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Consider the initial value problem
![\begin{displaymath}
\frac{d {\bf y}}{d t} = {\bf f}({\bf y}), \qquad {\bf y}(0) = {\bf y_0}.
\end{displaymath}](img4.png) |
(1) |
Suppose we write the Taylor exansion of the solution:
![\begin{displaymath}
{\bf y}(h) = {\bf y}(0) + \left.\frac{d {\bf y}}{d t} \right\vert _{t=0} h + \cdots
\end{displaymath}](img5.png) |
(2) |
Truncating and using (1), we obtain the formula for Euler's method
for the numerical solution of differential equations:
![\begin{displaymath}
{\bf y}(h) \approx {\bf y_0} + h {\bf f}({\bf y_0}).
\end{displaymath}](img6.png) |
(3) |
Of course, there is nothing special about
, so, letting
,
, we obtain
![\begin{displaymath}
{\bf y}_{n+1} \approx {\bf y}_n + h {\bf f}({\bf y}_n).
\end{displaymath}](img10.png) |
(4) |
By iterating, we find an approximation to the solution
of (1).
Here
is known as the stepsize.
Jeffrey M. Moehlis
2001-09-24