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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
