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POW-11: Preserving sums of squares October 26, 2008 in Elementary Math Problem Solving, Problem truth about enzyte of the Week (POW) | by Todd Trimble | 2 comments Time for another problem of the week! This one doesn’t necessarily require any number-theoretic knowledge, although a little bit might help point you in the right direction. As usual, let be the set of natural numbers, i.e., the set of nonnegative integers. Describe all functions such that for all . Please submit solutions to topological[dot]musings[At]gmail[dot]com by Sunday, November 2, 11:59 pm (UTC); do not submit solutions in Comments. Everyone with a correct solution will be inducted into our Hall of Fame! We look forward to your response. October 21, 2008 in Uncategorized | Tags: elementary function, integration | by Todd Trimble | 7 comments A reader brought up essentially this question: does anyone happen to know a proof that does not possess an elementary antiderivative? By “elementary”, I mean a member of the class of functions which contains all constants valued in the complex numbers, the identity function, the exponential and log functions, and closed under the four basic arithmetic operations and composition. October 17, 2008 in Uncategorized | Tags: Bonn, Germany, vishwanathan anand, vladimir kramnik, world chess championship 2008 | by Vishal Lama | 16 comments Perhaps most people are preoccupied with the global financial crisis right now, especially with people in the US much more focused on the upcoming US presidential election in November. So, for those who haven’t been following the news in chess closely I would like to bring their kind attention to the current World Chess Championship (2008) match (Bonn, Germany) between two supreme chess grandmasters, Vishwanathan Anand (India) and Vladimir Kramnik (Russia). Technically, Anand (pronounced Aa-nand and not A-naand; well, actually it is more like Aa-nundh) is the current world chess champion, but personally I think that his winning the World Chess Championship Mexico (2007) last year was a somewhat “unsatisfactory” accomplishment, if you will, given that he won the crown by winning a tournament and not a “classical” chess match. I fervently believe that a person should be crowned world chess champion (like Fischer, Kasparov, Capablanca, Kramnik, to name a few) only after he or she has won a “proper” world championship match played under “classical time controls” (remember the Fischer-Spassky match in 1972 and the Kasparov-Kramnik match in 2000?) Without that, the gravitas of the chess crown is somewhat diminished. So, here is Anand’s chance now to silence his critics, of which there are very few really, once and for all that he is indeed the undisputed world chess champion! And judging by the result of the third game, which he just won in a dramatic fashion (woohoo!), as well as the tremendous amount of home preparation it clearly seems he has done, there is no doubt that he is on a steady path to the crown. In all the first three games, Anand has demonstrated thus far that he is the superior player. Of course, there are nine more games left and the bets are not off by any means. After all, it was Kramnik who beat Kasparov convincingly in 2000 to win the crown. For analyses of the first two games, click here and here. Ok, so here is a poll that I invite our readers to participate in. (Obviously, if you choose the “wrong” answer, all your future comments on this blog will be deleted! So, think hard before you vote.) October 12, 2008 in Elementary Math Problem Solving, Problem of the Week (POW) | Tags: integration | by Todd Trimble | 8 comments The solutions are in! As is so often the case, solvers came up with a number of different approaches; the one which follows broadly represents the type of solution that came up most often. I’ll mention some others in the remarks below, some of it connected with the lore of Richard Feynman. Solution by Nilay Vaish: The answer to POW-10 is . Put
and integrate by parts: we have
The first term vanishes by a simple application of L’hôpital’s rule. We now have
where the second equation follows from , and the general elementary fact that Adding the last two displayed equations, we obtain
This gives
after a simple substitution. The last integral splits up as two integrals:
but these two integrals are equal, using the identity together with the general integration fact cited above. Hence the two sides of (3) equal
recalling equation (1) above. Substituting this for the last integral in equation (2), we arrive at
whence we derive the value of the desired integral . Remarks: 1. A number of solvers exploited variations on the theme of the solution above, which could be summarized as involving symmetry considerations together with a double angle formula. For example, Philipp Lampe and Paul Shearer in their solutions made essential use of the identity
in conjunction with the complementarity , and the general integration fact cited above. 2. Arin Chaudhuri (and Vishal in private email) pointed out to me that the evaluation of the integral
is actually fairly well-known: it appears for example in Complex Analysis by Ahlfors (3rd edition, p. 160) to illustrate contour integration of a complex analytic function via the calculus of residues, and no doubt occurs in any number of other places. 3. Indeed, Simon Tyler in his solution referred to this as an integral of Clausen type, and gave a clever method for evaluating it: we have
which works out to
The last integral can be expanded as a series
where the summands for odd vanish; the other terms can be collected and then resummed to yield
and by substituting this for the integral in (*), the original integral is easily evaluated. 4. Arin C. later described still another method which he says he got from an exercise in a book by Apostol — it’s close in spirit to the one I myself had in mind, called “differentiation under the integral sign”, famously referred to in Surely You’re Joking, Mr. Feynman!. As Feynman recounts, he never really learned fancy methods like complex contour integration, but he did pick up this method of differentiating under the integral sign from an old calculus book. In typical Feynman style, he writes, “It turns out that’s not taught very much in the universities; they don’t emphasize it. But I caught on how to use that method, and I used that one damn tool again and again. So because I was self-taught using that book [Wilson's Advanced Calculus], I had peculiar methods of doing integrals. The result was, when guys at MIT or Princeton had trouble doing a certain integral, it was because they couldn’t do it with the standard methods they had learned in school. If it was contour integration, they would have found it; if it was a simple series expansion, they would have found it. Then I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else’s, and they had tried all their tools on it before giving the problem to me.” |