An absorbing state $i$ has $p_{ii}=1$; once reached, the chain never leaves it. For a nonabsorbing initial state, condition on its first transition and use the Markov property:Here $h_1=1$ and $h_4=0$. The other equations areSolving givesFor absorption into either state one or four, let $e_i$ be the expected number of steps, with $e_1=e_4=0$. first-step analysis givesHence $e_2=2$ andThe expectation is finite because each transient two-step cycle has probability at most $(2/3)(1/4)=1/6$ of remaining transient; the survival tail is geometrically bounded. This also justifies solving the expectation equations without assuming finiteness in advance.

Uniform area distribution on the unit disk givesThe Jacobian determinant has absolute determinant $r$, soIntegrating out the other coordinate givesThe joint probability density function factors into these two marginal distribution, proving independence of $R$ and $\Theta$. FinallyThe angle at $R=0$ can be defined arbitrarily, since that event has probability zero.

Milne-Thomson's circle theorem states that if $f(z)$ is a complex potential of a two-dimensional potential flow whose singularities lie outside $|z|=a$, thenhas the circle as a streamline, with the original exterior singularities unchanged. Additive constants in the potential are immaterial.

Take $f(z)=\Gamma (2\pi i{)}^{-1}\log (z-z_1)$, using the convention that positive $\Gamma$ is positive circulation. Reflection in the circle gives, up to a constant,The reflected vortex at $a^2/{\bar{z}}_1$ has strength $-\Gamma$, while the central term has strength $+\Gamma$. Their combined circulation around the cylinder is zero; this is the circulation normalization implicit in the requested equation.

The vortex moves with the velocity induced by the other singularities, excluding its own singular term. Since complex velocity is $dw/dz=u-iv$, its value at the vortex is $d{\bar{z}}_1/dt$. ThusWriting $z_1=r{e}^{i\theta }$ and multiplying this equation by $z_1$ givesIts real part implies $\dot{r}=0$, and its imaginary part givesThe radius is constant and the angular velocity is constant on that orbit. For positive $\Gamma$ the orbit is clockwise with this convention. Retaining the conjugate on the time derivative is essential for the sign.

For a consistently oriented surface and boundary, Ampère's law isStokes theorem gives the left side as ${\int }_S(\nabla \times \mathbf{B})\cdot \mathbf{n}dS$, so the given curl equation proves the law.

On a circle of radius $r$ around the axis, the assumed azimuthal field is tangent and constant in magnitude. Thus the circulation is $2\pi r{B}_{\theta }(r)$. For $r\le a$, the enclosed electric current is $I{r}^2/a^2$; for $r\ge a$, it is $I$. ThereforeThe field is continuous at $r=a$ and vanishes regularly at the axis. Its direction follows the right-hand rule for positive electric current along ${\mathbf{e}}_z$. The radial component of the curl equation also gives ${\partial }_z{B}_{\theta }=0$, so the expression is valid at every height; no unmentioned axial dependence remains.

Autonomy gives conservation of energy, while the cyclic coordinate give conserved momenta:Here $p_{\psi }$ is the angular momentum along the body's symmetry axis, and $p_{\varphi }$ is its component along the fixed vertical. The time symmetry corresponds to $E$, not to an additional angle momentum.

With $u=\cos \theta$ and $-1<u<1$, solveInsert this and the conserved spin rate $n$ into the energy equation, multiply by $A$, and isolate $\dot{\theta }$:

For a fixed nonvertical angle, letThe energy is $A{\dot{\theta }}^2/2+V_{\mathrm{eff}}(\cos \theta )$. A constant angle in $(0,\pi )$ requires both $E=V_{\mathrm{eff}}(u_0)$ and $V_{\mathrm{eff}}^{\prime }(u_0)=0$, by the reduced Euler-Lagrange equation and $\sin {\theta }_0\ne 0$. Sinceboth $P(u_0)=0$ and $P^{\prime }(u_0)=0$, with $u_0\in (-1,1)$.

