From 41831ab06b44f17bc70914af90eb7794896bf1a8 Mon Sep 17 00:00:00 2001
From: Michael Wimmer <m.t.wimmer@tudelft.nl>
Date: Sun, 13 Sep 2020 21:42:03 +0200
Subject: [PATCH] fix some math
---
src/8_differential_equations_2.md | 74 +++++++++++++++----------------
1 file changed, 37 insertions(+), 37 deletions(-)
diff --git a/src/8_differential_equations_2.md b/src/8_differential_equations_2.md
index 96e9eaf..a94a60d 100644
--- a/src/8_differential_equations_2.md
+++ b/src/8_differential_equations_2.md
@@ -3,7 +3,7 @@
In the previous lecture, we focused on first order linear differential equations
as well as systems of such equations. In this lecture we switch focus to DE's
which involve higher derivatives of the function we would like to solve for. To
-facilitate this change we are going to change notation. In the previous lecture
+f`%acilitate this change we are going to change notation. In the previous lecture
we wrote differential equations for $x(t)$. In this lecture we will write DE's
of $y(x)$, where $y$ is an unknown function and $x$ is the independent variable.
For this purpose we make the following definitions,
@@ -57,64 +57,64 @@ $$y(x) = c_1 f_1 (x) + c_2 f_2 (x) + \cdots + c_n f_{n}(x). $$
To check that the $n$ solutions form a basis, it is sufficient to verify
$$ det \begin{bmatrix}
-f_1(x) & \hdots & f_{n}(x) \\
-f_1 ' (x) & \hdots & f_{n}'(x) \\
+f_1(x) & \cdots & f_{n}(x) \\
+f_1 ' (x) & \cdots & f_{n}'(x) \\
\vdots & \vdots & \vdots \\
-f^{(n-1)}_{1} (x) & \hdots & f^{(n-1)}_{n} (x) \\
+f^{(n-1)}_{1} (x) & \cdots & f^{(n-1)}_{n} (x) \\
\end{bmatrix} \neq 0.$$
The determinant in the preceding line is called the *Wronski determinant*. In
particular, to determine solutions, we need to find the eigenvalues of
-$$**A** = \begin{bmatrix}
-0 & 1 & 0 & \hdots & 0 \\
-0 & 0 & 1 & \hdots & 0 \\
-\vdots & \vdots & \vdots & \hdots & \vdots \\
--a_0 & -a_1 & -a_2 & \hdots & -a_{n-1} \\
+$$A = \begin{bmatrix}
+0 & 1 & 0 & \cdots & 0 \\
+0 & 0 & 1 & \cdots & 0 \\
+\vdots & \vdots & \vdots & \cdots & \vdots \\
+-a_0 & -a_1 & -a_2 & \cdots & -a_{n-1} \\
\end{bmatrix}.$$
It is possible to show that
-$$det(**A** - \lambda \mathbbm{1}) = -P(\lambda),$$
+$$det(A - \lambda I) = -P(\lambda),$$
-in which $P(\lambda)$ is the characteristic polynomial of the system matrix $**A**$,
+in which $P(\lambda)$ is the characteristic polynomial of the system matrix $A$,
$$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 \mathbbm{1}) = \begin{bmatrix}
-\lambda & -1 & 0 & \hdots & 0 \\
-0 & \lambda & -1 & \hdots & 0 \\
-\vdots & \vdots & \vdots & \hdots & \vdots \\
-a_0 & a_1 & a_2 & \hdots & a_{n-1} + \lambda \\
+$$- det(A - \lambda I) = \begin{bmatrix}
+\lambda & -1 & 0 & \cdots & 0 \\
+0 & \lambda & -1 & \cdots & 0 \\
+\vdots & \vdots & \vdots & \cdots & \vdots \\
+a_0 & a_1 & a_2 & \cdots & a_{n-1} + \lambda \\
\end{bmatrix} $$
-$$- det(**A** - \lambda \mathbbm{1}) = \lambda det \begin{bmatrix}
-\lambda & -1 & 0 & \hdots & 0 \\
-0 & \lambda & -1 & \hdots & 0 \\
-\vdots & \vdots & \vdots & \hdots & \vdots \\
-a_1 & a_2 & a_3 & \hdots & a_{n-1} + \lambda \\
+$$- 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 & \hdots & 0 \\
-\lambda & -1 & 0 & hdots & 0 \\
-\vdots & \vdots & \vdots & \hdots & \vdots \\
-0 & 0 & \hdots & \lambda & -1 \\
+-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 \mathbbm{1}) = \lambda det \begin{bmatrix}
-\lambda & -1 & 0 & \hdots & 0 \\
-0 & \lambda & -1 & \hdots & 0 \\
-\vdots & \vdots & \vdots & \hdots & \vdots \\
-a_1 & a_2 & a_3 & \hdots & a_{n-1} + \lambda \\
+$$- det(A - \lambda I) = \lambda det \begin{bmatrix}
+\lambda & -1 & 0 & \cdots & 0 \\
+0 & \lambda & -1 & \cdots & 0 \\
+\vdots & \vdots & \vdots & \cdots & \vdots \\
+a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
\end{bmatrix} + (-1)^{n+1} a_0 (-1)^{n-1}$$
-$$- det(**A** - \lambda \mathbbm{1}) = \lambda det \begin{bmatrix}
-\lambda & -1 & 0 & \hdots & 0 \\
-0 & \lambda & -1 & \hdots & 0 \\
-\vdots & \vdots & \vdots & \hdots & \vdots \\
-a_1 & a_2 & a_3 & \hdots & a_{n-1} + \lambda \\
+$$- 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 \mathbbm{1}) = \lambda (\lambda (\lambda \cdots + a_2) + a_1) + a_0$$
-$$- det(**A** - \lambda \mathbbm{1}) = P(\lambda).$$
+$$- 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
--
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