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You are not logged in. #76 20130331 11:38:27
Re: Contour integrationNope, that is how he solved them in the book. Do you think I invented this? Also, I check the correct answer and correct choice of poles by trying the others in combination. They do not work. 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 20130331 11:41:07
Re: Contour integrationCan you post how he exactly solved it? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #78 20130331 11:44:04
Re: Contour integrationI have! The only thing I am lacking is a coherent idea on which poles to take. That I did not write down because at the time I understood it better and it was self evident to me. 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 20130331 11:50:57
Re: Contour integrationWord to word? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #80 20130331 11:53:26
Re: Contour integrationYes, as I have written them down. Me and zetafunc did a few also. He came up with essentially the same method. The only problem is the ambiguity in explaining the choice of pole. For the contour I believe they always use the unit circle in these examples. 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 20130331 12:17:39
Re: Contour integrationNo, actually, the contour is chosen and the poles inside the contour are the ones you calculate the residue of. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #82 20130331 12:21:19
Re: Contour integrationOkay, choose some arbitrary contour and try to get the answer. This method gets the answer to every one of these problems. 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 20130331 12:24:39
Re: Contour integrationI did not say that the choice of contour is arbitrary. I said that the contours are chosen, not residues. In fact, sometimes, you may need to calculate the residue of more than one pole. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #84 20130331 12:33:39
Re: Contour integrationYes, we just did a problem where 2 poles were inside the unit circle. Sometimes you do all the poles. But as you can see from the drawing a circle is the simplest shape to go through the 3 points. It happens to be a unit circle. 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 20130331 12:38:33
Re: Contour integrationYou seem to be arbitrarily choosing which poles you are taking and which not. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #86 20130331 12:46:08
Re: Contour integrationNope, I have a rule. I just do not have any solid justification for it. 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 20130331 12:54:09
Re: Contour integrationWhat rule is that? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #88 20130331 12:56:14
Re: Contour integrationBecause we have intervals of 0 to infinity the bottom pole is out. Then I take the poles that have a positive imaginary part. That leaves the one pole I chose. 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. #89 20130331 13:05:08
Re: Contour integrationHm, okay. Then, how would you integrate log(x)/((1+x^2)^2) from 0 to infinity? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #90 20130331 13:09:29
Re: Contour integrationWe have not finished the other one. If we jump around we will get confused. We should calculate the residues for the other problem before moving on to the next one. 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. #91 20130331 13:11:46
Re: Contour integrationThat site is not correct. All sources I have found have it working on other types if functions as well. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #92 20130331 13:16:40
Re: Contour integrationI was going to say that but zetafunc shows a rather simple one where this method falls flat on its face. There must be some conditions when can use it and when you can not. 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. #93 20130331 13:19:45
Re: Contour integration
Actually, that was the point of the integral I gave you. As you will see, the method you will try on it will probably fail, but the Wiki article has no problem doing it. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #94 20130331 13:29:43
Re: Contour integrationThe residue method does not work on all cases. There are conditions. 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. #95 20130331 13:31:03
Re: Contour integrationYes, but the contour integration methid works on that one. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #96 20130331 13:36:20
Re: Contour integrationIt will fail on others. Either because of violations or impossibility to do. 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. #97 20130331 13:38:26
Re: Contour integrationThis was just to show that the method you have is not contour integration. But I do not know if it works. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #98 20130331 13:39:10
Re: Contour integrationWhy do you think it is not contour integration? 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. #99 20130331 13:45:40
Re: Contour integrationBecause I did understand some parts of what I redlad earlier, and some of those parts were not included in your method. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #100 20130331 13:48:19
Re: Contour integrationHi; 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. 