From cf201409a092b037371d09b4e09a16a0d7974898 Mon Sep 17 00:00:00 2001
From: Holden Rohrer <holden.rohrer@gmail.com>
Date: Wed, 20 Nov 2019 07:23:06 -0500
Subject: update

---
 tech-math/ws5.tex | 28 ++++++++++++++++++++++++++++
 1 file changed, 28 insertions(+)
 create mode 100644 tech-math/ws5.tex

(limited to 'tech-math/ws5.tex')

diff --git a/tech-math/ws5.tex b/tech-math/ws5.tex
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+++ b/tech-math/ws5.tex
@@ -0,0 +1,28 @@
+\def\pre#1{\leavevmode\llap{\hbox to \parindent{\hfil #1 \hfil}}}
+\noindent {\bf Q1} %#1
+
+\pre{a.} $(A^2-A-2)r(n) = 0$
+
+\pre{b.} $(A^4-3A^3+A^2+A)r(n) = 0$
+
+\pre{c.} $(A^3-5A+1)g(n) = 3^n$
+
+\pre{d.} $(A^3-A^2+2A-1)h(n-3) = 0$
+
+\pre{e.} $(A^5 - 4A^4 - A^2 - 3)r(n-5) = (-1)^n$
+
+\pre{f.} $(A^2 - A - 1)b(n-2) = 2^{n+1} - n^2$
+
+\noindent {\bf Q2)} %#3
+
+Equivalent to $$(A^2-3A+2)g(n) = 0.$$ $$A^2-3A+2=(A-1)(A-2).$$ Thus, $g(n) = c_1 + 2^n$.
+
+\noindent {\bf Q3)} %#5
+
+The Fibonacci formula is $a_{n+2} = a_{n+1} + a_n \to (A^2-A-1)a(n) = 0$. This has roots $1 + \sqrt{5} \over 2$ and $1 - \sqrt{5} \over 2$. Thus, its solution set is $$a(n) = c_1\left({1+\sqrt{5} \over 2}\right)^n + c_2\left({1+\sqrt{5} \over 2}\right)^n.$$
+
+\noindent {\bf Q4)} %#7
+
+This yields the solution set, by previously mentioned methods, $$f(n) = c_1 5^n + c_2 (-2)^n.$$ For this set, $$f(0)=2= c_1 + c_2$$ and $$f(1) = 10 = 5c_1 - 2c_2.$$ Solving as a system of equations, this gives $c_1 = 2$ and $c_2 = 0$. The solution is, therefore, $$f(n) = 2(5)^n$$.
+
+\bye
-- 
cgit 

