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[MUG] Simplification problem
| [MUG] Simplification problem |
|
Author: Carl Eberhart
Posted: 28/11/2000 16:58:03 GMT
|
>> From: Carl Eberhart
Help,
For some reason, I can't get Maple 6 to simplify the expression
f :=
1/2*(2*2^(1/3)*x^(1/3)-x)*2^(2/3)*sqrt(4-2^(2/3)*x^(2/3))/(x^(1/3));
to
fs := (2^(4/3)-x^(2/3))^(3/2);
even when I say
simplify(f-fs,assume=real);
Even when I substitute x = 2 into the difference, it does not simplify
to 0.
What's wrong?
Carl
--
Carl Eberhart, Department of Mathematics
University of Kentucky, Lexington, KY 40506-0027
(606) 257-1258
http://www.ms.uky.edu/~carl/
|
| [MUG] Re: Simplification problem |
|
Author: Maple Group
Posted: 01/12/2000 20:14:29 GMT
|
>> From: Maple Group
| >> From: Carl Eberhart
| For some reason, I can't get Maple 6 to simplify the expression
| f :=
| 1/2*(2*2^(1/3)*x^(1/3)-x)*2^(2/3)*sqrt(4-2^(2/3)*x^(2/3))/(x^(1/3));
| to
| fs := (2^(4/3)-x^(2/3))^(3/2);
| even when I say
| simplify(f-fs,assume=real);
| Even when I substitute x = 2 into the difference, it does not simplify
| to 0.
-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-
From: "HARALD PLEYM"
To:
Subject: Simplification problem
Date: Wed, 29 Nov 2000 19:10:16 +0100
Dear Carl,
This is the closest I can get:
> f :=
> 1/2*(2*2^(1/3)*x^(1/3)-x)*2^(2/3)*sqrt(4-2^(2/3)*x^(2/3))/(x^(1/3));
>
(1/3) (1/3) (2/3) (2/3) (2/3)
(2 2 x - x) 2 sqrt(4 - 2 x )
f := 1/2 ----------------------------------------------------
(1/3)
x
> radnormal(f,rationalized);
(2/3) (2/3) (2/3) (2/3)
- 1/2 sqrt(4 - 2 x ) (-4 + 2 x )
> g:=simplify(sqrt(%^2));
(2/3) (2/3) 3
g := 1/2 sqrt(-(-4 + 2 x ) )
> simplify(subs(x=2,f-g));
0
Cheers
Harald
-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-
Date: Wed, 29 Nov 2000 16:26:39 -0500
From: "Douglas B. Meade"
To:
Subject: Simplification problem
Carl,
I'm sure you'll get this from others, but your expression is not 0 when
x=2. I'm sure you meant to say x=4.
As for the problem of simplifying the expression, I do not have a
complete answer. You probably want to eliminate the possibility for
square roots of negative numbers. For this I believe you want to assume
abs(x)<=4. However, this is not enough to resolve your problem in a
simple manner.
Others will certainly give you more assistance.
Doug
-----------------------------------------------------------------------
Douglas B. Meade Phone: (803) 777-6183 FAX: (803) 777-6527
Department of Mathematics URL: http://www.math.sc.edu/~meade/
USC, Columbia, SC 29208 E-mail:
|
| [MUG] Re: Simplification problem |
|
Author: Helmut Kahovec
Posted: 04/12/2000 15:33:46 GMT
|
>> From: Helmut Kahovec
Carl Eberhart wrote:
| For some reason, I can't get Maple 6 to simplify the expression
| f :=
| 1/2*(2*2^(1/3)*x^(1/3)-x)*2^(2/3)*sqrt(4-2^(2/3)*x^(2/3))/(x^(1/3));
| to
| fs := (2^(4/3)-x^(2/3))^(3/2);
| even when I say
| simplify(f-fs,assume=real);
| Even when I substitute x = 2 into the difference, it does not simplify
| to 0.
| What's wrong?
Hello Carl,
The following maple6 session will show that your two expressions are
equal if x is positive.
> restart;
> expr:=1/2*(2*2^(1/3)*x^(1/3)-x)*2^(2/3)*sqrt(4-2^(2/3)*x^(2/3))/(x^(1/3));
(1/3) (1/3) (2/3) (2/3) (2/3)
(2 2 x - x) 2 sqrt(4 - 2 x )
expr := 1/2 ----------------------------------------------------
(1/3)
x
> rules:=1/x^(1/3)=1/xi,x^(1/3)=xi,x^(2/3)=xi^2,x=xi^3,
> 2^(1/3)=alpha,2^(2/3)=alpha^2,2=alpha^3,4=alpha^6:
> map(lprint,[rules]):
1/x^(1/3) = 1/xi
x^(1/3) = xi
x^(2/3) = xi^2
x = xi^3
2^(1/3) = alpha
2^(2/3) = alpha^2
2 = alpha^3
4 = alpha^6
> invrules:=op(map(u->rhs(u)=lhs(u),[rules])):
> map(lprint,[invrules]):
1/xi = 1/x^(1/3)
xi = x^(1/3)
xi^2 = x^(2/3)
xi^3 = x
alpha = 2^(1/3)
alpha^2 = 2^(2/3)
alpha^3 = 2
alpha^6 = 4
> subs({rules},expr);
4 3 2 6 2 2
(alpha xi - xi ) alpha sqrt(alpha - alpha xi )
1/2 --------------------------------------------------
xi
> simplify(%,assume=positive);
4 2 3 4 2
- 1/2 (-alpha + xi ) alpha sqrt(alpha - xi )
> combine(%);
4 2 4 2 3
(1/2 alpha - 1/2 xi ) sqrt(alpha - xi ) alpha
> subs({invrules},%);
(1/3) (2/3) (1/3) (2/3)
2 (2 - 1/2 x ) sqrt(2 2 - x )
> factor(%);
(1/3) (2/3) (3/2)
(2 2 - x )
With kind regards,
Helmut
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