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The symmetric power $S^{n}V$ is the quotient of $V^{\otimes n}$ identifying tensors which differ by permutation of the factors. The exterior power ${\Lambda }^{n}V$ is the alternating quotient, with a tensor having repeated equal factors set to zero. Both quotients inherit the given group action. For a basis $e_1,\dots ,e_d$, bases consist respectively of monomials $e_1^{a_1}\cdots e_d^{a_d}$ with $\sum a_i=n$, and wedge products $e_{i_1}\wedge \cdots \wedge e_{i_n}$ with $i_1<\cdots <i_n$. Thuswith the second dimension zero for $n>d$.

A finite-order group element is diagonalizable over $\mathbb{C}$, because its minimal polynomial divides $X^m-1$, which has no repeated roots. Choose an eigenbasis with eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$. The induced eigenvalues, including multiplicities, areThe zeroth powers are the one-dimensional trivial representation, with the empty product equal to $1$.

The binomial upper bound for a Ramsey number gives $R(3,t)\le \left(\genfrac{}{}{0ex}{}{t+1}{2}\right)=O(t^2)$. For the triangle-free Ramsey lower bound by alteration, a binomial random graph $G(n,p)$ has expected triangle count $\left(\genfrac{}{}{0ex}{}{n}{3}\right)p^3$ and expected independent $t$-set count $\left(\genfrac{}{}{0ex}{}{n}{t}\right)(1-p{)}^{\left(\genfrac{}{}{0ex}{}{t}{2}\right)}$. Choose a realization whose sum of these counts is at most its expectation. Delete one vertex from each surviving forbidden set, iterating through the originally listed sets. The remaining graph has neither a triangle nor an independent $t$-set. This random alteration method provesFor large $t$, set $n=\lfloor (t/(4\log t){)}^{3/2}\rfloor$ and $p=n^{-2/3}$. The triangle term is at most $n/6$. Using $\left(\genfrac{}{}{0ex}{}{n}{t}\right)\le (en/t{)}^t$ and $1-p\le e^{-p}$, the logarithm of the independent-set term divided by $t$ is at mostwhich tends to $-\infty$. That term is eventually below $1$. Thus $R(3,t)>5n/6-1$ and

Induction with Pascal's identity gives the binomial upper bound for a Ramsey number $R(s,t)\le \left(\genfrac{}{}{0ex}{}{s+t-2}{s-1}\right)$. Thus $R(s)\le \left(\genfrac{}{}{0ex}{}{2s-2}{s-1}\right)\le 4^{s-1}$, proving $R(s)=O(4^s)$.

For the lower bound, take a binomial random graph $G(n,1/2)$. The expected total number of $s$-cliques and independent $s$-sets isFor $n=\lfloor 2^{s/2}/2\rfloor$, this is at most $2^{1-s/2}/s!<1$ for all sufficiently large $s$. Some graph therefore has neither forbidden set, so $R(s)>n$. This proves $R(s)=\Omega (2^{s/2})$ by the probabilistic method.

The graph Ramsey number $R(s,t)$ is the least $N$ for which every graph on $N$ vertices has a clique of size $s$ or an independent set of size $t$; $R(s)=R(s,s)$ is the diagonal Ramsey number. The base cases are $R(2,t)=t$ and $R(s,2)=s$.

Assume the smaller parameters already have finite graph Ramsey numbers, and take $N=R(s-1,t)+R(s,t-1)$. For any vertex, either its neighbourhood has at least $R(s-1,t)$ vertices, or its nonneighbourhood has at least $R(s,t-1)$ vertices. In the first case it contains an independent $t$-set, or an $(s-1)$-clique which extends with the chosen vertex. In the second it contains an $s$-clique, or a $(t-1)$-independent set which extends with that vertex. Induction proves existence and

The recursive definition of ordinal addition is $\alpha +0=\alpha$, $\alpha +(\beta +1)=(\alpha +\beta )+1$, and $\alpha +\lambda ={\sup }_{\beta <\lambda }(\alpha +\beta )$ for a nonzero limit ordinal $\lambda$. The synthetic definition is the order type of the ordered disjoint union of a copy of $\alpha$ followed by a copy of $\beta$.

For the synthetic operation, adding the empty order does nothing, adding a successor appends one final point, and concatenation with a limit order is the increasing union of the initial concatenations with all its proper initial segments. Their order types have supremum ${\sup }_{\beta <\lambda }(\alpha +\beta )$. Thus the synthetic operation satisfies the recursive equations. Transfinite induction gives uniqueness of the recursively defined operation, proving equivalence.

The recursive definitions of ordinal multiplication and ordinal exponentiation areThe last equation in each line is for nonzero limit $\lambda$. In particular the convention $0^0=1$ belongs to this recursive definition.

The canonical momenta are $p_x=m\dot{x}$, $p_y=m\dot{y}+qBx$, $p_z=m\dot{z}$. The Legendre transform in mechanics givesHamilton's equations areThus $\ddot{x}=\Omega \dot{y}$, $\ddot{y}=-\Omega \dot{x}-qE/m$, and $\ddot{z}=0$, where $\Omega =qB/m$. The signs also follow from $\mathbf{E}=-E{\mathbf{e}}_y$ and $\mathbf{B}=B{\mathbf{e}}_z$.

