Solve Polynomial

Thomas Dean

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Jan 20, 2012, 11:43:53 PM1/20/12

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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

Nasser M. Abbasi

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Jan 21, 2012, 12:52:33 AM1/21/12

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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.

G. A. Edgar

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Jan 21, 2012, 9:35:15 AM1/21/12

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

<n...@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/

Thomas Dean

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Jan 21, 2012, 1:01:53 PM1/21/12

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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

acer

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Jan 21, 2012, 1:04:56 PM1/21/12

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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

Thomas Dean

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Jan 21, 2012, 1:09:31 PM1/21/12

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oops - I should have used -op(2,deq)

Peter Pein

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Jan 21, 2012, 1:50:13 PM1/21/12

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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