mm-4119 === Subject: Re: A =4 x4 matrix.Eigenvalues=? >Let A be the real matrix > / 0 a b c > A= / a 0 d e > b d 0 f / > c e f 0 / >where a^2+b^2+c^2+d^2+e^2+f^2 = 6 . >Consider that r_1 =< r_2 =< r_3 =< r_4 are eigenvalues of A. >It is known that r_4= 10. >It's true that all eigenvalues are integer numbers ? >If A is non-singular, which is the inverse matrix A^{-1} ? > It can't happen that r_4 = 10. The characteristic polynomial of > A is t^4 - (a^2+b^2+...+f^2) t^2 + c1 t + c0 for some c1 and c0. > So you'd need r1 + r2 + r3 + r4 = 0 and > -6 = r1 r2 + r1 r3 + r1 r4 + r2 r3 + r2 r4 + r3 r4 > = 1/2 ((r1 + r2 + r3 + r4)^2 - r1^2 - r2^2 - r3^2 - r4^2) > i.e. r1^2 + r2^2 + r3^2 + r4^2 = 12. In particular > r4 <= 2 sqrt(3). > Suppose we remove the requirement r_4=10. The only ways to > get 12 as a sum of four squares are 1^2 + 1^2 + 1^2 + 3^2 and > 0^2 + 2^2 + 2^2 + 2^2, so the only integer solutions to > r1+r2+r3+r4=0 and r1^2+r2^2+r3^2+r4^2=12 with r1<=r2<=r3<=r4 > are [-1,-1,-1,3] and [-3,1,1,1]. It is possible to get these > eigenvalues, e.g. with a=b=c=d=e=f=1 or a=b=c=d=e=f=-1 respectively. It looks to me (without being able to prove it) like the only case where 3 is an eigenvalue is a=b=c=d=e=f=1, i.e. a^2+...+f^2=6 and det(A-3I) = 0 imply a=b=...=f=1. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Is this simplifed multidimensional packing problem still NP-hard? Hi folks, It's known that the traditional multidimensional packing problem is NP-hard. The Traditional multidimensional packing problem is as follows: Given: vector w (1 x n), matrix A (m x n), vector b (m x 1) w, A, b are integer valued. find: vector y (n x 1) maximize: w * y subject to: Ay 1? Any hints or references would be appreciated. Bo === Subject: Demostration: Binomial Distribution mean and var. Need help Please!!!! Hi I need to demonstrate that the mean of a binomial distribution is np and the variance is np(1-p). Please, I need all your help. i'm a first year university student and I need this ASAP. === Subject: Re: Demostration: Binomial Distribution mean and var. Need help Please!!!! Very simple solution. Let X = number of trials you had successes in N total trials. Now X = I1 + I2 + ...+ IN where Ij = outcome of jth trial (it is distributed like Bernoulli(p) So the expectation of a sum is th sum of the expectations E(X) = p + p + ...+p = Np (there are N of them) I won't work out the formula for the variance, but the principle is the same (although here you need to know that the trials are independent in order to justify that the variance of a sum is the sum of the variances > Hi > I need to demonstrate that the mean of a binomial distribution is np > and the variance is np(1-p). > Please, I need all your help. i'm a first year university student and > I need this ASAP. === Subject: Re: Demostration: Binomial Distribution mean and var. Need help Please!!!! > Hi > I need to demonstrate that the mean of a binomial distribution is np > and the variance is np(1-p). > Please, I need all your help. i'm a first year university student and > I need this ASAP. The technique of indicator variables is easiest. Let k = the number of successes in n independent trials, where the probability of success on each trial is p. So k is a binomial random variable with parameters n, p. You want the mean and variance of k. Define a variable I_j = 1 if the j-th trial is a success, and 0 if it is a failure. Here are a few pertinent facts: 1. All the I's have the same distribution. 2. k = sum(I_j), j=1, n a. Therefore the mean of k is the mean of the sum of the I's, which is the same of the sum of the means. 3. What's the mean (expectation value) of I? It has two possible values, 0 and 1, so the expectation value is E[I] = 1*P(I=1) + 0*P(I=0). 4. You can also work out E[I^2] from the definition of expectation value (what possible values does I^2 have)? 5. Again noting that k is a sum of random variables, and that these variables are INDEPENDENT, then var(k) = sum[var(I_j)]. So you need to work out var(I_j), which I just gave you strong hints on how to do. The suggestion has been made to look in your university library. I suggest you do so. This proof will be found in many elementary probability texts, and likely many of them will use the indicator variable technique. I find that easiest, but there are other approaches. - Randy === Subject: Re: Demostration: Binomial Distribution mean and var. Need help Please!!!! > Hi > I need to demonstrate that the mean of a binomial distribution is np > and the variance is np(1-p). > Please, I need all your help. i'm a first year university student and > I need this ASAP. (1) Since you are a university student, a university library should be at a walking distance. And it must have a few books that present the proofs. Also, for first year university students the school usually provides teaching assistants (TA's) to help. (2) The proofs I saw start with the binomial expansion (u+v)^n = sum[k=0 to n] (n choose k) * u^k * v^(n-k) then one differentiates by u, multiplies by u, differentiates again by u, multiplies again by u, sets u = p, v = 1-p, does some extra algebra. You have seen V(X) = E(X^2) - (E(X))^2, I presume. Use it. === Subject: Re: Star Gate Topology of the Universe does anybody just look up a see the clouds blow by any more? holog === Subject: a problem of the determinate of a matrix B = [ b11, b12, b13, b14] [ b12, b22, b23, b24] [ b13, b23, b33, b34] [ b14, b24, b34, b44] BB = [ b11*b33-b13^2, b12*b33-b13*b23, b14*b33-b13*b34] [ b12*b33-b13*b23, b22*b33-b23^2, b24*b33-b23*b34] [ b14*b33-b13*b34, b24*b33-b23*b34, b44*b33-b34^2] It is known that B is positive definite, to determine whether BB is of full rank, that is det(BB)!=0. -- Liu Tong === Subject: Re: a problem of the determinate of a matrix > It is known that B is positive definite, to determine whether BB is of full > rank, that is det(BB)!=0. If B is positive definite, then det(B)>0. det(AB)=det(A)*det(B). M. K. Shen -------------------------------- http:://home.t-online.de/home/mok-kong.shen === Subject: Re: a problem of the determinate of a matrix >B = >[ b11, b12, b13, b14] >[ b12, b22, b23, b24] >[ b13, b23, b33, b34] >[ b14, b24, b34, b44] >BB = >[ b11*b33-b13^2, b12*b33-b13*b23, b14*b33-b13*b34] >[ b12*b33-b13*b23, b22*b33-b23^2, b24*b33-b23*b34] >[ b14*b33-b13*b34, b24*b33-b23*b34, b44*b33-b34^2] >It is known that B is positive definite, to determine whether BB is of full >rank, that is det(BB)!=0. According to Maple, det(BB) = det(B)*b33^2. To confirm this, do some row and column manipulations on B until you get BB. -- ARNOLD = Anagram of RONALD ENEGGER = Backwards mis-pronounced REAGAN This is a black -- I mean SCHWARZ -- period in California. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Re: Constant weight codes in GF(2^m)? > I hope that this is not too far off topic? I am interested in > constructing constant weight codes using elements from a Galois group > GF(2^m)? How can I construct such codes? In the literature that I > have found up to now, the authors only consider binary constant weight > codes. > Any comments, suggestions and/or pointers to literature will be > greatly appreciated > Jaco Versfeld Well, this old paper of mine gives a description of the cyclic, equidistant codes over GF(q). If you throw out 0 you have a constant weight code: Clark, W. Edwin Equidistant cyclic codes over ${rm GF}(q)$. Discrete Math. 17 (1977), no. 2, 139--141. It is proved that a cyclic $(n,k)$ code over $text{GF}(q)$ is equidistant if and only if its parity check polynomial is irreducible and has exponent $e=(q^k-1)/a$ where $a$ divides $q-1$ and $(a,k)=1$. The length $n$ may be any multiple of $e$. The proof of this theorem also shows that if a cyclic $(n,k)$ code over $text{GF}(q)$ is not a repetition of a shorter code and the average weight of its nonzero code words is integral, then its parity check polynomial is irreducible over $text{GF}(q)$ with exponent $n=(q^k-1)/a$ where $a$ divides $q-1$. === Subject: Re: JSH: Fun time > Does anyone really run a newsreader that executes javascript? If so, > why on earth would you want to? Sorry for the javascript. I thought the pop-up appropriate. > Heh-heh. > Nothing makes a donkey happier than a rut in the barnyard, as this > hee-hawing hoofer knows! > Nothing makes a donkey happier than a rut in the barnyard, as this > hee-hawing hoofer knows! > http://www.rubylane.com/ni/shops/viperswife/iteml/GW-684 http://www.cogsci.indiana.edu/farg/harry/bio/zoo/donkey.wav --John === Subject: Re: JSH: Fun time Johnboy, go **** yourself, or any other consenting parties, but leave the donkey alone. I'm a Democrat, after all! > /~~~ > / | /~~| `=` |/~~ > /_/ | `` | | `` |/~~ > / | | | | `` | > { <| | | | | > | | > | / . . ` / > | . . / > . / > . / > / > / > | | > | | --les ducs d'Enron! === Subject: Free the masses: Free the math! A science journal, instead of charging readers a subscription rate, I am all for this, except that the authors seem, to me, to be charged a huge amount. their public(non-student/faculty)-access to on-line journals subscribed to by these libraries. Here in my area, as an example, the university library has now cut off the public's access to its internet terminals, and consequently to member-only journal-searches. (Why?? ..You guessed it!... Budget cuts...) That is why I love the internet, for the most part. Math information (or information), whether peer-reviewed or not even close, is available in many cases for FREE (technically, but not always in practice). Hopefully, through PUBLIC libraries' books and internet terminals, even those of modest income will be forever able to learn some, at least, of the mysterious art of mathematics. (Gettin' socialistic about math...) ;) Leroy Quet === Subject: Re: Free the masses: Free the math! > That is why I love the internet, for the most part. Math information > (or information), whether peer-reviewed or not even close, is > available in many cases for FREE (technically, but not always in > practice). Hopefully, through PUBLIC libraries' books and internet > terminals, even those of modest income will be forever able to learn > some, at least, of the mysterious art of mathematics. > (Gettin' socialistic about math...) ;) Where's the central planning? -- http://hertzlinger.blogspot.com === Subject: Re: Free the masses: Free the math! > A science journal, instead of charging readers a subscription rate, this makes as much sense as having the inventors of the chicken pox vaccine pay for publishing their results. well, almost. math does not save lives, but rather ruins them in some cases. > I am all for this, except that the authors seem, to me, to be > charged a huge amount. > their public(non-student/faculty)-access to on-line journals subscribed > to by these libraries. > Here in my area, as an example, the university library has now cut off > the public's access to its internet terminals, and consequently to > member-only journal-searches. what is wrong with paying membership? > (Why?? ..You guessed it!... Budget cuts...) budget cuts? public univs and colleges seem to be sucking more and more money from the public now more than ever... > That is why I love the internet, for the most part. Math information > (or information), whether peer-reviewed or not even close, is > available in many cases for FREE (technically, but not always in > practice). Hopefully, through PUBLIC libraries' books and internet > terminals, even those of modest income will be forever able to learn > some, at least, of the mysterious art of mathematics. why should math be free? > (Gettin' socialistic about math...) ;) > Leroy > Quet === Subject: Re: Free the masses: Free the math! > why should math be free? Mathematicians are frequently absent-minded. If you have a forgetful functor, shouldn't you have a free functor to go with it? -- http://hertzlinger.blogspot.com === Subject: Re: Free the masses: Free the math! > A science journal, instead of charging readers a subscription rate, > I am all for this, except that the authors seem, to me, to be charged > huge amount. That is called vanity publishing. Few regard it as a Good Thing. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Free the masses: Free the math! > A science journal, instead of charging readers a subscription rate, > I am all for this, except that the authors seem, to me, to be charged > a > huge amount. > That is called vanity publishing. Few regard it as a Good Thing. But the very thing for JSH papers! === Subject: Re: Free the masses: Free the math! >A science journal, instead of charging readers a subscription rate, >I am all for this, except that the authors seem, to me, to be charged >huge amount. >That is called vanity publishing. Few regard it as a Good Thing. economics/finance has told me of first-tier journals that charge a fee when one submits a paper to be refereed. The journal keeps the fee, regardless of whether the paper is accepted or not. I don't know if the journal charges for the resubmission of a revision. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Free the masses: Free the math! >A science journal, instead of charging readers a subscription rate, >I am all for this, except that the authors seem, to me, to be charged >a >huge amount. >That is called vanity publishing. Few regard it as a Good Thing. > economics/finance has told me of first-tier journals that charge a fee > when one submits a paper to be refereed. The journal keeps the fee, > regardless of whether the paper is accepted or not. I don't know if the > journal charges for the resubmission of a revision. Sounds like a good scam. Yeah. Just submit your paper to our journal, and we will ,surely, certainly consider it for publication,..certainly, we REALLY will. You ONLY have to pay a *small* consideration-fee.... On those few occasions, when I have had a math-problem of mine published in either the Mathematics Magazine or the American Mathematical Monthly, my friends sometimes ask, So, are they paying you?... (of course not) I guess, or so it seems from this thread, I should actually be GLAD that they do not have ME pay THEM!... leroy Quet === Subject: Re: Free the masses: Free the math! > A science journal, instead of charging readers a subscription rate, > I am all for this, except that the authors seem, to me, to be charged > huge amount. > their > public(non-student/faculty)-access to on-line journals subscribed to > by > these libraries. > Here in my area, as an example, the university library has now cut off > the public's access to its internet terminals, and consequently to > member-only journal-searches. > (Why?? ..You guessed it!... Budget cuts...) > That is why I love the internet, for the most part. Math information > (or > information), whether peer-reviewed or not even close, is available > in > many cases for FREE (technically, but not always in practice). > Hopefully, through PUBLIC libraries' books and internet terminals, > even > those of modest income will be forever able to learn some, at least, > of > the mysterious art of mathematics. > (Gettin' socialistic about math...) ;) pay 1 cent per email and eliminate spam, pay per post, get used to the idea. ecommerce hasn't centralised yet, cent per image view or page view are about a decade away. I'll give you all premium accounts, they're 2c. don't fret about the library, avoiding $1 a day for an internet connection is mismanagement and a plea for funds not lack of them. Herc === Subject: Re: Free the masses: Free the math! > A science journal, instead of charging readers a subscription rate, > I am all for this, except that the authors seem, to me, to be charged > huge amount. I share your sentiment in many ways, but one thing that should be noted about the PLOS journals (right now at least) is that they are strictly biology and soon medicine. Big budget science. This kind of stuff just ends up taking up a whole lot of elbow grease and 'stuff'. The publication fee is relatively given the costs involved in these types of scientific endavour. Not to mention the fact, that their policy implies that if the work is of sufficient merit and the costs cannot be borne by the author, they will waive the fee. > their > public(non-student/faculty)-access to on-line journals subscribed to > by > these libraries. > Here in my area, as an example, the university library has now cut off > the public's access to its internet terminals, and consequently to > member-only journal-searches. > (Why?? ..You guessed it!... Budget cuts...) > That is why I love the internet, for the most part. Math information > (or > information), whether peer-reviewed or not even close, is available > in > many cases for FREE (technically, but not always in practice). > Hopefully, through PUBLIC libraries' books and internet terminals, > even > those of modest income will be forever able to learn some, at least, > of > the mysterious art of mathematics. > (Gettin' socialistic about math...) ;) > Leroy > Quet === Subject: Re: Free the masses: Free the math! Leroy Quet > A science journal, instead of charging readers a subscription rate, Six Hours a-day the young Students were employed in this Labour, and the Professor shewed me several Volumes in large Folio already collected, of broken Sentences, which he intended to piece together, and out of those rich Materials to give the World a compleat Body of all Arts and Sciences; which however might be still improved, and much expedited, if the Publick would raise a Fund for making and employing five hundred such Frames in Lagado, and oblige the Managers to contribute in common their several Collections. -- Jonathan Swift, _A Voyage to Laputa..._, 1726!! www.jaffebros.com/lee/gulliver/bk3/chap3-5.html LH === Subject: Re: Free the masses: Free the math! === >Subject: Free the masses: Free the math! >A science journal, instead of charging readers a subscription rate, >I am all for this, except that the authors seem, to me, to be charged >huge amount. >their >public(non-student/faculty)-access to on-line journals subscribed to >these libraries. >Here in my area, as an example, the university library has now cut off >the public's access to its internet terminals, and consequently to >member-only journal-searches. >(Why?? ..You guessed it!... Budget cuts...) >That is why I love the internet, for the most part. Math information >(or >information), whether peer-reviewed or not even close, is available >many cases for FREE (technically, but not always in practice). >Hopefully, through PUBLIC libraries' books and internet terminals, >even >those of modest income will be forever able to learn some, at least, >the mysterious art of mathematics. >(Gettin' socialistic about math...) ;) >Leroy >Quet (!) but thought they were mostly used to pad a vita. I saw no notice of peer review in your reference. Will acceptance for publication be strictly up to the editors? Marvin Marvin Sebourn osugeography@aol.com === Subject: Re: Factorial/Exponential Identity, Infinity > Splitting the > difference between almost all and almost none is around one half. Splitting the difference between almost all and almost none in a > real interval is almost anything in between. > Unless some highly specific splitting mechanism is guaranteed. What do you think about this? How can this disparity be resolved? Parity. Boyee. Anyways, we've seen from one method that an established asymptotic value for n choice n/2 approaches zero. Yet, binary sequences representing numbers normal to base two have equal densities of ones and zeros, and almost all numbers in the unit interval are absolutely normal, having finite measure. Proportion of sequences with equal zero-density and one-density according to: Combinatorics: approx. 0 Borel: approx. 1 Is Borel wrong? It reminds me of using input values dependent on the divergent variable in the Gamma function, except I haven't found that to actually disagree with anything besides diverging towards infinity. Each integer is finite. If there were two infinitely precise integers then their quotient might be irrational. In a consideration of splitting the real numbers in the unit interval into two (equal) parts, half of them have a zero as the first element in their binary expansion, half don't. The same holds true for each element of the seqence. I thought that half of the sequence had half ones and half zeros just because, it was a moment of insight. I'm beginning to think that it has to do with that half the elements of all the sequences are zeros and half the elements are ones. Unfortunately my space television antenna for receiving Galaxy 19 intro to reality programming went down because of blackbody interference from the Warpgate 93 spaceseal pods. Makes me want to haul out my alcohol-powered pro contractor graphite/chromolly seal club. All hail red meat. Damn terrorists. The universe is infinite. Where's Uncle Al when you need him? He'd have something to say. Would it offend half the people? Ross === Subject: Re: Factorial/Exponential Identity, Infinity Splitting the > difference between almost all and almost none is around one half. > Splitting the difference between almost all and almost none in a > real interval is almost anything in between. Unless some highly specific splitting mechanism is guaranteed. > What do you think about this? How can this disparity be resolved? Between what and what? === Subject: Re: Factorial/Exponential Identity, Infinity > Each integer is finite. If there were two infinitely precise integers > then their quotient might be irrational. Just what in hell is imprecise about an integer? === Subject: Re: Factorial/Exponential Identity, Infinity The dis-parity is about a combinatoric expression evaluating the proportion of binary sequences with density of ones being one half among every binary sequence as near zero and the measure-theoretic evaluation of the set of absolutely normal numbers between zero and one as having measure one or near one, where absolutely normal numbers have density of ones being one half. You know that. disrespect parity, yo. Parity appears to be a broadly applied scientific principle. About the integers, there isn't anything imprecise about an integer. There's imprecision in representing an irrational as a ratio of two integers. An infinite sum (sum of infinitely many) of ratios of integers may be an absolutely precise equal thing to an irrational, eg, zeta(3) is irrational, e = 1/0! + 1/1! + 1/2! + ..., etcetera. There are also infinite products of ratios of integers that equal irrational numbers, for example 2*(2*2/1*3)*(4*4/3*5)*(6*6/5*7)*... = pi. What shall we take 2*(2*4*6*...)^2 to be? Divide it by (3*5*7*...)^2, the quotient is pi. Agreeably, it's a bit more specific than that. It's like considering the sets of integers as a binary sequence with no radix, separator, decimal point. Ross === Subject: Re: Factorial/Exponential Identity, Infinity > The dis-parity is about a combinatoric expression evaluating the > proportion of binary sequences with density of ones being one half > among every binary sequence as near zero and the measure-theoretic > evaluation of the set of absolutely normal numbers between zero and > one as having measure one or near one, where absolutely normal numbers > have density of ones being one half. You know that. There is a disparity between measurements in grams and measurements in metres which is of equal relevance. > disrespect parity, yo. Parity appears to be a broadly applied > scientific principle. Parity with respect to binary sequences is a broadly applied scientific principle? > About the integers, there isn't anything imprecise about an integer. > There's imprecision in representing an irrational as a ratio of two > integers. I beg to differ. Any such representation exactly represents one and only one rational, and, though there are equivalent repressentations, the nature of that equivalence is preciesly known: a/b = c/d <==> a*d = b*c An infinite sum (sum of infinitely many) of ratios of > integers may be an absolutely precise equal thing to an irrational, > eg, zeta(3) is irrational, e = 1/0! + 1/1! + 1/2! + ..., etcetera. > There are also infinite products of ratios of integers that equal > irrational numbers, for example 2*(2*2/1*3)*(4*4/3*5)*(6*6/5*7)*... = > pi. What shall we take 2*(2*4*6*...)^2 to be? Divide it by > (3*5*7*...)^2, the quotient is pi. Agreeably, it's a bit more > specific than that. The issue with infinite sums and infinite products is their convergence, which is precisely defined. The value of a non-convergent expression for an infinite sum or infinite product is no more meaningful than the value of a quotient with zero as divisor, it is NaN! === Subject: Re: Factorial/Exponential Identity, Infinity > There's imprecision in representing an irrational as a ratio of two > integers. I'll treasure that :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Factorial/Exponential Identity, Infinity > There's imprecision in representing an irrational as a ratio of two > integers. Not when the two integers are both zero. At least, not precisely what I'd call imprecision. Lee Rudolph === Subject: Re: Web server question and Probability >At a Web server every 30 seconds arrive in average 3 requests. I want >to calculate the probability of waiting for more than 15 minutes so >that 80 requests arrive. Model it with a Poisson distribution. Doug === Subject: Re: Chi-square, Student T and f distribution > Are there any conditions under which the t, F and chi-square distributions > can yield the same results? Are there any conditions under which the t, F and chi-square distributions can all be appropriate? === Subject: Re: Please check probability answer... >Answer is 52/19 = 2.73682 probability. > Can a probability be greater than one? > Doug Most definitely! I have it on good authority from James Harris that probablities can be much greater than one. The problem is that mathematicians made an error in their definition of probability (just as they did in defining algebraic integers). This is the soon-to-be-entitled core error in probability and is certain to rock the mathematics community to the core (although they will of course no doubt not listen). Hope this clears up any confusion you might have had. JB === Subject: Re: Fundamental Reason for High Achievements of Jews If thine enemy smite thee on thy cheek, tear his head off and >down > his neck. [...] A real man > kills his enemies, takes his land and cattler, rapes his women, >enslaves > his children and contemplates how good life can be. In this evil world, if you > are not on top, some other part is and is humping you. The price of > liberty is efficiently killing the enemies of liberty. The rule is >very > simple. Support and help your friends. Kill your enemies. That is the > only way to survive. Do unto others, before they do unto you. Fascinating, even if it's off topic. Robert Kolker, who claims to be Jewish, advocates a philosophy > of life which sounds a lot like what we normally would associate > with Nazis and supremacists. If someone claimed the Jews believe what Kolker claims to > believe, that person would surely be labelled an anti-Semite. And yet, Kolker seems to be serious, and has received some support > from others in this thread. What gives? >Actually, the masses have been brainwashed >to give the German people a bad rap, >for trying to defend themselves from >the Bolsheviks.who were instigating >conflict and war, for power and riches in the 1900's. > Please give examples. Stalin was no angel but except for his defeat > in Finland never tried what the right wing Hitler did. All his crimes > were internial > Stalin did not invade Germany in June 1941, Hitler Invaded Russia, > Stalin did however cooperate with Hitler in the occupation of Poland. >As the Bolsheviks.were assassinating leaders and breaking >up the meetings of political rivals, > Please give examples including dates. For the most part it was the > Nazi's that that broke up and attacked rivals. >It is interesting to see that Thom does not comprehend, >that Germany, and many nations, >tried to control the class war instigations of the Bolsheviks >by many peaceful means, and by making treaties to >oppose the instigation of conflict and war, >and when all measures failed, they went after the Bolsheviks >in the nations they were using as bases of operations, just as >America went after real terrorists, and imagined terrorists in Iraq. >Note that the Germans were welcomed in most nations >as their local leaders had been bought or intimidated >by the Bolsheviks and were unable to maintain the >social, financial and cultural integrity of their nations. >Even France had little opposition to the Germans. >If FDR had joined with Germany, Spain, Japan, and other nations, >in going after the class war instigators, rather than backing them, >WWII would has lasted two months, and there would have been >no millions of deaths, no Cold War, no Korea, no Vietnam , >no nuclear weapons, and no instigation of global religious war. >Bush has also sold out to the religious war instigators, >and brought much harm to America and the world. >And it is interesting to see that Thom does not know >that the Bolsheviks were assassinating leaders, and assaulting >people, and breaking up meetings, and that the Black Shirts >and Nazi were reactions to this. >I suggest that he read some factual, concurrent history >of those times, rather than allow himself to be >brainwashed by present day spin motivated more by agenda, >than by facts or a respect for mankind. > I see our Holocaust defender is still spewing vile. It is interesting to see that Lloyd Parker considers it vile to post factual history, rather than parrot religious/ethnic propaganda; and to point out that the same group that instigated the class wars of the 1900's is instigating the religious wars of the 2000's. One has to wonder why some people do this. Are they brainwashed, or do they support the people who instigate conflict and war for power and riches????? -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews ... snip incredible nonsense from a dedicated anti-semite ... > One has to wonder why some people do this. > Are they brainwashed, > or do they support the people who instigate > conflict and war for power and riches????? > -- > Tom Potter http://tompotter.us Gee, I don't get this. How do I get into this power elite? I'm usually considered quite bright and I am Jewish. But I have never been in the company of any Jew who was in this power elite. My late brother-in-law was rich, Jewish, but not one of the power elite. How do I get there. Why have the other Jews kept this a secret from me and all my relatively poor ancestors? Even the atomic bomb had those who blabbed about it. Why haven't I heard about this grand scheme? I want in. I could contribute a lot. I have lots of Christian relatives who would provide cover for me. I know a few fairly powerful politicians and could influence them if it were worth my while. Why has all this been hidden from me for so many years? Could it be that Tom Potter is a troll who makes this all up out of whole cloth? Naw, not Tom Potter. Everybody knows he is honest to a fault. FK === Subject: Re: Fundamental Reason for High Achievements of Jews ... snip incredible nonsense from a dedicated anti-semite ... One has to wonder why some people do this. > Are they brainwashed, > or do they support the people who instigate > conflict and war for power and riches????? -- > Tom Potter http://tompotter.us > Gee, I don't get this. How do I get into this power elite? I'm usually > considered quite bright and I am Jewish. But I have never been in the > company of any Jew who was in this power elite. My late brother-in-law > was rich, Jewish, but not one of the power elite. How do I get there. > Why have the other Jews kept this a secret from me and all my relatively > poor ancestors? Even the atomic bomb had those who blabbed about it. Why > haven't I heard about this grand scheme? I want in. I could contribute a > lot. I have lots of Christian relatives who would provide cover for me. > I know a few fairly powerful politicians and could influence them if it > were worth my while. Why has all this been hidden from me for so many > years? Could it be that Tom Potter is a troll who makes this all up out > of whole cloth? Naw, not Tom Potter. Everybody knows he is honest to a > fault. > FK fkasner makes a good point! EVERYONE, Black, White, Brown, Yellow, Protestent, Catholic, Jewish, Muslim, Buddhists, male, female, gay, rich, poor, etc. who has ever dealt with Tom Potter, on a personal level, knows that he is honest to a fault. If honesty be a grevious fault, then greviously have I paid for it, and will pay for it. (With apologies to the Bard.) -- Tom Potter http://tompotter.us === Subject: Re: Fundamental Reason for High Achievements of Jews > ... snip incredible nonsense from a dedicated anti-semite ... Careful, Kaz, these days the ones who always cry anti-Semite are now perceived to be the real bigots and racists at heart. Reflect on that and ask yourself why this came about. Hint: 50+ years of incessant and loud Jew propaganda about Jewish moral and intellectual superiority, ......perhaps? All good things do come to an end, Kaz,.......ahahahaha.... > One has to wonder why some people do this. > Are they brainwashed, > or do they support the people who instigate > conflict and war for power and riches????? > Tom Potter I don't know, nor care, what post preceded this one, but since ever mankind existed some people of/in any tribe have always and will always instigate conflict and war for power and riches...... or what other reasons are there for killing each other?......and if the conditions were right, Tom, you'd be in the forefront doing the same, given the personality you exhibit on the Usenet......ahahahaha... > Gee, I don't get this. How do I get into this power elite? I'm usually > considered quite bright and I am Jewish. But I have never been in the > company of any Jew who was in this power elite. My late brother-in-law > was rich, Jewish, but not one of the power elite. How do I get there. > Why have the other Jews kept this a secret from me and all my relatively > poor ancestors? Even the atomic bomb had those who blabbed about it. Why > haven't I heard about this grand scheme? I want in. I could contribute a > lot. I have lots of Christian relatives who would provide cover for me. > I know a few fairly powerful politicians and could influence them if it > were worth my while. Why has all this been hidden from me for so many > years? Very astute and SEEMINGLY factual, Kaz, BUT..... See, Kaz, apparently you really don't get it. It's very simple. It is not so-called riches and power that gets Jews into deep periodically and epically. A lot of goys have far more of each, yet it normally does not get them into trouble. Most have learned how to handle it: Discretely, smoothly and QUIETLY. But not the Jews. With the Jews it is the opposite. It is their ing, incessant **loudness about themselves**, advocating that they are richer, better, smarter and more lovable BECAUSE THE ARE JEWISH, from private events to general media stuff, (like i.e. reporting on the front page or in headlines that it rained in Tel Aviv, but shoving a school bus accident report that killed 20 US-goy kids into the background......) that is what makes the goyim, which outnumber Jew almost a million fold, so irate and belligerent towards them. --- It is a simple as that. We won't even have to touch that for 50 years US taxpayers got forced to pay billions, annually, to the Jewish State... with NOTHING to show for in return! Will it change? No. Why not? Because that would mean for Jews to become Un-Jewish....ahahahaha.... Of course Potter will come along now and give recital of his Not all Jews are... song .........aahahahaha........ahahahaha......... > Could it be that Tom Potter is a troll who makes this all up out > of whole cloth? Naw, not Tom Potter. > Everybody knows he is honest to a fault. > FK ahahahaha............fault, Kaz, what fault?.........honesty and morality is Potterian by DEFAULT...........ahahahahaha........but then it becomes more evident with each post of yours that you, Kaz, are the Jewish mirror image of Potter.........ahahahah...... So, keep on singing.....All of you!.........It's is a beautiful choir.... ....accompanied by that clacking sound of shards of falling glass that comes from the stones we throw at each others houses of glass... ahahahaha......ahahahanson === Subject: BBP algorithm Does anyone know of a derivation for the BBP (Bailey-Borwein-Plouffe) Kayan === Subject: Re: BBP algorithm > Does anyone know of a derivation for the BBP (Bailey-Borwein-Plouffe) Yes. Write the sum as an integral and so the integral. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: What is 0/0 ? > The point is that 0/0 doesn't have meaning; don't you guys call > that undefined in the math biz? As I said in my first response in this thread, 0/0 is normally considered to be undefined. But another point is that 0/0 can be given meaning (and, IMO, in reasonable ways) if we in the math biz choose to do so. David === Subject: Re: What is 0/0 ? > The point is that 0/0 doesn't have meaning; don't you guys call > that undefined in the math biz? >As I said in my first response in this thread, 0/0 is normally considered >to be undefined. I remember. It doesn't hurt to say it often and say it loud ;-). /BAH Subtract a hundred and four for e-mail. === Subject: Re: What is 0/0 ? > The point is that 0/0 doesn't have meaning; don't you guys call > that undefined in the math biz? >As I said in my first response in this thread, 0/0 is normally considered >to be undefined. >But another point is that 0/0 can be given meaning >(and, IMO, in reasonable >ways) if we in the math biz choose to do so. Sure it would be interesting to see what kinds of algebras you could make, but that isn't going to apply in computing today. This propensity of doing something wrong just because there's a rule against it might be a way of establishing territorial imperative among teenage males but it sure as hell would create havoc in my bank account. /BAH Subtract a hundred and four for e-mail. === Subject: Re: What is 0/0 ? > The point is that 0/0 doesn't have meaning; don't you guys call > that undefined in the math biz? >As I said in my first response in this thread, 0/0 is normally considered >to be undefined. >But another point is that 0/0 can be given meaning >(and, IMO, in reasonable >ways) if we in the math biz choose to do so. > Sure it would be interesting to see what kinds of algebras you > could make, but that isn't going to apply in computing today. > This propensity of doing something wrong just because there's > a rule against it might be a way of establishing territorial > imperative among teenage males but it sure as hell would create > havoc in my bank account. The whole point is that it doesn't *have* to be undefined. The application determines what should be done with it. In bank accounts, divide by 0 should generate an error message. But in hydrodynamics or number theory or complex variable theory or other areas, it might (note: *might*) make perfect sense to simply remove all the removable singularities and not make a big deal about it. It all depends on the application. And that means that it isn't *always* wrong, and we don't need to *always* declare it to be undefined. Jon Miller === Subject: core error fixed > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. Stupid, evil definition... It's time to destroy it! No definition - no problem! That's the easiest way to fix everything. Let's never use the term algebraic integers again, let's just call them roots of monic polynomials with integer coefficient! All right, no core error from now on! === Subject: Prime number generating function Has anyone see this function: http://datashaping.com/Prime2.gif ? It's graph is shown at http://datashaping.com/cryptography.shtml . === Subject: Re: Prime number generating function > Has anyone see this function: http://datashaping.com/Prime2.gif ? I would say it's more of a prime-testing function than a prime-generating function. And it only tests primes that are <=(v**2)/2. > Its graph is shown at http://datashaping.com/cryptography.shtml . --Mike Amling === Subject: Re: Prime number generating function Hehe, looks nice. Have you done any timing analysis of the function? (I have too little knowledge of this to do that myself...) -Panic === Subject: Problem with Lp-norm I want to prove the triangle inequality for the Lp-norm b |f| = (S|f(x)|^p dx)^(1/p) a f(x) is a continous function on [a,b], and p is in [1,oo). I really don't get it, after writing several pages of inequalities. Can anyone give me a hint about the main idea of the proof? Rene. -- Ren.8e Meyer Student of Physics & Mathematics Zhejiang University, Hangzhou, China === Subject: Re: Problem with Lp-norm > I want to prove the triangle inequality for the Lp-norm > b > |f| = (S|f(x)|^p dx)^(1/p) > a > f(x) is a continous function on [a,b], and p is in [1,oo). I really > don't get it, after writing several pages of inequalities. Can anyone > give me a hint about the main idea of the proof? You can do it without Holder's inequality (for a reference see Royden's _Real Analysis_ text). Of course you will aim for |f + g|^p <= (|f| + |g|)^p. The main hint would be to use the convexity of the function f(t) = t^p. Hugh === Subject: Re: Problem with Lp-norm > I want to prove the triangle inequality for the Lp-norm > b > |f| = (S|f(x)|^p dx)^(1/p) > a > f(x) is a continous function on [a,b], and p is in [1,oo). I really > don't get it, after writing several pages of inequalities. Can anyone > give me a hint about the main idea of the proof? This is often derived from Holder's inequality; do you know the latter? === Subject: Re: An Ownership Puzzle > Peter, Paul, and Mary each want to own the bar of gold in their presence. > They all agree that the final owner of the gold bar will be the person who > is holding it at exactly one o'clock and they agree to the following rules. > At twelve o'clock Mary will hold the gold bar and, since twelve thirty is > half way between twelve o'clock and one o'clock, at twelve thirty Mary will > give the bar to Peter. Since twelve forty-five is half way between twelve > thirty and one o'clock, at twelve forty-five Peter will give the gold bar to > Paul. The half way time between twelve forty-five and one o'clock is twelve > fifty two and thirty seconds at which time Paul hands the gold bar back to > Mary. At the half way time between twelve fifty-two (and thirty seconds) > and one o'clock, Mary gives the gold bar to Peter and then at the next half > way time Peter gives it to Paul, and so on. In this puzzle each person can > pass the gold bar to the next person in no time at all. This passing of the > gold bar continues in the order Mary, Peter, Paul, Mary, Peter, Paul, ... > until exactly one o'clock and then stops. Who will be the final owner of > the gold bar? > Since the speed at which the bar is being passed increases without > limit as the time approaches one o'clock, the relativistic mass will > also increase without limit, so the environs of the gold bar will > collapse into a black hole before one o'clock. > This presumes the distances between Peter, Paul and Mary is non-zero > constant. That being the case, the bar will vaporise just before one. > Now Peter, Paul and Mary had better don asbestos mitts, very thick mitts > because when the speed increases until the melting point of the bar is > reached due to air friction, the recoil will great. Big enuf to knock > them over, to blow them away. > Since no information can escape from a black hole, we will never > know who got the gold bar. > As that black circumstance is precluded by other practical concers, we're > forced to accept that Peter, Paul and Mary are moving closer and closer > together. That eventually the bar is squizzed between three palms with no > one having time to grasp or hold the bar, that all three palms are > holding the bar at one o'clock. > Morrow of the story: > Peter, Paul and Mary best stay with Rosemary, Sage and Thyme. the puzzle, it would be like discussing the quantum uncertainty principle or various theories of quantisized space in response to a question about the truth of the continuum hypothesis. This puzzle is purely mathematical, it takes place in the ideal continuum space and time of mathematics. For the issue fishfry raised, whenever the gold bar is swapped (for example at twelve thirty), at that time the gold bar is in the hands of the previous holder, not the new holder. Because this puzzle is taking place in the idealized Euclidean space and idealized time (modeled by the real line), there are no relativistic or other physical constraints: Peter, Paul and Mary are not moving closer together at any time. So the question remains: who will be the final owner of the gold bar? Randy http://www.rlgerl.com === Subject: Re: An Ownership Puzzle Peter, Paul, and Mary each want to own the bar of gold in their > presence. > They all agree that the final owner of the gold bar will be the person > who > is holding it at exactly one o'clock and they agree to the following > rules. > At twelve o'clock Mary will hold the gold bar and, since twelve thirty > is > half way between twelve o'clock and one o'clock, at twelve thirty Mary > will > give the bar to Peter. Since twelve forty-five is half way between > twelve > thirty and one o'clock, at twelve forty-five Peter will give the gold > bar to > Paul. The half way time between twelve forty-five and one o'clock is > twelve > fifty two and thirty seconds at which time Paul hands the gold bar > back to > Mary. At the half way time between twelve fifty-two (and thirty > seconds) > and one o'clock, Mary gives the gold bar to Peter and then at the next > half > way time Peter gives it to Paul, and so on. In this puzzle each > person can > pass the gold bar to the next person in no time at all. This passing > of the > gold bar continues in the order Mary, Peter, Paul, Mary, Peter, Paul, > ... > until exactly one o'clock and then stops. Who will be the final owner > of > the gold bar? various theories of quantisized space in response to a question about the > truth of the continuum hypothesis. This puzzle is purely mathematical, it > takes place in the ideal continuum space and time of mathematics. For the > issue fishfry raised, whenever the gold bar is swapped (for example at > twelve thirty), at that time the gold bar is in the hands of the previous > holder, not the new holder. Because this puzzle is taking place in the > idealized Euclidean space and idealized time (modeled by the real line), > there are no relativistic or other physical constraints: Peter, Paul and > Mary are not moving closer together at any time. So the question remains: > who will be the final owner of the gold bar? This reminds me of the Brower map where every boundry point is the boundry of three countries. It is an infinite series but written in such a way that the map takes 1 year to complete so the implication is that the process has ended. Then there should have been a last country to extend its map (at least one would think) - so which was the last country to extend its border? If the answer is none (it is an infinite series), then is the map ever complete? === Subject: Re: An Ownership Puzzle <> Peter, Paul, and Mary each want to own the bar of gold in their >presence. <> They all agree that the final owner of the gold bar will be the person >who <> is holding it at exactly one o'clock and they agree to the following >rules. <> At twelve o'clock Mary will hold the gold bar and, since twelve thirty <> half way between twelve o'clock and one o'clock, at twelve thirty Mary >will <> give the bar to Peter. Since twelve forty-five is half way between >twelve <> thirty and one o'clock, at twelve forty-five Peter will give the gold >bar to <> Paul. The half way time between twelve forty-five and one o'clock is >twelve <> fifty two and thirty seconds at which time Paul hands the gold bar >back to <> Mary. At the half way time between twelve fifty-two (and thirty >seconds) <> and one o'clock, Mary gives the gold bar to Peter and then at the next >half <> way time Peter gives it to Paul, and so on. In this puzzle each >person can <> pass the gold bar to the next person in no time at all. This passing >of the <> gold bar continues in the order Mary, Peter, Paul, Mary, Peter, Paul, >... <> until exactly one o'clock and then stops. Who will be the final owner <> the gold bar? ....................... >the puzzle, it would be like discussing the quantum uncertainty principle or >various theories of quantisized space in response to a question about the >truth of the continuum hypothesis. This puzzle is purely mathematical, it >takes place in the ideal continuum space and time of mathematics. For the >issue fishfry raised, whenever the gold bar is swapped (for example at >twelve thirty), at that time the gold bar is in the hands of the previous >holder, not the new holder. Because this puzzle is taking place in the >idealized Euclidean space and idealized time (modeled by the real line), >there are no relativistic or other physical constraints: Peter, Paul and >Mary are not moving closer together at any time. So the question remains: >who will be the final owner of the gold bar? A similar problem occurs in the theory of Markov chains; there, instead of determinism, there is randomness. But the solution is still similar; the answer is not determined, and after one goes off the time scale in this manner, the process must be restarted, and this can be done in any manner whatever. Or to put it in another manner, without assuming some form of at least approximate smoothness at 1, knowing f(t) for all t < 1 tells us nothing about f(1). -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: An Ownership Puzzle Peter, Paul, and Mary each want to own the bar of gold in their > presence. > They all agree that the final owner of the gold bar will be the person > who > is holding it at exactly one o'clock and they agree to the following > rules. > At twelve o'clock Mary will hold the gold bar and, since twelve thirty > is > half way between twelve o'clock and one o'clock, at twelve thirty Mary > will > give the bar to Peter. Since twelve forty-five is half way between > twelve > thirty and one o'clock, at twelve forty-five Peter will give the gold > bar to > Paul. The half way time between twelve forty-five and one o'clock is > twelve > fifty two and thirty seconds at which time Paul hands the gold bar > back to > Mary. At the half way time between twelve fifty-two (and thirty > seconds) > and one o'clock, Mary gives the gold bar to Peter and then at the next > half > way time Peter gives it to Paul, and so on. In this puzzle each > person can > pass the gold bar to the next person in no time at all. This passing > of the > gold bar continues in the order Mary, Peter, Paul, Mary, Peter, Paul, > ... > until exactly one o'clock and then stops. Who will be the final owner > of > the gold bar? Since the speed at which the bar is being passed increases without > limit as the time approaches one o'clock, the relativistic mass will > also increase without limit, so the environs of the gold bar will > collapse into a black hole before one o'clock. This presumes the distances between Peter, Paul and Mary is non-zero > constant. That being the case, the bar will vaporise just before one. > Now Peter, Paul and Mary had better don asbestos mitts, very thick mitts > because when the speed increases until the melting point of the bar is > reached due to air friction, the recoil will great. Big enuf to knock > them over, to blow them away. > Since no information can escape from a black hole, we will never > know who got the gold bar. As that black circumstance is precluded by other practical concers, we're > forced to accept that Peter, Paul and Mary are moving closer and closer > together. That eventually the bar is squizzed between three palms with no > one having time to grasp or hold the bar, that all three palms are > holding the bar at one o'clock. > Morrow of the story: > Peter, Paul and Mary best stay with Rosemary, Sage and Thyme. > the puzzle, it would be like discussing the quantum uncertainty principle or > various theories of quantisized space in response to a question about the > truth of the continuum hypothesis. This puzzle is purely mathematical, it > takes place in the ideal continuum space and time of mathematics. For the > issue fishfry raised, whenever the gold bar is swapped (for example at > twelve thirty), at that time the gold bar is in the hands of the previous > holder, not the new holder. Because this puzzle is taking place in the > idealized Euclidean space and idealized time (modeled by the real line), > there are no relativistic or other physical constraints: Peter, Paul and > Mary are not moving closer together at any time. So the question remains: > who will be the final owner of the gold bar? humerous? huh Mary Herc === Subject: Re: An Ownership Puzzle what gets me is as the problem is stated : 1 o'clock WILL eventuate ONE of them will be holding the bar at 1 o'clock! but there does seem an instantaneous period of infinite swapping just up to 1 o'clock making it hard to determine. could the problem be multi choice with A/ Mary B/ Peter C/ Paul as options for the answers? and does it involve planch time intervals? Herc === Subject: Re: An Ownership Puzzle idealized Euclidean space and idealized time (modeled by the real line), > there are no relativistic or other physical constraints: Peter, Paul and > Mary are not moving closer together at any time. So the question > remains: who will be the final owner of the gold bar? Let p(n) be possessor at time t_n = 1 - 2^-n p(0) = Pete p(1) = Paul p(2) = Mary p(3) = Pete ... lim(n->oo) t_n = 1 lim(n->oo) p(n) doesn't exist because it's a cyclic sequence with frequency of 3. p(n) however, does has infinitely many subsequences that do converge. === Subject: Re: An Ownership Puzzle > address the puzzle. Because this puzzle is taking place in the > idealized Euclidean space and idealized time (modeled by the real line), > there are no relativistic or other physical constraints: Peter, Paul and > Mary are not moving closer together at any time. So the question > remains: who will be the final owner of the gold bar? > Let p(n) be possessor at time t_n = 1 - 2^-n > p(0) = Pete > p(1) = Paul > p(2) = Mary > p(3) = Pete > ... > lim(n->oo) t_n = 1 > lim(n->oo) p(n) doesn't exist > because it's a cyclic sequence with frequency of 3. > p(n) however, does has infinitely many subsequences that do converge. Yes, p(n) does not converge but the gold bar has to be somewhere at one o'clock. The question was, where is the gold bar at one o'clock? It has to be somewhere in the range of this problem, the range being the set consisting of the three participants. All three participants have agreed to stop passing the gold bar at one o'clock so the limit (n->oo) p(n) must be one of the three participants even though the subsequences do not converge. So the question remains: who will be the final owner of the gold bar? Randy http://www.rlgerl.com === Subject: Re: [JSH] Simple principle in core error proof ... Worthless dribble deleted! *** > Yup, call me crazy, but I think the bastards are out to get me!!! I'll go for calling you crazy. Does anyone second the motion. You are like a broken record, replaying the same old nonsense and you have lied to yourself for so long, you believe your own incompetence. Please go get some psychotropic drugs or jump off a bridge, you raving lunatic. How pathetic you have proven with your feeble attempts for fame, fortune and recognition. You are so sad and an example of schooling gone horribly wrong in the US (as much as it hurts to admit that). Perhaps you should renounce your degree and go back to grade school and see if people still think you are smart. Get a life! === Subject: Re: irreducible polynomial in R[X,Y] vanishing on finite set? >Given a finite subset S of R^2 (where R denotes the real numbers), is >there an irreducible polynomial in R[X, Y] which vanishes on S and >nowhere else? >Mike Yes. First observe that there exists a slope lambda such that any line with slope lambda passes through at most one point in S. This is because there are only finitely many slopes (y1 - y2)/(x1 - x2) where (x1, y1), (x2, y2) in S. By rotating S, we can assume no two points in S have equal x-coordinates. Let T consist of these x-coordinates. Find a polynomial f(x) such that S = {(x, f(x)) : x in T}. That is, f maps each x-coordinate to its corresponding y-coordinate. Set g(X, Y) = (Y - f(X))^2 + product (X - x)^2. x in T g is monic in Y, and has no factors involving only X. Nor does it have two linear factors in Y. -- ARNOLD = Anagram of RONALD ENEGGER = Backwards mis-pronounced REAGAN This is a black -- I mean SCHWARZ -- period in California. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Re: Why is x^y + y^x > 1 for 0 < x < 1, 0 < y < 1? > The inequality of the subject: header says it all. The inequality is > assuredly true; why? > --Ron Bruck The function is symmetric in x and y so we need only consider the upper triangle 0 < x <= y < 1. So we want to show for any fixed c in (0,1) that x^c + c^x > 1 for x in (0,c]. Because c^x is strictly convex in x, it stays above its tangent line at (0,1). So x^c + c^x > x^c + 1 + [ln(c)]x (*) for x > 0, and in particular for x in (0,c]. Let f(x) denote the RHS of (*). We have f(0) = 1 and f(c) = c^c + 1 + ln(c^c). As we know from infinitely many posts on sci.math, c^c >= (1/e)^(1/e). This implies f(c) > 1. But f is strictly concave down on (0,oo), which implies that on (0,c), f is above the secant line joining (0,f(0)) and (c,f(c)). This implies f > 1 on (0,c] and by (*) we're done. === Subject: Re: Why is x^y + y^x > 1 for 0 < x < 1, 0 < y < 1? > The inequality of the subject: header says it all. The inequality is > assuredly true; why? [convexity-based solution deleted] Here is my approach (involving a 'lovely' lemma right below): Observe first that x <= y is equivalent to x^(y-1) >= y^(x-1): indeed the latter is equivalent to (y-1)*lnx >= (x-1)*lny, which, via division of both sides by the *positive* quantity (y-1)*(x-1), is in turn equivalent to (lnx)/(x-1) > = (lny)/(y-1); the result follows now from the negativity of the derivative of (lnx)/(x-1) for 0 < x < 1, easily seen to be equivalent to x < 1 + x*lnx for 0 < x < 1 -- an inequality easily established by way of h(1) = 0 and h'(x) = lnx < 0 for 0 < x < 1, where h(x) = 1 + x*lnx - x. Assume WLOG x <= y and x^(y-1) >= y^(x-1). There are two cases to consider: Case 1: y > 1/e. Set f(x) = x^y + y^x, so that f'(x) = y*x^(y-1) + (lny)*y^x > y*x^(y-1) - y^x = y*[x^(y-1) - y^(x-1)] > 0; we conclude that f(x) > f(0) = 1. Case 2: y <= 1/e. Set g(y) = x^y + y^x, so that g'(y) = (lnx)*x^y + x*y^(x-1) <= (lny)*x^y + x*x^(y-1) <= -x^y + x^y = 0 and therefore g(y) >= g(1/e) = x^(1/e) + 1/(e^x) = x*x^(1/e-1) + 1/(e^x) >= x*[(1/e)^(x-1)] + 1/(e^x) = (e*x+1)/(e^x); with r(x) = (e*x+1)/(e^x) and r'(x) = (e-e*x-1)/(e*x), we see that x <= y <= 1/e < 1 - 1/e yields r'(x) > 0, hence g(y) >= r(x) > r(0) = 1. baloglouAToswego.edu === Subject: MKC Set Theories I dedicate this thread to JSH and to the memory of Pertti Lounesto --Every system of set theory that was published during the last century contains the standard axiom of extensionality, according to which collections are equal iff these are equi-membered.[1] AxAy[Az(z in x <-> z in y) -> x=y] Standard Extensionality --Nevertheless not all collections satisfy Standard Extensionality. It is a commonplace that *clubs* don't, for different clubs may be identical in membership. --Correy Extensionality differs from Standard Extensionality by allowing both for sets (collections which are determined by their members); and for non-sets (collections which are not determined by their members).[2] C4 AxAy[Az(z in x <-> z in y) -> {(set x & set y) <-> x = y}] Correy Extensionality --Their novel features notwithstanding (in my next posting I'll say what these are), from the git-go set theories which combine Correy Extensionality with the MK principle of Classification (I'll call these MKC theories), C3 EyAx[x in y <-> Set x & A] (with y not free in A) MK Classification have been ridiculed, scoffed at and misrepresented, not coincidentally by some of the same people who led a successful campaign to run the late Pertti Lounesto off of sci.math, and have led an unsuccessful campaign to silence JSH. Notes 1. Various theories of collection have been proposed since the 1900s. What they all share is the axiom of extensionality, which asserts that if x and y are collections then Az(z in x <-> z in y) -> x=y (Michael Potter, Different Systems of Set Theory, p. 1) 2. If {a} and {b} are determined by their members, then {a}={b} C4 AxAy[Az(z in x <-> z in y) -> {(set x & set y) <-> x = y}] Correy Extensionality In contrast, if {a} and {b} are not determined by their members, then ~({a}={b}) even when a=b, and it follows from C4 that {a} and {b} are non-sets. (Groupings that are like non-sets in this regard are the solitaire club and the chess club, which remain different clubs even when they have the same membership.) --John The road of excess leads to the palace of wisdom. William Blake (1757-1827) === Subject: Re: MKC Set Theories > I dedicate this thread to JSH and to the memory of Pertti Lounesto. To sum up so far: --There is one axiom that theories of collections proposed since the 1900s all share: Standard Extensionality. AxAy[Az(z in x <-> z in y) -> x=y] Standard Extensionality --The domain of any theory whose axioms include Standard Extensionality excludes collections that are 'not determined by their members'. (Recall, if {a} and {b} are 'determined by their members', then a=b <-> {a}={b}: this biconditional follows from C4 and first-order logic without identity. In contrast, if {a} and {b} are *not* determined by their members, then ~[a=b -> {a}={b}]. But a,b such that a=b & ~({a}={b}) are excluded from the domain of every theory of collections whose axioms include Standard Extensionality.) The admission of such collections thus marks a step in the direction of increased mathematical generality--notwithstanding the giggles, pouts and moues of David Ullrich, Robin Chapman and Lee Rudolph.[1] Notes 1. Perchance was it some such Trinity that inspired these lines of 'The Triple Fool'? Who are a little wise, the best fools be. --John Donne, 'The Triple Fool', st. 2 John Correy === Subject: how to do curve fitting for multiple variables? I post a message asking for help to do curve fiting for a single curve and just two variables x and y... and I got very helpful answers... Now I met with another more complex problem. It turns out that I have a bunch of those y=function(x) curves depending on the values of w and z, thus they seem like the curves y=function(x, w, z)... I do have the vectors containing data for y, x, w, and z... how to do this multiple variables curve fitting? -Walala === Subject: Re: how to do curve fitting for multiple variables? > I post a message asking for help to do curve fiting for a single curve and > just two variables x and y... and I got very helpful answers... > Now I met with another more complex problem. It turns out that I have a > bunch of those y=function(x) curves depending on the values of w and z, thus > they seem like the curves y=function(x, w, z)... > I do have the vectors containing data for y, x, w, and z... how to do this > multiple variables curve fitting? In the case where your model can be written as y = a1*f1(x,w,z) + a2*f2(x,w,z) + ... + an*fn(x,w,z) + noise that is, where the unknown parameters a1, a2, ..., an all appear linearly, then the techniques of linear least squares are easily generalized to do the curve fit in one step. (It requires a linear algebra package to be able to solve linear systems and, preferably, do matrix factorizations). The functions fk(x,w,z) can be anything at all so long as they have no free parameters that you are trying to estimate. In the case where your parameters occur nonlinearly, then you have a problem in nonlinear global minimization (of your error function), which requires iterative techniques. However, general nonlinear optimization packages are available on the web. - Randy === Subject: Re: how to do curve fitting for multiple variables? > I post a message asking for help to do curve fiting for a single curve and > just two variables x and y... and I got very helpful answers... > Now I met with another more complex problem. It turns out that I have a > bunch of those y=function(x) curves depending on the values of w and z, thus > they seem like the curves y=function(x, w, z)... > I do have the vectors containing data for y, x, w, and z... how to do this > multiple variables curve fitting? > In the case where your model can be written as > y = a1*f1(x,w,z) + a2*f2(x,w,z) + ... + an*fn(x,w,z) + noise > that is, where the unknown parameters a1, a2, ..., an all > appear linearly, then the techniques of linear least squares > are easily generalized to do the curve fit in one step. > (It requires a linear algebra package to be able to > solve linear systems and, preferably, do matrix > factorizations). The functions fk(x,w,z) can be anything > at all so long as they have no free parameters that you > are trying to estimate. > In the case where your parameters occur nonlinearly, then > you have a problem in nonlinear global minimization (of > your error function), which requires iterative techniques. > However, general nonlinear optimization packages are > available on the web. > - Randy I've attached the data: Could you please help me take a look at this? when z=1.7527e+004, w= 0.3536 0.0077 0.0072 0.0077 0.3536 0.0077 0.0072 0.0077 w is a parameter vector, the values here are used as a whole, they are not data points. maybe I should extract some feature to use one value to represent the whole vector in order to avoid parametrical vector curve fitting... oh, I do have a feature value for this parametric vector: 0.9314. x: 8.9289 3.8078 2.2907 1.5374 1.0899 0.8161 0.6354 for the above x(7 data points), w(one scalar feature value, or one parameter vector), z values(one scalar), we got y(7 data points): y: 0.0715 0.0176 -0.0078 -0.0225 -0.0350 -0.0430 -0.0496 ---------------------------------------------------------- when z=1.6135e+004, w=0.3536 0.0076 0.0072 0.0076 0.3536 0.0076 0.0072 0.0076 I also have a one-value feature extracted value for this parameter vector: 0.9827. x(7 data ponits)= 6.3795 1.8693 0.9923 0.6455 0.4612 0.3494 0.2753 we got y(7 data points): 0.0680 -0.0131 -0.0301 -0.0449 -0.0504 -0.0506 -0.0578 So you see the family curves (x, y) are dependent on values of w and z. And each (x, y) has very nice shape as I have attached in previous image. Now what mathematical approach should I take to model the relationship between y=functin(x, w, z)? -Walala === Subject: Re: how to do curve fitting for multiple variables? > I post a message asking for help to do curve fiting for a single curve and > just two variables x and y... and I got very helpful answers... > Now I met with another more complex problem. It turns out that I have a > bunch of those y=function(x) curves depending on the values of w and z, thus > they seem like the curves y=function(x, w, z)... > I do have the vectors containing data for y, x, w, and z... how to do this > multiple variables curve fitting? First question: Do you have *any* idea what your function should look like? Are you looking for some polynomial? Basically, the idea is you choose a category of functions, e.g. polynomials, with a number of unknown coefficients. For instance: Y = a*x + b*w + c*z. Then you must choose the coefficients a, b, and c, such that the difference between y (observed) and Y (fitted) is the least possible. To do that you form an expression with the sum of the squared difference, i..e. SUM (y-Y)^2 and minimize with respect to the unknown coefficients. You can include higher order terms (i.e. x^2 and x*w, etc.), but that is really grasping in thin air. What kind of relation do you expect? Without some additional information this curve fitting is basically meaningless. I mean, if you know that y is a bounded variable, then a polynomial relation is not a very good approximation. -Michael. === Subject: Re: how to do curve fitting for multiple variables? There is no simple answer. If you want fit with polynomials of several variables, see www.estlab.com Arto Huttunen > I post a message asking for help to do curve fiting for a single curve and > just two variables x and y... and I got very helpful answers... > Now I met with another more complex problem. It turns out that I have a > bunch of those y=function(x) curves depending on the values of w and z, > thus > they seem like the curves y=function(x, w, z)... > I do have the vectors containing data for y, x, w, and z... how to do this > multiple variables curve fitting? > First question: Do you have *any* idea what your function should look like? > Are you looking for some polynomial? > Basically, the idea is you choose a category of functions, e.g. polynomials, > with a number of unknown coefficients. For instance: Y = a*x + b*w + c*z. > Then you must choose the coefficients a, b, and c, such that the difference > between y (observed) and Y (fitted) is the least possible. To do that you > form an expression with the sum of the squared difference, i..e. SUM (y-Y)^2 > and minimize with respect to the unknown coefficients. > You can include higher order terms (i.e. x^2 and x*w, etc.), but that is > really grasping in thin air. What kind of relation do you expect? Without > some additional information this curve fitting is basically meaningless. I > mean, if you know that y is a bounded variable, then a polynomial relation > is not a very good approximation. > -Michael. === Subject: Sampel Data attached! Re: how to do curve fitting for multiple variables? > There is no simple answer. > If you want fit with polynomials of several variables, > see www.estlab.com > Arto Huttunen I post a message asking for help to do curve fiting for a single curve > and > just two variables x and y... and I got very helpful answers... Now I met with another more complex problem. It turns out that I have a > bunch of those y=function(x) curves depending on the values of w and z, > thus > they seem like the curves y=function(x, w, z)... I do have the vectors containing data for y, x, w, and z... how to do > this > multiple variables curve fitting? > First question: Do you have *any* idea what your function should look > like? > Are you looking for some polynomial? > Basically, the idea is you choose a category of functions, e.g. > polynomials, > with a number of unknown coefficients. For instance: Y = a*x + b*w + c*z. > Then you must choose the coefficients a, b, and c, such that the > difference > between y (observed) and Y (fitted) is the least possible. To do that you > form an expression with the sum of the squared difference, i..e. SUM > (y-Y)^2 > and minimize with respect to the unknown coefficients. > You can include higher order terms (i.e. x^2 and x*w, etc.), but