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 comp.soft-sys.math.maple  4344 articles. 3 followers. 9 Replies 1456 Views Similar Articles

[PageSpeed] 0

```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
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