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riovelo November 7th, 2016 05:51 AM

recurrence relation problem
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Hi, I would like to get help with the following recurrence relation problem.
Thanks in advance.

skipjack November 7th, 2016 06:03 AM

$P(k) = P(1)\dfrac{4^{k-1}}{k!}$

riovelo November 7th, 2016 06:13 AM

Thanks, but how is that explicit if the answer contains p(1)?

Benit13 November 7th, 2016 06:19 AM

I think the problem requires a boundary condition to be stated (P(1) or P(0)).

skipjack November 7th, 2016 07:09 AM

P(1) can have any value.

The formula I gave must also hold for k = 0, so it can be written as $P(k) = P(0)\dfrac{4^k}{k!}$, where P(0) can have any value.

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