mm-1084 === Subject: Re: Help with Integral > when I write the expansion of the integral in terms of bessel functions > int_0^oo exp(I*b*x)*exp(-a*sqrt(a^2+b^2)) dx= > 1/sqrt(pi)*sum_{k=0}^oo > (( Gamma(k+1/2)*2^k)/(2*k)! * (-b^2/a)^k*besselk(k+1,a) + > 1/sqrt(pi)*sum_{k=0}^oo > (( Gamma(k+1)*2^(k+1/2))/(2*k+1)! * (-b^2/a)^(k+1/2)*besselk(k+3/2,a) > I find it hard to believe, that the first sum is expressible as a simple besselk function, whereas the second should not be (the same form of expression, the index k is just 'shifted' by 1/2) ? > > By the way, if we take the first ( k=0) term of the sine part: > sqrt(2/(a*pi))*besselk(3/2,a), multiply it by b and replace besselk(3/2,a) with besselk(3/2,sqrt(a^2+b^2)), we get an even better limit for b->0 at > large a, but it is harder to connect to the limit b/sqrt(a^2+b^2) at small a. > > Anyway the solution is hidden somewhere in the solution space of > the Helmholtz equation. > > Andreas Hello Andreas, And sorry for the delay (well, I had nothing of value earlier so...). Let's change a little the form of your integral : int_0^oo e^(-a sqrt(1+x^2)) sin(b x) dx x = sinh(t) int_0^oo e^(-a cosh(t)) sin(b sinh(t)) cosh(t) dx Let's define f : f(a,b)= int_0^oo e^(-a cosh(t)) sin(b sinh(t)) dx so that your function is -d/da f with d/da the partial derivative relatively to a : -d/da f= int_0^oo e^(-a cosh(t)) sin(b sinh(t)) cosh(t) dx Let's integrate this by parts : = e^(-a cosh(t))(-cos(b sinh(t))/b)|_0^oo + 1/b (-a) int_0^oo e^(-a cosh(t)) sinh(t) cos(b sinh(t)) dt = e^(-a)/b - a/b d/db f so that f must be solution of the partial differential equation : a d/db f - b d/da f = e^(-a) a d/db f - b d/da f = 0 has the general solution g((a^2+b^2)/2) with 'g' a one parameter function to determinate... a special solution is int_1^x - e^(-t)/sqrt(a^2+b^2-t^2) dt the general solution of this PDE should be the sum : f(a,b)= int_1^a - e^(-t)/sqrt(a^2+b^2-t^2) dt + g((a^2+b^2)/2) we need f(a,0)= 0 so that int_1^a e^(-t)/sqrt(a^2-t^2) dt = g(a^2/2) or g(x)= int_1^sqrt(2 x) e^(-t)/sqrt(2 x - t^2) dt combining these : f(a,b)= int_1^a - e^(-t)/sqrt(a^2+b^2-t^2) dt + int_1^sqrt(a^2+b^2) e^(-t)/sqrt(a^2+b^2-t^2) dt f(a,b)= int_a^sqrt(a^2+b^2) e^(-t)/sqrt(a^2+b^2-t^2) dt = int_0^oo e^(-a cosh(t)) sin(b sinh(t)) dx (this looks numerically right) Of course we have again an integral without closed form and we can't even compute -d/da f(a,b) directly (result singular at the upper bound). But note the nice symmetry between a and b (broken at the bottom bound 'a' only...), the possibility of expanding 1/sqrt(r^2 - t^2) in powers of r^2=a^2+b^2 and so on. an example of rewriting ( not simpler :-( ) with t=sqrt(a^2+u^2*b^2) : b int_0^1 e^(-sqrt(a^2+u^2*b^2))/sqrt(a^2+u^2*b^2)*u/sqrt(1-u^2) du applying -d/da we get another integral for your problem : u(a,b)= a b int_0^1 du u*e^(-sqrt(a^2+b^2*u^2))/sqrt(1-u^2)*(sqrt(a^2+b^2*u^2)+1)/(a^2+b^2*u^2)^(3/ 2) using the previous change of variable again your initial integral becomes : u(a,b)= int_a^sqrt(a^2+b^2) a e^(-t) (t+1)/(t^2 sqrt(a^2+b^2-t^2)) dt I'll let you try other transformations eventually linked to Bessel functions (G. Watson's great book : 'A Treatise on the Theory of Bessel Functions' is available online : http://www.hti.umich.edu/cgi/t/text/text-idx?c=umhistmath;cc=umhistmath;sid= 3e0b8311798f39760e6af0eb9dd57793;q1=watson;view=toc;idno=acv1415.0001.001 ) Good luck! Raymond ---- resume of earlier results -------------------------------- A last look to my (corrected) equality for inspiration :-) : u(a,b)= exp(-a)/b sum_{n=1}^oo -(-(b/a)2)^n [ 1+a + sum_{k=1}^{n-1} ((2*k+1)*(n-k)+(n-2*k)*a) (a/2)^(2*k) (n+1-2*k)_{k-1} / [k! (3/2)_k (n-k+1/2)_k]] This looks rather complicated so let's show the first terms : u(a,b)= exp(-a)/b * [+b^2/a^2 * ( 1 + a ) -b^4/a^4 * ( 1 + a + 1/3*a^2 ) +b^6/a^6 * ( 1 + a + 2/5*a^2 + 1/5*a^3/3 ) -b^8/a^8 * ( 1 + a + 3/7*a^2 + 2/7*a^3/3 + 1/105*a^4 ) +b^10/a^10 * ( 1 + a + 4/9*a^2 + 3/9*a^3/3 + 1/63*a^4 +1/945*a^5 ) -b^12/a^12 * ( 1 + a +5/11*a^2 +4/11*a^3/3 + 2/99*a^4 +1/495*a^5+ 1/10395*a^6) +...] e^(-a)*b/(a^2+b^2)*(1+a) may be obtained by collecting the two first columns inside the parenthesis. ( by the way a fine approximation of u(a,b) is e^(-a)*b/(a^2+b^2)*(1+a-b^2/3)*(1+b^4/(3*5*a^2)) ) === Subject: Re: New Zealand Mathematics Magazine > Continued fractions in a search for odd perfect numbers. > New Zealand Math. Mag. 19 (1982/83), no. 2, 63--69. > Does anyone have access to this who would be able to help out? .... The N.Z. Mathematics Magazine is a modest publication mainly for school teachers, but I happen to have the complete set. The Heyworth just send me your snail-mail address. Ken Pledger. === Subject: e^x - {1+(x/n)}^n hello.......doctor~ let be 0 e^x (Uniformly convergence) because f_n_(x) is increasing function on [a,b], then f_n_(b) is maximum. so, M_n = sup|f_n_(x) - e^x| = e^b - {1+(b/n)}^n --> 0 thank you very much for your advice. === Subject: Re: e^x - {1+(x/n)}^n Nntp-Posting-Host: apps.cwi.nl >hello.......doctor~ > >let be 0 >i need the fact that >f_n_(x) = e^x - {1+(x/n)}^n is >increasing function on [a, b]. > >but i can't show that. > >so, i need your advice. > >from this, i can show that f_n_(x) => e^x >(Uniformly convergence) >because f_n_(x) is increasing function on [a,b], >then f_n_(b) is maximum. > >so, M_n = sup|f_n_(x) - e^x| = e^b - {1+(b/n)}^n >--> 0 > >thank you very much for your advice. You need some more hypotheses. If, for example n = 1, so f_1_(x) = e^x - 1 - x, then f(-1) ~= 0.37 whereas f(0) = 0. For b > a >= 0, show f_n_'(x) >= f_n_(x) for all x >= 0. Try to show that this, together with f_n_(0) = 0, implies f_n_(x) >= 0 for all x and hence f_n_(x) >= 0 for all x >= 0. -- If we don't protect our forests, the bushlands will become Bushlands. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA === Subject: Re: e^x - {1+(x/n)}^n OpenPGP: id=31C8E9F7 > hello.......doctor~ let be 0 f_n_(x) = e^x - {1+(x/n)}^n is > increasing function on [a, b]. > Show that the Taylor series of f_n has positive coefficients, only, by using n^k >= n!/(n-k)! (for 0 <= k <= n). J. === Subject: Re: Is equal chatetas square triangle proved ...?? > Ecsouse me I forgot to tell PITAGORA's THEOREM.. > > You can find proofs of Pythagoras' Theorem all over the place, > but I don't know what chatetas are. Ecsouse -------> Excuse > PITAGORA's -------> Pythagoras' > chatetas -------> Katheten (german, plural of Kathete) In English there is only the word sides for the shorter > two sides of a rectangular triangle. But in German as well > as in some other European languages we use Kathete or some- > thing like that. It seems as if the Greeks did so first :-) .... The English word is cathetus (singular), catheti (plural), stright from Latin. However, these words are now so old-fashioned that most people have never met them. Ken Pledger. === Subject: Re: Is equal chatetas square triangle proved ...?? > > > Ecsouse me I forgot to tell PITAGORA's THEOREM.. > > You can find proofs of Pythagoras' Theorem all over the place, > but I don't know what chatetas are. > > Ecsouse -------> Excuse > PITAGORA's -------> Pythagoras' > chatetas -------> Katheten (german, plural of Kathete) > > In English there is only the word sides for the shorter > two sides of a rectangular triangle. But in German as well > as in some other European languages we use Kathete or some- > thing like that. It seems as if the Greeks did so first :-) .... > > > The English word is cathetus (singular), catheti (plural), > stright from Latin. However, these words are now so old-fashioned that > most people have never met them. I met the second one, and she lets me call her Cathy. === Subject: Re: Paranormal Test Proposal for ASTROLOGY >> The current protocol is to >> 1/ NOT LOOK AT WHAT THE PSYCHICS ARE DOING >> 2/ DEFINE A WATERTIGHT CONTRACT BEFORE ANYTHING IS SEEN >> 3/ SETUP A **SINGLE TEST** FOR THE CLAIMANT >> 4/ USE YOU GOT 5/10 1ST GO WHICH IS IN AVERAGE RESULTS >> SO YOU CANT EVER BE TESTED AGAIN. >> What a total bullshit, however rigorous looking, contractual offer. Actually this is bollocks. The challenge allows you to demonstrate > your amazing powerz (not that you have any) in any way you please. If > you wished to do so in a series of tests, you could do that, subject > to the usual stuff about having to being able to do something which is > highly unlikely by chance alone. In fact that's exactly how homeopathy > and other claims have been tested before - using hundreds of samples & > double blind testing. > IF any of you want to claim you have PROVEN any of my claims are >> unsubstantiated PUT FORWARD YOUR STATISTICALLY SIGNIFICANT RESULTS... >> BECAUSE YOU DONT HAVE ANY. If you have some claim about the >> paranormal then make it significant like we have to. You want us to disprove a negative now? Do you want us to disprove the existence of sentient road signs while > you're at it? Wait a second! I have proof that happens. Those road signs smell a drunken driver and lure them into hitting them all the time. What's jHerc's excuse? -- Cujo - The Official Overseer of Kooks and Trolls in dfw.*, alt.paranormal, alt.astrology and alt.astrology.metapsych. Colonel of the Fanatic Legion. FL# 555-PLNTY Motto: ABUNDANCE!. Charter Member - Digital Brownshirts and Library Gestapo. At the moment, half the spool of this group is subsequent to things that === Subject: Re: Help with exercise in Apostol > And yet, the benefits above notwithstanding, I think an exercise of > this kind relates better to a Modern Algebra course than to a Calculus > course. > > So I say, forget this silly exercise. Proceed onwards to the Calculus! quasi the exercise despite getting bogged down a bit. Your comment that it relates better to a Modern Algebra course got me thinking that I might want to start studying in that area as well. Can you recommend a text for self study that takes the same kind of ground up approach? Louis === Subject: Re: Help with exercise in Apostol > ... >the exercise despite getting bogged down a bit. Your comment that it >relates better to a Modern Algebra course got me thinking that I might >want to start studying in that area as well. Can you recommend a text >for self study that takes the same kind of ground up approach? Modern Algebra? Now you're talking about a math that's abstract by design (but lots of fun), whereas for Calculus, the abstraction was introduced as a necessity to put it on a rigorous foundation. As far as a recommended self study text for Modern Algebra, I'm not sure. 2 very widely used and respected texts are: Topics In Algebra Herstein A First Course in Abstract Algebra Fraleigh (I used Herstein's Topics In Algebra when I first learned). As a prerequisite to the study of Modern Algebra, I strongly recommend studying some elementary number theory -- not a complete course, just some of the basics -- for example, the first few chapters of a typical number theory text. The text I used when I learned Number Theory was: An Introduction to the Theory of Numbers Niven & Zuckerman and if you use that text, just do the first 2 chapters, then proceed to Modern Algebra. You can also try doing them in parallel. Another often recommended prerequisite to the study of Modern Algebra is Linear Algebra, but I think this would slow things down too much and with insufficient gain. In fact, I think it should be the opposite -- learn the basics of Modern Algebra at an introductory level first, then study Linear Algebra. Linear Algebra is made clearer when the concepts which are emphasized in Modern Algebra such as sets, relations, functions, injections, surjections, bijections, operations, inverses, generators, homomorphisms and isomorphisms have been developed. However Linear Algebra is an important prerequisite to the study of then next level of Modern Algebra, and is also absolutely essential to much of modern math, so should be studied right after an introductory Modern Algebra course, if not before. quasi === Subject: Re: Clarification on definition of limits > >>>Yeah, I'm trying to sneak up on topology from behind. Shh! >> >>I highly recommend this text: >> >>General Topology >>Stephen Willard >> >>It's so well written that, in my opinion, it's in a class by itself. > > If you try to work through Willard, since I have a copy, I'd be glad to help with hints and explanations. It would give me an excuse to review the material I once knew, and tackle the parts I never started. quasi === Subject: Re: Mathematical concepts >> I do not believe that a concept can be defined, and >> is rarely precise. It is, however, something which >> can be used, and can only be tested by whether it >> can be used. I can put into words the cardinal and >> ordinal concepts of numbers, but the concepts are >> not understood unless the person can use them. > >I disagree. Concepts can be defined. > > This is the opinion of the educationists, who have done > a good job in destroying education. I don't believe educationists know anything about concepts. All they know about is their drills and behavioral tests, but not tests for concepts and in general little that will aid the student in his concept-forming process. Evolutionary epistemology is the name of this field. >A concept is nothing more than a cognitive capacity to >discriminate. > >It won't be hard to see, for instance, if a student >has the concept of a number. For that, he merely has >to distinguish numbers from not-numbers. A sufficient >precision in successive discrimination tasks will demonstrate >the existence of the concept in the student. > > This is utter nonsense. There are dozens of number > concepts, some of which are applicable to the same > numbers. The counting numbers have both the > cardinal and ordinal concepts, both of which generalize > further. In addition, there are several other types > of numbers within which these can be embedded. Dozens of number concepts. I would certainly love to hear about that. If you mean that there is no singular cognitive process related to discriminating the number-ness of a piece of information, I think you are making the mistake of the mathematician, instead of the educationist, in assuming that there is some kind of reality to number theory which indeed houses so many conceptions of number