From 62cf45a5551e40ff35336a889f500ef227db009a Mon Sep 17 00:00:00 2001
From: Michael Wimmer <m.t.wimmer@tudelft.nl>
Date: Tue, 15 Sep 2020 21:56:27 +0200
Subject: [PATCH] move proof to info box
---
src/8_differential_equations_2.md | 80 ++++++++++++++++---------------
1 file changed, 42 insertions(+), 38 deletions(-)
diff --git a/src/8_differential_equations_2.md b/src/8_differential_equations_2.md
index 3db758c..a3e1fff 100644
--- a/src/8_differential_equations_2.md
+++ b/src/8_differential_equations_2.md
@@ -104,40 +104,44 @@ in which $P(\lambda)$ is the characteristic polynomial of the system matrix $A$,
$$P(\lambda) = \lambda^n + a_{n-1} \lambda^{n-1} + \cdots + a_0.$$
-As we demonstrate below, the proof relies on the co-factor expansion technique
-for calculating a determinant.
-$$- \det(A - \lambda I) = \begin{bmatrix}
-\lambda & -1 & 0 & \cdots & 0 \\
-0 & \lambda & -1 & \cdots & 0 \\
-\vdots & \vdots & \vdots & \cdots & \vdots \\
-a_0 & a_1 & a_2 & \cdots & a_{n-1} + \lambda \\
-\end{bmatrix} $$
-$$- \det(A - \lambda I) = \lambda \det \begin{bmatrix}
-\lambda & -1 & 0 & \cdots & 0 \\
-0 & \lambda & -1 & \cdots & 0 \\
-\vdots & \vdots & \vdots & \cdots & \vdots \\
-a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
-\end{bmatrix} + (-1)^{n+1}a_0 \det \begin{bmatrix}
--1 & 0 & 0 & \cdots & 0 \\
-\lambda & -1 & 0 & cdots & 0 \\
-\vdots & \vdots & \vdots & \cdots & \vdots \\
-0 & 0 & \cdots & \lambda & -1 \\
-\end{bmatrix}$$
-$$- \det(A - \lambda I) = \lambda \det \begin{bmatrix}
-\lambda & -1 & 0 & \cdots & 0 \\
-0 & \lambda & -1 & \cdots & 0 \\
-\vdots & \vdots & \vdots & \cdots & \vdots \\
-a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
-\end{bmatrix} + (-1)^{n+1} a_0 (-1)^{n-1}$$
-$$- \det(A - \lambda I) = \lambda \det \begin{bmatrix}
-\lambda & -1 & 0 & \cdots & 0 \\
-0 & \lambda & -1 & \cdots & 0 \\
-\vdots & \vdots & \vdots & \cdots & \vdots \\
-a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
-\end{bmatrix} + a_0$$
-$$- \det(A - \lambda I) = \lambda (\lambda (\lambda \cdots + a_2) + a_1) + a_0$$
-$$- \det(A - \lambda I) = P(\lambda).$$
+!!! info "Proof of $\det(A - \lambda I) = -P(\lambda)$"
+
+ As we demonstrate below, the proof relies on the co-factor expansion
+ technique for calculating a determinant.
