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Divide the second-order difference equation by $h^2$ and set $x=nh$. The first quotient is a central finite difference approximation to $y^{\prime \prime }(x)$; the quotient $(y_{n+1}-y_{n-1})/(2h)$ approximates $y^{\prime }(x)$. Their Taylor series errors are $O(h^2)$ for a sufficiently smooth function. This is therefore a finite difference method for the second-order linear differential equation in part (a).

At a fixed $x\ge 0$, take $h\to 0$ with $n=x/h$ integral. Since $\log r_{-}=-2h+O(h^3)$, the exact bounded discrete solution satisfiesLikewise $(1-2h{)}^{x/h}\to e^{-2x}$. The decay selected by boundedness in the linear recurrence relation becomes exactly the decay selected by boundedness in the ordinary differential equation.

Classification and characteristic exponents. For a normalized second-order linear differential equation, an ordinary point at zero means $p$ and $q$ are real analytic functions there. A regular singular point is a singular point where $xp$ and $x^{2}q$ nevertheless extend as real analytic functions to zero. Write $p_0={\lim }_{x\to 0}xp(x)$ and $q_0={\lim }_{x\to 0}{x}^{2}q(x)$. The lowest-power coefficient in the Frobenius method gives the indicial equationIts roots are the characteristic exponents at a regular singular point. If their difference is not an integer, there are two linearly independent solutions $x^{{\sigma }_j}{A}_j(x)$, with $A_j$ real analytic and $A_j(0)\ne 0$. For distinct real roots ${\sigma }_1>{\sigma }_2$ with integer difference, a solution for the larger root exists, and a second has the form $x^{{\sigma }_2}B(x)+c{y}_1\log x$; the logarithmic coefficient can vanish. For a repeated root, the second solution has the form $y_1\log x+x^{{\sigma }_1}B(x)$ after normalization. This is a Logarithmic Frobenius solution. Complex conjugate roots $a\pm ib$ give real solutions whose leading behaviour is $x^a\cos (b\log x)$ and $x^a\sin (b\log x)$, with analytic-series corrections. All these descriptions are on $x>0$, or on a fixed complex branch.

For the present equation, division by $x$ gives $p=1+(1-m)/x$ and $q=(1-m)/x$. Therefore $xp=x+1-m$ and $x^{2}q=(1-m)x$ are real analytic. Zero is regular singular when $m\ne 1$; when $m=1$, it is ordinary after removing the common factor $x$. In that exceptional case the normalized equation is $y^{\prime \prime }+y^{\prime }=0$.

For the Frobenius method substitution, the lowest coefficient and the general linear recurrence relation areThese equations come respectively from the coefficients of $x^{\sigma -1}$ and $x^{n+\sigma -1}$. At $m=1$, the same formal exponents $0,1$ describe the two possible leading powers at an ordinary point, rather than singular behaviour.

For the exponent $\sigma =m$, the linear recurrence relation reduces to $(m+n)a_n=-a_{n-1}$. The excluded nonpositive integers ensure that no denominator is zero. Normalize $a_0=1$ and iterate:The ratio test gives an infinite radius of convergence for the power series factor. Hence a solution isFor nonintegral $m$, choose instead the exponent $\sigma =0$. Since $n-m\ne 0$ for every positive integer $n$, its linear recurrence relation gives $n{a}_n=-a_{n-1}$. ThusThe distinct nonintegral leading exponents establish linear independence. More explicitly, $W(e^{-x},y_m)=m{e}^{-x}{x}^{m-1}$ by the Abel identity and the leading coefficient at zero, so the Wronskian is nonzero. The factor $x^m$ uses the real branch on $x>0$.

For positive integral $m$, the exponent-zero linear recurrence relation fixes $a_n=(-1{)}^n{a}_0/n!$ for $1\le n<m$, but at $n=m$ its factor $n-m$ vanishes and leaves $a_m$ free. Choose $a_0=1$ and $a_m=0$. All later coefficients then vanish by the same linear recurrence relation.

A polynomial solution of degree $m-1$ isFor $m=1$ this is the constant solution $1$, in agreement with the ordinary point exception. The terminating Frobenius series arises because the coefficient equation at the resonant index becomes $0=0$, permitting truncation rather than forcing a logarithm.

Write $K=k+1$ and $h_j={\mathbb{E}}_j{D}_j$. The expected duration of symmetric gambler's ruin satisfies $h_0=h_K=0$ and, by conditioning on the first step,These expected values are finite: from any interior state, $K$ consecutive right steps ensure absorption, an event of probability at least $2^{-K}$. Blocks of $K$ steps therefore bound the survival probability by a decreasing geometric sequence. The candidate $h_j=j(K-j)$ has the required boundary values and satisfies $h_{j+1}-2h_j+h_{j-1}=-2$. HenceFor completeness, the difference of two solutions of this linear recurrence relation has zero second difference and is affine in $j$; its two zero boundary values force it to vanish.

Subtract successive terms and use a telescoping series:Thus the linear recurrence relation has solution

The original PDF contains this subpart; the local TeX has omitted it. Write $a_n$ for the leading coefficient of $P_n$. The Legendre polynomial recurrence relation and the starting polynomials give $a_0=a_1=1$, and induction givesIndeed, $x{P}_n$ has degree of a polynomial $n+1$, while $P_{n-1}$ has smaller degree of a polynomial and cannot cancel its leading term. HenceAt $x=1$, the same linear recurrence relation gives $(n+1)P_{n+1}(1)=(2n+1)P_n(1)-n{P}_{n-1}(1)$. Starting from $P_0(1)=P_1(1)=1$, induction yields

The homogeneous linear recurrence relationhas characteristic polynomial $(r-2{)}^2$, soFor a particular solution, substitute $y_n=an+b$. The left-hand side becomesso matching $n$ gives $a=1$ and $b=2$. HenceThe conditions $y_0=1$ and $y_1=0$ give $C=-1$ and $D=-1/2$. Therefore

The characteristic polynomial of the linear recurrence relation isThe repeated-root rule therefore givesThe condition $x_0=0$ gives $A=0$. The other two conditions giveso $B+C=2$ and $B+2C=3$. Thus $B=C=1$, and

PutThe initial values are $S_1=S_2=1$. Using Pascal's identity and the convention that an out-of-range binomial coefficient is zero,Thus $(S_n)$ has the same initial values and linear recurrence relation as the Fibonacci numbers. Mathematical induction gives

A terminating Frobenius series is a Frobenius method solution whose analytic power series factor has only finitely many nonzero coefficients. Its coefficient linear recurrence relation must allow the first omitted coefficient and all following coefficients to be zero. At an integer separation of characteristic exponents at a regular singular point, a resonant coefficient equation can become $0=0$, permitting such a choice rather than forcing a logarithmic term. With exponent zero, termination produces a polynomial solution.

A variable-coefficient affine recurrence is a linear recurrence relation with coefficients allowed to depend on the index. If every $a_n$ is nonzero, a discrete integrating factor reduces it to a telescoping series.