To satisfy everybody there is the following way. I defined two functions.
"|" and nth_roots(), | is a infix operator which performs multiplication
when both arguments are lists. Here is the output.
(%i1) n_roots(-3,3);
5/6 1/3 5/6 1/3
3 %i - 3 3 %i + 3 1/3
(%o1)
[- --------------, --------------, - 3 ]
2 2
(%i2) display2d:false;
(%o2) false
(%i3) n_roots(2,2);
(%o3) [-sqrt(2),sqrt(2)]
(%i4) n_roots(3,5) | n_roots(-2,3);
(%o4)
[-2^(1/3)*3^(1/5),3^(1/5)*(2^(1/3)*sqrt(3)*%i+2^(1/3))/2,-3^(1/5)*(2^(1/3)*sqrt(3)*%i-2^(1/3))/2,-2^(1/3)*3^(1/5)*%e^-(2*%i*%pi/5),3^(1/5)*(2^(1/3)*sqrt(3)*%i+2^(1/3))*%e^-(2*%i*%pi/5)/2,
-3^(1/5)*(2^(1/3)*sqrt(3)*%i-2^(1/3))*%e^-(2*%i*%pi/5)/2,-2^(1/3)*3^(1/5)*%e^-(4*%i*%pi/5),3^(1/5)*(2^(1/3)*sqrt(3)*%i+2^(1/3))*%e^-(4*%i*%pi/5)/2,
-3^(1/5)*(2^(1/3)*sqrt(3)*%i-2^(1/3))*%e^-(4*%i*%pi/5)/2,-2^(1/3)*3^(1/5)*%e^(4*%i*%pi/5),3^(1/5)*(2^(1/3)*sqrt(3)*%i+2^(1/3))*%e^(4*%i*%pi/5)/2,-3^(1/5)*(2^(1/3)*sqrt(3)*%i-2^(1/3))*%e^(4*%i*%pi/5)/2,
-2^(1/3)*3^(1/5)*%e^(2*%i*%pi/5),3^(1/5)*(2^(1/3)*sqrt(3)*%i+2^(1/3))*%e^(2*%i*%pi/5)/2,-3^(1/5)*(2^(1/3)*sqrt(3)*%i-2^(1/3))*%e^(2*%i*%pi/5)/2]
(%i5)
Here is the code. The multiplication or addition of all of the roots of the
two numbers can be a very complicated expression. In this case the list
length = 5*3 = 15. That is why you need |. You can't overload * directly,
I tried but got an error about * is a built in operator. You would need to
handle + to when both sides are lists.
Here is the code, it is very simple. It allows for multiple answers to a
root calculation and comes very close to the definition that x^(1/n) =
solve([y^(1/n)=x], y);
n_roots(__c,__n):=
block(
[_x],
map(lambda([_z], rhs(_z)), solve(_x^__n=__c,_x))
);
infix("|", 135, 135);
infix("++", 101,101);
"|"(_l1,_l2):=
block
(
[_ans:[]],
if listp(_l1) and listp(_l2) then
(
for _p in _l1 do (
for _q in _l2 do (
_ans : cons(_p*_q, _ans)
)
)
)
elseif listp(_l1) and not listp(_l2) or listp(_l2) and not listp(_l1)
then
(
_ans : _l2 * _l1
)
else
(
_ans : _l2 * _l1
),
_ans
)$
"++"(_l1,_l2):=
block
(
[_ans:[]],
if listp(_l1) and listp(_l2) then
(
for _p in _l1 do (
for _q in _l2 do (
_ans : cons(_p+_q, _ans)
)
)
)
elseif listp(_l1) and not listp(_l2) or listp(_l2) and not listp(_l1)
then
(
_ans : _l2 + _l1
)
else
(
_ans : _l2 + _l1
),
_ans
)$
This is possible to do in Maxima 5.23.2 so you don't need to worry about
sqrt(x^2) = abs(x) if you use n_roots(x^2,2). I am not sure I like the name
n_roots(), I was calling it nth_root() at some point but that is not so
great either. The multitude of all possible answers is explicit in this
case, you can get very long lists when multiplication or addition is
involved.
n_roots(x^2,2);
[- x, x]
Rich
----Original Message-----
From: Richard Fateman
Sent: Friday, February 11, 2011 10:10 AM
To: Ren? K?lin
Cc: maxima at math.utexas.edu
Subject: Re: [Maxima] sqrt(x)*sqrt(x)
On 2/9/2011 11:50 PM, "Ren? K?lin" wrote:
> Hello
>
> I tried this (with Maxima 5.21.1):
>
> domain:real$
> t:x^(1/2)*x^(1/2);
> assume(x<0);
> t;
> forget(x<0)$
> assume(x>=0);
> t;
>
> with the output:
>
> (%o2) x
> (%o3) [x<0]
> (%o4) x
> (%o6) [x>=0]
> (%o7) x
>
> I only want to consider some real numbers. Because the squareroot of x is
> not defined for x<0, %o4 is wrong, isn't it?
>
> The Problem seems to be that t:x^(1/2)*x^(1/2) is simplified immediately
> to x in any case.
>
> (I sent this post 2 days ago with a wrong title. Sorry if you received it
> twice.)
It seems hopeless to point out that x>0 does not mean that sqrt(x)>0,
mathematically.
There are 2 square roots. For example sqrt(16) is {-4, 4}, even though
16>0.
The square root of x, for x<0 IS defined. It just happens to be imaginary.
There is no simple way of making Maxima forget about complex numbers,
even if you decide that certain of the inputs represent real numbers.
What you seem to be insisting on is
not that x>0, but that sqrt(x)>=0.
Perhaps you should
define a new function positive_sqrt(x):= .... if (x<0) then error else
....
Yes, I know that in Maxima sqrt(z^2) comes out abs(z), but if you do
assume(z>0), then abs(z) comes out as z.
This first step, sqrt(z^2) to abs(z) is, in general, wrong, so there
you are, with the subsequent step propagating the questionable result.
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