that seem formally incommensurable. Is it clear what I mean by that? However, note that I say nothing about how concepts can be instilled. This is a hard problem, and I don't think educationists have addressed it. They definitely haven't addressed it in my country. >In particular, I think we can say that mathematics is translucent >enough to allow for a Turing Test for mathematical concepts. > > If you try to come up with one, a machine given the > limited conceptual capacity of someone with your > attitude might pass, but a first-class mathematician > might not. I'm glad that you can infer my conceptual capacity from my attitude. So, your words are to take precedence about cognitive science since you deem yourself to be a first-class mathematician, and since you have asserted me not to be a first-class mathematician, or neither have the capacity to be one, well everything ought to be clear. I don't believe you even understood what I mean above. (See, this is a test for a mathematical concept after all...) I will be patient enough to explain once more. I am merely telling you that the whole purpose of making mathematics is to be symbolic enough so that every concept can be eventually put to test. I wouldn't normally discuss this without first discussing what separates mathematics from other fields of inquiry. At any rate, let us proceed. Show me an example of a *mathematical* concept that cannot be tested first, and then I may start taking your comments seriously. This ought to be a quite basic concept, like set, space, etc., so that we can discuss it briefly. Complex concepts would be missing the point, since few people truly understand those. If this is a fundamental property, it ought to be *independent* of complexity, which is merely a matter of difficulty in perceiving, explaining, etc. Turing himself had a much stronger claim about this, he implies in his computational intelligence paper that every aspect of intelligence, *and not just making mathematics, which is just one aspect* can be detected with proper questions prepared by the judge. Now, I'm going to assume that Turing is a first-class mathematician. If we are to take Turing seriously (and although he has the wrong attitude, we must believe him since he's first class) then mathematical concepts which are merely components of mathematical intelligence could be tested by a competent judge in a Turing test. So, how does Turing's view stack up against yours? Or are you a 0-class mathematician? To disagree with Turing would take more power than a cheap attempt at trying to discredit a USENET poster who did the mistake of responding to you. If you haven't already read Turing's paper: it's all over the web. http://www.abelard.org/turpap/turpap.htm But do read it. Remember. Turing is a first-class mathematician. His opinion MUST count. -- Eray === Subject: Re: Addition and subtraction of Rational Expressions > Hey everyone, i've got a question for ya. For this problem Add or subtract > (10/(b-9)) + (5/(b^2 - 81)) I end up with (5(2b+19))/(b-9)(b+9) as my answer. The book says the answer is (10b+95)/(b-9)(b+9) Which I guess are basically the same answer, but why doesn't the book > do the last step I do that factors out what's on top in the fraction? Jon If one is using standard NG notation, the correct result should be (10b+95)/(((b-9)(b+9)) or even (10*b+95)/(((b-9)*(b+9)) because in standard NG notation (10b+95)/(b-9)(b+9) = (10b+95)(b+9)/(b-9) === Subject: Re: Addition and subtraction of Rational Expressions what does NG stand for? === Subject: Re: Addition and subtraction of Rational Expressions > what does NG stand for? Notational Goof-ups. === Subject: Re: Addition and subtraction of Rational Expressions > what does NG stand for? News Group. === Subject: Re: Addition and subtraction of Rational Expressions > Hey everyone, i've got a question for ya. For this problem Add or subtract > (10/(b-9)) + (5/(b^2 - 81)) I end up with (5(2b+19))/(b-9)(b+9) as my answer. The book says the answer is (10b+95)/(b-9)(b+9) Which I guess are basically the same answer, but why doesn't the book > do the last step I do that factors out what's on top in the fraction? Because if they did then someone would post to sci.math to say that he ended up with (10b+95)/(b-9)(b+9) while the book says the answer is (5(2b+19))/(b-9)(b+9) and why didn't the book multiply out what's on top in the fraction? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Addition and subtraction of Rational Expressions So the both answers are correct? It's just hard to read the mind of the person grading the test. === Subject: Re: Addition and subtraction of Rational Expressions > So the both answers are correct? It's just hard to read the mind of > the person grading the test. It's hard to read the mind of a person posting a followup to some message when that person posts nothing of the message to which she is following up. But I think I know what you're talking about. In which case, you've changed the subject: you made believe you were interested in why some book gave the answer in a form differing from yours, but what you are really interested in is passing a test. That is, you're not actually interested in the mathematics, just in the credential you hope to obtain. There's nothing wrong with that, but it's more ethical to be above board with your motivations when you ask other people for help. Now that we've got that settled, here are some guides to reading the mind of the person grading the test: If somewhere on the test sheet it says, leave your answers in factored form, well, that's a pretty big hint as to what the person grading the test wants. If the person grading the test is the same as the person teaching the class, keep your eyes open during the semester to see what form she usually chooses as the final form of the answers in the examples she gives, and do thou likewise. If all else fails, write something like this on your test: The answer is 2(b + 17), which is the same thing as 2 b + 34, and I haven't been able to figure out which of these two forms you think is simpler. I can't imagine any teacher giving anything less than full marks for this (unless, of course, the correct answer was 3 q - 47). -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Addition and subtraction of Rational Expressions Gerry, You're right....you caught me.....I am so sneaky......look at me.....Mr. sneaky, I asked for the correct answer to a problem and what i really meant was what's the correct answer to the problem? Jon === Subject: Re: Addition and subtraction of Rational Expressions > Gerry, You're right....you caught me.....I am so sneaky......look at > me.....Mr. sneaky, I asked for the correct answer to a problem and what > i really meant was what's the correct answer to the problem? Ah, but *why* did you want the correct answer to the problem? You made as if it arose from mathematical concern over the discrepancy between your answer and the one in the book, but in fact you wanted the correct answer so you could score well on a test. As I've already said, there's nothing wrong with that - I just prefer to see people being open & above board on these matters. By the way, did you get beyond the part of my message that you've responded to here, and read my attempt to answer your question? If so, did you find it at all useful? I only ask because in such circumstances it's considered polite to show gratitude, rather than sarcasm. But perhaps I wasted my time in attempting to help you, and shouldn't bother in the future, if there is one. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Addition and subtraction of Rational Expressions >So the both answers are correct? It's just hard to read the mind of >the person grading the test. Yes, both are correct (unless your teacher has a strong bias towards one form or the other). Simplification of an expression just means to rewrite it in another form which is simpler in some sense, but what's simpler is relative to a given problem. So for example, what simpler x^2+x or x(x+1)? About the same, but one form might be more useful than the other in a given context. So the choice is optional, either one is ok. On the other hand some simplifications are mandatory, and hence, automatic: x + x should be automatically be simplified to 2x x - x should be automatically be simplified to 0 x * x should automatically be simplified to x^2 But what about x^6-1? Is it already simplified or should it be factored? Well, x^6-1 is brief and simple looking as it is, and if you factor it completely, you get (x+1)(x-1)(x^2+x+1)(x^2-x+1), which, you have to admit, makes it look less simple. So should it be factored? It depends on the situation. If the factored form helps you in the next steps of the problem, then yes, factor it, but otherwise, why not leave it alone? quasi === Subject: Re: Addition and subtraction of Rational Expressions > >> Hey everyone, i've got a question for ya. For this problem >> Add or subtract >> (10/(b-9)) + (5/(b^2 - 81)) >> I end up with >> (5(2b+19))/(b-9)(b+9) >> as my answer. The book says the answer is >> (10b+95)/(b-9)(b+9) >> Which I guess are basically the same answer, but why doesn't the book >> do the last step I do that factors out what's on top in the fraction? > >Because if they did then someone would post to sci.math >to say that he ended up with (10b+95)/(b-9)(b+9) >while the book says the answer is (5(2b+19))/(b-9)(b+9) >and why didn't the book multiply out what's on top in >the fraction? Haha, but so true. What that example illustrates is that simplification is, in many cases, not an absolute concept. It's not cast in stone (except when it is). In other words: Simplicity, like beauty, is in the eye of the beholder. (quote from some great philosopher ... ok, me) quasi === Subject: Re: Addition and subtraction of Rational Expressions > Hey everyone, i've got a question for ya. For this problem Add or subtract > (10/(b-9)) + (5/(b^2 - 81)) I end up with (5(2b+19))/(b-9)(b+9) as my answer. The book says the answer is (10b+95)/(b-9)(b+9) Which I guess are basically the same answer, but why doesn't the book > do the last step I do that factors out what's on top in the fraction? Jon Perhaps they felt that any subsequent operation would have to start by multiplying out 5(2b+19) to get (10b+95) again. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Multivariable calculus help? > Given a function A(y) and the distribution function of y, f(y), I can > find the expectation value of A(y) by the formula: > > Integral A(y) f(y) dy > > Now, if y is itself a function of another variable x with its own > distribution function g(x), the above is also equal to: > > Integral A(y(x)) g(x) dx > > So to get f(y), given g(x) and y(x), I would say: > > f(y) dy = g(x) dx > > So you're telling me a function of y is equal to a function of x? This > can only happen if the function is constant. > > so: > > f(y) = g(x) / (dy/dx) > > Am I right? > > No. I think you're trying to do a substitution for > Integral A(y) f(y) dy, where y = y(x). Not really.... It's more like I'm aware of the form of that substitution, and I'm trying to solve for f(y) based on it. I realize that I mistakenly called f(y) and g(x) the distribution functions. I meant probability density functions. Does that make my problem more clear? === Subject: Re: Solving a summation > > I need help with a problem I've run into while developing a piece of > software. The software basically calculates a series of numbers of the > form: > > 1 / (1 + ix) > > Where I is an integer. So the resulting series looks like: > > 1/(1+x) + 1/(1+2x) + 1/(1+3x) ... > > The input to the algorithm is the number of terms, n. The output of the > algorithm is x. The constraint is that the sum of all the terms must be > some predetermined value, v. The only restriction on x is that it is > real, positive, non-zero. > > So the equation looks like this (sorry for the bad ASCII art): > > n > --- > 1 > / ------ = v > --- 1 + ix > i=1 > > So what I need is a way to solve that summation to get x in terms of n > and v. I've spent several days banging my head against this, and I just > can't find any way to get x out of the summation, or to split the > fraction into something that I can work with. > > Any advice would be appreciated. Two useful closed-form approximations for x will be given. Which of the two approximations should be used seems to depend, roughly, on whether v/n is close to 0 or not. 1. For v/n close to 0: I will use generalized harmonic numbers, although the approximation could also be expressed in terms of related functions. See , for example. HarmonicNumber[n, p] represents the sum, from i = 1 to n, of i^p. For a given value of n, we will need to compute HarmonicNumber[n, 1], HarmonicNumber[n, 2], and HarmonicNumber[n, 3], which for brevity I shall denote as H1, H2, and H3, resp. Then, using the first three terms of a series expansion, x is approximately H1/v - H2/H1 + (H1*H3 - H2^2)/H1^3 * v. Example 1a: For reference, note that if n = 100 and x = 10, then v = 0.503491... Now suppose that we were given n = 100 and v = 0.503491... and asked to approximate x. Since v/n = 0.005... is close to 0, we use the first approximation. The resulting approximate value for x is 10.0005 . Example 1b: For reference, note that if n = 10 and x = 1, then v = 2.01987... Now suppose that we were given n = 10 and v = 2.01987... and asked to approximate x. Is v/n = 0.2... close enough to 0 so that using this approximation is reasonable? It seems so, since the resulting approximate value for x is 1.0098 . 