       Re: Legendre transform

• To: mathgroup at smc.vnet.net
• Subject: [mg64315] Re: [mg64312] Legendre transform
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Sat, 11 Feb 2006 03:32:45 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```Off[Solve::ifun];
Off[InverseFunction::ifun];

Clear[legendreTransform];

legendreTransform[f_, x_Symbol, y_Symbol]:=
Module[{d=D[f, x], transform},
transform/;
(transform=(-f+x*d)/.Solve[y==d, x]//Last;
FreeQ[transform,Solve]&&
FreeQ[transform,InverseFunction])]/;!(x===y);

legendreTransform[Exp[x], x, y]

y*Log[y] - y

legendreTransform[%, y, x]

E^x

Bob Hanlon

>
> From: Tobin Fricke <fricke at ocf.berkeley.edu>
To: mathgroup at smc.vnet.net
> Subject: [mg64315] [mg64312] Legendre transform
>
> As an exercise in learning Mathematica, and also as a computational and
> pedagogical tool for myself, I'm attempting to implement a function to
> compute the Legedre transform of an expression.
>
> The procedure is, given a function f(x):
>
>    1. compute y(x) = f'(x) and solve for x(y)
>    2. the legendre transform is g(y) = -f(x(y)) + x(y) y
>
> Here's my first attempt:
>
>    legendreTransform[f_, x_, y_] := Block[{},
>        solutions = Solve[y == D[f, x], x];
>        x[k_] := x /. First[solutions];
>        -(f /. x -> x[y]) + x[y] y]
>
> For instance, it correctly gives the transform of xLog[x]-x as e^x:
>
>    Simplify[legendreTransform[x Log[x] - x, x, y],
>             Assumptions -> {Element[y, Reals]}]
>
> This seems to work, though (1) things go horribly wrong if x==y, and (2)
> errors aren't handled at all--for instance, it should detect whether the
> Solve[] fails, etc.
>
> Another attempt is written in a such a way as to operate on *functions*
> rather than expressions.  To me this seems like the preferred way, but
> maybe it isn't:
>
>    legendreTransform[f_] = Block[{x, y},
>        x[y_] = x /. First[Solve[y == f'[x], x]];
>        Function[y, -f[x[y]] + x[y] y]]
>
> Any hints/critiques/etc appreciated.  I'd like to end up with a version
> that will let me take the legendreTransform with respect to an arbitrary
> argument of a function with arbitrarily many arguments.
>
> thanks,
> Tobin
>
>  http://en.wikipedia.org/wiki/
Legendre_transform#Legendre_transformation_in_one_dimension
>
>

```

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