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Solve Polynomial

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  Solve Polynomial Thomas Dean <tomdean@speakeasy.net> - 2012-01-20 20:43 -0800
    Re: Solve Polynomial "Nasser M. Abbasi" <nma@12000.org> - 2012-01-20 23:52 -0600
      Re: Solve Polynomial "G. A. Edgar" <edgar@math.ohio-state.edu.invalid> - 2012-01-21 07:35 -0700
    Re: Solve Polynomial Thomas Dean <tomdean@speakeasy.net> - 2012-01-21 10:01 -0800
      Re: Solve Polynomial Thomas Dean <tomdean@speakeasy.net> - 2012-01-21 10:09 -0800
      Re: Solve Polynomial Peter Pein <petsie@dordos.net> - 2012-01-21 19:50 +0100
    Re: Solve Polynomial acer <maple@rogers.com> - 2012-01-21 10:04 -0800

#322 — Solve Polynomial

Maple 15 does not produce a solution to diff(eq,x)=0;  But, one exists.

eqn := (2*x^2-5*x+2)/(5*x^2-7*x-6);
deq:=diff(eqn,x);
solve(deq=0,x);
# returns no solution
eq2 := op(1, deq) = op(2, deq);
solve(eq2,x,AllSolutions);
returns two solutions as I expect.

numer(eqn) is zero at x=2 and x=1/2
denom(eqn) is zero at x=2 and x=-3/5

eqn is zero at x=1/2 and undefined at x=2 and x=-3/5.  I assume 0/0 is 
undefined.

limit(eqn,x=2) is 3/13.

How can I use solve to do produce a solution for diff(eq,x)=0?

Tom Dean

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#323

On 1/20/2012 10:43 PM, Thomas Dean wrote:
> Maple 15 does not produce a solution to diff(eq,x)=0;  But, one exists.
>
> eqn := (2*x^2-5*x+2)/(5*x^2-7*x-6);
> deq:=diff(eqn,x);
> solve(deq=0,x);
> # returns no solution

I think if you plot  deq as function of x you'll see why
there is no solution. deq is positive for all x.

> eq2 := op(1, deq) = op(2, deq);
> solve(eq2,x,AllSolutions);
> returns two solutions as I expect.
>
> numer(eqn) is zero at x=2 and x=1/2
> denom(eqn) is zero at x=2 and x=-3/5
>
> eqn is zero at x=1/2 and undefined at x=2 and x=-3/5.  I assume 0/0 is
> undefined.
>
> limit(eqn,x=2) is 3/13.
>
> How can I use solve to do produce a solution for diff(eq,x)=0?
>
> Tom Dean

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#324

In article <jfdjr1$smo$1@speranza.aioe.org>, Nasser M. Abbasi
<nma@12000.org> wrote:

> On 1/20/2012 10:43 PM, Thomas Dean wrote:
> > Maple 15 does not produce a solution to diff(eq,x)=0;  But, one exists.
> >
> > eqn := (2*x^2-5*x+2)/(5*x^2-7*x-6);
> > deq:=diff(eqn,x);
> > solve(deq=0,x);
> > # returns no solution
> 
> I think if you plot  deq as function of x you'll see why
> there is no solution. deq is positive for all x.

and, simplified, deq = 11/(5*x+3)^2, so we can see it is never zero.
However, it has a pole at x=-3/5, maybe that is what you want to find?

> 
> > eq2 := op(1, deq) = op(2, deq);
> > solve(eq2,x,AllSolutions);
> > returns two solutions as I expect.
> >
> > numer(eqn) is zero at x=2 and x=1/2
> > denom(eqn) is zero at x=2 and x=-3/5
> >
> > eqn is zero at x=1/2 and undefined at x=2 and x=-3/5.  I assume 0/0 is
> > undefined.
> >
> > limit(eqn,x=2) is 3/13.
> >
> > How can I use solve to do produce a solution for diff(eq,x)=0?
> >
> > Tom Dean
>

-- 
G. A. Edgar                              http://www.math.ohio-state.edu/~edgar/

