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#76 2011-07-22 07:13:26

bobbym
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Re: Tricky integral of a rational function

Hi;

I will provide something in a few minutes.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#77 2011-07-22 07:17:22

Au101
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Re: Tricky integral of a rational function

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

Last edited by Au101 (2011-07-22 07:17:41)

#78 2011-07-22 07:23:54

bobbym
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Re: Tricky integral of a rational function

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.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#79 2011-07-22 07:35:29

Au101
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Re: Tricky integral of a rational function

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.

#80 2011-07-22 07:38:50

bobbym
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Re: Tricky integral of a rational function

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.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#81 2011-07-22 07:56:46

Au101
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Re: Tricky integral of a rational function

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

#82 2011-07-22 08:00:25

bobbym
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Re: Tricky integral of a rational function

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


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#83 2011-07-22 08:20:03

Au101
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Re: Tricky integral of a rational function

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

#84 2011-07-22 08:23:11

bobbym
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Re: Tricky integral of a rational function

Hi;

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


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#85 2011-07-22 14:00:39

gAr
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Re: Tricky integral of a rational function

Hi,

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





"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense"  - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

#86 2011-07-22 14:25:52

bobbym
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Re: Tricky integral of a rational function

Hi gAr;

That is two quadratics for the numerators also. Because



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


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#87 2011-07-22 14:36:05

gAr
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Re: Tricky integral of a rational function

Hi bobbym,

Oh, okay!


"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense"  - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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