# Finding Legendre Transform of f(x) = (|x|+1)2

• Niles
In summary, the conversation is discussing the Legendre transform of the function f(x) = (|x|+1)^2. The process of finding the Legendre transform is explained and the solutions for the different intervals of x are given. The validity of these solutions for the different intervals of p is also questioned.
Niles
Hi guys

I am looking at f(x) = (|x|+1)2. I write this as

$$f(x) = \left\{ {\begin{array}{*{20}c} {x^2 + 1 + 2x\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,x > 0} \\ {x^2 + 1 - 2x\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,x < 0} \\ {1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,x = 0} \\ \end{array}} \right.$$

I want to find the Legendre transform of this function. For x>0 I get the Legendre transform

f*(p) = p2/4-p-1/2.

I am a little unsure of how this works. Because I need to find the Legendre transform of f for x<0 and x=0. But how do these solutions get "patched" together?

I don't get that at all. For x> 0, f is differentiable so f*(p) is just -f(x) where x is such that f'(x)= p. Here, f'(x)= 2x+ 2= p when x= (p-2)/2. So f*(p)= -f((p-2)/2)= -((p-2)2/4+ 1+ 2(p-1)/2)= -((1/4)p2- p+ 1+ 1+ p- 2)= -(1/4)p^2.

Similarly, for x< 0, f'(x)= 2x- 2= p when x= (p+2)/2. So F*(p)= -f((p+2)/2)= -)(p+2)2/4+ 1+ 2(p+2)/2)= -((1/4)p2+ p+ 1+ 1+ p+ 2)= -(1/4)p2- 2p- 4.

f is not differentiable at x= 0 so have to go back to the basic definition: f*(0)= max(0(x)-f(x))= max -f(x). Of course, that maximum is 0: f*(0)= 0. Since f is not differentiable at 0, it Legendre transform is not continuous there and they cannot be "patched" together.

Thanks. Ok, I've done it a slight different way. First, I must admit that I have not heard of a theorem stating that if f is differentiable, then f*(p) is just -f(x).

I used that since f is strictly convex (we look at x>0 for now), then f*(p)=xp-f(x(p)), where x is such that f'(x)= p.But OK: Let us look at the solutions found in #2 (I am mostly trying to learn the method). So

for x>0: f*(p) = -(1/4)p^2
for x<0: f*(p) = -(1/4)p^2- 2p- 4
x=0 : f*(p) = 0 (by the way, shouldn't this be -1 rather than 0?)What solutions are "valid" for the different intervals of p? I mean, we have 3 solutions, one for each interval of x. But how do I find out what intervals of p they correspond to?

Last edited:

## 1. What is a Legendre transform?

A Legendre transform is a mathematical operation that transforms a function of one variable into a related function of a different variable. It is commonly used in physics and mathematics to simplify calculations and solve problems in a different way.

## 2. How do you find the Legendre transform of a function?

To find the Legendre transform of a function, you first need to take the derivative of the function with respect to its variable. Then, use the inverse function to solve for the variable in terms of the derivative. This resulting function is the Legendre transform of the original function.

## 3. What is the purpose of finding the Legendre transform of a function?

The Legendre transform is useful for simplifying complicated functions and solving optimization problems. It also helps to transform a function into a different form that may be easier to work with mathematically.

## 4. How do you find the Legendre transform of f(x) = (|x|+1)2?

To find the Legendre transform of f(x) = (|x|+1)2, we first take the derivative with respect to x, which is 2(|x|+1). Then, we use the inverse function to solve for x in terms of the derivative, which is x = ±√(y/2 - 1). Therefore, the Legendre transform of f(x) is g(y) = ±√(y/2 - 1) - (y/2 - 1).

## 5. What are some common applications of the Legendre transform?

The Legendre transform has many applications in physics, mathematics, and engineering. It is commonly used in thermodynamics to transform between different energy and entropy variables. It is also used in optimization problems, statistical mechanics, and quantum field theory.

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