Legendre Transformation of f(x) = x^3

In summary: Thanks for taking the time to look at this. I am going to go through the example in the book and see if there is anything I can understand.
  • #1
RJLiberator
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Homework Statement


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

Homework Equations


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

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.
 
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  • #2
RJLiberator said:

Homework Statement


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

Homework Equations


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

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.

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.
 
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  • #3
I am following an example laid out by the book, where am I going wrong? Is it due to the concave vs. convex nature?

and you need another variable p as the argument of the transformed function.

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]
 
  • #4
RJLiberator said:
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]

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.
 
  • #5
I will continue to look into the problem and with your added information I'll try to make sense of it all.
apparently your un-named book does it differently.
Statistical and Thermal Physics:
With Computer Applications
Harvey Gould & Jan Tobochnik

 

Related to Legendre Transformation of f(x) = x^3

What is the Legendre transformation?

The Legendre transformation is a mathematical operation that transforms a function of one variable into a different function of the same variable. It is used to find the conjugate function of a given function.

How is the Legendre transformation calculated?

The Legendre transformation of a function f(x) is calculated by taking the derivative of f(x) with respect to x and setting it equal to the variable y. The resulting equation, y = f'(x), is then solved for x in terms of y to get the transformed function.

Why is the Legendre transformation important?

The Legendre transformation is important because it allows us to convert between different representations of a function. It is used in thermodynamics, physics, and other fields to simplify and solve complex problems.

What is the physical interpretation of the Legendre transformation?

The physical interpretation of the Legendre transformation is that it represents a change in the coordinate system. The original function f(x) is represented in terms of the variable x, while the transformed function is represented in terms of its conjugate variable y.

Can the Legendre transformation be applied to any function?

No, the Legendre transformation can only be applied to convex functions, which have a positive second derivative and a curvature that is always increasing. This is because the transformation relies on the convexity of the function to produce a unique solution.

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