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 · 41m ago
Yes, there's a similar diagram on the Wikipedia page for Integration by Parts.
gitroom · 40m ago
haha this stuff twists my brain up but i love the visual angle. makes me wonder if breaking things down more like this really changes the way i remember them long-term. stuff like diagrams actually help or am i just tricking myself?
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):
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.