From rusin@math.niu.edu Tue Jan 25 09:18:41 2000 Return-Path: Received: from vesuvius.math.niu.edu (vesuvius.math.niu.edu [131.156.3.93]) by mail.math.niu.edu (8.9.1a/8.9.1) with ESMTP id JAA27966; Tue, 25 Jan 2000 09:18:41 -0600 (CST) From: Dave Rusin Received: (from rusin@localhost) by vesuvius.math.niu.edu (8.9.3/8.8.5) id JAA06834; Tue, 25 Jan 2000 09:18:39 -0600 (CST) Date: Tue, 25 Jan 2000 09:18:39 -0600 (CST) Message-Id: <200001251518.JAA06834@vesuvius.math.niu.edu> To: lklosinski@math.scu.edu Subject: Seeking background information in re Putnam problem Cc: rusin@math.niu.edu Status: RO Dear Prof Klosinski, One of the problems from this year's Putnam exam (B4) turns out to have some interesting mathematics behind it: it may be both strengthened and generalized. This is not my usual research area and so I am not really sure how best to analyze the problem but I would like to learn more about it. Could you direct me to the author of this problem? (I know the problem authors are usually not known to the public, and I would respect that confidence; indeed, I usually shepherd the discussions of the exam in the sci.math newsgroup, and have in that capacity exchanged email with the proposers of other problems while maintaining confidentiality. See e.g. http://www.math.niu.edu/~rusin/problems-math/index.html Perhaps you could simply invite the problem's author to contact me?) The "right" statement of B4 seems to be that for sufficiently smooth functions on the real line, the conditions f>0, f'>0, f^(k) < f^(l) for specific k > l (e.g k=3 and l=0 in the Putnam problem) often but not always imply f^(m) <= f^(n) for all n < m < k. Such a conclusion follows for all the easily-constructed examples of functions meeting the hypotheses (namely any Stieltjes transform f(x) = \int_{[0,1]} { P(t) exp(tx) dt } where P is a positive function on [0,1]), but this class of functions is not broad enough to include all functions under consideration. I think I can prove this statement with (k,l,m,n)=(3,0,1,0), which strengthens the Putnam result, and I understand the (easier) cases with k=2, but that's not enough examples for me even to conjecture the general case. Thank you for your assistance. dave Prof. David Rusin Director of Undergraduate Studies Department of Mathematical Sciences Northern Illinois University DeKalb, Illinois, 60115 USA Email rusin@math.niu.edu Web http://www.math.niu.edu/~rusin/ Telephone 1-815-753-6739 Fax 1-815-753-1112 From rusin@math.niu.edu Thu Jan 20 16:31:16 2000 Return-Path: Received: from vesuvius.math.niu.edu (vesuvius.math.niu.edu [131.156.3.93]) by mail.math.niu.edu (8.9.1a/8.9.1) with ESMTP id QAA11524; Thu, 20 Jan 2000 16:31:16 -0600 (CST) From: Dave Rusin Received: (from rusin@localhost) by vesuvius.math.niu.edu (8.9.3/8.8.5) id QAA00905; Thu, 20 Jan 2000 16:31:15 -0600 (CST) Date: Thu, 20 Jan 2000 16:31:15 -0600 (CST) Message-Id: <200001202231.QAA00905@vesuvius.math.niu.edu> To: grubb@math.niu.edu Subject: That stupid Putnam problem. Cc: rusin@math.niu.edu Status: R Why do I waste my time like this? :-) You may recall we were considering functions known only to satisfy f>0 f'>0 f'''0, f'>0, f^(k) {f^(m) < f^(n)} In a few cases I could draw the desired inference. In others, not. My analysis looks at h = f'/f ; we assume that h > 0 and that one of these is greater than another, and then we see what others among the set must be ordered: 1 h h' + h^2 h'' + 3 h h' + h^3 h''' + 4 h h'' + 3 h'^2 + 6 h' h^2 + h^4 etc. Well, here's an interesting example: if h=(2/3-1/4*sin(x/3)-1/13*sin(2*x/3)) then the first four of these line up, but the fifth does not, that is, we can cook up a function with f>0, f'>0, and f''' < f'' < f' < f but f^(iv) isn't even less than f'', much less f''' . In particular, it's certainly not expressible as an integral \int P(u) exp(ut) as above. dave From rusin@math.niu.edu Thu Jan 20 09:21:43 2000 Return-Path: