enter image description here


I perhaps want to expand in taylor series about $x_n$.

$$ y_{n+1} = y_n + h y'_n + h^2/2 y''_n $$

$$ y_{n-1} = y_n - h y'_n + h^2/2 y''_n $$

$$ f(y_{n+1} ) = y'_{n+1} = y'_n + h y_n'' + ... $$

$$ f(y_{n-1} ) = y'_{n-1} = y'_n - h y_n'' + ... $$

So we ahve

$$ T_k = 2h y'_n - h/3 ( 6 y'_n + O(h^2) ) = O(h^3) $$

So it is ${\bf third}$ order accurate. Is this correct?

  • $\begingroup$ Are you sure that the higher degree Taylor terms do not cancel? At the moment you have only computed that the method is at least second order ($p=2$ with local error $O(h^{p+1})$). This would be not enough to make this a named method. $\endgroup$ – LutzL May 13 at 17:51
  • $\begingroup$ should I expand up to h^3? $\endgroup$ – ILoveMath May 13 at 18:26
  • $\begingroup$ The Simpson rule is global 4th order, which means that you would have to consider terms up to degree 5. $\endgroup$ – LutzL May 13 at 20:24

It's probably easier to just derive the order of local error of Simpson's rule for integration. We can derive it by applying this rule to polynomial base functions $$ f_{m}(x) \in \left\lbrace 1, x - x_{n - 1}, (x - x_{n-1})^{2}, ..., (x - x_{n-1})^{m}, ... \right\rbrace $$ and we see that for $f_{4}(x) = (x - x_{n-1})^{4}$ this integration rule is no longer accurate $$ \frac{(2h)^{5}}{5} = \frac{h}{3} \left(0 + 4h^{4} + (2h)^{4} \right) + C \cdot f^{(4)}(\xi) $$ so we can calculate the constant in the local error \begin{align*} 4! \cdot C &= \frac{32 h^{5}}{5} - \frac{20 h^{5}}{3} \\ C &= - \frac{h^{5}}{90}. \end{align*}

Then we can derive this method for solving ordinary differential equations in the following way. Given equation $$ y' = f(x, y) $$ integrate it $$ \int_{x_{n-1}}^{x_{n+1}} y'(x)dx = \int_{x_{n-1}}^{x_{n+1}} f(x, y(x))dx \\ $$ and apply Simpson's rule on the right hand side $$ y(x_{n+1}) - y(x_{n-1}) = \frac{h}{3} \left( f(x_{n-1}, y_{n-1}) + 4 f(x_{n}, y_{n}) + f(x_{n+1}, y_{n+1})\right) - \frac{1}{90} h^{5} f^{(4)}(\xi) $$ disregard the error term and replace the exact values with approximations $y_{n+1} \approx y(x_{n+1})$ and $y_{n-1} \approx y(x_{n-1})$. Note that we have obtained the method and so the order of local error equals 5.

  • $\begingroup$ Thanks for your reply! I have a question, for stability, what would be the stability region? $\endgroup$ – ILoveMath May 13 at 21:48

$y(x+h)-y(x-h)$ is twice the odd part of (the Taylor expansion of) $y(x+h)$. The same way, $y'(x+h)+y'(x-h)$ is twice the even part of (the Taylor expansion of) $y'(x+h)$. Thus \begin{alignat}1 y(x+h)-y(x-h)&=2\sum_j\frac{y^{(2j+1)}(x)h^{2j+1}}{(2j+1)!}&=2y'(x)h+2\frac{y'''(x)h^3}{3!}+2\frac{y^{(5)}(x)h^{5}}{5!}+...\\ \frac{h}3(y'(x+h)+4y'(x)+y'(x-h))&=\frac{2h}3\left(2y'(x)+\sum_j\frac{y^{(2j+1)}(x)h^{2j}}{(2j)!}\right)&=2y'(x)h+\frac23\frac{y'''(x)h^3}{2!}+\frac23\frac{y^{(5)}(x)h^{5}}{4!}+... \\[1.5em]\hline \text{in the difference }y(x+h)-y(x-h)&-\frac{h}3(y'(x+h)+4y'(x)+y'(x-h))&=-\frac43\frac{y^{(5)}(x)h^{5}}{5!}+... \end{alignat} so that in general, the truncation error is $O(h^5)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.