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I want to ask for a hint in solving the following problem from Berkeley Problems in Mathematics:

Let $h>0$ be given. Consider the linear difference equation $$\frac{y((n+2)h)-2y((n+1)h)+y(nh)}{h^{2}}=-y(nh),n\in \mathbb{Z}^{*}$$ 1) Find the general solution of the equation by trying suitable exponential subsititions.

2) Find solutions with $y(0)=0$ and $y(h)=h$. Denote it by $S_{h}(nh)$.

3). Let $x$ be fixed and $h=\frac{x}{n}$. Show that $$\lim_{n\rightarrow \infty}S_{\frac{x}{n}}(\frac{nx}{n})=\sin[x]$$

I am having trouble even using the first step. I tried $y=e^{P(t)}Q(t)$ but I feel it is unlikely to generate the required relation. So I am stuck. I think I need a hint as the book does not have a solution to this problem.

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I guess the first equation should $...=0$? – Frenzy Li Jul 21 '12 at 9:13
Sorry for the mistake, corrected. – Bombyx mori Jul 21 '12 at 9:27
up vote 1 down vote accepted

The equation:

$$\frac{y((n+2)h)-2y((n+1)h)+y(nh)}{h^{2}}=-y(nh),n\in \mathbb{Z}^{*}\,,$$

is nothing but a difference equation with constant coefficients coming from a second order differential equation with constant coefficients. Now, you need to solve this difference equation. Assume your solution $y(nh) = r^n $ (by the way that's the suitable exponential substitution) and substitute back in your equation and solve the resulting polynomial for $r$. Note that you can write the equation in the form.

$$ y \left( n+2 \right) -2\,y \left( n+1 \right) + \left( 1+{h}^{2} \right) y \left( n \right) =0 $$

with keeping in mind that $ h=\frac{x}{n}\,. $ Your general solution is going to have the form $y(n)=y(nh) = C_{1}r_{1}^n + C_{2}r_{2}^n $ which is going to be a function in $h$. Then you can advance with 2 which is determining the constants $C_{1}$ and $C_{2}$ subject to the initial conditions you are given. I hope it is clear now.

Substituting $ y(n) = r^n $ in the difference equation gives

$$ r^{n+2} - 2 r^{n+1} + (1+h^2)r^n =0 \Rightarrow r^2 - 2 r +(1+h^2)=0 \,.$$

Solve the above polynomial and construct your general solution.

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you mean $r$ as a function of $h$? I feel confused. – Bombyx mori Jul 21 '12 at 10:42
Yes, see the edit. – Mhenni Benghorbal Jul 21 '12 at 13:29

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