mm-4519 === Subject: Re: Textbook comments request (Hungerford's Algebra) Has the expected level of competence at the undergraduate level decreased? I was not a mathematics major, but took algebra out of Herstein, and considered it par for the course, as did everyone else. By the way, it was one of the best elective classes I ever took for the short and long term benefits conferred. As a programmer I was always puzzled at the antipathy to mathematics manifested by other programmers. -- Michael Press === Subject: On the level of undergraduate training on mathematics (was: Re: Textbook comments request (Hungerford's Algebra)) I don't know if it has decreased or not, but one thing that was mentioned by a friend of mine that has taught at the University of Ohio and University of Tennessee is that while our (brazilian) undergraduate courses cover deeper aspects than what is covered by undergraduate courses in the US, people usually make graduate students sweat a lot to get their degree, basically nullifying the difference in the degrees obtained at a Ph.D. level. Oh, BTW, one cultural difference that I see is that a Master's degree in the US is mostly taken as a consolation prize for those that can't get a Ph.D. degree. Is that impression of mine correct? Here, a Master's degree is almost *required* to enter a Doctorate level program and going the Doctorate level is really not the norm here (quite a few exceptions I'd add) and it can even touch quite a bit of research (at least, mine did). Indeed, I'd say that Abstract Algebra is absolutely essential for developing the formal, *mathematical thinking* that one should have for writing elegant programs. You can't really imagine how many headaches I get when I try to give a stronger mathematical bias when I'm teaching Analysis of Algorithms or Elementary Graph Theory. And that ruins all the beauty of intelligent, elegantly devised algorithms. :-( -- Rog.8erio Brito : rbrito@ime.usp.br : http://www.ime.usp.br/~rbrito Homepage of the algorithms package : http://algorithms.berlios.de Homepage on freshmeat: http://freshmeat.net/projects/algorithms/ === Subject: Re: IM style (off topic) and built it into the program . SAVED (dot saved), a kind of shell. It built on supervisor primitives provided by Bob Fenichel; we provided the user interface. (The SDC timesharing system had a similar feature allowing communication between the user and operator but I don't know if it worked for user to user.) We had abbreviation conventions, nothing like ROFL, but people would say (over) to indicate they were waiting for a reply. later this was shortened to -o- or similar, and o/o meant over and out as in radio. === Subject: Re: Estimating the number of great white sharks Given any value of n, it is taken that the sum of probabilities (expected value) is equal to one when a tagged shark is caught. Of course this is the theoretical mean of a probability distribution. For example: say n = 7. Then 0/N + 1/N + 2/N + 3/N + 4/N + 5/N + 6/N = 1 21/N = 1 N = 21 This is equal the sum of the numerators of the progression which can be written as Sum = 7^2/2 - 7/2 = 24.5 - 3.5 = 21 Hence N = n^2/2 - n/2 seems to work fine. Phil H === Subject: Re: Reciprocals of the natural numbers Yeah, that's my proof too, but a tad messy. I was looking for a neat -- if slightly sophisticated -- proof. === Subject: Re: Reciprocals of the natural numbers With H n := 1 + 1/2 + 1/3 + ... 1/n, write 2*3...n + 1*3...n + 1*2*4...n + 1*2...(n-1) H n = ------------------------------------------- n! Let p be the largest prime less than or equal to n. Then p divides n! and also every term in the numerator with the single exception of === Subject: Re: Reciprocals of the natural numbers Yeah, Betrand's postulate does give a proof.(the desired p lies between N and N/2) But again, Bertrand's postulate is proved rather computationally. I'm looking for a proof that's algebraic, maybe something using arithmetic functions or even something sophisticated like group theory but at any rate avoiding the use of the order property. The first proof given also depends on the fact that 2 is the smallest prime. I want to avoid using the order property. === Subject: Re: Reciprocals of the natural numbers [ ... ] [ ... ] I thought the basic idea of the proof can be expressed as: If the set {1,2,...,n} contains two numbers divisible by 2^k, then it contains a number divisible by 2^(k+1); with conclusion that given n, there is an exponent k such that exactly one of the numbers {1,2,...,n} is divisible by 2^k. I'd say, this proof does not rely on the order property, but rather on the fact that 1+1=2. (Really!) === Subject: Re: Reciprocals of the natural numbers Okay -- the way I interpreted it when I proved it was that if 2^k divides another number in the given range the number is of the form a*(2^k)where a is at least 2. Thus the number is not in the given range, i.e between 2^k and 2^(k+1). Actually the proof works for any prime