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chap1.html
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<html>
<head>
<script src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML" type="text/javascript">
</script>
</head>
<body>
<h2>
Exercise 1.2
</h2>
<h3>
Recurrent Relation:
<h3/>
<p>
\(A(0)=0 \\
A(n)=3A(n-1)+2, \text{for }n\ge0\)
</p>
<h3>
Closed Form Hypothesis:
</h3>
<p>
\(A(n)=3^n-1\)
</p>
<h3>
Inductive Proof:
</h3>
<p>
\(A(0)=3^0-1=0 \\
A(n)=3(3^{n-1}-1)+2=3^n-1\)
</p>
<h2>
Exercise 1.2 Alternative - Generating Functions
</h2>
<p>
\(\text{Recurrent relation} \\
A_0=0 \\
A_{n+1}=3A_n+2, \text{for }n\ge0 \\
\text{Define the generating function} \\
G(x)=\sum_{n\ge0}A_nx^n \\
\text{Express }A_{n+1}\text{ in terms of }G(x) \\
\sum_{n\ge0}A_{n+1}x^n=A_1+A_2x+a_3x^2+...=\{A_1x+A_2x^2+a_3x^3+...\}/x \\
\sum_{n\ge0}A_{n+1}x^n=G(x)/x \\
\text{Express }3A_{n}+2\text{ in terms of }G(x) \\
\sum_{n\ge0}(3A_{n}+2)x^n=3(\sum_{n\ge0}A_{n}x^n)+(\sum_{n\ge0}2x^n)\\
\sum_{n\ge0}(3A_{n}+2)x^n=3G(x)+\frac2{1-x},\text{ for }x<1 \\
\text{Solve }G(x) \\
G(x)/x=3G(x)+\frac2{1-x},\text{ for }x<1 \\
G(x)-3xG(x)=\frac{2x}{1-x}=(1-3x)G(x) \\
G(x)=\frac{2x}{(1-x)(1-3x)} \\
\text{Partial Fractions Expansion} \\
\frac{2x}{(1-x)(1-3x)}=\frac{y}{1-x}+\frac{z}{1-3x} \\
2x=(1-3x)y+(1-x)z\\
(1-3(\frac13))y+(1-\frac13)z=\frac23z=\frac23 \\
2x=(1-3x)y+(1-x)\\
y=\frac{-1+3x}{1-3x}=-1 \\
G(x)=\frac{1}{1-3x}-\frac{1}{1-x},\text{ for }x<1 \\
\text{Expand }G(x)\text{ into a power series} \\
G(x)=\sum_{n\ge0}(3x)^n - \sum_{n\ge0}x^n=
\sum_{n\ge0}3^nx^n - \sum_{n\ge0}x^n \\
[x^n]G(x)=A_n=3^n-1
\)
</p>
<h2>
Exercise 1.6
</h2>
<h3>
Recurrent Relation:
</h3>
<p>
\(B(2)=0 \\
B(n)=B(n-1)+n-2,\text{ for }n>2\)
</p>
<h3>
Unroll The Recurrence:
</h3>
<p>
\(B(n)=B(n-2)+(n-1-2)+(n-2) \\
B(n)=0+(3-2)+...+(n-1-2)+(n-2) \\
B(n)=0+(n-2)+...+(4-2)+(3-2) \\
2B(n)=(n+3-4)(n-2)\)
</p>
<h3>
Closed Form Hypothesis:
</h3>
<p>
\(B(n)=\frac{(n-1)(n-2)}2,\text{ for }n\ge2\)
</p>
<h3>
Inductive Proof:
</h3>
<p>
\(\frac{(2-1)(2-2)}2=0 \\
\frac{(n-1)(n-2)}2=\frac{(n-2)(n-3)}2+n-2 \\
\frac{n^2-3n+2}2=\frac{n^2-3n+2}2\)
</p>
<h2>
Exercise 1.8
</h2>
<p>
\(n=5m+l,\text{ where }0\le l\le4\text{ and } m \in \Bbb N \\
Q(5m+l)=P(l) \\
P(0)=\alpha \\
P(1)=\beta \\
P(2)=\frac{1+\beta}\alpha \\
P(3)=\frac{1+\beta+\alpha}{\alpha\beta} \\
P(4)=\frac{1+\alpha}\beta\)
</p>
<h2>
Exercise 1.11a
</h2>
<h3>
Recurrent Relation:
<h3/>
<p>
\(D(0)=0 \\
D(2n)=2D(2n-2)+2, \text{for }n\ge0\)
</p>
<h3>
Closed Form Hypothesis:
</h3>
<p>
\(D(2n)=2^{n+1}-2\)
</p>
<h3>
Inductive Proof:
</h3>
<p>
\(D(2(0))=2^{0+1}-2=0 \\
s = 2n-2 \\
D(s) = 2^{(s/2)+1}-2 \\
D(2n)=2D(s)+2=2^{(s/2)+2}-4+2=2^{n+1}-2\)
</p>
<h2>
Exercise 13
</h2>
<h3>
Recurrent Relation:
</h3>
<p>
\(Z(0)=1 \\
Z(n)=Z(n-1)+9n-8,\text{ for }n>0\)
</p>
<h3>
Unroll The Recurrence:
</h3>
<p>
\(Z(n)=Z(n-2)+(9(n-1)-8)+(9n-8) \\
Z(n)=1+(9(1)-8)+(9(2)-8)+...+(9n-8) \\
Z(n)=1+(9n-8)+(9(n-1)-8)+...+(9-8) \\
2Z(n)=2+n(9(n+1)-16)\)
</p>
<h3>
Closed Form Hypothesis:
</h3>
<p>
\(Z(n)=\frac{n(9n-7)}2+1,\text{ for }n>0\)
</p>
<h3>
Inductive Proof:
</h3>
<p>
\(Z(1)=\frac{1(9(1)-7)}2+1=2 \\
Z(n)=\frac{(n-1)(9(n-1)-7)}2+1+9n-8 \\
Z(n)=\frac{(n-1)(9n-16)}2+9n-7 \\
Z(n)=\frac{9n^2-7n+2}2 \\
Z(n)=\frac{n(9n-7)}2+1\)
<h2>
Exercise 1.15
</h2>
<h3>
Recurrent Relation:
</h3>
<p>
\(I(1)=0 \\
I(2)=2 \\
I(2n)=2I(n)-1,\text{ for }n>0 \\
I(2n+1)=2I(n)+1,\text{ for }n>0\)
</p>
<h3>
Closed Form Hypothesis:
</h3>
<p>
\(n=3(2^m)+l,\text{ where }0\le l<3(2^m)\le n \\
I(n)=2l+1,\text{ for }n>2\)
</p>
<h3>
Inductive Proof:
</h3>
<p>
Basis:\(I(3)=I(3(2^0)+0)=2(0)+1=1\)<br/>
Even:\(I(n)=2(2(l/2)+1)-1=2l+1,\text{ for }n>2\)<br/>
Odd:\(I(n)=2(2(l-1)/2+1)+1=2l+1,\text{ for }n>2\)<br/>
</p>
</body>
<html>