2. For v/n not close to 0: x is approximately (5/3 + Sqrt(2(1 - v/n * e^(1 - v/n))))/v - (1 + 2/3 * e^(1 - v/n))/n . Example 2a: For reference, note that if n = 100 and x = 0.01, then v = 69.0653... Now suppose that we were given n = 100 and v = 69.0653... and asked to approximate x. Since v/n = 0.69... is not close to 0 (at least for our purpose here), we use the second approximation. The resulting approximate value for x is 0.01002 . Example 2b: Referring to example 1b, suppose that we were given n = 10 and v = 2.01987... but decided to approximate x using the second approximation. The resulting approximate value for x is 1.0969, and so is not as good as using our first approximation. Although I have presented the transition between the two approximations as being based on whether v/n is close to 0 or not, I am not entirely sure that that is the best way to determine the transition. Further examination may be useful... David W. Cantrell === Subject: Re: well- versus regular ordering of R (was, a crank thread) >I have very little feel for what >a well-ordered uncountable set looks like, except that >it's well-ordered (duh). Which is exactly the right insight. Such sets are constructed by uncountably many arbitrary choices as to what comes next, and it makes no difference what you choose. > I don't even know if it's >possible to describe a well-ordering on the reals >in any useful way, since I've only seen a proof that >such a thing exists, and it uses the axiom of choice >in a pretty big way. Again, that's the right intuition. You can start the well-ordering any way you want, just write a countable sequence of your choice as the starting part of the well-order. For example you could choose some sequential ordering of the rationals as the starting order, in which case, you would then already have a dense set, so any new terms would have to fill in holes all over the place. Any concrete description of such a well-orderering of the reals would only be able to account for countably many terms, so for the rest, you basically just have to imagine it as a kind of arbitrary continuation of the choice process (unless you have special glasses fitted with Zorn's lenses in which case you can see the whole thing clearly). quasi === Subject: Re: well- versus regular ordering of R (was, a crank thread) > Robert Low , speaking sweet reason to > one of the usual suspects, incidentally raises a question in > my (cranky, because of the heat; but I hope, not crankish) > mind. > >A well-ordering of the reals is a completely >different beast from the usual ordering. When you try to construct >this uncountable 'sequence' of nested intervals by going to the >'next real up' according to the well-ordering you will sometimes >be getting a smaller (in the usual sense) real, so that the >ordering isn't compatible with the nesting. > > Is it possible to say anything interesting about how different > the usual ordering of the reals is from a well-ordering (where > a well-ordering might be *any*, or *every*, according to taste)? > For instance, what are the possible types (order types, > homeomorphism types, what have you; even cardinalities, in > case the continuoum hypothesis is not assumed, or is denied) > of sets of pairs of reals (x,y) such that, for a given well-ordering > W, x < y but x W y? What is the simplest such set of pairs? > The smallest? The largest? The nastiest? Etc., etc. > Here's a thought to start (warning: this is based on my ill-tested understanding of set theory): Given a well-order W on the reals, let W(x) be the set of all elements that are less than x, in the well-order. It seems to me that there is exactly one x (0) with W(x) = {}; a countable number of reals x having W(x) finite; and an uncountable number of them having W(x) countable. (And I /think/ that's all of it - there is no largest element in the well-order, so every W(x) is at most countable - is that correct, or am I just plain wrong?) That would tend to imply that for every real x, there are always an uncountable number of y's with y < x in the usual order, but x < y in the well-order. Then the probability that than any pair x,y has the same usual order as the well-order is 0, yes? I'm kinda talking out of my hat here while I'm wildly waving my hands here, and (as long as I'm mixing my metaphors) probably sticking my foot in my mouth... === Subject: Re: well- versus regular ordering of R (was, a crank thread) >Then the probability that than any pair x,y has the same usual order as >the well-order is 0, yes? No, the probablity that a random pair (x,y) (for any reasonable concept of random') has the same usual order as the well order is, by symmetry, 1/2, not 0. quasi === Subject: Re: well- versus regular ordering of R (was, a crank thread) > I don't even know if it's possible to describe a well-ordering on the > reals in any useful way, since I've only seen a proof that such a > thing exists, and it uses the axiom of choice in a pretty big way. One does not need the axiom of choice to show that there exists an uncountable well-ordered set. Defining a cardinal as an ordinal whence there exists no bijection to a smaller ordinal, one can show without choice that there exists a cardinal strictly greater than any given ordinal. See, for example, the first chapter of Kunen's set theory text. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: well- versus regular ordering of R (was, a crank thread) > I don't even know if it's possible to describe a well-ordering on the > reals in any useful way, since I've only seen a proof that such a > thing exists, and it uses the axiom of choice in a pretty big way. One does not need the axiom of choice to show that there exists an > uncountable well-ordered set. Non-responsive. RL wanted a proof that a well-ordering exists -> on the reals <- not just a proof that there exists a well-ordered uncountable set. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: well- versus regular ordering of R (was, a crank thread) > >One does not need the axiom of choice to show that there exists an >uncountable well-ordered set. You need something. Are you saying that the ordinary axioms of set theory are sufficient to prove the existence of an uncountable well-ordered set? I agree that you can get an uncountable set, for example, P(N), but how can you prove, without some possibly weakened form of the axiom of choice, that P(N) can be well-ordered? quasi === Subject: Re: well- versus regular ordering of R (was, a crank thread) > >One does not need the axiom of choice to show that there exists an >uncountable well-ordered set. You need something. Are you saying that the ordinary axioms of set theory are sufficient > to prove the existence of an uncountable well-ordered set? The set of countable ordinals is uncountable & well-ordered, no? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: well- versus regular ordering of R (was, a crank thread) > >> >>One does not need the axiom of choice to show that there exists an >>uncountable well-ordered set. >> You need something. >> Are you saying that the ordinary axioms of set theory are sufficient >> to prove the existence of an uncountable well-ordered set? > >The set of countable ordinals is uncountable & well-ordered, no? Sure, but to prove the existence of the set of all coutable ordinals, how do you prove that without invoking some form of the axiom of choice? quasi === Subject: Re: well- versus regular ordering of R (was, a crank thread) >> >>> >>>One does not need the axiom of choice to show that there exists an >>>uncountable well-ordered set. >>>> You need something. >>>> Are you saying that the ordinary axioms of set theory are sufficient >>> to prove the existence of an uncountable well-ordered set? >> >>The set of countable ordinals is uncountable & well-ordered, no? >Sure, but to prove the existence of the set of all coutable ordinals, >how do you prove that without invoking some form of the axiom of >choice? Given a set X, take the set A of all well-orderings of subsets of X (i.e. A is a subset of P(P(X^2)), and B is an element of A iff B is a subset of P(X^2) and B is a well-ordering of a subset of X). For each element of A, replace that element by the corresponding order-type, which is an ordinal (using the Axiom of Replacement). The resulting set of ordinals is itself an ordinal, and there do not exist a subset Y of X and well-ordering of Y such that the order-type is this ordinal. Specifically, there is no one-to-one mapping from this ordinal to X - the ordinal is too big. At no point was the Axiom of Choice, or any of its equivalents, used in the construction of this ordinal. When you do this for X = N, the ordinal that is produced by the above contruction is the smallest uncountable ordinal, omega_1 or aleph_1. A subset Z of P(Y^2) is a well-ordering of Y iff (1) for all x, y, and z in Y, ((x,y) in Z and (y,z) in Z) => (x,z) in Z; (2) for all x, y in Y, ((x,y) in Z and (y,x) in Z) => x = y; (3) for all y in Y, (y,y) in Z; (4) for all nonempty C in P(Y), there exists x in C, for all y in C, (x,y) in Z. (1) is transitivity and (2) is antisymmetry, so that Z is an order. (3) is reflexivity, so that Z is weak (or reflexive) order. (4) is the statement that every nonempty subset of Y has a smallest element (i.e. Z is a well-ordering). (4) also implies that Z is a linear (or total) order. ----- === Subject: Re: well- versus regular ordering of R (was, a crank thread) quasi a .8ecrit : >> >> >>> >>>>One does not need the axiom of choice to show that there exists an >>>>uncountable well-ordered set. >>> >>>You need something. >>> >>>Are you saying that the ordinary axioms of set theory are sufficient >>>to prove the existence of an uncountable well-ordered set? >> >>The set of countable ordinals is uncountable & well-ordered, no? > Sure, but to prove the existence of the set of all coutable ordinals, > how do you prove that without invoking some form of the axiom of > choice? > Existence of a set of orders (on A) is a consequence of axioms of comprehension and power set > quasi === Subject: Re: well- versus regular ordering of R (was, a crank thread) > Sure, but to prove the existence of the set of all coutable ordinals, > how do you prove that without invoking some form of the axiom of choice? Why do you keep ignoring my previous post? > There exists a cardinal strictly greater than any ordinal. As I > stated before, a cardinal is by defintion an ordinal whence there > exsits no bijection to a smaller ordinal. Here is the proof from Kunen. > > Let a be an infinite ordinal. Let W = {R in P(a x a): R > well-orders a}. > sup({order-type(a,R): R in W}) is a cardinal greater than a. Thus, there exists an uncountable cardinal k. {a in k: a is countable} (which is called aleph-1 or omega-1) is the set of all countable ordinals. It exists by the above proposition and the axiom of comprehension. BTW, Gerry M. is correct: While it is true that one does not need AC to show the existence of uncountable well-ordered sets, one does need some version of choice to construct a well-ordering of the reals. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: well- versus regular ordering of R (was, a crank thread) > > >>One does not need the axiom of choice to show that there exists an >>uncountable well-ordered set. >> >> > >You need something. > >Are you saying that the ordinary axioms of set theory are sufficient >to prove the existence of an uncountable well-ordered set? > >I agree that you can get an uncountable set, for example, P(N), but >how can you prove, without some possibly weakened form of the axiom of >choice, that P(N) can be well-ordered? > > I did not claim one can order P(N). I said that there exists a cardinal strictly greater than any ordinal. As I stated before, a cardinal is by defintion an ordinal whence there exsits no bijection to a smaller ordinal. Here is the proof from Kunen. Let a be an infinite ordinal. Let W = {R in P(a x a): R well-orders a}. sup({order-type(a,R): R in W}) is a cardinal greater than a. Or, as some would like to phrase it, even without choice, there exists aleph-a for each ordinal a. card(aleph-a) < card(aleph-b) for a < b. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: well- versus regular ordering of R (was, a crank thread) > >> >> >>>One does not need the axiom of choice to show that there exists an >>>uncountable well-ordered set. >>> >>> >> >>You need something. >> >>Are you saying that the ordinary axioms of set theory are sufficient >>to prove the existence of an uncountable well-ordered set? >> >>I agree that you can get an uncountable set, for example, P(N), but >>how can you prove, without some possibly weakened form of the axiom of >>choice, that P(N) can be well-ordered? > >I did not claim one can order P(N). I said that there exists a cardinal >strictly greater than any ordinal. Um ... I think you need the axiom of choice, to define the ordinals, unless you want to settle for only countably many ordinals. quasi === Subject: Re: well- versus regular ordering of R (was, a crank thread) > Um ... I think you need the axiom of choice, to define the ordinals, > unless you want to settle for only countably many ordinals. Not at all. The proof which you omitted proves the opposite. The axioms of inifinity and power set (and the others, but not choice) give you what you need. Study your set theory. I already gave you one reference, i.e., Kunen. