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January 4th, 2016, 12:57 PM  #1 
Newbie Joined: Dec 2014 From: Brooklyn Posts: 10 Thanks: 0  Solve this Recurrence Relation?
Can Anyone provide a function satisfying: f(1) = 1 f(x) = 1 + x/f(x) This isn't homework or anything, just something I stumbled across. 
January 4th, 2016, 03:14 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,464 Thanks: 2038 
Did you type the equations correctly? They don't specify a recurrence relation. The second equation implies f(1) = 1 + 1/f(1), which isn't true if f(1) = 1. 
January 4th, 2016, 05:11 PM  #3 
Newbie Joined: Dec 2014 From: Brooklyn Posts: 10 Thanks: 0 
Wow, my bad. Here is the correct version: f(1) = 1 f(x) = 1 + x/f(x1) for x > 1 Does that make more sense? 
January 4th, 2016, 06:33 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,464 Thanks: 2038 
Is f(x) defined only when x is a positive integer?

January 4th, 2016, 06:37 PM  #5 
Newbie Joined: Dec 2014 From: Brooklyn Posts: 10 Thanks: 0 
Yes

January 4th, 2016, 09:28 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,464 Thanks: 2038 
How did you "stumble across" this?

January 5th, 2016, 02:37 AM  #7 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,150 Thanks: 730 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Perhaps you mean: $\displaystyle x_{n+1} = 1 + \frac{n}{x_n}$ $\displaystyle x_1 = 1$ ? 
January 5th, 2016, 03:04 AM  #8 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,150 Thanks: 730 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Assuming the iteration scheme is as outlined by my previous post, I calculated the first 1000 values of the function (in Python) and it appears to be very slowly diverging as n increases. Here's a plot of the first 100:
Last edited by Benit13; January 5th, 2016 at 03:08 AM. 
January 5th, 2016, 04:51 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,464 Thanks: 2038 
For very large values of $n$, $x_n$ is approximately 1/2 + √$n$.


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