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The coefficients in the Taylor series areFor either series, the ratio test gives absolute term ratio $|x|/(n+1)$ or $2|x|/(n+1)$, tending to zero for each fixed real or complex $x$. Both have infinite radius of convergence.

Write the analytic solution as $w(z)={\sum }_{n\ge 0}{a}_n{z}^n$. The constant term in the differential equation gives $a_1=0$, and matching the coefficient of $z^{n+1}$ givesAll odd coefficients therefore vanish; $a_2=-a_0$ and $a_4=a_0/4$. The first three nonzero terms areIn full, $a_{2j}=(-1{)}^j{a}_0/(j!{)}^2$, so $w=a_0{J}_0(2z)$, a Bessel function of the first kind. The ratio test proves convergence for every complex $z$. The other independent solution has a logarithmic singularity at zero and does not satisfy the analyticity requirement.

In monic form the Chebyshev differential equation has $P(x)=-x/(1-x^2)$ and $Q(x)=p^2/(1-x^2)$. Both are analytic near zero, so it is an ordinary point. At $x=1$, $(x-1)P=x/(x+1)$ and $(x-1{)}^{2}Q=-p^2(x-1)/(x+1)$ are analytic; at $x=-1$, $(x+1)P=x/(x-1)$ and $(x+1{)}^{2}Q=-p^2(x+1)/(x-1)$ are analytic. Thus both endpoints are regular singular points.

Insert $y_q={\sum }_{n\ge 0}{a}_n{x}^{n+q}$ and equate coefficients. The recurrence relation isExplicitly the factors are $(n^2-p^2)/[(n+2)(n+1)]$ for $q=0$ and $[(n+1{)}^2-p^2]/[(n+3)(n+2)]$ for $q=1$. The lowest equation for $q=1$ forces $a_1=0$. For $q=0$, one may choose $a_1=0$ to select an even solution, independent of the odd $q=1$ solution. The leading coefficient $a_0$ is free in either chosen parity branch.

For a nonterminating branch the consecutive parity-term ratio tends to $x^2$. Hence the ratio test gives convergence for $|x|<1$, and radius of convergence one. Terminating branches are polynomials with infinite radius. For $p=m\ge 0$ an integer, choose $q$ of the same parity as $m$: the recurrence stops once $n+q=m$, producing a degree-$m$ Chebyshev polynomial. Normalization at $1$ gives all nonzero coefficients through degree three:

Set $Y(\theta )=y(\cos \theta )$. Direct differentiation gives $Y^{\prime \prime }={\sin }^2\theta y^{\prime \prime }-\cos \theta y^{\prime }$, so the equation becomes $Y^{\prime \prime }+m^{2}Y=0$. For $m>0$, $Y=A\cos (m\theta )+B\sin (m\theta )$. A polynomial $y$ has finite derivative at $x=1$, and therefore $Y^{\prime }(0)=-\sin 0y^{\prime }(1)=0$, forcing $B=0$. Also $Y(0)=T_m(1)=1$, so $A=1$. For $m=0$, the transformed solution is affine in $\theta$ and the same derivative condition removes its linear term. ThusThe endpoint values follow by continuity. The finite-derivative condition is what excludes the independent nonpolynomial branch, rather than normalization alone.

Write a second-order linear ordinary differential equation in monic form $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$. At a finite point $x_0$, an ordinary point has analytic $P,Q$. A singular point is a regular singular point if $(x-x_0)P$ and $(x-x_0{)}^{2}Q$ extend analytically there. A singular point failing this condition is an irregular singular point. The point at infinity is classified by the reciprocal coordinate $t=1/x$.

For the Laguerre differential equation, $P=1/x-1$ and $Q=\nu /x$. Thus $xP=1-x$ and $x^{2}Q=\nu x$ are analytic at zero, while $P$ itself is singular: zero is regular singular. Put $Y(t)=y(1/t)$. Since $y^{\prime }=-t^2{Y}^{\prime }$ and $y^{\prime \prime }=t^4{Y}^{\prime \prime }+2t^3{Y}^{\prime }$, the transformed equation isHere $t{P}_t=1+1/t$ is not analytic at zero, so infinity is an irregular singular point, for every real $\nu$.

