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

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The discussion centers on finding a closed-form solution for a complex difference equation involving nonlinear Bell polynomials. The equation expresses s_n as a function of previous terms s_k, with the challenge being to eliminate the dependence on these previous terms. Initial conditions are provided, and the complexity of the sum increases with n, complicating the derivation of a closed form. Participants emphasize the need for relationships between the Bell polynomials to simplify the equation, as the current formulation leads to an exponential number of terms. The conversation highlights the difficulty in establishing recurrence identities for the multivariate Bell polynomials necessary for solving the problem.
phoenixthoth
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While working on another problem, the following difference equation came up.

s_{n}=\frac{1}{r_{1}-r_{1}^{n}}}\sum_{k=1}^{n-1}s_{k}B_{n,k}\left(<br /> r_{1},r_{2},...,r_{n-k+1}\right)

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!
 
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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).
 
Here is a recurrence I found in the literature which might help.

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(<br /> r_{1},...,r_{j-k+2}\right)
 
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