Received: from olympus.math.niu.edu (olympus.math.niu.edu [131.156.3.4]) by mail.math.niu.edu (8.9.1a/8.9.1) with ESMTP id JAA09214; Thu, 20 Jan 2000 09:21:42 -0600 (CST) From: Dave Rusin Received: (from rusin@localhost) by olympus.math.niu.edu (8.9.3/8.8.5) id JAA13764; Thu, 20 Jan 2000 09:21:42 -0600 (CST) Date: Thu, 20 Jan 2000 09:21:42 -0600 (CST) Message-Id: <200001201521.JAA13764@olympus.math.niu.edu> To: bellout@math.niu.edu Subject: That Putnam problem Cc: rusin@math.niu.edu Status: R Hamid, I'm curious about this approach you mentioned yesterday in the coffee room. Recall that the Putnam exam question asked, if f : R -> R has continous 3rd derivative, and if we know f > 0, f' > 0, and f''' < f, then f' < c f for c=1.66 . (Actually the Putnam question only asked for a proof with c=2, and they gave some extra hypotheses, but I have a proof which uses only the information shown. On the other hand, I think one should be able to give a proof for any c>1.) One line of reasoning I have is incomplete, but it suggests that any set of conditions of the form { f>0, f'>0, f^(k) < f^(l) } for some k > l ought to be enough to prove f^(m) <= f^(n) for every m > n; the problem I set above is the case (k,l,m,n) = (3,0,1,0). I can prove some easier cases like (k,l,n,m) = (2,1,1,0) by more direct means, but that sort of approach does not appear to generalize well. Now, this technique you were suggesting -- can you suggest a paper where it has been used on a similar problem? I thought I understood what the general approach would be, but when I tried to work it out I realized there must be some points I'm missing. Thanks dave From bellout@math.niu.edu Thu Jan 20 09:26:26 2000 Return-Path: Received: from pike.math.niu.edu (pike.math.niu.edu [131.156.3.83]) by mail.math.niu.edu (8.9.1a/8.9.1) with ESMTP id JAA09250 for ; Thu, 20 Jan 2000 09:26:26 -0600 (CST) From: Hamid Bellout Received: (from bellout@localhost) by pike.math.niu.edu (8.9.3/8.8.5) id JAA01707 for rusin@math.niu.edu; Thu, 20 Jan 2000 09:26:25 -0600 (CST) Message-Id: <200001201526.JAA01707@pike.math.niu.edu> Subject: Re: That Putnam problem To: rusin@math.niu.edu (Dave Rusin) Date: Thu, 20 Jan 2000 09:26:24 -0600 (CST) In-Reply-To: <200001201521.JAA13764@olympus.math.niu.edu> from "Dave Rusin" at Jan 20, 2000 09:21:42 AM X-Mailer: ELM [version 2.5 PL2] MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Content-Transfer-Encoding: 7bit Status: R > > > Hamid, > > I'm curious about this approach you mentioned yesterday in the coffee room. > Recall that the Putnam exam question asked, > if f : R -> R has continous 3rd derivative, and if we know > f > 0, f' > 0, and f''' < f, then f' < c f for c=1.66 . > (Actually the Putnam question only asked for a proof with c=2, and > they gave some extra hypotheses, but I have a proof which uses only > the information shown. On the other hand, I think one should be able > to give a proof for any c>1.) > > One line of reasoning I have is incomplete, but it suggests that any > set of conditions of the form { f>0, f'>0, f^(k) < f^(l) } for some k > l > ought to be enough to prove f^(m) <= f^(n) for every m > n; the > problem I set above is the case (k,l,m,n) = (3,0,1,0). I can prove some > easier cases like (k,l,n,m) = (2,1,1,0) by more direct means, but that > sort of approach does not appear to generalize well. > > Now, this technique you were suggesting -- can you suggest a paper where > it has been used on a similar problem? I thought I understood what the Better than a paper here is a book which tries to make it more genera AUthor : R. Sperb TItle: **** maximum principle **** (I do not recall the exact title but it has maximum priciple somwhere in it) It is in our library. look up chapter about uses of what he calls P-functions > general approach would be, but when I tried to work it out I realized there > must be some points I'm missing. > > Thanks > > dave > >