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Let $L$ be the last vertex visited for the first time, and assume $N\ge 2$. Every vertex is eventually painted with probability one: from any position, $N-1$ consecutive clockwise steps visit the entire cycle, and the probability of such a block is $2^{-(N-1)}$. Thus the probability that coverage has not occurred after $m$ successive blocks is at most $(1-2^{-(N-1)}{)}^m$, which tends to zero.

Fix $j\ne 0$ and cut the cycle at $j$. Represent it as the path with positions $0,1,\dots ,N$, where the two endpoints both represent $j$ and the original starting vertex is at position $N-j$. Up to the first visit to $j$, the walk is a simple symmetric random walk on this path, killed at either endpoint. The event $L=j$ is precisely that it visits both $1$ and $N-1$ before being killed: travelling between these two positions visits all intermediate vertices, while visiting all non-$j$ vertices necessarily includes these two.

For $N\ge 3$, first wait until the walk reaches one of $1,N-1$. This happens almost surely before it can reach an absorbing endpoint. If it reaches $1$ first, it must next reach $N-1$ before zero. The gambler's ruin probability is $1/(N-1)$: the hitting probability $h(i)$ on $0\le i\le N-1$ satisfiesso its successive differences are constant and $h(i)=i/(N-1)$. Starting at $1$ gives the stated probability. If the first endpoint reached is $N-1$, reflection of the path gives the same probability of reaching $1$ before $N$. The Strong Markov property at the first such hit therefore gives $\mathbb{P}(L=j)=1/(N-1)$, independently of which endpoint was hit first and of the start position.

Consequently the last visited vertex of a simple random walk on a cycle has distributionFor $N=2$ the same formula follows directly since the only other post must be last. For $N=1$, painting the initial post already completes the task, so there is no subsequent last unpainted post.

The time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\Psi =E\Psi$. For the hydrogen atom, separate variables as $\Psi =g(r)Y_{\ell m}(\theta ,\phi )$. The spherical-coordinate Laplacian has radial part $r^{-2}{\partial }_r(r^2{\partial }_r)$ and angular part $r^{-2}{\Delta }_{S^2}$. A spherical harmonic obeys ${\Delta }_{S^2}{Y}_{\ell m}=-\ell (\ell +1)Y_{\ell m}$. The first displayed term is therefore radial kinetic energy; the second is the attractive Coulomb's law potential energy $-e^2/(4\pi {\epsilon }_{0}r)$ of electron and proton; the third is the angular kinetic energy, also called the centrifugal potential. Regular single-valued angular states have $\boxed{\ell =0,1,2,\dots }$, with orbital angular momentum squared ${\hslash }^2\ell (\ell +1)$.

For $g=r^{\alpha }{e}^{-\beta r}$, direct differentiation givesSubstitution into the Radial Schrodinger equation for the hydrogen atom and equality of the coefficients of $r^{-2}$, $r^{-1}$ and the constant term giveThe regular radial solution selects $\alpha =\ell$, rather than the singular branch $-\ell -1$. At $\ell =0$, $r^{-1}$ is locally square integrable but is still inadmissible as a regular Coulomb eigenfunction: its Laplacian has a point-source contribution at the origin. Bound-state decay requires $\beta >0$. Thus, with the Bohr radius $a_0=4\pi {\epsilon }_0{\hslash }^2/(m{e}^2)$,These are the nodeless states with principal quantum number $n=\ell +1$. Increasing $\ell$ raises the energy towards zero, so emission is from the state labelled $\ell +1$ to the one labelled $\ell$. Using the Planck constant $h=2\pi \hslash$ and photon energy $h\nu$, the frequency is

For variations $y+ϵ\eta$ with $\eta (x_1)=\eta (x_2)=0$, integration by parts gives $\delta I={\int }_{x_1}^{x_2}[f_y-(d/dx)f_{y^{\prime }}]\eta dx$. Since this vanishes for every such variation, the Euler-Lagrange equation isWhen $f$ is independent of $y$, it has the first integral $\boxed{\partial f/\partial y^{\prime }=C}$.

For a ray represented as a graph traversed with increasing $x$, the travel time is $T={\int }_{x_1}^{x_2}\sqrt{1+y^{\prime 2}}/c(x)dx$, with $c(x)>0$. By Fermat principle, use $f=\sqrt{1+y^{\prime 2}}/c(x)$. The first integral becomesThe sign of $y^{\prime }$ is constant, and $y_2>y_1$ makes it positive. Write $C=1/k$ with $k>0$ and solve to obtainA finite slope requires $k>c(x)$ in the interior; a limiting vertical tangent is possible at an endpoint.

For $c(x)=ax$ and $k=a{x}_2$, positive speed on the whole interval and $k\ge c(x)$ select $a>0$ and $0<x_1<x_2$. The slope is $x/\sqrt{x_2^2-x^2}$. Integrating backwards from $(x_2,y_2)$ givesso $\boxed{(y_2-y{)}^2=x_2^2-x^2}$, with the branch $y\le y_2$. The endpoint integral is finite despite the vertical tangent. The specified value $k=a{x}_2$ fits the first endpoint precisely when $y_2-y_1=\sqrt{x_2^2-x_1^2}$; for other endpoint data the preceding integral determines a different $k$.

The contraction mapping theorem states that a contraction mapping $T:X\to X$ on a nonempty complete metric space has a unique fixed point. Here contraction means that $d(Tx,Ty)\le qd(x,y)$ for all $x,y$, with one constant $0\le q<1$. Choose $x_0\in X$ and define $x_{n+1}=T{x}_n$. Successive distances satisfy $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for every $m\ge 1$,Thus $(x_n)$ is a Cauchy sequence. Completeness gives $x_n\to x_{\ast }$. Since a contraction mapping is a continuous function, $T{x}_{\ast }=\lim T{x}_n=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ were another fixed point, $d(x_{\ast },y_{\ast })=d(T{x}_{\ast },T{y}_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing distance zero. This proves both existence and uniqueness.

Now suppose $T=f^k$ is a contraction mapping, where $k$ is a positive integer. Let $y$ be its unique fixed point. Since $f$ commutes with its own iterate, $T(fy)=f(Ty)=fy$. Thus $fy$ is also a fixed point of $T$, so $fy=y$. Any fixed point of $f$ is a fixed point of $T$, and must equal $y$. A contractive iterate therefore gives a unique fixed point of $f$, without assuming continuity of $f$ itself.

The space $X=C([0,1],\mathbb{R})$ is complete for the uniform norm. Indeed, a uniformly Cauchy sequence converges pointwise by completeness of $\mathbb{R}$; passing to the limit in its uniform Cauchy bound shows uniform convergence, and the uniform limit of continuous functions is a continuous function. The operator $F$ maps this space to itself: $s\mapsto \varphi (h(s),s)$ is a continuous function, and its integral plus the continuous $g$ is a continuous function.

For $n=0$ the desired bound is the definition of $\Vert h-k{\Vert }_{\infty }$. Suppose it holds for $n$. The Lipschitz condition and the fact that the same $g$ cancels giveThis proves the estimate by induction. Taking the supremum over $0\le t\le 1$, the Lipschitz constant of $F^n$ is at most $M^n/n!$. These constants tend to zero, since the ratio of successive terms is $M/(n+1)$. Some iterate is consequently a contraction. The preceding result gives a unique fixed point of $F$, which is exactly $\boxed{\text{the unique continuous solution of the integral equation on }[0,1]}$. No assumption $M<1$ is needed.