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: well- versus regular ordering of R (was, a crank thread) > >> Um ... I think you need the axiom of choice, to define the ordinals, >> unless you want to settle for only countably many ordinals. > > >Not at all. The proof which you omitted proves the opposite. The >axioms of inifinity and power set (and the others, but not choice) give >you what you need. Study your set theory. I already gave you one >reference, i.e., Kunen. Study my set theory? Kunen? Ok, I'll go to the library right now to look that up. Wait, the the library is closed, so maybe tomorrow. So what you're saying is that the power set axiom, together with the axiom of infinity are sufficient to define the ordinals up to any given cardinal? If that's the case, I was unaware of it. I was under the impression that the following statements are equivalent: (1) there exists an ordinal of any given cardinality (2) any set can be well-ordered (just make a bijection to the appropriate ordinal) (3) the axiom of choice holds quasi === Subject: Re: well- versus regular ordering of R (was, a crank thread) quasi a .8ecrit : >> >> >>>Um ... I think you need the axiom of choice, to define the ordinals, >>>unless you want to settle for only countably many ordinals. >> >> >>Not at all. The proof which you omitted proves the opposite. The >>axioms of inifinity and power set (and the others, but not choice) give >>you what you need. Study your set theory. I already gave you one >>reference, i.e., Kunen. > Study my set theory? Kunen? Ok, I'll go to the library right now to > look that up. Wait, the the library is closed, so maybe tomorrow. So what you're saying is that the power set axiom, together with the > axiom of infinity are sufficient to define the ordinals up to any > given cardinal? No, he didn't say that. He said that it is possible to get cardinals strictly greater than any given ordinal. So from Axiom of infinity, you ger w_0 (aka aleph_0), then the set of all well-orderings of w_0 exists (by power set and selection) and is well ordered (of order type w_1, aka aleph_1) etc. up to aleph_(w_0) (defined by union), then... up to *any* aleph_alpha, as long as alpha is a defined ordinal (so you get huge monstruosities like aleph_aleph_42, etc. ) It is *still* possible that card(P(IN)) be greater than *all* those (in fact, it can be weakly inaccessible), so there is no way to use this to get a well-ordering on IR. Last, without choice, other sets can be without well-orderings, both very large ones and also very small, like Dedekind-finite sets... If that's the case, I was unaware of it. I was under > the impression that the following statements are equivalent: (1) there exists an ordinal of any given cardinality (2) any set can be well-ordered (just make a bijection to the > appropriate ordinal) (3) the axiom of choice holds quasi === Subject: Re: well- versus regular ordering of R (was, a crank thread) Is it possible to say anything interesting about how different > the usual ordering of the reals is from a well-ordering (where > a well-ordering might be *any*, or *every*, according to taste)? > For instance, what are the possible types (order types, > homeomorphism types, what have you; even cardinalities, in > case the continuoum hypothesis is not assumed, or is denied) > of sets of pairs of reals (x,y) such that, for a given well-ordering > W, x < y but x W y? What is the simplest such set of pairs? > The smallest? The largest? The nastiest? Etc., etc. Lee Rudolph Well, it is known that there is no analytic well-ordering of the reals (a set is analytic if it is the continuous image of a Borel set, so what I'm saying is that no continuous function from R^n [for any n] to the plane carries a Borel set onto the graph of a well-ordering W); for details consult Jech's _Set Theory_, section 41. So any well-ordering of the reals is in some sense (i.e. in the sense of descriptive set theory) complex, whereas the graph of < is open in the plane, and so dead simple. On the other hand, assuming the axiom of constructibility there is a sigma^1_2 well-ordering of the reals, i.e. a well-ordering whose graph is the projection onto the first two coordinates of a subset of R^3 whose complement is analytic. So in this case we have a fairly tight bound on how complex a well-ordering is required to be. Hope that satisfies your curiosity somewhat. Bob Beaudoin === Subject: Re: A problem on probability >dies (basically when we run out of 'B' molecules). Probably has >>no correlation to reality. >> >>for A=10000 B=20000 C=100000 >> >>0 1 2 3 4 5 6 >> >>0 A's left when B gone >> >>('AC', 199) >>('AAC', 6) >>('BC', 18273) >>('C', 69401) >>('AAAC', 2) >>('BAC', 420) >>('BAAC', 35) >> >>More chains of length 1 and 2, but still no 3's. I wonder what >>would happen if I could run 10**23? > Why is #BC + #BAC + #BAAC != 20000 ??? Anyway, to the original problem: Since B is the limiting resource, no B's will remain unpaired. For a given > B, there are 10 times as many C's as there are A's. Therefore, the > probability of ending up with an empty BC chain is 10/11. The prabability of > a chain starting with BA is similary 1/11. The probability of a BA chain > pairing up with a C is again 10/11. In general, if Pn is the probability of > a chain with exactly n A's, then Pn = 10/11 * (1/11)^n -Michael. I think your argument gives the right answer (though I'm not sure I understand why it is correct, since it isn't true that the free A,B,C stay in the same proportion throughout the reaction). At least, under certain assumptions the original problem gives rise to Pn as above. This is the model where you pick a right hand edge (RH of free A or B) uniformly at random and join it to a left hand edge (LH of free A or C) picked uniformly at random (so long as that doesn't form a loop). Keep going until there is nothing left to join. I don't know if this is a physically reasonable model but if it is I think it should correspond to chemistry at absolute zero. The number of chains (A^r)C should be about 80000*10/11*(1/11)^r, and the number of chains B(A^r)C should be about 20000*10/11*(1/11)^r. I'm not sure if mensanator's simulation corresponds to the above model. It agrees fairly well with this formula, but there is a slight discrepancy for AAC. This is a result of doing a simulation using the above model for A=100000000, B=2A, C=10A. r (A^r)C Theoretical B(A^r)C Theoretical 0 727263492 727272727.27 181826681 181818181.82 1 66126909 66115702.48 16520246 16528925.62 2 6008258 6010518.41 1502231 1502629.60 3 546662 546410.76 136980 136602.69 4 49708 49673.71 12642 12418.43 5 4551 4515.79 1106 1128.95 6 383 410.53 104 102.63 7 33 37.32 10 9.33 8 3 3.39 0 0.85 9 1 0.31 0 0.08 10 0 0.03 0 0.01 To derive the above distribution, it is OK to forget the process of molecule construction and just count the number of final configurations (this is a fact about continuous time Markov chains on graphs). Label each original molecule and count the final states, not forgetting that a set of identical final chains doesn't come with an order. For example if there was 1 A, 2 Bs and 10 Cs to start with, then the number of ways of getting the final state {AC BC BC C C C C C C C} is 10*9*8/2 because there are 10*9*8 ways of choosing three Cs from ten to pair with A,B,B, and you divide by two because the final BCs are indistinguishable. The number of ways of getting the final state {BAC BC C C C C C C C C} is 10*9. In general, let's say there are A As, B Bs and C Cs to start with and C>=A+B. The final chains will be (A^r)C and B(A^r)C. Let C_r be the number of final (A^r)C and D_r be the number of final B(A^r)C. The number of ways of labelling this final state is N = A!B!C!/(C_0!C_1!C_2!...D_0!D_1!D_2!...) and the constraints are sum(D_r)=B, sum(C_r)=C-B, sum(r(C_r+D_r))=A. If A,B,C are large, then this distribution on the set of possible {C_0 C_1 ... D_0 D_1 ...} will be very concentrated, and so we may maximise N. This could be done by a Stirling approximation which corresponds to an entropy calculation, so we need to extremise sum [ C_rlog(C_r)+D_rlog(D_r)-mu*D_r-nu*C_r-lambda*r(C_r+D_r) ] where mu,nu,lambda are Lagrange multipliers. This results in C_r=(C-B)*p^r*q, D_r=B*p^r*q where p=A/(A+C) and q=C/(A+C), as above. --- You could also consider introducing a non-zero temperature. Now the process has bonds forming at rate R for each possible new bond and broken at rate R' for each existing bond (this assumes each bond has the same energy). You wait until the mixture is in equilibrium. Then you need to consider the final states (A^r) (say A_r of these for r>=1) and B(A^r) (say B_r of these) too, and you need to introduce a temperature parameter. The corresponding calculation is to extremise sum [ A_rlog(A_r)+B_rlog(B_r)+C_rlog(C_r)+D_rlog(D_r) -mu*(B_r+D_r)-nu*(C_r+D_r)-lambda*r(A_r+B_r+C_r+D_r) ] and the solution is A_r = (A'-(B+C)+BC/A')p^r*q (r>=1) B_r = B(1-C/A')p^r*q C_r = C(1-B/A')p^r*q D_r = (BC/A')p^r*q where A'=Aq/p, q=1-p, and p is a free parameter which can be chosen in the range 0 to A/(A+C), and C>=A+B. p parametrises temperature, with 0 corresponding to infinite temperature and A/(A+C) corresponding to absolute zero (the original calculation). --- Alex === Subject: Re: A problem on probability > Here is my solution. Please correct me if you think there is a mistake > somewhere. > > Recall that [A] = x, [B] = 2x and [C] = 10x. Let me use '|' to denote > a link separating two compounds. (Please do not confuse with OR > operator.) As you put it correctly, for a particular link, > > P(A|A) = ([A]/([A]+[B]))*([A]/([A]+[C])) = 1/33 > P(B|A) = ([B]/([A]+[B]))*([A]/([A]+[C])) = 2/33 > P(A|C) = ([A]/([A]+[B]))*([C]/([A]+[C])) = 10/33 What does that mean???? > > Probability of forming a chain of length n is: > P(B(|A|)^nC) = P(all |A|'s are linked together) * (P(A|A))^(n-1) * > P(B|A) * P(A|C) > > Let us try to find P(all |A|'s are linked together) first. For the > sake of explanation, let us take an example with n=3. The chain we are > looking for here is B|A|A|A|C. > > Let us enumerate possible link configurations (=2^3) for |A|A|A| where > n=3. Note that there are four links with the two end-links referring > to the links with B and C. > > |A|A|A| > |A A|A|A| > |A|A A|A| > |A|A|A A| > |A A|A A|A| > |A A|A|A A| > |A|A A|A A| > |A A|A A|A A| > > Thus, in general, one out of 2^n link configurations is |A|A|...|A|. > Consequently, P(all |A|'s are linked together) = 2^n > > Therefore, P(B(|A|)^nC) = 1/(2^n) * (P(A|A))^(n-1) * P(B|A) * P(A|C) > = 1/(2^n) * 1/33^(n-1) * 2/33 * 10/33 > = 5/(2^(n-2) * 33^(n+1)) > > Just to cross-check, if n = 0, we get P(B|C) = 20/33. Again, what does that mean? > > We also know that, > P(B|C) = ([B]/([A]+[B]))*([C]/([A]+[C])) = 20/33, which matches with > above. > > Please let me know what you think. I am quite dubious. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: A problem on probability Hi James, also who took time to give me various directions to solve this > problem. Here is my solution. Please correct me if you think there is a > mistake somewhere. I suspect I would agree with your analysis if I understood it, but I don't understand how your link configurations are chosen, what they have to do with BAAAC. Putting your notation in my own terms this looks like 1, 2, 2, 2, 3, 3, 3, and 4 molecules respectivly. > |A|A|A| BAAAC > |A A|A|A| BA,AAAC > |A|A A|A| BAA,AAC > |A|A|A A| BAAA,AC > |A A|A A|A| BA,AA,AAC > |A A|A|A A| BA,AAA,AC > |A|A A|A A| BAA,AA,AC > |A A|A A|A A| BA,AA,AA,AC But I don't think my understanding of that is too important. I think now that my earlier analysis could be improved. What I didn't include was the probability that a particular innie is not mated with an outies. The simplest method would be to assume that all outies find innies (still randomly), although that would not be physical for thermodynamic reasons -- the entropy of that would be very low. But still, it serves as a good toy model to see how to work things out. If there are [A}+[B] = 3x outies mated, then there are 3/11 of the innies mated, or 3/11 x of A and 30/11 x of C (randomly). 80/11 x of the >C remains unmated. 10/11 x produces BC 20/11 x has at least one A, ?AC: (8/11)(20/11) x of that remains unmated, >AC (1/11)(20/11) x finds a B at that point, BAC (2/11)(20/11) x finds another A, ?AAC: ... (2/11)(2/11)(20/11) x finds another A. ?AAAC: ... (2/11)^(n-1)(20/11) x, at least n A's, ?A^nC (1/11)(2/11)^(n-1)(20/11) x, BA^nC So, the quantity of BA^nC, in the case where all the A and B outies bond, is q(BA^nC) = (1/11)(2/11)^(n-1)(20/11) x = (10/11)(2/11)^n x But that would mean a very low entropy, since there is only one way to select all of A and B. It may be, though, that the bond energy is high enough that you can neglect the unbonded fraction of the outies. In my own case, I'm not familiar enough with all this to make that call. What I will do later is find out what percentages of [B], [C], and [A] (innies and outies) find mates. The calculation starting with C will probably look very similar with some adjustment for fewer A and B bonding. The calculation could also be run starting with the B and working in the other direction -- this would be a good check; all the q(BA^nC) should agree when calculating from either the left or the right. For this model, I think we can let the number of accessible states n(r,s,t,u) = C( [B]N, [B]N*r )*C( [A]N, [A]N*s )* C( [A]N, [A]N*t )*C( [C]N, [C]N*u ) and the energy U(r,s,t,u) = ( [B]N*r + [A}N*s )*E = ( [A]N*t + [C]N*u )*E where C(n,m) = n!/(m!(n-m)!), N is Avagaodro's number, r is the fraction of B bonding, s of A's outies, t of A's innnies, and u of C's Yours, Jim Burns > Recall that [A] = x, [B] = 2x and [C] = 10x. Let me use '|' to denote a > link separating two compounds. (Please do not confuse with OR operator.) > As you put it correctly, for a particular link, P(A|A) = ([A]/([A]+[B]))*([A]/([A]+[C])) = 1/33 > P(B|A) = ([B]/([A]+[B]))*([A]/([A]+[C])) = 2/33 > P(A|C) = ([A]/([A]+[B]))*([C]/([A]+[C])) = 10/33 Probability of forming a chain of length n is: > P(B(|A|)^nC) = P(all |A|'s are linked together) * (P(A|A))^(n-1) * > P(B|A) * P(A|C) Let us try to find P(all |A|'s are linked together) first. For the sake > of explanation, let us take an example with n=3. The chain we are > looking for here is B|A|A|A|C. Let us enumerate possible link configurations (=2^3) for |A|A|A| where > n=3. Note that there are four links with the two end-links referring to > the links with B and C. |A|A|A| > |A A|A|A| > |A|A A|A| > |A|A|A A| > |A A|A A|A| > |A A|A|A A| > |A|A A|A A| > |A A|A A|A A| Thus, in general, one out of 2^n link configurations is |A|A|...