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

• RJLiberator
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.
RJLiberator
Gold Member

## 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.

RJLiberator said:

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

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:

$$g(m) = mx-f(x)= x(3x^2)-x^3$$
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.

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

## 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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