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You are not logged in. #1 20100809 02:29:36
Tricky integral of a rational functionHi all! None of the standard methods work because of the irreducibility of both the numerator and the denominator (the denominator can be factored out as a product of two polynomials of sixth degree, but that's not very helpful). On the other hand, Mathematica seems to have no problem evaluating the integral. So I must be missing something. It seems that the coefficients in the denominator are carefully chosen because changing any one of them gives Mathematica a hard time and it doesn't evaluate the integral at all. Any suggestions will be greatly appreciated #2 20100809 07:37:50
Re: Tricky integral of a rational functionHi Heirot; 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. #3 20100809 17:11:15
Re: Tricky integral of a rational functionI doubt that Mathematica runs any algorithm because it takes as much time to do this integral as it takes to calculate integral of x. Perhaps this integral if from the tables. How could I check that? #4 20100809 18:07:11
Re: Tricky integral of a rational functionHi; 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. #5 20100809 19:49:05
Re: Tricky integral of a rational functionI agree with you that the indefinite integral would be more challenging to calculate. Because of the limits of integration one can use methods of contour integration or something of the sort that would greatly simplify the calculation. I was hoping that the special choice of the coefficients in the denominator makes that choice possible. E.g. expand the lower limit to minus infinity and then use the semicircle contour. It would pick up singularities in the upper halfplane (which are zeros of the denominator with positive imaginary part), summing their residues to give the integral. Now, since all of the poles are simple ones, I was hoping to express the sum of their residues in the terms of the coefficients of the denominator via Viete formulas. Do you think that could work? #6 20100810 04:54:06
Re: Tricky integral of a rational functionHi Heirot; 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. #7 20100814 22:55:53
Re: Tricky integral of a rational functionHi You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #8 20100815 00:16:38
Re: Tricky integral of a rational functionHi bob; 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. #9 20100816 06:37:54
Re: Tricky integral of a rational functionHi Last edited by bob bundy (20100816 06:39:36) You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #10 20100818 21:45:25
Re: Tricky integral of a rational functionMy son wants the complete complex factorisation for each x^6 polynomial. If mathematica can do this, please would you put it to work for us. Last edited by bob bundy (20100818 21:50:09) You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #11 20100818 23:35:26
Re: Tricky integral of a rational functionHi bob bundy; 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. #12 20100823 20:21:18
Re: Tricky integral of a rational functionHi bobbym You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #13 20110716 21:42:19
Re: Tricky integral of a rational functionhi bobbym, gAr and anyone else who can maybe help. where P and Q are real. Lacking any other way to proceed I set up three such quartics 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: 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 my 1000th post. 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. If someone could test that, I’d be grateful. Then I started the partial fractions. 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) Last edited by bob bundy (20110716 21:42:44) You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #14 20110716 22:04:48
Re: Tricky integral of a rational functionHi; 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. #15 20110716 22:14:57
Re: Tricky integral of a rational functionhi bobbym, Last edited by bob bundy (20110716 22:15:17) You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #16 20110716 22:20:57
Re: Tricky integral of a rational functionHi all, "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." #17 20110716 22:57:47
Re: Tricky integral of a rational functionHi;
Those coefficients are quite close. But the exact, closed form for them Mathematica can not find. 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. #18 20110716 23:06:57
Re: Tricky integral of a rational functionhi bobbym Last edited by bob bundy (20110716 23:07:17) You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #19 20110716 23:28:22
Re: Tricky integral of a rational functionHi; Here are my equations. P=.776376329716313367678396729764340830599615711886 Q=.38281829417943271655606084094051556034456756849056 R=.6084053916280706468434233044383803346185796004872 S=.1083620659490676250704100110401129566047127759067 T=8.615218278655615985478179965797278834781804687627 U=24.106271297265356993763009485782695345494455926176 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. #20 20110716 23:44:22
Re: Tricky integral of a rational functionhi bobbym, You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #21 20110717 00:16:38
Re: Tricky integral of a rational functionHi Bob; 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. #22 20110717 00:27:23
Re: Tricky integral of a rational functionEven this didn't find any. "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." #23 20110717 00:33:06
Re: Tricky integral of a rational functionHi gAr; 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. #24 20110717 00:50:21
Re: Tricky integral of a rational functionHi bobbym, "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." #25 20110717 01:08:15
Re: Tricky integral of a rational functionThis page tells about few methods: http://mpec.sc.mahidol.ac.th/radok/phys … /sec41.htm "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." 