|A|. > Consequently, P(all |A|'s are linked together) = 2^n Therefore, P(B(|A|)^nC) = 1/(2^n) * (P(A|A))^(n-1) * P(B|A) * P(A|C) > = 1/(2^n) * 1/33^(n-1) * 2/33 * 10/33 > = 5/(2^(n-2) * 33^(n+1)) Just to cross-check, if n = 0, we get P(B|C) = 20/33. We also know that, > P(B|C) = ([B]/([A]+[B]))*([C]/([A]+[C])) = 20/33, which matches with above. Please let me know what you think. Partha. > I'm going to make a very naive model, in hopes that it will > inspire someone to tell me what's wrong with it. That may > move you tolerably close to your goal. > > Let us say that each B has one outie, each C has one > innie and each A has one outie and one innie. Let us > suppose that each link is formed by drawing at random > from the pool of outies and the pool of innies. > Each pool is approximately infinite from the POV of > any individual molecule. > > Then, for a particular link, the chance that it is AA is > p(AA) = ( [A]/([A]+[B]) )*( [A]/([A]+[C]) ), > and likewise, > p(BA) = ( [B]/([A]+[B]) )*( [A]/([A]+[C]) ), > p(AC) = ( [A]/([A]+[B]) )*( [C]/([A]+[C]) ), > p(BC) = ( [B]/([A]+[B]) )*( [C]/([A]+[C]) ). > > For the formation of the A-less BC we have > p(BC) = ( [B]/([A]+[B]) )*( [C]/([A]+[C]) ), > and > p(BAC) = p(BA)*p(AC) = p(BC)*p(AA) > p(BAAC) = p(BA)*p(AA)*p(AC) = p(BC)*p(AA)^2 > ... > p(BA^nC) = p(BC)*P(AA)^n, > ... > > Because I'm not sure of what I mean when I say the > probability that the bond formed is AA, I'm going to normalize > over the p(BA^nC). I guess that means the normalized > p'(BA^nC) is the probability that a random molecule is > BA^nC, and so may be a prediction of molarity, or maybe > molarity ratios. > > p'(BA^nC) = p(BA^nC)/(p(BC) + p(BAC) + ...) > = p(BC)*p(AA)^n/(p(BC) + p(BC)*p(AA) + ...) > = p(AA)^n/( 1 + p(AA) + p(AA)^2 + ...) > = p(AA)^n*( 1 - p(AA) ) > > So the probability of a particular molcule being BA^nC > p'(BA^nC) = p(AA)^n*( 1 - p(AA) ) > where > p(AA) = ( [A]/([A]+[B]) )*( [A]/([A]+[C]) ). > > I can already see there are problems with this. For example, > it doesn't seem to make sense if [B] or [C] -> 0. Maybe > this will help; if not, at least I had fun playing with > your problem. > > Jim Burns === Subject: Re: a ESP test (statistical problem) > >> A ESP test (extra sensory perception) is carried like this: >the >> conductor of the experiment and a person. The cards were laid on the >table >> with the back side upwards in such a way that only the conductor of the >> experiment knew the sequence of the cards. The conductor chosed a card >and >> the person should then say which card it was. In a sequence of 100 >> experiments the person answered correctly 37 times. >> Is there reason to believe that this person has special supernatural >powers? >> >> > >No. >A Blind Monkey would make at least 20 out of 100. >Your guy missed 53% of the time. 37 is a lot more than 20, so if a monkey gets 37 out of 100, that is one amazing monkey. Find me that monkey so I can take it with me to the race track. quasi === Subject: Re: a ESP test (statistical problem) >A ESP test (extra sensory perception) is carried like this: >conductor of the experiment and a person. The cards were laid on the table >with the back side upwards in such a way that only the conductor of the >experiment knew the sequence of the cards. The conductor chosed a card and >the person should then say which card it was. In a sequence of 100 >experiments the person answered correctly 37 times. >Is there reason to believe that this person has special supernatural powers? Well, there are some non-supernatural explanations, for example: (1) the person ia cheating -- peeking somehow (2) the deck is partially marked (but the experimenter is unaware of it) (3) the person being tested is able to get some indication as to which card by looking at the experimenter's facial expression, in the same way that top poker players, when playing against a novice, can almost always tell whether the novice has a great hand or is bluffing. There are other non-supernatural explanations, but some explanation is needed since a score of 37 is very high. If after each guess, the cards are reshuffled, then the probability of getting a score of 37 out of 100 is: C(100,37)*(1/5)^37*(4/5)^63 which equals approximately 1/27000. Of course, if you have 27,000 monkees being tested, it wouldn't be surprising if one of them scored that high. quasi === Subject: Re: change of variable in multiple integrations... > >> Hi all, >> >> I understand the change of variables in bi-varite integration: >> >> For example, if you have a> >> Integrate(Integrate(f(x1, x2), w.r.t x2 from x1 to b), w.r.t x1 from a to >> b) >> >> If I change order of integration, it becomes >> >> Integrate(Integrate(f(x1, x2), w.r.t x1 from a to x2), w.r.t x2 from a to >> b) >> >> This I understand well. >> >> How about more variables: for example, a> >> Original ordering: (first of all, is this a correct one)? >> >> Integrate(Integrate(Integrate(Integrate(f(x1, x2, x3, x4), w.r.t x4 from >> x3 >> to b), w.r.t x3 from x2 to b, w.r.t x2 from x1 to b, w.r.t x1 from a to >> b) >> >> Then change the order of x2 and x4: (is the following correct?) >> >> Integrate(Integrate(Integrate(Integrate(f(x1, x2, x3, x4), w.r.t x2 from >> x1 >> to x3), w.r.t x3 from x1 to x4, w.r.t x4 from x1 to b, w.r.t x1 from a to >> b) > > Yes, but I think your answer will be off by a sign. When you change > variables, you need to multiply f(...) by the determinant of a certain > matrix. This is why when you do integration with polar coordinates, > you have the extra r factor. In short: > > Change from (x,y) to (r,T) [T for theta]: > x = r cos T > y = r sin T > > dx/dr = cos T, dx/dT = -r sin T > dy/dr = sin T, dy/dT = r cos T > > | cos T -r sin T | > | sin T r cos T | > > = r cos^T - (-r) sin^T = r (cos^T + sin^T) = r. > > Thus: int (f(x,y) dy dx) = int(f(r cos T, r sin T) r dr dT). > >> As you can see, these kind things can be very confusing when number of >> variables become larger... is there any general rules governing such >> techniques? > > Yes. Look up Jacobian in a 3-semester calculus textbook, or > http://mathworld.wolfram.com/Jacobian.html . > > > > P.S. This seems correct; if anyone sees any obvious mistakes, please > let me know, because I'll be teaching this in a month or so and haven't > touched Calc III since my undergraduate days ... > > P.P.S. No, it was my senior year in high school, Fall 1988. > > > Hi Chris, > > I understood the Jacobian and change of polar coordinates etc. But I still > did not see where did I miss a sign? > > Could you please elaborate directly on this problem... You're swapping variables x2 and x4. This is like using the following transformation: y1 = x1 y2 = x4 y3 = x3 y4 = x2 Now you need to calculate partial derivatives. The Jacobian becomes | 1 0 0 0 | | 0 0 0 1 | | 0 0 1 0 | | 0 1 0 0 | The determinant of this matrix is -1, so the new integral is of f(x1,x4,x3,x2) times -1, not just f(x1,x4,x3,x2). (Multiplying by -1 changes the sign.) === Subject: Re: Unique element in cyclic subgroups >Hi folks,I have this problem I'm trying to prove. I feel its easy but >nothing in abstract algebra is easy for me. I just started memorizing >the countless theorems and corollaries. Problem: >Let G be a group and suppose a belongs to group G which generates a >cyclic subgroup of order 2 and is the unique such element. Show that >ax=xa for all x belongs to G. I started off trying to use xax^(-1) >where x^(-1) is the inverse but I I got confused within the proof and >went nowhere. plz help. Hint: If a has order 2 and x is an arbitrary element of G, what is the order of xax^(-1)? quasi === Subject: Re: Unique element in cyclic subgroups > Hi folks,I have this problem I'm trying to prove. I feel its easy but > nothing in abstract algebra is easy for me. I just started memorizing > the countless theorems and corollaries. Problem: > Let G be a group and suppose a belongs to group G which generates a > cyclic subgroup of order 2 and is the unique such element. Show that > ax=xa for all x belongs to G. I started off trying to use xax^(-1) > where x^(-1) is the inverse but I I got confused within the proof and > went nowhere. plz help. > I think you need to be clear in your assumptions. Is a supposed to be the only element of order 2? If not, the theorem is false. (Consider G=S_3 and a some transposition). You might try seeing what the order of xax^(-1) is. Once you find that, the uniqueness of a should get you somewhere. === Subject: Re: Smallest Graph with Chromatic Number = 5 > [...] > Does this mean that I cannot use the mce as a basis for a proof of the > 4CT? > There were no 5-chroma planar graphs when H & A proposed the mce! They > were > allowed to use it! Why can't I? > > Because they're trying to prove there are NO mce's. Basically, they > prove that the following statement is true: > > (1) If there's a mce to the 4CT, then (some contradiction). > > Since (some contradiction) is a false statement, and (1) is true, then > the antecedant (there's a mce) is also false. Hence, there is no mce, > hence there is no counterexample, hence the 4CT is true. > > If you're trying to prove the 4CT, you can use this outline as well. > But if you can find an explicit mce, then the 4CT is automatically > false. However, it has been explicitly shown that there are no > counterexamples with fewer than 42 vertices. (This was done _before_ > A&H, btw.) > > > If I were attempting to prove the 4CT, could I say, > Let there be a 5-chroma planar graph with at least 42 vertices!? Yes you can. The base case is settled (for n = 1, 2, 3, ..., 42, and, according to Myerson, even higher numbers). === Subject: Re: Smallest Graph with Chromatic Number = 5 > If I were attempting to prove the 4CT, could I say, > Let there be a 5-chroma planar graph with at least 42 vertices!? Yes you can. The base case is settled (for n = 1, 2, 3, ..., 42, and, > according to Myerson, even higher numbers). Don't take my word for it. Two relevant items from Math Reviews: MR0398880 (53 #2731) Stromquist, Walter The four-color theorem for small maps. J. Combinatorial Theory Ser. B 19 (1975), no. 3, 256--268. 05C15 Each time that the list of known four-color reducible configurations increases significantly, the result usually follows that the 4CC (four color conjecture) holds for a class of maps with a higher bound than before on the number of faces. Dually, it is possible to color planar graphs with more vertices. Recently O. Ore and J. Stemple aroused interest with an announcement of a proof that the bound could be lifted to 39 [J. Combinatorial Theory 8 (1970), 65--78; MR0265215 (42 #127)]; soon after, J. Mayer gave a proof for a bound of 47 [J. Combinatorial Theory Ser. B 19 (1975), no. 2, 119--149]; and now the present author obtains a bound of 51. A set of some 40 reducible configurations is exhibited, drawn from many old and newly published results, and such that each planar triangulation having fewer than 52 vertices and obeying the usual degree and connectivity restriction will contain at least one instance of one configuration of the set. The proof uses a concept of value function based on the Euler polyhedral formula. This is closely parallel to the Heesch discharging principal [H. Heesch, ibid. 13 (1972), 46--55; MR0304224 (46 #3359)]. The exposition is for the most part concise and clear, giving enough details of critical cases to allow the reader to finish the other cases. Further improvements along the same line by other authors will follow, including the surprise proof of the 4CC by K. Appel and W. Haken [Illinois J. Math. 20 (1976), no. 2, 218--297; MR0392641 (52 #13458)], which uses a discharge method, and a list of over 1,900 reducible configurations. This latter will on appearance make the current paper technically obsolete, yet the reviewer recommends this paper and its exposition quite strongly as a stepping stone to the result of Appel and Haken. Reviewed by F. Bernhart MR0389645 (52 #10476) Sachs, H. On some investigations in the theory of the four-colour problem. Infinite and finite sets (Colloq., Keszthely, 1973; dedicated to P. Erd.9as on his 60th birthday), Vol. III, pp. 1227--1234. Colloq. Math. Soc. Janos Bolyai, Vol. 10, North-Holland, Amsterdam, 1975. 05C15 the 4-colour problem, in the light of recent developments. One often expects such writing to be a compound of propaganda, opinion, and hope. informative discussion by a detached and yet knowledgeable observer. His most personal conclusion, which serves as leitmotif throughout, is clearly stated at the outset: He sees the arguments for and against the 4-colour conjecture in a state of mutual counterbalance. {The statement that every planar graph with fewer than 45 vertices can be 4-colored [G. A. Donec, A study of questions on the coloring of planar graphs (Russian), Candidate's Dissertation, Inst. Kyberneti, Akad. Nauk, Ukrain. SSR, Kiev, 1970] has since been improved. J. Mayer with the help of recent computer work on reducibility, has obtained a bound of 96 (Probl.8fme des quatre couleurs: un contre-exemple doit avoir au moins 96 sommets, J. Combinatorial Theory Ser. B., to appear), overtaking 45 and also values of 48, 52, and 58 [Mayer, J. Combinatorial Theory Ser. B. 19 (1975), 119--149; W. Stromquist, ibid. 19 (1975), 256--268; H.-J. Presia, Math. Nachr. 65 (1975), 223--234; ibid. 67 (1975), 181--198; MR0376416 (51 #12592a)].} A side glance at a difficult solved problem in number theory illustrates the suggestion that should a non-4-colourable graph exist, its discovery may well turn on a deeper structural insight, or more systematic construction methods. Several partial results illustrate another point. Namely, a promising line of investigation will frequently yield many results in favor of the 4-colour conjecture, but past a certain point the method begins to fail, and fail in principle. {For the entire collection see MR0360076 (50 #12526).} Reviewed by F. Bernhart -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Smallest Graph with Chromatic Number = 5 > If you're trying to prove the 4CT, you can use this outline as well. > But if you can find an explicit mce, then the 4CT is automatically > false. However, it has been explicitly shown that there are no > counterexamples with fewer than 42 vertices. (This was done _before_ > A&H, btw.) A buddy of mine, Walter Stromquist, pushed that 42 up a bit, I think to 52, in a doctoral dissertation he got accepted just before A & H announced their results. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: geometric series calculation Hello I need help in summing the geometric series below, it is the result of the following problem workout. ******************************problem****************************** Assume X is a random variable with ensemble Ex={0,1,2,...} and P(X=k+1) = a.P(X=k), k is element of Ex, where 0 Hello > > I need help in summing the geometric series below, it is the result of > the following problem workout. > > ******************************problem****************************** > Assume X is a random variable with ensemble Ex={0,1,2,...