Mark's question at Yahoo Answers (Linear recurrence relation)

In summary, the solution to the 2nd order homogeneous linear recurrence is given by x_n=C_1\left( \dfrac{1}{3}\right)^n+C_2\left(\dfrac{-4}{3}\right)^n where . The sequence is bounded if and only if .
  • #1
Fernando Revilla
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Question - Find the general solution to the 2nd order homogeneous linear recurrence below, and give a necessary and sufficient condition on u0 and u1 such that the sequence defined by the recurrence is bounded.

2*x subscript(n + 1) + 3*x subscript (n) -2*x subscript (n-1) = 0

I've found the general solution using the auxiliary equation, but I'm not sure how to prove it's bounded. I know that if a sequence converges, it means that it is bounded, but I have no clue how to show whether a recurrent sequence converges. Any help will be greatly appreciated!

Here is a link to the question:

2nd order homogeneous linear recurrence? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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  • #2
Hello Mark,

From we get and so, the general solution is As that is, is bounded. Taking into account that , the sequence is bounded if and only if , as a consequence is bounded if and only if . Now, for and for

But if and only if (necessary and sufficient condition on and such that the sequence defined by the recurrence relation is bounded).

Edit: See the following posts.
 
Last edited:
  • #3
Thank you So much for your reply, I've perfectly understood how to solve it now! Just one quick question though, shouldn't the roots of the auxiliary equation be -2 and 1/2? It's not a big deal as the concept remains the same, but I just wanted to confirm whether it was a calculation mistake on your part or have I solved the auxiliary equation wrong.

Thank you very much once again!
 
  • #4
You are correct, the characteristic roots are indeed:

 
  • #5
TheAvenger said:
Thank you So much for your reply, I've perfectly understood how to solve it now! Just one quick question though, shouldn't the roots of the auxiliary equation be -2 and 1/2? It's not a big deal as the concept remains the same, but I just wanted to confirm whether it was a calculation mistake on your part or have I solved the auxiliary equation wrong.

Thank you very much once again!

All right, the concept remains the same. Out of curiosity, my mistake: I looked at instead of !

P.S. Does anyone know about a quadratic equation's tutorial? :)
 

FAQ: Mark's question at Yahoo Answers (Linear recurrence relation)

What is a linear recurrence relation?

A linear recurrence relation is a mathematical equation that describes a sequence of numbers, where each term is a linear combination of previous terms. In other words, the next term in the sequence is a function of the previous terms.

How is a linear recurrence relation different from a regular recurrence relation?

A linear recurrence relation is a special type of recurrence relation where the coefficients of the terms are constants. In a regular recurrence relation, the coefficients can be any mathematical expression. This makes linear recurrence relations easier to solve and analyze.

What is the importance of linear recurrence relations in mathematics?

Linear recurrence relations are important in many areas of mathematics, including number theory, combinatorics, and computer science. They are used to model and solve various problems, such as finding the number of ways to arrange objects or determining the growth rate of a population.

How do you solve a linear recurrence relation?

To solve a linear recurrence relation, you can use various methods such as substitution, iteration, and generating functions. The specific method used will depend on the complexity of the relation. In some cases, the solution can be found using algebraic manipulations, while in others, a computer program may be necessary.

Can you provide an example of a linear recurrence relation?

One example of a linear recurrence relation is the Fibonacci sequence, where each term is the sum of the two previous terms (e.g. 0, 1, 1, 2, 3, 5, 8, 13, 21, ...). The relation for this sequence is expressed as F(n) = F(n-1) + F(n-2), where n represents the term number. This relation can be used to generate any term in the sequence.

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