I Derive local truncation error for the Improved Euler Method

[oceanix]

TL;DR Summary

Can't figure out how to find the local truncation error for the Improved Euler method

I'm trying to find the local truncation error of the autonomous ODE: fx/ft = f(x).

I know that the error is |x(t1) − x1|, but I can't successfully figure out the Taylor expansion to get to the answer, which I believe is O(h^3).

Any help would be greatly appreciated!

 

[BvU]

Did you try googling " local truncation error for the Improved Euler " ? I got lots of good hits.

 

[pasmith]

Is that <br /> \begin{align*}<br /> \bar x_{i+1} &amp;= x_i + hf(x_i) \\<br /> x_{i+1} &amp;= x_i + \tfrac12 h(f(x_i) + f(\bar x_i))\mbox{?}\end{align*} The method is to expand the right-hand side of the second equation in a Taylor series about x_i,
<br /> \begin{align*}<br /> x_{i+1} &amp;= x_i + \tfrac12 h \left(f(x_i) + f(x_i + hf(x_i))\right) \\<br /> &amp;= x_i + \tfrac12 h \left(2f(x_i) + hf(x_i)f&#039;(x_i) + \tfrac12h^2f(x_i)^2f&#039;&#039;(x_i) + O(h^3) \right)<br /> \end{align*}<br />
and compare it to the Taylor series of the correct value, <br /> \begin{align*}<br /> x(t_i + h) &amp;= x(t_i) + h\dot x(t_i) + \tfrac12 h^2 \ddot x(t_i) + \tfrac16 h^3 x^{(3)}(t_i) + O(h^4) \\<br /> &amp;= x_i + hf(x_i) + \tfrac12h^2 f&#039;(x_i)f(x_i) + \tfrac16 h^3 f(x_i)(f(x_i)f&#039;&#039;(x_i) + f&#039;(x_i)^2) + O(h^4).<br /> \end{align*}