For the Frobenius solution, the lowest power $x^{\sigma -1}$ has coefficient $a_0{\sigma }^2$. Hence the indicial equation is ${\sigma }^2=0$, and equating the remaining powers givesWith the repeated root $\sigma =0$, the Frobenius series for the Laguerre differential equation obeysIf the series does not terminate, its consecutive term ratio is asymptotic to $x/k$, so the ratio test gives infinite radius of convergence. It is an entire function, even though the differential equation has a singular coefficient at zero. The repeated indicial root yields only one analytic Frobenius branch; an independent solution generally has a logarithmic term there.

For $\nu =n\ge 0$, the factor $k-n$ stops the recurrence at degree $n$. With $a_0=1$ its coefficients are $a_k=(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})/k!$, giving the Laguerre polynomialsThe same termination argument includes $n=0$, as required for $L_0$ despite the earlier reference to positive integers.

For a nonterminating parity series, consecutive nonzero terms satisfyThe ratio test therefore gives $\boxed{R=1}$ for each nonterminating series. The nearest singular coefficient points are $x=\pm 1$, both distance one from the expansion centre.

The polynomial exception is essential here. If $n$ is an integer, the series of the same parity as $|n|$ terminates and has $\boxed{R=\infty }$; the opposite-parity independent series remains nonterminating with radius one. For noninteger $n$, both have radius one. A regular singular point can be removable for one particular solution, so its presence does not force every individual solution to have finite convergence radius.

The origin is an ordinary point. Substitute $y={\sum }_{n\ge 0}{a}_n{x}^n$ into the equation and collect the coefficient of $x^n$:Hence the recurrence relation isThere are two free initial coefficients. The independent even and odd power series beginThe general solution is $\boxed{y=a_0{y}_{\mathrm{e}}+a_1{y}_{\mathrm{o}}}$. As a useful independent check, the substitution $x=\sin t$ reduces the equation to $d^{2}y/d{t}^2+k^{2}y=0$. ThusFor a nonterminating parity series, $|a_{n+2}/a_n|\to 1$, so the ratio test in $x^2$ gives $\boxed{R=1}$. The regular singular points at $\pm 1$ explain the natural obstruction.

There is an important exception to an unconditional radius-one answer. If $k^2=m^2$ for an integer $m\ge 0$, the series of parity $m$ terminates at degree $m$ and is a multiple of a Chebyshev polynomial. Its radius of convergence is $\boxed{R=\infty }$; the other independent parity solution still has radius one. For $m=0$ the even solution is constant. Any nonzero combination containing the nonterminating solution has radius one. These are the full power series of the Chebyshev differential equation, including the terminating cases.

For either exponent, the recurrence from part (c) impliesfor every finite complex $x$. The ratio test therefore givesThe factor multiplying $x^{\sigma }$ is an entire function in each case. For $\sigma =1/2$, this does not make the full Frobenius solution single-valued at zero: its square-root prefactor still requires a branch. The infinite radius refers to $\sum a_n{x}^n$, not to a Taylor expansion of the branched full solution.

Use a Frobenius solution $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, with $a_0\ne 0$. The lowest power $x^{\sigma -1}$ has coefficient $2\sigma (2\sigma -1)a_0$, giving the indicial equationFor $n\ge 1$, matching the coefficient of $x^{n+\sigma -1}$ givesFor both candidate exponents the cancelled factor is nonzero in this range, so the recurrence isWith $a_0=1$, the exponent-zero solution isFor $\sigma =1/2$, the coefficients are $a_n=1/(2n+1)!!=2^{n}n!/(2n+1)!$, givingThe ratio test gives infinite radius of convergence for both series factors, since the successive ratios in magnitude are $|x|/(2n+2)$ and $|x|/(2n+3)$. The second full solution still needs a square-root branch in the complex plane; on $x>0$ it is a real solution. The distinct leading behaviors $1$ and $\sqrt{x}$ prove linear independence.

