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Looking for Equation Solver

 
 
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      09-05-2005
Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> On 03 Sep 2005 22:51:13 GMT, http://www.velocityreviews.com/forums/(E-Mail Removed) <(E-Mail Removed)> wrote:
> > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> >> I wrote one
> >>
> >> http://www.algebra.com/services/rend...simplifier.mpl
> >>
> >> Besides simplifying, it shows work and plots math formulas. It is not
> >> available publicly though.

> >
> > Also, it doesn't give the right answer.
> >
> > "Text form: ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
> > simplifies to (x-3)/(x+2)"
> >
> > No, it doesn't.

>
> I will appreciate corrections and suggestions. What's the right answer
> there?


I didn't solve it fully; I just set x to zero which was enough to show the
simplification was wrong.

It appears that the correct answer is (x-3)^2/(x-2)/(x+2). It seem like
your program realizes that (x-3) is a common factor in two of the quadratic
groups, but then "cancels out" both of those groups in their entirety,
rather than just the factor of x-3.

Xho



>
> i


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Ignoramus25850
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      09-07-2005
Thanks guys (all of you). You got it. I corrected the solver, it was a
programming mistake. Thank you!

i

On Sun, 4 Sep 2005 21:29:30 +0200, Dave <(E-Mail Removed)> wrote:
>
> "Ignoramus14363" <ignoramus14363@NOSPAM.14363.invalid> wrote in message
> news:bdwSe.98947$(E-Mail Removed).. .
>> On Sat, 3 Sep 2005 22:47:50 -0700, Keith Keller
>> <(E-Mail Removed)-francisco.ca.us> wrote:
>>> On 2005-09-03, Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid>
>>> wrote:
>>>>
>>>> On 03 Sep 2005 22:51:13 GMT, (E-Mail Removed) <(E-Mail Removed)>
>>>> wrote:
>>>>>
>>>>> Also, it doesn't give the right answer.
>>>>>
>>>>> "Text form: ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
>>>>> simplifies to
>>>>> (x-3)/(x+2)"
>>>>>
>>>>> No, it doesn't.
>>>>
>>>> I will appreciate corrections and suggestions. What's the right answer
>>>> there?
>>>
>>> If you can't figure it out on paper, why are you writing a program to do
>>> it? Sheesh, I might as well go write an OS! :-\

>>
>> Let me know when you have something useful and substantial to say,
>> okay?
>>
>> (hint, I did solve it on paper)
>>
>> If you can explain why the above simplification is incorrect, I would
>> appreciate hearing that.
>>
>> i
>>

>
> The error in the explanation this example is at the line:
> 'Canceled out common factors (x-(3)),(x-(3))'
>
> Where (1) is left in each case instead of (x-3) and (x-2) in numerator and
> denominator respectively.
>
> The explantion would be clearer if you showed the factorisation before the
> cancellation, and this error would have been more obvious too.
>
> Hope this helps
>
> Dave
>
>



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Peyton Bland
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      11-14-2005

Hi,

I'm coming to the party a bit late, but you guys are scaring me, so I
had to reply (!)...

In article <20050904211634.239$(E-Mail Removed)>, <(E-Mail Removed)>
wrote:

> Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > On 03 Sep 2005 22:51:13 GMT, (E-Mail Removed) <(E-Mail Removed)> wrote:
> > > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > >> I wrote one
> > >>
> > >> http://www.algebra.com/services/rend...simplifier.mpl
> > >>
> > >> Besides simplifying, it shows work and plots math formulas. It is not
> > >> available publicly though.
> > >
> > > Also, it doesn't give the right answer.
> > >
> > > "Text form: ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
> > > simplifies to (x-3)/(x+2)"
> > >
> > > No, it doesn't.

> >
> > I will appreciate corrections and suggestions. What's the right answer
> > there?

>
> I didn't solve it fully; I just set x to zero which was enough to show the
> simplification was wrong.


I did this, too, and got -1 when x = 0. What did you get, Xho?

> It appears that the correct answer is (x-3)^2/(x-2)/(x+2). It seem like
> your program realizes that (x-3) is a common factor in two of the quadratic
> groups, but then "cancels out" both of those groups in their entirety,
> rather than just the factor of x-3.


Have another look... I get (x+2)/(x-2) . Hopefully, no one coerced
the software in question to yield any of these other answers given
earlier.

Cheers,
Peyton Bland
 
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Paul Lalli
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      11-14-2005
Peyton Bland wrote:
> I'm coming to the party a bit late, but you guys are scaring me, so I
> had to reply (!)...


It may have been wise to figure out what you're talking about before
reviving a 3 month old thread.

> In article <20050904211634.239$(E-Mail Removed)>, <(E-Mail Removed)>
> wrote:
>
> > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > > On 03 Sep 2005 22:51:13 GMT, (E-Mail Removed) <(E-Mail Removed)> wrote:
> > > >
> > > > Also, it doesn't give the right answer.
> > > >
> > > > "Text form: ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
> > > > simplifies to (x-3)/(x+2)"
> > > >
> > > > No, it doesn't.
> > >
> > > I will appreciate corrections and suggestions. What's the right answer
> > > there?

> >
> > I didn't solve it fully; I just set x to zero which was enough to show the
> > simplification was wrong.

>
> I did this, too, and got -1 when x = 0. What did you get, Xho?