that is > really grasping in thin air. What kind of relation do you expect? Without > some additional information this curve fitting is basically meaningless. I > mean, if you know that y is a bounded variable, then a polynomial relation > is not a very good approximation. > -Michael. Could you please help me take a look at this? when z=1.7527e+004, w= 0.3536 0.0077 0.0072 0.0077 0.3536 0.0077 0.0072 0.0077 w is a parameter vector, the values here are used as a whole, they are not data points. maybe I should extract some feature to use one value to represent the whole vector in order to avoid parametrical vector curve fitting... oh, I do have a feature value for this parametric vector: 0.9314. x: 8.9289 3.8078 2.2907 1.5374 1.0899 0.8161 0.6354 for the above x(7 data points), w(one scalar feature value, or one parameter vector), z values(one scalar), we got y(7 data points): y: 0.0715 0.0176 -0.0078 -0.0225 -0.0350 -0.0430 -0.0496 ---------------------------------------------------------- when z=1.6135e+004, w=0.3536 0.0076 0.0072 0.0076 0.3536 0.0076 0.0072 0.0076 I also have a one-value feature extracted value for this parameter vector: 0.9827. x(7 data ponits)= 6.3795 1.8693 0.9923 0.6455 0.4612 0.3494 0.2753 we got y(7 data points): 0.0680 -0.0131 -0.0301 -0.0449 -0.0504 -0.0506 -0.0578 So you see the family curves (x, y) are dependent on values of w and z. And each (x, y) has very nice shape as I have attached in previous image. Now what mathematical approach should I take to model the relationship between y=functin(x, w, z)? -Walala === Subject: Primitive Recursion and Language Theory Quick question: What are the language theoretic approaches *primitive* recursive languages? Recursively enumerable languages have a nice non-algorithmic characterization in terms of phase-structure (type-0) grammars. However, I haven't seen anything similar to characterize languages whose characteristic function is primitive recursive. Are there charaterizations that avoid referring to classes of functions? Any thoughts/references would be appreciated. Rex Butler === Subject: Re: Primitive Recursion and Language Theory Distribution: inet > Quick question: What are the language theoretic approaches > *primitive* recursive languages? Recursively enumerable languages > have a nice non-algorithmic characterization in terms of > phase-structure (type-0) grammars. > However, I haven't seen anything similar to characterize languages > whose characteristic function is primitive recursive. Are there > charaterizations that avoid referring to classes of functions? Any > thoughts/references would be appreciated. Try looking up the Grzegorczyk hierarchy: it's a stratification of primitive recursive functions based on the depth of nesting of the recursions; it's purely syntactical. You can get a (nearly) identical stratification by looking at the depth of nesting of bounded loops in restricted programming languages. Dennis === Subject: Re: Determining temperature in a tank while adding water > I have a tank of water with a known temperature. I am going to add water of a different known temperature.The flow rate of the water entering the tank is the same as the water leaving the tank. How do I set up the differential equation so that I can determine the temperature in the tank at any time t. I am assuming a uniform mixture of the water in the tank and neglecting heat gain or loss due to the surrounding environment. The temperature of the water leaving the tank is also the temperature of the water in the tank, i.e. T(t). The temperature of the water entering the tank is T1. The heat added to the container is the temperature difference multiplied by the amount of water exchanged (flow rate) and the specific heat of water (4.2 J/gK). The total heat of the container is the temperature times the total amount of water times the specific heat. Hope that helps. -Michael. === Subject: Re: Determining temperature in a tank while adding water > I have a tank of water with a known temperature. I am going to add water > of a different known temperature.The flow rate of the water entering the > tank is the same as the water leaving the tank. How do I set up the > differential equation so that I can determine the temperature in the tank at > any time t. I am assuming a uniform mixture of the water in the tank and > neglecting heat gain or loss due to the surrounding environment. > The temperature of the water leaving the tank is also the temperature of the > water in the tank, i.e. T(t). The temperature of the water entering the tank > is T1. The heat added to the container is the temperature difference > multiplied by the amount of water exchanged (flow rate) and the specific > heat of water (4.2 J/gK). The total heat of the container is the temperature > times the total amount of water times the specific heat. Or consider two solutions of salt in water at different concentrations instead of two different temperatures. Then you can deal directly with the amounts of salt in the tank as time passes. === Subject: Re: How Euclid did not prove the Pythagorean Theorem's converse >In http://www.oswego.edu/~baloglou/math/euclid-1.48.html I present a >geometrical argument that *at the same time* proves the Pythagorean >Theorem and illustrates how it fails for non-right angles (thus proving >its converse): this geometrical argument is a straightforward extension >of Euclid's proof of the Pythagorean Theorem, and argument for a right >angle, see http://www.oswego.edu/~baloglou/math/euclid-1.47.html, to the >case of an obtuse angle, the argument for an acute angle being similar. >As you may (not) recall, Euclid's original proof of the converse of the >Pythagorean Theorem employs the Pythagorean Theorem itself, something that >I do not like that much*: my efforts to locate other proofs of the converse >in the literature led nowhere -- save for 'my' proof above, that is! Of course the converse of the Pythagorean Theorem is implicit in Euclid's equivalents of the law of cosines (propositions II.12 (obtuse angle) and II.13 (acute angle)). It is interesting that proofs of these propositions in the spirit of Euclid's proof of the Pythagorean Theorem are late: Heath (v. I, p. 404) is not aware of any such proof before Gregory of St. Vincent's Opus geometricum quadraturae circuli et sectionum coni (1647). baloglouAToswego.edu === Subject: Re: How Euclid did not prove the Pythagorean Theorem's converse > ... Euclid's original proof of the converse of the > Pythagorean Theorem employs the Pythagorean Theorem itself, something that > I do not like that much ... Why not? Seems perfectly reasonable to me. It's not like he's proving the inverse. ~ Chris > In other words, you do not accept ordinary logic. In that case, you > better formulate the logical principles you do accept. For all anyone > knows, it cannot be proved in your logic, whatever it is. On the > other hand, it is likely that not much can be proved in your logic. I > certainly will not waste time on it (more than I have). Designing a > logic is not for the faint of heart, nor for the unsophisticated. Of course I accept ordinary logic. The converse of a theorem is independent of the original theorem. Once you prove the theorem, then you can use it to prove the converse. === Subject: Re: Equivalent Binomial Random Variables > Unless I misunderstood your point, the mean and variance will remain > unchanged. That is, X1, X2 and X3 will have approximately the same > mean and variance, but is enough to let me state that they are > equivalent? If not, what are the conditions in order to state so? X1, X2, and X3 will have approximately the same mean, variance, skewness, kurtosis, etc; that is, their PMFs will be approximately the same (except, as you have noted) in the tails. How close they must be before you declare them equivalent is a subjective matter: equivalence of this sort is in the eye of the beholder. > Are you making a reference to the relation between > the Poisson and the Binomial distribution? Yes. === Subject: Re: Equivalent Binomial Random Variables Hi Mr. Koopman, > And my point also remains unchanged: that you should start by considering > what happens to the mean and variance when you increase n and decrease p > in a way that leaves n*p constant, which is what your values approximate. Are you making a reference to the relation between the Poisson and the Binomial distribution? If that's the case, there are two interesting sites about this fact: http://mathworld.wolfram.com/PoissonDistribution.html http://www.math.uah.edu/statold/poisson/poisson6.html Daniel Sadoc === Subject: Re: Core error, mathematicians, complex numbers > Mr. Ullrich used a quotation from De Morgan on the topic of replying > to crackpots to the effect that doing so is necessary to prevent the > gullible public at large from believing them. That is Arturo Magidin, not Mr. Ullrich. Gib === Subject: Re: how about A1*X*B1+A2*X*B2=X? > Can you say something about A1*X*B1+A2*X*B2=X? Ah, that's interesting. Considering again X = e_{jk}, we find that (with @ = tensor product) A1 @ B1 + A2 @ B2 = I @ I. Now I claim this implies for some scalars c and d, either A1 = c I, A2 = d I, c B1 + d B2 = I or B1 = c I, B2 = d I, c A1 + d A2 = I Note that if, say, A1 is not a scalar multiple of I, there must be a vector u such that u and v = A1 u are linearly independent. Take a vector x such that x.u = 0 and x.v = 1. Then for any vector w, B1 w + (x. A2 u) B2 w = (x.u) w = 0 i.e. B1 w = - (x. A2 u) B2 w so B1 = - (x.A2 u) B2. So A1 @ B1 + A2 @ B2 = (-(x.A2 u) A1 + A2) @ B2 = I @ I which implies B2 and B1 are multiples of I. > For the above problem, can I say it equals to > (A1+A2)*X*(B1+B2)=X... No, of course not. That would say A1 X B1 + A1 X B2 + A2 X B1 + A2 X B2 = X. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: finite groups >Is it true that in every finite simple group G >all isomorphic maximal subgroups >are Aut(G)-conjugate? > No, but it is not easy to find examples. I was about to give up, but > then I found one! > Believe it or not, the alternating group A_{253} has two maximal subgroups > isomorphic to PSL(2,23). One of them has dihedral D_{24} as point > stabilizer, and the other has S_4, so they cannot be conjugate in > S_{253}. >Hi Derek, >I have one question about your example. >The dihedral subgroup D_{24} of PSL(2,23) is extended to a subgroup >of PGL(2,23) of the same index 253. This means that PGL(2,23) >also has a permutational representation of degree 253. >Could it be that the first of your subgroups is not maximal >but a subgroup of index 2 in a larger subgroup isomorphic to PGL(2,23)? No, the extension to PGL(2,23) lies in S_{253} but not in A_{253}. I just checked that directly, but I have a computer list available with the maximal subgroups of A_n for n <= 1000, which is how I found the example. Incidentally, I am fairly certain that there are infinitely many examples, although I have not proved it completely. There are many families of simple groups with non-isomorphic maximal subgroups of the same order. For example the groups PSp(4,p) for odd primes p have two non-isomorphic maximal subgroups of order p^3(p-1)|PSL(2,p)|. If you form the permutation representations on the cosets of those two subgroups, then I would expect to get isomorphic maximal subgroups of the appropriate alternating group, not conjugate in the symmetric group. Derek Holt. === Subject: Re: finite groups >Is it true that in every finite simple group G >all isomorphic maximal subgroups >are Aut(G)-conjugate? No, but it is not easy to find examples. I was about to give up, but > then I found one! Believe it or not, the alternating group A_{253} has two maximal subgroups > isomorphic to PSL(2,23). One of them has dihedral D_{24} as point > stabilizer, and the other has S_4, so they cannot be conjugate in > S_{253}. >Hi Derek, >I have one question about your example. >The dihedral subgroup D_{24} of PSL(2,23) is extended to a subgroup >of PGL(2,23) of the same index 253. This means that PGL(2,23) >also has a permutational representation of degree 253. >Could it be that the first of your subgroups is not maximal >but a subgroup of index 2 in a larger subgroup isomorphic to PGL(2,23)? >No, the extension to PGL(2,23) lies in S_{253} but not in A_{253}. I just >checked that directly, but I have a computer list available with the >maximal subgroups of A_n for n <= 1000, which is how I found the example. >Incidentally, I am fairly certain that there are infinitely many examples, >although I have not proved it completely. There are many families of >simple groups with non-isomorphic maximal subgroups of the same order. >For example the groups PSp(4,p) for odd primes p have two non-isomorphic >maximal subgroups of order p^3(p-1)|PSL(2,p)|. If you form the permutation >representations on the cosets of those two subgroups, then I would expect >to get isomorphic maximal subgroups of the appropriate alternating group, >not conjugate in the symmetric group. Hmmm! After looking through Jim Heckman's proposed list, I see that this family does not yield maximal subgroups for p=3,5,7, so maybe it never does! Derek Holt === Subject: Re: finite groups > ...... >For example the groups PSp(4,p) for odd primes p have two non-isomorphic >maximal subgroups of order p^3(p-1)|PSL(2,p)|. If you form the permutation >representations on the cosets of those two subgroups, then I would expect >to get isomorphic maximal subgroups of the appropriate alternating group, >not conjugate in the symmetric group. > Hmmm! After looking through Jim Heckman's proposed list, I see that this > family does not yield maximal subgroups for p=3,5,7, so maybe it never does! We could also consider the family of two parabolic subgroups of G_2(q) corresponding to the short and long roots of the Dynkin diagram. They are maximal, have the same order and, if I am not mistaken are non-isomorphic unless q is a power of 3. But of course the hardest part would be to verify whether the corresponding permutational embeddings are maximal. I don't see how this can be done in general. Anvita === Subject: Re: finite groups >Is it true that in every finite simple group G >all isomorphic maximal subgroups >are Aut(G)-conjugate? > No, but it is not easy to find examples. I was about to give up, but > then I found one! > Believe it or not, the alternating group A_{253} has two maximal > subgroups > isomorphic to PSL(2,23). One of them has dihedral D_{24} as point > stabilizer, and the other has S_4, so they cannot be conjugate in > S_{253}. > Hi Derek, > I have one question about your example. > The dihedral subgroup D_{24} of PSL(2,23) is extended to a subgroup > of PGL(2,23) of the same index 253. This means that PGL(2,23) > also has a permutational representation of degree 253. > Could it be that the first of your subgroups is not maximal > but a subgroup of index 2 in a larger subgroup isomorphic to PGL(2,23)? Good catch! How about these, then? Hmm, let's see... 1) A_{40} has 2 classes of maximal subgroups isomorphic to PSU(4,2), such that their nomalizers in S_{40} have point stabilizers isomorphic to 3^{1+2}_+ : 2A_4 and 3^3 : S_4, respectively. 2) A_{63} -> PSU(3,3) -> 4.S_4 and 4^2.S_3 . 3) A_{156} -> PSp(4,5) -> 5^{1+2}_+ : (4 x A_5) and 5^3 : (2 x A_5).2 . 4) A_{253} -> M_{23} -> PSL(3,2^2) : 2 and 2^4 : A_7 . 5) A_{400} -> PSp(4,7) -> 7^{1+2}_+ : (6 x PSL(2,7)) and 7^3 : (3 x PSL(2,7)).2 . 6) A_{495} -> M_{12} -> 2^{1+4}_+.S_3 and 4^2 : D_{12} . 7) A_{820} -> PSp(4,3^2).2 -> (3^{1+2}_+)^2 : (8 x PSL(2,3^2)) and 3^6 : (4 x PSL(2,3^2)).2 . (Actually, I'm not sure these 2 classes of maximal subgroups are isomorphic. There's more than 1 extension PSp(4,3^2).2 .) Unless I've missed some, that appears to exhaust the a priori possibilities for alternating groups < A_{1000}. -- Jim Heckman === Subject: Re: finite groups >Is it true that in every finite simple group G >all isomorphic maximal subgroups >are Aut(G)-conjugate? > No, but it is not easy to find examples. I was about to give up, but > then I found one! > Believe it or not, the alternating group A_{253} has two maximal > subgroups > isomorphic to PSL(2,23). One of them has dihedral D_{24} as point > stabilizer, and the other has S_4, so they cannot be conjugate in > S_{253}. > Hi Derek, > I have one question about your example. > The dihedral subgroup D_{24} of PSL(2,23) is extended to a subgroup > of PGL(2,23) of the same index 253. This means that PGL(2,23) > also has a permutational representation of degree 253. > Could it be that the first of your subgroups is not maximal > but a subgroup of index 2 in a larger subgroup isomorphic to PGL(2,23)? >Good catch! How about these, then? Hmm, let's see... But my example works - both copies of PSL(2,23) are maximal. >1) A_{40} has 2 classes of maximal subgroups isomorphic to >PSU(4,2), such that their nomalizers in S_{40} have point >stabilizers isomorphic to 3^{1+2}_+ : 2A_4 and 3^3 : S_4, >respectively. Only one of those is maximal in A_{40} >2) A_{63} -> PSU(3,3) -> 4.S_4 and 4^2.S_3 . Only one is maximal >3) A_{156} -> PSp(4,5) -> 5^{1+2}_+ : (4 x A_5) and >5^3 : (2 x A_5).2 . Neither maximal. >4) A_{253} -> M_{23} -> PSL(3,2^2) : 2 and 2^4 : A_7 . Neither maximal >5) A_{400} -> PSp(4,7) -> 7^{1+2}_+ : (6 x PSL(2,7)) and >7^3 : (3 x PSL(2,7)).2 . Neither maximal >6) A_{495} -> M_{12} -> 2^{1+4}_+.S_3 and 4^2 : D_{12} . Neither maximal >7) A_{820} -> PSp(4,3^2).2 -> (3^{1+2}_+)^2 : (8 x PSL(2,3^2)) >and 3^6 : (4 x PSL(2,3^2)).2 . (Actually, I'm not sure these 2 >classes of maximal subgroups are isomorphic. There's more than 1 >extension PSp(4,3^2).2 .) Neither maximal >Unless I've missed some, that appears to exhaust the a priori >possibilities for alternating groups < A_{1000}. I expect my example is the only one. My stataments were all just based on consulting known lists of alternating groups up to degree 1000. If you want me to justify any of them, I will need to find a larger subgroup I suppose. Of course, lists like that often contain the odd error. Derek Holt. >Jim Heckman === Subject: Re: finite groups Bleh, ignore previous response. Hit send too quick. in message : >Is it true that in every finite simple group G >all isomorphic maximal subgroups >are Aut(G)-conjugate? > No, but it is not easy to find examples. I was about to give up, but > then I found one! > Believe it or not, the alternating group A_{253} has two maximal > subgroups > isomorphic to PSL(2,23). One of them has dihedral D_{24} as point > stabilizer, and the other has S_4, so they cannot be conjugate in > S_{253}. > Hi Derek, > I have one question about your example. > The dihedral subgroup D_{24} of PSL(2,23) is extended to a subgroup > of PGL(2,23) of the same index 253. This means that PGL(2,23) > also has a permutational representation of degree 253. > Could it be that the first of your subgroups is not maximal > but a subgroup of index 2 in a larger subgroup isomorphic to PGL(2,23)? >Good catch! How about these, then? Hmm, let's see... > But my example works - both copies of PSL(2,23) are maximal. OK. Is there an easy way to see that the PGL(2,23) in the one case lies in S_{253} but not in A_{253}? And the analogous question for the following cases? >1) A_{40} has 2 classes of maximal subgroups isomorphic to >PSU(4,2), such that their nomalizers in S_{40} have point >stabilizers isomorphic to 3^{1+2}_+ : 2A_4 and 3^3 : S_4, >respectively. > Only one of those is maximal in A_{40} Ah, OK. I see now that all I really knew was that PSU(4,2) and its automorphism group have non-permutation-isomorphic primitive reps of degree 40. Again, is there an easy way to see that only one of the classes is maximal? >2) A_{63} -> PSU(3,3) -> 4.S_4 and 4^2.S_3 . > Only one is maximal And again... >3) A_{156} -> PSp(4,5) -> 5^{1+2}_+ : (4 x A_5) and >5^3 : (2 x A_5).2 . > Neither maximal. OK, but then don't both lie in Aut(PSp(4,5)) = PSp(4,5).2, which would then satisfy the maximal condition? Or are there even larger primitive groups of degree 156 in which at least one of them lies? Hmm..., if so, it looks like they would have to be normal extensions of PSL(4,5), right? >4) A_{253} -> M_{23} -> PSL(3,2^2) : 2 and 2^4 : A_7 . > Neither maximal So again if I understand correctly, at least one of these must lie in a larger primitive group of degree 253, which it looks like must thus be among normal extensions of A_{22} or PSL(2,23). Hmm..., must be the latter, since M_{23} isn't a subgroup of A_{22}, and so this probably reduces to your example. But wait a minute, M_{23} is too big to lie in PSL(2,23) either. I'm confused. :-( >5) A_{400} -> PSp(4,7) -> 7^{1+2}_+ : (6 x PSL(2,7)) and >7^3 : (3 x PSL(2,7)).2 . > Neither maximal And here PSL(4,7)? (Or maybe even a primitive group with product action and composite socle A_5 x A_5?) >6) A_{495} -> M_{12} -> 2^{1+4}_+.S_3 and 4^2 : D_{12} . > Neither maximal And here A_{12} or PSO+(10,2) (more likely the former)? >7) A_{820} -> PSp(4,3^2).2 -> (3^{1+2}_+)^2 : (8 x PSL(2,3^2)) >and 3^6 : (4 x PSL(2,3^2)).2 . (Actually, I'm not sure these 2 >classes of maximal subgroups are isomorphic. There's more than 1 >extension PSp(4,3^2).2 .) > Neither maximal And here A_{41}, PSL(2,41) or PSL(4,3^2) (likely the latter)? >Unless I've missed some, that appears to exhaust the a priori >possibilities for alternating groups < A_{1000}. > I expect my example is the only one. > My stataments were all just based on consulting known lists of alternating > groups up to degree 1000. Are any of these known lists available in books? I'm sure you've guessed I'm working from the appendices of Dixon and Mortimer's _Permutation Groups_. > If you want me to justify any of them, I will > need to find a larger subgroup I suppose. Of course, lists like that > often contain the odd error. Well, it would be cool if you could find the larger subgroups. :-) I've given some suggestions above. -- Jim Heckman === Subject: Re: finite groups >Bleh, ignore previous response. Hit send too quick. >in message : >Is it true that in every finite simple group G >all isomorphic maximal subgroups >are Aut(G)-conjugate? No, but it is not easy to find examples. I was about to give up, but > then I found one! Believe it or not, the alternating group A_{253} has two maximal > subgroups > isomorphic to PSL(2,23). One of them has dihedral D_{24} as point > stabilizer, and the other has S_4, so they cannot be conjugate in > S_{253}. Hi Derek, I have one question about your example. > The dihedral subgroup D_{24} of PSL(2,23) is extended to a subgroup > of PGL(2,23) of the same index 253. This means that PGL(2,23) > also has a permutational representation of degree 253. > Could it be that the first of your subgroups is not maximal > but a subgroup of index 2 in a larger subgroup isomorphic to PGL(2,23)? >Good catch! How about these, then? Hmm, let's see... > But my example works - both copies of PSL(2,23) are maximal. >OK. Is there an easy way to see that the PGL(2,23) in the one >case lies in S_{253} but not in A_{253}? And the analogous >question for the following cases? Let x be an element of order 2 in PGL(2,23) - PSL(2,23). Then x has two fixed points and 11 2-cycles on the 24 points. It lies in exactly 11 subgroups in the conjugacy class of maximal subgroups isomorphic to D_{48}, one for each of its 11 2-cycles. So, when we move to the action on 253 points, it fixes exactly 11 points and hence has 121 2-cycles, and is an odd permutation. of the maximal >1) A_{40} has 2 classes of maximal subgroups isomorphic to >PSU(4,2), such that their nomalizers in S_{40} have point >stabilizers isomorphic to 3^{1+2}_+ : 2A_4 and 3^3 : S_4, >respectively. > Only one of those is maximal in A_{40} >Ah, OK. I see now that all I really knew was that PSU(4,2) and >its automorphism group have non-permutation-isomorphic primitive >reps of degree 40. >Again, is there an easy way to see that only one of the classes >is maximal? I can only go by the lists I have. They are basically compiled from the using list of primitive groups in Dixon & Mortimer, and then checking for containment, which is painstaking but do-able. The maximal primitive subgroups of A_40 are PSU(4,2) (with the 3^3 : S_4 stabilizer) and PSL(4,3), which must contain the other PSU(4,2) (which is isomorphic to PSp(4,3)). >2) A_{63} -> PSU(3,3) -> 4.S_4 and 4^2.S_3 . > Only one is maximal A_63 has maximal primitive subgroups PSU(3,3), PSU(3,3).2 and PSL(6,2) - so that explains what is happening here. >And again... >3) A_{156} -> PSp(4,5) -> 5^{1+2}_+ : (4 x A_5) and >5^3 : (2 x A_5).2 . > Neither maximal. >OK, but then don't both lie in Aut(PSp(4,5)) = PSp(4,5).2, which >would then satisfy the maximal condition? Or are there even >larger primitive groups of degree 156 in which at least one of >them lies? Hmm..., if so, it looks like they would have to be >normal extensions of PSL(4,5), right? The only maximal primitive subgroup is PSL(4,5).2 here, which must contain everything. >4) A_{253} -> M_{23} -> PSL(3,2^2) : 2 and 2^4 : A_7 . > Neither maximal >So again if I understand correctly, at least one of these must >lie in a larger primitive group of degree 253, which it looks >like must thus be among normal extensions of A_{22} or >PSL(2,23). Hmm..., must be the latter, since M_{23} isn't a >subgroup of A_{22}, and so this probably reduces to your >example. But wait a minute, M_{23} is too big to lie in >PSL(2,23) either. I'm confused. :-( Three maximal primtiive subgroups, A_23, and the two PSL(2,23)'s. The A_23 (acting on pairs) contains both M_23's. >5) A_{400} -> PSp(4,7) -> 7^{1+2}_+ : (6 x PSL(2,7)) and >7^3 : (3 x PSL(2,7)).2 . > Neither maximal >And here PSL(4,7)? (Or maybe even a primitive group with product >action and composite socle A_5 x A_5?) Yes, both contained in PSL(4,7). A_400 also has maximal primitive subgroup isomorphic to S_20 wr C_2. >6) A_{495} -> M_{12} -> 2^{1+4}_+.S_3 and 4^2 : D_{12} . > Neither maximal >And here A_{12} or PSO+(10,2) (more likely the former)? No, the only max. prim. subgp. of A_495 is PSO-(10,2). >7) A_{820} -> PSp(4,3^2).2 -> (3^{1+2}_+)^2 : (8 x PSL(2,3^2)) >and 3^6 : (4 x PSL(2,3^2)).2 . (Actually, I'm not sure these 2 >classes of maximal subgroups are isomorphic. There's more than 1 >extension PSp(4,3^2).2 .) > Neither maximal >And here A_{41}, PSL(2,41) or PSL(4,3^2) (likely the latter)? Yes, I am sure it is the last - actually PSL(4,3^2).2^2 >Unless I've missed some, that appears to exhaust the a priori >possibilities for alternating groups < A_{1000}. > I expect my example is the only one. > My stataments were all just based on consulting known lists of alternating > groups up to degree 1000. >Are any of these known lists available in books? I'm sure you've >guessed I'm working from the appendices of Dixon and Mortimer's >_Permutation Groups_. As I said, they come originally mainly from Dixon and Mortimer - I don't know of any other written source. A few mistakes have been discovered in the Dixon and Mortimer lists. I believe that there is now a very high degree of confidence that the lists of primitive groups in GAP and Magma (both returned by the PrimitiveGroup function) are correct. They have been independentyl computed with extreme care. Derek Holt. === Subject: Re: finite groups in message : >Is it true that in every finite simple group G >all isomorphic maximal subgroups >are Aut(G)-conjugate? > No, but it is not easy to find examples. I was about to give up, but > then I found one! > Believe it or not, the alternating group A_{253} has two maximal > subgroups > isomorphic to PSL(2,23). One of them has dihedral D_{24} as point > stabilizer, and the other has S_4, so they cannot be conjugate in > S_{253}. > Hi Derek, > I have one question about your example. > The dihedral subgroup D_{24} of PSL(2,23) is extended to a subgroup > of PGL(2,23) of the same index 253. This means that PGL(2,23) > also has a permutational representation of degree 253. > Could it be that the first of your subgroups is not maximal > but a subgroup of index 2 in a larger subgroup isomorphic to PGL(2,23)? >Good catch! How about these, then? Hmm, let's see... > But my example works - both copies of PSL(2,23) are maximal. OK. Is there an easy way to see that the PGL(2,23) in the one case lies in S_{253} but not in A_{253}? >1) A_{40} has 2 classes of maximal subgroups isomorphic to >PSU(4,2), such that their nomalizers in S_{40} have point >stabilizers isomorphic to 3^{1+2}_+ : 2A_4 and 3^3 : S_4, >respectively. > Only one of those is maximal in A_{40} Ah, OK. Is there an easy way to see this? (Same question for subsequent cases where this occurs >2) A_{63} -> PSU(3,3) -> 4.S_4 and 4^2.S_3 . > Only one is maximal >3) A_{156} -> PSp(4,5) -> 5^{1+2}_+ : (4 x A_5) and >5^3 : (2 x A_5).2 . > Neither maximal. OK, but then don't both lie in Aut(PSp(4,5)) = PSp(4,5).2, which would then satisfy the condition? >4) A_{253} -> M_{23} -> PSL(3,2^2) : 2 and 2^4 : A_7 . > Neither maximal >5) A_{400} -> PSp(4,7) -> 7^{1+2}_+ : (6 x PSL(2,7)) and >7^3 : (3 x PSL(2,7)).2 . > Neither maximal >6) A_{495} -> M_{12} -> 2^{1+4}_+.S_3 and 4^2 : D_{12} . > Neither maximal >7) A_{820} -> PSp(4,3^2).2 -> (3^{1+2}_+)^2 : (8 x PSL(2,3^2)) >and 3^6 : (4 x PSL(2,3^2)).2 . (Actually, I'm not sure these 2 >classes of maximal subgroups are isomorphic. There's more than 1 >extension PSp(4,3^2).2 .) > Neither maximal >Unless I've missed some, that appears to exhaust the a priori >possibilities for alternating groups < A_{1000}. > I expect my example is the only one. > My stataments were all just based on consulting known lists of alternating > groups up to degree 1000. If you want me to justify any of them, I will > need to find a larger subgroup I suppose. Of course, lists like that > often contain the odd error. > Derek Holt. >Jim Heckman -- Jim Heckman === Subject: Re: finite groups in message : >Is it true that in every finite simple group G >all isomorphic maximal subgroups >are Aut(G)-conjugate? No, but it is not easy to find examples. I was about to give up, but > then I found one! Believe it or not, the alternating group A_{253} has two maximal > subgroups > isomorphic to PSL(2,23). One of them has dihedral D_{24} as point > stabilizer, and the other has S_4, so they cannot be conjugate in > S_{253}. > Hi Derek, > I have one question about your example. > The dihedral subgroup D_{24} of PSL(2,23) is extended to a subgroup > of PGL(2,23) of the same index 253. This means that PGL(2,23) > also has a permutational representation of degree 253. > Could it be that the first of your subgroups is not maximal > but a subgroup of index 2 in a larger subgroup isomorphic to PGL(2,23)? > Good catch! [...] Or not. On further thought, just because the relevant permutation rep of PSL(2,23) is embedded in one of PGL(2,23) in S_{253} doesn't mean it's so embedded in A_{253}. Offhand, I don't know an easy way to tell, but probably Derek will. -- Jim Heckman === Subject: what type of equation is this ? My apologies if this is a newbie type question. The Equation of the form Ut + H( Ux, x) = 0 is the Hamilton-Jacobi Equation in 1-D. What are equations of the following type called? Ut + H( U, Ux, x) = 0 and what are the numerical techniques used to solve it ? across a shock ? === Subject: Re: what type of equation is this ? >My apologies if this is a newbie type question. >The Equation of the form Ut + H( Ux, x) = 0 >is the Hamilton-Jacobi Equation in 1-D. >What are equations of the following type called? >Ut + H( U, Ux, x) = 0 >and what are the numerical techniques used >to solve it ? across a shock ? sometimes called Hamilton's principle or characteristic function in this context. You're saying , essentially, that the Hamiltonian is explicitly dependent on U itself. Just how is H dependent on U? I.e., what is H (U, Ux, x)? === Subject: Re: Limit of sequence Never use indirect proof when it's not necessary. You easily find that a-e0. Just let e->0 and you're done. -gs- >I'm having difficulty proving the following proposition on the limit >of a sequence xn: >if a <= xn <= b for every n and xn -> x, then a <= x <= b >Since xn -> x then I know that there is some N such that |xn - x| < e >whenever n > N. So I have >xn - e < x < xn + e. >By the initial condition I also have >a - e <= xn - e <= b - e, and >a + e <= xn + e <= b + e >Now I try to combine into >a - e <= xn - e < x < xn + e <= b + e. But that just tells me that >a < x + e < b + 2e. >Close, except I have a couple of e's that I don't really want :-) >Any suggestions on how to prove this would be greatly appreciated! >Tom > How about indirect proof? If the theorem is not true, then either > x> b or x< a. > If x> b, what happens if you pick e= (x-b)/2? === Subject: isomorphism on abelian groups if G is a finite abelian group of order n and phi: G->G is defined by phi(a)=a^m for all a in G, what's a necessary and sufficient condition for phi such that it's an isomorphism from G onto itself ? === Subject: Re: isomorphism on abelian groups >if G is a finite abelian group of order n and phi: G->G is defined by >phi(a)=a^m for all a in G, what's a necessary and sufficient condition >for phi such that it's an isomorphism from G onto itself ? An isomorphism is a homomorphism which is bijective. For phi to be a homomorphism (also called automorphism when the domain and range are the same) you need *) phi(1) = 1 *) phi(x^(-1)) = phi(x)^(-1) for all x in G *) phi(xy) = phi(x) phi(y) for all x, y in G [The definition in your textbook may omit some of these. For example, the third with x = y = 1 gives phi(1) = phi(1)^2, from which we can derive 1 = phi(1) after multiplying by phi(1)^(-1).] For phi to be a bijection you need *) phi(x) = phi(y) implies x = y (injective) *) For all y in G there exists x in G with phi(x) = y (surjective) Play with these five conditions, determining the restrictions on m such that all are satisfied. Some of the five are satisfied for all m, but others are non-trivial. -- ARNOLD = Anagram of RONALD ENEGGER = Backwards mis-pronounced REAGAN This is a black -- I mean SCHWARZ -- period in California. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Re: isomorphism on abelian groups > if G is a finite abelian group of order n and phi: G->G is defined by > phi(a)=a^m for all a in G, what's a necessary and sufficient condition > for phi such that it's an isomorphism from G onto itself ? It's surjective iff it's injective, so the question is, when does it have a trivial kernel? I'll let you think about this :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html === Subject: Re: isomorphism on abelian groups G is defined by > phi(a)=a^m for all a in G, what's a necessary and sufficient condition > for phi such that it's an isomorphism from G onto itself ? > It's surjective iff it's injective, so the question is, > when does it have a trivial kernel? > I'll let you think about this :-) Any finite abelian group is isomorphic to a direct product Z_p1^n1 x Z_p2^n2 x...x Z_pj^nk, for primes pj. of cyclic groups of prime power order. Any two such decompositions have the same number of factors of each order. I'd surmise that m would have to be coprime to every pj. ---- === Subject: Re: Andrica's Conjecture > Well, that takes care of that, but there are still some problems. > Here is the most important example: > Let n = 22. Then p_n and p_(n+2) can be 2, as you say, but then > we can let p_(n+3) = 5 and p_(n+5) = 3; however, you state that > p_(n+5) can't equal 3. In other words, I have switched the 3 and > the 5 in the divisor list. The example here is kind of silly, because > for very small numbers there are very small intervals between prime > numbers. However, as numbers get larger, this is not the case at all, > and I could possibly switch the list so that p_(n+11) = 5. In this > case, n + 1 = (n + 11) - 10, and so p_(n+1) = 5. Note that this kind > of switching could happen basically anywhere on the list; for example, > if p_(n+7) = 3, then so does p_(n+1), or if p_(n+57) = 5, then so does > p_(n+37). In this way we can continue re-using the same primes over > and over again. Your argument that p_m and p_(m+P) must either be > equal or different and unequal to P does not take into account the > fact that the relationship between p_(m + p1) and p_(m + p2) is a > difference of a composite number. > You may claim, Okay, well obviously that won't happen. You may > be able to swap around a couple of the prime numbers, but you can't > move them *that* far. Well, if so, prove it. Also, notice that if > you try to redefine p_(n+k) as the smallest prime that divides n+k but > doesn't divide n+j for j < > Maybe I _should_ have written: > hcf(pm, p(m+M)) | hcf(pm, M) rather than hcf(pm, p(m+M))=hcf(pm, M) > The rest of the argument still follows anyway... (in fact the rest of the > argument becomes slightly more immediate) > J Hi again Daniel, however it seems to be incomplete - was there a bit you accidentally left off? [it could be a problem with _my_ viewing of it thru Google groups??, if so, pls let me know]. is unimportant - I'm only using the pigeon-hole principle, effectively, [since every one has to be coprime to ALL the others] and showing there's an overflow James === Subject: Re: Andrica's Conjecture > Hi again Daniel, > however it seems to be incomplete - was there a bit you accidentally > left off? [it could be a problem with _my_ viewing of it thru Google > groups??, if so, pls let me know]. Nope. My fault. Sorry. > is unimportant - I'm only using the pigeon-hole principle, > effectively, [since every one has to be coprime to ALL the others] and > showing there's an overflow The problem is, you can't pigeonhole because your uniqueness argument is broken. Relative primality is NOT an equivalence operator -- just because a and b are relatively prime and b and c are relatively prime doesn't mean than a and c are relatively prime. This is the problem -- you start by saying that several of the p_n are relatively prime to the first one, and then you claim that they're different from each other. That's untrue; they can be the same, because after all, for P1, P2 primes > 2, p_(m + P2) = p_(m + P1 + C) for some _composite_ number, and if p_(m+P1) | c, then we can let p_(m + P1) = p_(m + P2). > James === Subject: Re: Andrica's Conjecture You're (sort of) correct in what you say, though incorrect that what you correctly observe, negatively affects my proof. Let me explain why: There will exist factors with the properties you've found (presumably by searching for specific (numerical) examples - which I'm always in favour of! :-), however you're missing a crucial facet of the proof. It's that old chestnut again I call the any/all issue. I only need there to exist ANY prime factors with the characteristics I need - you're looking at another set - included in ALL the relevant prime factors - though with extra members to my set, which extra members are therefore irrelevant to my proof. It's these extra members which have the property you demonstrate. This shows the limitation of numerically examining a specific (mis-)interpretation of the proof [and the all/any issue is a common one] J === Subject: Re: cross-section > How well can a 3-dimensional solid be specified if we know its > cross-sections, i.e., intersections with planes? > Example: Given that every cross-section is a triangle, then the solid > is a tetrahedron. Proof? > What if every cross-section is a square? Does the solid exist? I doubt it. I think that the only solid with all its cross-sections of the same shape, is the sphere. Or, more generally, the ellipsoid, if you admit affine-equivalent cross-sections. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: cross-section > How well can a 3-dimensional solid be specified if we know its > cross-sections, i.e., intersections with planes? > Example: Given that every cross-section is a triangle, then the solid > is a tetrahedron. Proof? > What if every cross-section is a square? Does the solid exist? > phil If M and M' are the middpoints of opposite edges, any plan perpendicular to hte segment MM' produce a rectangular cross-section.Particularly, the mean plane of MM' produce a square cross-section. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: cross-section > Example: Given that every cross-section is a triangle, then the solid > is a tetrahedron. Proof? Nonsense. A tetrahedron will have cross-sections that are not triangular. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: cross-section > Example: Given that every cross-section is a triangle, then the solid > is a tetrahedron. Proof? > Nonsense. A tetrahedron will have cross-sections that are not > triangular. In particular, consider the plane parallel to a pair of opposite edges and equidistant from them, and how this plane intersects the tetrahedron. === Subject: Re: cross-section > Example: Given that every cross-section is a triangle, then the solid > is a tetrahedron. Proof? > Nonsense. A tetrahedron will have cross-sections that are not > triangular. > In particular, consider the plane parallel to a pair of opposite > edges and equidistant from them, and how this plane intersects the > tetrahedron. I believe this sort of question is generally dealt with under that phrase will probably lead you to many interesting points. I am fairly sure that there are some very counter-intuitive examples. -Mike === Subject: Re: cross-section > Example: Given that every cross-section is a triangle, then the solid > is a tetrahedron. Proof? > Nonsense. A tetrahedron will have cross-sections that are not > triangular. > In particular, consider the plane parallel to a pair of opposite > edges and equidistant from them, and how this plane intersects the > tetrahedron. >I believe this sort of question is generally dealt with under >the topic geometric topology I think that belief is wrong. I usually at least glance at, and occasionally read with close attention, the papers posted at the ArXiv under the rubric Geometric Topology, and I very much doubt that any example of this sort of question is dealt with in any of the papers there. (The closest approximation I can think of is in some papers describing generalizations of normal surfaces in triangulated 3-manifolds.) Yet I think it would be generally agreed, by those in the field, that math.GT does, in fact, represent Geometric Topology (which I have seen defined as topology which is interested in concrete topological spaces and maps, particularly homeomomorphisms, more than in homotopy types, homotopy equivalences, and so on) pretty well. >that phrase will probably lead you to many interesting points. >I am fairly sure that there are some very counter-intuitive examples. The problem seems to me very untopological, but if a search string with topology must be used, I think combinatorial topology might work better. I think it's a purely geometric question, though. Lee Rudolph === Subject: Re: Primes as number of solutions of a diphantine equation. >For example, n is prime iff the curve y = n/x only passes to two >integer points: (1,n) and (n,1). So, why we don't call what are the >numbers that the parabola only passes to two integer points, or passes >to any integer points?. >For example, what are the numbers n such that y=nx^2 has integer >solutions? > Hint: any integer x will do. Yes, I'm in agreement with you that all integers x have this propierty (there are integers y, n such that y = n x^2), but I asked for n: what are the integers n such that there are x, y integers such that y = nx^2? Because we see n prime as the integer such that the curve y = f(x,n) has only two integer solutions ((x,y)=(n,1) and (1,n)), then we could interest about the number of integer solutions of a curve. And I started with f(x,n) = nx^2 a parabola. Xan. === Subject: Re: Primes as number of solutions of a diphantine equation. > I think there is merit in pursuing your idea, however in the form you have > presented it, there is not much you can do once you write n=xy but if you > use a mapping by which you tranform the original problem into a similar one, > you may make some progress. > I will give you this hint: find the mapping to tranform your useless > original equation into something more useful. then instead of working with > n=x*y, you work in the transformed space. You will find that the function > that represents the product x*y in the transformed space has many > interesting properties and has the form: > alpha=x*(C-x)/(6*x+1) where C is a constant that depends on N. There is only > one such function for every N. What is alpha? What do you do?. Can you explain me? === Subject: Re: WHAT IS THE COINCIDENCE OF THIS was Re: To prove psychic powers, someone has to sit this test! for you to play on LInk is above and in my signature. > signed up, waiting for confirm, the issue is the new topic! Herc. I've sent you two invitations to rejoin the group and to inspect the method and conditions of our proposed Trial and have not had a response. These were sent to both your Yahoo email and the Ozemail one. Have you lost interest in attempting to demonstrate your powers? As you say in the initial subject of this thread: To prove psychic powers, someone has to sit this test! That'll be you. We cannot proceed without your participation, and I am not going to further invite others to participate if you are not going to attempt this. -- Eric Hocking (moderator) You're invited to test of the paranormal at: http://www.twofromoz.freeserve.co.uk/sceptic/herctest/ e_hocking@yahoo.co.uk === Subject: Re: WHAT IS THE COINCIDENCE OF THIS was Re: To prove psychic powers, someone has to sit this test! > Herc. I've sent you two invitations to rejoin the group and to > inspect the method and conditions > of our > proposed Trial and have not had a response. These were sent to both > your Yahoo email and the Ozemail one. You don't need to contact him via email! He's psychic. He should know to contact you, and what your contact information is without you lifting a finger to inform him. LOL! > Have you lost interest in attempting to demonstrate your powers? As > you say in the initial subject of this thread: > To prove psychic powers, someone has to sit this test! What powers? Psychics are predatory quacks looking for little old ladies to conn out of their life's savings for a feel-good bull session. > That'll be you. We cannot proceed without your participation, and I > am not going to further invite others to participate if you are not > going to attempt this. He isn't going to participate because psychics are frauds. Why do you think they rely on newsgroups and 900 numbers? Why do you think the law requires them to have For entertainment purposes only tatooed to their every venue? So people will realize that they are just spinning new-aged feel-good bull. Your sister has been trying to contact you... or perhaps it's an aunt or niece... She has good news... and so forth. -- Felony case 02-CR-0617 9/1/03: Oregon Department of Justice V. Raymond Ronald Karczewski, Defendant. The defendant's name is NOT copyrighted. === Subject: Re: WHAT IS THE COINCIDENCE OF THIS was Re: To prove psychic powers, someone has to sit this test! > Herc. I've sent you two invitations to rejoin the group and to > inspect the method and conditions > of our > proposed Trial and have not had a response. These were sent to both > your Yahoo email and the Ozemail one. > You don't need to contact him via email! He's psychic. He should know to > contact you, and what your contact information is without you lifting a > finger to inform him. LOL! he foresaw the future, and knew he'd fail the test. === Subject: Re: ref. for infinite series bmqtm4$q5kj6$1@ID-143665.news.uni-berlin.de... > David Petry a .8ecrit dans le message de > I'm looking for a reference for the proof that the infinite series > There is an entirely elementary proof of this relation, > which I suspect is well known but I've never seen it. > Factor (1+x)^N - (1-x)^N, for N an odd integer, as > product ( w^k(1+x) - w^{-k}(1-x) ), k=-(N-1)/2..(N-1)/2 > where w = exp( i pi / N ) > Then find the coefficient of x^3 for that expression, and > compare it to the same coefficient using the Binomial Theorem > for the first expression, and take a limit. > I find : Binomial(n,3)=n*Sum((cotan(k*Pi/n))^2,k=1..(n-1)/2) sin(x)<=x<=tan(x) donc (cotan(x))^2<=1/x^2<=1/(sin(x))^2=1+(cotan(x))^2 then Sum((cotan(k*Pi/n))^2,k=1..(n-1)/2)<=Sum(1/(k*Pi/n))^2,k=1..(n-1)/2)<=(n-1)/ 2+ Sum((cotan(k*Pi/n))^2,k=1..(n-1)/2) then (Binomial(n,3)/(n^3))*Pi^2<=Sum(1/(k^2,k=1..(n-1)/2)<=(Pi/n)^2*((n-1)/2)+(Bi nomial(n,3)/(n^3))*Pi^2 and limit (Binomial(n,3)/(n^3))),n=oo)=1/6 and limit((Pi/n)^2*((n-1)/2),n=00)=0 === Subject: Re: ref. for infinite series David Petry a .8ecrit dans le message de > David Petry a .8ecrit dans le message de I find : > n*Binomial(n,3)=Sum((cotan(k*Pi/n))^2,k=1..(n-1)/2) > and .... > Please give me an hint for continue... > limit n->oo cotan(s/n)*(1/n) = 1/s > Also, the left hand side of your equation is almost, but > not quite, right. I find : Binomial(n,3)=n*Sum((cotan(k*Pi/n))^2,k=1..(n-1)/2) === Subject: Re: online math handbook web-based symbolic calculator, math handbook and computer algebra system. www.mathHandbook.com === Subject: Re: Data analysis software > It plots and analyses any x-y data for peak location, peak height, > peak > width, semi-derivative, derivative, integral, semi-integral, > convolution, > deconvolution, curve fitting, and separating overlapped peaks and > background. www.chemSoftware.com === Subject: Re: Two coin flip/ clarification for C Bond > [...] > Now consider the following: > [i] Let x = 3. > [ii] Let y = 3. > Do you understand that these are not necessarily equivalent because I > haven't stated what the rest of the question is after this? >We could say that x = y. >We could say they are different in that x has two lines that cross; y >has two lines that don't. >Mathematically speaking, they are equal. >Are they equivalent? Depends on the definition of equivalent. > Maybe you're trying to confuse us by pretending to be dense? > Yes, _if_ [i] and [ii] are both true _then_ x = y. That wasn't > the question. The question is whether [i] and [ii] are > equivalent statements. They obviously are not. > For example, suppose [i]. It follows that x^2 = 9. > Now suppose [ii]. Does that imply that x^2 = 9? > Of course it doesn't. >This >argument obvuscates the real argument. > There's more than one argument in the universe. You got me > replying when you said that I'd said that At least one was a head > is the same statement as At least one was a tail. I never said > anything so ridiculous. Okay, there is a lot of ambiguity in arguments. Not much in our question. There is a relationship between heads and tails which is unique. They are on opposite sides of the coin. a)A coin was flipped, it landed heads. a')A coin was flipped, it landed tails. There is a relationship between a and a' which is unique. Equivalent? Complimentary? Equal? I don't know. When you flip a coin, one happens, _OR_ the other. b)Two coins were flipped and they landed HH. b')Two coins were flipped and they landed TT. Same same b and b', except they each happen .25. When we say two coins were flipped, there is a relationship between 4 (four) outcomes. I don't know what it's called, but there are four events tied together. If this relationship isn't defined (if it doesn't have a name) maybe we can define it. When we flip two coins, HH happens, _OR_ TT happens, _OR_ HT happens, _OR_ TH happens. We toss them thousands of times, and any one of the happenings could have been the first toss. Suppose that b' happened first and: Two coins were flipped and at least one is a tail. What are the chances for two tails?, _OR_ Suppose that b happened first and: Two coins were flipped and at least one is a head. What are the chances for two heads? Bill and Jane only hear the statements. All they have to do to make the proper bet is answer the questions correctly. On b' Bill bet there are two tails. Jane bet for one of each. On b Bill bet for two heads. Jane bet for one of each. Bill will win with either of these occurrances. What odds should they bet? Would both be the same, or would they be different? Eldon > ************************ > David C. Ullrich === Subject: Re: Less symbols, core error proof > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. >What part of the *definition* is inconsistent? For it to be >inconsistent would suggest that there are no algebraic integers. No, if the definition showed there were no algebraic integers there would be nothing inconsistent about that (until we found another proof that there _is_ at least one algebraic integer). There's simply no way that this _definition_ can possibly be inconsistent. As always James is not saying exactly what he means - he means that the definition, together with other facts that he imagines are true, is inconsistent. ************************ David C. Ullrich === Subject: Re: Less symbols, core error proof >1. First the problematic definition: >Algebraic integers are defined to be roots of monic polynomials with >integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where >monic refers to the leading coefficient. >My assertion is that the over hundred year old definition excludes >numbers that have to be included to keep from having contradiction >i.e. mathematical inconsistency. >What part of the *definition* is inconsistent? For it to be >inconsistent would suggest that there are no algebraic integers. > No, if the definition showed there were no algebraic integers > there would be nothing inconsistent about that (until we found > another proof that there _is_ at least one algebraic integer). True. Of course, the fact that x+2 has solution -2 trivially shows that there is at least one algebraic integer. > There's simply no way that this _definition_ can possibly be > inconsistent. As always James is not saying exactly what he > means - he means that the definition, together with other > facts that he imagines are true, is inconsistent. Unfortunately, James has yet to take a class in which he finds out how mathematicians move from definition to properties. He seems to be convinced that we decide on the properties, and then construct the definition to fit. While that happens sometimes, as often we start with the definition and *then* explore the properties. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Core error argument objection refuted, short Adjunct Assistant Professor at the University of Montana. [.some quick comments.] >I was going through it in an e-mail reply to some questions Nora >sent. You can get it from the prime factorization of f and some Galois >Theory applied to the splitting field, at least when the polynomial is >irreducible. >What follows is almost certainly harder than it needs to be... That's >my usual m.o. What followed was my argument that f will be factored as f=u1*u2*u3, with u1, u2, u3 pairwise coprime, and each of a1, a2, a3 will be divisible by u2*u3, u1*u2, and u1*u2, respectively. Some notes on that argument: (1) Although I did not state so explicitly, it should be clear that I assuming that f was a rational prime throughout. The argument will go through, word for word, if we replace Q by the field of definition of the polynomial and f by a prime element of its ring of integers (that is, f|ab -> f|a or f|b); so, for example, it applies to James's example of f=sqrt(2), m=1, because sqrt(2) is a prime element of Z[sqrt(2)], the ring of integers of Q(sqrt(2)), the field of definition of the polynomial. (2) The argument should also work if f is a product of distinct primes, by analysing each prime separately. (3) The argument should also work if the ideal (f) is a product of distinct prime ideals, replacing divisibiilty of elements by divisibility of ideals throughout. (4) I believe the argument will also go through if we replace f with f^j for j>0, f a prime element; but there are some steps I used with j=1 which would no longer hold. For example, I discarded the possibility that (f)=p1^2*p2^2, because then there would be no way to split off (f)^2 into three elements; but if we are considering, say, f^3, then that would not be a problem. I think these difficulties can be dealt with, but I haven't thought it through. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Question on algebras >Let A be an any algebra over the field of real numbers. >Choose x,y in A. >Call a finite sequence (x,x_1,x_2, ...,x_n) a (finite) directed path >from x to y if x*x_1*...*x_n=y and x_j in A forall 1 <= j <=n. >Let P_{xy} be the set of all directed paths from x to y. >Let P'_{xy} subset P_{xy} be the set of all directed paths from >x to y without y being a member of any of the finite sequences >in P_{xy}. >Question: Does there exist a set A which is a (possibly non-divisor) >algebra and has two elements x, y such that P'_{xy}, or >even P_{xy} = empty?. >I have found an example algebra A' for which P_{xy} = not empty >forall x, y in A' No you haven't. If x = 0 and y <> 0 then P_{xy} is empty. >(to say the least) seemingly needs proving >-but, of course, why prove it if its already part of some more >general theorem? >Warning: Please go easy on me if I've overlooked something obvious >as I am a physics student and not a mathematician. I'm sure path, >for example, probably has another name. >C. Dement ************************ David C. Ullrich === Subject: Re: Question on algebras > What is an example of an algebra that does not have the following > property: forall x, y in A exists z in A: x * z = y. Every algebra A which contains more than one element is an example. Just let x = 0 and y be any non-zero element. For example, let A be the algebra of all real numbers. === Subject: Power series simplyfication of MoebiusMu expression F is a finite field with q elements. I have learned that the number of monic irreducible polynomials of degree t for a field with q elements is I(t,q) = 1/t * the summation for divisors d of t of MoebiusMu(t/d)*q^d. Hence, the number of monic pairs of factors in F multiplying to a polynomial of degree t is: q^t - 1/t * the summation for divisors d of t of MoebiusMu(t/d)*q^d. Can someone help me simplyfy the power series expression: The summation as t goes from 0 to infinity of q^t - 1/t * the summation for divisors d of t of MoebiusMu(t/d)*q^d ? Diana -- God made the integers, all else is the work of man. L. Kronecker, Jahresber. DMV 2, S. 19. === Subject: Re: Power series simplyfication of MoebiusMu expression >.... > Can someone help me simplyfy the power series expression: > The summation as t goes from 0 to infinity of > q^t - 1/t * the summation for divisors d of t of MoebiusMu(t/d)*q^d ? > Diana I believe that you want t to go from 1, not 0, right? For at t = 0, all right of the minus is undefined. Now, sum{t= 1 to infinity} 1/t^r sum{d|t} mu(t/d) q^d = polylog(q,r) /zeta(r), where polylog(q,r) = sum{t=1 to infinity} q^t/t^r (for those r and q where the sum converges), and where I have otherwise shortened your notation. (And zeta(r) = polylog(1,r), of course {for r > 1}.) So, sum{t= 1 to infinity} (q^t - 1/t sum{d|t} mu(t/d} q^d) = q/(1-q) - polylog(q,1)/zeta(1). But 1/zeta(r) -> 0 as r -> 1 from above. And, for -1 <= q < 1, polylog(q,1) = -log(1-q) (but this ends up not mattering). So, *IF* -1 <= q < 1, your sum (if t is from 1) is simply q/(1-q). I have avoided being rigorous, however, and I may have misinterpreted your question. Leroy Quet === Subject: Re: Power series simplyfication of MoebiusMu expression Glad math is a better subject for me than spelling: That would be simplify and simplification... > F is a finite field with q elements. I have learned that the number of monic > irreducible polynomials of degree t > for a field with q elements is > I(t,q) = 1/t * the summation for divisors d of t of MoebiusMu(t/d)*q^d. > Hence, the number of monic pairs of factors in F multiplying to a polynomial > of degree t is: > q^t - 1/t * the summation for divisors d of t of MoebiusMu(t/d)*q^d. > Can someone help me simplyfy the power series expression: > The summation as t goes from 0 to infinity of > q^t - 1/t * the summation for divisors d of t of MoebiusMu(t/d)*q^d ? > Diana > -- > God made the integers, all else is the work of man. > L. Kronecker, Jahresber. DMV 2, S. 19. === Subject: Re: ordered pairs >I was curious as to why an ordered pair (x,y) would be defined as >{{x}{x,y}}. I understand that a set has no order but therefore, wouldn't >that be the same as {{x,y} {x}}. > Yes, it would. But the point is that {{x,y},{x}} = {{z,w},{z}} if and > only if x=z and y=w. So that (x,y) = (z,w) if and only if x=z and y=w, > so that (x,y)=(y,x) if and only if x=y. So it's not that the set is > now ordered, is that the set {{x}, {x,y}} can be used to say which of > x and y goes first and which goes second. Could a *variant* of Axiom of Choice be said to be assumed here, we only know the ordering by issuing a IS_SINGLETON function on the 2 members. Its almost a contextual ordering relying on the data itself, I've devised other schemes that tend to rely on type checking. Herc === Subject: Re: ordered pairs Adjunct Assistant Professor at the University of Montana. >I was curious as to why an ordered pair (x,y) would be defined as >{{x}{x,y}}. I understand that a set has no order but therefore, wouldn't >that be the same as {{x,y} {x}}. > Yes, it would. But the point is that {{x,y},{x}} = {{z,w},{z}} if and > only if x=z and y=w. So that (x,y) = (z,w) if and only if x=z and y=w, > so that (x,y)=(y,x) if and only if x=y. So it's not that the set is > now ordered, is that the set {{x}, {x,y}} can be used to say which of > x and y goes first and which goes second. >Could a *variant* of Axiom of Choice be said to be assumed here, we only know >the ordering by issuing a IS_SINGLETON function on the 2 members. No, no variant of the Axiom of Choice is being assumed. Given two sets, x and y, there is a set containing x and y (Axiom of Pairing). Therefore, there exists a set that contains exactly both of them, {x,y} (Axiom of specification), and one that contains exactly x, {x} (Axiom of specification). Applying the Axiom of Pairing again, you get that there is one set that contains the set {x,y} and the set {x}. The Axiom of specification gives you the set { {x,y}, {x} }. The fact that this is then used to ->define<- an ordered pair has nothing to do with ordering, and we DO NOT check to see if which one is the singleton. We ->define<- the first element of the ordered pair A={ {x,y},{x} } to be the unique element of intersection{ {x,y},{x} }. This intersection exists by applying first the axiom of of unions, which gives you the set S = {x: x is in some element of A} and then the axiom of specification, which gives you the set I = {x in S: x is in all elements of A}. Then the unique element of I is the first element of the ordered pair, and the second element of the ordered pair is defined to be either the unique element of S-I, if one exists, or else the unique element of x, if none exists. If x is equal to y, then the ordered pair becomes A = { {x,y}, {x} } = { {x,x},{x}} = {{x},{x}} = { {x} }. So S = {x}, and I={x}; x is the first element, x is the second element. If x is different from y, then S = {x,y} and I={x}; so the first element is the unique element of I, and the second element is the unique element of S-I = {y}. No need to test to see if something is a singleton. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Quick Math Guide to core error issues > last couple of days, partly out of EXTREME FRUSTRATION at my > situation. I've found that I can have fun with postings, which makes > me feel better. Translation: weird mood = drunk and/or in the manic phase of (one of) your disorders > Oh well, maybe someday you'll have reason to give me your name, but > it's not a big deal. In any event, unlike with Nora Baron, I won't > put quotes around your name. How chivalrous of you. So you're going to be slightly less obnoxious in your dealings with Penny than with Nora -- at least until it's clear you can't snow her into agreeing with you. Then you'll be calling her a liar, too. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Funktor What is that? What is the precise definition? Why is it said that fundamental field (Puankare's field) is a functor? And what is the result? === Subject: Re: Funktor > What is that? Looks like a Germanic spelling of functor. >What is the precise definition? In category theory a functor F from a category C to a category D assigns to each object A of C an object FA of D, and to each arrow f: A -> B of C an arrow Ff fA -> FB of D such that identities and composition are preserved by F. > Why is it said that fundamental field (Puankare's field) What? > is a functor? And > what is the result? Result? Like for instance Forfar 4, East Fife 5? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Funktor > Why is it said that fundamental field (Puankare's field) >What? > is a functor? And Puankare' is probably meant to be Poincare'. I don't know what Poincare's field is, but the original poster might mean the fundamental group of a topological space, which was (according to MathWorld) introduced by Poincare'. Under all these assumptions: when someone says that the fundamental group is a functor, he/she means that not only there exists a fundamental group pi(X) for every topological space, but that there actually is a functor pi from the category of topological spaces to the category of groups such that pi(X) for every topological space X is the fundamental group of X. So, in particular, a homomorphism f : X -> Y of topological spaces gives rise to a group homomorphism pi(f) : pi(X) -> pi(Y) and this behaves nicely with respect to function composition and identities. -- Peter van Rossum, | Universal law of linearity: for all Dept. of Mathematics, New Mexico | f : R -> R and for all x, y in R: State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y) === Subject: Re: Funktor >What is the precise definition? > In category theory a functor F from a category C to a category D > assigns to each object A of C an object FA of D, and to each arrow > f: A -> B of C an arrow Ff fA -> FB of D such that identities > and composition are preserved by F. >Why is it said that fundamental field (Puankare's field) > What? Cannot answer the original question, but I think this is phonetic transcription of Poincare -- Vale ! Christianus Auriocus === Subject: Re: Need help for this integral >I'd like to calculate the value of the integral >int_{-infty} ^infty exp(ikx) Ln(1/Abs(x-x')) dx. >Is it possible to calculate the integral analytically? >Can anyone give me hints for how to solve such case? Can you give me any sense in which the integral exists? The integrand is going to infinity as x does. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Need help for this integral >I'd like to calculate the value of the integral >int_{-infty} ^infty exp(ikx) Ln(1/Abs(x-x')) dx. >Is it possible to calculate the integral analytically? >Can anyone give me hints for how to solve such case? > Can you give me any sense in which the integral exists? > The integrand is going to infinity as x does. How about limit a->0 integral (x=-oo..oo) exp(ikx - a|x-x'|) ln(1/|x-x'|) dx === Subject: Re: Need help for this integral >I'd like to calculate the value of the integral >int_{-infty} ^infty exp(ikx) Ln(1/Abs(x-x')) dx. >Is it possible to calculate the integral analytically? >Can anyone give me hints for how to solve such case? > Can you give me any sense in which the integral exists? > The integrand is going to infinity as x does. Let's try x'=0, k=1. We have convergence at 0, so how about a principal value where we integrate from -X to X, then let X increase to infinity? Here is what I get with Maple: > k := 1; k := 1 > fp := exp(I*k*x)*log(1/x); fm := exp(I*k*x)*log(1/(-x)); /1 fp := exp(I x) ln|-| x/ / 1 fm := exp(I x) ln|- -| x/ > fint := int(fp,x=0..X)+int(fm,x=-X..0); fint := Pi + I exp(I X) ln(X) + I Ei(1, -I X) - I exp(-I X) ln(-X) - Pi exp(-I X) - I Ei(1, I X) > simplify(%) assuming X>0; -2 ln(X) sin(X) + 2 Si(X) > limit(Si(X),X=infinity); 1 - Pi 2 So the 2 Si(X) term converges, and we are left with bigger and bigger oscillations. So not even this principal value exists. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Tough Probability question... >Anyone know how to attack the following? Mensa is a club for people >with >high IQ scores. To qualify, your IQ must be at least 132, putting you >the >top 2% of the general population. If a group of 10 people are chosen at >random, what is the probability that at least two of them qualify for >Mensa? > IQ tests are allegedly designed so that > (1) distribution of IQ scores may be treated as normal, > Why normal? Log-normal would be more appropriate. >No it wouldn't. The scoring of the tests is designed so that the scores are >symmetric about the mean (subject to measurement error). Log-normal has an >upward drift over time (oops, I'm sampling over time rather than population >here.) >I really don't think we mere mathematicians want to get bogged down in the >quagmire of the right way to measure intelligence. >Unless, of course, someone comes up with a compelling argument that it >exists (or doesn't). >Jon Miller Why is there any concern over the distribution of IQ? This is a straight binomial problem, and it makes no difference what the distribution of IQ happens to be. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Tough Probability question... > Why is there any concern over the distribution of IQ? > This is a straight binomial problem, and it makes no > difference what the distribution of IQ happens to be. I have the bad habit of answering questions asked rather than the questions that should have been asked. And of wandering off on tangents. OP's question has been answered many posts back (Randy Poe's post, plus another one that even gives more detail) and the rest is just playing games. Which (IMHO) is allowed, even in a math ng. I am just assuming that the other posts are offered in the same vein. Jon Miller === Subject: Re: Tough Probability question... This problem can be solved with the Bernoulli experiment. think of the problem as hit / no-hit or 0 and 1 n = total number of tries k = number of hits p = possiblity for a hit nCr(n,k) = combinations of k in n nCr -> on my TI-89 this function is called that way nCr(n,k) * p^k * (1-p)^(n-k) for your problem: P(A) = at least 2 P(NOT A) = none or exactly 1 P(A) = 1 - P(NOT A) 1 - (0.98^10) + (10 * (0.02^1 * (1-0.02)^(10-1))) = 0.0161776407 approx. 