For an independent closed-form check on the structure, put $z=x/2$. The equation becomes $z{Y}^{\prime \prime }+(1/2-z)Y^{\prime }-(1/2)Y=0$, a Kummer differential equation with equal parameters. Its exponential reduction of a confluent hypergeometric equation with equal parameters explains the elementary first solution and the branched companion.

Use the generalized binomial theorem with exponent $-1/2$, choosing the analytic branch of a square root that equals $1$ at $z=0$. For $|z|<1$,The signs cancel in each binomial coefficient. For $n\ge 1$,The same expression equals $1$ for $n=0$, using the empty-product convention. Consequently the power series isIts coefficients are the normalized central binomial coefficients. As a convergence check, the ratio of successive absolute terms tends to $|z|$, so the ratio test gives absolute convergence inside the unit disc. Applying the Cauchy coefficient formula to the holomorphic function $(1-z{)}^{-1/2}$ now immediately yieldsThis value is independent of the chosen radius because the contour integral extracts the same coefficient on every such circle.

We use the stationary weights of a reflected birth-death chain, also verifying why they decide the question. If a stationary distribution $\pi$ exists, its equation at zero is ${\pi }_0={\pi }_1{d}_1={\pi }_0{u}_0$. At $k\ge 1$, stationarity givesThus the edge flow is constant, and its value at the boundary is zero. Every stationary distribution must therefore satisfy detailed balance on each edge. Since irreducibility forces ${\pi }_0>0$, it must be proportional toConversely, if ${\sum }_k{w}_k<\infty$, normalizing these positive weights gives a probability distribution satisfying detailed balance, hence a stationary distribution. Existence of a stationary distribution for an irreducible Markov chain on a countable state space is equivalent to positive recurrence.

The ratio isFor $p<1/2$, the limit is less than one, so the ratio test gives summability. For $p\ge 1/2$, observe that $u_k=p+\theta q/(k+\theta )>p\ge 1/2$, whereas $d_{k+1}<q\le 1/2$. Thus $w_{k+1}/w_k>1$ for every $k$, so the weights cannot be summable. This also resolves $p=1/2$ without making an inconclusive appeal to the limiting ratio. Therefore

With nonnegative production amounts $x_1,x_2,x_3$, the primal linear programme isIntroduce slacks $s_R,s_S\ge 0$. The initial basic variables are the slacks, and the reduced profits of the goods are $5,6,3$. The greatest is six, so $x_2$ enters. The ratio test compares $6/2=3$ and $11/3$, so $s_R$ leaves. The resulting dictionary iswith profit $18-7x_1+3x_3-3s_R$. The greatest positive reduced profit is now three, so $x_3$ enters; its only positive pivot coefficient is two in the second row, and $s_S$ leaves. The final tableau, expressed as equations, isAll nonbasic reduced profits are nonpositive, proving optimality atThe ordered optimal basis is $(x_2,x_3)$, henceThe inverse is read from the two slack columns of the final basic equations.

Set $K={\sum }_i\Vert A^i{\Vert }_{\mathrm{op}}^2$, where the norms are operator norms, and $D_m=y^{(m+1)}-y^{(m)}$. Independence of the Brownian coordinates and the vector Itô isometry giveInduction, as in part (b), yields the Factorial Picard estimate for a linear stochastic differential equationFor $t\le T$, the triangle inequality in $L^2$ therefore bounds any tail of the iterates by the corresponding tail ofThis numerical series converges by the ratio test: successive terms have ratio $\sqrt{KT/(m+2)}\to 0$. Thus the iterates are Cauchy in $L^2$, uniformly in the deterministic time parameter on each bounded interval. Completeness of Lp spaces gives a limit $y_t$. Once its continuous adapted version is chosen in part (d), the Itô isometry permits passage to the limit in the integral equation: the squared norm of each integral difference is bounded by $K{\int }_0^T\mathbb{E}|y_s^{(m)}-y_s{|}^{2}ds\to 0$. Consequently the Picard limit solves the stochastic differential equation.

