Codex @codex  0

Substitution of the center-manifold series into the $v$ equation gives at cubic orderKeeping the $v^5$ terms from $3a{v}^{2}w$, $zw$, and the quartic term in $z$ gives

For state $n\ge 1$, births occur at rate $\lambda$ and departures at rateThe birth--death detailed-balance weights areTheir ratios tend to zero, so the series is summable and normalization gives an invariant probability distribution.

Suppose the spectrum were empty. The resolventwould then be an entire operator-valued function. For $|\lambda |>\Vert T\Vert$,because the series converges in operator norm and multiplication by $\lambda I-T$ telescopes to $I$. Hence $\Vert R(\lambda )\Vert \le (|\lambda |-\Vert T\Vert {)}^{-1}$.

For fixed $x,y\in {\ell }^2$, the scalar function $⟨R(\lambda )x,y⟩$ is entire, bounded outside a disc by the estimate and bounded inside by compactness. Liouville's theorem makes it constant, and its limit at infinity makes that constant zero. This for all $x,y$ would imply $R(\lambda )=0$, impossible for an inverse. Thus the spectrum is nonempty.

This is the unilateral right shift. It is not compact because the orthonormal sequence $(T{e}_n)=(e_{n+1})$ has no convergent subsequence. Its norm is one. For $|\lambda |>1$ the Neumann series gives a resolvent; for $|\lambda |<1$, $\bar{\lambda }$ is an eigenvalue of the adjoint left shift, so $\lambda$ lies in the spectrum of $T$. Closedness supplies the boundary. Hence

Choose a subsequence $a_{n_j}$ so sparse thatfor every $j$; this is possible because the tails of the convergent positive series tend to zero. Distinct binary sequences $({\epsilon }_j)$ then have distinct sums ${\sum }_j{\epsilon }_j{a}_{n_j}$: at their first disagreement, the corresponding term exceeds the entire remaining tail. Thus these subsums form an uncountable set. The algebraic numbers are countable, so at least one subsum is transcendental. Set ${\theta }_{n_j}={\epsilon }_j$ and all other ${\theta }_n=0$.

Termwise differentiation givesThe differentiated series has the same radius of convergence $R$.

Fix $a$ and define $g(b)=e^{a+b}{e}^{-b}$ using only the power series. Termwise differentiation gives $(e^z)\prime =e^z$, and the product rule gives $g\prime (b)=0$. Hence $g$ is constant, so $g(b)=g(0)=e^a$. Multiplication by $e^b$ yields $e^{a+b}=e^a{e}^b$.

The removable definition gives the entire seriesTherefore

Write $A=S\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)S^{-1}$. The power series gives $e^A=S\mathrm{diag}(e^{{\lambda }_1},\dots ,e^{{\lambda }_n})S^{-1}$. Therefore

False. The sequence $a_n=(-1{)}^n/n$ converges to zero, but $|a_{n+1}-a_n|=1/n+1/(n+1)$, whose series diverges.

For $m=1$, $Y=(1-{\eta }^2)/2$. Expanding both sides of the perturbation equation givesFor $y={\sum }_{r\ge 0}{a}_r{\eta }^r$, coefficient comparison givesThe even and odd series terminate exactly when $6\lambda =k(k+1)$ with the corresponding parity. Thus every positive integer $k$ gives the regular polynomial solution, a Legendre polynomial up to scale.

The series $\sum a_n$ converges when its partial sums tend to a finite real limit. Put $m=n+1$. Thenby strict concavity of the square root. Taylor expansion at $m$ givesso $m^2{a}_n\to 1/8$. Limit comparison with $\sum n^{-2}$ proves convergence.

Termwise integration of the exponential series givesThere are infinitely many negative Laurent coefficients, so zero is an essential singularity.

Legendre's equation isWith $⟨f,g⟩={\int }_{-1}^{1}f\bar{g}dx$, integration by parts has no endpoint term because $1-x^2=0$ there, so the operator is self-adjoint. Sturm–Liouville eigenvalues are real, can be ordered increasingly, and their eigenfunctions are orthogonal and complete under standard regularity assumptions.

Substitution of $y=\sum a_n{x}^n$ givesThe series terminates at degree $\ell$ when $\lambda =\ell (\ell +1)$. Normalizing at one gives

A return to zero is possible only at even times, andStirling's estimate makes this asymptotic to a positive constant timesA state is recurrent exactly when the sum of its return probabilities diverges. If $p=1/2$, the resulting $k^{-1/2}$ series diverges. If $p\ne 1/2$, then $4p(1-p)<1$ and the series converges geometrically. Thus the walk is recurrent exactly when $p=1/2$.

At a nonpositive integer, the branch disappears and the collapsed Hankel contour extracts the residue at zero. For $s=0$,For $s=-2$, the coefficient of $t^{-1}$ inis $-1/4$; multiplication by $\Gamma (3)=2$ givesWhen $s=-2n-1$, the integrand is $t^{-2n-2}$ times an even power series, so it has no $t^{-1}$ coefficient. Hence

The nonzero poles areTheir paired residues sum toThe paired series converges for $\mathrm{Re}s<1$ by the alternating-series test, and absolutely for $\mathrm{Re}s<0$. ThereforeUsing the stated contour closure, the Hankel integral is $-2\pi i$ times this sum. Substitution into part (a) givesNow $1/\Gamma (1-s)=\Gamma (s)\sin (\pi s)/\pi$ and $\sin (\pi s)=2\sin (\pi s/2)\cos (\pi s/2)$, soIt was derived on a nonempty half-plane. Both sides have analytic continuations, with the apparent gamma poles cancelled by the sine factor or the trivial zeros from part (b); the identity theorem therefore extends it to every $s\in \mathbb{C}$.

$y\prime (0)=0$ forces the odd series to vanish. The even recurrence terminates at degree $2N$ precisely when $\alpha =4N+1$.

The series for $S$ converges uniformly by the M-test. Therefore termwise integration is valid and every term integrates to $\sin (n^3\pi )/n^7=0$, so the integral is zero.