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November 7th, 2016, 05:51 AM  #1 
Newbie Joined: Nov 2016 From: Israel Posts: 7 Thanks: 0  recurrence relation problem
Hi, I would like to get help with the following recurrence relation problem. Thanks in advance. 
November 7th, 2016, 06:03 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,929 Thanks: 2205 
$P(k) = P(1)\dfrac{4^{k1}}{k!}$

November 7th, 2016, 06:13 AM  #3 
Newbie Joined: Nov 2016 From: Israel Posts: 7 Thanks: 0 
Thanks, but how is that explicit if the answer contains p(1)?

November 7th, 2016, 06:19 AM  #4 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 732 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
I think the problem requires a boundary condition to be stated (P(1) or P(0)).

November 7th, 2016, 07:09 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,929 Thanks: 2205 
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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