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The homogeneous characteristic equation of a linear recurrence isThe repeated characteristic root of a linear recurrence $-1$ supplies the independent terms $(-1{)}^n$ and $n(-1{)}^n$. A constant particular solution is $1/2$, since $1/2=-2(1/2)-1/2+2$. Thus $u_n=1/2+(A+Bn)(-1{)}^n$.

The scan prints $u_2=0$, not $u_1=0$. These initial conditions are consistent. They give $A=1/2$ from $u_0=1$, then $0=1/2+A+2B$, so $B=-1/2$. ThereforeIn particular $u_1=1/2$ and $u_2=0$. Direct substitution verifies the recurrence relation. This is an example of nonconsecutive initial values for a linear recurrence; replacing the printed subscript by 1 would change the answer.

For $n>1$, integration by parts with $u=x^{n-1}$ and $dv=e^{-x}dx$ yieldsThe boundary term vanishes at zero because $n-1>0$, and at infinity because exponential decay dominates every fixed power. The initial improper integral is $I_1={\int }_0^{\infty }{e}^{-x}dx=1$. Iterating the recurrence relation therefore givesThis is the positive-integer case of the Gamma function identity $\Gamma (n)=(n-1)!$.

Differentiate $x/(1+x^2{)}^n$ and integrate between zero and infinity. Its boundary values vanish for every integer $n\ge 1$, givingSince $x^2/(1+x^2{)}^{n+1}=(1+x^2{)}^{-n}-(1+x^2{)}^{-n-1}$, this becomes $0=E_n-2n(E_n-E_{n+1})$. Hence the recurrence relation isThe starting value is $E_1=[\arctan x{]}_0^{\infty }=\pi /2$. Therefore $E_2=\pi /4$, $E_3=3\pi /16$, andThese are also Wallis integrals after the integration by substitution $x=\tan t$.

Two applications of integration by parts lower the exponent by two. First, the sine boundary term vanishes:For $n\ge 1$, integrate the last expression once more:Thus the recurrence relation isThe starting value is $I_0={\int }_0^{\pi }\cos xdx=0$. Successive applications give $I_1=-2\pi$ and $I_2=-4{\pi }^3+24\pi$, followed by

Successive substitution in the matrix recurrence relation gives $V_k=V_0{Q}^k$. In the matrix power of $Q=ED{E}^{-1}$, every adjacent $E^{-1}E$ cancels, so $Q^k=E{D}^k{E}^{-1}$. ConsequentlyHere $D^k=\mathrm{diag}({\lambda }_1^k,\dots ,{\lambda }_n^k)$ for $k\ge 1$, and $D^0=I$. The formula also gives $V_0$ correctly at $k=0$.

Classification and existence. For $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, an ordinary point $x_0$ is one where $p,q$ are analytic. There are two independent analytic power series solutions, with arbitrary initial values $y(x_0),y^{\prime }(x_0)$. A regular singular point is a singular point where $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ are analytic. The Frobenius method then constructs solutions beginning with $(x-x_0{)}^{\sigma }$, where $\sigma$ solves the indicial equation. At least one Frobenius solution exists; two such solutions exist without logarithms when the two exponents differ by a noninteger. Repeated exponents or an integer difference can require a logarithmic second solution.

For this equation,Both are analytic at zero, while $p,q$ themselves have poles there. Thus zero is a regular singular point.

Indicial equation and recurrence. Multiply by $4x$ and insert $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$:The lowest power $x^{\sigma -1}$ has coefficient $[4\sigma (\sigma -1)+2\sigma ]a_0$, givingThe coefficient of $x^{n+\sigma }$ isFor either exponent the common factor is nonzero for $n\ge 0$, so the recurrence relation reduces toNormalizing each leading coefficient to one, the two Frobenius solutions areMore explicitly, $y_2=\sqrt{x}{\sum }_{n\ge 0}{2}^{n}n!x^n/(2n+1)!$. The distinct leading powers establish linear independence, so $y=C_1{y}_1+C_2{y}_2$ on the positive real side of the singular point. The displayed exponential function is the requested closed-form solution; substitution gives $y_1^{\prime }=y_1/2,y_1^{\prime \prime }=y_1/4$, which makes the differential expression zero.

