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For a linear recurrence relation with two distinct characteristic equation of a linear recurrence roots $r_1,r_2$, the solution is $u_n=A{r}_1^n+B{r}_2^n$. If $|r_1|<1<|r_2|$, boundedness for $n\to \infty$ forces $B=0$, and $u_0$ fixes $A$. This selects the decaying discrete mode in a finite difference method for a boundary problem on an unbounded interval. Roots on the unit circle require separate treatment; a repeated root there can contribute an unbounded factor $n$.

Collecting the shifted terms gives a second-order difference equation with coefficients $1-h/2$, $-(2+6h^2)$ and $1+h/2$. Substitute $y_n=r^n$ to obtain its characteristic equation of a linear recurrence:For real $h\ne 0,2$, the roots and general solution areIn the small-step regime, indeed for $0<h<2$, the polynomial is positive at $r=0$, negative at $r=1$ and positive for sufficiently large positive $r$. Thus $0<r_{-}<1<r_{+}$. The bounded solution of a second-order constant-coefficient recurrence discards the growing root, and $y_0=1$ fixes its remaining coefficient:The Taylor series establishes the displayed approximation to the root. Raising that approximation to the $n$th power is a leading approximation, rather than a uniform relative approximation for arbitrarily large $n$: its relative error is controlled when $n{h}^2$ is small. For completeness, the degenerate values give $y_n=A+Bn$ when $h=0$, and $y_n=C13^{-n}$ when $h=2$.

Let $h_k$ be the probability that the simple random walk started at $k$ hits $n$ before zero. The first-step analysis recurrence and boundary conditions areIf $p\ne q$, the characteristic equation of a linear recurrence has roots $1$ and $q/p$, so the boundary conditions giveIf $p=q=1/2$, the repeated root gives the limiting formula $\boxed{h_m=m/n}$. This is the gambler's ruin probability.

For a fixed integer stake $a\le N$, the fact that $a$ divides $N!$ lets us measure Patricia's wealth in units of £$a$. She starts at $M=N!/a$ and succeeds upon reaching $2M$. The formula above factors asFor $p=18/37$, $r=19/18>1$, so this probability increases as $M$ decreases. She should therefore choose the largest stake, $\boxed{a=N}$. For $p=19/37$, $r=18/19<1$, so the probability increases as $M$ increases; she should choose the smallest stake, $\boxed{a=1}$.

The characteristic equation of a linear recurrence isIts repeated root givesSince $t_n=nh$, this can be written $y_n=A+(B/h)t_n$, exactly the discrete form of the continuous solution $y(t)=C+Dt$ of $y^{\prime \prime }=0$.