Talk:Geometric integrator

Constant acceleration

This page seems a little misleading. The reason that the midpoint rule gives an accurate answer is that the acceleration is constant. The midpoint is actually the analytical solution when the acceleration is constant. However, this is not generally applicable to systems where the acceleration is not constant. Mattopia 11:37, 19 October 2006 (UTC)

What makes you think that the acceleration is constant? Isn't acceleration = q'' = −sin q. -- Jitse Niesen (talk) 12:30, 19 October 2006 (UTC)

OK, I see my mistake. The article states that the acceleration due to gravity is g=1, but actually the acceleration itself is not constant. (OK, so its a simple pendulum and it should be obvious...I guess I didn't look hard enough). Mattopia 16:05, 19 October 2006 (UTC)

Exact trajectories are missing.

The text says:

>We plot, in $\mathbb {R} ^{2}$, the exact trajectories and the numerical solutions of the system.

, but I don't see the exact trajectories on the picture. 89.135.16.141 (talk) 05:07, 10 July 2014 (UTC)

I think the thin lines (closed curves around the equilibria) are the exact trajectories. The two thick lines are the numerical solutions. -- Jitse Niesen (talk) 13:50, 10 July 2014 (UTC)

Oh, yes, this is evident (already), thank you. Now I've also discovered, that there is a bigger version of this pictore too here: [[1]] 86.101.236.13 (talk) 16:09, 10 July 2014 (UTC)

Assessment comment

The comment(s) below were originally left at Talk:Geometric integrator/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Article only talks about symplectic integrators, but geometric integration is more: preservation of symmetries, reversibility, first integrals, volume, etc. -- Jitse Niesen (talk) 13:34, 2 August 2007 (UTC)

Last edited at 22:56, 7 August 2007 (UTC). Substituted at 02:09, 5 May 2016 (UTC)