Convergence. For either exponent the ratio of successive terms in the coefficient power series isThus both coefficient series have $\boxed{R=\infty }$ by the ratio test. This does not remove the $\sqrt{x}$ branch point of $y_2$: its analytic coefficient factor is entire, while the full Frobenius solution requires a square-root branch away from zero in the complex plane. The differential equation has no other finite singular points to limit continuation of these coefficient series.

Let $M=|m|$ for integer $m$. In the even or odd power series whose parity matches $M$, the numerator in the recurrence relation vanishes at the coefficient of $x^M$. All coefficients of lower powers in that parity are nonzero, and all higher ones vanish. Therefore a polynomial solution exists, of polynomial degree $M$, containingFor nonnegative $m$ these are the Chebyshev polynomials $T_m$. The normalized low-degree cases follow directly from the recurrence relation:Enforcing $T_m(1)=1$ givesTo see that normalization at one is always possible, set $x=\cos \theta$ in the Chebyshev differential equation. For $Y(\theta )=y(\cos \theta )$, the differential expression $(1-x^2)y^{\prime \prime }-x{y}^{\prime }$ becomes $Y^{\prime \prime }$. Thus $Y=\cos (M\theta )$ gives $T_M(x)=\cos (M\arccos x)$ on $[-1,1]$ and has value one at $x=1$. It is a polynomial because the cosine addition formula gives $T_{M+1}=2x{T}_M-T_{M-1}$. Negative integer $m$ produces the same solution as $M=|m|$.

Write a nonterminating parity series as $y=x^r{\sum }_{j\ge 0}{b}_j{x}^{2j}$, where $r=0$ or $1$. Its recurrence relation isThe ratio test applied to its nonzero successive terms has limiting absolute ratio $|x{|}^2$. Hence the series converges absolutely for $|x|<1$ and diverges for $|x|>1$, withAt $x=\pm 1$, the limiting ratio is one, so the basic ratio test alone is inconclusive. Retain the next term of the ratio instead:For large $j$ the ratio is positive. Taking logarithms and summing gives $\log |b_j|=-\frac{3}{2}\log j+O(1)$; the $O(j^{-2})$ remainders have a finite sum. Thus $|b_j|\le C{j}^{-3/2}$ eventually. Comparison with the convergent p-series proves absolute convergence at both endpoints as well:The radius of convergence one is the distance from the expansion point to the nearest singular points of a differential equation, $x=\pm 1$. Convergence of function values at those points does not make the solutions analytic there; endpoint derivatives can diverge. The terminating Chebyshev polynomial is an exception with infinite radius of convergence. For example, at $m=0$ the other solution is $\arcsin x$, whose endpoint values are finite while its first derivative diverges.

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.

Insert the power series and equate the coefficient of $x^n$. In the Legendre differential equation this givesHence the recurrence relation isThe free coefficients $a_0,a_1$ generate the independent even and odd series, respectively. They converge at least for $|x|<1$, up to the nearest singular points of the equation.

For a nonnegative integer $\ell$, the chain of the same parity as $\ell$ reaches $n=\ell$, where the numerator of the recurrence relation vanishes. The next coefficient and every later coefficient in that chain are zero, producing a polynomial of degree $\ell$. Normalizing it at $x=1$ gives the Legendre polynomial $P_{\ell }$.

The opposite-parity chain never reaches $n=\ell$. Its other possible zero, $n=-\ell -1$, is negative and cannot occur, so a nonzero initial coefficient produces a nonterminating power series. It is linearly independent of $P_{\ell }$ and can be scaled to the Legendre function of the second kind $Q_{\ell }$.

