My Math Forum Solve this Recurrence Relation?

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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(x-1) 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
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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:
Attached Images
 newval.jpg (7.4 KB, 5 views)

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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