MHB How to derive a recurrence relation from explicit form

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To derive a recurrence relation from an explicit formula, one can start by substituting the closed form into the recurrence relation. The explicit form provided is P_N = (A^N/N!)/(Σ(A^x/x!)), while the recurrence relation is P_i = (A P_{i-1})/(i + A P_{i-1}) for i=1 to N. It is suggested that the recurrence can be expressed as 1/P_i = (i/(A P_{i-1})) + 1, which may simplify the derivation. The discussion indicates that algebraic manipulation can be sufficient to establish the recurrence relation. Overall, the thread emphasizes the importance of understanding both forms to facilitate the derivation process.
find_the_fun
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I am given a formula in explicit form and as a recurrence relation. It is asked to derive the recurrence relation from the explicit form. How is this done?
 
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What is the closed, or explicit, form?
 
closed: [math]P_N=\frac{\frac{A^N}{N!}}{\sum\limits_{x=0}^N \frac{A^x}{x!}}[/math]

recurrence: [math]P_i = \frac{A p_{i-1}}{i+A p_{i-1}}[/math] for i=1 to N
 
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I was hoping you would post a closed form solution in the form:

$$A_n=\sum_{k=0}^m\left(c_kr_k^n\right)$$

Leading to a linear recurrence.

I will have to defer here to someone more knowledgeable in this field. :D
 
find_the_fun said:
recurrence: [math]P_i = \frac{A p_{i-1}}{i+A p_{i-1}}[/math] for i=1 to N
I assume $P_{i-1}$ should be written with a capital $P$ in the right-hand side.

I'll have to think more how to derive the recurrence formula, but it is easy to prove it once it is known: just substitute the closed formula into it. It is probably easier to work with
\[
\frac{1}{P_i}=\frac{i}{AP_{i-1}}+1.
\]
 
Can this be done using nothing but algebraic manipulation? If so, I think I found a way.
 

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