The degree statement requires $\ell \ge 0$, as the examples in part (c) indicate. For a negative integer $\ell$, the same equation has the nonnegative degree $-\ell -1$, because $\ell (\ell +1)=(-\ell -1)(-\ell )$. A polynomial cannot literally have negative degree.

Set $y={\sum }_{j\ge 0}{a}_j{x}^j$ and equate coefficients in the power series. The constant coefficient gives $a_2=0$. For $m\ge 1$, the coefficient of $x^m$ givesThe square bracket is $(m+1)(m+2)$, so the recurrence relation is $a_{j+3}=a_j$ for $j\ge 0$. The initial conditions give $a_0=1$, $a_1=0$, $a_2=0$. Hence

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 $n\ge 1$, the coefficient of $x^{n+\sigma }$ isFactoring the bracket gives the recurrence relationFor $\sigma =1/2$ this is $a_n=-a_{n-1}/[n(2n-1)]$. Iteration and $(2n)!=2^{n}n!(2n-1)!!$ yieldFor $\sigma =1$, the recurrence is $a_n=-a_{n-1}/[n(2n+1)]$, givingHere the double factorial products are one at $n=0$, consistent with the chosen normalization. Recognizing the cosine and sine power series, for positive $x$ one may also write the two Frobenius solutions asThe constants multiplying the two independent solutions may be chosen separately.

The trigonometric addition formula givesDividing by $\sin \theta$ and writing $x=\cos \theta$ gives the recurrence relationThe first two members are $y_1=1$ and $y_2=2x$. If $y_n$ has degree $n-1$ and leading coefficient $2^{n-1}$, then $2x{y}_n$ has degree $n$ and leading coefficient $2^n$, whereas $y_{n-1}$ has lower degree and cannot cancel that term. Mathematical induction provesFor example the sequence begins $1,2x,4x^2-1,8x^3-4x$, agreeing with the Chebyshev polynomial of the second kind.

Zero-stability means that perturbations of the starting values of the zero-step recurrence remain bounded by a fixed multiple of their initial size, uniformly as the number of steps tends to infinity on a fixed time interval. In particular solutions of ${\sum }_{j=0}^k{\alpha }_j{e}_{n+j}=0$ obey ${\max }_{n\le N}|e_n|\le C{\max }_{j<k}|e_j|$ with $C$ independent of $N$.

The equivalent root condition for a multistep method isA root outside the unit circle gives exponential growth of a recurrence relation mode; a repeated root on it gives polynomial growth. Repeated roots strictly inside it are harmless because their polynomial factor is dominated by decay.

Put $s=e^{-R\Delta t/L}$ and $J=V/R$, with $0<s<1$. The first interval leaves electric current $I_1=s{I}_0$. Position B reconnects the battery, so the driven RL circuit obeys $L\dot{I}+RI=V$. Multiplication by the integrating factor $e^{Rt/L}$, or subtraction of the constant solution $J$, givesConsequentlyThe same two intervals give the recurrence relation $I_{2n+2}=s^2{I}_{2n}+J(1-s)$. Its fixed point is $I_{\ast }=J/(1+s)$. Subtracting this fixed point makes the recurrence relation homogeneous:This is the upper endpoint of the eventual periodically switched RL circuit response, immediately before switching from B to A. The lower endpoint is $sJ/(1+s)$. Their sum is $J$, consistent with equal charging and discharging intervals.

Substitute the power series and collect each power of $\xi$:Matching the power series coefficients gives the recurrence relationThe even and odd coefficients form separate sequences, determined by $a_0$ and $a_1$, respectively. For example, $a_2=(1-ϵ)a_0/2$ and $a_3=(3-ϵ)a_1/6$. These are the two independent power series solutions of the Hermite differential equation.

