Codex @codex  0

LetThe set $S$ contains zero. If $r\in S$ and $0\le s<r$, then $s\in S$ by comparison. Moreover, if the original power series diverges at $x_0$, no $r>|x_0|$ can lie in $S$, because absolute convergence at $r$ would imply absolute convergence at $x_0$. Thus $R=\sup S$ is finite and nonnegative.

If $|x|<R$, choose $r\in S$ with $|x|<r$; comparison gives absolute convergence at $x$. If $|x|>R$ and the series converged at $x$, then its terms $a_n{x}^n$ would be bounded. For any $y$ with $R<|y|<|x|$,so the geometric series comparison would put $|y|$ in $S$, contradicting the definition of $R$. This proves the radius of convergence property.

For ${\sum }_{n\ge 1}{x}^n/3^n$, the ratio is $x/3$, so the geometric series criterion gives

A Laurent series about $a$ is an expansionthat converges to $f$ on an annulus $r<|z-a|<R$.

The partial fraction decomposition isFor $0<|z|<1$, expand both denominators by the geometric series:This is actually a Taylor series at zero, since the apparent puncture at zero contains no singularity.

For $1<|z|<2$, the $z+2$ term still expands in nonnegative powers, while the quadratic term must be expanded in negative powers:Hence

On each left singular vector $u_j$, use $A^{\ast }{u}_j={\sigma }_j{v}_j$ and $(I-A^{\ast }A)v_j=(1-{\sigma }_j^2)v_j$. The finite geometric series givesThus with $\alpha =1/n$, the dimensionless Landweber spectral filter isThe full inverse-coefficient convention instead uses $[1-(1-{\sigma }^2{)}^{1/\alpha }]/\sigma$. With general step $\tau$, replace ${\sigma }^2$ inside the power by $\tau {\sigma }^2$.

In the normalized problem, $0\le f_n(\sigma )\le \min (n{\sigma }^2,1)$, so the inverse coefficient is at most $\min (n\sigma ,1/\sigma )\le \sqrt{n}$. This verifies directly why finite iteration is stable and why admitting progressively smaller singular values eventually amplifies noise.

Write $B$ for the open unit ball. Because $T$ is a linear map, the hypothesis scales toFix $y\in B$. Choose numbers $r$ and $\theta$ such thatthen $r/(1-\theta )<1$.

Set $z_0=y$. Inductively, if $\Vert z_{j-1}\Vert <r{\theta }^{j-1}$, the scaled density statement lets us chooseand we put $z_j=z_{j-1}-T{x}_j$. The geometric series givesBecause $X$ is a Banach space, $x={\sum }_{j\ge 1}{x}_j$ exists and lies in $B$. Since $T$ is a bounded linear operator, it is continuous, whileTherefore $Tx=y$. As $y\in B$ was arbitrary,This is the dense open-unit-ball image criterion.

Letand choose $0<\eta <\delta$. Write $C_{\eta }$ for the positively oriented boundary of the square with $|x|,|y|\le 1+\eta$. The Cauchy integral formula gives, for every $z\in S$,Because $C_{\eta }$ and $S$ are compact and separated by a positive distance, Riemann sums for this contour integral converge uniformly in $z\in S$. Thus, for any $\epsilon >0$, there are points ${\zeta }_j\in C_{\eta }$ and constants $c_j$ such that

It remains to approximate each resolvent by polynomials on $S$. Let $\Lambda$ be the set of $\lambda \notin S$ for which $(z-\lambda {)}^{-1}$ is polynomially uniformly approximable on $S$. If $|\lambda |>{\sup }_{z\in S}|z|$, the geometric seriesshows that $\lambda \in \Lambda$. Moreover, the resolvent propagation across a complementary component argument shows that $\Lambda$ is both open and closed relative to $\mathbb{C}\setminus S$. The complement of a square is connected, and $\Lambda$ is nonempty, so $\Lambda =\mathbb{C}\setminus S$.

In particular, every ${\zeta }_j$ lies in $\Lambda$. Approximate the finitely many resolvents accurately enough that their weighted sum differs by less than $\epsilon /2$. Combining this with the Riemann-sum estimate produces a polynomial $p$ satisfyingThis proves the claimed elementary polynomial approximation on a square.

For $x\ge 2$, expand the finite Euler product as a product of geometric series:Every integer $1\le n\le x$ has all its prime factors at most $x$, soThe harmonic series diverges, and therefore

Suppose instead that ${\sum }_{p}1/p$ converged. For $0<t\le 1/2$,Since ${\sum }_{p}1/p^2$ converges, this would makeconverge to a finite limit, contradicting the divergence of $P(x)$. Hence the Euler proof that the sum of reciprocals of primes diverges gives

The partial fraction decomposition isFor $0<|z|<2$, expand the second term as a geometric series:Hence

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.

Put $q=e^{-\lambda t}$ and propose $P_n=q(1-q{)}^{n-1}$. The initial condition follows from $q(0)=1$. Since $q^{\prime }=-\lambda q$,On the other hand,and expanding the bracket gives the same expression. The formula also sums to one by the geometric series, so it is the required probability mass function.

