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The product rule givesApply the divergence theorem, with outward unit normal $\mathbf{n}$, and use the vanishing Laplacian of the harmonic function $g$:The boundary value of $f$ is one. Another application of the divergence theorem therefore givesThis is the gradient pairing with a boundary-constant function and a harmonic function, obtained directly from Green's first identity. The harmonic function must be regular throughout the volume; an interior singularity can supply an additional boundary flux.

The tensor contraction $C_{mkl}={ϵ}_{ijm}{T}_{ijkl}$ is an isotropic third-rank tensor under proper rotations. It therefore equals $\nu {ϵ}_{mkl}$. Cyclically permuting three indices is even, so ${ϵ}_{mkl}={ϵ}_{klm}$ and ${ϵ}_{ijm}{T}_{ijkl}=\nu {ϵ}_{klm}$.

For the given Kronecker delta expression, the $\alpha$ term vanishes by antisymmetry, while the $\beta$ and $\gamma$ terms give respectively $\beta {ϵ}_{klm}$ and $\gamma {ϵ}_{lkm}=-\gamma {ϵ}_{klm}$. HenceThese computations also illustrate isotropic contractions of a fourth-rank tensor.

The tensor contraction $B_{kl}={\delta }_{ij}{T}_{ijkl}$ is an isotropic second-rank tensor, because the Kronecker delta and $T$ are invariant under rotations. The classification proved above yields $B_{kl}=\mu {\delta }_{kl}$ for some scalar $\mu$.

For the given three-term tensor, contraction over $i,j$ givesThus $\mu =3\alpha +\beta +\gamma$.

Both the Levi-Civita symbol and $T$ are invariant under proper rotations. Their tensor contractionis consequently an isotropic vector. The introductory proof shows that it must vanish, so ${ϵ}_{ijk}{T}_{ijkl}=0$.

For the displayed combination of Kronecker deltas, the three terms areIn each term the tensor contraction repeats two indices of the Levi-Civita symbol; each term is zero. This verifies the identity for arbitrary $\alpha ,\beta ,\gamma$.

During the addition interval, subtracting the formulas in (ii) and (iii) cancels the common unforced temperature:This is zero precisely when $e^{-a(t-1)}=1/(1+a)$, soThe inequalities $0<\log (1+a)<a$ give $1<t^{\ast }<2$, so using the active-interval formula was justified. The difference has derivative $(1+a)e^{-a(t-1)}>0$, making this crossing unique within that interval. Immediately after time one it is minus one; after time two it is strictly positive by (iv). Hence there is no other equality at any finite time after time one. Both temperatures approach the same ambient limit, but that limiting equality is not an additional crossing.

At time two, continuity of the second temperature givesThe forcing then turns off, so Newton's law of cooling again gives exponential decay:Subtracting this from the first temperature and expressing both corrections with the same exponential factor givesBecause $e^a>1+a$ for $a>0$, this difference is positive. Thus the cup cooled instantaneously is warmer after the addition interval: its cooling occurred earlier and has undergone more subsequent exponential attenuation.

On the active interval the excess temperature satisfies ${\theta }_2^{\prime }+a{\theta }_2=-1$. Multiplication by the integrating factor $e^{at}$ and integration from time one givesThere is no impulse in this equation, so continuity supplies ${\theta }_2(1)=(T_0-T_{\infty })e^{-a}$. ThereforeThe equivalent correction to the common unforced temperature is $-(1-e^{-a(t-1)})/a$, which vanishes continuously at the beginning of the interval.

Integrate the first equation over $(1-\epsilon ,1+\epsilon )$. The integral of the ordinary cooling term tends to zero as $\epsilon \downarrow 0$, while the Dirac delta function integrates to one. Thus the jump condition for an impulse isAfter the jump there is no further forcing, so Newton's law of cooling givesEquivalently,The last term has jump minus one at time one and then decays at the same exponential rate as the unforced temperature.

Before time one, both forcing terms vanish. The separable differential equation ${\theta }_i^{\prime }=-a{\theta }_i$ gives ${\theta }_i(t)=C_i{e}^{-at}$. The initial values give $C_1=C_2=T_0-T_{\infty }$, and henceIn particular the common left-hand limit at time one is $T_{\infty }+(T_0-T_{\infty })e^{-a}$.

This is a gambler's ruin hitting problem. For $0<p<1$, let $h_j$ be the probability of reaching $+3$ before $-3$ when the current difference is $j$. Conditioning on the next independent game givesStopping occurs almost surely: conditional on being inside the barriers, three consecutive wins by either player force absorption within the next three games. This event has probability $p^3+q^3>0$, so the chance of surviving $m$ such blocks is at most $(1-p^3-q^3{)}^m$, which tends to zero.

If $p\ne q$, the characteristic roots of the interior recurrence are $1$ and $r=q/p$, so $h_j=A+B{r}^j$. Imposing the two boundary values yieldsIn particular,For $p=q=1/2$ the recurrence instead has affine solutions, and the boundary values give $h_j=(j+3)/6$, hence $h_0=1/2$. The deterministic cases $p=0,1$ give zero and one respectively. The formula covers all of them continuously:The stopping argument and the boundary recurrence show why the answer concerns the first barrier reached rather than the win difference after a predetermined number of games.

