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For a Markov chain invariant measure, $xP=x$, and hence $x{P}^n=x$. The irreducible Markov chain property supplies, for each $j$, an $n$ with $(P^n{)}_{kj}>0$. Nonnegativity gives $x_j\ge x_k(P^n{)}_{kj}>0$. Every coordinate of a nonzero invariant measure is positive.

Start the Markov chain at $k$, and let $T_k=\inf \{n\ge 1:X_n=k\}$ be its first return time. Each product in the series is the probability of a path with no intermediate visit to $k$. ThusThis is the expected number of visits to $j$ in an excursion, including the return endpoint and excluding the starting point. Because the chain is recurrent Markov chain, $T_k<\infty$ almost surely, and exactly one visit to $k$ is counted: ${\gamma }_k^k=1$. Counting the starting point instead of the return endpoint gives the equivalent usual excursion occupation measure.

For the lower bound, let $J=I\setminus \{k\}$, $Q=(p_{ij}{)}_{i,j\in J}$, and $r=(p_{kj}{)}_{j\in J}$. The stationarity equations on $J$, with $x_k=1$, are $x_J=r+x_{J}Q$. Iterating and using nonnegativity yieldsThe $j$th coordinate on the right is precisely the partial excursion series for $j\ne k$. Taking its increasing limit gives $x_j\ge {\gamma }_j^k$; for $j=k$ equality follows from $x_k={\gamma }_k^k=1$. The iteration is valid also on a countable state space because all summands are nonnegative.

For distinct nodes, define the divided difference recursively byOnly the function values at the nodes are needed. Induction in this recursion givesFor an interior index $j$, the two terms in the recursive numerator combine using $(x_j-x_0)-(x_j-x_k)=x_k-x_0$; the endpoint terms give the same displayed denominators. This proves the formula, including its symmetry in the nodes.

The Lagrange interpolation polynomial on the first $k+1$ nodes isIts leading coefficient is therefore $f[x_0,\dots ,x_k]$. Meanwhile $p_k-p_{k-1}$ vanishes at $x_0,\dots ,x_{k-1}$ and has degree at most $k$. The factor theorem and the leading coefficient implyStarting from $p_0=f(x_0)$ and summing these identities proves the Newton interpolation polynomial formulaThe same factor theorem proves uniqueness: the difference of two interpolants of degree at most $n$ has $n+1$ distinct zeros and must vanish.

The grounded conducting plane supplies the induced charge needed for a uniform field between the plates and zero field above the upper plate. Gauss's law gives $E_z=-\sigma /{ϵ}_0$ in the gap. With electric potential zero at the grounded plate,Integrating the electrostatic energy density ${ϵ}_0|\mathbf{E}{|}^2/2$ across the gap gives the capacitor energy per unit areaFor an isolated upper plate, charge is fixed, so increasing $d$ to $(1+\alpha )d$ gives exactly $\Delta u=\frac{1}{2}\sigma V\alpha$, with $V$ its initial value. The positive increase equals the mechanical work done against attraction.

At fixed voltage, the new charge density is $\sigma /(1+\alpha )$, and the energy becomes $u/(1+\alpha )$. Consequently $\Delta u=-\frac{1}{2}\sigma V\alpha +O({\alpha }^2)$. Charge $\sigma \alpha +O({\alpha }^2)$ per unit area flows back to the voltage source, which receives energy $\sigma V\alpha +O({\alpha }^2)$. The externally supplied mechanical work is still $\frac{1}{2}\sigma V\alpha +O({\alpha }^2)$: the field loses half this returned electrical energy and the mechanical work supplies the other half. Thus the fixed-voltage capacitor energy balance includes both field and source energy.

Applying the angular momentum operator directly givesFor an eigenvalue $\lambda$, comparison with $\lambda (f(x)+iy)z$ gives $\hslash x=\lambda f(x)$ and $\hslash f^{\prime }(x)=\lambda$. The first identity excludes $\lambda =0$; differentiating it and using the second gives ${\lambda }^2={\hslash }^2$. No additive constant in $f$ is permitted. The possibilities are $f(x)=x$, $\lambda =\hslash$, and $f(x)=-x$, $\lambda =-\hslash$.