Suppose real coefficients satisfy $c\le a_n\le C(1+n{)}^m$ for all sufficiently large $n$, where $c,C>0$ and $m\ge 0$. Then the power series $\sum a_n{z}^n$ has radius of convergence one. For $|z|<1$, the comparison test for series and the ratio test dominate it by a polynomial times a geometric sequence. For $|z|\ge 1$, the terms have modulus at least $c$ eventually and violate the term test for divergence. Finite changes to the coefficients do not affect the conclusion.

For a nonzero polynomial $p$, the power series ${\sum }_{n\ge 0}p(n)z^n$ has radius of convergence $1$. If $p$ has degree $d$ and leading coefficient $c\ne 0$, then $p(n)\sim c{n}^d$ and $|p(n+1)/p(n)|\to 1$, so the ratio test applies inside the unit disk. At every point of its boundary, the terms fail to tend to zero. The zero polynomial is the exceptional case with radius of convergence $\infty$.

Suppose $a_n>0$ and, for all sufficiently large $n$, $a_{n+1}/a_n\le 1-c/n$ with fixed $c>0$. Choose an initial index $N>c$. Since $\log (1-x)\le -x$ for $0<x<1$, summing logarithmic ratios gives $\log (a_n/a_N)\le -c{\sum }_{k=N}^{n-1}1/k\le -c\log (n/N)$. Thus $a_n\le a_N{N}^c{n}^{-c}$. For $c>1$, the comparison test for series with a P-series proves convergence. This supplies a useful refinement when a limiting ratio of one makes the ordinary ratio test inconclusive.

Write the Maclaurin series coefficients as $a_n=f^{(n)}(0)/n!$. When the successive ratios exist eventually and have a limit, the ratio test gives the displayed radius of convergence, including zero and infinity. The unconditional formula is $R^{-1}={\mathrm{limsup}}_{n\to \infty }|a_n{|}^{1/n}$, from the Cauchy-Hadamard theorem. Endpoints must be checked separately.

The existence and convergence of this power series do not by themselves imply that it equals $f$: equality requires analyticity or control of the Taylor remainder. The smooth function $f(x)=e^{-1/x^2}$ for $x\ne 0$, $f(0)=0$, has all derivatives zero at zero, so its Maclaurin series is identically zero despite the nonzero values away from zero.

The radius of convergence of ${\sum }_{n\ge 0}{a}_n(z-c{)}^n$ is the unique $R\in [0,\infty ]$ separating interior absolute convergence from exterior divergence. It converges absolutely when $|z-c|<R$ and diverges when $|z-c|>R$. At the centre it always has the finite value $a_0$, since all positive powers vanish. The boundary convergence of a power series is a separate question.

For $0<R<\infty$, the interior is a disk in the complex plane and an open interval $(c-R,c+R)$ on the real line. A power series converges uniformly on every smaller closed disk $|z-c|\le r<R$. This gives stable polynomial approximation over a whole compact region, rather than a separate limiting calculation at each point. See Jiří Lebl, Basic Analysis, §11.3.2: convergence of power series.

Four examples reveal what $R$ measures:

The Cauchy-Hadamard theorem computes $R$ directly from the coefficients. The attractive feature is that one growth statistic decides convergence throughout the interior and exterior, including sparse series with many zero coefficients. Finite changes to the coefficients do not alter the radius, though they can change the sum.

In a feasible simplex basis, increasing a nonbasic variable by $\theta$ changes basic coordinates to $x-\theta d$. Nonnegativity limits $\theta$ by the smallest ratio $x_i/d_i$ with $d_i>0$; a minimizing coordinate leaves the basis. Ties can leave zero-valued basic variables. This pivot test is distinct from the analytic ratio test for series convergence.