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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