponens, and the maximal-consistency properties: a maximal consistent set contains exactly one of $A,\neg A$, is closed under derivability, and satisfies $A\to B\in \Phi$ exactly when $A\notin \Phi$ or $B\in \Phi$. These follow from the deduction theorem and the classical schemas; in particular adjoining a missing formula to a maximal set must yield a contradiction. Define $v(p)=T$ iff $p\in \Phi$. Induction on formula construction now provesFor negation use the exactly-one property, and for implication use its displayed membership rule, which is precisely the Boolean truth table. Thus every member of $\Phi$ is true under $v$, providing the required valuation.

For $0<\beta <1$, the upright angle has $\ddot{\theta }=(gl/I_1)(1-\beta )\theta$ and is unstable. There is one nonzero equilibrium in $0<\theta <\pi$,Differentiate the right side of the nutation equation at this angle. Since the bracket vanishes there, the derivative is $-2(gl/I_1)\sin {\theta }_{\ast }\tan ({\theta }_{\ast }/2)=-4(gl/I_1){\sin }^2({\theta }_{\ast }/2)$. Therefore it is stable, withThis equilibrium represents steady precession, not a top with all Euler angles stationary. The equality case $\beta =1$ is excluded by the two cases in the question and requires a nonlinear marginal analysis.

Put $\beta =q^2/(4I_{1}gl)>1$. There is no nonzero equilibrium angle because ${\cos }^4(\theta /2)\le 1<\beta$. The upright angle $\theta =0$ is stable: its linearized equation is $\ddot{\theta }=-(gl/I_1)(\beta -1)\theta$. HenceThese are small nutational oscillations of the rapidly spinning top. No downward equilibrium belongs to the momentum level under consideration.

Split the real contour at zero and rotate its positive part to angle $\theta \in (0,\pi /4)$ and its negative part to angle $\pi -\theta$. Rotate the two rays of $\partial D^{+}$ to these same angles, preserving orientation. The initial-transform pole at $k=i$ is not crossed by either real-contour rotation, while ${\hat{u}}_0(-k)$ has its pole at $-i$, outside the upper half-plane. The apparent poles in ${\tilde{g}}_0$ are removable.

On the rotated rays, both $\mathrm{Im}k>0$ and $\mathrm{Re}{k}^2>0$. The initial-data factor has Gaussian decay $|e^{ikx-k^{2}t}|=e^{-x\mathrm{Im}k-t\mathrm{Re}{k}^2}$. For the boundary term, retain the integral formwhich is $O(|k{|}^{-2})$ there. Multiplication by $e^{ikx}$ gives exponential decay as $|k|\to \infty$ for every fixed $x>0$. Thus all integrands are exponentially damped on the deformed contours; the endpoint boundary value at $x=0$ is taken afterwards, rather than claiming uniform exponential decay there.

Use the half-line transform $\hat{u}(k,t)={\int }_0^{\infty }{e}^{-ikx}u(x,t)dx$ and let $g_j(t)={\partial }_x^{j}u(0,t)$. Integrating the heat equation twice by parts givesHere ${\hat{u}}_0(k)=(1+ik{)}^{-2}$ andwith removable singularities at the apparent denominator zeros. Let $D^{+}=\{\mathrm{Im}k>0,\mathrm{Re}{k}^2<0\}$, and orient its boundary from infinity on the $3\pi /4$ ray to zero and then out on the $\pi /4$ ray. Transform inversion and contour deformation giveTo eliminate the unknown boundary derivative, evaluate the transform identity at $-k$: ${\tilde{g}}_1={\hat{u}}_0(-k)+ik{\tilde{g}}_0-e^{k^{2}t}\hat{u}(-k,t)$. The resulting extra contour integral of $e^{ikx}\hat{u}(-k,t)$ vanishes by analyticity and decay in the upper half-plane. This proves that the boxed expression involves only the prescribed data. It is intended for $x,t>0$, with boundary and initial values obtained as limits.