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The polynomial Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $p$ such that ${\sup }_{z\in K}|p(z)-f(z)|<\epsilon$.

For an explicit approximation on the left semicircle $K=\{z:|z|=1,\mathrm{Re}z\le 0\}$, defineThe two power-series expansionsproduce this formula. The first is a geometric series whose ratio has modulus at most $4/\sqrt{17}<1$ on $K$. The second converges uniformly on the whole unit circle; the elementary bound $\left(\genfrac{}{}{0ex}{}{j+k}{k}\right)\le 2^{j+k}$ shows that truncating it at $k=n^2$ gives a total error tending to zero even after summing over $0\le j\le n$. Consequently $p_n\to 1/z$ uniformly on $K$. This is an explicit polynomial approximation of the reciprocal on the left semicircle.

No such sequence exists on the punctured unit circle $S=\{z:|z|=1,z\ne 1\}$. If polynomials $p_n$ converged uniformly there, they would be Uniformly Cauchy. Continuity givesso they would converge uniformly on the entire unit circle. The uniform limit theorem would force the value at the missing point to be $1$, and hence the limit would be $1/z$ on the full circle. But the Cauchy integral theorem gives $\oint p_n(z)dz=0$ for every $n$, while uniform convergence and contour integration would implya contradiction. This is the polynomial approximation obstruction on a punctured circle.

A renewal-reward process consists of i.i.d. regenerative cycles $({\xi }_n,R_n)$, where ${\xi }_n>0$ is the $n$th cycle length and $R_n$ is the reward accumulated during that cycle. If $0<\mathbb{E}{\xi }_1<\infty$ and $\mathbb{E}|R_1|<\infty$, the renewal-reward theorem states that the cumulative reward $Z(t)$ satisfiesalmost surely, with the usual negligible-remainder condition for an unfinished cycle.

For the machine, let $X\sim \mathrm{Exp}(\lambda )$ be its working lifetime after a repair. The next repair occurs at the first inspection following failure, so one repair-to-repair cycle has lengthwhile its working-time reward is $R=X$. Since an exponential distribution of rate $\lambda$ has mean $1/\lambda$,By the tail-sum formula for the positive integer-valued variable $\lceil X/m\rceil$,Thereforeand the long-run proportion of time for which the machine is working is