How to derive a recurrence relation from explicit form

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SUMMARY

The discussion focuses on deriving a recurrence relation from an explicit formula in the context of probability distributions. The explicit form provided is \( P_N = \frac{\frac{A^N}{N!}}{\sum\limits_{x=0}^N \frac{A^x}{x!}} \), while the corresponding recurrence relation is \( P_i = \frac{A P_{i-1}}{i + A P_{i-1}} \) for \( i=1 \) to \( N \). Participants suggest that algebraic manipulation can be employed to derive the recurrence from the explicit form, specifically using the transformation \( \frac{1}{P_i} = \frac{i}{A P_{i-1}} + 1 \). The conversation emphasizes the importance of understanding both forms for effective application.

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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
 
Last edited:
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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