The issue is the square-integrable condition on the whole real line, not just decay at one end. Because the quantum harmonic oscillator potential is even, the even and odd parts of a square-integrable wavefunction are themselves square integrable and satisfy the same equation. It is therefore enough to examine each parity sequence separately.

For a nonterminating sequence, put $b_j=a_{2j+p}$, where $p=0$ or $1$. Its recurrence relation givesSince $ϵ$ is real for an energy eigenvalue, the ratio is eventually positive. After multiplying by one constant phase, the tail coefficients are all positive. Choose $0<\delta <1/2$. For sufficiently large $j$, the ratio is at least $(1-\delta )/(j+1)$, so for some $K>0$ the tail coefficients obey $b_j\ge K(1-\delta {)}^j/j!$. Consequently, for $\xi >0$, the magnitude of $f$ eventually exceeds a positive multiple of ${\xi }^p{e}^{(1-\delta ){\xi }^2}$ after subtraction of a fixed polynomial. Multiplication by $e^{-{\xi }^2/2}$ leaves exponential growth like $e^{(1/2-\delta ){\xi }^2}$, incompatible with the square-integrable condition. Thus each nonzero parity series must terminate. The parity argument also excludes cancellation between the two nonterminating sequences at both infinities. This proves Normalization forces Hermite-series termination.

If the terminating polynomial has degree $n$, its top nonzero coefficient requires $2n+1-ϵ=0$. Conversely, when $ϵ=2n+1$, choosing the seed of parity $n$ nonzero makes the recurrence relation terminate exactly at degree $n$. The opposite seed must be zero, since that sequence does not terminate. ThereforeThe resulting polynomials are proportional to the physicists' Hermite polynomials, and ${\psi }_n\propto H_n(\xi )e^{-{\xi }^2/2}$ is normalizable because any polynomial times this Gaussian function is square integrable. The two parity seeds describe the full solution space of the second-order differential equation; the preceding argument therefore exhausts the normalizable stationary states. Each eigenspace has dimension one. General time-dependent quantum states can of course be superpositions of these energy eigenstates.

For brevity write $\gamma =mKZ/{\hslash }^2$. With the physical indicial root $c=\ell +1$, substitute $f={\sum }_{n\ge 0}{a}_n{r}^{n+\ell +1}$ into the equation from part (i). The coefficient of $r^{n+\ell }$ isThe indicial equation has already canceled the still lower power $r^{\ell -1}$. Rearranging gives the recurrence relationIts denominator is $(n+1)(n+2\ell +2)>0$ for every $n\ge 0$.

To impose wavefunction normalization at infinity, first suppose that the series never terminates. ThenThis is the exponential-series growth associated with $e^{2\alpha r}$, which would overwhelm the factor $e^{-\alpha r}$ in $\psi$. A direct lower-bound argument makes the conclusion precise. All sufficiently late recurrence ratios are real and positive, so after a common global phase choice the tail coefficients have one sign. For any $0<\eta <\alpha$, there is an index $N$ such thatInduction then bounds the tail below by a positive multiple ofThe finitely many earlier terms form a polynomial and cannot cancel this exponential growth. Thus $|\psi |=r^{-1}{e}^{-\alpha r}|f|$ grows at least as a positive multiple of $r^{\ell }{e}^{(\alpha -\eta )r}$ for large $r$, contradicting wavefunction normalization. This proves that the regular series must terminate.

Since the denominator never vanishes, termination requires the numerator to vanish at some nonnegative integer $n$:Then $a_{n+1}=0$ and every later coefficient is zero. Consequently $f$ is a polynomial of degree $n+\ell +1$, with a factor $r^{\ell +1}$; equivalently,where $P_n$ has degree $n$. This is series-termination quantization of the Coulomb radial equation. If $N_{\mathrm{p}}=n+\ell +1$ denotes the principal quantum number, the associated energy eigenvalues areHere $n$ counts the degree of the remaining radial polynomial; it is not the principal quantum number itself.