How Can a Closed Form for This Complex Difference Equation Be Determined?

In summary, the conversation was about a difference equation that came up while working on another problem. The equation involves polynomials and an initial condition, and the goal is to find a closed form formula that does not involve previous values. The conversation also mentioned the use of Bell polynomials and a recurrence identity that might be helpful in simplifying the formula.
  • #1
phoenixthoth
1,605
2
While working on another problem, the following difference equation came up.

[tex]s_{n}=\frac{1}{r_{1}-r_{1}^{n}}}\sum_{k=1}^{n-1}s_{k}B_{n,k}\left(
r_{1},r_{2},...,r_{n-k+1}\right)[/tex]

The B_{n,k} are (nonlinear) polynomials in the variables r_{1}, ..., r_{n,k-n+1} that don't involve the s's. (In fact they are Bell polynomials though I'm not sure it is necessary to know B's to get a formula for s_{n}.)

Initial condition: s_{1}=1.

You can see that s_{n} is equal to some function of the previous s_{k} for k running from 1 through n-1.

I want a closed form formula for s_{n} that does not involve any other s_{k}. I've taken it up to 5 to look for a pattern and it seeeeems like there is some function of the B's that's being iterated to get the result of simplifying and substituting previous values of s_{k}.

Actually, I'm not sure this is a difference equation because the sum is not fixed but increases in complexity as n increases.

Any feedback or inquiries as to how this problem arose are quite welcome and appreciated!
 
Mathematics news on Phys.org
  • #2
phoenixthoth said:
though I'm not sure it is necessary to know B's to get a formula for s_{n}

You'll need to use some kind of relationship between the B's to simplify the formula, otherwise s_n will have 2^(n-2) terms (this can be shown by induction).
 
  • #3
  • #4
Here is a recurrence I found in the literature which might help.

[tex]B_{n,k}\left( r_{1},r_{2},...,r_{n-k+1}\right) =\frac{1}{k}\sum_{j=k-1}^{n-1}\binom{n}{j}r_{n-j}B_{j,k-1}\left(
r_{1},...,r_{j-k+2}\right) [/tex]
 

FAQ: How Can a Closed Form for This Complex Difference Equation Be Determined?

1. What is a difference equation?

A difference equation is a mathematical equation that describes the relationship between the current value of a variable and its previous values. It is used to model and predict the behavior of dynamic systems in various fields such as physics, economics, and engineering.

2. How is a difference equation different from a differential equation?

While both equations involve the use of mathematical functions to describe the relationship between variables, the main difference is in the way they are solved. Differential equations involve continuous functions, whereas difference equations involve discrete functions. This means that differential equations are solved using calculus, while difference equations are solved using algebra.

3. What are the applications of difference equations?

Difference equations are used to model and predict the behavior of various systems, such as population growth, chemical reactions, and stock market fluctuations. They are also used in control systems to regulate and optimize the performance of machines and processes. In economics, difference equations are used to study the effects of different policies on the economy.

4. Can difference equations be solved analytically?

Yes, some difference equations can be solved analytically by using algebraic manipulation and mathematical techniques such as generating functions and characteristic equations. However, more complex difference equations may require numerical methods for solution, such as iteration or simulation.

5. How are difference equations used in computer science?

Difference equations are used in computer science to analyze and design algorithms, as well as to model and simulate the behavior of computer systems. They are also used in digital signal processing to filter and process discrete-time signals, and in coding theory to detect and correct errors in digital communication systems.

Similar threads

Back
Top