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by aria
4210 days ago
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In practice you only need to do $H^{-1} g$; L-BFGS stores the {s_k} and {y_k} vectors which allow you to do the $H^{-1} g$ directly rather than needing to ever form the hessian or its inverse. There are techniques that aren't BFGS-based which approximate the hessian rather than the inverse and in that case you'd be better off solving. |
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I know this is irrelevant to the main part of the post, which is to explain LBFGS, I'm just genuinely interested if there are applications where its better to pass the inverse of H rather than H itself for some reason.