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[quote=bob bundy]hi bobbym, gAr and anyone else who can maybe help.

(with apologies for the long post)

It was almost a year ago that this post first appeared.  No one has posted a solution yet.

The proposal was:

Factorise the order 12 polynomial into 12 complex factors and use Cauchy.

But how to factorise?

After puzzling on this for some time, I devised an alternative strategy.

Factorise into 3 quartic factors, split into partial fractions with numeric numerators, and integrate each (without Cauchy).

As the polynomial has real  and even coefficients each quartic must have four complex factors of the form (a + bi), (a – bi), (-a + bi) and ( -a –bi) and so each quartic will have the form:

[math](x^4 + Px^2 + Q)[/math]

where P and Q are real.

Lacking any other way to proceed I set up three such quartics

[math] (x^4 + Px^2 + Q) (x^4 + Rx^2 + S)(x^4 + Tx^2 + U)[/math]

and multiplied this out to obtain 7 simultaneous equations in 6 unknowns.

I set up the equations in Excel and used goal seek to try and number crunch a solution.

To show if I was approaching a solution I took the 6 trial values for P, Q ...U, and computed how far distant from the target coefficients my solution lay.

My initial distance was over 100 units away, but by goal seeking each variable in turn I got down to about 18.  But I couldn’t get closer.  Eventually, I worked out this is because of the way goal seek works.  It kept ‘hopping’ over the desired solution because it’s steps were to large.

So I switched to decimal search and got to within 0.6 of target.

After a pause I wrote a computer program to do the decimal searching for me and, within a few minutes, got to within 10^-16 of target.  I don’t think I can number crunch to a closer solution.

The factorisation is then:
[math] (x^4 + Px^2 + Q) (x^4 + Rx^2 + S)(x^4 + Tx^2 + U)[/math]
with

P = -0.776376326
Q = 0.382818292
R = -0.608405391
S = 0.108362067
T = -8.61521826
U = 24.1062714

It would be better if I could arrive at exact values by some analytic method but I cannot.

Feeling I was getting close, I decided to go all out with the rest of the solution, planning to post it as [u][b]my 1000th post.[/b][/u]

But I’ve hit a snag.:(

Firstly, I don’t know if mathematica can test out that factorisation by multiplying it out, back to the power 12 poly.  
[b]If someone could test that, I’d be grateful.[/b]

Then I started the partial fractions.

[math]  \frac{A}{x^4 + Px^2 +Q} + \frac{B}{x^4 + Rx^2 +S} +\frac{C}{x^4 + Tx^2 +U}[/math]

I used the coefficients for x^6, x^4 and x^2 to make three simultaneous equations and solved them:
I have
A = -1.23654966
B = 1.045292468
C = 4.0356581263

But these are not consistent with the x^8 and x^0 equations.

Furthermore, I cannot see how to integrate the 1/quartic expressions anyway.

So I’m stuck. :(

Any help would be gratefully received.

Bob (post 987)[/quote]

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Topic review (newest first)

gAr: 2011-07-22 14:36:05

Hi bobbym,

Oh, okay!

bobbym: 2011-07-22 14:25:52

Hi gAr;

That is two quadratics for the numerators also. Because

That is why I went right to 2 quadratics in the numerators.

gAr: 2011-07-22 14:00:39

Hi,

Shouldn't the partial fraction we expect be of the form:

bobbym: 2011-07-22 08:23:11

Hi;

Because in the "fit" for the coefficients a,b,c,d,e,f it gets taken care of all by itself.

Au101: 2011-07-22 08:20:03

Hmmm, yes I see, but ought we not account for it. I tried taking the four outside, although perhaps that was not correct. What I am interested in, though, is why we don't have to worry about it when computing a,b,c,d,e and f

bobbym: 2011-07-22 08:00:25

I think that is just some factor that Mathematica pulled out, for what reason... I do not know.

Au101: 2011-07-22 07:56:46

Oooh yes, that's very good - I'm still confused about that pesky four though

bobbym: 2011-07-22 07:38:50

Hi;

to tell that our numerators would be quadratics

One way is to say the numerator is an 8th degree poly, a quadratic times by a six degree
poly is an 8th degree poly.

Au101: 2011-07-22 07:35:29

Oooh thanks bobbym that's perfect, my only question would be where we've accounted for the fact that the product of our denominators is four times the original denominator. I also wonder if it would be possible to tell that our numerators would be quadratics if we hadn't had the answer to begin with - more out of curiosity than anything else.

bobbym: 2011-07-22 07:23:54

Hi;

Now what you do is form a 6 x 6 set of simultaneous linear equations. This is done by substituting x = -3,-2,-1,0,1,2 in the above. That wipes out the x and you are left with:

Which is exactly what we expected.

Au101: 2011-07-22 07:17:22

Oh of course - take your time - I hadn't meant for this thread to cause you - or anybody else - even more hard work.

bobbym: 2011-07-22 07:13:26

Hi;

I will provide something in a few minutes.

Au101: 2011-07-22 06:56:54

Hmmm, I can definitely do 2) in theory - although that's not to say that I might not have some trouble with doing a large, difficult, practical example - although I should think I would probably enjoy trying and, well, as for 1) I not only know the form, I know what they are, since I wish only to verify it, as much as I would like to be able to derive something like this - not being a computer - or at least an extremely experienced and intelligent professor of advanced mathematics - I think that to do so may well be somewhere beyond me.

Where I am stuck is with regards to where to go with these 2 pieces of knowledge.

bobbym: 2011-07-22 06:50:55

Hi;

The way I would do it would be by solving a set of simultaneous linear equations.

You would need two things:

1) To know the form for A and B beforehand.
2) The ability to solve a 6 x 6 set of linear eqautions.

Au101: 2011-07-22 06:44:55

Oh, well, what I did was - mainly for my own satisfaction - to expand

Which of course gives a massive expansion, if it's of interest:

Which, when you add it all together gives:

Which, of course, is:

So I had satisfied myself with the denominators and I tried using my standard partial fractions method and said

But then I got:

And gave up, because I knew that I must have been on the wrong track. And, well, that's when I asked you lol

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