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You seem really confused about something. The article is talking about SDFs that do not have that property, which it is then trying to force to have that property via a hack of sorts (f/|del f| apparently), because once the property holds it is very useful for making the algorithm for more efficient.

The arbitrary scaling is not arbitrary if the whole algorithm is defined around K=1. The Lipschitz property in analysis is "bounded by any K". The algorithm is specifically using "bounded by K=1".

Also: for the purpose of numerical methods, it does not matter if a function "has a derivative" in a classical sense or not. Certainly distributions, step functions, etc are fine. Numerical methods cannot, after all, tell if a function is discontinuous or not. The article is using some known closed-form functions as examples, but it is trying to illustrate the general case, which is not going to be a closed form but just an arbitrary SDF that you constructed from somewhere---which it will be very useful to enforce the K=1 Lipschitz condition on.



I think there are probably different populations commenting here with different backgrounds, and consequently different assumptions and different blind spots. For example, for Newton's Method, it does very much matter whether the function has a derivative in the classical sense everywhere in the relevant interval; the quadratic convergence proof depends on it. Discontinuous functions can break even binary chop.


This is about differentiability.

>which it is then trying to force to have that property via a hack of sorts (f/|del f| apparently)

The functions he talks about do not have a derivative. So that hack is nonsensical.


Max, min, and the L1 norm? Those have derivatives (gradients) almost everywhere. The derivatives are undefined at a set of points of measure zero, and the functions are continuous across those gaps.


And that means, by definition, that they do not have a derivative.

If you don't believe me open a real analysis textbook.


You seem to be under the mistaken and misguided impression that the only kind of derivative is the one in a real analysis textbook.

I bet you think delta functions aren't real either, lol.


Are Dirac deltas hyperreal? Asking for a friend.


Evidently they do? Maybe there is some domain knowledge you're missing here as you belligerent protest that it's nonsense.




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