} and > P(X=k+1) = a.P(X=k), k is element of Ex, where 0 a) find the probability mass distribution for X. > b) Evaluate P(A) where A=[X is an odd number]. > > > the probability mass distribution function should be > P(X=k) = a^k P(X=0) > and since the sum of all probabilities add to 1. then > > > inf > P(X=0) = 1 - sum P(X=0) a^k > k=1 > > using the sum of infinite geometric series we have > > a > = 1 - P(X=0) ------- > 1-a > 1-a > = -------- > a > > am I missing something, because the book says it should be (1-a)? Let's see: P(X=0) = 1 - P(X=0) * a/(1-a) P(X=0) * (1 + a/(1-a)) = 1 P(X=0) * (1-a+a)/(1-a) = 1 P(X=0) /(1-a) = 1 P(X=0) = 1-a. Hmm. Looks right. === Subject: Re: Zariski Topology >Here is one example. The Cayley-Hamilton theorem says that a matrix >satisfies its own characteristic polynomial and elementary proofs are >fairly unpleasant. So here is a proof using the Zariski topology. >Matrices with distinct eigenvalues form a non-empty open (therefore >dense) set in the Zariski topology on R^{n^2}, R being the base field. >That is because the discrimant of that polynomial is a polynomial in >the entries and the non-vanishing of that discriminant is equivalent to >the distinctness of the roots. Assuming the field is algebraically >closed, any matrix with distinct eigenvalues is diagonalizable and the >C-H theorem is evidently true for them. If p_A is the characteristic >polynomial of A, its coefficients are polynomials in the entries, as >are each of the entries of p_A(A). The Zariski topology is the weak >topology for all the polynomial functions and therefore any polynomial >function that vanishes on a Zariski-dense set is 0. > >Since nothing changes under field extension, it is trivial that this >argument is valid over an arbitrary field. You don't really need >algebraically closed; it is sufficient to extend the field by the roots >of the polynomial. > >Here is a simpler one. If A and B are n x n matrices, then AB and BA >have the same characteristic polynomial. This is obvious if A or B is >invertible since then AB and BA are conjugate. But the non-singular >matrices are dense (non-vanishing of the determinant) and thus p_{AB} - >p_{BA} vanishes on this dense set and hence identically 0. Great examples. quasi === Subject: Re: a fair die problem... > Hi all, > > Suppose I have a fair die: I roll it untill an even number -- either 2, 4, > or 6, occur 10 times. Then stop. > > What should be the total number of rolls until this event occur? > > > A nasty answer follows. I don't know if/how it can be simplified much. > > I make the probability of stopping on the n'th roll to be given by > > Pr(n) = 1/2 * S_i S_j [(n-1)!/((n-1-(9+i+j))! * 9! * i! * j!) > * (1/6)^(9+i+j) * (1/2)^(n-1-(9+i+j))] > > where S_i S_j signifies a double (nested) sum, over i = 0 to 9, j = 0 > to 9, with the condition 9+i+j <= n-1 > > (To speed up calculation you can obviously take constants out of the > sums. E.g. a factor of 1/9! can be taken to the front, and a factor of > (n-1)!/i! can be taken out of the inner sum. I left it like it is for > clarity. OTOH, if you're doing it on a computer then it scarcely > matters much; if you're doing it by hand then very good luck to you!) > > This looks suspiciously like an approximation to a normal distribution, > with the mean, median and mode all around 44 or 45. In other words, the > average number of rolls until you stop, the most likely number of rolls > until you stop, and the point at which the probability of getting this > far falls below 1/2, are all around 44 or 45 rolls. The variance is > quite high though, so this number doesn't tell you very much about what > actually WILL happen. > > Explanation: > > To end on the n'th roll, the n'th roll must be a 2, 4 or 6. For the > moment assume it's a 2. In the previous n-1 rolls there must have been > exactly 9 2's, and there must have been i 4's and j 6's where 0 <= i < > 10 and 0 <= j < 10. Also we must have 9+i+j <= n-1. The expression > inside the sum is the probability of getting 9 2's, i 4's and j 6's in > those n-1 rolls, and the sums are taken over the appropriate ranges of > i and j. > > The probability of the n'th roll being a 2 is 1/6, so we should > multiply the double sum by 1/6. However, by symmetry, the end-on-4 and > end-on-6 scenarios have identical probability, so we multiply instead > by 3*(1/6), or 1/2. Of course! (Essentially, it boils down to calculating the number of nonnegative solutions to the equation x1 + x2 + x3 + x4 + x5 + x6 = n, with the additional conditions that x2 = 10, x4 < 10, and x6 < 10.) === Subject: Re: a fair die problem... > Hi all, > > Suppose I have a fair die: I roll it untill an even number -- either 2, 4, > or 6, occur 10 times. Then stop. > > What should be the total number of rolls until this event occur? > A nasty answer follows. I don't know if/how it can be simplified much. I make the probability of stopping on the n'th roll to be given by Pr(n) = 1/2 * S_i S_j [(n-1)!/((n-1-(9+i+j))! * 9! * i! * j!) > * (1/6)^(9+i+j) * (1/2)^(n-1-(9+i+j))] where S_i S_j signifies a double (nested) sum, over i = 0 to 9, j = 0 > to 9, with the condition 9+i+j <= n-1 (To speed up calculation you can obviously take constants out of the > sums. E.g. a factor of 1/9! can be taken to the front, and a factor of > (n-1)!/i! can be taken out of the inner sum. I left it like it is for > clarity. OTOH, if you're doing it on a computer then it scarcely > matters much; if you're doing it by hand then very good luck to you!) This looks suspiciously like an approximation to a normal distribution, > with the mean, median and mode all around 44 or 45. In other words, the > average number of rolls until you stop, the most likely number of rolls > until you stop, and the point at which the probability of getting this > far falls below 1/2, are all around 44 or 45 rolls. The variance is > quite high though, so this number doesn't tell you very much about what > actually WILL happen. Here's what I get in Mathematica: In[1]:= pr[n_] := 1/2 Sum[Multinomial[n-1-(9+i+j),9,i,j] * (1/6)^(9+i+j) * (1/2)^(n-1-(9+i+j)), {i,0,9}, {j,0,9}] In[2]:= Sum[pr[n], {n,1,Infinity}] Out[2]= 1 In[3]:= Sum[n*pr[n], {n,1,Infinity}] 113594395286860 Out[3]= --------------- 2541865828329 In[4]:= N[%] Out[4]= 44.6894 Note that the condition 9+i+j <= n-1 didn't have to be specified, because Multinomial[] returns 0 if any of its arguments are negative integers. Scott -- Scott Hemphill hemphill@alumni.caltech.edu This isn't flying. This is falling, with style. -- Buzz Lightyear === Subject: Re: a fair die problem... > >>> Suppose I have a fair die: I roll it untill an even number ... >>> ... occur 10 times. Then stop. >> >> Maybe I did not specify clearly: > > You could say that. > >> The stopping criteria is 2, 2, 2, 2, ... 2, 2, 2... occur 10 times (not >> neccessarily consequtively), >> or 4, 4, 4...., occur 10 times (not neccessarily consequtively), or 6, 6, >> 6, >> occur 10 times (not neccessarily consequtively) >> >> whichever pattern comes first... > > This is a somewhat more complicated problem, yes. I don't see any > obvious shortcuts off the top of my head, but if something occurs I > will post. > > Glen > Yes, it is hard. That's why I asked... === Subject: Re: linear regression with truncated data Tom === Subject: Re: infinity > > > So here's what I think about it. It's obvious that the number of balls in > the vase keeps growing as you approach noon. > However, the algorithm which describes the filling of the vase with balls > never reaches noon. Therefore, I think it is not even correct to ask the > original question 'how many balls at noon' (do mathemathicians call that > ill-defined, -posed or something like that?). There is some sense to what you say. There is still some room IMO for claiming that the question itself is > ill-defined. It's certainly not physically possible for such an > operation to take place in the real world to start with; so the > question of what would happen at noon? is really best interpreted as > what would be a logically consistent answer to the question, assuming > that it /could/ be done in practice? The basic idea I had was that we can make a function B(n) which describes the number of balls in the vase from one minute to noon to every time less than noon. But the description of the experiment does not allow us to reach noon. It does not have a sense of a continuously progressing time which is required to reach noon. So I would say that the experiment directly implies that 'noon' is not within the domain of B(n), and therefore the function, or description of the experiment, cannot give an answer to the question. *** For nitpickers :) , as B(n) is a function of variable 'n', statements about the domain should be based on 'n' only, I guess. But every 'n' implies a certain time t_n, so I think you get the idea... I hope... else beat me again :( *** infinite balls in the vase at noon. Different answers while the function B(n) that describes the number of balls in the vase at any 'n' is the same for both... The question is a physical one (number of balls), and for that reason the answer cannot depend on how the balls are numbered or what color they have. Suppose instead of your experiment 1, we put a ball into the vase on > odd steps, and then take it out the even steps. Then it seems obvious that there is no single well-defined answer to > how many balls are there at noon?. Neither empty nor containing a > ball seems to be correct - there is no answer that is somehow > /logically preferable/ to all others; and that is the basic sense of > not well-defined. For the same reason as I described above: for each n we can find the number of balls in the vase, but there is no n which equals 'noon'. Therefore, because the number of balls change at every step n, we should not want to force an answer. We can just say that no answer is possible because 'noon' is not in the domain of the function B(n) which describes the number of balls in the vase. On the other hand, if you did /nothing/ at each step, that there would > somehow be a ball in the vase at noon seems obviously wrong. The vase > would obviously be empty, because we intuitively require that there > can't be a ball in the vase unless it was put there at some finite step > n. Because there are no steps defined where the number of balls in the vase changes, we don't need to stick to those discrete moments t_n, but we can directly describe the number of balls in 'continuous time': nothing happens. So we are not bound by a function B(n) in which 'noon' is not element of the domain. Similarly, it seems intuitive that if at step 1, you put ball 1 in > the vase, and then at no later step do you remove it, then that ball > must be in the vase at noon. These two intuitions are the assumptions we are making when we ask > assuming it is possible in the real world, what balls are in the vase > at noon?; and a well-defined answer would be one which is consistent > with these intuitions, and also be the /only/ one consistent with these > intutions. Again, IMHO, forgetting that the answer must be given by a function which describes the number of balls in the vase, and that 'noon' is not in the domain of that function. We don't _need_ to give a number as an answer. In this experiment, we try to connect some logical reasoning (about putting balls with number in the vase and getting some out) with a physical entity (number of balls in the vase). That connection cannot be made because we have no description or function which can make the connection! Thus, in experiment 1, we can agree that, if at some definite step t_n, > we remove the ball labelled n, and then there is no later step that > we put the ball back in, then it is perfectly reasonable to insist > that, /whatever/ we mean by the balls in the vase at noon, it can't > mean that ball n is in the vase at noon. Since, in your construction, the above is true for every n, the only > /possible/ answer consistent with our assumptions is that there are no > balls in the vase at noon. So your explanation is that for each ball n in the vase, there is a step t_n where that ball is taken out. So each ball is taken out eventually which results in an empty vase at noon. This is based on _physical_ observations: every ball put in the vase is taken out at some time. Another _physical_ observation I made in my original reply is: at every 'n' the number of balls in the vase increases by 9. So there is no interval between one minute to noon and noon where the number of balls in the vase decreases. So how can the vase be empty at noon? IMO, both reasonings are OK, but they _seem_ to end up with different answers. Reason for this contradiction is that we're not allowed to formulate a numeric answer to the original question, because noon is not in the domain of the experiment. And since that /is/ the single, consistent answer to the question, we > say the vase is empty at noon is a well-defined answer to the > question. > === Subject: Re: infinity > For unnumbered balls the outcome -as far as I can tell - is also an empty > vase! This is not my invention however, I read this here: > http://www.cut-the-knot.org/Probability/infinity.shtml > (scroll down a bit for the relevant part, although you might like to read it > all) I found his argument not /quite/ so convincing as the argument for numbered balls. His argument seems to be essentially that, assuming that the ball to be removed at each step is selected with a uniform probability from the set of balls in the vase at that step, that therefore the /probability/ that any /particular/ ball will still be in the vase tends to 0 as the number of steps tends toward infinity. Therefore there is probability 1 that the vase is empty at noon. I have two (slight) problems with this. First, the assumption that the balls are distinguishable but selected at random is quite reasonable but not neccessary; we could equally well say that the balls are merely chosen by some /unknown/ method, not neccessarily with random distribution (for example, maybe the balls at the bottom of the vase are naturally chosen with somewhat less frequently than the balls at the top of the pile). If we only assume that some method is used and that the balls are essentially indistinguishable, I think a better answer is a countably infinite number of balls are in the vase at noon. But even accepting the premise, the fact that an event has probability 0 of occuring is not (in general) the same as saying that it is logically /impossible/ for that event