[PDF] Q1: Exercise 2.3-3 (10 points) Use mathematical induction to show

CS 161 Lecture 3 - 1 Solving recurrences

Last class we introduced recurrence relations such as T(n) = 2T(?n/2?) + n. Typically Here a = 9

Discrete Mathematics II (Spring 2015) - 8.2 Solving Linear

Determine what is the degree of the recurrence relation. • Need to know the general solution equations. • Need to find characteristic equation. • Need to find 

Recurrences

n = 2k for some constant k and leaving out base case and constant in ?). Methods for solving recurrences. 1. Substitution method. 2. Iteration method.

Q1: Exercise 2.3-3 (10 points) Use mathematical induction to show

recurrence. T(n) =2 ifn=2. 2T(n/2) + n if n = 2k for k > 1 is T(n) = n logn. . Base Step: If n = 2

Master Theorem: Practice Problems and Solutions

The Master Theorem applies to recurrences of the following form: 2. T(n)=4T(n/2) + n2. 3. T(n) = T(n/2) + 2n. 4. T(n)=2nT(n/2) + nn.

GENERATING FUNCTIONS AND RECURRENCE RELATIONS

A recurrence recurrence relation is a set of equations Linear Recurrence. Fibonacci Sequence an = an?1 + an?2 n ? 2. ... n=2 n(n ? 1)xn +.

Solving Recurrences

For example the following recurrence (written in two different but standard ways) describes the identity function f (n) = n:.

1.1 Time complexity and Big-Oh notation: exercises

2. A quadratic algorithm with processing time T(n) = cn2 spends T(N) To have the well-defined recurrence assume that n = km with the integer.

Sample Induction Proofs

(?1)nn(n + 1). 2 . Base case: When n = 1 the left side of (1) is (?1)12 = ?1

3. Recurrence 3.1. Recursive Definitions. To construct a recursively

Example 3.2.1. The factorial function f(n) = n! is defined recursively as follows: 1. Initial Condition: f(0) = 1. 2. Recursion: f(n +1)=(n + 1)f(n).

Recurrence Relations - Hong Kong University of Science and

an= 2 n¡1; n ‚1: Given a recurrence relation for a sequence with initial conditions Solving the recurrence relation means to ?nd a formula to express the general termanof the sequence 2 Homogeneous Recurrence Relations Any recurrence relation of the form

Solving Recurrence Relations - Texas A&M University

n 2 and the base cases of the induction proof (which is not the same as the base case of the recurrence!) are n= 2 and n= 3 (We are allowed to do this because asymptotic notation only requires us to prove our statement for n n 0 and we can set n 0 = 2 ) We choose n= 2 and n= 3 for our base cases because when we expand the recurrence

Recurrence Relations - University of Ottawa

Many sequences can be a solution for the same recurrence relation a n = 2a n 1 a n 2; for n 2 The following sequences are solutions of this recurrence relation: I a n = 3n for all n 0 I a n = 5 for all n 0 The initial conditions for a sequence specify the terms before n 0 (before the recurrence relation takes e ect)

Solving Recurrence Relations - Princeton University

n 2 to be C0 and C1 Since the sequences fang and f 1rn 1 + 2r n 2g satisfy the degree 2 recurrence and agree on the rst two values they must be identical Example: Find the solution to the recurrence relation an = an 1 +an 2 with initial condi-tions a0 = 2 and a1 = 7 Solution: The characteristic equation is r2 r 2 = 0 i e (r 2)(r +1) = 0

Recurrence Relations - University of Illinois Urbana-Champaign

2 Recall that 1=nk= nk and we have rn 2(r25r + 6) = 0 Then we can factor the polynomial into rn 2(r 2)(r 3) = 0 We assumed that r is non-zero so we can divide by the rst factor rn 2 which leaves (r 2)(r 3) = 0 Thus the characteristic function (also: polynomial) remains on the left-hand side

Searches related to 2^n n^2 recurrence PDF

n = 2n for every nonnegative integer n is a solution of the recurrence relation a n = 2a n-1 - a n-2 for n = 234 Assume a 0=1 and a 1=2 Solution: Check initial conditions a 0= 20 = 1 a 1= 2 1 = 2 Check if {2n} satisfies a n = 2a n-1 - a n-2 2a n-1 - a n-2 = 2(2n-1) - 2n-2 = 2n-2(4-1) = 3(2n-2) 2n So {2n} is not a solution for a n = 2a

What is the solution of the recurrence relation?

.Thenasequence (a n ) is a solution of the recurrence relation a n= c 1a n?1+ c 2a n?2 if and only if a n= ? 1r n 1+ ? 2r n 2 for n ? 0 for some constants ? 1,? 2 Idea of the Proof The proof proceeds exactly as in the case of the Fibonacci numbers. Try to prove it yourself!

What is the solution of recurrence relation an = 10an-1 - 25an-2?

The solution of the recurrence relation an = 10an-1 - 25an-2, is: (a0 = 1, a1 = 15) Q8. Solution to recurrence relation T (n) = T (n - 1) + 4 is given by, where n > 0 and T (0) = 3.

What is a 2 nd-order linear recurrence?

This is a subset of the type of 2 nd -order recurrence formulas used by MCS . The Fibonacci number s are the best-known example of a 2 nd -order linear recurrence.

What is the initial condition for a recurrence relation?

So (a_n = 3a_ {n-1} + 2) is our recurrence relation and the initial condition is (a_0 = 1 ext {.}) We are going to try to solve these recurrence relations.