Mathematica 9 is now available
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: [mg64739] Re: Limit
  • From: "Scout" <Scout at nodomain.com>
  • Date: Thu, 2 Mar 2006 06:47:23 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

"Scout" <Scout at nodomain.com>
> "Sinan Kapçak" <sinankapcak at yahoo.com>
>>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,
 you could try to solve the recurrence equation that defines your continued
 fraction:
 i.e.
 f[x_]:= x+(x+1) / f[x+2];
 with x=1.
 Also try to solve separately the 2 recurrence equations:
 RSolve[{A[k + 1] == (2k - 1) A[k] + 2k A[k - 1], A[0] == 1, A[1] == 1},
 A[k], k]

 RSolve[{B[k + 1] == (2k - 1) B[k] + 2k B[k - 1], B[0] == 0, B[1] == 1},
 B[k], k]

 and the limit of their ratio exists and it is the value of the continued
 fraction:

 Limit[A[n] / B[n] , n->Infinity].

 You are now busy little job ;-)

    ~Scout~


  • Prev by Date: Help: running external program on Mac OS X
  • Next by Date: Re: Lists of all values of a two-variable function
  • Previous by thread: Re: Limit
  • Next by thread: Re: Limit