3 ms·
Doesn't continuous time basically mean "this is what we expect for sufficiently small time steps"? Very similar to how one would for example take the first orde
by phreeza 6mo ago
Doesn't continuous time basically mean "this is what we expect for sufficiently small time steps"? Very similar to how one would for example take the first order Taylor dynamics and use them for "sufficiently small" perturbations from equilibrium. Is there any other magic to continuous time systems that one would not expect to be solved by sufficiently small time steps?
- measurablefunc 6mo agoYou should look into condition numbers & how that applies to numerical stability of discretized optimization. If you take a continuous formulation & naively discretize you might get lucky & get a convergent & stable implementation but more often than not you will end up w/ subtle bugs & instabilities for ill-conditioned initial conditions.
- phreeza 6mo agoI understand that much, but it seems like "your naive timestep may need to be smaller than you think or you need to do some extra work" rather than the more fundamental objection from OP?
- measurablefunc 6mo agoThe translation from continuous to discrete is not automatic. There is a missing verification in the linked analysis. The mapping must be verified for stability for the proper class of initial/boundary conditions. Increasing the resolution from 64 bit floats to 128 bit floats doesn't automatically give you a stable discretized optimizer from a continuous formulation.
- phyalow 6mo agoOr you can just try stuff and see if it works
- measurablefunc 6mo agoPoint still stands, translation from continuous to discrete is not as simple as people think.
- phreeza 6mo agoNumerical issues totally exist but the reason has nothing to do with the fact that Cauchy sequences don't exist on a computer imo.
- measurablefunc 6mo agoThe abstract formulation is different from the concrete implementation. It is precisely b/c the abstractions do not exist on computers that the abstract analysis does not automatically transfer the necessary analytical properties to the digital implementation. Cauchy sequences & Dedekind cuts are abstract & do not exist on digital computers.
- heyethan 6mo ago[flagged]
- tsimionescu 6mo agoInfinity has properties that finite approximations of it just don't have, and this can lead to serious problems for certain theorems. In the general case, the integral of a continuous function can be arbitrarily different from the sum of a finite sequence of points sampled from that function, regardless of how many points you sample - and it's even possible that the discrete version is divergent even if the continous one is convergent. I'm not saying that this is the case here, but there generally needs to be some justification to say that a certain result that is proven for a continuous function also holds for some discrete version of it. For a somewhat famous real-world example, it's not currently known how to produce a version of QM/QFT that works with discrete spacetime coordinates, the attempted discretizations fail to maintain the properties of the continuous equations.