For $f=x$, the polynomial $\psi =(x+iy)z$ is homogeneous of degree two and satisfies $\Delta \psi =0$. Expanding the three squared orbital angular momentum operators gives, with $D=x{\partial }_x+y{\partial }_y+z{\partial }_z$,Here $D\psi =2\psi$ and $D^2\psi =4\psi$. Therefore ${\hat{L}}^2\psi =6{\hslash }^2\psi$. This is the degree-two case of angular momentum of a homogeneous harmonic polynomial; its angular labels are $l=2$, $m=1$.

Put $u=ct-x$ and $v=ct+x$. The chain rule turns the wave equation into $4y_{uv}=0$. Integrating first in $v$ and then in $u$ gives the D'Alembert formulaAt the left endpoint, $g(u)=-f(u)$ for $u>0$. At the right endpoint, $f(ct-L)+g(ct+L)=0$, henceThus the travelling waves can be represented by $g=-f$, with $f$ and $g$ both $2L$-periodic. More precisely, the positive-time boundary conditions constrain only $f$ on $(-L,\infty )$ and $g$ on $(0,\infty )$; their unused arguments may be filled in by this periodic extension. They do not constrain arbitrary pre-existing choices outside those ranges. This is periodic reflection for a fixed-end string.

Using $g=-f$, differentiation gives $y_t=c(f^{\prime }(ct-x)-f^{\prime }(ct+x))$ and $y_x=-f^{\prime }(ct-x)-f^{\prime }(ct+x)$. The cross terms in the wave energy cancel, soThis integrates a $2L$-periodic function over exactly one period. $E(t)$ is constant. Equivalently, the wave equation gives $E^{\prime }(t)=[y_t{y}_x{]}_0^L=0$, because the fixed endpoint values imply $y_t=0$ there.

Rouché's theorem states that if $f,g$ are holomorphic functions on a neighbourhood of a simple closed contour and its interior, and $|g|<|f|$ on the contour, then $f$ and $f+g$ have the same number of zeros inside, counted with multiplicity.

On $|z|=1$, use $f=6z^2$: the other terms have absolute value at most $1+3+1=5<6$. Hence there are two zeros inside the unit circle. On $|z|=2$, use $f=3z^7$: its modulus is $384$, whereas the other terms have total modulus at most $256+24+1=281$. Hence there are seven zeros inside the larger circle. The strict inequalities also exclude zeros on either boundary. There are $7-2=5$ roots in the annulus, counted with multiplicity.

For differentiable maps $F$ at $x$ and $G$ at $F(x)$, the chain rule assertsWrite $F(x+h)=F(x)+Ah+r(h)$ with $\Vert r(h)\Vert =o(\Vert h\Vert )$, and $G(F(x)+u)=G(F(x))+Bu+s(u)$ with $\Vert s(u)\Vert =o(\Vert u\Vert )$. Since $u=Ah+r(h)=O(\Vert h\Vert )$, substitution givesBoth remaining terms are $o(\Vert h\Vert )$, proving differentiability and the stated chain rule. The estimate for $s$ also holds when its argument is zero, by setting $s(0)=0$.

For the circle constraint take $H(u)=u_1^2+u_2^2$. The composite $H\circ F$ is constant, so the chain rule givesSince $\Vert F(x)\Vert =1$, this is a nonzero row vector annihilating $D{F}_x$. Thus $D{F}_x$ cannot be invertible: $\det D{F}_x=0$ at every $x$. Geometrically, the derivative of a map into a level set takes values in the tangent space to the circle.

The conjugacy class of $x$ is its orbit under the conjugation action, and its stabilizer is the centralizer $C_G(x)$. The orbit-stabilizer theorem givesBy Lagrange's theorem, the order of the centralizer divides $p^n=|G|$, so the index is a power of $p$. A conjugacy class has size one exactly when its element belongs to the center of a group.

Separate these singleton classes in the class equation:where the sum takes one representative of each nonsingleton conjugacy class. Every summand is divisible by $p$. Thus $p\mid |Z(G)|$, and the identity ensures $|Z(G)|>0$. Consequently $|Z(G)|\ge p$. This is the nontrivial center of a finite p-group property.