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Title: Infinite system of equations Post by Eigenray on Jun 4th, 2005, 12:32pm Determine all solutions, in nonnegative real numbers, of the following infinite system of equations: a1 + b1 = 2 a2 + 2a1b1 + b2 = 4 a3 + 3a2b1 + 3a1b2 + b3 = 8 ... an + (nC1) an-1b1 + ... + (nCk)an-kbk + ... + bn = 2n ... |
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Title: Re: Infinite system of equations Post by Michael_Dagg on Jun 5th, 2005, 9:08am Looks easy. |
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Title: Re: Infinite system of equations Post by Eigenray on Jun 5th, 2005, 2:15pm Well, finding the solutions is easy, but proving those are the only ones maybe less so. If we drop the nonnegative real condition, the ai may be chosen arbitrarily. Perhaps a better question is what's the weakest condition we can replace it by and still have just the one family of solutions. |
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Title: Re: Infinite system of equations Post by Deedlit on Jun 5th, 2005, 3:18pm Well, assuming any deviations force larger and larger deviations until the values go negative, then presumably any bound on Ai and Bi, or both upper and lower bounds on one or the other, would eventually be violated. I suppose a single bound on one sequence of variables might not be enough, since we could send the values in the other direction. We could then examine the rate of divergence - but I should probably solve the original problem first... |
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Title: Re: Infinite system of equations Post by Eigenray on Jul 15th, 2005, 10:18pm I guess I'll give a hint: I found this in a [hide]complex analysis[/hide] textbook. |
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Title: Re: Infinite system of equations Post by raprap on Jul 16th, 2005, 4:39am 8)The second thing that struck me was a binomial expansion' a1 + b1 = 2 (a1+b1)^2=2^2 . . . (a1+b1)^n=2^n if so if a1=a1 b1=b1 then a2=a1^2 b2=a1^2 . . . an=a1^n bn=b1^n The family of solutions would be contingent upon initial values of a1 or b1. The first thing that struck me was an infinite square triangular matrix, mostly because if it existed the determinant would be the product of the diagonal, and as long as ai or bi <>0 a solution would exist. But then I noticed that it was missing a row for the first element. and the matrix wasn't square. Oh Well! Rap |
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Title: Re: Infinite system of equations Post by Eigenray on Aug 15th, 2005, 1:28pm Indeed, raprap, for any 0<=a1<=2, a solution is given by an=a1n, bn=(2-a1)n. The surprising result is that these are the only solutions in nonnegative real numbers. I probably shouldn't have put this in medium, but I was actually wondering if someone would find an elementary solution. The proof I know requires some nontrivial results about [hide]entire functions[/hide]. |
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Title: Re: Infinite system of equations Post by Eigenray on Jan 21st, 2006, 7:50pm *Bumpity* I guess this should have gone in the putnam section ([hide]the complex analysis section would have been a giveaway, I think[/hide]). To continue the previous hint, [hide]entire functions of finite order[/hide]. |
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Title: Re: Infinite system of equations Post by Eigenray on Feb 10th, 2006, 9:36pm Set a0=b0=1, and let f(z) = [sum]n=0oo anzn/n!, g(z) = [sum]n=0oo bnzn/n!, so that the given equations may be expressed as f(z)g(z) = [sum]n=0oo 2nzn/n! = e2z. Since an, bn are all nonnegative, they must also be bounded by 2n, so f and g are entire functions; moreover |f(z)| < [sum] 2n|z|n/n! = e2|z|, and similarly for g, so they are of finite order 1. Since f(z)g(z)=e2z, neither f nor g has any zeros, so by the Hadamard factorization theorem, f(z) = er+sz, and therefore g(z) = e-r+(2-s)z, for some constants r,s. Since a0=1, we have r=0, so f(z) = esz = [sum] sn/n! zn, and therefore an = sn = a1n for all n, and similarly bn = (2 - a1)n. (This problem is an example in Complex Analysis, by Bak and Newman.) |
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