Put $q=1-p$ and track the difference $S_n$ between the numbers of wins. It is a biased random walk with increments $+1$ and $-1$, stopped on first reaching $3$ or $-3$.

To finish at $+3$ on game five, there must be four wins for $A$ and one for $B$, with the final game won by $A$. The unique loss for $A$ can be in position one, two or three. If it were in position four or five, the first three games would already have stopped the contest. Each of the three allowed paths remains strictly inside the barriers until the last step and has probability $p^{4}q$. By interchanging players, there are also three paths ending at $-3$, each with probability $p{q}^4$. ThereforeFor a fair game this is $3/16$. Counting all five-game sequences with final win difference three would overcount paths on which the contest ended earlier.

For any fixed $r>0$, $\log y=o(y^{1/r})$ as $y\to \infty$. For example, l'Hôpital's rule applied to $\log y/y^{1/r}$ gives the limit zero. With $y=\log n$, this implies $(\log \log n{)}^r<\log n$ eventually. Thereforefor all sufficiently large $n$. The comparison series diverges by the integral test for convergence, sinceThe comparison test for series givesOne may obtain the same growth bound by extending the given logarithmic estimate to real $y$ through $\lceil y\rceil$ and choosing any exponent smaller than $1/r$; no special convergence test for iterated logarithms is needed.

The comparison test for series says that an eventual upper bound by a convergent positive series proves convergence, while an eventual lower bound by a divergent positive series proves divergence. The p-series $\sum n^{-s}$ converges exactly when $s>1$.

If $p>1$, then $(\log n{)}^q\ge 1$ for all sufficiently large $n$, soand the series converges. If $0<p<1$, choose $\eta =(1-p)/2>0$. The allowed logarithmic bound, applied with exponent $\eta /q$, gives $(\log n{)}^q<n^{\eta }$ eventually. Thusand the series diverges.

For $p=1$, use the integral test for convergence: for an eventually positive decreasing function $g$, $\sum g(n)$ and $\int g(x)dx$ either both converge or both diverge. Here $g(x)=1/[x(\log x{)}^q]$ is decreasing for $x>1$, and the substitution $u=\log x$ givesThis integral converges exactly when $q>1$. Hence the polynomial-logarithmic series convergence classification isAll other allowed positive $p,q$ give divergence.

The parametric equation of a straight line through $a$ with unit direction $\hat{t}$ isIf the two lines meet, then $a_1+s{\hat{t}}_1=a_2+t{\hat{t}}_2$ for some real $s,t$, so $a_1-a_2$ lies in the linear span of their direction vectors. Its dot product with their cross product is consequently zero:This is not sufficient without excluding parallel lines. For instance $a_1=(0,0,0)$, $a_2=(0,1,0)$ and ${\hat{t}}_1={\hat{t}}_2=(1,0,0)$ satisfy it, but the lines are distinct and never meet. For nonparallel lines the condition is sufficient, because the two directions span the plane perpendicular to their nonzero cross product.

For nonparallel skew lines, put $n=({\hat{t}}_1\times {\hat{t}}_2)/|{\hat{t}}_1\times {\hat{t}}_2|$. Every joining vector is $a_1-a_2+s{\hat{t}}_1-t{\hat{t}}_2$, with fixed component $(a_1-a_2)\cdot n$ along $n$. Its length is at least the absolute value of that component. Its perpendicular component can be canceled by a unique choice of $s,t$, since the two independent directions span $n^{\perp }$. This lower bound is attained, proving the distance between skew lines:

Adding the two expanded unitary matrix identities instead cancels the commutator and gives $2(A^2+B^2)=2I$. HenceTogether with part (i), this characterizes the commuting Hermitian parts of a unitary matrix.

By the unitary matrix identities,Subtracting the two equations yields $2i(AB-BA)=0$. ThereforeThe subtraction is important: entrywise real and imaginary parts cannot generally be used to separate these terms, since the Hermitian matrices need not have real entries.

For $|z|=r<1$, the inequality $n!\ge n$ for $n\ge 1$ givesThe first $r$ comes from $0!=1$; both initial terms have exponent one. Thus the factorial-gap power series converges absolutely inside the unit circle. For $|z|>1$, its terms have magnitudes $|z{|}^{n!}\to \infty$, so the term test for divergence proves divergence and $R=1$.

At every point with $|z|=1$, each term has magnitude one, so the terms cannot tend to zero, regardless of their phases. HenceThe large gaps between exponents make this a lacunary power series; cancellation cannot overcome the necessary condition that the terms of a convergent series tend to zero.

For $z\ne 0$, the ratio of successive absolute terms isThe ratio test therefore gives absolute convergence for $|z|<1$ and divergence for $|z|>1$, while at zero the series is trivially convergent. Hence $R=1$. On the boundary, $|z^n/n^2|=1/n^2$, and the convergent p-series proves