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Legendre Transformation of f(x) = x^3

  1. Aug 24, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data

    Find the Legendre Transformation of [tex]f(x)=x^3[/tex]

    2. Relevant equations
    [tex]m(x) = f'(x) = 3x^2[/tex]
    [tex] x = {\sqrt{\frac{m(x)}{3}}}[/tex]
    [tex]g = f(x)-xm[/tex]

    3. The attempt at a solution

    I am reading a quick description of the Legendre Transformation in my required text and it has the example giving for the function [tex]f(x) = \frac{1}{2}e^{2x} [/tex]

    Thus, I am trying to follow through with it as follows:

    [tex] g=f(x)-xm = x^3-x(3x^2)[/tex]

    The problem for me exists in the next step. For them, they had an easy simplification for their example problem. In my problem, m does not equal x^3, m is equal to 3x^2.

    Can I do this to solve the transformation:

    Since [tex] x = {\sqrt{\frac{m(x)}{3}}}[/tex]

    We have
    [tex]({\sqrt{\frac{m(x)}{3}}})^3- {\sqrt{\frac{m(x)}{3}}}m = g(m)[/tex]

    And that is the Legendre transformation of x^3.

    Note* First time ever being exposed to Legendre Transformation.
     
  2. jcsd
  3. Aug 24, 2016 #2

    Ray Vickson

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    The Legendre Transform ##f^*(p)## of the convex function ##f(x)## with domain ##x \geq 0## is
    $$f^*(p) = \max_{x \geq 0} [p x - f(x)] $$
    (or with ##\sup## replacing ##\max## if necessary..

    Note that you SUBTRACT ##f(x)##, not add it, and you need another variable ##p## as the argument of the transformed function.

    Everything would fall apart if you tried to maximize ##f(x) - px##; that is, if you added ##f## instead of subtracting it, the whole theory would fail. There are good reasons for that, but never mind them for now.
     
  4. Aug 24, 2016 #3

    RJLiberator

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    I am following an example laid out by the book, where am I going wrong? Is it due to the concave vs. convex nature?

    Would my "m" variable be equal to your "p" variable?

    So, if I follow the path of mx-f(x), I get:

    [tex] g(m) = mx-f(x)= x(3x^2)-x^3 [/tex]
    And so:
    [tex] g(m) = {\sqrt{\frac{m(x)}{3}}}m-({\sqrt{\frac{m(x)}{3}}})^3[/tex]
     
  5. Aug 24, 2016 #4

    Ray Vickson

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    Yes, my "p" is your "m".

    Some people use ##f(x) - mx## and minimize it instead of maximizing, but the max form is pretty standard. I think it is often done differently in Physics than in Mathematics, and apparently your un-named book does it differently.

    I cannot make any sense of what you did; your formula for the Lagrange-transformed function ##g(m)## should not have an "##x##" in it, and the final answer should be negative.
     
  6. Aug 24, 2016 #5

    RJLiberator

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    I will continue to look into the problem and with your added information I'll try to make sense of it all.


    Statistical and Thermal Physics:
    With Computer Applications
    Harvey Gould & Jan Tobochnik

     
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