Fix \(h>0\) and consider \[ \frac{y((n+2)h)-2y((n+1)h)+y(nh)}{h^2}=-y(nh), \qquad n=0,1,2,\dots. \]
- Find the general solution by exponential substitution.
- Find the solution satisfying \[ y(0)=0, \qquad y(h)=h, \] and denote it by \(S_h(nh)\).
- For fixed \(x\), put \(h=x/n\). Prove that \[ \lim_{n\to\infty}S_{x/n}(x)=\sin x. \]
Solution
Put \[ u_n\coloneqq y(nh). \]
1. The difference equation is equivalent to\[ u_{n+2}-2u_{n+1}+(1+h^2)u_n=0. \]
Multiply the given equation by \(h^2\) and move the term \(-h^2y(nh)\) to the left.
2. The characteristic equation of the recurrence in step 1 is\[ r^2-2r+(1+h^2)=0, \]with distinct roots\[ r_\pm=1\pm ih. \]
Substituting \(u_n=r^n\) into the recurrence gives \[ r^n\left(r^2-2r+1+h^2\right)=0. \] Since \(h>0\), the two roots \(1\pm ih\) are distinct.
3. The general solution is\[ \boxed{ y(nh) = A(1+ih)^n+B(1-ih)^n }, \]where \(A,B\in\CC\) are arbitrary.For real-valued solutions, the coefficients are exactly those satisfying\(B=\overline A\).
Each characteristic root from step 2 yields a solution, and the two solutions are linearly independent because the roots are distinct. A second-order recurrence is uniquely determined by \(u_0\) and \(u_1\), so their span is the full complex solution space. Since the two basis solutions are complex conjugates, their linear combination is real for every \(n\) exactly when \(B=\overline A\).
4. The solution satisfying \(y(0)=0\) and \(y(h)=h\) is\[ \boxed{ S_h(nh) = \frac{(1+ih)^n-(1-ih)^n}{2i} }. \]
Step 3 and \(u_0=0\) give \[ A+B=0. \] Thus \(B=-A\). The condition \(u_1=h\) then gives \[ h = A\left((1+ih)-(1-ih)\right) = 2ihA, \] so \(A=1/(2i)\) and \(B=-1/(2i)\).
5. For every fixed \(x>0\),\[ \lim_{n\to\infty}S_{x/n}(x)=\sin x. \]
The standing hypothesis \(h>0\) makes the substitution \(h=x/n\) literal when \(x>0\). By step 4, \[ S_{x/n}(x) = \frac{ \left(1+\frac{ix}{n}\right)^n - \left(1-\frac{ix}{n}\right)^n }{2i}. \] Using the standard exponential limit, \[ \left(1+\frac{ix}{n}\right)^n\longrightarrow e^{ix}, \qquad \left(1-\frac{ix}{n}\right)^n\longrightarrow e^{-ix}. \] Therefore \[ \lim_{n\to\infty}S_{x/n}(x) = \frac{e^{ix}-e^{-ix}}{2i} = \sin x. \] The same closed formula extends continuously to \(x=0\) and, if the recurrence is allowed for nonzero negative step size, to negative \(x\) as well.
6. Q.E.D.
Steps 3, 4, and 5 answer parts 1, 2, and 3 respectively.