The source's unqualified final sentence omits the nonvertical restriction. A vertical axis has $u=\pm 1$, where the displayed reduction divides by zero and the common root need not be interior. For example, the stationary upright top has $n=h=0$, $E=mga$, andIts only repeated root is $u=1$, so it has no common root with its derivative in $(-1,1)$. Thus the interior double-root conclusion holds for nonvertical steady precession. The vertical configurations require separate endpoint treatment. This is nonvertical qualification in the steady-top double-root condition.

Let $\eta$ be a smooth variation with $\eta ={\eta }^{\prime }=0$ at both endpoints, and consider $y_{ϵ}=y+ϵ\eta$. stationary point gives the first variationIntegrate the middle term once by parts and the last term twice. All boundary terms vanish because both $\eta$ and ${\eta }^{\prime }$ vanish there. Assuming the derivatives needed for the classical formula, the given fundamental lemma of the calculus of variations yieldsWithout assuming those classical derivatives initially, the same calculation with compactly supported variations states this equation weakly; sufficient regularity then gives the classical equation.

For the specific integrand, $F_y=-2y$, $F_{y^{\prime }}=0$, $F_{y^{\prime \prime }}=2y^{\prime \prime }$, so the equation is $y^{(4)}-y=0$. Its weak solutions here are smooth, and its general classical solution isPut $\ell =\pi /2$, $c=\cosh \ell$, $s=\sinh \ell$. The conditions at zero give $C=-A$ and $D=1-B$. Those at $\ell$ giveUsing $c^2-s^2=1$, solve to obtain $A=c/2$, $B=(1-s)/2$, $C=-c/2$, $D=(1+s)/2$. Thus an extremal isSubstitution checks all four boundary values. The additive term $x^2$ in the integrand contributes no variation in $y$, so it does not alter this equation.

The required condition is $D_{2}F(a,b)\ne 0$. The implicit function theorem then gives a unique local differentiable branch $y=f(x)$ with $f(a)=b$ andThe denominator remains nonzero after shrinking the neighborhood by continuity.

For the specified smooth function,At $(0,0)$ both equal one, so the condition holds and $f^{\prime }(0)=-1$. Differentiate $F(x,f(x))=0$ twice:At zero, $F_{xx}=F_{yy}=0$, $F_{xy}=1$, and $F_y=1$. ConsequentlyThe additional derivatives exist because this particular $F$ is smooth; the general first-derivative theorem only requires the stated continuous first partial derivatives.

A dominated differentiation under the integral sign suffices. Suppose $\varphi (\cdot ,t)$ is measurable, $\varphi (\cdot ,t_0)$ is integrable at one parameter value, and for almost every $x$ the map $t\mapsto \varphi (x,t)$ is continuously differentiable on an interval. If $|{\partial }_t\varphi (x,t)|\le H(x)$ throughout that interval for an integrable $H$, then the parameter integral is differentiable, andOne direct sufficient version assumes ${\partial }_t\varphi (x,t)$ is continuous in $t$ almost everywhere and has that common domination; dominated convergence theorem then proves the assertion.

Here $|f(x)e^{itx}|=|f(x)|$, so the complex-valued function $h$ is integrable for every real $t$. For a real increment $u\ne 0$, the elementary bound $|e^{iux}-1|\le |ux|$ givesThe right side is integrable by hypothesis and the integrand converges pointwise to $ixf(x)e^{itx}$. dominated convergence theorem, applied to real and imaginary parts, provesThis supplies the domination explicitly rather than merely differentiating a formal Fourier transform.

Euler's theorem states that if $\gcd (a,m)=1$, then $a^{\varphi (m)}\equiv 1(\mathrm{mod}m)$, where $\varphi$ is Euler's totient function. This includes Fermat's little theorem when $m$ is prime.

For each prime-power factor $p_i^{e_i}$, the integer $\varphi (p_i^{e_i})$ divides the defined least common multiple $\lambda (m)$. Euler's theorem modulo that prime power therefore givesThe prime powers are pairwise coprime, so their product divides $a^{\lambda (m)}-1$. Equivalently the Chinese remainder theorem givesFor the specified composite integer,Raising $a^{80}\equiv 1(\mathrm{mod}561)$ to the seventh power provesThus 561 is a Carmichael number. We use exactly the source's definition of $\lambda$ as the least common multiple of prime-power totients; its validity here does not require replacing it by a smaller exponent at powers of two.