f



x^(3/2)+x^(5/2) = (x+x^2)*x^(1/2)

Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to
(x+x^2)*x^(1/2).
I am willing to assume(x>0).

Maple does recognize these as being equal:

> p:=x^(3/2)+x^(5/2):
> q:=(x+x^2)*sqrt(x):
> simplify(p-q);

                          0

Here's a clumsy way to do it:

> p:=x^(3/2)+x^(5/2):
> assume(t>0):
> subs(x=t^2,p);
> simplify(%,{t^2=x});
> collect(%,t);

Is there something simpler?

(This has undoubtedly come up before, but I cannot find it anywhere.)

--Edwin




0
Edwin
12/29/2003 10:02:34 PM
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If you do not insist upon your incomplete factorization:
 > x^(3/2) + x^(5/2);

                             (3/2)    (5/2)
                            x      + x

 > factor(%);

                              (3/2)
                             x      (1 + x)


All the best    Jurgen

Edwin Clark schrieb:
> Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to
> (x+x^2)*x^(1/2).
> I am willing to assume(x>0).
> 
> Maple does recognize these as being equal:
> 
> 
>>p:=x^(3/2)+x^(5/2):
>>q:=(x+x^2)*sqrt(x):
>>simplify(p-q);
> 
> 
>                           0
> 
> Here's a clumsy way to do it:
> 
> 
>>p:=x^(3/2)+x^(5/2):
>>assume(t>0):
>>subs(x=t^2,p);
>>simplify(%,{t^2=x});
>>collect(%,t);
> 
> 
> Is there something simpler?
> 
> (This has undoubtedly come up before, but I cannot find it anywhere.)
> 
> --Edwin
> 
> 
> 
> 

0
J
12/30/2003 12:51:42 AM
I wrote:

> Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to
> (x+x^2)*x^(1/2).
> I am willing to assume(x>0).

------
>
> Here's a clumsy way to do it:
>
> > p:=x^(3/2)+x^(5/2):
> > assume(t>0):
> > subs(x=t^2,p);
> > simplify(%,{t^2=x});
> > collect(%,t);

When I copied this bit of code in for some reason the last line was not
copied. It should include:

> subs(t=sqrt(x),%);


>
> Is there something simpler?
>
> (This has undoubtedly come up before, but I cannot find it anywhere.)
>
> --Edwin




0
Edwin
12/30/2003 1:10:10 AM
Edwin Clark wrote:

>| Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to
>| (x+x^2)*x^(1/2).
>| I am willing to assume(x>0).


Well, you may do the following:

> restart;
> p:=x^(3/2)+x^(5/2);

                               (3/2)    (5/2)
                         p := x      + x

> series(p/sqrt(x),x)*sqrt(x);

                                  2   1/2
                            (x + x ) x

> whattype(%);

                                  *


Helmut
0
Helmut
12/30/2003 4:04:09 AM
"Edwin Clark" <eclark@math.usf.edu> wrote in message news:<_v1Ib.153184$b01.3400427@twister.tampabay.rr.com>...
> Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to
> (x+x^2)*x^(1/2).
> I am willing to assume(x>0).
> Is there something simpler?
> --Edwin

Why not x^(3/2)+x^(5/2)=x^(3/2)*(1+x)?
> x^(3/2)+x^(5/2):
  %=factor(%,sqrt(x)) assuming x>0;
            x^(3/2)+x^(5/2)=x^(3/2)*(1+x)

Ken
0
maplemath
12/30/2003 4:10:45 AM
Helmut Kahovec wrote:

>| Edwin Clark wrote:
>| 
>| >| Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to
>| >| (x+x^2)*x^(1/2).
>| >| I am willing to assume(x>0).
>| 
>| Well, you may do the following:
>| 
>| > restart;
>| > p:=x^(3/2)+x^(5/2);
>| 
>|                                (3/2)    (5/2)
>|                          p := x      + x
>| 
>| > series(p/sqrt(x),x)*sqrt(x);
>| 
>|                                   2   1/2
>|                             (x + x ) x
>| 
>| > whattype(%);
>| 
>|                                   *


Sorry! I forgot converting the series object back to a polynomial:

> restart;
> p:=x^(3/2)+x^(5/2);

                               (3/2)    (5/2)
                         p := x      + x

> series(p/sqrt(x),x)*sqrt(x);

                                  2   1/2
                            (x + x ) x

> map(convert,%,polynom);

                                  2   1/2
                            (x + x ) x

Helmut
0
Helmut
12/30/2003 4:31:19 AM
"Helmut Kahovec" wrote

> Edwin Clark wrote:
>
> >| Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to
> >| (x+x^2)*x^(1/2).
> >| I am willing to assume(x>0).
>
>
> Well, you may do the following:
>
> > restart;
> > p:=x^(3/2)+x^(5/2);
>
>                                (3/2)    (5/2)
>                          p := x      + x
>
> > series(p/sqrt(x),x)*sqrt(x);
>
>                                   2   1/2
>                             (x + x ) x
>
Thanks Helmut,

I notice that you can also get the result by expand(p/sqrt(x))*sqrt(x).

But both of these run into trouble if p has a constant term say
p=x^(3/2)+x^(5/2) + 1;

I was also thinking of things like a + b*x^(3/2) + c*y^(5/2) +
d*x^(7/2)*y^(5/2). In general, how does one convert all terms of the form
z^((2*n+1)/2) to z^n*sqrt(z) and maintain this form while collecting
coefficients of sqrt(z) . Maple does this automatically if x, y , z are
positive integers. For example:

> 2^(5/2) + 3^(7/2)*5^(5/2);

                           1/2        1/2  1/2
                        4 2    + 675 3    5

--Edwin





0
Edwin
12/30/2003 4:57:14 AM
"Ken Lin" wrote
> "Edwin Clark" wrote
> > Is there a simple way to get Maple to convert x^(3/2) + x^(5/2) to
> > (x+x^2)*x^(1/2).
> > I am willing to assume(x>0).
>
> Why not x^(3/2)+x^(5/2)=x^(3/2)*(1+x)?

