Prev: Unsolvability of the Entscheidungsproblem as a Corollary of Gödel’s 2nd Theorem
Next: about squarefree integers-
From: lloyd on 31 May 2010 15:46
Hi - simple question: does anyone know a website where I can enter a
number in symbolic form and get a list of semiconvergents? It's the
kind of thing I thought WolframAlpha would be able to do, but
apparently it understands continuedFraction(whatever) but not
semiconvergents(whatever). Failing that, if someone wants to reply
with a list of semiconvergents to 3^(6/13), that would be appreciated.
From: Glenn on 13 Jun 2010 11:41
On May 31, 12:46 pm, lloyd <lloyd.hough...(a)gmail.com> wrote:
> Hi - simple question: does anyone know a website where I can enter a
> number in symbolic form and get a list of semiconvergents? It's the
> kind of thing I thought WolframAlpha would be able to do, but
> apparently it understandscontinuedFraction(whatever) but not
> semiconvergents(whatever). Failing that, if someone wants to reply
> with a list of semiconvergents to 3^(6/13), that would be appreciated.

http://wims.unice.fr/wims/wims.cgi?lang=en&module=tool/number/contfrac.en&cmd=new
This should do the trick for you.
From: Glenn on 13 Jun 2010 12:12
http://wims.unice.fr/wims/wims.cgi
Then click on "Contfrac" to get the program.
You'll see the following menu:
==============================
Number to : n =
Number of terms in the expansion: 10203050100200300500

Use n = 3^(6/13) and select "30" terms for the expansion.
When you click "expand" you'll get the following output:
=============================================
Continued fraction expansion of n = 3^(6/13):

1.66038885600108665650778885021671080466 = 1 + 1/1+ 1/1+ 1/1+ 1/17+
1/30+ 1/1+ 1/9+ 1/1+ 1/2+ 1/3+ 1/1+ 1/1+ 1/1+ 1/1+ 1/2+ 1/1+ 1/42+
1/10+ 1/5+ 1/2+ 1/6+ 1/47+ 1/25+ 1/8+ 1/3+ 1/1+ 1/2+ 1/1+ 1/2+ . . .

With javascript, placing the mouse over a denominator will show you
the convergent of the corresponding term (limited precision):
=======================================
Clicking on any of terms 2 through 27 will give you the corresponding
convergent.
Only the first 16 significant digits are accurate,
so the last three "convergents" contain roundoff errors.

Hope that helps!
From: lloyd on 29 Jun 2010 17:30
Thank you Glenn, that is perfect!
 | 
Pages: 1
Prev: Unsolvability of the Entscheidungsproblem as a Corollary of Gödel’s 2nd Theorem
Next: about squarefree integers-