Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2006
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2006

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Limit

  • To: mathgroup at smc.vnet.net
  • Subject: [mg64693] Re: Limit
  • From: Peter Pein <petsie at dordos.net>
  • Date: Tue, 28 Feb 2006 01:49:25 -0500 (EST)
  • References: <dtrvmt$lv0$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Sinan Kapçak schrieb:
> i want to find the value of the limit 
> 
> 1+(2/(3+(4/(5+(6/(7+... 
> 
> how can i do that with Mathematica?
> 
> Tnx..
> 

Hi Sinan,

First write a function to approximate your continued fraction:

cf[n_] := 1 + Fold[(#2 - 1)/(#1 + #2) & , 2*n + 2, Reverse[2*Range[n] + 1]]

(*Test*)
cf[3] - (1 + 2/(3 + 4/(5 + 6/(7 + 8))))
--> 0

Let's have a look, if it converges (and how fast):
N[cf /@ Range[10]]
-->
{1.2857142857142858, 1.5945945945945945, 1.5346534653465347, 1.5422045680238332,
   1.5414334169814963, 1.5414984960673195, 1.541493802752502, 1.5414940982569976,
   1.5414940817435092, 1.5414940825731342}

With the help of http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html one could guess
N[1/(Sqrt[E] - 1)]
-->1.5414940825367982
is the value of the c.f.

At least the first 150 Digits are the same:
N[cf[100], 150] == 1/(Sqrt[E] - 1)
--> True

If I only had paid more attention to the number theory lectures... :-\

Peter


  • Prev by Date: Re: finding the position of a pattern in list
  • Next by Date: code efficiency
  • Previous by thread: Re: Limit
  • Next by thread: Re: Limit