Find the solution to an=7 an-2+6 an-3 with a0=9 a1=10, and a2=32 .

Name: Problem 13, Chapter 8: Discrete Mathematics and its Applications
Uploaded: 2020-02-27T01:44:38-08:00
Duration: 8 min 44 s
Channel: Bryan Lynn
Description: Find the solution to an=7 an-2+6 an-3 with a0=9 a1=10, and a2=32 .

Find the solution to an=7 an-2+6 an-3 with a0=9 a1=10, and...

Key Concepts

Linear Recurrence Relations

A linear recurrence relation defines each term in a sequence as a linear combination of previous terms, usually with constant coefficients. This concept helps in breaking down a sequence into manageable parts and forms the foundation for finding patterns or closed-form solutions.

Homogeneous Recurrence Relations

A homogeneous recurrence relation is characterized by the fact that all terms of the recurrence are dependent on previous terms without any additional non-homogeneous (or forcing) term added. This allows for the application of specific methods, such as the characteristic equation method, to find its general solution.

Characteristic Equation

The characteristic equation is derived by substituting a trial solution, often of the form r^n, into the homogeneous linear recurrence relation. Solving this polynomial equation yields the roots that determine the structure of the general solution, including cases of distinct or repeated roots.

Initial Conditions

Initial conditions provide the specific values at the beginning of the sequence, which are necessary to determine the unique constants in the general solution of a linear recurrence. They tailor the general solution to fit the particular sequence being studied.

Transcript

00:01 Alrighty, so for this one, our given equation, a .n, is equal to seven times a .n minus two, so seven times the answer we got two times ago, two iterations ago, plus six times a n minus three.

00:21 And so it'd be the solution from three iterations ago.

00:26 And in this instance, we have a0 is equal to 9, a1 is equal to 10, and a2 is equal to 32.

00:39 And so, of course, in this case, n must be greater than are equal to 3 for this to really make any sense.

00:46 So we're going to start with our roots characteristic equation.

00:50 So here we'll let a .n be equal to r cubed.

00:58 A n minus 1 is going to be equal to r squared.

01:08 A n minus 2 is going to be equal to r.

01:12 And a n minus 3 will be equal to 1.

01:19 So what's happening is as each of these indexes decreases by 1, so does each of the powers on the r? so once we make that substitution, we end up with r cubed.

01:35 Is equal to 7r minus 6.

01:42 Okay, so if we subtract 7r minus 6 from both sides, this is equal to r cubed minus 7r, and i apologize should be a plus 6 here, so it be minus 6 is equal to 0.

01:59 And we can just factor this.

02:03 So this is gonna end up being equal to r minus 3 times r plus 1, times r plus two is equal to zero.

02:16 Okay, that gives us our roots, right? so now we know that r must be equal to three, negative one, negative two.

02:25 For these roots to be equal to zero.

02:29 So we can use those to start building our solution to the recurrence relation.

02:35 So the solution is gonna be of the form, a .n is equal to, say, alpha one, times negative 2 to the power of n plus some other constant alpha 2 times negative 1 to the power of n and then plus another constant alpha 3 times 3 n.

03:01 So the roots that we found in the previous section are what we're raising to the power of n.

03:08 And now we will use the initial conditions to figure this out.

03:15 So we know that a not, which is equal to 9, if we put a 0 in for all these ends, we will get is equal to alpha 1 plus alpha 2 plus alpha 3.

03:32 And a 1, which was equal to 10, if we put a 1 in where we have an n above, we'll get a negative 2 alpha 1 minus.

03:50 Alpha 2 plus 3 alpha 3.

03:58 Okay, and then lastly, a2, which is equal to 32, we put a 2 in, we're gonna get 4 times alpha 1 plus alpha 2 plus 9 alpha 3.

04:22 Okay, so now what we wanna do is we're first going to add the first two equations.

04:29 So we're going to add alpha not plus, or a not plus a 1 together...

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