From bd13537ee2fb704b02b961b5d06dd4f406f19a71 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 21 Oct 2023 14:21:59 +0200
Subject: Improve knowledge base

---
 source/know/concept/legendre-transform/index.md | 31 ++++++++++++++-----------
 1 file changed, 17 insertions(+), 14 deletions(-)

(limited to 'source/know/concept/legendre-transform/index.md')

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)}
-- 
cgit v1.2.3