Offset between non-homogeneous and homogeneous recurrence sequences

In summary, the offset between a linear, non-homogeneous recurrence sequence and its corresponding homogeneous sequence can be determined by finding the steady state value of the recurrence rule. This offset is independent of the initial values of the sequence and can be calculated by setting up an equivalence between three successive terms of the sequence.
  • #1
dodo
697
2
I have a question; help is welcome.

Let sn be a linear, non-homogeneous recurrence sequence, and let hn be a corresponding homogeneous sequence (with initial values to be determined).

As it turns out, the offset between the two (sn - hn) is given by the steady state value of sn, if the initial values of hn are offset from those of sn by the same amount. Precisely what is the reason for that?

This steady state value bears no relation to the initial values of the sequence sn; more properly, it should be called the steady state value of the recurrence rule. I believe it is clear that a linear recurrence rule will have exactly one steady state value, neither none nor multiple (as the steady state is given by the root of a first-degree polynomial). And the steady state of hn is, of course, 0 (the root of a first-degree polynomial through the origin -- d'oh!). Therefore, if the desired offset does not depend on the initial values of sn (the offset was hand-set on the initial values, but who says it will stay that way further on in the sequences?), then the difference of the two steady states (thus the steady state of sn, as the one of hn is zero) should do. But why is the part in bold true?

Thanks--
 
Physics news on Phys.org
  • #2
To start with a simpler question, how would you get the closed form of this sequence? (And why?)[tex]
G_n = G_{n-1} + G_{n-2} + 3
[/tex]with initial values[tex]
G_0 = 0, \quad G_1 = 1
[/tex](Not a homework problem; change the sequence to any other that has a constant term, if you wish.)

- - -

P.S. The answer is that you look for the closed form of[tex]
F_n = F_{n-1} + F_{n-2}
[/tex]with initial values[tex]
F_0 = 3, \quad F_1 = 4
[/tex]and then subtract 3 from that; obtaining, in the end,[tex]
G_n = \frac {(3+\sqrt 5) \phi^n + (3-\sqrt 5) \psi^n} 2 - 3
[/tex]with the usual values for[tex]
\phi = \frac {1 + \sqrt 5} 2, \quad \psi = \frac {1 - \sqrt 5} 2
[/tex]
but the question is why is the first step valid.

- - -

P.P.S.: Hmm, maybe the proof by induction that you'd use for this example can be extended to the generic case. I'll try and be back.
 
Last edited:
  • #3
Dodo said:
I have a question; help is welcome.

Let sn be a linear, non-homogeneous recurrence sequence, and let hn be a corresponding homogeneous sequence (with initial values to be determined).

As it turns out, the offset between the two (sn - hn) is given by the steady state value of sn, if the initial values of hn are offset from those of sn by the same amount. Precisely what is the reason for that?

This steady state value bears no relation to the initial values of the sequence sn; more properly, it should be called the steady state value of the recurrence rule. I believe it is clear that a linear recurrence rule will have exactly one steady state value, neither none nor multiple (as the steady state is given by the root of a first-degree polynomial). And the steady state of hn is, of course, 0 (the root of a first-degree polynomial through the origin -- d'oh!). Therefore, if the desired offset does not depend on the initial values of sn (the offset was hand-set on the initial values, but who says it will stay that way further on in the sequences?), then the difference of the two steady states (thus the steady state of sn, as the one of hn is zero) should do. But why is the part in bold true?

Thanks--

I think the answer is to set up the following equivalence

s2 = A(s1) + B(s0) + C = h2 - Offset

= A*h1 + B*h0 - Offset= A*(s1 + Offset) + B*(s0 +Offset) - Offset

As you can see the terms A(s1) and B(s0) on both the right and left hand sides of the equation cancel out.
 
Last edited:
  • #4
Yes, ramsey, I think that is helpful. Not just for the first three terms, but for any three successive terms, if an offset was to have these properties, it would have to comply with an equation where only A,B,C appear. (Assuming A+B is not 1, because then the whole method breaks -- but that's another story.)

It is better to define the offset with the opposite sign (so that sn = hn + Offset, and not - Offset) because then the equation simplifies to Offset = C/(1 - A - B) instead of C/(A + B - 1). The former value can be shown to be the steady state of the sequence rule:

A ( C/(1 - A - B) ) + B ( C/(1 - A - B) ) + C = (AC + BC + C(1 - A - B)) / (1 - A - B) = C/(1 - A - B)

(that is, for this value of "Offset", if sn-2 and sn-1 are both equal to Offset, then sn will be equal to Offset too).
 
  • #5


The offset between non-homogeneous and homogeneous recurrence sequences is an interesting concept and has practical applications in various fields of science and mathematics. The reason for the steady state value of sn being the offset between the two sequences is due to the nature of linear recurrence rules.

Firstly, it is important to understand that a linear recurrence rule will always have a steady state value, which is the root of a first-degree polynomial. This steady state value is independent of the initial values of the sequence sn, as it is solely determined by the recurrence rule itself.

In the case of a non-homogeneous sequence sn and its corresponding homogeneous sequence hn, the initial values of hn are set to be offset by the same amount as the initial values of sn. This means that the two sequences will have the same behavior at the beginning, but as the sequences progress, the non-homogeneous sequence will deviate from the homogeneous one.

However, the steady state value of hn will always be 0, as it is the root of a first-degree polynomial through the origin. This means that the difference between the steady state values of sn and hn will be equal to the steady state value of sn. In other words, the offset between the two sequences is determined by the steady state value of sn, which is independent of the initial values of the sequence.

In conclusion, the reason for the steady state value of sn being the offset between non-homogeneous and homogeneous recurrence sequences is due to the nature of linear recurrence rules and the fact that the steady state value is independent of the initial values of the sequence. I hope this helps to clarify the concept.
 

1. What is the difference between non-homogeneous and homogeneous recurrence sequences?

A non-homogeneous recurrence sequence is one where the coefficients of the terms are not constant, while a homogeneous recurrence sequence has constant coefficients. This means that the terms in a non-homogeneous sequence may have varying factors or variables, while those in a homogeneous sequence will have the same factor or variable throughout.

2. How is the offset calculated in non-homogeneous and homogeneous recurrence sequences?

The offset in a non-homogeneous recurrence sequence is calculated by finding the difference between the highest degree of the terms and the degree of the recurrence relation. In a homogeneous recurrence sequence, the offset is simply the degree of the recurrence relation.

3. What does the offset represent in these types of sequences?

The offset in both non-homogeneous and homogeneous recurrence sequences represents the number of initial terms that are needed to fully determine the sequence. This means that the offset is the number of known terms that are required to find the unknown terms in the sequence.

4. How can the offset affect the behavior of these sequences?

The offset can affect the behavior of non-homogeneous and homogeneous recurrence sequences by determining the number of initial terms needed to find the unknown terms. If the offset is large, it may take more known terms to determine the sequence, making it more difficult to predict its behavior.

5. Are there any real-world applications of these types of sequences and their offset?

Yes, recurrence sequences and their offset have many applications in fields such as computer science, engineering, and economics. They can be used to model population growth, financial investments, and algorithmic processes. Determining the offset can help in predicting future behavior and making informed decisions based on the given sequence.

Similar threads

  • Differential Equations
Replies
2
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
18
Views
2K
  • General Math
Replies
11
Views
1K
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
2K
  • Topology and Analysis
Replies
10
Views
2K
  • Thermodynamics
Replies
7
Views
2K
Replies
2
Views
2K
  • Quantum Physics
Replies
1
Views
621
  • Precalculus Mathematics Homework Help
Replies
1
Views
3K
Back
Top