to occur. Thus, the statement that the vase is empty is not the same as the vase is empty, with probability 1 (the latter being all that is asserted at the link you gave). Still, it is pretty slick... :) === Subject: Re: infinity > >> For unnumbered balls the outcome -as far as I can tell - is also an empty >> vase! This is not my invention however, I read this here: >> http://www.cut-the-knot.org/Probability/infinity.shtml >> (scroll down a bit for the relevant part, although you might like to read it >> all) > I found his argument not /quite/ so convincing as the argument for > numbered balls. > His argument seems to be essentially that, assuming that the ball to be > removed at each step is selected with a uniform probability from the > set of balls in the vase at that step, that therefore the /probability/ > that any /particular/ ball will still be in the vase tends to 0 as the > number of steps tends toward infinity. Therefore there is probability 1 > that the vase is empty at noon. The probability of any particular ball being left in the vase at noon does not tend to 0. The probability of any particular ball being left in the vase at noon is exactly 0. And since there are countably many balls, it follows by countable additivity of probability measure that the probability of the vase being nonempty at noon is also 0. > I have two (slight) problems with this. > First, the assumption that the balls are distinguishable but selected > at random is quite reasonable but not neccessary; we could equally well > say that the balls are merely chosen by some /unknown/ method, not > neccessarily with random distribution (for example, maybe the balls at > the bottom of the vase are naturally chosen with somewhat less > frequently than the balls at the top of the pile). If we only assume > that some method is used and that the balls are essentially > indistinguishable, I think a better answer is a countably infinite > number of balls are in the vase at noon. If the balls are merely chosen by some unknown method, then for all we know, the balls may be removed in last-in, first-out order, which guarantees that infinitely many balls will be left at noon. That's why more information is needed. The assumption of a uniform distribution at each time step is sufficient, but not necessary, to conclude that the vase is emptied with probability 1. > But even accepting the premise, the fact that an event has probability > 0 of occuring is not (in general) the same as saying that it is > logically /impossible/ for that event to occur. Nobody has claimed otherwise. > Thus, the statement that the vase is empty is not the same as the > vase is empty, with probability 1 (the latter being all that is > asserted at the link you gave). > Still, it is pretty slick... :) -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: infinity > > > >> For unnumbered balls the outcome -as far as I can tell - is also an empty >> vase! This is not my invention however, I read this here: >> http://www.cut-the-knot.org/Probability/infinity.shtml >> (scroll down a bit for the relevant part, although you might like to read it >> all) > > I found his argument not /quite/ so convincing as the argument for > numbered balls. > > His argument seems to be essentially that, assuming that the ball to be > removed at each step is selected with a uniform probability from the > set of balls in the vase at that step, that therefore the /probability/ > that any /particular/ ball will still be in the vase tends to 0 as the > number of steps tends toward infinity. Therefore there is probability 1 > that the vase is empty at noon. > > The probability of any particular ball being left in the vase at noon > does not tend to 0. The probability of any particular ball being left > in the vase at noon is exactly 0. > > And since there are countably many balls, it follows by countable > additivity of probability measure that the probability of the vase being > nonempty at noon is also 0. > > > I have two (slight) problems with this. > > First, the assumption that the balls are distinguishable but selected > at random is quite reasonable but not neccessary; we could equally well > say that the balls are merely chosen by some /unknown/ method, not > neccessarily with random distribution (for example, maybe the balls at > the bottom of the vase are naturally chosen with somewhat less > frequently than the balls at the top of the pile). If we only assume > that some method is used and that the balls are essentially > indistinguishable, I think a better answer is a countably infinite > number of balls are in the vase at noon. > > If the balls are merely chosen by some unknown method, then for all we > know, the balls may be removed in last-in, first-out order, which > guarantees that infinitely many balls will be left at noon. That's why > more information is needed. Yes, well I suppose I shouldn't just mumble my thoughts. :) The tie-the-knot site uses explicitly numbered balls, just as the OP's first example did; but most recently, the topic seems to have changed to essemtially unlabelled balls. If the balls are chosen by some unknown method, but the balls are indistinguishable (which is what I essentially said), then it seems to me that we are within our rights to relabel the balls at each step, since at each step all we /really/ know about the balls in the vase is the cardinality of the set of those balls. The sequence is not then some function from the naturals to the naturals indicating what which ball is removed at that step, as distingushed by its labelling. Instead we have a function from the naturals to some set of balls, about which all we know is its resultant cardinality (i.e., that they /can be/ labelled distinctly in a certain way). This sequence would seem to have as its limit a countable set of balls, not an empty or finite one. > The assumption of a uniform distribution at > each time step is sufficient, but not necessary, to conclude that the > vase is emptied with probability 1. > > But even accepting the premise, the fact that an event has probability > 0 of occuring is not (in general) the same as saying that it is > logically /impossible/ for that event to occur. > > Nobody has claimed otherwise. > The poster to whom I was replying said: >> For unnumbered balls the outcome -as far as I can tell - is also an empty >> vase! To me, saying the outcome is an empty vase is not the same as saying that thye probability that the vase would be empty is 1. === Subject: Re: infinity > If we only assume > that some method is used and that the balls are essentially > indistinguishable, I think a better answer is a countably infinite > number of balls are in the vase at noon. > > If the balls are merely chosen by some unknown method, then for all we > know, the balls may be removed in last-in, first-out order, which > guarantees that infinitely many balls will be left at noon. That's why > more information is needed. > > Yes, well I suppose I shouldn't just mumble my thoughts. :) > > The tie-the-knot site uses explicitly numbered balls, just as the OP's > first example did; but most recently, the topic seems to have changed > to essemtially unlabelled balls. > > If the balls are chosen by some unknown method, but the balls are > indistinguishable (which is what I essentially said), then it seems > to me that we are within our rights to relabel the balls at each > step, since at each step all we /really/ know about the balls in the > vase is the cardinality of the set of those balls. > > The sequence is not then some function from the naturals to the > naturals indicating what which ball is removed at that step, as > distingushed by its labelling. > > Instead we have a function from the naturals to some set of balls, > about which all we know is its resultant cardinality (i.e., that they > /can be/ labelled distinctly in a certain way). > > This sequence would seem to have as its limit a countable set of balls, > not an empty or finite one. > No, no, no. The problem statement requires that there be a total of a countable number of balls which can ever be in the urn at some step n in the sequence (if we had a step where we added an uncountable number of balls, then the original vase is empty at noon argument wouldn't hold). Therefore there is a bijection betwen the set of naturals and the balls which allows us, in principle, to say that they are distinguishable (or at least labelled in some definite way). Whether I know what this bijection is or not, it certainly exists; call it T. Then over all possible functions which at step n select a ball from the subset of N represented by the balls in the vase at step n, there are an equal number (cardinality) of such functions which select one of these balls as any another; that translates into select a ball from the set of balls in the vase, with uniform distribution, and then the original probability argument holds. I stand corrected (even if by myself). === Subject: Re: Update: Objections to Cantor's Theory >>>Anyways, with uncountably many nested intervals, each sharing at most >>>one endpoint with its parent interval, there are uncountably many >>>disjoint intervals. >>It takes more that saying it three times to make it true. > You might say that because as an opaque statement it would imply the > rationals uncountable, and make inconsistent claims you see as true, > but it does not address the simple deduction that as described > uncountably many nested intervals assures the existence of uncountably > many disjoint intervals. Last time round the loop, Ross. For a guy with a 180+ IQ you seem to have trouble noticing when somebody provides you with what you ask for. You start off with an uncountable set of nested intervals of reals. You then pick out an uncountable sequence of them where the left hand endpoints are strictly increasing, according to a well-ordering. You then claim that by doing this you get an uncuntably infinite set of pairwise disjoint intervals. But this would only work if the left hand endpoint were increasing in the usual sense; the well-ordering is not compatible with the standard ordering. You have not found an inconsistency in the usual picture of the reals, with a countable dense set of rationals, and an uncountable (also dens) set of irrationals. You have simply produced a fallacious argument. > If you actually have a line of reasoning why contrary to as described > that is so, I'd be interested to learn of it. If you were interested to learn of it, you'd make an attempt to understand it. But as a prerequisite, you'd have to understand what you write, and I see little to no evidence of that. === Subject: Re: Update: Objections to Cantor's Theory On 3 Aug 2005 16:41:20 -0700, Ross A. Finlayson >> Anyways, with uncountably many nested intervals, each sharing at most >> one endpoint with its parent interval, there are uncountably many >> disjoint intervals. >> >> It takes more that saying it three times to make it true. >> >> Yes, obviously you can find uncountably many pairs of >> intervals with no common element. All the intervals >> in (0,1) are disjoint from all those in (2,3). >> >> But there is no uncountable set of intervals such that >> every pair in the set is disjoint. > >You might say that because as an opaque statement it would imply the >rationals uncountable, and make inconsistent claims you see as true, >but it does not address the simple deduction that as described >uncountably many nested intervals assures the existence of uncountably >many disjoint intervals. You go from your uncountable nested intervals and show that there exists a set of disjoint intervals. You then claim (without proof) that there are uncountable sets of disjoint intervals. It's the lack of proof that's the problem. Especially when there are proofs the other way (You've actually hidden one in the thread). >If you actually have a line of reasoning why contrary to as described >that is so, I'd be interested to learn of it. I discuss a variety of >considerations thereupon myself, perhaps in those you might see a line >of reasoning to avoid the otherwise inescapable conclusion. The line of reasoning to escape your conclusion is that you haven't proven it yet. Martin ************************* CALL FOR PAPERS EvoOpt 2006 Special Track on Evolutionary Optimization at the 19th International FLAIRS Conference (in cooperation with the American Association for Artificial Intelligence) 11th-13th May, 2006, Melbourne, Florida, USA ---------------------------------------------------------------------------- ---- Submission Deadline: November 21, 2005 EvoOpt 2006 website: http://evoopt2006.dei.uc.pt FLAIRS 2006 website: http://www.indiana.edu/~flairs06/ ---------------------------------------------------------------------------- ---- EvoOpt-2006 will focus on the application of Evolutionary Algorithms (EAs) to complex optimization problems. In the past few years, these algorithms have been successfully applied to a large number of optimization problems. Some of the most relevant examples belong to the class of combinatorial optimization problems such as the traveling salesperson, scheduling, packing, planning or routing. 