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#325

On 01/20/12 20:43, Thomas Dean wrote:
> eq2 := op(1, deq) = op(2, deq);
> solve(eq2,x,AllSolutions);

eqn := (2*x^2-5*x+2)/(5*x^2-7*x-6);
                                        2
                                     2 x  - 5 x + 2
                              eqn := --------------
                                        2
                                     5 x  - 7 x - 6

deq:=diff(eqn,x);
                                           2
                         4 x - 5       (2 x  - 5 x + 2) (10 x - 7)
               deq := -------------- - ---------------------------
                         2                      2           2
                      5 x  - 7 x - 6        (5 x  - 7 x - 6)

solve(deq=0,x);
eq2 := op(1, deq) = op(2, deq);
                                            2
                        4 x - 5         (2 x  - 5 x + 2) (10 x - 7)
              eq2 := -------------- = - ---------------------------
                        2                        2           2
                     5 x  - 7 x - 6          (5 x  - 7 x - 6)

solve(eq2,x,AllSolutions);
                                    1/2           1/2
                           37   2649     37   2649
                           -- - -------, -- + -------
                           80     80     80     80

evalf(%);
                           -0.1808554616, 1.105855462

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#327

On 01/21/12 10:01, Thomas Dean wrote:
> On 01/20/12 20:43, Thomas Dean wrote:
>> eq2 := op(1, deq) = op(2, deq);
>> solve(eq2,x,AllSolutions);
>
> eqn := (2*x^2-5*x+2)/(5*x^2-7*x-6);
> 2
> 2 x - 5 x + 2
> eqn := --------------
> 2
> 5 x - 7 x - 6
>
> deq:=diff(eqn,x);
> 2
> 4 x - 5 (2 x - 5 x + 2) (10 x - 7)
> deq := -------------- - ---------------------------
> 2 2 2
> 5 x - 7 x - 6 (5 x - 7 x - 6)
>
> solve(deq=0,x);
> eq2 := op(1, deq) = op(2, deq);
> 2
> 4 x - 5 (2 x - 5 x + 2) (10 x - 7)
> eq2 := -------------- = - ---------------------------
> 2 2 2
> 5 x - 7 x - 6 (5 x - 7 x - 6)
>
> solve(eq2,x,AllSolutions);
> 1/2 1/2
> 37 2649 37 2649
> -- - -------, -- + -------
> 80 80 80 80
>
> evalf(%);
> -0.1808554616, 1.105855462
oops - I should have used -op(2,deq)

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#328

Am 21.01.2012 19:01, schrieb Thomas Dean:
> On 01/20/12 20:43, Thomas Dean wrote:
>> eq2 := op(1, deq) = op(2, deq);
>> solve(eq2,x,AllSolutions);
>
> eqn := (2*x^2-5*x+2)/(5*x^2-7*x-6);
> 2
> 2 x - 5 x + 2
> eqn := --------------
> 2
> 5 x - 7 x - 6
>
> deq:=diff(eqn,x);
> 2
> 4 x - 5 (2 x - 5 x + 2) (10 x - 7)
> deq := -------------- - ---------------------------
> 2 2 2
> 5 x - 7 x - 6 (5 x - 7 x - 6)
>
> solve(deq=0,x);
> eq2 := op(1, deq) = op(2, deq);
> 2
> 4 x - 5 (2 x - 5 x + 2) (10 x - 7)
> eq2 := -------------- = - ---------------------------
> 2 2 2
> 5 x - 7 x - 6 (5 x - 7 x - 6)
>
> solve(eq2,x,AllSolutions);
> 1/2 1/2
> 37 2649 37 2649
> -- - -------, -- + -------
> 80 80 80 80
>
> evalf(%);
> -0.1808554616, 1.105855462

If you want deq=0, you should look at
eq2 := op(1, deq) = -op(2, deq);
(please note the sign of op(2,deq);

Cheers,
Peter

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#326

Did you not forget a minus sign, when you wrote, 

   eq2 := op(1, deq) = op(2, deq);

for `deq` a sum of two operands?


acer

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