# Legendre Transformation of f(x) = x^3

1. Aug 24, 2016

### RJLiberator

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

Find the Legendre Transformation of $$f(x)=x^3$$

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

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

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.

2. Aug 24, 2016

### Ray Vickson

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.

3. Aug 24, 2016

### 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?

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:
$$g(m) = {\sqrt{\frac{m(x)}{3}}}m-({\sqrt{\frac{m(x)}{3}}})^3$$

4. Aug 24, 2016

### Ray Vickson

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. Aug 24, 2016

### RJLiberator

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