Legendre Transformation of f(x) = x^3

[RJLiberator]

Homework Statement

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

Homework Equations

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

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 f(x) = \frac{1}{2}e^{2x}

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

g=f(x)-xm = x^3-x(3x^2)

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 x = {\sqrt{\frac{m(x)}{3}}}

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

And that is the Legendre transformation of x^3.

Note* First time ever being exposed to Legendre Transformation.

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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

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 f(x) = \frac{1}{2}e^{2x}

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

g=f(x)-xm = x^3-x(3x^2)

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 x = {\sqrt{\frac{m(x)}{3}}}

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

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.

 

[RJLiberator]

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:

g(m) = mx-f(x)= x(3x^2)-x^3
And so:
g(m) = {\sqrt{\frac{m(x)}{3}}}m-({\sqrt{\frac{m(x)}{3}}})^3

 

[Ray Vickson]

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:

g(m) = mx-f(x)= x(3x^2)-x^3
And so:
g(m) = {\sqrt{\frac{m(x)}{3}}}m-({\sqrt{\frac{m(x)}{3}}})^3

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.

 

[RJLiberator]

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