+
+ $$- \det(A - \lambda I) = \begin{bmatrix}
+ \lambda & -1 & 0 & \cdots & 0 \\
+ 0 & \lambda & -1 & \cdots & 0 \\
+ \vdots & \vdots & \vdots & \cdots & \vdots \\
+ a_0 & a_1 & a_2 & \cdots & a_{n-1} + \lambda \\
+ \end{bmatrix} $$
+ $$- \det(A - \lambda I) = \lambda \det \begin{bmatrix}
+ \lambda & -1 & 0 & \cdots & 0 \\
+ 0 & \lambda & -1 & \cdots & 0 \\
+ \vdots & \vdots & \vdots & \cdots & \vdots \\
+ a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
+ \end{bmatrix} + (-1)^{n+1}a_0 \det \begin{bmatrix}
+ -1 & 0 & 0 & \cdots & 0 \\
+ \lambda & -1 & 0 & \cdots & 0 \\
+ \vdots & \vdots & \vdots & \cdots & \vdots \\
+ 0 & 0 & \cdots & \lambda & -1 \\
+ \end{bmatrix}$$
+ $$- \det(A - \lambda I) = \lambda \det \begin{bmatrix}
+ \lambda & -1 & 0 & \cdots & 0 \\
+ 0 & \lambda & -1 & \cdots & 0 \\
+ \vdots & \vdots & \vdots & \cdots & \vdots \\
+ a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
+ \end{bmatrix} + (-1)^{n+1} a_0 (-1)^{n-1}$$
+ $$- \det(A - \lambda I) = \lambda \det \begin{bmatrix}
+ \lambda & -1 & 0 & \cdots & 0 \\
+ 0 & \lambda & -1 & \cdots & 0 \\
+ \vdots & \vdots & \vdots & \cdots & \vdots \\
+ a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
+ \end{bmatrix} + a_0$$
+ $$- \det(A - \lambda I) = \lambda (\lambda (\lambda \cdots + a_2) + a_1)
+ + a_0$$
+ $$- \det(A - \lambda I) = P(\lambda).$$
In the second last line of the proof we indicated that the method of co-factor
expansion demonstrated is repeated an additional $n-2$ times. This completes the
@@ -434,7 +438,7 @@ This equation looks exactly like, and turns out to be, an eigenvalue equation!
Recall that a space of functions can be transformed into a Hilbert space by
equipping it with a inner product,
- $$\langle f, g \rangle = \int^{L}_{0} dx f*(x) g(x) $$
+ $$\langle f, g \rangle = \int^{L}_{0} dx f^*(x) g(x) $$
Use of this inner product also has utility in demonstrating that particular
operators are *Hermitian*. The term hermitian is precisely defined below.
@@ -455,16 +459,16 @@ Hilbert space). Denote this Hilbert space $\mathcal{H}_{0}$ and consider let
$f, \ g \ \epsilon \ \mathcal{H}_0$ denote two functions from the Hilbert space.
Then, we can investigate
-$$\langle f, L g \rangle = \frac{- \hbar^2}{2m} \int^{L}_{0} dx f*(x) \frac{d^2}{dx^2}g(x).$$
+$$\langle f, L g \rangle = \frac{- \hbar^2}{2m} \int^{L}_{0} dx f^*(x) \frac{d^2}{dx^2}g(x).$$
As a first step, it is possible to do integration by parts in the integral,
-$$\langle f, L g \rangle = \frac{+ \hbar^2}{2m} ( \int^{L}_{0} dx \frac{d f*}{dx} \frac{d g}{dx} - [f*(x)\frac{d g}{dx}] \big{|}^{L}_{0} )$$
+$$\langle f, L g \rangle = \frac{+ \hbar^2}{2m} ( \int^{L}_{0} dx \frac{d f^*}{dx} \frac{d g}{dx} - [f^*(x)\frac{d g}{dx}] \big{|}^{L}_{0} )$$
The boundary term vansishes, due to the boundary conditions $f(0)=f(L)=0$,
-which directly imply $f*(0)=f*(L)=0$. Now, intergrate by parts a second time
+which directly imply $f^*(0)=f^*(L)=0$. Now, intergrate by parts a second time
-$$\langle f, L g \rangle = \frac{- \hbar^2}{2m} (\int^{L}_{0} dx \frac{d^2 f*}{dx^2} g(x) - [\frac{d f*}{dx} g(x)] \big{|}^{L}_{0} ).$$
+$$\langle f, L g \rangle = \frac{- \hbar^2}{2m} (\int^{L}_{0} dx \frac{d^2 f^*}{dx^2} g(x) - [\frac{d f^*}{dx} g(x)] \big{|}^{L}_{0} ).$$
As before, the boundary term vanishes, due to the boundary conditions
$g(0)=g(L)=0$. Upon cancelling the boundary term however, the expression on
--
GitLab