Lecture 30: Solving Recurrences

We looked at deriving recurrences for counting last lecture. We also started looking into solving the very simplest of recurrences. This lecture, we will look at some more interesting ways to tackle recurrences. Especially, we will introduce generating functions as an interesting way of solving recurrences.

Some Basic Summation Facts

We recall the following basic summation facts:

$begin{array}{rclcl} sum_{i=0}^n i &=& 0+1 + 2 + 3 + cdots + n &=& frac{n(n+1)}{2} sum_{i=0}^n a^i &=& a^0 + a^1 + a^2 + cdots + a^n &=& frac{a^{n+1} -1 }{a-1}, mbox{provided} a not= 1 sum_{i=0}^{infty} a^i &=& 1 + a+ a^2 + a^3 + cdots &=& frac{1}{1 -a }, mbox{provided} -1 < a < 1 end{array}$

Given a knowledge of the closed forms of the summations above, we can calculate other kinds of summations. Let us take an example.

Example-1

Find the value of $sum_{i=0}^n (i+1) a^i = 1. a^0 + 2 . a^1 + 3. a^2 + ldots + (n+1) . a^{n}$ for some constant , where a not= 1 .

Answer

Let represent the value of the summation above.

$begin{array}{r|lclclclclclclcl} (1) rightarrow & S &=& 1 &+& 2 a &+& 3 a^2 &+& 4 a^3 &+& cdots &+& (n+1) a^n (2) rightarrow & a S &=& & & a &+& 2 a^2 &+& 3 a^3 &+& cdots &+& n a^n &+& (n+1) a^{n+1} hline mbox{Subtract (2) from (1)} rightarrow &S - aS &=& 1 &+& a &+& a^2 &+& a^3 &+& cdots &+& a^n &-& (n+1) a^{n+1} end{array}$

From the reasoning above, we get

$(1-a) S = frac{1 - a^{n+1} }{1 - a } - (n+1) a^{n+1} ,.$

Since a not= 1 , we conclude

$S = frac{1 - a^{n+1} }{(1-a)^2} - frac{ (n+1) a^{n+1} }{ 1 - a } ,.$

Solving Recurrences: Basics

We consider recurrences of the form: $a_n = K a_{n-1} + L, n > 0$ with base case a_0=C fixed. K,L,C are constants.

Example-1

Take the recurrence: $a_n = 2 a_{n-1} +1, n > 0$ and a_0 = 1 . Here is the derivation for the closed form.

Answer

$begin{array}{rclcl} a_n &=& 2 a_{n-1} +1 &=& 2 (2 a_{n-2} +1) + 1 &=& 4 a_{n-2} + 2 + 1 &=& 4 (2 a_{n-3} +1) + 2 + 1 &=& 8 a_{n-3} + 4 + 2 + 1 & =& cdots &vdots & &=& 2^n a_{n-n} + 2^{n-1} + cdots + 2 + 1 &=& 2^n a_0 + sum_{i=1}^{n-1} 2^{n-1} &=& 2^n + frac{(2^n -1)}{2-1} &=& 2^{n+1} -1 end{array}$

Observation

From the example above, we can proceed to solve recurrences of the form $a_n = K a_{n-1} + L, n > 0$ as follows:

Answer

$begin{array}{rclclcl} a_n &=& K a_{n-1} + L &=& K( K a_{n-2} + L) + L &=& K^2 a_{n-2} + KL + L &=& K^2 (K a_{n-3} + L) + KL + L & =& K^3 a_{n-3} + L( K^2 + K + 1) &=& K^4 a_{n-4} + L (K^3 + K^2 + K + 1) &vdots & &=& K^n a_0 + L ( K^{n-1} + cdots + K + 1) &=& K^n a_0 + L frac{K^n -1}{K -1 } end{array}$

Second-Order Recurrences

We will now consider recurrences that are second-order of the form:

$a_n = K a_{n-1} + L a_{n-2}, n geq 2$

with base cases a_0, a_1 .

Well-Known examples of such recurrences include:

Fibonacci numbers: and .
Lucas Numbers: and .
Neganacci numbers: and .

Quadratic Equation

For the second order recurrence

$a_n = K a_{n-1} + L a_{n-2}, n geq 2$

We setup a quadratic equation

x^2 - K x - L = 0

Let r,s be two different roots to the quadratic equation above. Then the closed form solution to the recurrence has the form

a_n = A r^n + B s^n

for some coefficients A,B .

If the equation has a single repeated root r=s , then the solution is given by

a_n = (A+Bn) r^n

Example # 1

Consider the recurrence:

$G_0 = 1, G_1 = 2, G_n = - G_{n-1} + 2 G_{n-2}, n geq 2$

Answer

We write out some terms of this recurrence

G_2 = 0, G_3 = 4, G_4 = -4, G_5 = 12, cdots .

Here we have K = -1 and L = 2 . Therefore we solve the quadratic equation:

x^2 + x - 2 = 0

Its roots are: $x = frac{-1 pm sqrt{1 + 8}}{2} = frac{-1 pm 3}{2} = 1, -2$ .

Therefore, the solution is of the form

G_n = A 1^n + B ( -2 )^n

Given G_0 = 1 , we obtain A + B = 1 , and G_1 = 2 = A - 2 B . Solving, we obtain the solution

$A = frac{4}{3}, B = - frac{1}{3}$

Therefore,

$G_n = frac{4}{3} - frac{1}{3} left( -2 right)^n = frac{4 + (-1)^{n+1} 2^n}{3}$ .

Example # 2

Consider the Fibonacci recurrence

$F_{0} = 0, F_1= 1, F_{n} = F_{n-1} + F_{n-2}, n geq 2$ .

Answer

Here K = L = 1 . The characteristic polynomial is

x^2 - K x - L = 0 ;; x^2 -x -1 = 0 .

Solving this equation, we obtain the roots

$x = frac{1 pm sqrt{ 1 + 4} }{2} = frac{1 + sqrt{5}}{2}, frac{1 - sqrt{5}}{2}$ .

Therefore, the closed form is

$F_n = A left( frac{1 + sqrt{5}}{2} right)^n + B left(frac{1 - sqrt{5}}{2} right)^n$ .

Plugging, in n = 0 , we obtain,

A + B = F_0 = 0 .

For n=1 ,

$A frac{1 + sqrt{5}}{2} + B frac{1 - sqrt{5}}{2} = F_1 = 1$ .

Writing $phi = frac{1 + sqrt{5}}{2}$ and $psi = frac{1 - sqrt{5}}{2}$ , we obtain the equations:

A + B = 0, phi A + psi B = 1

Solving them we obtain, (psi - phi) A = -1 or $A = frac{-1}{psi - phi} = frac{1}{sqrt{5}}$ . Likewise, $B = -frac{1}{sqrt{5}}$ . The overall closed form is

$F_n = frac{1}{sqrt{5}}left( phi^n - psi^n right)$ .

A special note: usually, we have Fibonacci numbers starting as F_0 = 1, F_1 = 1 . The closed form for the sequence with this base case will be

$tilde{F_n} = frac{1}{sqrt{5}}left( phi^{n+1} - psi^{n+1} right)$ .