
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: 20,476 Thanks: 2039 
$P(k) = P(1)\dfrac{4^{k1}}{k!}$

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: 2,150 Thanks: 730 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: 20,476 Thanks: 2039 
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. 

Tags 
problem, recurrence, relation 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Recurrence Relation and Closed Form Relation  uniquegel  Algebra  4  September 8th, 2014 04:18 PM 
Recurrence relation  Joselynn  Real Analysis  2  September 14th, 2013 12:52 AM 
recurrence relation fn+4  fe phi fo  Applied Math  3  December 4th, 2011 09:22 AM 
Recurrence relation  kec11494  Applied Math  6  December 17th, 2010 11:57 PM 
Recurrence Relation  roguebyte  Advanced Statistics  1  December 31st, 1969 04:00 PM 