Find the solution to an=5 an-2-4 an-4 with a0=3 a1=2, a2=6, and a3=8

Name: Problem 14, Chapter 8: Discrete Mathematics and its Applications
Uploaded: 2019-09-04T05:19:29-08:00
Duration: 5 min 40 s
Channel: James Chok
Description: Find the solution to an=5 an-2-4 an-4 with a0=3 a1=2, a2=6, and a3=8

Find the solution to an=5 an-2-4 an-4 with a0=3 a1=2, a2=6,...

Key Concepts

Linear Homogeneous Recurrence Relations

This concept refers to a recurrence relation where each term is defined as a linear combination of previous terms with constant coefficients and without any additional non-homogeneous terms. Such recurrences can be solved systematically by translating the recurrence into a polynomial equation whose roots determine the form of the general solution.

Characteristic Equation

The characteristic equation is derived by assuming a solution of the form r^n for the recurrence relation. Substituting this form into the relation transforms it into a polynomial equation in terms of r. The roots of this polynomial, including their multiplicities, dictate the structure of the closed-form solution of the recurrence.

Initial Conditions

Initial conditions are the starting values provided for the recurrence relation, which are essential for determining the unique particular solution. They are used to compute the specific coefficients in the closed-form solution once the general solution, based on the characteristic equation, is obtained.

Closed-Form Solution

A closed-form solution is an explicit expression that defines the nth term of the sequence without requiring the computation of previous terms. It is derived from the combination of the roots of the characteristic equation, adjusted by the initial conditions to fit the specifics of the recurrence.

Transcript

00:01 Okay, so in this question we want to find a solution to this recurrence equation.

00:05 So first off, the recurrence, the characteristic equation is going to be r4 is equal to 5r squared minus 5 .4.

00:15 So moving everything to one side, you have r to 4 minus 5, 5r squared plus 5 4 is equal to 0.

00:25 So now the question is, how do you factorize this? so the first thing that you can do is, is, you know, is the substitute r is equal to plus or minus one.

00:35 And then you'll notice that this thing would factorize, well, r equals to plus and minus one will make this left -hand side equals to zero.

00:43 Because this will become one minus by five, plus by four, so that's zero.

00:48 What you will also find out is plus and minus two also makes this equation zero.

00:57 So the factors of this is then, you can factorize this as r minus, minus 2, and by r plus 2, r minus 1, r plus 1 is equal to 0.

01:12 And so then your roots are basically these ones up here.

01:15 R is equal to plus or minus 1, plus or minus 2.

01:21 So now we want to solve the recurrence relation.

01:25 So our relation is an an is going to be some constant times by, let's say, minus 2 to the n plus by alpha 2 times by minus 1 to the n plus by alpha 3 times by 1 to the n plus by alpha 4 times by 2 to the n so then we'll use our initial conditions which is a 0 is equal to 3 which is also equal to when we put n into these powers so this will become 1 1 1 and 1 is equal to then alpha 1 plus by alpha 2 plus by alpha 3 plus alpha 4 we have a 1 is going to be equal to 2 because that's the intro condition and then minus 2 to 1 is just minus 2 alpha 1 minus 1 to 1 is just minus 1 so minus alpha 2 1 to 1 to plus alpha 2 and finally plus by 2 alpha 4.

02:38 So a2 given by the initial condition is going to be equal to 6.

02:43 So then this is equal to 4 alpha 1 plus by alpha 2 plus by alpha 3 plus by 4 alpha 4.

02:56 And finally a3 which is equal to 8 which is going to be equal to minus 2 to the 3 is minus 8.

03:05 So minus 8 alpha 1, minus 1 cube is just minus 1, so minus alpha 2, plus by alpha 3 plus by 8 alpha 4.

03:19 So we have a system of equations, in particular these sets of equations...

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