For $Bq\ne 0$, put $v_d=-E/B$, $a=v_{x0}-v_d$, $b=v_{y0}$. The motion isDifferentiation verifies both equations and all initial conditions. The centre of the circular motion has the electric drift velocity $-E{\mathbf{e}}_x/B$. For $B=0$, take $x=x_0+v_{x0}t$, $y=y_0+v_{y0}t-qE{t}^2/(2m)$ and the same $z$. For $q=0$, the particle moves freely.

The canonical momentum is $p=L_{\dot{q}}$, and a regular Legendre transform in mechanics defines $H(q,p,t)=p\dot{q}-L(q,\dot{q},t)$ after solving for $\dot{q}$. Its differential isbecause the $d\dot{q}$ terms cancel. With the Euler-Lagrange equation $\dot{p}=L_q$, this gives Hamilton's equationsIn particular $H_q=-L_q$ at corresponding velocities, so a cyclic coordinate for the Lagrangian is also cyclic for the Hamiltonian.

A Hamming code decoder succeeds on a seven-bit block exactly when there are zero or one errors. Its radius-one decoding regions are disjoint, so an error of weight at least two cannot decode back to the original codeword. ThusNumerically $Q\approx 0.97929$, so only about $1.0212$ transmissions are expected. Even accounting for seven rather than four bits per letter, expected bits sent are $7N/Q\approx 7148$, compared with $4N/P\approx 218830$. The Hamming method is substantially better under the stated error detection and independent retransmission assumptions.

Independence of individual bit errors gives $P=(1-p{)}^{4N}$. Independent retransmissions have a geometric distribution with success probability $P$, so the expected total number of transmissions, including the successful one, is $1/P$. At the supplied values, $P\approx 0.018279$ and $1/P\approx 54.71$.

The principle of stationary action says that the physical path makes the action stationary under smooth variations $q+\eta h$ with fixed endpoints $h(t_1)=h(t_2)=0$. It need not minimize the action. Differentiating and integrating by parts givesThe fundamental lemma of the calculus of variations then gives the Euler-Lagrange equation

The Gershgorin circle theorem says that every eigenvalue of a square matrix lies in at least one disc centred at a diagonal entry with radius the sum of the absolute values of the off-diagonal entries in that row.

For interior unknowns write the update as $u^{n+1}=B{u}^n$. The matrix is real symmetric and tridiagonal, with diagonal $1-\mu (a_{m-1/2}+a_{m+1/2})$ and adjacent entries $\mu a_{m+1/2}$. Each Gershgorin radius is at most $\mu (a_{m-1/2}+a_{m+1/2})$. Its real interval is therefore contained inFor $0<\mu \le 1/(2a_{+})$, every eigenvalue lies in $[-1,1]$. Symmetry gives an orthogonal diagonalization, so $\Vert B^n{\Vert }_2\le 1$ for every $n$: errors in initial data cannot grow. This is stability in the discrete ${\ell }^2$ norm, also in its $h$-weighted version. One can additionally see maximum-norm stability directly: the update weights are nonnegative with row sums at most one once the zero boundary unknowns are removed. Hence $\Vert B{\Vert }_{\infty }\le 1$.

Let all spatial derivatives be evaluated at $x=mh$, $t=nk$. Expand the two adjacent flux contributions symmetrically. The bracket in the scheme isFor clarity, the $h^2$ terms are $a{u}_{xx}+a^{\prime }{u}_x$; odd orders cancel under $h\mapsto -h$. Under four spatial derivatives the next coefficient is $a{u}_{xxxx}/12+a^{\prime }{u}_{xxx}/6+a^{\prime \prime }{u}_{xx}/8+a^{(3)}{u}_x/24$. Time expansion gives $u(x,t+k)=u+k{u}_t+O(k^2)$. Since $u_t=(a{u}_x{)}_x$ and $k=\mu h^2$,This stronger smooth-data residual implies the requested $O(h^3)$ consistency bound. Even a less detailed Taylor expansion with an $O(h^3)$ spatial remainder would suffice for the printed order. Uniform bounds on the needed derivatives over the region under consideration make the estimate uniform in the interior mesh index; the zero boundary values enter the scheme exactly.

For constant viscosity, differentiating $\sigma =-pI+\mu (\nabla u+\nabla u^T)$ gives $\nabla \cdot \sigma =-\nabla p+\mu {\nabla }^{2}u$, since $\nabla \cdot u=0$. This is zero by the body-force-free Stokes equations.

Apply the divergence theorem to the annular fluid. Its outer normal is $+n_R$, but its inner normal is minus the sphere's outward normal. Therefore $0={\int }_{S_R}\sigma n_{R}dS-{\int }_{S_a}\sigma n_{a}dS$. The second integral is the hydrodynamic force on the inner sphere, $-F$, because that sphere exerts $F$ on the fluid. Thus again $G=-F$, without differentiating the far field. Local momentum conservation carries the same total force through every enclosing surface.

Put $Q=F\cdot x$. Differentiate the given Stokeslet velocity:Symmetrizing and subtracting the pressure term cancels the diagonal terms, leaving the Cauchy stress tensorOn the radius-$R$ surface with outward sphere normal $n=x/R$, the force exerted by the exterior fluid on the sphere is the tractionUse $dS=R^{2}d\Omega$ and ${\int }_{S^2}{n}_i{n}_{j}d\Omega =(4\pi /3){\delta }_{ij}$ to findThe $R^2$ factors cancel, proving independence of radius. This calculation uses the printed far-field Stokeslet; higher multipoles of the exact sphere flow have zero net force through enclosing surfaces.