3 ms·
Upvoted for the cute proof- without-words geometrical diagram of the Legendre transform, but the fact that you defined the inverse map as (x, y) to (\hat y, \ha
by crdrost 1y ago
Upvoted for the cute proof- without-words geometrical diagram of the Legendre transform, but the fact that you defined the inverse map as (x, y) to (\hat y, \hat x) I found impossible to keep my head straight. Probably it's easy if I slow down and stop skimming the article.
IMO the easier derivation — may just be personal tastes as someone more on the engineering side — is just another integration by parts.
So with f(g(x)) = g(f(x)) = x, define y = g(x) for U-substitution with x = f(y):
∫ g(x) dx = ∫ g(f(y)) f'(y) dy
= ∫ y f'(y) dy
= y f(y) – ∫ f(y) dy
= g(x) x – F(g(x)) + C
The more interesting thing is that this is a really basic integration by parts which means that this diagram of yours that I like, is more universal than it appears at first? I'd have to think about that a bit more, how you can maybe graphically teach integration by parts that way, is there always a u substitution so that you can get u f(u) or so and get this nice pretty rectangle in a rectangle... hmm.
- messe 1y agoYes, there's a similar diagram on the Wikipedia page for Integration by Parts.
- gowld 1y agoAnd it links to the more specific page on Integral of inverse functions: https://en.wikipedia.org/wiki/Integral_of_inverse_functions https://en.wikipedia.org/wiki/Integral_of_inverse_functions
- 3abiton 1y agoI also got confused until I read your feedback.