↑ stuart clark:
Hi.
Skrytý text:
Since
[mathjax]\begin{pmatrix} 1 & \frac{x}{n} \\ -\frac{x}{n} & 1 \end{pmatrix}^n = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ -i & i \end{pmatrix}\begin{pmatrix} \left(1-\frac{ix}{n}\right)^n & 0 \\ 0 & \left(1+\frac{ix}{n}\right)^n \end{pmatrix}\begin{pmatrix} 1 & i \\ 1 & -i \end{pmatrix}[/mathjax], there holds
[mathjax]\lim_{n\rightarrow \infty}\begin{pmatrix}1 & \frac{x}{n}\\ -\frac{x}{n} & 1 \end{pmatrix}^n = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ -i & i \end{pmatrix}\begin{pmatrix} \mathrm{e}^{-ix} & 0 \\ 0 & \mathrm{e}^{ix} \end{pmatrix}\begin{pmatrix} 1 & i \\ 1 & -i \end{pmatrix} =\begin{pmatrix} \cos(x) & \sin(x) \\ -\sin(x) & \cos(x) \end{pmatrix} = \mathrm{e}^{\mathbb{A}x}[/mathjax],
where [mathjax]\mathbb{A}=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} [/mathjax]
Hence, when solving system of ODE [mathjax] \boldsymbol{y}'=\mathbb{A}\boldsymbol{y}[/mathjax] with the exact solution, [mathjax]\boldsymbol{y}(x)=\mathrm{e}^{\mathbb{A}x}\boldsymbol{y}(0)[/mathjax], we can compute its approximation by considering
[mathjax]\boldsymbol{y}_n(x) = \left(\mathbb{I}+\frac{x}{n}\mathbb{A}\right)^n\boldsymbol{y}(0) \; \longrightarrow \;\mathrm{e}^{\mathbb{A}x}\boldsymbol{y}(0)[/mathjax].
for all 