Solving recurrences (CSCI 2824, Spring 2015)

Topic covered:

Solving Recurrences: some basic ideas on generating functions for solving interesting recurrences (Section 5.6 of the book).

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 aneq 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 Linear 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 2

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

Suppose that Kneq 1 . Then

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

If K=1 , then

$a_n=a_{n-1}+L=a_{n-2}+2L = cdots = a_0+nL$ .

Linear Recurrences with Two Initial Terms

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 .

Solving the likes of Fibonacci sequence by setting up a quadratic equation

For the recurrence

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

We setup a quadratic equation

x^2 = K x + L ,

or equivalently,

x^2 - K x - L = 0 ,

whose real roots are given by

$r = frac{K+sqrt{K^2+4L}}{2},quad s = frac{K-sqrt{K^2+4L}}{2}.$

If rneq s , then the closed form solution to the recurrence has the form

a_n = A r^n + B s^n

for some coefficients A,B determined by the initial values a_0, a_1 .

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

a_n = (A+Bn) r^n

for some coefficients A,B determined by the initial values a_0, a_1 .

Example 3

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 4

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

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