How do they do that?

[Mikito Harakiri]

> simplify((sqrt(5)+2)^(1/3)-(sqrt(5)-2)^(1/3));

1

I enabled trace level 50, or so, and was overwhelmed with information. It
seems like Maple tries to find a numerical approximation, first. Why
numerical approximation matters?

(I have read about this problem that roots in this expression are resistant
to any straightforward application of power operation; one have to consider
a cubic equation root of which this expression is).

[Richard Fateman]

Maybe they started by reading the literature, see, for
example,
http://citeseer.nj.nec.com/landau93note.html

As for why you might want a numerical approximation,
sometimes you want to know if expression > 0.
If you can determine that expression is between 1.5 and 1.6
but you don't know even the next digit, you can still answer
the question.

RJF

[Robert Israel]

In article <HFIIb.6$Mf4...@news.oracle.com>,

Mikito Harakiri <mikha...@iahu.com> wrote:
>> simplify((sqrt(5)+2)^(1/3)-(sqrt(5)-2)^(1/3));

> 1

>I enabled trace level 50, or so, and was overwhelmed with information. It
>seems like Maple tries to find a numerical approximation, first. Why
>numerical approximation matters?

I suspect that the fact that sqrt(5)-2 > 0 may be important here, in
deciding which branch of some multivalued function to choose. This
can be tricky.

Somewhat more generally, I believe
(sqrt(b^3+a^2)+a)^(1/3) - (sqrt(b^3+a^2)-a)^(1/3)
is one of the roots of z^3 + 3 b z - 2 a if b > 0 and a is real,
but not necessarily in other cases.

Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2

[steve_H]

"Mikito Harakiri" <mikha...@iahu.com> wrote in message news:<HFIIb.6$Mf4...@news.oracle.com>...
> > simplify((sqrt(5)+2)^(1/3)-(sqrt(5)-2)^(1/3));
>
> 1
>
>

This is funny, I can get 4/3 not 1. here is how.

using the relation,

exp( (1/3)*ln(x) ) = x^1/3

then write ( sqrt(5)+2 )^1/3 ---> exp( (1/3)*ln( sqrt(5)+2 ) )

then let x=5, and use series expansion for

exp( (1/3)*ln( sqrt(x)+2 ) )

and

exp( (1/3)*ln( sqrt(x)-2 ) )

and subtract the difference. here it is:

> a:=series( (1/3)*exp(ln(sqrt(x)+2)),x);

1 (1/2) 2
a := - x + -
3 3

> b:=series( (1/3)*exp(ln(sqrt(x)-2)),x);

1 (1/2) 2
b := - x - -
3 3

> a-b;

4
-
3

and not 1 . may be I made a mistake somewhere.

[Carl Devore]

On 31 Dec 2003, steve_H wrote:
> "Mikito Harakiri" <mikha...@iahu.com> wrote in message news:<HFIIb.6$Mf4...@news.oracle.com>...
> > > simplify((sqrt(5)+2)^(1/3)-(sqrt(5)-2)^(1/3));
> > 1

> exp( (1/3)*ln( sqrt(x)+2 ) )

> > a:=series( (1/3)*exp(ln(sqrt(x)+2)),x);

> and not 1 . may be I made a mistake somewhere.

Look at the different placement of the 1/3 in the two lines above.