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Cambridge, UK +44 (0)7849748462 === Subject: ACS Short Course in Computational Chemistry and Drug Design Precedence: list iQCVAwUBQu5Zj6CIqsWvyF5FAQG0swP/avX4jjcr0Azvy/qwkMDi+IKLB95OKE64 Vo6oVh3JGcMG6MLOPSJO7qCviSNasJhRroz5mTd4CDkmJjjBnsQt9FkgqwRX3PcE vOnPFp4sS8Oyi2H053grzndjCOJLUfQ0ZHARZjgkV3R8Vnpc9rrDlEDBuqwHf1Bc CV0YMLAdNCQ= =/qT0 ACS Short Course Computational Chemistry and Computer-Assisted Drug Design: Practical Approaches 230th ACS National Meeting Washington Convention Center, Washington, DC Friday-Saturday, August 26-27, 2005 This introductory level course is designed for organic chemists, pharmaceutical chemists, and biochemists who are interested in learning more about computational and combinatorial methods, or scientists who need to develop a working knowledge of the fundamentals and need to understand the concepts and terminology of this rapidly developing area. Program Overview of Computational Chemistry and Computer-Assisted Drug Design Molecular Mechanics: Background, Development, Concepts, Force Fields Conformational Searching Molecular Dynamics Simulations: Background, Development, Concepts, and Applications Protein Structure Prediction Overview of Quantum Chemistry Methods and Its Application to Drug Design DNA and Protein Sequence and Structure Analysis Drug Design Methods and Pharmacophore Design QSAR and Property Prediction Methods 3D Database and 3D Searching Examples of Pharmacophore Perception and 3D Searching Combinatorial Chemistry and Chemical Diversity Concepts Faculty Phillip Bowen (UNCG), Osman Guner (Accelrys Inc.), Robert Pearlman (UT-Austin), Christopher Lipinski (Pfizer), Alexander Tropsha (UNC), Iosif Vaisman (GMU). The course will be taught from 9:00 a.m. to 5:00 p.m. on both days. If you have questions about the course, contact Dr. Bowen at 336-334-5714 or at jpbowen@uncg.edu Registration Web: http://www.chemistry.org/portal/a/c/s/1/acsdisplay.html?DOC=education%5Cprof essional%5Cscbp12.html Email: shortcourses@acs.org Phone: 800-227-5558, ext. 4508, or 202-872-4508. === Subject: A question about implementing MM-PBSA method by CHARMM Precedence: list iQCVAwUBQu5ZrKCIqsWvyF5FAQFxGwP/S5is+OyxmAFyo5STvZmR190uYOsg7yft QQR0nQlaMllSaa7opqGjFaGTp3auve/EwJg1BUfZ3wvH9atLFhPiRByD5KiZjAal n90govvic6dN8NcBPvGGYaKRWEX5qj6t5pB+Wh11ft/HK7z9gifjSE0eiKDHlQ9J FD7ignWz3UU= =pkWu hi, I'm trying to using MM-PBSA method to estimate changes in binding free energies of protein-protein complexes as a result of mutating one or compute the molecular mechanic terms and using DELPHI, with PARSE radii and Cornnel charges, to compute electrostatic contribution to solvation free energy. When I using CHARMM to do the same job, the results becomes quite different from those published ones (e.g. Irina Massova & Peter protein dielectric constant, but the result is still not encouraging. appreciate if any one can share his/her experience in this aspect. p.s. It is the first time I post to bionet groups. please tell me if there is anything improper. -- Cean === Subject: KDD-2005 Final Call for Participation Precedence: list iQCVAwUBQvJZlqCIqsWvyF5FAQGK2wP/aW4yBGr/+gIF7rI+YbAJ97YdyTBpzZTs S6sbrybStm9Dkp+P8uMHwV/gmFIvBtgp0liNumA4z2x7vgKQNFI+kWkU1JnZo8Di ImkSr8sSbvAV/7S2Zskl1oOWtOKUJtK9HhxihPdmPGq6jzzy8z/tA+7HCKKz6ajc 9vg+vmdjoQY= =m8ff ************************************** **** FINAL CALL FOR PARTICIPATION **** ************************************** K D D - 2 0 0 5 *************************************** Knowledge Discovery and Data Mining *************************************** The Eleventh ACM SIGKDD International Conference on Knowledge Discovery and Data Mining will take place in Chicago, Illinois, USA, from August 21 through 24, 2005. Visit the conference web site at http://www.kdd2005.com or http://www.acm.org/sigkdd/kdd2005/ KDD is the premier international conference on knowledge discovery and data mining. The conference will provide a forum for academic researchers and industry and government innovators to share in their results and experience. The program will include keynote presentations, oral paper presentations, poster presentations, workshops, tutorials, and panels, as well as the KDD Cup competition. CONFERENCE HIGHLIGHTS Keynotes - Prabhaker Raghavan. Head of Reasearch at Yahoo! and professor of Computer Science at Stanford. - Gian Fulgoni, chariman and co-founder of comScore Networks, Mining the Internet: The Eighth Wonder of the World. KDD Cup - Winners of the KDD Cup data mining competition will be annnounced. Technical papers - Research track: 40 presentations and 36 posters - Industry/Gov. track: 14 presentations and 11 posters Full Day Workshops - Data Mining Methods for Anomaly Detection - Workshop on Open Source Data Mining (OSDM 2005) - Workshop on Utility-Based Data Mining (UBDM-2005) - 4th Workshop on Multi-Relational Data Mining (MRDM 2005) - 5th Workshop on Data Mining in Bioinformatics (BIOKDD 2005) - Workshop on Data Mining Standards, Services and Platforms (DM-SSP 2005) - WebKDD'05: Taming Evolving, Expanding and Multi-faceted Web Clickstreams - Workshop on Link Discovery: Issues, Approaches and Applications (LinkKDD-2005) Half Day Workshops - Data Mining Success Stories and KDD Practice Prize (KDDSS-2005) - 6th International Workshop on Multimedia Data Mining: Mining Integrated Media and Complex Data Tutorials - Introduction to Logistic Regression - Data Visualization and Mining using the GPU - Randomized Algorithms for Matrices and Massive Data Sets - Principles and Applications of Probabilistic Learning CONFERENCE VENUE Come and join us at KDD and enjoy your stay in Chicago. The conference will take place at the Hyatt Regency Chicago Hotel. Area attractions include the easy-to-navigate Museum Campus; lively Navy Pier; a resurrected North Loop theater district; and a who's who of luxury shopping destinations along the city's fabled Magnificent Mile. A busy convention trade has sparked hotel construction, and the city's eclectic mix of restaurants has gained an international reputation for excellence. Chicago Air and Water Show The oldest and biggest exhibition of its kind in the country, the Air and Water Show displays daredevil acrobatics both in the air and water. It can be viewed from many lakeside locations, but the main hub is above and beyond North Avenue Beach. Both civilian and military demonstrations take place and electrifying stunts are performed, the highlight being the US Navy Blue Angels' jet formation. INFORMATION Please visit our web site for more details about the program. http://www.acm.org/sigkdd/kdd2005/ === Subject: Help with a bioinformatics problem posting-account=6fRz6wwAAAB-ItAxaRFJm0M_J4ygwiJW Status: RO Content-Length: 332 Precedence: list iQCVAwUBQvOf26CIqsWvyF5FAQFDfQP8C/tdQLY8RFCk7QbttP5jUjGC4bFGzfsO wzO/rV1fd4LeE2SjrPen6mrbacmXTFSXKxy9kAGb+sv2OTLej5U3jhp4rkBSSOef EY92sGo2kdt6FcAX+bUOH+ism04p4wu1OWdPffP9I9HVYcQhQ0O2on0iaL4Rvgre V2P7Va+ry/g= =7p6H hi.... i'm trying to do a genome (i think) comparison between A. agrarius, R novegius, C glareolus and being new to bioinformatics, i'm not exactly sure what would be the best nucleotide or genome part that i should select is doing a multiple sequence analysis. can someone point me to the correct thing i should do? joe === Subject: Re: Help with a bioinformatics problem Precedence: list iQCVAwUBQvZTJqCIqsWvyF5FAQEWowP9HRVc7iY3zKWxBrwuzS6XA0iam9cXg2ex BXYmn5amUZSDF83lkDxRdcMHPJx5nQmK1WPejzjoUAsnWZAiYvRnYyj5+3BCYGoO XeIfC5O7vKRglkTElKaBy9OwWD6Qx+zkGC9hk3pVQmDxMkjumivFD2KxTFQL0Rwf YELykvOzo54= =fG1J > i'm trying to do a genome (i think) comparison between A. agrarius, R > novegius, C glareolus and being new to bioinformatics, i'm not exactly > sure what would be the best nucleotide or genome part that i should > select is doing a multiple sequence analysis. can someone point me to the correct thing i should do? You should perhaps describe in more detail what you intend to compare. While the chromosomes of a genome may be available as single sequences, it's usually only single genes which are subjected to multiple alignment. Thomas Jahns -- Computers are good at following instructions, but not at reading your mind. D. E. Knuth, The TeXbook, Addison-Wesley 1984, 1986, 1996, p. 9 === Subject: Re: Help with a bioinformatics problem Precedence: list iQCVAwUBQvbK1aCIqsWvyF5FAQFqRgP8DS06PjYldLGe+SsH73NjE169GIpIO9QC +cTnm/KVAy9//ZoeLUHc6Xvyt9WFSjf976Fx3abny5wPyMVjYZczNYl0nAEFQRiq 8prvJTaqogM4NP5sHMdyEgaIC/uhtWLLGQWgPIUe+u6OUppRHwG+ONOyZN6dzu/T xhNzleOJeSw= =bIWg >> i'm trying to do a genome (i think) comparison between A. agrarius, R >> novegius, C glareolus and being new to bioinformatics, i'm not exactly >> sure what would be the best nucleotide or genome part that i should >> select is doing a multiple sequence analysis. >> can someone point me to the correct thing i should do? > > You should perhaps describe in more detail what you intend to > compare. While the chromosomes of a genome may be available as single > sequences, it's usually only single genes which are subjected to > multiple alignment. Not quite true. There are alignments of whole genomes done for finding syntenic blocks. These can be particularly useful when looking for operons in bacterial genomes. For computational reasons, they are done as pairwise alignments, but they are sometimes interpreted as multiple alignments. Doing synteny mapping for large genomes is a bit tricky, as it is easy to be mislead by repeats. The UCSC genome browser group has done several whole-genome alignments, including rat, mouse, human, chicken, ... , but not A. agrarius or C. glareolus. http://genome.ucsc.edu/ Their tools are available---you probably want to use BLAT, by Jim Kent, though there may be other relevant tools. ------------------------------------------------------------ Kevin Karplus karplus@soe.ucsc.edu http://www.soe.ucsc.edu/~karplus Professor of Biomolecular Engineering, University of California, Santa Cruz Undergraduate and Graduate Director, Bioinformatics (Senior member, IEEE) (Board of Directors, ISCB) life member (LAB, Adventure Cycling, American Youth Hostels) Effective Cycling Instructor #218-ck (lapsed) Affiliations for identification only. === Subject: Re: Help with a bioinformatics problem 6jcKWVKF0EPuW20XA/71FwFIC6vuXkTXchG6UKEOFZLgKCQvL3 Precedence: list iQCVAwUBQvPdjKCIqsWvyF5FAQE7NAP/TL+jdp2ZN7KTakR13xOTvz9TZc2fJqlV oCOmj7jglindcEAz7k00UCOLyBuRuO5/2TVHD7eaRCQq0FGBB3FvNufiajVknd2u PjGqhvbfZN4X0MClf+9/5/LREAn+q+rOEdtV1DNiv7+tNNhs28MFcfBPJ7jiJCrX tfZ9/v46NCU= =T6Hy > i'm trying to do a genome (i think) comparison between A. agrarius, R > novegius, C glareolus and being new to bioinformatics, i'm not exactly > sure what would be the best nucleotide or genome part that i should > select is doing a multiple sequence analysis. > can someone point me to the correct thing i should do? It's hard to give any useful advice without knowing, what you have in mind and why you want to compare these genomes. I suggest you elaborate a bit on your goals and I'm sure you'll get some good hints. cu Philipp -- Dr. Philipp Pagel Tel. +49-89-3187-3675 Institute for Bioinformatics / MIPS Fax. +49-89-3187-3585 GSF - German National Research Center for Environment and Health http://mips.gsf.de/staff/pagel === Subject: KDD-2005 Travel Announcement Precedence: list iQCVAwUBQvZTRaCIqsWvyF5FAQHWHQP+P/QfUeWMJW3iFML8aLW7l+m0io1OCOKB VT939mThQbd4ZtQ8gzCjbf1+Jfwq4OwPdGL8Ws2PgK9qm7zMdy7S5dUEdpF4qcbI SceZ8nrNyWwT4R+KVQh/GM/G9UjH0nqmGbG6F+/zyvqPiaJd8iRThvfZNzE+H3ms x+JweBR7wdI= =8EjN ******* TRAVEL ANNOUNCEMENT ********** K D D - 2 0 0 5 *************************************** Knowledge Discovery and Data Mining *************************************** The organizers of the KDD 2005 conference are pleased to announce a special offer for KDD attendees. TravelZoo.com announced a special offer from Hyatt Regency Chicago with room prices starting at $89 per night. Rooms must be booked by August 8th. When making reservations, please specify special offer code 89SPC and mention TravelZoo. This offer was made available by TravelZoo.com and not KDD. Hyatt Regency Chicago 151 East Wacker Drive Chicago, Illinois, USA. 60601 Telephone: 312 565 1234 *************************************** Conference Details *************************************** The Eleventh ACM SIGKDD International Conference on Knowledge Discovery and Data Mining will take place in Chicago, Illinois, USA, from August 21 through 24, 2005. Visit the conference web site at http://www.kdd2005.com or http://www.acm.org/sigkdd/kdd2005/ KDD is the premier international conference on knowledge discovery and data mining. The conference will provide a forum for academic researchers and industry and government innovators to share in their results and experience. The program will include keynote presentations, oral paper presentations, poster presentations, workshops, tutorials, and panels, as well as the KDD Cup competition. CONFERENCE HIGHLIGHTS Keynotes - Prabhaker Raghavan. Head of Reasearch at Yahoo! and professor of Computer Science at Stanford. - Gian Fulgoni, chariman and co-founder of comScore Networks, Mining the Internet: The Eighth Wonder of the World. KDD Cup - Winners of the KDD Cup data mining competition will be annnounced. Technical papers - Research track: 40 presentations and 36 posters - Industry/Gov. track: 14 presentations and 11 posters Full Day Workshops - Data Mining Methods for Anomaly Detection - Workshop on Open Source Data Mining (OSDM 2005) - Workshop on Utility-Based Data Mining (UBDM-2005) - 4th Workshop on Multi-Relational Data Mining (MRDM 2005) - 5th Workshop on Data Mining in Bioinformatics (BIOKDD 2005) - Workshop on Data Mining Standards, Services and Platforms (DM-SSP 2005) - WebKDD'05: Taming Evolving, Expanding and Multi-faceted Web Clickstreams - Workshop on Link Discovery: Issues, Approaches and Applications (LinkKDD-2005) Half Day Workshops - Data Mining Success Stories and KDD Practice Prize (KDDSS-2005) - 6th International Workshop on Multimedia Data Mining: Mining Integrated Media and Complex Data Tutorials - Introduction to Logistic Regression - Data Visualization and Mining using the GPU - Randomized Algorithms for Matrices and Massive Data Sets - Principles and Applications of Probabilistic Learning CONFERENCE VENUE Come and join us at KDD and enjoy your stay in Chicago. The conference will take place at the Hyatt Regency Chicago Hotel. Area attractions include the easy-to-navigate Museum Campus; lively Navy Pier; a resurrected North Loop theater district; and a who's who of luxury shopping destinations along the city's fabled Magnificent Mile. A busy convention trade has sparked hotel construction, and the city's eclectic mix of restaurants has gained an international reputation for excellence. Chicago Air and Water Show The oldest and biggest exhibition of its kind in the country, the Air and Water Show displays daredevil acrobatics both in the air and water. It can be viewed from many lakeside locations, but the main hub is above and beyond North Avenue Beach. Both civilian and military demonstrations take place and electrifying stunts are performed, the highlight being the US Navy Blue Angels' jet formation. INFORMATION Please visit our web site for more details about the program. http://www.acm.org/sigkdd/kdd2005/