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The cross product $x\times y$ is the vector perpendicular to $x$ and $y$, of magnitude $|x||y|\sin \alpha$ and orientation fixed by the right-hand rule. In suffix notation,where ${\epsilon }_{ijk}$ is the Levi-Civita symbol. ThereforeUsinggivesHence the vector triple product identity is

Similarly,Thus, for $0<\theta <\pi$,At $\theta =0$, $1-z=0$, whose complex argument is undefined. Geometrically, $|1-z|$ is the chord joining the two unit-circle points $z$ and $1$. The isosceles triangle with vertex angle $\theta$ has half-chord $\sin (\theta /2)$, and the chord direction is perpendicular to the radius through its midpoint, giving the stated argument.

By Euler's formula, $z=e^{i\theta }$. Factoring out the half-angle givesSince $0\le \theta <\pi$, the cosine is positive. ThereforeOn the Argand diagram, the points $0,1,z,1+z$ form a rhombus. Its diagonal from $0$ to $1+z$ bisects the angle between the unit vectors $1$ and $z$, so its direction is $\theta /2$; resolving either side along the diagonal gives its length $2\cos (\theta /2)$.

We first prove by induction thatIt holds for $n=1$, and if it holds at $n$, thenConsequentlyPart (c) and the squeeze theorem now give

For $u_n=2^{-n}{n}^k$,Thus for all sufficiently large $n$ the ratio is, for example, at most $3/4$. Comparison with a decaying geometric sequence, equivalently the ratio test, yields

If $s\le 0$, the terms $n^{-s}$ do not tend to zero, so the series diverges by the term test for divergence. If $s>0$, the terms form a nonnegative decreasing sequence, and the Cauchy condensation test gives the condensed seriesThis geometric series converges exactly when $1-s<0$. Hence the p-series criterion is

For each $k\ge 0$, monotonicity gives the dyadic-block boundsSumming the upper bounds proves convergence of $\sum x_n$ whenever $\sum 2^k{x}_{2^k}$ converges. Summing the lower bounds gives, up to the first term,so convergence of $\sum x_n$ forces convergence of the condensed series. This is the Cauchy condensation test:

The upper bound follows immediately from convexity, sinceFor the lower bound, writeApplying convexity and rearranging givesChoose a fixed $r_0>0$ with $[y_0-r_0,y_0+r_0]\subset (a,b)$. Applying these bounds with $r=r_0$ and $r=-r_0$ shows that the difference $f(y_0+h)-f(y_0)$ is trapped between quantities of order $|h|/r_0$. Both tend to zero, so every convex function on an open interval is continuous.

Now suppose $c$ is a local minimum but some $x$ satisfies $f(x)<f(c)$. For sufficiently small $\lambda >0$, the point $z=(1-\lambda )c+\lambda x$ lies in the neighbourhood on which $c$ is minimal, whereas convexity givesa contradiction. ThusDifferentiability at the minimizer is unnecessary: $f(x)=|x-c|$ is convex, has its global minimum at $c$, and is not differentiable there.

The function $f$ is continuous at $y_0$ when for every $\epsilon >0$ there is a $\delta >0$ such thatSuppose $f$ has a local minimum at $c$ and is differentiable there. For sufficiently small $h>0$,Both one-sided limits equal $f^{\prime }(c)$, so it is both nonnegative and nonpositive. Therefore

A finite irreducible Markov chain is positive recurrent. Its mean recurrence time to state $i$, starting at $i$ and counting the first positive return, is $1/{\pi }_i$. This identity comes from the long-run fraction of time spent at that state: successive return cycles each contain one visit at their start, so the fraction is the reciprocal of the mean cycle length. It remains valid for a periodic chain.

Here ${\pi }_0=2^{-k}$, henceFor $k=1$ the return is deterministically after two minutes, which checks the formula.

If there are $i$ balls in box A, a ball currently in B is selected with probability $(k-i)/k$, increasing the count, and a ball in A is selected with probability $i/k$, decreasing it. For $k\ge 1$,with every other transition zero. This is the Ehrenfest urn model.

The stationary distribution isIt is normalized by the binomial theorem, and it obeys the detailed balance:All other pairs have zero flux in both directions. These identities prove stationarity and show that this is a reversible Markov chain in equilibrium. The finite chain is irreducible and has period two; reversibility does not require aperiodicity.

Expand the second stage by the Taylor theorem, writing $f=f(t_n,y_n)$ and $D_{y}f$ for the derivative in the state variable:The update therefore agrees with the exact solution's Taylor series:Thus the one-step local truncation error is $O(h^3)$, and the method has order at least two under the usual smoothness and Lipschitz continuity assumptions.

For $f=\lambda y$, the two-stage Runge-Kutta method givesOn the real axis, $R(z)=\frac{1}{2}(z+1{)}^2+\frac{1}{2}\ge \frac{1}{2}$, so $|R(z)|\le 1$ is equivalent to $(z+1{)}^2\le 1$. HenceThe decay convention gives $(-2,0)$. Since negative real values such as $z=-3$ lie outside the linear stability domain, the method is not A-stable. Its order is exactly two, since $R(z)$ lacks the cubic term of $e^z$.

Apply the numerical method to the scalar test equation $y^{\prime }=\lambda y$, with $z=h\lambda$. Its linear stability domain consists of those $z\in \mathbb{C}$ for which the discrete solution stays bounded as the step count tends to infinity. For a one-step method with stability function $R$, this isIf asymptotic decay is required, use $|R(z)|<1$ instead; this only changes the boundary conventions in part (b). A method is A-stable if its stability domain contains the left half-plane, so every decaying linear test equation is stable for every positive step size. For a multistep method, the corresponding definition uses the root condition for a multistep method for every characteristic root, including simplicity of unit-modulus roots.