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From: Robert Israel on 13 Jan 2010 19:03
"Achava Nakhash, the Loving Snake" <achava(a)hotmail.com> writes:

> On Jan 13, 4:00=A0am, Ross <rossclem...(a)gmail.com> wrote:
> > PS: I think I've just properly understood. The Planet Math equations
> > are correct, but produce complex numbers in their solution, as said.
> > This causes difficulty to me, as I'm creating a computational example,
> > and upgrading the code to handle complex numbers would be too
> > difficult and computationally expensive. Since sqrt( a ) =3D sqrt( -1 * -
> > a ) =3D i sqrt( -a ), I could do the algebra to handle the case where
> > the operand of the sqrt is negative, by solving the equation to remove
> > i, then using that resulting equation when the operand is negative.
> > And of course, the other "more complicated looking" solutions use
> > "discriminants" to branch in the solution structure.
> >
> > So, I'd guess that the lengthier solution that does work for me is a
> > similar, or even the same, solution method. Just that more steps have
> > been done so that no complex arithmetic is performed. In which case, I
> > have a solution, and it's probably as simple as any I'm going to get.
> >
> > This is similar to designing digital filters by direct pole/zero
> > placement on an Argand diagram. When the poles or zeroes are not on
> > the real axis, they must be in conjugate pairs (r,omega), (r,-omega),
> > so that when the filter different equation is derived, the imaginary
> > parts drop out of the working, and final equation.
>
> Are you sure that you want to use closed-form solutions for solving
> these equations on the computer? A glance at the formula for solving
> quadratic equations shows you that it can be very sensitive to small
> changes in the coefficients, and my practical experience with solving
> quadratic equations on the computer showed that the result were often
> quite inaccurate. I finally ditched that approach and used the Newton
> method. I would not be a bit surprised if that approach is far more
> stable and accurate for cubic equations as well, particularly if you
> know in advance that you only get one root.

The problem with quadratics is that in the usual quadratic formula
(-b (+/-) sqrt(b^2 - 4 a c))/(2 a)
the root close to 0 suffers from "catastrophic cancellation" if |b^2| is
large compared to |4 a c|. That can be cured by using the alternative
formula

-2 c/(b (+/-) sqrt(b^2 - 4 a c))

for that root.

I don't know if the cubic suffers from similar problems, or if there is a
similar cure.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Gib Bogle on 14 Jan 2010 00:38
Ross wrote:
> Are the formulae for findin the roots of cubic equations on Wikipedia
> correct? I've been through it over and over, and simply don't get
> roots.
>
> I've gone to the original source, which is:
> http://planetmath.org/?op=getobj&from=objects&name=CubicFormula
>
> Just looking at the first equation to calculate r1, then if I have the
> cubic x^3 -2x^2 -11x + 12, then I get NaNs when I try to find r1.
>
> Specifically, if I set a1=-2, b1=-11, and c1=12, then the value of
> which the square root is taken:
> (2*a1^3 - 9*a1*b1 + 27*c1)^2 + 4*(-(a1^2)+3*b1)^3
>
> works out to be negative, leading to the NaN result.
>
> Here's a cut and paste from "bc" where I first confirm that my
> variables are set properly, then work out the operand of the square
> root.
>
> a1
> -2
> b1
> -11
> c1
> 12
> (2*a1^3 - 9*a1*b1 + 27*c1)^2 + 4*(-(a1^2)+3*b1)^3
> -190512
>
> Negative!
>
> I've also tried the modified form of these equations from the
> Wikipedia page, but I get the same result. NaNs.
>
> I've really tried to find the error in my calculations or reasoning,
> but I just can't see any. Are these equations correct? Am I misusing
> them?
>
> I can successfully use the method in this page: http://www.1728.com/cubic2.htm.
> However, I'd still like to know what I'm doing wrong with the
> Wikipedia method.
>
> Any hints/help?

I've just been reading A History of Mathematics by Boyer (and Merzbach).
The cubic gave the old-timers plenty to think about, and solving it
was a great feat. As David Ullrich mentioned, the occurrence of square
roots of negative numbers in a cubic with three real roots was a real
head-scratcher.
From: David Bernier on 14 Jan 2010 01:04
Gib Bogle wrote:
> Ross wrote:
[...]

