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Topic: cube root of a given number
Replies: 112   Last Post: Jan 10, 2013 1:39 PM

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Iain Davidson

Posts: 1,173
Registered: 12/12/04
Re: cube root of a given number
Posted: Aug 11, 2007 7:11 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On 11 Aug, 22:10, rich burge <r3...@aol.com> wrote:
> On Aug 11, 11:53?am, "sttscitr...@tesco.net" <sttscitr...@tesco.net>
> wrote:
>
>
>

> > Finding solutions to the cubic Pell is easy (use Pari/gp).
>
> > Do you mean systematically using generalizations of continued
> > fractions or
> > other methods for finding cubic units or just intelligent
> > trial and error ?

>
> What I had in mind was something like this:
>
> (12:39) gp > ?bnfinit
> bnfinit(P,{flag=0},{tech=[]}): compute the necessary data for future
> use in
> ideal and unit group computations, including fundamental units if they
> are not
> too large. flag and tech are both optional. flag can be any of 0:
> default, 1:
> insist on having fundamental units, 2: do not compute units, 3: small
> bnfinit,
> which can be converted to a big one using bnfmake. See manual for
> details
> about tech.
>
> (12:39) gp > a=bnfinit(X^3-25,1);
>
> (12:39) gp > a.fu
> %34 = [Mod(4/5*X^2 - 2*X - 1, X^3 - 25)]
>
> (12:41) gp > p3(k,x,y,z)=x^3+k*y^3+k^2*z^3-3*k*x*y*z
>
> (12:41) gp > p3(25,-1,-2,4/5)
> %35 = -1


I'm not familiar with Pari/gp, but there are methods using
the regulator and cycles of ideals that always find
fundametal units. Aren't they decribed in Cohen's book ?
What's the basic idea ? Isn't this a long way from (vector)
continued fraction algorithms ?

> I also have a homebrewed method for finding solutions that is pretty
> fast (it finds a 30000+ digit solution for k=1000700 in less than a
> minute) but comes without any guarantees about the solution being
> fundamental.


Yes, you have mentioned this before, but were not specific.
Didn't you say it was a probabilistic version of Jacobi -erron ?
> > Yes, I saw your recent post and Israel's interesting answer.
> > Have you found some way of estimating an upper bound
> > for the power of the unimodular matrix so that you can
> > find the fundamental unit after a small number of trials ?

>
> No, not really. Moveover, given any reasonable definition of "a small
> number of trials" I believe it is possible to show the approach I had
> in mind does not work. I have not played with it much, however.
>
> Here is a link to an article on this subject you may find interesting:
>
> Determining the Fundamental Unit of a Pure Cubic Field Given Any Unit
> N. S. Jeans; M. D. Hendy
> Mathematics of Computation, Vol. 32, No. 143. (Jul., 1978), pp.
> 925-935.


Thanks, that looks very interesting.






Date Subject Author
4/20/04
Read cube root of a given number
vsvasan
4/20/04
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Iain Davidson
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7/26/07
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Iain Davidson
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Iain Davidson
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Iain Davidson
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Iain Davidson
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8/5/07
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8/11/07
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