Find the solution to an=2 an-1+an-2-2 an-3 for n=3,4,5, …, with a0=3, a1=6, and a2=0

Name: Problem 12, Chapter 8: Discrete Mathematics and its Applications
Uploaded: 2019-08-31T03:14:22-08:00
Duration: 6 min 43 s
Channel: James Chok
Description: Find the solution to an=2 an-1+an-2-2 an-3 for n=3,4,5, …, with a0=3, a1=6, and a2=0

Find the solution to an=2 an-1+an-2-2 an-3 for n=3,4,5, …,...

Key Concepts

General Solution

Once the roots of the characteristic equation are obtained, they are used to construct the general solution of the recurrence relation. Depending on whether the roots are distinct, repeated, or complex, the general solution is expressed as a linear combination of terms formed by the roots raised to the power n. This representation encompasses all possible solutions to the recurrence before any initial conditions are applied.

Initial Conditions

Initial conditions are specific values assigned to the first few terms of the sequence, and they are essential for determining the unique constants in the general solution. After obtaining the general solution, these conditions are used to solve a system of linear equations that provide the exact values for the constants, thereby tailoring the solution to the problem at hand.

Linear Homogeneous Recurrence Relations

This concept involves sequences defined by relations where each term is a linear combination of previous terms, with constant coefficients. The recurrence is called homogeneous if it equates to zero when subtracting the terms, meaning no additional functions of n are added. Understanding these relations is crucial to analyze sequences and predict their long?term behavior using a systematic approach.

Characteristic Equation

For linear homogeneous recurrence relations with constant coefficients, the characteristic equation is formed by substituting a trial solution of the form r^n into the recurrence relation. This transforms the recurrence into an algebraic equation in r, whose roots dictate the behavior of the sequence. The process of solving the characteristic equation simplifies the recurrence problem to finding polynomial roots.

Transcript

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

00:05 So the characteristic equation, we let a sub n be equal to r to the 3.

00:12 So then you have r3 is equal to 2 a or r squared plus by r minus by 2.

00:22 Then moving everything to the left side, you have r cube minus by 2r squared minus by r plus 2 is equal to 0.

00:35 So now we want to factorize this or we can just simply try to find roots to this.

00:41 So because this is a cubic where there's no general way to factorize this, what you have to do is you have to substitute values for this function.

00:49 R cubed minus 5 2 r squared minus r plus 2.

00:54 So you just substitute random numbers for r into this equation until you find until you get 0 and then you can simplify from that.

01:00 So if you you do do that, what you'll find out is f of 1, 1 minus by 2, minus by 1 plus by 2, so that's equal to 0.

01:11 You also find that f of minus 1, so minus 1 cube is minus 1, minus 1, minus 2 times by minus 1 squared, so that's this be minus 2, negative minus minus minus 1, so that's plus 1 and then plus 2, so that's also equal to 0.

01:27 And finally, you also find out that f of 2 is also 0.

01:30 I'll just let you do the calculations.

01:32 So this is you just have to guess and check for these numbers, but you'll find that these three are your roots.

01:39 So then this thing, this thing right here, can be factorized into r minus 1, r plus by 1, and r minus by 2 is equal to 0.

01:51 So then your roots are r is equal to 1, minus 1, and 2.

01:57 So your general equation, your characteristic equation is an a sub n is a, is a, and equal to c sub 1, this is 2.

02:09 Let's say 2 to the n plus by c sub 2 minus 1 to the n plus by c sub 3, 1 to the n.

02:19 Now 1 to the n is just 1.

02:22 This is c sub 1 to the n plus c sub 2 minus 1 to the n plus by c sub 3.

02:31 Now we have, using the integer conditions, a sub 0 is equal to 3, substituting n equals to 0 into this equation, you have 3 is going to be equal to c sub 1 plus by c sub 2 plus c sub 3.

02:50 And then you have a sub 1 is equal to 6.

02:53 So 2 to 1 gives you 2.

02:56 So 2, c sub 1, minus 1 to 1 gives you minus 1.

03:00 So minus c sub 1.

03:01 So minus c sub 1.

03:02 Sub 2 plus by c sub 3 and finally we have a sub 2 is equal to 2 squared or 8 2 is equal to 0 sorry and 2 squared gives you 4 so 4 c sub 1 minus 1 squared gives you 1 so that's plus c sub 2 plus c sub 3 okay now let's just write these out better so you have 3 is equal to a sub 1 plus by a sub 2 plus by a sub 3...

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