>> 12
>> (2*a1^3 - 9*a1*b1 + 27*c1)^2 + 4*(-(a1^2)+3*b1)^3
>> -190512
>>
>> Negative!
>>
>> I've also tried the modified form of these equations from the
>> Wikipedia page, but I get the same result. NaNs.
>>
>> I've really tried to find the error in my calculations or reasoning,
>> but I just can't see any. Are these equations correct? Am I misusing
>> them?
>>
>> I can successfully use the method in this page:
>> http://www.1728.com/cubic2.htm.
>> However, I'd still like to know what I'm doing wrong with the
>> Wikipedia method.
>>
>> Any hints/help?
>
> I've just been reading A History of Mathematics by Boyer (and Merzbach).
> The cubic gave the old-timers plenty to think about, and solving it was
> a great feat. As David Ullrich mentioned, the occurrence of square
> roots of negative numbers in a cubic with three real roots was a real
> head-scratcher.

It would be interesting to know which came first:
- trigonometric and inverse trigonometric functions used to solve
casus irreducibilis equations or:
- the ( Argand) complex plane .

David Bernier
From: Adam on 16 Jan 2010 14:03
On Wed, 13 Jan 2010 04:51:18 -0600, David C. Ullrich wrote:

>On Wed, 13 Jan 2010 00:56:36 -0800 (PST), Ross <rossclement(a)gmail.com>
>wrote:
>
>>Are the formulae for findin the roots of cubic equations on Wikipedia
>>correct? I've been through it over and over, and simply don't get
>>roots.
>>
>>I've gone to the original source, which is:
>>http://planetmath.org/?op=getobj&from=objects&name=CubicFormula
>>
>>Just looking at the first equation to calculate r1, then if I have the
>>cubic x^3 -2x^2 -11x + 12, then I get NaNs when I try to find r1.
>>
>>Specifically, if I set a1=-2, b1=-11, and c1=12, then the value of
>>which the square root is taken:
>>(2*a1^3 - 9*a1*b1 + 27*c1)^2 + 4*(-(a1^2)+3*b1)^3
>>
>>works out to be negative, leading to the NaN result.
>
>No, the square root of a negative number is a complex number.
>Complex numbers come into the formula even when the roots
>of the original cubic turn out to be real.
>
>This has a lot to do with the history of complex numbers.
>Way back then nobody was interested in saying that
>x^2 = -1 had a solution because it obviously didn't.
>But these funny "imaginary" numbers did arise in the
>solution to cubics with _real_ roots - hence they couldn't
>be just ignored. At first they were just "imaginary"
>temporary things that luckily went away when we got the
>final answer - people gradually decided they were interesing
>in themselves.

Imaginary numbers are the "dark matter" of math?

They outnumber real numbers.

From: Don Redmond on 16 Jan 2010 14:47
On Jan 14, 12:04 am, David Bernier <david...(a)videotron.ca> wrote:
> Gib Bogle wrote:
> > Ross wrote:
>
> [...]
>
>
>
> >> 12
> >> (2*a1^3 - 9*a1*b1 + 27*c1)^2 + 4*(-(a1^2)+3*b1)^3
> >> -190512
>
> >> Negative!
>
> >> I've also tried the modified form of these equations from the
> >> Wikipedia page, but I get the same result. NaNs.
>
> >> I've really tried to find the error in my calculations or reasoning,
> >> but I just can't see any. Are these equations correct? Am I misusing
> >> them?
>
> >> I can successfully use the method in this page:
> >>http://www.1728.com/cubic2.htm.
> >> However, I'd still like to know what I'm doing wrong with the
> >> Wikipedia method.
>
> >> Any hints/help?
>
> > I've just been reading A History of Mathematics by Boyer (and Merzbach).
> > The cubic gave the old-timers plenty to think about, and solving it was
> > a great feat. As David Ullrich mentioned, the occurrence of square
> > roots of negative numbers in a cubic with three real roots was a real
> > head-scratcher.
>
> It would be interesting to know which came first:
> - trigonometric and inverse trigonometric functions used to solve
> casus irreducibilis equations or:
> - the ( Argand) complex plane .
>
> David Bernier

Viete, at least, in 1500's did the trig soln of cubics. Argand and
Wessel
did the Argand diagram in the 1700's. It was spread by Gauss.

Don
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