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November 7th, 2016, 06:51 AM   #1
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recurrence relation problem

Hi, I would like to get help with the following recurrence relation problem.
Thanks in advance.
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November 7th, 2016, 07:03 AM   #2
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$P(k) = P(1)\dfrac{4^{k-1}}{k!}$
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November 7th, 2016, 07:13 AM   #3
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Thanks, but how is that explicit if the answer contains p(1)?
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November 7th, 2016, 07:19 AM   #4
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I think the problem requires a boundary condition to be stated (P(1) or P(0)).
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November 7th, 2016, 08:09 AM   #5
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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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