My guess would be the correct answer that we all gave three months ago.
-9/4

((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))


(0^2-6*0+9)/(0^2-3*0-10)
------------------------
(0^2-5*0+6)/(0^2-8*0+15)

9 / -10
------
6 / 15

9 15
- -- * --
10 6

135 27 9
- --- = - -- = - -
60 12 4


(make sure you view the above in a fixed-width font)

> > It appears that the correct answer is (x-3)^2/(x-2)/(x+2). It seem like
> > your program realizes that (x-3) is a common factor in two of the quadratic
> > groups, but then "cancels out" both of those groups in their entirety,
> > rather than just the factor of x-3.

>
> Have another look... I get (x+2)/(x-2) .


Yes, well, you're wrong. See previous responses for how and why.

> Hopefully, no one coerced
> the software in question to yield any of these other answers given
> earlier.


Yes, improving the buggy software would certainly be worse than using a
working one to begin with.

Paul Lalli

 
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      11-14-2005
Peyton Bland <(E-Mail Removed)> wrote:
> Hi,
>
> I'm coming to the party a bit late, but you guys are scaring me, so I
> had to reply (!)...


That being the case, you might want to spend some more time explaining
what it is you are trying to say. It is hard to just pick up a thread
after 3 months.

> In article <20050904211634.239$(E-Mail Removed)>, <(E-Mail Removed)>
> wrote:
>
> > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > > On 03 Sep 2005 22:51:13 GMT, (E-Mail Removed) <(E-Mail Removed)>
> > > wrote:
> > > > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > > >> I wrote one
> > > >>
> > > >> http://www.algebra.com/services/rend...simplifier.mpl
> > > >>
> > > >> Besides simplifying, it shows work and plots math formulas. It is
> > > >> not available publicly though.
> > > >
> > > > Also, it doesn't give the right answer.
> > > >
> > > > "Text form: ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
> > > > simplifies to (x-3)/(x+2)"
> > > >
> > > > No, it doesn't.
> > >
> > > I will appreciate corrections and suggestions. What's the right
> > > answer there?

> >
> > I didn't solve it fully; I just set x to zero which was enough to show
> > the simplification was wrong.

>
> I did this, too, and got -1 when x = 0. What did you get, Xho?


Well, for the original full form, I got -9/4. For the original incorrect
simplification I got -3/2. For the correct simplification I got
(obviously) -9/4. For your new incorrect simplification I got -1.


>
> > It appears that the correct answer is (x-3)^2/(x-2)/(x+2). It seem
> > like your program realizes that (x-3) is a common factor in two of the
> > quadratic groups, but then "cancels out" both of those groups in their
> > entirety, rather than just the factor of x-3.

>
> Have another look...


Are you trying to tell me to double check my work, or are you telling me
that the bug has been fixed and I should go look at the original web page
again?


> I get (x+2)/(x-2) .


Which is obiously wrong, as (setting x=0) -9/4 != -1.

> Hopefully, no one coerced
> the software in question to yield any of these other answers given
> earlier.


I have no idea what that means.

Xho

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Peyton Bland
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      11-15-2005
In article <(E-Mail Removed). com>, Paul
Lalli <(E-Mail Removed)> wrote:

> Peyton Bland wrote:
> > I'm coming to the party a bit late, but you guys are scaring me, so I
> > had to reply (!)...

>
> It may have been wise to figure out what you're talking about before
> reviving a 3 month old thread.


My sincere apologies to the group. I carelessly mis-read the
expression (the ratio of polynomials).

> > In article <20050904211634.239$(E-Mail Removed)>, <(E-Mail Removed)>
> > wrote:
> >
> > > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > > > On 03 Sep 2005 22:51:13 GMT, (E-Mail Removed) <(E-Mail Removed)>
> > > > wrote:
> > > > >
> > > > > Also, it doesn't give the right answer.
> > > > >
> > > > > "Text form: ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
> > > > > simplifies to (x-3)/(x+2)"
> > > > >
> > > > > No, it doesn't.
> > > >
> > > > I will appreciate corrections and suggestions. What's the right answer
> > > > there?
> > >
> > > I didn't solve it fully; I just set x to zero which was enough to show the
> > > simplification was wrong.

> >
> > I did this, too, and got -1 when x = 0. What did you get, Xho?

>
> My guess would be the correct answer that we all gave three months ago.
> -9/4
>
> ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
>
>
> (0^2-6*0+9)/(0^2-3*0-10)
> ------------------------
> (0^2-5*0+6)/(0^2-8*0+15)
>
> 9 / -10
> ------
> 6 / 15
>
> 9 15
> - -- * --
> 10 6
>
> 135 27 9
> - --- = - -- = - -
> 60 12 4
>
>
> (make sure you view the above in a fixed-width font)
>
> > > It appears that the correct answer is (x-3)^2/(x-2)/(x+2). It seem like
> > > your program realizes that (x-3) is a common factor in two of the
> > > quadratic
> > > groups, but then "cancels out" both of those groups in their entirety,
> > > rather than just the factor of x-3.

> >
> > Have another look... I get (x+2)/(x-2) .

>
> Yes, well, you're wrong. See previous responses for how and why.
>
> > Hopefully, no one coerced
> > the software in question to yield any of these other answers given
> > earlier.

>
> Yes, improving the buggy software would certainly be worse than using a
> working one to begin with.
>
> Paul Lalli


Peyton Bland
 
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