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You are not logged in. #26 20131218 06:55:38
Re: Getting the coefficient.Hi bobbym for the coefficient of ... Last edited by anonimnystefy (20131218 06:55:51) 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 #27 20131218 06:57:16
Re: Getting the coefficient.What are you getting for the coefficient of x^50000? 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. #28 20131218 07:30:15
Re: Getting the coefficient.Sorry, that is the coefficient of 50000. 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 #29 20131218 07:42:50
Re: Getting the coefficient.That is not what I am getting. See you in a bit. 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. #30 20131218 07:56:52
Re: Getting the coefficient.My recurrence is: Last edited by anonimnystefy (20131218 09:36:16) 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 #31 20131218 09:26:47
Re: Getting the coefficient.The recurrence should only be for the numerator. 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. #32 20131218 09:30:16
Re: Getting the coefficient.It does not matter. The answer will be the same. 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 #33 20131218 09:31:09
Re: Getting the coefficient.I suggest you try it first before saying that. 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. #34 20131218 09:45:04
Re: Getting the coefficient.The results are the same. 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 #35 20131218 09:49:41
Re: Getting the coefficient.Could be because your recurrence is wrong. Code:RecurrenceTable[{a[n] == 1/(10000 + n) (60005 a[5 + n]  n a[5 + n] + 50004 a[4 + n]  n a[4 + n] + 40003 a[3 + n]  n a[3 + n] + 30002 a[2 + n]  n a[2 + n] + 20001 a[1 + n]  n a[1 + n]), a[10000] == 1, a[10001] == 10000, a[10002] == 50005000, a[10003] == 166716670000, a[10004] == 416916712502500}, a, {n, 10000, 50000}] // Last; 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. #36 20131218 09:53:33
Re: Getting the coefficient.I know exactly what's wrong. 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 #38 20131218 10:00:34
Re: Getting the coefficient.I am using the wrong GF! 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 #39 20131218 10:04:22
Re: Getting the coefficient.Well, that will certainly cause a few problems. What will you do about that? 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. #41 20131221 06:05:24
Re: Getting the coefficient.Part 3 You should recognize that are the coefficients with no restrictions. Now Only the terms x^6 to x^39996 could possibly contribute to the coefficient of x^40000, making the final answer: The whole process looks a lot like a PIE so it can be generalized to the following series. For, with In Mathematica speak: Code:r = 40000; n = 10000; m = 5; ans = Binomial[r + n  1, n  1] + Sum[(1)^ k Binomial[n, k] Binomial[(r + n  1)  (m + 1) k, n  1], {k, 1, Floor[r/(m + 1)]}]; To get the probability of the sum being 50000: Which agrees with Part 2's answer. We can now safely say that the result in Part 1 is incorrect. Okay, what went wrong?
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. 