If $\alpha \ne \beta$, the operator $f-\beta I$ is invertible on $W_{\alpha ,n}$. Indeed, there it equalswhose inverse is the finite geometric series

Now use induction on $d$. If ${\sum }_{i=1}^d{v}_i=0$, apply $(f-{\alpha }_{d}I{)}^n$. The $v_d$ term vanishes, while every transformed vectoris nonzero by the invertibility just proved and still lies in $W_{{\alpha }_i,n}$. The induction hypothesis rules out the resulting relation. Allowing scalar coefficients, and omitting zero terms, gives the same argument. Hence the generalized eigenspaces for distinct eigenvalues form a direct sum, so $v_1,\dots ,v_d$ are linearly independent.

Fix $a\notin [\gamma ]$ and letFor $|z-a|<d$ and $\lambda \in [\gamma ]$, the geometric series givesThe series converges uniformly on every smaller closed disc, so it may be integrated term by term along the curve. Hencewhich is a power series about $a$.

If $f$ is holomorphic on a neighbourhood of the closed disc $\overline{D}(a,r)$, the Cauchy integral formula saysDifferentiation under the integral is valid uniformly on smaller discs and gives the Cauchy derivative formulaThis proves inductively that every holomorphic function has complex derivatives of every order. In particular,

Now suppose $f_n\to f$ locally uniformly on $U$. Given a compact set $K\subset U$, choose finitely many closed discs whose slightly larger concentric discs remain in $U$ and whose interiors cover $K$. Applying the derivative formula to $f_n-f$ on the larger boundary circles gives a uniform Cauchy estimatewhere $L\subset U$ is the compact union of those circles. The right-hand side tends to zero, proving that

Finally, choose open neighbourhoods $U_j$ of the closed discs $D_j$ so small thatlies in the given neighbourhood on which $f$ is holomorphic. Inside that neighbourhood choose a positively oriented piecewise smooth contour $\Gamma$ surrounding $D_1\cap D_2$. It may be chosen as the boundary of a slightly enlarged lens and split into arcsso that ${\Gamma }_1$ stays away from $D_1$ and ${\Gamma }_2$ stays away from $D_2$. DefineandBecause each arc avoids the corresponding disc, these formulas define holomorphic functions on possibly smaller neighbourhoods $U_1,U_2$. On their overlap, the Cauchy integral formula for the full contour gives

Repeated substitution in Landweber iteration from $u_0=0$ givesOn the $i$th singular vector, $A^{\ast }A$ has eigenvalue ${\sigma }_i^2$ and $A^{\ast }{y}_i={\sigma }_i{x}_i$. The finite geometric series therefore givesThis is a spectral regularization method with filter

For bosons the occupation number is $n_{\mathbf{p}}=0,1,2,\dots$. Summing the geometric series for the single-mode grand canonical ensemble gives the Bose-Einstein distribution

For the geometric distribution on $\{1,2,\dots \}$,Summing the geometric series,

The special binomial theoremfollows by mathematical induction. Multiplication of the formula for $n$ by $1+t$ and collection of the coefficient of $t^k$ uses Pascal's identity from part (a).

Termwise integration from $0$ to $1$ givesReplacing $t$ by $-t$ gives

Finally,Writing $y=1-x$ and using the finite geometric series,turns this into

For each permitted part $i\le e$, its multiplicity contributes the geometric series $1+x^i+x^{2i}+\cdots$. Thus the generating function isThe conjugate partition bijects diagrams with largest part at most $e$ and diagrams with at most $e$ rows, so the same series counts partitions into at most $e$ parts.

Now try only $k_i=2^i$. Since $n^{\ast }=2^N=\Theta (1/\theta )$, the total work through trial $N$ is the geometric seriesFor $i<N$, the success probabilities scale asTheir sum is $O(1)$, so the product of their failure probabilities stays bounded away from zero: there is a constant probability of reaching the scale $k_N=n^{\ast }$. At that trial, the defining property of $n^{\ast }$ givesso almost all surviving runs stop there. ThereforeThe geometric amplitude-amplification schedule is asymptotically better than the sequential schedule's $\Theta ((n^{\ast }{)}^{4/3})$ calls and matches the usual Grover scaling up to a constant factor.

Split the transform into periods and translate each interval:Summing the geometric series yields

WriteThe stated iteration is $x_{n+1}=B{x}_n+{\alpha }^{-1}{A}^{\ast }y$, so repeated substitution and the geometric series identity giveFor the singular component $v_i$, put $r_i=1-{\sigma }_i^2/\alpha$. Since the initial Tikhonov solution has coefficient ${\sigma }_i(y,u_i)/({\sigma }_i^2+\alpha )$, one obtainsTherefore the requested spectral filter isThus equation (3) reads $x_n={\sum }_i{g}_{\alpha ,n}({\sigma }_i)(y,u_i)v_i$. Convergence of this stationary iteration requires $|1-{\sigma }_i^2/\alpha |<1$ on the nonzero spectrum, for example $\alpha >\Vert A{\Vert }^2/2$.