Because I want to express everything in terms of sqrt(x). As I mentioned in
my reply to Helmut I really would like all subexpressions like x^(2*n+1) to
be
converted to x^n*sqrt(x) where there could be several such in an expression
and several variables not just x.

The following works except for subexpressions like y^(7/2)*x^(5/2):

> p:= 1+x^(3/2) + x^(7/2) + 2*y^(7/2)*x^(5/2) + y^(9/2)+3*y^(5/2):
> assume(t>0,r>0):
> subs({x=t^2,y = r^2},p):
> simplify(%,{t^2=x,r^2=y}):
> collect(%,[t,r], distributed):
> subs({t=sqrt(x),r=sqrt(y)},%);

               (7/2)  (5/2)     3       1/2     4      2   1/2
        1 + 2 y      x      + (x  + x) x    + (y  + 3 y ) y

--Edwin



0
Edwin
12/30/2003 5:48:47 AM
Edwin Clark wrote:

>| I notice that you can also get the result by expand(p/sqrt(x))*sqrt(x).
>| 
>| But both of these run into trouble if p has a constant term say
>| p=x^(3/2)+x^(5/2) + 1;
>| 
>| I was also thinking of things like a + b*x^(3/2) + c*y^(5/2) +
>| d*x^(7/2)*y^(5/2). In general, how does one convert all terms of the form
>| z^((2*n+1)/2) to z^n*sqrt(z) and maintain this form while collecting
>| coefficients of sqrt(z) . Maple does this automatically if x, y , z are
>| positive integers. For example:
>| 
>| > 2^(5/2) + 3^(7/2)*5^(5/2);
>| 
>|                            1/2        1/2  1/2
>|                         4 2    + 675 3    5


Well, I do not think that this will be possible, since Maple automatically
rewrites expressions like x*sqrt(x) as x^(3/2). It is analogous to
automatically expanding 2*(x+y) to 2*x+2*y.

However, you may try the following:

> restart;
> p:=1+x^(3/2)+x^(7/2)+2*y^(7/2)*x^(5/2)+y^(9/2)+3*y^(5/2);

              (3/2)    (7/2)      (7/2)  (5/2)    (9/2)      (5/2)
    p := 1 + x      + x      + 2 y      x      + y      + 3 y


You may assign the order term of the series objects any sufficiently large
number:

> N:=1000:

Remember that expressions within parentheses are of type series:

> subsindets(
    p,
    `^`,
    u->series(u/op(1,u)^(1/2),op(1,u),N)*op(1,u)^(1/2)
  );

           1/2     3   1/2       3   1/2   2   1/2     4   1/2
  1 + (x) x    + (x ) x    + 2 (y ) y    (x ) x    + (y ) y

               2   1/2
         + 3 (y ) y

Transforming back to polynomials triggers Maple's automatic simplifier:

> subsindets(
    %,
    series,
    u->convert(u,polynom)
  );

           (3/2)    (7/2)      (7/2)  (5/2)    (9/2)      (5/2)
      1 + x      + x      + 2 y      x      + y      + 3 y


Helmut
0
Helmut
12/31/2003 3:09:15 AM
Edwin Clark wrote:

>| The following works except for subexpressions like y^(7/2)*x^(5/2):
>| 
>| > p:= 1+x^(3/2) + x^(7/2) + 2*y^(7/2)*x^(5/2) + y^(9/2)+3*y^(5/2):
>| > assume(t>0,r>0):
>| > subs({x=t^2,y = r^2},p):
>| > simplify(%,{t^2=x,r^2=y}):
>| > collect(%,[t,r], distributed):
>| > subs({t=sqrt(x),r=sqrt(y)},%);
>| 
>|                (7/2)  (5/2)     3       1/2     4      2   1/2
>|         1 + 2 y      x      + (x  + x) x    + (y  + 3 y ) y


Well, as I indicated in my previous message, you may alternatively do the
following:

> restart;
> p:=1+x^(3/2)+x^(7/2)+2*y^(7/2)*x^(5/2)+y^(9/2)+3*y^(5/2);

              (3/2)    (7/2)      (7/2)  (5/2)    (9/2)      (5/2)
    p := 1 + x      + x      + 2 y      x      + y      + 3 y

> subsindets(
    p,
    `^`,
    u->series(u/op(1,u)^(1/2),op(1,u),1000)*op(1,u)^(1/2)
  );

           1/2     3   1/2       3   1/2   2   1/2     4   1/2
  1 + (x) x    + (x ) x    + 2 (y ) y    (x ) x    + (y ) y

               2   1/2
         + 3 (y ) y

> collect(%,[sqrt(x),sqrt(y)],distributed);

               3    1/2      4        2    1/2
  1 + ((x) + (x )) x    + ((y ) + 3 (y )) y

               3   1/2   2   1/2
         + 2 (y ) y    (x ) x

> subsindets(
    %,
    series,
    u->convert(u,polynom)
  );

                  3   1/2     4      2   1/2      (7/2)  (5/2)
        1 + (x + x ) x    + (y  + 3 y ) y    + 2 y      x


Helmut
0
Helmut
12/31/2003 3:24:27 AM
Reply:

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