Solve the nonhomogeneous recurrence relation hn=6 hn-1-9 hn-2+2 n, (n ≥ 2) h0=1

Solve the nonhomogeneous recurrence relation hn=6...

Key Concepts

Linear Recurrence Relations

A linear recurrence relation defines each term of a sequence as a linear combination of previous terms, typically with constant coefficients. Understanding these relations is fundamental in mathematics because they appear in various contexts, such as algorithm analysis, financial modeling, and counting problems in combinatorics.

Homogeneous vs Nonhomogeneous Recurrences

A homogeneous recurrence relation has a right-hand side equal to zero (or, more generally, does not include an additional term), whereas a nonhomogeneous recurrence includes an extra term independent of the sequence. The nonhomogeneous part requires finding a particular solution that addresses this additional term, in addition to solving the associated homogeneous equation.

Characteristic Equation

The characteristic equation is a polynomial derived from a homogeneous linear recurrence relation and is crucial for finding its general solution. By solving this polynomial equation for its roots, one can determine the form of the solution depending on whether the roots are distinct, repeated, or complex.

Method of Undetermined Coefficients

This method is used to find a particular solution to a nonhomogeneous linear recurrence relation by guessing a form of the solution based on the nonhomogeneous term. The unknown coefficients are then determined by substituting the guess into the recurrence, making it a systematic way to handle additional nonhomogeneous terms.

Initial Conditions

Initial conditions provide the starting values of the sequence and are essential for determining the unique solution of a recurrence relation. They allow for the elimination of arbitrary constants found in the general solution, ensuring that the resulting sequence exactly matches the defined criteria from the outset.

Transcript

00:01 So here we have that t, our function t is going to be equal to n times t squared, evaluated at n over 2.

00:13 And initial condition, t1 is equal to 6.

00:21 So first we're going to substitute n equals 2 to the power of k.

00:27 And so this becomes t of 2 to the power of k is going to be equal to 2, 2, 2, to the power of k times t squared at two to the power of k over two, which of course is going to be equal to two to the power of k times t squared, referred at two, k minus one.

01:00 And then we're going to make another substitution here, where we let a k equal log 2, t, two to the power of k.

01:14 And when we do that, we get that this is equal to log 2, 2 to the power of k times t squared to k minus 1.

01:35 And we can use some of our log rules.

01:40 So we get log 2 times 2 to the k plus log 2 of 2 to the k, plus log 2 of t to the k to the k minus 1, squared.

02:14 Okay, and this is going to be equal to k plus 2, log 2, t times 2 to the k minus 1 power, and then this is equal to k plus 2 a subk minus 1.

02:52 Okay, so that's going to be a lot easier to work with, not we've made those substitutions.

03:04 And, for, simplicity, we can really write this as like a -n, a -k would be a -n is equal to n plus 2 -a -n -minus -1, right? so it'd be in the same form that we're used to.

03:22 I'm going to translate our initial conditions as well, so we would get, so we're going to rewrite to get a -n is equal to 2 -a -n -1 plus n, and then we would have, a knot is equal to log 2 t to the 2 to the power of 0 right which is going to be just one to the power becomes t1 which we know from what we're given is that that is really just six okay so then we want to find the root's characteristic equation here and when we do that we get that our our homogeneous portion is going to be equal to.

04:36 So we would say r squared is equal to 2.

04:40 I say r is equal to 2 when we set a .n equal to r and a .n minus 1 equal to 1.

04:46 So it would be alpha times 2 to the power of n.

04:53 And then we want to move on to the particular solution.

04:58 In this case, our function is just equal to n, which is the same as 1.

05:11 Multiplying that by 1 to the power of n and 1 is not a root.

05:19 So we'll take that form of our particular solution, which should be equal to p1 times n plus p .0 times 1 to 1 to the power of n, which is just p1 times n plus p .0.

05:44 And then now we need to solve for this and make sure it satisfies our recurrent relation.

05:52 Here and so we will rewrite it.

05:58 So we have a .n equals two times a .n minus 1 plus n.

06:04 We'll put in our particular solution, p1 times n plus p0, in for a .n.

06:11 We'll put it in for a .n minus one as well.

06:13 So we get two times p1 times n minus 1 plus p knot plus n.

06:30 We can extend that, so it's going to be 2p1 times n minus 2 p1 plus 2p0 plus n.

06:49 And we can subtract from the left -hand side and the right, so we would get 0 is equal to p -1 times n, minus 2p1 plus p not plus n.

07:08 So we get up the n's and the knot ends...