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-rw-r--r--source/know/concept/legendre-transform/index.md31
1 files changed, 17 insertions, 14 deletions
diff --git a/source/know/concept/legendre-transform/index.md b/source/know/concept/legendre-transform/index.md
index c4fdeb4..d09613f 100644
--- a/source/know/concept/legendre-transform/index.md
+++ b/source/know/concept/legendre-transform/index.md
@@ -10,48 +10,49 @@ layout: "concept"
The **Legendre transform** of a function $$f(x)$$ is a new function $$L(f')$$,
which depends only on the derivative $$f'(x)$$ of $$f(x)$$,
-and from which the original function $$f(x)$$ can be reconstructed.
+and from which the original $$f(x)$$ can be reconstructed.
The point is that $$L(f')$$ contains the same information as $$f(x)$$,
just in a different form,
analogously to e.g. the [Fourier transform](/know/concept/fourier-transform/).
Let us choose an arbitrary point $$x_0 \in [a, b]$$ in the domain of $$f(x)$$.
Consider a line $$y(x)$$ tangent to $$f(x)$$ at $$x = x_0$$,
-which must have slope $$f'(x_0)$$, and intersects the $$y$$-axis at $$-C$$:
+which has slope $$f'(x_0)$$ and intersects the $$y$$-axis at $$y = -C$$:
$$\begin{aligned}
y(x)
- = f'(x_0) (x - x_0) + f(x_0)
- = f'(x_0) \: x - C
+ &= f'(x_0) (x - x_0) + f(x_0)
+ \\
+ &= f'(x_0) \: x - C
\end{aligned}$$
Where $$C \equiv f'(x_0) \: x_0 - f(x_0)$$.
-We now define the Legendre transform $$L(f')$$ such that
-for all $$x_0 \in [a, b]$$ we have $$L(f'(x_0)) = C$$
+We now define the *Legendre transform* $$L(f')$$,
+such that for all $$x_0 \in [a, b]$$ we have $$L(f'(x_0)) = C$$
(some authors use $$-C$$ instead).
Renaming $$x_0$$ to $$x$$:
$$\begin{aligned}
L(f'(x))
- = f'(x) \: x - f(x)
+ &= f'(x) \: x - f(x)
\end{aligned}$$
We want this function to depend only on the derivative $$f'$$,
but currently $$x$$ still appears here as a variable.
-We fix this problem in the easiest possible way:
+We solve this problem in the easiest possible way:
by assuming that $$f'(x)$$ is invertible for all $$x \in [a, b]$$.
If $$x(f')$$ is the inverse of $$f'(x)$$, then $$L(f')$$ is given by:
$$\begin{aligned}
\boxed{
L(f')
- = f' \: x(f') - f(x(f'))
+ = f' \: x(f') - f\big(x(f')\big)
}
\end{aligned}$$
The only requirement for the existence of the Legendre transform is thus
the invertibility of $$f'(x)$$ in the target interval $$[a,b]$$,
-which can only be true if $$f(x)$$ is either convex or concave,
+which is only satisfied if $$f(x)$$ is either convex or concave,
meaning its derivative $$f'(x)$$ is monotonic.
The derivative of $$L(f')$$ with respect to $$f'$$ is simply $$x(f')$$.
@@ -72,9 +73,11 @@ To show this, let $$g(L')$$ be the Legendre transform of $$L(f')$$:
$$\begin{aligned}
g(L')
- = L' \: f'(L') - L(f'(L'))
- = x(f') \: f' - f' \: x(f') + f(x(f'))
- = f(x)
+ &= L' \: f'(L') - L(f'(L'))
+ \\
+ &= x(f') \: f' - f' \: x(f') + f(x(f'))
+ \\
+ &= f(x)
\end{aligned}$$
Moreover, a Legendre transform is always invertible,
@@ -84,7 +87,7 @@ so a proof is:
$$\begin{aligned}
L''(f')
- = \dv{}{f'} \Big( \dv{L}{f'} \Big)
+ &= \dv{}{f'} \Big( \dv{L}{f'} \Big)
= \dv{x(f')}{f'}
= \dv{x}{f'(x)}
= \frac{1}{f''(x)}