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I'm trying to find the legendre transformation of

$$ f(x)=x^3 $$

I have calculated it using the approach we learned in class:

1 - Find the derivative of function => $y(x) = f'(x)$

2 - Take the inverse of the derivative => $x(y)$

3 - (I) Integrate this inverse to find the legendre transformation

3 - or (II) use $g(y)=-f(x(y))+x(y)y$

I did this and my Legendre transformation turned out to be

$$ g(y)=\frac{2}{3}y\sqrt{\frac{y}{3}} $$

Which I think is correct. However, the next assignment is then

Check the relation $f''g''=1$.

Which isn't true:

$f''=6x$ and $g''=\frac{1}{2\sqrt{3y}}$

I don't understand why this should work. I could not find any information online about a relation between the second derivatives of a function with it's legendre transform.

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HINT: Write $g''$ in terms of $x$, or $f''$ in terms of $y$.

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  • $\begingroup$ Just wanted to add, in general it's a good idea to write out things like $f''$ explicitly if you're running into a wrong answer, since if you'd written it out as $\frac{df^2}{dx^2} \frac{dg^2}{dx^2} = 1$ I think you would've seen the error. $\endgroup$ – CuriousHegemon Nov 19 '18 at 18:17

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