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Exactly. Automatic differentiation works because we have the chain rule (f . g)'(x) = f'(g(x)) g'(x), telling us how to differentiate functions which are compo
by IngoBlechschmid 6y ago
Exactly.
Automatic differentiation works because we have the chain rule (f . g)'(x) = f'(g(x)) g'(x), telling us how to differentiate functions which are composed from simpler building blocks. But there is no analogous rule for integration.
- a1369209993 6y agoTechnically there is: D f = f' <=> S f' = f # inverse # chain rule D (f . g) x = D f (g x) * D g x # divide by D g x D (f . g) x / D g x = (D f) (g x) # S vs D inverse D (S f' . g) x / D g x = (f' . g) x # commute = and integrate both sides S (f' . g) x = S \x( D (S f' . g) x / D g x ) x # rename and drop argument S (f . g) = S \x( D (S f . g) x / D g x ) The problem is that the RHS is, in the general case, not any less complex (and often much more) than the LHS. Edit: Note that (S f . g) is integrating only f, then composing with g, which is what you want; it's the S \x(...) that causes problems.
- IngoBlechschmid 6y agoNice observation, thank you for sharing!