From 56b740565fd6467564be28b5465ad24cccf71109 Mon Sep 17 00:00:00 2001 From: Holden Rohrer Date: Thu, 16 Jan 2020 21:24:46 -0500 Subject: moved comb hw --- tech-math/ws5.tex | 28 ---------------------------- 1 file changed, 28 deletions(-) delete mode 100644 tech-math/ws5.tex (limited to 'tech-math/ws5.tex') diff --git a/tech-math/ws5.tex b/tech-math/ws5.tex deleted file mode 100644 index ff21fa2..0000000 --- a/tech-math/ws5.tex +++ /dev/null @@ -1,28 +0,0 @@ -\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