1.6 % === Subject: Re: Tough Probability question... > Anyone know how to attack the following? Mensa is a club for people with > high IQ scores. To qualify, your IQ must be at least 132, putting you in the > top 2% of the general population. If a group of 10 people are chosen at > random, what is the probability that at least two of them qualify for > Mensa? > So far I have: Approx pop of US is 292,000000 but Im not sure how to > proceed.Now do I figure out the chances of 2 people being 132 or above from > With this type of problem, it is often simpler to find the probability > of the opposite case. So, what is the probability that NONE of the > group qualify for Mensa, and what is the probability that ONE person > qualifies. These two answers, combined with the one you want, cover > all the possibilities... > I don't think that the population of the US is relevant to the > answer...Isn't mensa an international organization anyway? To this hint and my own, I'll add a pedantic note: Technically this is sampling without replacement in the sense that after you take one person out of the pool, the probabilities are slightly changed. However, nobody would solve it that way. Treat it as sampling with replacement in the sense that the probability of any given person having IQ 132 (or above) is exactly 2%. The pool is large enough to be considered infinite. - Randy === Subject: Re: Tough Probability question... > Anyone know how to attack the following? Mensa is a club for people > with > high IQ scores. To qualify, your IQ must be at least 132, putting you > in > the top 2% of the general population. If a group of 10 people are > chosen > at random, what is the probability that at least two of them qualify > for > Mensa? > So far I have: Approx pop of US is 292,000000 Did the question say anything about the USA? There are other countries in the world, you know. > Well obviously us furreiners are too stupid to qualify for Mensa. Duh. > It is largely true that youse furriners is too stupid to score above 131 on > ours IQ tests, but I think Mensa allows other methods of intelligence > estimation. So you might be able to make it in, anyways. If'n ya wanna, > o'course. > Further question. What is the probability that a randomly selected > president of the US qualifies for Mensa? Of course, does he really have to > be that intelligent, or can he just act the part? We'll go a little further... Due to the system of elections, the selection test is obviously conducted and evaluated by the population. What does that say about the intellegence of them? > Jon Miller Jeroen Boschma === Subject: Re: Tough Probability question... >Anyone know how to attack the following? Mensa is a club for people with >high IQ scores. To qualify, your IQ must be at least 132, putting you in >the >top 2% of the general population. If a group of 10 people are chosen at >random, what is the probability that at least two of them qualify for >Mensa? > IQ tests are allegedly designed so that > (1) distribution of IQ scores may be treated as normal, > Why normal? Log-normal would be more appropriate. Ask the designers of the tests, not me! > (1) the mean IQ score is 100, > (3) the standard deviation of IQ scores is 15. === Subject: pls help arith prob! please help me with the following simple problem that is frustrating me immensely!: An old car has to travel a 2-mile route, uphill and down. Because it is so old, the car can climb the first mile-the ascent-no faster than an average speed of 15mph. How fast does the car have to travel the 2nd mile (the descent-where it can go faster) in order to achieve an average speed of 30 mph for the entire trip? thank you very very much! === Subject: Re: pls help arith prob! >please help me with the following simple problem that is frustrating me >immensely!: >An old car has to travel a 2-mile route, uphill and down. Because it is so >old, the car can climb the first mile-the ascent-no faster than an average >speed of 15mph. How fast does the car have to travel the 2nd mile (the >descent-where it can go faster) in order to achieve an average speed of 30 >mph for the entire trip? >thank you very very much! Complete solution e-mailed to the address you had the integrity to provide. === Subject: Re: pls help arith prob! > please help me with the following simple problem that is frustrating me > immensely!: > An old car has to travel a 2-mile route, uphill and down. Because it is so > old, the car can climb the first mile-the ascent-no faster than an average > speed of 15mph. How fast does the car have to travel the 2nd mile (the > descent-where it can go faster) in order to achieve an average speed of 30 > mph for the entire trip? > thank you very very much! What have you done so far? If s is the speed during the second half, can you find a formula for the total time of the whole trip? === Subject: Re: pls help arith prob! > please help me with the following simple problem that is frustrating me > immensely!: > An old car has to travel a 2-mile route, uphill and down. Because it is so > old, the car can climb the first mile-the ascent-no faster than an average > speed of 15mph. How fast does the car have to travel the 2nd mile (the > descent-where it can go faster) in order to achieve an average speed of 30 > mph for the entire trip? > thank you very very much! Is this a Newtonian or an Einsteinian problem? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === decoding algorithm for correcting both erasures and errors of Reed-Solomon codes (appeared in IEEE Transactions on Communications, In section IV, the authors present an algorithm, (Algorithm 2), to decode both erasures and errors, using an algorithm based on the Euclidean algorithm together with Berlekamp-Massey algorithm. I have tried to implement the pseudocode in MATLAB, and used Example 1 to verify if my implementation works. The output of my implementation does not correspond at all with the output given in the example. My first problem is that the polynomial OMEGA^(b) is not computed according to the pseudocode. In the pseudocode, OMEGA^(b)(x) is multiplied by x, in line 41, but when I look at the output given in Example 1, the code is not multiplied by x. Did anyone implement this algorithm? Any suggestions would be greatly appreciated Jaco Versfeld === === Subject: Inverting the Laplace trasform I have some problems of inverting the Laplace transform of this function F(s)=sh(k*s)/s^2 ; k=const sh(x) is hyperbolic sinus Obviously, F(s) has a simple pole at s=0. Now by definition to recover f(t) we need to take the residue at s=0 from this expression: F(s)*e^(s*t) Result is k. But this is not correct. Where is my error? === Subject: Re: Inverting the Laplace trasform >I have some problems of inverting the Laplace transform of this function >F(s)=sh(k*s)/s^2 ; k=const >sh(x) is hyperbolic sinus >Obviously, F(s) has a simple pole at s=0. Now by definition to recover f(t) >we need to take the residue at s=0 from this expression: >F(s)*e^(s*t) >Result is k. But this is not correct. Where is my error? Do you have any reason to think that this F(s) should be a Laplace transform? Note that if f(t) is an exponentially bounded function (the kind of function for which the Bromwich inversion formula is valid) its Laplace transform F(s) satisfies |Lf(s)| <= A/(Re(s)-a) for some constants a and A when Re(s) > a. Even if you allow distributions involving a finite number of derivatives of such functions, the result will be bounded by a polynomial in |s|. But your F has no such bound. So I doubt it's the Laplace transform of anything sensible. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: minimization problem I have the following problem and I would appreciate any help. g(t,c_i, f_i(t)), i=1..n where c_i is a parameter non-linear to g and f_i(t) known functions. I want to find c_i so that g(t)=g(t,c_i,f_i(t))=:g_i(t) for all i What I have tried so far is to minimize integral (sum_i,j (g_i(t)-g_j(t))^2) dt or equivalently minimize integral ((sum_i g_i(t)^2)-(1/n)*(sum_i g_i(t))^2) dt So Ive differentiated the above term with respect to c_j and set it to 0: int (g_j(t)) dt = 1/n * int (sum_i g_i(t)) dt. But this is not very helpful since I cant solve for the cs. Any suggetions? Or any literature that might help? Lydia === Subject: Re: Boolean Algebra - Arithmetic Relationship > Where do axioms come from? Are these simply highly reliable patterns It depends on what you want to deduce from them. It seems to me that the axioms and definitions are chosen so that interesting theorems may be deduced. History works back to them, logic works forward from them. > that emerge out of empirical reality? > Can axiom be thought of as a pattern or template, that given one form > of truth can imply another? > Can an axiom be modeled therefore as an edge in a directed graph? > Given some truth (state) and an applicable axiom it moves me to > another truth (state)? I think of trees (upside ones with the leaves at the top) with nodes like this A B C | / | / * | D At * a rule has been applied to three previous theorems A, B, C to produce a new theorem D. Here the number three is not significant. If it's 0, D as an axiom. 1 and 2 are common. > If this is true, doesn't it imply that all transformations can be > modeled by a Turing machine? A Turing Machine can confirm that a deduction in a formal system is correct (if it is) but it cannot discover all of the correct deductions. > -Steve -- G.C. === Subject: Re: Boolean Algebra - Arithmetic Relationship > I think of trees (upside ones with the leaves at the top) with nodes > like this > A B C > | / > | / > * > | > D > At * a rule has been applied to three previous theorems A, B, C to > produce a new theorem D. Here the number three is not significant. If > it's 0, D as an axiom. 1 and 2 are common. Are rules, operations, and axioms essentially different forms of the same thing? Is it possible to define a grammar where any syntactically correct statement is also semantically meaningful and 'true'? Where I can't write a gramatically correct statement that is is false? -Steve === Subject: Re: Considering study of mathematics or physics after age 50 >I'm in my mid-fifties. Will my age be (perceived as) an impediment to >either undergraduate or graduate studies? By whom? The only place where it would seem relevant is in finding a good thesis advisor,which might not be relevant for a masters. >- how one chooses between/among programs in applied mathematics and >physics? Based on your own aptitudes and interests. >- level of effort/time required? Substantial. But if you have the interest, money, talent and time, I don't see why not. Certainly if I had the money I would seriously consider going back to school, first for some review, second to get up to date and third for a PhD. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: homoeomorphims between different dimensions? <3f8f75ef$3$fuzhry+tra$mr2ice@news.patriot.net> at 12:10 PM, Robin Chapman said: >It's isomorphic to Z. Now is that a homework question? No, it's an obvious counterexample. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: homoeomorphims between different dimensions? > at 12:10 PM, Robin Chapman It's isomorphic to Z. Now is that a homework question? > No, it's an obvious counterexample. Counterexample to what? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: homoeomorphims between different dimensions? > at 12:10 PM, Robin Chapman said: >It's isomorphic to Z. Now is that a homework question? >No, it's an obvious counterexample. > Counterexample to what? I think he missed the adjective local, or may not know about the concept of homology at a point. Just to complete the thought: The i'th homology group of the space X at the point p is defined to be the relative group H_i(X, X {p}). If X is locally contractible at p, then one can replace this group by the following: H_i(U, dU) (two steps: first is via excision, to H_i(U, U {p}) for U any neighborhood of p, and the second is the isomorphism between H_i(U, U{p}) and H_i(U, dU) [dU = boundary of U] for U a contractible neighborhood). This yields local homology for open subsets of R^n to be Z in dimension n, zero everywhere else. Dale === Subject: Re: Naive Q: Set theory, logic - which comes first? <1xUcb.14254$O85.6040@pd7tw1no> <3f79e264$7$fuzhry+tra$mr2ice@news.patriot.net> <3f82ebb8$3$fuzhry+tra$mr2ice@news.patriot.net> at 06:10 PM, Robert J. Kolker said: >However, Goedel numbering schemes are somewhat arbitrary. They are >merely artifacts for self reference, hence any theorems proven about >a particular set of numbers used as Goedel encoding are unlikely to >be of any fundemental importance in number theory. Fundamental importance is a separate issue. An unimportant statement about natural numbers is still a statement about natural numbers. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Naive Q: Set theory, logic - which comes first? <1xUcb.14254$O85.6040@pd7tw1no> <3f79e264$7$fuzhry+tra$mr2ice@news.patriot.net> <3f82ebb8$3$fuzhry+tra$mr2ice@news.patriot.net> <3f8c413c$18$fuzhry+tra$mr2ice@news.patriot.net> <3f8fd096$5$fuzhry+tra$mr2ice@news.patriot.net> at 05:22 PM, Torkel Franzen said: > Different how? The general usage of the term true has nothing to do with constructing models. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Naive Q: Set theory, logic - which comes first? Cc: nobody > The general usage of the term true has nothing to do with > constructing models. What does constructing models have to do with the observation or assumption that an arithmetical sentence is true? === Subject: Re: Naive Q: Set theory, logic - which comes first? <3f79e264$7$fuzhry+tra$mr2ice@news.patriot.net> <3f82ebb8$3$fuzhry+tra$mr2ice@news.patriot.net> <3f8c413c$18$fuzhry+tra$mr2ice@news.patriot.net> <3f8fd096$5$fuzhry+tra$mr2ice@news.patriot.net> at 03:05 PM, mattias_wikst71@hotmail.com (Mattias Wikstr?m) said: >Here is a definition which I hope will satisfy you: The natural >numbers are the things obtainable from the two operations (the first >of which is actually a constant) 0:N and s:N->N. That's still not a definition. >This is an example of an inductive definition. No. You have neither defined 0 and s not given any axioms for them. >I have (see above). above is some numenclature from Peano and the claim This is an example of an inductive definition. >I hope we agree that the Peano Postulates are consistent. I hope that they are ;-) >Here you have an example of a statement about natural numbers >which we agree is true, but which cannot be proven from the >Peano Postulates. My hoping that they are consistent does not make them consistent. My agreeing that they are consistent would not make them consistent. >I will show how the Peano Postulates can be proven from the >following definition: >The natural numbers are the things obtainable from the two >operations 0:N and s:N->N. That's too imprecise to be a definition, and you would need more machinery to get Peano's Postulates. >This definition is complete if we understand operation the right >way. Namely, there should be no sence of equality for the things we >obtain by theese operations other than that two expressions built up >from signs denoting theese operations denote the same thing iff they >are equal. This is vaguely formulated, but I hope it is clear what I >mean. I know what you mean, but how are you going to formulate it rigorously without presupposing something equivalent to Peano's Postulates? -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Open set definition >1.- The two concepts agree when the topology involved is that of all >open subsets of R. That is any open set in the topology is open in >the usual, elementary sense, that is the set is a neighborhood of all >its points. That's always true; you can define a topolog either in terms of neighborhoods or in terms of open sets. Don't confuse the concept of neighborhood witht he special case of open interval. >2.- For other topologies (say the collection of all subsets of R)a >set may be open according to one definition, closed according to the >other, or even both open and closed. The 2 definitions are >therefore unrelated. No. If you have two topologies on R, each has its own neighborhoods. The open sets of any topology are defined by its neighborhoods, and vice versa. If you are using a nonstandard topology for R then a neighborhood of a real x might not contain any open interval around x, and an open interval might not contain any neighborhood of its points. >There must be some general utility in defining open sets as the >elements of a properly defined topology. Sure. There are lots of topological spaces of interest besides the real line. Although properly defined topology is redundant; if it's not properly defined then it is not a topology. >Can someone please outline the reason? The most obvious example is to look at the plane rather than the line. You can no longer use open intervals as the basis for the standard topology. You could use products of intervals, but it is often convenient to use distance instead, which leads to the next example. If you have a metric space, you can define a topology where the open sets are unions of open balls. Frequently it is convenient to define a topology on a space of functions. See any book on Functional Analysis for examples. These spaces are very important for both Mathematics and Physics, and have been used to prove important results in the theory of differential equations. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Minimum polynomials Well as I understand it, the resolvent cubic has roots aa'+bb', ab+a'b' and ab'+a'b. It should be a cubic since any automorphism that permutes the roots of the quartic, a, a', b, b' will permute the three values above. Bill. > Could you please give a few more details. I don't see why the root of the > resolvent is equal to aa'+bb'. > -Michael. >OK, someone has pointed out the solution to me. I feel a bit stupid for >not knowing it. I should. >In a case like the one I mention, the cubic resolvent equation has a >rational root, which is easily calculated. This leads to the evaluation >I'm after, since that root is precisely the value aa' + bb' where the >a's and b' are the conjugate pairs of the original roots of my quartic. >Bill. === Subject: Re: foundations of calculus > what texts, if any, are available if I want to study the foundations, > ie how newton and leibniz actually conceived calculus, You can read Newton himself. The Principia, for example, is available in English. Leibniz? I don't know. Generally, see E Hairer, G Wanner Analysis by Its History Springer. -- G.C. === Subject: Re: foundations of calculus >what texts, if any, are available if I want to study the foundations, >ie how newton and leibniz actually conceived calculus, > You can read Newton himself. The Principia, for example, is available > in English. > Leibniz? I don't know. > Generally, see E Hairer, G Wanner Analysis by Its History Springer. Look at Dover's The History of the Calculus and Its Conceptual Development by Carl B. Boyer Origins in antiquity, medieval contributions, work of Newton, Leibniz, rigorous formulation. Treatment is verbal. Price $11.95 url: http://store.yahoo.net/doverpublications/by-subject-science-and-mathematics- mathematics-history-of-mathematics-1.html or, for just the book, http://store.yahoo.net/doverpublications/0486605094.html Martin Cohen === Subject: Re: foundations of calculus at 12:34 PM, a_cjones@hotmail.com (cdj) said: >sheesh - i was hoping that the rest of the world wasn't going to >following america's >descent-into-idiocy-by-reading-secondary-or-abridged-sources.... just >read the damn Principia - sheesh.... And I was hoping that the rest of the world wasn't going to follow America's descent into idiocy by confusing origins with publicity. The secondary sources can be very useful in telling you where to look, and you would have done well to consult them before making a fool of yourself. The roots of Calculus go back well before Newton. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: foundations of calculus > at 12:34 PM, a_cjones@hotmail.com (cdj) said: >sheesh - i was hoping that the rest of the world wasn't going to >following america's >descent-into-idiocy-by-reading-secondary-or-abridged-sources.... just >read the damn Principia - sheesh.... > And I was hoping that the rest of the world wasn't going to follow > America's descent into idiocy by confusing origins with publicity. The > secondary sources can be very useful in telling you where to look, and > you would have done well to consult them before making a fool of > yourself. The roots of Calculus go back well before Newton. rofl And I quote the OP: ##what texts, if any, are available if I want to study the foundations, ##ie how newton and leibniz actually conceived calculus I made no alethic errors in my post. Yours was completely irrelevant to the OP. (a) the OP did not ask for the roots of Calculus which, as you astutely notice, go back well before Newton. The OP didn't ask about origins. The OP didn't ask about publicity. To find out what the OP _did_ ask for, either read the OP, or see (b), below. (b) the OP specifically asked for how Newton/Leibniz thought of their topic. I stand by my response: to understand what a person thought of their chosen topic, there is no place better to go than the person himself. (c) almost unfailingly, secondary sources are crap. This, combined with the increasing practice of 2ndary sources being the _only_ source used exacerbates the idiocy problem I mentioned initially. (d) wrt Newton specifically, the only good secondary source I've seen is by (if I recall the name correctly - it's been awhile) Dana Densmore. It will be clear to anyone at all familiar with this book that my recognition of this as a good secondary source is completely consonant with (c), above. confuse origins with publicity? lol - good one cdj === === Subject: Re: pls help arith prob! === >Subject: pls help arith prob! >please help me with the following simple problem that is frustrating me >immensely!: >An old car has to travel a 2-mile route, uphill and down. Because it is so >old, the car can climb the first mile-the ascent-no faster than an average >speed of 15mph. How fast does the car have to travel the 2nd mile (the >descent-where it can go faster) in order to achieve an average speed of 30 >mph for the entire trip? >thank you very very much! Or, is it even possible??? Here's a similar problem: If I can only eat one donut per hour, can I eat two donuts in an hour? adam === Subject: Applying ergodic theorem to Brownian Motion I am reading a paper that applies the ergodic theorem where the measure in question is the Wiener measure, and the transformation in question is w(t) --> (1/sqrt(a))w(at), where w refers to a continuous path in Wiener space. Now, I can see that this transformation is measure preserving, but why is it ergodic? Intuitively it's very believable somehow, but I don't see the proof. Anyone know? Greg === Subject: Re: Applying ergodic theorem to Brownian Motion > I am reading a paper that applies the ergodic theorem where the > measure in question is the Wiener measure, and the transformation in > question is w(t) --> (1/sqrt(a))w(at), where w refers to a continuous > path in Wiener space. Now, I can see that this transformation is > measure preserving, but why is it ergodic? Intuitively it's very > believable somehow, but I don't see the proof. Anyone know? Call the above transformation T_a, a>0. Let F be a bounded non-negative measurable function on the Wiener path space that is T_a invariant for all positive a. Let F_t be path-continuous version of the martingale E_t[F], where E_t is the conditional expectation with respect to the history of the Wiener process up to time t. Using the T_a-invariance of Wiener measure it is easy to check that F_t(T_a) = F_{at} for all positive a and t. Therefore, for z>0, E[exp(-z*F_t)]= E[exp(-z*F_t(T_a))]= E[exp(-z*F_{at})], and the latter converges to E[exp(-z*F_{0})]=exp(-z*E[F]) as a --> 0+, by the continuity of t--> F_t and the fact that F_0=E[F]. That is, for z>0,t>0 E[exp(-z*F_t)]=exp(-z*E[F}). This means that F_t is almost surely equal to its mean value E[F]. Since $F_t converges to F almost surely as t-->infinity, we conclude that F=E[F] almost surely. So invariant functions are constant, so the group of transformations {T_a, a>0} is ergodic. It seems to me that the same argument, with a confined to the multiplicatice subgroup {b^n: n in Z} of positive real numbers generated by a fixed b>0, shows that each T_b is ergodic. -- A. === Subject: Re: Applying ergodic theorem to Brownian Motion > I am reading a paper that applies the ergodic theorem where the > measure in question is the Wiener measure, and the transformation in > question is w(t) --> (1/sqrt(a))w(at), where w refers to a continuous > path in Wiener space. Now, I can see that this transformation is > measure preserving, but why is it ergodic? Intuitively it's very > believable somehow, but I don't see the proof. Anyone know? > Greg Paradoxically named, the ergodic theorem applies to transformations that are not necessarily ergodic. === Subject: Re: Applying ergodic theorem to Brownian Motion This reminds me of something that happened when I was an undergraduate. My adviser, Ethan Coven, taught a graduate course entitled Ergodic Theory. Coven had an evaluation of that course posted inside his office. In response to the instruction to evaluate the course content (as opposed to the quality of teaching), one student had written something like, The course content was interesting but somewhat disappointing. Then again, I did misread the course catalogue entry. I was expecting a course on Erotic Theory. BTW, while I was still an undergraduate, someone swiped this evalution from Coven's bulletin board. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Convergence of infinite sum; can anyone solve this? > I am trying to prove the convergence of the following series; > The limit as N approaches infinity of; > (1/(ln(N)) times (sum from 1 to N) of |mu(N)|/N, [as noted, the sum here should be sum(|mu(n)|/n, n= 1..N). I e-mailed you this reply earlier, but as I received no acknowledgement and as you have posted since then in this thread, I assume there's a good chance you never saw it.] As Robin Chapman pointed out, if we set F(s) = sum(|mu(n)|/n^s), then F(s) = zeta(s)/zeta(2s) for Re(s) > 1. Now F is a Dirichlet series with nonnegative coefficients. It converges for Re(s) > 1, and (as follows from the above expression in terms of zeta) is analytic for Re(s) > 1/2 except for a simple pole at s=1 with residue 1/zeta(2) = 6/pi^2. So by the Wiener-Ikehara Tauberian theorem (see, e.g., Murty's problem book on analytic number theory) we have S(x):= sum(|mu(n)|, n <= x) = 6x/pi^2 + R(x), where R(x)/x -> 0 as x -> oo. Partial summation says that the sum you want is S(N)/N + int_1^N{S(t)/t^2 dt}. The term S(N)/N is bounded as N -> oo. In estimating the integral, 6/pi^2 * log N pops out as the contribution from the main term in the estimate for S(x). The fact that R(x)/x -> 0 implies int_1^N{R(t)/t^2 dt} = o(log N), so the limit you are after is 6/pi^2. One can also attack this problem elementarily. The key observation is that |mu(n)| = sum(mu(d), d^2 | n). Then reverse the order of summation to obtain sum(|mu(n)|/n, n=1 .. N) = sum(mu(d)/d^2 * sum(1/k, k<= N/d^2), d <= sqrt(N)). The inner sum can be estimated as log(N/d^2) + gamma + O(d^2/N). After a little manipulation, one finds sum(|mu(n)|/n, n=1..N) = log N * sum(mu(d)/d^2, d<= sqrt(N)) + O(1). The sum here is almost 1/zeta(2); we have omitted the tail consisting of the terms with d> sqrt(N), which incurs an error which is O(sum(1/t^2, t > sqrt(N))) = O(1/sqrt(N)). This implies the first term on the right hand side above is 6 log N/pi^2 + O(log(N)/sqrt(N)). Hence the entire sum is 6 log N/pi^2 + O(1), which again shows the limit you're after is 6/pi^2. With a little more care about the error terms, the same method implies sum(|mu(n)|/n, n= 1..N) - 6 log N/pi^2 in fact tends to a limit. Hope this helps, Paul === Subject: Re: Convergence of infinite sum; can anyone solve this? >Liz > I am trying to prove the convergence of the following series; > The limit as N approaches infinity of; > (1/(ln(N)) times (sum from 1 to N) of |mu(N)|/N, > Presumably meaning 1/ln(N) sum_{n=1}^N |mu(n)|/n > where |mu(N)| is the absolute value of the Mobius function, and ln(N) > is the natural log of N. > |mu(N)| is 1 whenever N is square-free, and 0 otherwise. >We know that >(1/(ln(N)) times (sum from 1 to N) of 1/N >has a limit at infinity, from the def'n of Euler's constant. So, we are down >to showing that this has a limit: >(1/(ln(N)) times (sum from 1 to N) of k(N)/N >where k(N)=1 if N is divisible by a square, else k(N)=0. >But that expression must go to zero, because sum(1/N^2) is finite. > No, that expression doesn't go to 0. For example, k(4 j) = 1 for any > positive integer j, so > 1/ln(N) sum_{n=1}^N k(n)/n >= 1/ln(N) sum_{j=1}^floor(N/4) 1/(4j) > ~= ln(N/4)/(4 ln(N)) -> 1/4 > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 6/pi^2 does seem to be where the series is heading; see Matlab code below. sum = 0; jj = 1; mag(1) = 0; narray = 1:1:100000; for n=1:1:100000, f = factor(n); a = length(f); ff = unique(f); b = length(ff); if (a == b) %%% if you are here, mu(n) == 1 sum = sum + (1/n); end if (n > 1) magg(jj) = (1/log(n))*sum; end jj = jj + 1; end plot(narray,magg) === Subject: Re: Quanta and Cakes > bn0qo5$qjh6e$1@ID-186372.news.uni-berlin.de>... > ThereI[Hyphen]as a misspelling: > The authorI[Hyphen]as name is Teller, not Heller! > OK, it came to me like Heller, I'm sorry. > ThatI[Hyphen]as all right. :-) > [Teller's homepage: http://www-philosophy.ucdavis.edu/teller/] > Oh the man is professor. > Yes indeed. > But still, his example with the box with two > some better example, with only two quantum states available (I can not > find no realistic one). > The question II[Hyphen]am now having is why the apparently well-known author > of > An Interpretative Introduction to Quantum Field Theory (Princeton: > Princeton University Press. 1995) > has chosen an example that turns out to be bad, as you maintain! > Well, probably thatI[Hyphen]as due to a metaphysical bias of his. > Unfortunately, II[Hyphen]am not a physicist and so I cannot comment > competently on your criticism of TellerI[Hyphen]as quantum statistical > example. > Nevertheless it seems to me that thereI[Hyphen]as reason to suppose that the > example chosen by Teller does not back his conclusion (Quantum objects > are entities without haecceities!) the way he thinks it does. May be, I've misunderstand Mr. Teller's example. With his example he may mean not real box but something like this: Suppose to have system, where only two states can be occupied by Everything else than is OK. Or maybe I'm complete idiot and I'm missing some point from the beginning :) > Anyway, the metaphysical issue whether quanta are endowed with > haecceities or not (if that concept makes any real sense at all ...) > cannot be decided by quantum physics itself, because the latterI[Hyphen]as > philosophical interpretation is underdetermined by the empirical data. > I agree with van Fraassen that quantum mechanics does not force the > issue. > BTW, you might find the following text interesting: > French, Steven (2000). /Identity and Individuality in Quantum Theory/. > In Stanford Encyclopedia of Philosophy. Retrieved from > http://plato.stanford.edu/entries/qt-idind/] I just scrolled through, but nice. Funny what are people over the world working on. Palo > PH === Subject: Re: Quanta and Cakes > bn0qo5$qjh6e$1@ID-186372.news.uni-berlin.de>... > ThereI[Hyphen]as a misspelling: > The authorI[Hyphen]as name is Teller, not Heller! > OK, it came to me like Heller, I'm sorry. ThatI[Hyphen]as all right. :-) > [Teller's homepage: http://www-philosophy.ucdavis.edu/teller/] > Oh the man is professor. Yes indeed. > But still, his example with the box with two > some better example, with only two quantum states available (I can not > find no realistic one). The question II[Hyphen]am now having is why the apparently well-known author of An Interpretative Introduction to Quantum Field Theory (Princeton: Princeton University Press. 1995) has chosen an example that turns out to be bad, as you maintain! Well, probably thatI[Hyphen]as due to a metaphysical bias of his. Unfortunately, II[Hyphen]am not a physicist and so I cannot comment competently on your criticism of TellerI[Hyphen]as quantum statistical example. Nevertheless it seems to me that thereI[Hyphen]as reason to suppose that the example chosen by Teller does not back his conclusion (Quantum objects are entities without haecceities!) the way he thinks it does. Anyway, the metaphysical issue whether quanta are endowed with haecceities or not (if that concept makes any real sense at all ...) cannot be decided by quantum physics itself, because the latterI[Hyphen]as philosophical interpretation is underdetermined by the empirical data. I agree with van Fraassen that quantum mechanics does not force the issue. BTW, you might find the following text interesting: French, Steven (2000). /Identity and Individuality in Quantum Theory/. In Stanford Encyclopedia of Philosophy. Retrieved from http://plato.stanford.edu/entries/qt-idind/] PH === Subject: Re: So much fun Nice try James But the real JSH doesn't put the JSH in the title message. YOU you stupid LOSER!!! HELL YEAH!!! I am the MAX. So bow down before me worm. The real JSH tries to sound erudite and sophisticated. So, please, leave the cranking to the pros. GREG > Most of my life I've had this label: smart. > But here on this newsgroup people made fun of me, and kept trying to > argue with me. I could come out here and not be at the top of the > heap but be considered on the bottom. > I had people to play with. > Usually, when people figure out what I can do they quit playing with > me. > They get tired of being beaten. > But so many of you kept playing, and I've had a lot of fun. > Now though I wonder if it's about over, so while I have the chance, > I'll have even more fun. > Now I call you names and curse. > It's now much more fun. > How long will you still play with me? Let's see. > James Harris === Subject: Re: So much fun Discussion, linux) > Nice try James > But the real JSH doesn't put the JSH in the title message. JSH has often put JSH in the title message. He does not consistently do so, however. > YOU you stupid LOSER!!! > HELL YEAH!!! > I am the MAX. So bow down before me worm. > The real JSH tries to sound erudite and sophisticated. > So, please, leave the cranking to the pros. Lately, there have been many crude posts putatively by James Harris. Some of these are almost certainly by James -- most of the post is in his typical style with his typical arguments and then they devolve to curses at the end. Others, like this one, seem a bit more questionable, but James has not disputed that they are from him. Of course, there are many occasions in which posters have forged JSH posts and James has not spoken up, so his silence on this post is not strong evidence one way or the other. Nonetheless, I tentatively think that this post is from the real McCoy, until stronger evidence to the contrary comes. -- Jesse Hughes LOL. How arrogant you are. Now when you realize that I DID prove Goldbach's conjecture and that I proved Fermat's Last Theorem as well, how are you going to feel then? -- James Harris === Subject: Formulae for Latin squares of size 2^n For size 2^n, an obvious type of Latin squares has its element in row i and column j (counting from 0) given by s(i,j) = i + j mod 2^n (We consider those that result from row and/or column permutations of the Latin square to belong to the same type.) One can easily verify that there is another type whose elements may be described as follows: t(i,j) = 2*i*j + i + j mod 2^n Are there additional types having elements that could be in advance. M. K. Shen ----------------------------------- http://home.t-online.de/home/mok-kong.shen === Subject: Re: Formulae for Latin squares of size 2^n > t(i,j) = 2*i*j + i + j mod 2^n >Are there additional types having elements that could be >in advance. Any two-input multipermutation works, although one could consider this to be a circular definition. But anyway, t(i, j) = i XOR j works. While a multipermutation operator doesn't necessarily form a group over the N elements, any group operator over N elements must form a multipermutation... Greg. -- Greg Rose 232B EC8F 44C6 C853 D68F E107 E6BF CD2F 1081 A37C Crypto Mini-FAQ: http://www.schlafly.net/crypto/faq.txt Qualcomm Australia: http://www.qualcomm.com.au === Subject: Re: Formulae for Latin squares of size 2^n > t(i,j) = 2*i*j + i + j mod 2^n >Are there additional types having elements that could be >in advance. > Any two-input multipermutation works, although one > could consider this to be a circular definition. > But anyway, t(i, j) = i XOR j works. While a > multipermutation operator doesn't necessarily form > a group over the N elements, any group operator > over N elements must form a multipermutation... is also connected to the fact that a Latin square can serve as a combiner of bit groups as described in Terry Ritter's webpage.) I hope to be able to gather more types that are susceptible to simple computational formulations. M. K. Shen === Subject: Re: Formulae for Latin squares of size 2^n > For size 2^n, an obvious type of Latin squares has its > element in row i and column j (counting from 0) given by > s(i,j) = i + j mod 2^n > (We consider those that result from row and/or column > permutations of the Latin square to belong to the same type.) > One can easily verify that there is another type whose > elements may be described as follows: > t(i,j) = 2*i*j + i + j mod 2^n > Are there additional types having elements that could be > in advance. The multiplication table of any group (lots of these of order 2^n as evidenced in another recent thread). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Formulae for Latin squares of size 2^n > The multiplication table of any group (lots of these > of order 2^n as evidenced in another recent thread). Are there general ways to simply formulate such tables like those I gave? (Given i,j, I want to compute the element in row i and column j numerically.) Could you give a key word of the said thread? (I look only M. K. Shen === Subject: Sylow Group Hi I am learning algebra course I am learning sylow group, on the wikipedia website, it says: The Sylow theorems of group theory form a partial converse to the theorem of Lagrange, which states that if H is a subgroup of a finite group G, then the order of H divides the order of G. The Sylow theorems guarantee, for certain divisors of the order of G, the existence of corresponding subgroups, and give information about the number of those subgroups. I remember if G is a finite cyclic group of order n, then G has exactly one subgroup of order d for each divisor d of n, and G has no other subgroups. Can I generalize that for any finite group G of order n, G has exactly one subgroup of order d for each divisor d of n, and G has no other subgroups === Subject: Re: Sylow Group > .... > Can I generalize that for any finite group G of order n, > G has exactly one subgroup of order d for each divisor d of n, and G > has no other subgroups.... The alternating group A_4 has order 12. Has it exactly one subgroup of order 3, and exactly one of order 6? Ken Pledger. === Subject: Re: Sylow Group Adjunct Assistant Professor at the University of Montana. >Hi I am learning algebra course >I am learning sylow group, on the wikipedia website, it says: >The Sylow theorems of group theory form a partial converse to the >theorem of Lagrange, which states that if H is a subgroup of a finite >group G, then the order of H divides the order of G. The Sylow >theorems guarantee, for certain divisors of the order of G, the >existence of corresponding subgroups, and give information about the >number of those subgroups. >I remember if G is a finite cyclic group of order n, then G has >exactly one subgroup of order d for each divisor d of n, and G has no >other subgroups. Yes. >Can I generalize that for any finite group G of order n, >G has exactly one subgroup of order d for each divisor d of n, and G >has no other subgroups No, you cannot. That is exactly why Sylow's Theorems are only a partial converse, and yet they are important. If you could generalize, then Sylow's Theorems would be pretty uninteresting! To see the exactly one clause is wrong, consider the Klein 4-group, which is the product of two cyclic groups of order 2. It is of order 4, but it has THREE subgroups of order 2: {(0,0), (1,0)}, {(0,0), (0,1)}, and {(0,0),(1,1)}. And as for at least one subgroup of order d, that one is a bit harder to come up with counterexamples; there are certainly smaller examples, but the alternating group on 5 elements, A_5, has 60 elements and has no subgroup of order 30. Sylow's Theorems state that if p is a prime, p^n divides the order of G, and p^{n+1} does NOT divide the order of G, then G has at least one subgroup of order p^n, that any two such subgroups are conjugate, and that the total number of subgorups of order p^n is a divisor of G and is congruent to 1 modulo p. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Sylow Group > I remember if G is a finite cyclic group of order n, then G has > exactly one subgroup of order d for each divisor d of n, and G has no > other subgroups. > Can I generalize that for any finite group G of order n, > G has exactly one subgroup of order d for each divisor d of n, and G > has no other subgroups No. Take a census of the subgroups of some noncyclic groups, e.g., the Klein 4-group V_4, the symmetric groups S_3 and S_4 and the alternating group A_4 to get a feeling for how wrong this is. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: How do you solve this? Any ideas? % - Mod function k - Constant k%n = (n+1)/2 So if k is 8, possible solutions are 5 and 3. Whereas, k is 13, the solution is 5. The real problem is, trying to do a mathematical equivalent of a mod function without using the cosine function and without creating an external variable. === Subject: Re: How do you solve this? Any ideas? >% - Mod function >k - Constant >k%n = (n+1)/2 >So if k is 8, possible solutions are 5 and 3. or 15. >Whereas, k is 13, the solution is 5. or 25. In other words, k = (n+1)/2 + y n for some integer y. Or 2 k - 1 = (2y+1) n So n can be any factor (> 1) of 2k-1, and 2y+1 = (2k-1)/n. Of course n=2k-1 always works. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: How do you solve this? Any ideas? >% - Mod function >k - Constant >k%n = (n+1)/2 >So if k is 8, possible solutions are 5 and 3. > or 15. >Whereas, k is 13, the solution is 5. > or 25. > In other words, k = (n+1)/2 + y n for some integer y. > Or 2 k - 1 = (2y+1) n > So n can be any factor (> 1) of 2k-1, and 2y+1 = (2k-1)/n. > Of course n=2k-1 always works. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 That is not very helpful. Y is an external variable. You had to create y in order to solve the problem. I already did this except I used 'A'. If that is the best Canada can produce, God help us all. Just kidding (laugh). === Subject: Re: How do you solve this? Any ideas? > In other words, k = (n+1)/2 + y n for some integer y. > Or 2 k - 1 = (2y+1) n > So n can be any factor (> 1) of 2k-1, and 2y+1 = (2k-1)/n. > Of course n=2k-1 always works. >That is not very helpful. Y is an external variable. You had to create y >in order to solve the problem. I already did this except I used 'A'. I don't see what's unhelpful about it. You don't actually need the y. You wanted to know how to find the n's that work, and I told you: n can be any factor (> 1) of 2k-1. >If that is the best Canada can produce, God help us all. Just kidding >(laugh). Who do you want, Wayne Gretzky? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Sum of unique prime squares? Is there a theorem that states any square can be written as a sum (including negative terms) of only squares of primes and one so that no term is repeated twice? 16 = 25 - 9 36 = 49 - 9 - 4 64 = 81 - 49 - 25 - 9 + 4 - 1 100 = 81 + 49 - 25 - 4 - 1 144 = 81 + 49 + 9 + 4 + 1 196 = 169 + 49 - 25 + 4 - 1 ... === Subject: Re: Sum of unique prime squares? >Is there a theorem that states any square can be written as a sum >(including negative terms) of only squares of primes and one so that >no term is repeated twice? >16 = 25 - 9 >36 = 49 - 9 - 4 >64 = 81 - 49 - 25 - 9 + 4 - 1 I don't think you need the 1. 64 = 13^2 - 11^2 + 5^2 - 3^2 81 = 11^2 - 7^2 + 3^2 100 = 11^2 - 5^2 + 2^2 144 = 13^2 - 5^2 196 = 17^2 - 11^2 + 7^2 - 5^2 + 2^2 225 = 13^2 + 11^2 - 7^2 - 5^2 + 3^2 256 = 13^2 + 7^2 + 5^2 + 3^2 + 2^2 Actually it seems that all positive integers at least up to 1000 can be written as differences of sums of squares of distinct primes. I wouldn't be surprised if this was true for all positive integers, but I don't immediately see a way to prove it. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Sum of unique prime squares? > Is there a theorem that states any square can be written as a sum > (including negative terms) of only squares of primes and one so that > no term is repeated twice? > 16 = 25 - 9 > 36 = 49 - 9 - 4 > 64 = 81 - 49 - 25 - 9 + 4 - 1 ^^ square of a prime? [...] Hugo === Subject: Re: Sum of unique prime squares? > Is there a theorem that states any square can be written as a sum > (including negative terms) of only squares of primes and one so that > no term is repeated twice? > 16 = 25 - 9 > 36 = 49 - 9 - 4 > 64 = 81 - 49 - 25 - 9 + 4 - 1 > ^^ > square of a prime? Replace with: 64 = 49 + 25 - 9 - 1 100 = 121 - 25 + 4 === Subject: If G is a group of order p^mr, where p is not divided by r Let X be the set of all subsets of G of size p^m. Then the textbook says |X| = C_{p^m}^{p^mr} === Subject: Re: If G is a group of order p^mr, where p is not divided by r Adjunct Assistant Professor at the University of Montana. You should always include the hypothesis in the text, not just the header or subject. So, Let G be a group of order p^m*r. It seems to me that you mean where p does not divide r, rather than where p is not divided by r, since presumably p is a prime number? >Let X be the set of all subsets of G of size p^m. Then >the textbook says |X| = C_{p^m}^{p^mr} What is C_{p^m}^{p^mr}? A choice coefficient (p^m*r choose p^m, perhaps)? Because my first impulse was to say that that did not make sense, since C is the centralizer. Anyway: If you have a set with R elements, then the number of subsets with S elements it has is R choose S; that's the definition of R choose S: the number of ways of choosing S elements out of R possible you can prove that R choose S is equal to R!/(S!)(R-S)!. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Higgs Field and Conformal Group? bcc On Charles A. Muses (1919-2000) and Hyperspace: gave at the Parapsychology Research Group in San Francisco, December 1, 1986. The Parapsychology of Time In 1967 quantum physicist and Nobel laureate Eugene P. Wigner predicted a convergence of psychology and physics, in his book *Symmetries and Reflections*. For some years Dr. Charles Muses has investigated both consciousness and higher dimensionality. Specifically, he has worked with sphere-packing in higher dimensions by means of hyper number, This would seem to be relevant of lattice models for the bosonic dimensions, what about the super-symmetry fermionic dimensions of M-theory? some properties of which he will discuss in the parapsychological context. He has done experimental work in states of higher awareness and has proposed that the voluntary calling forth of such such states is a keystone in coming stages of human evolutionary advance. Dr. Muses received his Ph.D. in philosophy from Columbia University. He has also studied mathematics, cybernetics, and physics. He has travelled widely in his investigations of shamanistic and other spiritual practices. He edited the *Journal for the Study of Consciousness* (1968-1973) and was co-editor with Arthur Young of the book, *Consciousness and Reality* (1972). In 1985 he published the books *Destiny and Control in Human Systems* and *The Lion Path*. I met Charles Muses in October 1973 at the Institute for the Study of Consciousness in Berkeley, CA. This was when the Institute was first being set up by Arthur Young, who had asked me to be his research associate. See The Star Gate Conspiracy by Picknett & Prince. Again one cannot take all in that book on face value uncritically. Young probably knew Eugene McDermott - see my book Destiny Matrix. Did they know Leo Strauss at the University of Chicago? Charles was very keen on the hyperspace idea. Arthur was not so happy with going beyond the 4 dimensions of spacetime. Arthur encouraged me to read Eddington's works and finite group theory. In particular, Arthur wanted me to work out the symmetry group of the tetrahedron. I did this in early 1974, and found that this group T has 12 elements of rotations, but when reflections are allowed this is enlarged to the full 24 element group consisting of all permutations of 4 things (also called the symmetric-4 group S4). This is equivalent to the Octahedral group O of rotations of the octahedron (and its dual the cube). In process of generating the group table for O (= S4), I found that 6 cosets of K4 (the Klein-4 group) within S4, seemed to map perfectly in S4 (the number of elements in the classes being 1, 3, 8, 6, and 6). The first 3 classes correspond to the T (=A4) subgroup of 12 even permutations Moreover the odd permutations consisting of the two remaining classes map onto 3 families of fermions, with two quarks and two leptons in each family). I played a catalytic role here. I forget exactly what happened then? Saul-Paul please put in the details. Was I playing off your idea of the Klein group and some wacky Cabala analogy having to do with Carlo Suares back then in the 70's. I recall I was playing with the Young Diagrams from H. Weyl's Theory of Groups in Quantum Mechanics and his point about the finite group Sn skeleton to all the continuous Lie Groups. I saw that the column diagrams were obviously fermions, the rows bosons and in-between suggested parastatistics. You saw that the cosets of S4 fit the standard model with 3 generations. That was a nice insight. This was all finite group modeling, yet I was very intrigued by Muses's work on hyperspace, especially his paper, The Amazing 24th Dimension, published in the *Journal for the Study of Consciousness* (which he edited for Arthur Young), Vol. 5, No. 1 (1972), pp. 82-24. There Muses describes the packing of spheres in 24 dimensional space, via the Leech Lattice. There are 196,560 spheres packed around a central sphere in this maximally dense packing. Since S4 is a 24 element group, I thought there might be a connection between this group and the magical 24-d space. So my next move was to construct the 24-d group algebra of S4. This is done by regarding the 24 elements of S4 as vectors in a vector space. Moreover, if we make this a complex vector space, then via various theorems (such as Wedderburn's) we can see that C[S4] = M1 + M2 + M3 + M3 + M1 where C[S4] is the complex group algebra of S4, and Mn is the total matrix algebra of n x n complex matrices. Then the unitary elements within C[S4] is of necessity the 24-d unitary Lie group: U = U(1) x U(2) x U(3) x U(3) x U(1) which we can rewrite as T4 x U(1) x SU(2) x SU(3) x SU(3), and thus we and extra SU(3) which may function as a hyperweak gauge group acting on the three family labels -- thus breaking CP symmetry -- but this is speculative. Remind me what is T4? We are still a long way from the Leech lattice in 24-d vector space that Charles Muses described as important for both physics and consciousness. You were working for the consciousness side from Piaget's idea that K4 described some of the functional aspects of learning in kids. Is that correct? In that case, the group structure was pure phenomenology from the empirical study of certain psychological tests. Correct? I mean no physics model here as yet - except by some kind of formal analogy. There is, of course, one force missing in the above unification -- gravity. There you need Diff(4) from breaking the 4-parameter translation subgroup of the 10 parameter Poincare group. Gravity as curved spacetime is a broken symmetry of global translational invariance. The Poincare group, in turn, is from breaking I suppose a 5-parameter subgroup of the 15 parameter conformal group of the light cone used in Penrose's twistors. One parameter is dilation. I forget - what are the other 4? What is the physical interpretation of the 5 additional generators of the Conformal group Lie algebra? 4-momentum from translation sub-group and space-time rotations from 6-parameter Lorentz sub-group. What are the others? Shipov in his torsion theory seems to want to break the symmetry of the 6 Lorentz group parameters to get the torsion twists generalization of space-time rotations! That makes sense actually. What about those 5 conformal generators? Starting from 15 parameter conformal group of the light cone, C(15) = ?(5) T(4)O(1,3) Is ?(5) = D(1)?(4)? D(1) a one parameter dilation whose breaking of symmetry introduces the rest mass m into E^2 = (mc2)^2 + (pc)^2 mass shell What are the 4 missing generator charges to add to 4-momentum and space-time rotations? Breaking symmetry in this context means that the vacuum does not have the symmetry of the dynamical action. I mean More is different spontaneous symmetry breaking not dynamical symmetry breaking by adding weird terms to the action of the local field theories. When you locally gauge you restore the symmetry of the action, though not the vacuum. 4-momentum local gauging -> Einstein gravity compensating dynamical field that also spontaneously breaks global flatness of the vacuum, i.e. tidal tensor curvature not locally eliminated the way the non-tensor g-force connection for parallel transport can be. Lorentz group local gauging -> Shipov torsion compensating field adds the dislocation twist to space-time. What about the dilatation and the 4 other things? What are they physically? They should already be there in global special relativity without gravity hidden as rest masses of quantum fields in some way? OK in my theory, gravity comes from ripples in Goldstone phase, and exotic vacuum dark energy/matter come from ripples in Higgs field where Higgs Field(x) e^i(Goldstone Phase(x)) Higgs Field = |PSI| So dilation is |PSI| -> constant |PSI| This is physical here since |PSI|^2 is the density of virtual electron-positron bound states in the vacuum condensate. Dilation symmetry is broken here. The 4 other charge generators of the conformal group, must, therefore, be the 4-gradient of the Higgs Field! That sort of makes physical sense. Will there be a compensating gauge field? to be continued === Subject: (sample data attached)Re: how to do data mining/curve fitting to explore underlying math model for a bunch of variables? > What I am trying to do is not just polynomial curve. > I attach one curve as image file in this email. That curve shows a > relatioship between y and x, at fixed w and z. If I vary w and z, I got a > family of such curves. They are so beautiful that use high-order polynomial > to fit them may be not enough. I guess the curves can be described a better, > clearer, more intuitive, and more elegant math model. > The problem is how to find such underlying math model elegantly? That's why > I put in a bunch of varables y=function(x, w, z), and let an algorithm to > find their elegant relationship for me. But the question is: is there such a > data-mining kind of algorithm? > -Walala Could you please help me take a look at this? when z=1.7527e+004, w= 0.3536 0.0077 0.0072 0.0077 0.3536 0.0077 0.0072 0.0077 w is a parameter vector, the values here are used as a whole, they are not data points. maybe I should extract some feature to use one value to represent the whole vector in order to avoid parametrical vector curve fitting... oh, I do have a feature value for this parametric vector: 0.9314. x: 8.9289 3.8078 2.2907 1.5374 1.0899 0.8161 0.6354 for the above x(7 data points), w(one scalar feature value, or one parameter vector), z values(one scalar), we got y(7 data points): y: 0.0715 0.0176 -0.0078 -0.0225 -0.0350 -0.0430 -0.0496 ---------------------------------------------------------- when z=1.6135e+004, w=0.3536 0.0076 0.0072 0.0076 0.3536 0.0076 0.0072 0.0076 I also have a one-value feature extracted value for this parameter vector: 0.9827. x(7 data ponits)= 6.3795 1.8693 0.9923 0.6455 0.4612 0.3494 0.2753 we got y(7 data points): 0.0680 -0.0131 -0.0301 -0.0449 -0.0504 -0.0506 -0.0578 So you see the family curves (x, y) are dependent on values of w and z. And each (x, y) has very nice shape as I have attached in previous image. Now what mathematical approach should I take to model the relationship between y=functin(x, w, z)? -Walala === Subject: (Sample data attached!)Re: (image attached: curve fitting) how do you curve-fit the following image/curve? > Please help me to use as simple math model to model the following curve? > (please see attached images) > How to do that? > -walala > Much more useful would be a table of those (x,y) points that the > curve is expected to pass through, one table for each of the two > cases. > There are standard statisitical methods for finding the best of a > family of curves passing through a given set of points. I've attached some data, could you please take a look for me? Could you please help me take a look at this? when z=1.7527e+004, w= 0.3536 0.0077 0.0072 0.0077 0.3536 0.0077 0.0072 0.0077 w is a parameter vector, the values here are used as a whole, they are not data points. maybe I should extract some feature to use one value to represent the whole vector in order to avoid parametrical vector curve fitting... oh, I do have a feature value for this parametric vector: 0.9314. x: 8.9289 3.8078 2.2907 1.5374 1.0899 0.8161 0.6354 for the above x(7 data points), w(one scalar feature value, or one parameter vector), z values(one scalar), we got y(7 data points): y: 0.0715 0.0176 -0.0078 -0.0225 -0.0350 -0.0430 -0.0496 ---------------------------------------------------------- when z=1.6135e+004, w=0.3536 0.0076 0.0072 0.0076 0.3536 0.0076 0.0072 0.0076 I also have a one-value feature extracted value for this parameter vector: 0.9827. x(7 data ponits)= 6.3795 1.8693 0.9923 0.6455 0.4612 0.3494 0.2753 we got y(7 data points): 0.0680 -0.0131 -0.0301 -0.0449 -0.0504 -0.0506 -0.0578 So you see the family curves (x, y) are dependent on values of w and z. And each (x, y) has very nice shape as I have attached in previous image. Now what mathematical approach should I take to model the relationship between y=functin(x, w, z)? -Walala === Subject: Re: (image attached: curve fitting) how do you curve-fit the following image/curve? walala (please see attached images) If you have a recent version of Maple you can easily obtain several different descriptions of the curves. But you must provide a list of pairs (x,y) which lie on or near the curve. If you send me a list of such pairs I can compute the equations for you if you don't have maple. If you have Maple 8 or 9 execute the commands: with(CurveFitting); Interactive([[x1,y1],..., [xn,yn]]); (where [x1,y1],..., [xn,yn] are points on the curve. This will bring up a window which will give you several options for fitting the curve. --Edwin === Subject: Not the words 1:2 --> 1/2=0,5--> density on 1, corret 0,1/0,2= 0.5 density 0n 0,1, not corret 0,1/0,2=0,05, density on 1, corret 0,1/2 =0,05, density on 0,1, not corret 0,1/2=0,005, density on 1, corret. -- Andrea Sorrentino, libero Ricercatore Web: http://digilander.libero.it/socratis === 3 formats are now online for my talk re: John Archibald Wheeler's ideas and the New Cosmology What Is The Universe Made Of?. The version I uploaded a few minutes ago is 99% of what I will present Friday Oct 24 from Sirag. http://qedcorp.com/APS/StarGate1.mov a self running movie in Quick Time. Use your Quick Time Viewer to directly access the URL for instant streaming on DSL- also you can open it to 2x screen size. http://qedcorp.com/APS/AustinTx102403.mov is point and click interactive slide show vesion in Quick Time http://qedcorp.com/APS/WheelerAustin.ppt is Power Point made in MAC but I think Windows will read it? Photos of Wheeler and Feynman never seen before are included. === Subject: symmetries of a rectangle Is the set of symmetries (rotations, reflections, with the identity) of a non-square rectangle in the plane a group under composition? I think it's an abelian group (just by inspection)... === Subject: Re: symmetries of a rectangle > Is the set of symmetries (rotations, reflections, with the identity) of a > non-square rectangle in the plane a group under composition? The set of symmetries of any object whatsoever forms a group under composition. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: symmetries of a rectangle > Is the set of symmetries (rotations, reflections, with the identity) of a > non-square rectangle in the plane a group under composition? > I think it's an abelian group (just by inspection)... No, the result of rotation by 180 degrees (pi radians) and reflection about a diagonal don't commute. Norm === Subject: Re: symmetries of a rectangle > Is the set of symmetries (rotations, reflections, with the identity) of a > non-square rectangle in the plane a group under composition? > I think it's an abelian group (just by inspection)... > No, the result of rotation by 180 degrees (pi radians) and reflection > about a diagonal don't commute. > Norm A non-square rectange is not symmetric under reflection through a diagonal. It can only be reflected horizontally or vertically or both, which is the same as pi-rotation. So the symmetry group should be isomorphic to (Z/2Z)+(Z/2Z). Have a tolerable existence. Eli Have a tolerable existence. Eli === Subject: Re: treatment of Lebesgue integral There are many ways to develop Lebesgue integral, and one way is as > follows: f is lebesgue integrable if there exists a sequence {f_n} of step >function > such that (a) sum int |f_n| < infty > (b) f(x) = sum f_n(x) for every x such that sum |f_n(x)| < infty Then the integral of f is defined as int f = sum int f_n My question is: does anyone know any book that introduces Lebesgue > integral in this way? >No it doesn't look right. The function: >f(x) = 1 if 0f(x) = 0 otherwise >is lebesgue integrable. > And this function satisfies the condition above: > Let the rationals in (0,1) be r_1, r_2, ... . Let f_n(x) = 0 > except for f_n(r_j) = 1, 1 <= j <= n. Then f_n is a step > function, >Is it? I would take the following or something very similar as a definition >of a step function. A function f on (0,1) is a step function if there are a >partition P = {0and constants r_n in R n=0,1,...N such that >f(s)=r_0 for s<=t_1 >f(s)=r_n for t_nf(s)=r_N for s>t_N >In otherwords it's caglad or of course it could be cadlag if you alter the >inequality. Either way you must have continuity on either the right or left. Well, that's not what _I_ would take as the definition of step function... This all seemed very familiar, so I took the drastic step of looking it up when I got to the office. In the textbook for a class I took many years ago, Asplund & Bungart A First Course in Integration, the definition of step function includes the things I called step functions above. And the fact that a function is Lebesgue integrable if and only if it satisfies (a) and (b) above is exactly Lemma 2.1.10 (they say it's due to Mikusinski). ************************ David C. Ullrich === Subject: Re: treatment of Lebesgue integral There are many ways to develop Lebesgue integral, and one way is as > follows: f is lebesgue integrable if there exists a sequence {f_n} of step >function > such that (a) sum int |f_n| < infty > (b) f(x) = sum f_n(x) for every x such that sum |f_n(x)| < infty Then the integral of f is defined as int f = sum int f_n My question is: does anyone know any book that introduces Lebesgue > integral in this way? >No it doesn't look right. The function: >f(x) = 1 if 0f(x) = 0 otherwise >is lebesgue integrable. > And this function satisfies the condition above: > Let the rationals in (0,1) be r_1, r_2, ... . Let f_n(x) = 0 > except for f_n(r_j) = 1, 1 <= j <= n. Then f_n is a step > function, >Is it? I would take the following or something very similar as a definition >of a step function. A function f on (0,1) is a step function if there are a >partition P = {0and constants r_n in R n=0,1,...N such that >f(s)=r_0 for s<=t_1 >f(s)=r_n for t_nf(s)=r_N for s>t_N >In otherwords it's caglad or of course it could be cadlag if you alter the >inequality. Either way you must have continuity on either the right or left. > Well, that's not what _I_ would take as the definition of step > function... Well, constant of half open intervals is probably a less clumsy way of putting it. > This all seemed very familiar, so I took the drastic step of looking > it up when I got to the office. In the textbook for a class I took > many years ago, Asplund & Bungart A First Course in Integration, > the definition of step function includes the things I called > step functions above. I've never come across a definition of step function which would allow your example. Interestingly I've just come across http://www.nmt.edu/~iavramid/notes/lebesg.pdf which appears to be using the OP definitions, with my definition of step function! The counterexample I gave is Lebesgue integrable, yet I'm sure there's no way to get it as a (countable) sum of step functions. Even if there was I could choose an uncountable Lebesgue measurable set rather than the rationals, too complex to be obtained via the same method (Cantor set maybe). The cardinality of the set of all Borel sets is less than the cardinality of the set of all Lebesgue measurable sets (c and 2^c respectively). This is what makes me think intuitively that (b) can't work, the cardinality of all step functions is c (I think) and of all Lebesgue measurable functions its 2^c so not every Lebesgue measurable function can be expressed as a sum of a countable number of step functions. Something is wrong here, I wonder, are they calling it Lebesgue integrable when they mean Riemann integrable? >And the fact that a function is Lebesgue > integrable if and only if it satisfies (a) and (b) above is exactly > Lemma 2.1.10 (they say it's due to Mikusinski). > ************************ > David C. Ullrich === Subject: Re: treatment of Lebesgue integral >There are many ways to develop Lebesgue integral, and one way is as >follows: >f is lebesgue integrable if there exists a sequence {f_n} of step function >such that >(a) sum int |f_n| < infty >(b) f(x) = sum f_n(x) for every x such that sum |f_n(x)| < infty >Then the integral of f is defined as int f = sum int f_n >My question is: does anyone know any book that introduces Lebesgue >integral in this way? It's not quite the way the integral is _introduced_, but the approach you have in mind appears in Asplund & Bungart A First Course in Integration: The characterization of integrable functions above is exactly Lemma 2.1.10 in that book (credited to Mikusinski). ************************ David C. Ullrich === Subject: Re: treatment of Lebesgue integral > Tien-Chien Chen > There are many ways to develop Lebesgue integral, and one way is as > follows: > f is lebesgue integrable if there exists a sequence {f_n} of step function > such that > (a) sum int |f_n| < infty > (b) f(x) = sum f_n(x) for every x such that sum |f_n(x)| < infty > Then the integral of f is defined as int f = sum int f_n > My question is: does anyone know any book that introduces Lebesgue > integral in this way? > Do you mean sup rather than sum? > I don't know a book that starts from there, but I think it can be worked out > by going through the dominated convergence theorem, and a couple of related > results, in reverse. I don't think so. The function: f(x) = 1 if 0 hello every one > i need to find information about Monte Carlo simulation for protein folding > can anyone give me the algorithem === Subject: Re: monte carlo algorithm Ken Dill, Skolnick, James MacCammon, Hagai Merovitch, Harold Shegara, Martin Karplus, and Michael Levitt are a few of the top people in the field of computational protein folding analysis that have used Monte-Carlo algorithms. > hello every one > i need to find information about Monte Carlo simulation for protein folding > can anyone give me the algorithem === Subject: Re: monte carlo algorithm you can start with these two books which are well known Monte Carlo Methods in statistical physics editor K Binder and computer simulations of liquids by M P Allen and D J Tildesley they are yours if you want them but I really don't want to pay shipping costs and I am not joking, it's your lucky day today I guess. So if you want them, find a way to absorb the cost of shipping mahfoud errifi@golden.net > hello every one > i need to find information about Monte Carlo simulation for protein folding > can anyone give me the algorithem === Subject: Re: monte carlo algorithm > hello every one > i need to find information about Monte Carlo simulation for protein folding > can anyone give me the algorithem Protein folding is one of the hardest problems around. If you mean you want to be part of the group that lets their computer be used as part of a multi-computer simulation go here: http://www.stanford.edu/group/pandegroup/folding/ Bill === Subject: Julian day formula Hi everyone I was wondering if anyone out there could walk me through a formula for calculating the Julian day (JD) number for a particular Gregorian days in the year + leap year every 4 years) and that it is the number of days since Jan 1 4713 BC. I have seen a few formulae on the net but I couldn't follow them properly. I tried my own method of taking the difference in years, subtracting one (b/c there was no year zero) and multiplying by 365.25 but this is consistently out by ~60 days. Any help would be gladly appreciated. This has been driving me nuts for days! === Subject: Re: Julian day formula > Hi everyone > I was wondering if anyone out there could walk me through a formula > for calculating the Julian day (JD) number for a particular Gregorian > days in the year + leap year every 4 years) and that it is the number > of days since Jan 1 4713 BC. Actually, that's a misconception. The Julian day number is not the same as the Julian date, and has no particular connection to it. A Julian date is a date on the Julian calendar. Similarly, a Gregorian date is a date on the Gregorian calendar. Either type of date can be used to compute a Julian day number; there is a slight adjustment at one step in the algorithm. By the way, your question is incompletely specified, since you did not > I have seen a few formulae on the net but I couldn't follow them > properly. I tried my own method of taking the difference in years, > subtracting one (b/c there was no year zero) and multiplying by 365.25 > but this is consistently out by ~60 days. It's much more complicated than that. Here is a function that I have used. double julian_day(struct tm *clock) { int a, b, month, year; double day; month = clock->tm_mon + 1; year = clock->tm_year + 1900; if (month < 3) { month += 12; year--; } if (is_gregorian(year,month,clock->tm_mday)) { a = year / 100; b = 2 - a + a/4; } else b = 0; day = clock->tm_mday + dayfraction(clock); return (1461*(year+4716))/4 + (153*(month+1))/5 + day + b - 1524.5; } This code depends on a function is_gregorian which determines whether the Gregorian calendar or the Julian calendar should be applied. There is also a function dayfraction which corrects for the time of day, correspond to 0.5, because half the day has elapsed. In case you are not familiar with the C library, clock->tm_mday is the day of the month (1-31). You can ignore the part about clock->tm_mon and clock->tm_year, and just use month (1-12) and year as indicated. > Any help would be gladly appreciated. This has been driving me nuts > for days! See if your library has a copy of _Astronomical Algorithms_ by Jean Meeus. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Julian day formula > Hi everyone > I was wondering if anyone out there could walk me through a formula > for calculating the Julian day (JD) number for a particular Gregorian > days in the year + leap year every 4 years) and that it is the number > of days since Jan 1 4713 BC. > I have seen a few formulae on the net but I couldn't follow them > properly. I tried my own method of taking the difference in years, > subtracting one (b/c there was no year zero) and multiplying by 365.25 > but this is consistently out by ~60 days. > Any help would be gladly appreciated. This has been driving me nuts > for days! http://aa.usno.navy.mil/ with explanation. Jaap === Subject: Re: tangential quadrilateral First, Ptolemy's theorem on cyclic quadrilaterals goes in the wrong direction. It assumes that there is a circle passing thru the three vertices of the quadrilateral. Brian Hutchings proof seems to be correct and is very nice, though it is not obvious to me why the following is true: If three angle bisectors of a quadrilateral, D, intersect in a point, P, then a circle can be inscribed in D. Of course P would be the center of such a circle. > If a circle can be inscribed in a convex quadrilateral, R, in the xy > plane, then R is called a tangential quadrilateral. It is well known, > and not hard to show, that if R is a tangential quadrilateral, then a + > c = b + d, where a and c, and b and d denote the lengths of opposite > sides of R. It is noted in John W. Harris and Horst Stocker: Handbook of > Mathematics and Compuitational Science, Springer, New York, 1998, but > not proved, that a + c = b + d is also sufficient for R to be a > tangential quadrilateral. I might be able to grind out a proof myself, > but I was wondering if anyone knows if the details of a proof have been > published anywhere. === Subject: Re: tangential quadrilateral > First, Ptolemy's theorem on cyclic quadrilaterals goes in the wrong > direction. It assumes that there is a circle passing thru the three vertices > of the quadrilateral. Brian Hutchings proof seems to be correct and is very > nice, though it is not obvious to me why the following is true: If three > angle bisectors of a quadrilateral, D, intersect in a point, P, then a > circle can be inscribed in D. Of course P would be the center of such a > circle. I think you confuse Brian's message with mine. Or perhaps I son't see any message. In any case, if three angle bisectors of a quadrilateral, D, intersect in a point 'I', then 'I' is at the smae distance from the four edges, and it is possible to inscribe a circle in the quadrilateral. Remember that the points of an angle bisector are at the same (perpendicular) distance from both sides of the angle. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com > If a circle can be inscribed in a convex quadrilateral, R, in the xy > plane, then R is called a tangential quadrilateral. It is well known, > and not hard to show, that if R is a tangential quadrilateral, then a + > c = b + d, where a and c, and b and d denote the lengths of opposite > sides of R. It is noted in John W. Harris and Horst Stocker: Handbook of > Mathematics and Compuitational Science, Springer, New York, 1998, but > not proved, that a + c = b + d is also sufficient for R to be a > tangential quadrilateral. I might be able to grind out a proof myself, > but I was wondering if anyone knows if the details of a proof have been > published anywhere. === Subject: Re: Kepler's Equation (was: How do you do this integral?) > Now you can see why an attempt to solve for theta, > even when reduced to an attempt to solve for u, > is just as hard as solving Kepler's equation which has been studied in great detail. >Has anybody ever done detailed numerical studies of it directly, and not >via the brute force methods of pushing f=ma and dh/dt=0 ( h = angular >momentum )? Have the solutions ever been tabulated and treated as known >functions? > If you insist on a formula for E in terms of > M and e, (0 < e < 1), The Encyclopedic > Dictionary of Mathematics offers a Fourier series > E = M + sum[n=1 to inf] (2/n)*J_n(n*e) * sin(n*M) where J_n is the Bessel function of order n. > In practice, one uses numerical methods to > approximate E. >Has anybody ever done a study of how quickly this expansion converges, as a >function of eccentricity? See the book Solving Kepler's Equation over Three Centuries by Peter Cowell (1993). -- John E. Prussing University of Illinois at Urbana-Champaign Department of Aerospace Engineering http://www.uiuc.edu/~prussing === Subject: norm equivalence is it true that l_p(N), and l_q(N) norms on f:N->R are not equivalent? as in, sup ||f||_p/||f||_q over f, pis it true that l_p(N), and l_q(N) norms on f:N->R are not equivalent? as in, >sup ||f||_p/||f||_q over f, p This is a little like what I discuss at these threads: > Let A(0) = {1} = {a(0)}; > Let A(n+1) = {A(n),-a(n),A(n)}, > for all n >= 0, > where a(n) is the n_th term in each sequence A where a(n) is defined. > (the A's are each a sequence formed by concatenation.) > So, we have: > A(4) = > 1, -1, 1, 1, 1, -1, 1, -1, 1, -1, 1, 1, 1, -1, 1, -1, 1, -1, 1, 1, 1, > -1, 1, -1, 1, -1, 1, 1, 1, -1, 1 > (For each m, A(m) has 2^(m+1)-1 terms, a(0) to a(2^(m+1)-2).) > So, if we let > F_m(y) = sum{k=0 to 2^(m+1)-2} a(k) y^k > (and F(y) = limit{m-> oo} F_m(y)), > then: > F_m(y) = > F_{m-1}(y) (1 + y^(2^m)) - a(m-1) y^(2^m -1), > and so > F(y) - 1 = > sum{k=1 to oo} (F_{k-1}(y) y - a(k-1)) y^(2^k -1), > and therefore > F_{2^m-1}(1) + F{m-1}(1) - 1 = > sum{k=0 to 2^m-2) F_k(1). > (all of above right?) > So, what exactly is the closed-form for F(y)?? I believe (but am not absolutely certain) that a(m-1) = -(-1)^c(m), where c(m) = the number of e()'s where: e(e(e(...e(m)...))) = 0, and e(m) is a nonnegative integer such that: 2^e(m) is the highest power of 2 dividing m. I discuss this c-sequence in the first link I give above. (Is there anything related to these sequences yet in the Encyclopedia of Integer Sequences?) Leroy Quet === Subject: Eigenfunctions of the Fourier transform I heard an interesting talk today from a guy who is searching (so far unsuccessfully) for a new class of eigenfunctions of the Fourier Transform, of a certain type. I was aware that gaussians transform to gaussians, so that exp(-ax^2) transforms to itself for suitable a. What I didn't realize was that functions of the form (Hermite polynomial)*(exp(-ax^2/2)) are also eigenfunctions. Apparently that is a consequence of a certain differential operator (d^2/dx^2 - x^2) commuting with the Fourier Transform operator. My questions for the newsgroup: Is anyone aware of other classes of eigenfunctions of the Fourier Transform? Are there other known operators that commute with the Fourier Transform? Especially is there a characterization of any such operator? - Randy === Subject: Re: Eigenfunctions of the Fourier transform >I heard an interesting talk today from a guy who is searching (so far >unsuccessfully) for a new class of eigenfunctions of the Fourier >Transform, of a certain type. I was aware that gaussians transform to >gaussians, so that exp(-ax^2) transforms to itself for suitable a. >What I didn't realize was that functions of the form (Hermite >polynomial)*(exp(-ax^2/2)) are also eigenfunctions. >Apparently that is a consequence of a certain differential operator >(d^2/dx^2 - x^2) commuting with the Fourier Transform operator. Known to physicists as the Hamiltonian of the harmonic oscillator. >Is anyone aware of other classes of eigenfunctions of the Fourier >Transform? The Fourier transform F on L^2(R) (with the normalization that makes it unitary) satisfies F^4 = I, so its eigenvalues are fourth roots of unity. For any f in L^2, g1(x) = f(x) + F(f)(x) + f(-x) + F(f)(-x), g2(x) = f(x) - i F(f)(x) - f(-x) + i F(f)(-x), g3(x) = f(x) - F(f)(x) + f(-x) - F(f)(-x), g4(x) = f(x) + i F(f)(x) - f(-x) - i F(f)(-x) are eigenfunctions for 1, i, -1 and -i respectively. >Are there other known operators that commute with the Fourier >Transform? Especially is there a characterization of any such >operator? All operators that leave the four eigenspaces invariant. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Racist Nonsense <3F82F182.992FEC19@hate.spam.net> <3F832520.71D6778@hate.spam.net> <3f89f614$40$fuzhry+tra$mr2ice@news.patriot.net> <1482GpA5WNj$EwIC@aglow.demon.co.uk> <3f8fd502$6$fuzhry+tra$mr2ice@news.patriot.net> Shmuel (Seymour J.) Metz What would you call it? The ONLY claims for high IQ I have seen where >particular jewish groups have been mentioned have been for Askenasi >Jews. >That say more about you than anything else. For a long time the >cultural centers of the Jewish world were Spain and the Babylon. The >Ashkenazi communities were considered to be backwaters. My post was not about ancient history, but what seemed to be happening in newsgroups now. It`s not about old arguments, it`s about hopefully making people think before they start new ones. >But, using the kiss principle: >The KISS principle in Biology lead to a lot of incorrect taxonomy. Simplicity is a goal in all research, hence eg the search for a unified theory. The most complex theory is susceptible to one anomalous piece of data. Seems to me that`s what I`ve provided. > There seems to be agreement here that the scale of those changes is >the result of forced evolution. >If true, the same arguments would apply for all three communities. Are you disagreeing with forced evolution? Things would go a bit quicker if we agree on what is common ground and what isn`t, if we have to discuss it, and I don`t, particularly. I have said pretty much all I wanted to say. >Where there are problems is that the `selective bell curve culling` >theory delibrately ignores the craniofacial data. >How so? An influx of new genomes does not stop the forces of natural >selection. The new genes are as subject to culling as the old ones. Yes, but even here if the `culling` (how i hate that word) was significantly selective (not at all easy to show, I suspect) then it might tend to select those with more host genes. >One has to account for the apparent startling quoted differences in >IQ between the Askenasim and other jews. >Whom are you quoting? I'm not aware of any scientific study showing a >difference in IQ between Ashkenazi Jews and either Sephardim or >Oriental Jews. Read my original message - it is the wording of some claims in current newsgroups I am complaining about, and the very claims themselves. Going into the IQ`s of various groups in the way it`s being done in these newsgroups is what I object to. Elitism is poisonous. It killed millions within living memory. Particularily as I have seen no evidence to support, eg, that jews have any significant IQ advantage over other groups. If you have any decent evidence handy to support these elitist claims (and they ARE elitist, and breed anti-Semitism), eg that there is a significant gap between the Jews and the Germans and the Poles and the people of Connecticut, well, ok, we could look at it. I didn`t get any replies to my request for this in my original message. >Skin color would be pretty >far down on the list of what I'd consider relative. Definite agreement on that. -- Jan === Subject: How I Divided by Zero, and Lived to Tell About It! 2001. The only significant change is that two now-nonfunctioning links in the original have been replaced with functioning links.] Contents: Introduction and History A Construction of a Nonnegative Extended Real Number System A Relation Between Arithmetic on [0,+oo] and Floating-Point Arithmetic Should +oo Be Considered to Be a Number? Avoiding Paradoxes Uses of Division by Zero, the Projective Extension of R, etc. Introduction and History: In this newsgroup, we often have question about division of a nonzero quantity by zero. Consideration of this question goes back at least to Aristotle (in his _Physics_). The idea that a nonzero quantity divided by zero yields infinity was perhaps first made explicit in 1152 by Bhaskara. The infinity symbol, resembling an 8 turned sideways, which is most commonly used nowadays was introduced in 1657 by John Wallis. (Not having that symbol available here, I shall use oo as a substitute.) Wallis stated plainly that 1/0 = oo. This idea is alive today in the minds of many people, some of whom regard oo as a number, albeit a most unusual one. On the other hand, taking a/b to be that number which, upon multiplication by b, yields a, Martin Ohm concluded in 1828 that, If a is not zero, but b is zero, then the quotient a/b has no meaning. This opposing idea, that division of a nonzero quantity by zero must be meaningless, is also alive today in the minds of many people. Among them, it is common to assert that oo is not a number and that, when oo is used in analysis, it should never be taken to indicate anything more than a limiting process. notion of division can be extended so that the quotient of a positive quantity by zero has meaning, that quotient being equal to +oo, which will be defined as a specific mathematical object. A Construction of a Nonnegative Extended Real Number System: by zero can be meaningfully interpreted in the context of a system of actually have been A Construction of [0,+oo] from Q+ Using Equivalence Classes of Cauchy Sequences.) I shall assume that readers are familiar with the construction of R from Q using equivalence classes of Cauchy sequences and, as such, will give only a bare bones presentation of the construction of [0,+oo] here. (Interested readers who are unfamiliar with such constructions may consult appropriate texts in real analysis. Unfortunately, the information on this topic at the FAQ site for this newsgroup is too meager to be of much use.) We shall assume that the system Q of rationals, together with appropriate relations and operations, has already been developed. (Of particular interest will be reciprocation as a unary operation on Q+, the positive rationals.) If we were then to proceed to develop equivalence classes of Cauchy sequences, as is often done in constructing R from Q, but instead using only Cauchy sequences of _positive_ rationals, the construction could be considered as giving the nonnegative reals, [0,+oo). For our purpose, however, we must have exactly one more equivalence class, that corresponding to +oo itself. There are at least two ways that this can be accomplished. The naive method would be simply to adjoin, to the existing set of equivalence classes of Cauchy sequences of positive rationals, another equivalence class, +oo, consisting of those positive rational sequences which increase without bound. Obviously, such sequences are not Cauchy -- in the usual metric, that is. This thought, however, leads us to a more sophisticated method of accomplishing our goal: Consider a different metric, such as d(x,y) =| x/(1+x) - y/(1+y) |, under which there is exactly one more equivalence class, +oo, of Cauchy sequences of positive rationals than under the usual metric. Although it is of no great significance whether we use the naive or the more sophisticated method, we shall use the latter, if for no other reason, so that we may legitimately refer to all of the sequences of positive rationals in our equivalence classes as being Cauchy. Thus, nonnegative extended reals will be regarded here as equivalence classes of Cauchy sequences of positive rationals. The binary operation of addition on [0,+oo] is easily defined: If (a_n) and (b_n) are members of equivalence classes x and y, resp., then x + y is the equivalence class having (a_n + b_n) as a member. Note in particular that it follows that x + (+oo) = +oo = +oo + x for all x in [0,+oo]. Although multiplication is also easily defined, it is, unfortunately, not a binary operation on [0,+oo]: Let (a_n) and (b_n) be members of equivalence classes x and y, resp. If the set of all (a_n * b_n) is an equivalence class in [0,+oo], then x * y is that equivalence class; otherwise, x * y is undefined in our system. Not surprisingly, the only undefined products are 0 * (+oo) and +oo * 0. Also note in particular that it follows that, for all x in (0,+oo], x * (+oo) = +oo = +oo * x, a property bearing a nice symmetry with the property that, for all x in [0,+oo), x * 0 = 0 = 0 * x. Addition is associative and commutative; and multiplication (when defined) is associative and commutative, and it distributes over addition. [Cf. the brief discussion of Arithmetic in [0,oo] by Walter Rudin in his _Real and Complex Analysis_ (New York: McGraw-Hill, 1966), third edition, 1987.] Reciprocation is easily defined as a unary operation on [0,+oo]: If (a_n) is a member of equivalence class x, then 1/x is the equivalence class having (1/a_n) as a member. Note in particular that, just as in Q+, reciprocation is equivalent to multiplicative inversion for nonzero finite x in our system. Furthermore, it follows that 1/0 = +oo and 1/(+oo) = 0, and reciprocation is an involution on [0,+oo]. Recalling that the products of 0 and +oo are undefined, these two elements do not have multiplicative inverses. Thus, for the zero and infinite elements of our system, reciprocation does not happen to be equivalent to multiplicative inversion. Division may then be defined in terms of multiplication and reciprocation: The quotient x/y is the product x * (1/y) if that product is defined; otherwise, the quotient is undefined. Note that, as expected, it follows that 0/0 and +oo/(+oo) are undefined in our system. All other quotients are defined; in particular, x/0 = +oo and 0/x = 0 for all x in (0,+oo], and x/(+oo) = 0 and +oo/x = +oo for all x in [0,+oo). Negation is out of the question since we are dealing only with nonnegative values. Nonetheless, subtraction can still be defined when minuend exceeds subtrahend: Given x and y in [0,+oo] with x > y, let (a_n) and (b_n) be members of the equivalence classes x and y, resp. Since x > y, there exists N such that, for all n > N, a_n > b_n. Then the difference x - y is that equivalence class containing (c_n) where c_n = a_(n+N) - b_(n+N). We have, in particular, x - 0 = x for all positive x and +oo - x = +oo for all finite x. [Note that this definition of subtraction cannot be extended naively to cover the case when minuend equals subtrahend. It is perhaps debatable whether x - x should be undefined or defined to be 0 for finite x.] A Relation Between Arithmetic on [0,+oo] and Floating-Point Arithmetic: If floating-point arithmetic were restricted to nonnegative values, it would be similar to arithmetic on [0,+oo]. In particular, the floating-point number +0.0 closely resembles the zero in our construction of [0,+oo]. It could perhaps be argued that neither should be considered as a zero, but rather that both should be considered as positive infinitesimals. [Concerning the note at the end of the preceding paragraph: In designing floating-point arithmetic, it was an essentially arbitrary choice that, for finite x, x - x should be evaluated as +0.0, rather than as -0.0.] Should +oo Be Considered to Be a Number?: The word infinity is often used to indicate a rather broad concept. This fact, however, should not preclude considering the specific mathematical object +oo, as defined above, to be a number. It is, after all, an object of the same type as other elements of [0,+oo]. Furthermore, although the behavior of +oo is certainly unusual, it is no more unusual that of 0. Since 0 is considered to be a number, the behavior of +oo should not preclude considering it to be a number as well. [Also, for comparison, it might be noted that, according to the internationally accepted standard for floating- point arithmetic, its +Infinity is specified to be a number.] Of course, the properties of +oo are what are truly important, rather than the relatively unimportant issue of whether it is to be called a number or not. If preferred, +oo could be considered to be the sole nonnumeric element of [0,+oo]. The author, however, prefers not to make such a distinction, deeming it to be inelegant. (Perhaps a reasonable compromise stance would be to consider, symmetrically, both 0 and +oo as improper elements.) Avoiding Paradoxes: Statements to the effect that division by zero must lead to paradoxes are often encountered in the literature. However, such paradoxes typically arise when thinking, mistakenly, that division by zero must be equivalent to cancellation of factors of zero. But certainly such cancellation is not allowable in arithmetic on [0,+oo]. Cancelling factors of zero and treating 0 and 1/0 (or +oo) as being multiplicative inverses are equivalently incorrect. Understanding this, paradoxes can be avoided easily. Uses of Division by Zero, the Projective Extension of R, etc.: It is not uncommon, even for nonmathematicians in everyday life, to encounter division of a positive quantity by zero in a context in which all quantities must be nonnegative. An interesting example is discussed in the thread Percent Increase Test . In such cases, arithmetic on [0,+oo] is appropriate. In this newsgroup, questions such as Is 1/0 = oo ? are often met with derisive responses, stating, without qualification, that division by zero is impossible. On the other hand, I cannot remember a question such as Is sqrt(-1) = i ? being met with a similar response, stating bluntly that it is impossible to take square roots of negatives. Yet the two questions are quite similar. Neither 1/0 nor sqrt(-1) exists in the real number system, of course; but, merely by the fact that the questioners mentioned oo or i, it would be reasonable to presume that the questions were intended for broader contexts than just that of the reals. Extended real number systems are useful, possibly even more so, for lay calculations, than the complex number system. There are, of course, many instances in which it is inadequate to deal with just nonnegative values. Although the affine extension of R, having two signed infinities, -oo and +oo, does not allow -- in any reasonably nonproblematic way, at least -- for division by its unsigned zero, the projective extension, R U {oo}, having a single unsigned infinity, does allow for division of nonzero values by zero. This extension could be developed much as we developed [0,+oo] here, although some of the details would be slightly messier. Some applications of this extension are mentioned in . Thoughtful comments will be welcomed, especially suggestions for David W. Cantrell === Subject: Re: Basic Probability It is true even for some uncountable sets. The middle third Cantor set is an example of an uncountable set with measure zero (which means that it does not affect a proability by its removal). > If I have an experiment where I randomly draw a number from the interval > [0,1], is the probability that I pick a number in the interval [0,1) equal > to one? > Likewise, if I remove a countable number of points, will the probability of > picking in the interval be equal to one? === Subject: Re: division by zero...?? > A (potentially) stupid question :) How I Divided by Zero, and Lived to Tell About It! I hope that it will answer your question, at least in part. David W. Cantrell > What would happens if someone discovers a self-consistent new algebra > of numbers in which we could divide by zero?? revolutionnary? > interesting but not useful? === Subject: Probability Question Got an interesting question from a friend--any thoughts? Dave and Mike are going from destionation A to B. They are going their own way and methods are independent of each other, Length of time it takes Dave to reach B is assumed to have an exponential distribution with mean 1/x=5 hr. Mike, exponential distribution with mean 1/y=4 hr. a) what is the probablility Mike reaches B last, b) Given Mike reaches B last, what is the expected length of Dave' journey. c) Given Mike reaches B first, what is the expected length of Mike's journey. d) Given Mike reaches B last, what is the expected length of Mike's journey. Interesting... thoughts? === Subject: Re: Probability Question >Got an interesting question from a friend--any thoughts? I think it's an interesting homework exercise. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Antipodal Points and S^1 Let's say I have a continuous map f : S^1 ---> R (where R is the real numbers). What is the proof that there always exists an element x of S^1 such that f(x) = f(-x) regardless of the continuous map f? Justin === Subject: Re: Antipodal Points and S^1 Intermediate value theorem > Let's say I have a continuous map f : S^1 ---> R (where R is the real > numbers). What is the proof that there always exists an element x of S^1 > such that > f(x) = f(-x) regardless of the continuous map f? > Justin === Subject: Re: Simple principle, core error proven Good, with a cubic in this thread. ... > 2. The important tool I use is a polynomial: > P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) Q(m) = 49((4 m^2 - 3m) 5^3 - 3(-1 + m)5 + 7) > where m varies in the ring of algebraic integers. > Some may find it looks odd. But the entire point of that form is to > do something a little different than anyone else apparently has done > before, which is to get non-polynomial factors. > And the factorization with those factors is > P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) Q(m) = (5 c1 + 7)(5 c2 + 7)(5 c3 + 7) > where the a's are roots of the following cubic: > a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m). c^3 + 3(-1 + m)c^2 - 49(4 m^2 - 3m). > So you can see that the a's are functions of m, and if you really want > headaches you can go ahead and solve the cubic to get a view of them. > However, I can move on without needing their explicit form. > g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7) (Still that typo for a_3.) h1 = (5 c1 + 7), h2 = (5 c2 + 7), h3 = (5 c3 + 7) > and I can find those terms that are free of m by setting m=0, just > like with any polynomial, you know like with S(m) = 2(m^2 + 2m + 1), > S(0) = 2, which you can just look at and see here, but with my P(m) > it's just a little harder, so I set m=0, which gives me > P(0) = 49(3(5) + 7), Q(0) = 49(3(5) + 7), > which fits with the cubic as at m=0 it gives > a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3, c^3 - 3c^2 = 0, so c1 = c2 = 0, c3 = 3, > to show that at m=0, the three factors are > g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22. h1 = 7, h2 = 7, h3 = 3(5) + 7 = 22. > Now dividing P(m) by 49 gives > P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7 Q(m)/49 = (4 m^2 - 3m) 5^3 - 3(-1 + m)5 + 7 > and the question is what happens to the g's, but look now at P(0)/49 > as that is > P(0)/49 = 3(5) + 7 Q(0)/49 = 3(5) + 7 > as two factors of 7, each 7, have beeen divided off, which is easy to > see. > The only way that can happen is if > P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) Q(m)/49 = (5 c1/7 + 1)(5 c2/7 + 1)(5 c3 + 7) > where only two of the a's have 7 as a factor, where the idea is almost > trivially simple as consider a polynomial like > S(m) = 7(m^2 + 2m + 1) = (b_1 m + 7)(b_2 m + 1) Skipping a red herring. Here a polynomial in m is factored in polynomials in m. That is not the case with factoring P (or Q) above). > It's the same basic idea while what I have is more complicated as I'm > showing you an over hundred year old flaw, and it turns out that it > takes a somewhat complicated expression to show it which > P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7 > is, but the same principle works as just like with S(m), with P(m), > dividing out 49, affects the independent or constant terms, revealing > factors of the a's. > That is, from the distributive property, factors that are 7 must > divide through. > So now I know that the correct factorization is > P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) Q(m)/49 = (5 c1/7 + 1)(5 c2/7 + 1)(5 c3 + 7) > which is like > S(m)/7 = (b_1 m/7 + 1)(b_2 m + 1). > 3. So *two* of the a's *should* have 7 as a factor, and in fact they > do, in a proper ring, but the ring of algebraic integers has problems, > so that for certain values of m, they won't. Let's recapitulate. I have Q(m) = 49((4 m^2 - 3m) 5^3 - 3(-1 + m)5 + 7), I factorise as Q(m) = (5 c1 + 7)(5 c2 + 7)(5 c3 + 7) where the c's are roots of c^3 + 3(-1 + m)c^2 - 7^2(4 m^2 - 3m), exactly two of the c's must be divisible by 7. We check for m=0 and get c1 = c2 = 0, c3 = 3, so it is correct. We check for m=1 and get: c1 = cbrt(49) c2 = cbrt(49) * (-1 + sqrt(-3))/2 c3 = cbrt(49) * (-1 - sqrt(-3))/2 Which two are divisible by 7? Note that (-1 +- sqrt(-3))/2 are units. If c1 is divisible by 7 in a ring, cbrt(49)/7 is in that ring, but that is cbrt(1/7). So the cube of that (1/7) is also in the ring and 7 is a unit. As (-1 +- sqrt(-3))/2 are units, they are coprime to 7. So when c2 is divisible by 7, so is cbrt(49), and so is c1 (and c3). So when m=1 either all of the c's are divisible by 7, or none is. The majority prefers the latter, and remains in the ring of algebraic integers. It gets more interesting when we start with Q(m) = 49((2 m^2 - m) 5^3 - 3(-1 + m)5 + 7) the c-polynomial is: c^3 + 3(-1 + m)c^2 - 49(3 m^2 - 2m). For m = 0 and m = 1 the roots are as given above, for m = 2, the roots are: c1 = 7 c2 = sqrt(8).[-sqrt(2) + sqrt(-5)] c3 = sqrt(8).[-sqrt(2) - sqrt(-5)] Note that for c2 and c3 the latter factors are algebraic integers that are factors of 7. So we see that the distribution of the algebraic integer factors of 7 critically depends on the value of m. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Simple principle, core error proven > Luckily for me the mathematical argument that proves that I've been > correct all along can be further simplified by the use of *numbers* > instead of variables, as while algebra as an idea is well-founded, so > letter symbols should do, when mathematicians are lying about the > mathematics, you need to use what you can, and pray. > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. No, the definition just doesn't include enough to make things work the way you should. If the definition had an error of the type you suggest, the contradiction would lie in one of the theorems *about* algebraic integers, not in the definition. For example, the theorem that the algebraic integers form a ring. > 2. The important tool I use is a polynomial: > P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) > where m varies in the ring of algebraic integers. > Some may find it looks odd. But the entire point of that form is to > do something a little different than anyone else apparently has done > before, which is to get non-polynomial factors. > And the factorization with those factors is > P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) > where the a's are roots of the following cubic: > a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m). > So you can see that the a's are functions of m, and if you really want > headaches you can go ahead and solve the cubic to get a view of them. > However, I can move on without needing their explicit form. > g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7) > and I can find those terms that are free of m by setting m=0, or by inspection you could simply rewrite a_1 as a_1(m) etc, and realize that the 7's are the constant terms, and therefor *independent* of m. > just > like with any polynomial, you know like with S(m) = 2(m^2 + 2m + 1), > S(0) = 2, which you can just look at and see here, but with my P(m) > it's just a little harder, so I set m=0, which gives me > P(0) = 49(3(5) + 7), > which fits with the cubic as at m=0 it gives > a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3, > to show that at m=0, the three factors are > g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22. > Now dividing P(m) by 49 gives > P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7 > and the question is what happens to the g's, but look now at P(0)/49 > as that is > P(0)/49 = 3(5) + 7 > as two factors of 7, each 7, have beeen divided off, which is easy to > see. So far, so good. > The only way that can happen is if > P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) > where only two of the a's have 7 as a factor, Wrong. Factors of 7 will be distributed as determined by the values of each of the a's for each value of m. The factorization depends on m because the a's depend on m. This has been demonstrated with other polynomials through explicit computations. I don't recall if Arturo has done the computations for this particular one or not. > where the idea is almost > trivially simple as consider a polynomial like > S(m) = 7(m^2 + 2m + 1) = (b_1 m + 7)(b_2 m + 1) > and notice that S(0) = 7, while S(m)/7, gives you > S(m)/7 = m^2 + 2m + 1, > which means that > S(m)/7 = (b_1 m/7 + 1)(b_2 m + 1) = m^2 + 2m + 1. This is a different situation, as it is a reducible polynomial. P(m) is, in general, irreducible. > It's the same basic idea while what I have is more complicated as I'm > showing you an over hundred year old flaw, and it turns out that it > takes a somewhat complicated expression to show it which > P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7 > is, but the same principle works as just like with S(m), with P(m), > dividing out 49, affects the independent or constant terms, revealing > factors of the a's. Ok, you've supposedly shown us a flaw. What is it supposed to be? > That is, from the distributive property, factors that are 7 must > divide through. > So now I know that the correct factorization is > P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) Wrong for the reasons above. > which is like > S(m)/7 = (b_1 m/7 + 1)(b_2 m + 1). Only superficially. > 3. So *two* of the a's *should* have 7 as a factor, and in fact they > do, in a proper ring, but the ring of algebraic integers has problems, > so that for certain values of m, they won't. Where do you get should? Also, you seem to think that the fact that they don't for certain values of m indicates a problem with the algebraic integers instead of your proof. When you see an apparent contradiction, start questioning your work. Also, what is a proper ring? Or are you claiming that the algebraic integers do not form a ring? > It's that *inconsistency* which shows you there's a problem because > mathematics isn't about being wishy-washy, where sometimes something > works and then other times it doesn't. Which indicates there is a problem, but until you can lock down the holes in your proof (which you can't, you are saying the polynomial behaves in a way you later say it doesn't), you have nothing. Also, you would do better to locate the exact source of the problem, rather than claim that something is broken. > That error has sat in mathematics--the body of discoveries commonly > called mathematics--for over a *hundred* years. > Since I found the error I should probably get rich and famous from it, > but so far mathematicians I've contacted seem more interested in > denying or hiding the error than in telling the truth. However, that > means there is this error, which may sink lots of proofs over the > past hundred years in an area of mathematics called algebraic number > theory. If you knew much math history, you would know that you might get famous, but it's unlikely that wealth would follow. Mathematics is not the field for wealth. > And in fact, mathematicians may be engaging in a bigger fraud because > they may *know* by now just how big the problem is, and it may be a > bigger embarrassment than even I am aware of at this point. Or most may be either ignorant or unconcerned. > They may see themselves as having no choice but to hide it to keep > their money, their prestige, and their history as it is known to most > as of now. Have you done any research into math society/finances *besides* this newsgroup? Have you asked anyone what their day-to-day lives are like? > Some of them may be trying to hide it partly out of envy or jealousy > of my discovery as well. Mathematicians can be VERY vicious for petty > and childish reasons I've found. > If you are a math student, you probably will want to stay out of the > area of algebraic number theory, or consider carefully things your > professors supposedly prove in the area. Or just pay attention during abstract algebra and learn what James has refused to. > While mathematicians behave this way, you have to wonder now about > what they teach you. > James Harris > ------------------------- > Intellectual laziness is about deciding > ahead of time what you wish to believe, > and daring God to be different. > http://lostincomment.blogspot.com/ This is a quote that says volumes about you. Have you used it as a mirror lately? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Simple principle, core error proven > Luckily for me the mathematical argument that proves that I've been > correct all along can be further simplified by the use of *numbers* > instead of variables, as while algebra as an idea is well-founded, so > letter symbols should do, when mathematicians are lying about the > mathematics, you need to use what you can, and pray. 1. First the problematic definition: Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. > No, the definition just doesn't include enough to make things work the > way you should. If the definition had an error of the type you suggest, > the contradiction would lie in one of the theorems *about* algebraic > integers, not in the definition. For example, the theorem that the > algebraic integers form a ring. I'm merely stating the problem and giving my assertion, so there's no reason to get into a debate here. > 2. The important tool I use is a polynomial: P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) Oh yeah, that's P(m) = 14706125 m^3 - 900375 m^2 + 899640 m + 22 and I think it's easier to work with as I give it originally. Notice that what I do is creatively break the polynomial up, so that I can factor it in a special way. But it's still just a polynomial. where m varies in the ring of algebraic integers. Some may find it looks odd. But the entire point of that form is to > do something a little different than anyone else apparently has done > before, which is to get non-polynomial factors. And the factorization with those factors is P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) where the a's are roots of the following cubic: a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m). So you can see that the a's are functions of m, and if you really want > headaches you can go ahead and solve the cubic to get a view of them. However, I can move on without needing their explicit form. > g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7) and I can find those terms that are free of m by setting m=0, > or by inspection you could simply rewrite a_1 as a_1(m) etc, and realize > that the 7's are the constant terms, and therefor *independent* of m. That's wrong. Posters in the past have made a similar claim, but readers should note that the reality is that the term independent of m with ONLY TWO of the factors is 7, while for the third it is 3(5) + 7, that is 22. Now here's a good place for me to emphasize to readers to read *carefully* as the problem I'm facing is that mathematicians appear to be trying to protect themselves from an over one hundred year old problem, so they will just say wrong things to you, hoping you'll just trust. > just > like with any polynomial, you know like with S(m) = 2(m^2 + 2m + 1), > S(0) = 2, which you can just look at and see here, but with my P(m) > it's just a little harder, so I set m=0, which gives me P(0) = 49(3(5) + 7), which fits with the cubic as at m=0 it gives a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3, to show that at m=0, the three factors are g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22. Now dividing P(m) by 49 gives P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7 and the question is what happens to the g's, but look now at P(0)/49 > as that is P(0)/49 = 3(5) + 7 as two factors of 7, each 7, have beeen divided off, which is easy to > see. > So far, so good. The math is rather basic, and like I said, it's easy to see. The only way that can happen is if P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) where only two of the a's have 7 as a factor, > Wrong. Factors of 7 will be distributed as determined by the values of > each of the a's for each value of m. The factorization depends on m > because the a's depend on m. This has been demonstrated with other > polynomials through explicit computations. I don't recall if Arturo has > done the computations for this particular one or not. That's not how it works, but the problem is that for many a_1, a_2 and a_3 are mysteries, so I'll help out by asking some what if questions. What if, a_1 has sqrt(m) as a factor? It has some factor of m, since it equals 0, when m is a factor. So what then? What if that factor could be m, would you believe Will Twentyman then? Some of you may be confused by a_3, but what if it were m+3? Let's look at that possibility with x, y and z algebraic integers introduced for any remaining factors: Then you'd have P'(m) = (5xm + 7)(5ym + 7)(5z(m+3) + 7) and how many of you NOW would accept that 7 divides off *dependent* on the value of m? I hope none. What works with the what if scenario is P'(m) = (5(7)m + 7)(5(7)m + 7)(5(m+3) + 7) and hopefully some of you can now see how simple it really is, and how diabolical mathematicians in arguing against my result have been. They've convinced many of you that independent, constant terms can vary based on the value of m. Here you can see how it can happen that only two of the factor have 7 as a factor while for the last it's blocked. Now the a's aren't quite that simple in one sense, but they are in another as you can see at m=0, that P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) gives P(0) = (7)(7)( 5(3) + 7). Now finally, can any of you come up with a *rational* way for 49 to divide off P(m), such that the independent terms vary based on the value of m? But, they're independent terms, so they can't--rationally. I'm telling you that mathematicians are trying to hide an error in their discipline. As they are mathematicians they KNOW HOW TO LIE TO YOU, and you can see that they're quite willing to do it. Their lies are messing up my life, and can't be helping world society. Meanwhile we, the public, can't know just how bad the problem is, though the longer mathematicians lie about the over hundred year old error, the more likely it seems that it is a meltdown. James Harris === Subject: Re: Simple principle, core error proven > Luckily for me the mathematical argument that proves that I've been > correct all along can be further simplified by the use of *numbers* > instead of variables, as while algebra as an idea is well-founded, so > letter symbols should do, when mathematicians are lying about the > mathematics, you need to use what you can, and pray. 1. First the problematic definition: Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. No, the definition just doesn't include enough to make things work the > way you should. If the definition had an error of the type you suggest, > the contradiction would lie in one of the theorems *about* algebraic > integers, not in the definition. For example, the theorem that the > algebraic integers form a ring. > I'm merely stating the problem and giving my assertion, so there's no > reason to get into a debate here. As a principle, we should always note the first error in a proof and stop investigating the proof until that error is fixed. Your first error is that you believe a definition could lead to a contradiction. Until you fix that error it would be pointless to read any further. === Subject: Re: Simple principle, core error proven > Luckily for me the mathematical argument that proves that I've been > correct all along can be further simplified by the use of *numbers* > instead of variables, as while algebra as an idea is well-founded, so > letter symbols should do, when mathematicians are lying about the > mathematics, you need to use what you can, and pray. 1. First the problematic definition: Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. No, the definition just doesn't include enough to make things work the > way you should. If the definition had an error of the type you suggest, > the contradiction would lie in one of the theorems *about* algebraic > integers, not in the definition. For example, the theorem that the > algebraic integers form a ring. > I'm merely stating the problem and giving my assertion, so there's no > reason to get into a debate here. > 2. The important tool I use is a polynomial: P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) > Oh yeah, that's > P(m) = 14706125 m^3 - 900375 m^2 + 899640 m + 22 That should be P(m) = 14706125 m^3 - 900375 m^2 + 17640 m + 1078 as I forgot to multiply the last one by 49 and the Window's calculator wasn't on scientific for the other. My apologies for the error. The point is that I'm just using a polynomial, but I'm doing a special factorization. > and I think it's easier to work with as I give it originally. > Notice that what I do is creatively break the polynomial up, so that I > can factor it in a special way. > But it's still just a polynomial. where m varies in the ring of algebraic integers. Some may find it looks odd. But the entire point of that form is to > do something a little different than anyone else apparently has done > before, which is to get non-polynomial factors. And the factorization with those factors is P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) There's something neat about that factorization that I'd want to emphasize, which I'll do below. where the a's are roots of the following cubic: a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m). So you can see that the a's are functions of m, and if you really want > headaches you can go ahead and solve the cubic to get a view of them. However, I can move on without needing their explicit form. > g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7) and I can find those terms that are free of m by setting m=0, or by inspection you could simply rewrite a_1 as a_1(m) etc, and realize > that the 7's are the constant terms, and therefor *independent* of m. > That's wrong. Posters in the past have made a similar claim, but > readers should note that the reality is that the term independent of m > with ONLY TWO of the factors is 7, while for the third it is 3(5) + 7, > that is 22. > Now here's a good place for me to emphasize to readers to read > *carefully* as the problem I'm facing is that mathematicians appear to > be trying to protect themselves from an over one hundred year old > problem, so they will just say wrong things to you, hoping you'll just > trust. > just > like with any polynomial, you know like with S(m) = 2(m^2 + 2m + 1), > S(0) = 2, which you can just look at and see here, but with my P(m) > it's just a little harder, so I set m=0, which gives me P(0) = 49(3(5) + 7), which fits with the cubic as at m=0 it gives a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3, to show that at m=0, the three factors are g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22. And notice that is the factorization of the last coefficient. Remember P(m) = 14706125 m^3 - 900375 m^2 + 17640 m + 1078 and 1078 = 7(7)(22). Using P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) is just a way to factor into non-polynomial factors. I keep emphasizing that as unfortunately mathematicians are *lying* to try and hide an over hundred year old error, and I need to help you all to see that so that you understand that the people confusing you and muddying the waters are mathematicians. After all, if I can get past them I get famous, and hopefully then, rich, so I have LOTS of motivation to help you understand. They have the opposite motivation. I must say that mathematicians are remarkably evil, and quite diabolical in their misuse of your trust. > Now dividing P(m) by 49 gives P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7 and the question is what happens to the g's, but look now at P(0)/49 > as that is P(0)/49 = 3(5) + 7 as two factors of 7, each 7, have beeen divided off, which is easy to > see. So far, so good. > The math is rather basic, and like I said, it's easy to see. The only way that can happen is if P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) where only two of the a's have 7 as a factor, Wrong. Factors of 7 will be distributed as determined by the values of > each of the a's for each value of m. The factorization depends on m > because the a's depend on m. This has been demonstrated with other > polynomials through explicit computations. I don't recall if Arturo has > done the computations for this particular one or not. > That's not how it works, but the problem is that for many a_1, a_2 > and a_3 are mysteries, so I'll help out by asking some what if > questions. > What if, a_1 has sqrt(m) as a factor? It has some factor of m, since > it equals 0, when m is a factor. So what then? > What if that factor could be m, would you believe Will Twentyman > then? > Some of you may be confused by a_3, but what if it were m+3? > Let's look at that possibility with x, y and z algebraic integers > introduced for any remaining factors: > Then you'd have > P'(m) = (5xm + 7)(5ym + 7)(5z(m+3) + 7) > and how many of you NOW would accept that 7 divides off *dependent* on > the value of m? > I hope none. > What works with the what if scenario is > P'(m) = (5(7)m + 7)(5(7)m + 7)(5(m+3) + 7) > and hopefully some of you can now see how simple it really is, and how > diabolical mathematicians in arguing against my result have been. > They've convinced many of you that independent, constant terms can > vary based on the value of m. Yup, I do something that is to get non-polynomial factors and mathematicians use that against me, so that I have to remind you how factorization works, and remind you of that interesting thing called the distributive property. Distributive property: a(b+c) = ab + ac They're making *fools* of you all!!! The mathematicians are making you all look like fools to the extent that I have to remind you of the distributive property!!! > Here you can see how it can happen that only two of the factor have 7 > as a factor while for the last it's blocked. Now the a's aren't quite > that simple in one sense, but they are in another as you can see at > m=0, that > P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) gives > P(0) = (7)(7)( 5(3) + 7). > Now finally, can any of you come up with a *rational* way for 49 to > divide off P(m), such that the independent terms vary based on the > value of m? > But, they're independent terms, so they can't--rationally. > I'm telling you that mathematicians are trying to hide an error in > their discipline. As they are mathematicians they KNOW HOW TO LIE TO > YOU, and you can see that they're quite willing to do it. > Their lies are messing up my life, and can't be helping world society. > Meanwhile we, the public, can't know just how bad the problem is, > though the longer mathematicians lie about the over hundred year old > error, the more likely it seems that it is a meltdown. You see, mathematicians may be trying to hide a situation at the limits of the worst case. James Harris === Subject: Re: Simple principle, core error proven Adjunct Assistant Professor at the University of Montana. [.snip.] >I keep emphasizing that as unfortunately mathematicians are *lying* to >try and hide an over hundred year old error, and I need to help you >all to see that so that you understand that the people confusing you >and muddying the waters are mathematicians. >After all, if I can get past them I get famous, and hopefully then, >rich, so I have LOTS of motivation to help you understand. Nonsense, and nonsense. But, in any case, you have claimed that I am one of the people trying to confuse and muddying the waters. I have offered to literally stop being in your way. You've refused the offer. If you are so interested in getting past us, why did you refuse the offer? I'm not saying I will endorse your work (I cannot: it's just plain wrong), but I am willing to stop telling you why it is wrong. I am willing to stop speaking on the subject. But you don't want me to stop, do you? >They have the opposite motivation. >I must say that mathematicians are remarkably evil, and quite >diabolical in their misuse of your trust. You are twisted, and your vision of the world reflects your twisted-ness. [.snip.] Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Simple principle, core error proven [...] >I keep emphasizing that as unfortunately mathematicians are *lying* to >try and hide an over hundred year old error, and I need to help you >all to see that so that you understand that the people confusing you >and muddying the waters are mathematicians. >After all, if I can get past them I get famous, and hopefully then, >rich, so I have LOTS of motivation to help you understand. > Nonsense, and nonsense. > But, in any case, you have claimed that I am one of the people trying > to confuse and muddying the waters. I have offered to literally stop > being in your way. You've refused the offer. > If you are so interested in getting past us, why did you refuse the > offer? I'm not saying I will endorse your work (I cannot: it's just > plain wrong), but I am willing to stop telling you why it is wrong. I > am willing to stop speaking on the subject. > But you don't want me to stop, do you? Of course he doesn't. He's just calling you evil and a liar, so that you become somehow more motivated in answering his queries and sometimes perhaps just because he thinks it's fun to do it. You happily play his game for some reason I don't fully understand. He needs and borrows your maths knowledge in order to get his crappy proofs right, while still hoping to make a discovery that will make him famous... I guess you two have achieved a symbiotic relationship :-) The problem is though, James is more on the parasite side. While he may provide some entertainment, I don't see anything else of even of little value comming from him. On the other side, I think you invest quite some time in order to comment the errors in James' proofs, doing his homework because he's not able to. Hmm, how about teaching James to fish, instead of feeding him daily? Already tried that? -- Edgar === Subject: Re: Simple principle, core error proven > Hmm, how about teaching James to fish, instead of feeding him daily? Already >tried that? I like Terry Pratchett's version - Give a man a fire and he's warm for a day. Set a man on fire and he's warm for the rest of his life. -- Richard -- Spam filter: to mail me from a .com/.net site, put my surname in the headers. FreeBSD rules! === Subject: Re: Simple principle, core error proven << I like Terry Pratchett's version - Give a man a fire and he's warm for a day. Set a man on fire and he's warm for the rest of his life. -- Richard Amig te melegedsz igy, nosza. A baj csak akkor van ha masokat kivansz igy melengetni === Subject: Re: Simple principle, core error proven << << I like Terry Pratchett's version - Give a man a fire and he's warm for a day. Set a man on fire and he's warm for the rest of his life. -- Richard Amig te melegedsz igy, nosza. A baj csak akkor van ha masokat kivansz igy melengetni Sorry, this was a mistake to send here. === Subject: Re: Simple principle, core error proven > << > << > I like Terry Pratchett's version - Give a man a fire and he's warm > for a day. Set a man on fire and he's warm for the rest of his life. > -- Richard > Amig te melegedsz igy, nosza. A baj csak akkor van ha masokat kivansz > igy melengetni > Sorry, this was a mistake to send here. Oh, a bit Hungarian on the way does not matter so very much. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Simple principle, core error proven > 2. The important tool I use is a polynomial: P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) > Oh yeah, that's > P(m) = 14706125 m^3 - 900375 m^2 + 899640 m + 22 Is it? (Hint: WRONG!) Don't you think you should learn to *multiply* before you attempt something more advanced? > James Wacky Harris (quoted term added for clarity) By the way, when you replace a variable in a single variable expression with a constant, such as you did when you replaced 'm' with 0 in your expression for P(m), you are *evaluating* P(m) at m=0, not eliminating terms dependent on 'm'. Replacement of 'm' with any constant will reduce P(m) to a number. For a polynomial in 'm', such as P(m), the value of P(0) is simply the coefficient of m^0, i.e. the constant term. The best you can say is that you have eliminated all terms which are multiples of 'm', not 'independent of m'. Replacing 'm' by any number reduces the expression to a constant -- what's the point? -- A fool and his proof are soon refuted. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Simple principle, core error proven Adjunct Assistant Professor at the University of Montana. > Luckily for me the mathematical argument that proves that I've been > correct all along can be further simplified by the use of *numbers* > instead of variables, as while algebra as an idea is well-founded, so > letter symbols should do, when mathematicians are lying about the > mathematics, you need to use what you can, and pray. > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. > No, the definition just doesn't include enough to make things work the > way you should. If the definition had an error of the type you suggest, > the contradiction would lie in one of the theorems *about* algebraic > integers, not in the definition. For example, the theorem that the > algebraic integers form a ring. >I'm merely stating the problem and giving my assertion, so there's no >reason to get into a debate here. There is no debate. Your opinion about where the alleged problem is quite simply incorrect; what you state is logical nonsense. > 2. The important tool I use is a polynomial: > P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) >Oh yeah, that's >P(m) = 14706125 m^3 - 900375 m^2 + 899640 m + 22 >and I think it's easier to work with as I give it originally. No, it is much harder. The reason it is harder is that you are HIDING what you are doing; rather than clarifying the situation, you are obfuscating it. The reason you obfuscate it is that the way you factor P(m) has little to do with the specific values you are using. The factorization of P(m) is not a factorization as a polynomial in m, and it is not a numeric factorization, it is a factorization as a polynomial in a DIFFERENT variable, and by hiding that variable you are inviting miscommunication and misconception. [.snip.] >That's not how it works, but the problem is that for many a_1, a_2 >and a_3 are mysteries, so I'll help out by asking some what if >questions. >What if, a_1 has sqrt(m) as a factor? It has some factor of m, since >it equals 0, when m is a factor. So what then? What if it does not? What then? >What if that factor could be m, would you believe Will Twentyman >then? What if that factor is not m? If my grandmother had wheels, she would be a bicycle. But she doesn't. So what is she? >Some of you may be confused by a_3, but what if it were m+3? But it's not. It's not a POLYNOMIAL in m. That's one of the big parts of the problem. But if you really want to stop talking about stupid what ifs, here is what a_1, a_2, and a_3 are (assuming I did not make a mistake with the Cardano formulas): a1 = m*f^2-1 + (1/2^{1/3})*(cuberoot (D-2+sqrt(C))+cuberoot(D-2-sqrt(C))) a2 = m*f^2-1 + [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C))) + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))) a3 = m*f^2-1 + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C))) + [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))). where C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8} + 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^2. D = m^3*f^6 - 3*m^2*f^4 + 3*m*f^2 In this case, you have set f=7. So C= -41523861603m^6 + 5084554482m^5 - 259416054m^4 + 6823642m^3 - 93639m^2 + 588 m D = 117649m^3 - 7203m^2 + 147 and a1 = 49m - 1 + (1/2^{1/3})*(cuberoot(D-2+sqrt(C))+cuberoot(D-2-sqrt(C))) a2 = 49m - 1 + [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C))) + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))) a3 = 49m - 1 + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C))) + [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))). So they aren't polynomials in m. So your what ifs are red herrings and irrelevant. >Let's look at that possibility with x, y and z algebraic integers >introduced for any remaining factors: >Then you'd have >P'(m) = (5xm + 7)(5ym + 7)(5z(m+3) + 7) >and how many of you NOW would accept that 7 divides off *dependent* on >the value of m? >I hope none. >What works with the what if scenario is >P'(m) = (5(7)m + 7)(5(7)m + 7)(5(m+3) + 7) You kidding? You are ASSUMING what you want: you are assuming that x and y are multiples of 7, but that z is coprime to 7. So, yes, amazing: if we assume your conclusion is correct, then it follows that your conclusion is correct. All hail the conquering hero! But you might note, that even in your silly what ifs, if m=4, then 7 factors out of the LAST term as well, even though it did not factor out of it the first time. But it is irrelevant: you are assuming your conclusion; that's called a circular argument. [.snip.] >P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) gives >P(0) = (7)(7)( 5(3) + 7). >Now finally, can any of you come up with a *rational* way for 49 to >divide off P(m), such that the independent terms vary based on the >value of m? It's not the independent terms that vary, it's the a_1, a_2, and a_3. In order to factor out 7 from 5a_1+7, you need ->a_1<- to be a multiple of 7. It has nothing to do with the independent terms, that's just your red herring. Plug in the value of m into the CORRECT formulas, and note that it is NOT true that a_3 is coprime to 7 for each value of m, that it is NOT true that a_1 is a multiple of 7, and that it is NOT true that a_2 is a multiple of 7. For each value of m, you will get a factorization of 7 into three algebraic integers, 7=w_1(m)*w_2(m)*w_3(m), which depend on m, pairwise coprime, and a_1 will be a multiple of w_2(m)*w_3(m), a_2 will be a multiple of w_1(m)*w_3(m), and a_3 will be a multiple of w_1(m)*w_2(m). It just so happens that at m=0, you get w_1=7, w_2=1, w_3=1. >But, they're independent terms, so they can't--rationally. But that's irrelevant, so you bringing it up is irrational. [.snip.] Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Simple principle, core error proven > Luckily for me the mathematical argument that proves that I've been > correct all along can be further simplified by the use of *numbers* > instead of variables, as while algebra as an idea is well-founded, so > letter symbols should do, when mathematicians are lying about the > mathematics, you need to use what you can, and pray. > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. > No, the definition just doesn't include enough to make things work the > way you should. If the definition had an error of the type you suggest, > the contradiction would lie in one of the theorems *about* algebraic > integers, not in the definition. For example, the theorem that the > algebraic integers form a ring. >I'm merely stating the problem and giving my assertion, so there's no >reason to get into a debate here. > There is no debate. Your opinion about where the alleged problem is > quite simply incorrect; what you state is logical nonsense. > 2. The important tool I use is a polynomial: > P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) >Oh yeah, that's >P(m) = 14706125 m^3 - 900375 m^2 + 899640 m + 22 Should be P(m) = 14706125 m^3 - 900375 m^2 + 17640 m + 1078 as I forgot to multiply the last, 22, by 49, and just messed up the third coefficient. >and I think it's easier to work with as I give it originally. > No, it is much harder. The reason it is harder is that you are HIDING > what you are doing; rather than clarifying the situation, you are > obfuscating it. Well, it looks simpler to me. > The reason you obfuscate it is that the way you factor P(m) has little > to do with the specific values you are using. The factorization of > P(m) is not a factorization as a polynomial in m, and it is not a > numeric factorization, it is a factorization as a polynomial in a > DIFFERENT variable, and by hiding that variable you are inviting > miscommunication and misconception. > [.snip.] What variable? >That's not how it works, but the problem is that for many a_1, a_2 >and a_3 are mysteries, so I'll help out by asking some what if >questions. >What if, a_1 has sqrt(m) as a factor? It has some factor of m, since >it equals 0, when m is a factor. So what then? > What if it does not? What then? >What if that factor could be m, would you believe Will Twentyman >then? > What if that factor is not m? > If my grandmother had wheels, she would be a bicycle. But she > doesn't. So what is she? You're involving your grandmother??!!! >Some of you may be confused by a_3, but what if it were m+3? > But it's not. It's not a POLYNOMIAL in m. That's one of the big parts > of the problem. > But if you really want to stop talking about stupid what ifs, here > is what a_1, a_2, and a_3 are (assuming I did not make a mistake with > the Cardano formulas): Well what's wonderful about what I do is *simplification* and it's incredible that I can show something about the a's when they're so massively complicated to look at explicitly. It's a testimony to the power of algebra, and a simple idea, in my case, shifting things around to get a powerful factorization. > a1 = m*f^2-1 + (1/2^{1/3})*(cuberoot > (D-2+sqrt(C))+cuberoot(D-2-sqrt(C))) > a2 = m*f^2-1 + [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C))) > + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))) > a3 = m*f^2-1 + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C))) > + > [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))). > where > C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8} > + 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^2. > D = m^3*f^6 - 3*m^2*f^4 + 3*m*f^2 > In this case, you have set f=7. So > C= -41523861603m^6 + 5084554482m^5 - 259416054m^4 + 6823642m^3 > - 93639m^2 + 588 m > D = 117649m^3 - 7203m^2 + 147 > and > a1 = 49m - 1 + (1/2^{1/3})*(cuberoot(D-2+sqrt(C))+cuberoot(D-2-sqrt(C))) > a2 = 49m - 1 + [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C))) > + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))) > a3 = 49m - 1 + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C))) > + > [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))). > So they aren't polynomials in m. So your what ifs are red herrings > and irrelevant. I never said they were polynomials, but I *did* point out simple principles. Now then since *two* of the a's go to 0, when m=0, there's some factor of m in there, but can anyone see it? Anyone see the factor of m revealed by using a simple but powerful technique? Arturo Magidin is like the guy who jumps up and yells that it's TOO HARD. I'm the can-do guy who did. >Let's look at that possibility with x, y and z algebraic integers >introduced for any remaining factors: >Then you'd have >P'(m) = (5xm + 7)(5ym + 7)(5z(m+3) + 7) >and how many of you NOW would accept that 7 divides off *dependent* on >the value of m? >I hope none. >What works with the what if scenario is >P'(m) = (5(7)m + 7)(5(7)m + 7)(5(m+3) + 7) > You kidding? You are ASSUMING what you want: you are assuming that x > and y are multiples of 7, but that z is coprime to 7. So, yes, > amazing: if we assume your conclusion is correct, then it follows that > your conclusion is correct. All hail the conquering hero! Yeah, and I *said* I was doing the what if thing. Readers should note, Arturo Magidin is trying to snow you with complexity. But I found ways to slash through the complexity. > But you might note, that even in your silly what ifs, if m=4, then 7 > factors out of the LAST term as well, even though it did not factor > out of it the first time. But it is irrelevant: you are assuming your > conclusion; that's called a circular argument. > [.snip.] And notice the attempt to fool you yet again readers by focusing on a specific case. The reality is that given P(m) = 14706125 m^3 - 900375 m^2 + 17640 m + 1078 which has a factor of 49, if you divide P(m) by 49 you get 22 as the constant term. That means that your factorization will not have 7 as a factor for any of the factors of the constant term as 22 is coprime to 7. Notice the simplicity in the argument. Look at what I have next... >P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) gives >P(0) = (7)(7)( 5(3) + 7). >Now finally, can any of you come up with a *rational* way for 49 to >divide off P(m), such that the independent terms vary based on the >value of m? > It's not the independent terms that vary, it's the a_1, a_2, and > a_3. In order to factor out 7 from 5a_1+7, you need ->a_1<- to be a > multiple of 7. It has nothing to do with the independent terms, > that's just your red herring. The *independent* terms do not vary with m, which is why I use them as I do. That P(0)/49 = 22 tells the tale. There's no way that *any* of the independent terms can have 7 as a factor because they're all factors of 22. Remember the polynomial is P(m) = 14706125 m^3 - 900375 m^2 + 17640 m + 1078 and it's really not so complicated. > Plug in the value of m into the CORRECT formulas, and note that it is NOT > true that a_3 is coprime to 7 for each value of m, that it is NOT true > that a_1 is a multiple of 7, and that it is NOT true that a_2 is a > multiple of 7. For each value of m, you will get a factorization of 7 > into three algebraic integers, 7=w_1(m)*w_2(m)*w_3(m), which depend on > m, pairwise coprime, and a_1 will be a multiple of w_2(m)*w_3(m), a_2 > will be a multiple of w_1(m)*w_3(m), and a_3 will be a multiple of > w_1(m)*w_2(m). It just so happens that at m=0, you get w_1=7, w_2=1, > w_3=1. Arturo Magidin is a mathematician who was educated at Berkeley University, which I verified by going to the Berkeley website. He's now *lying* to you and trying to confuse with a lot of talk, when I'm talking straight and simple. I use simple idea, and basic algebra, while you can see what Arturo Magidin is doing. Why would he do it? Because mathematicians are hiding an over one hundred year old error and probably figure that the way to hide it is to confuse you by throwing a lot of stuff at you. It's about power, prestige and *money*. Now then, what do you gain by letting mathematicians lie, even if you're a math student? >But, they're independent terms, so they can't--rationally. > But that's irrelevant, so you bringing it up is irrational. > [.snip.] I use simple ideas and can trace out EVERYTHING I do, and explain in detail. Here you see this mathematician coming in to try and make you think it's all so complicated. Why? Well even if you have doubts about him, more than likely you'll side with mathematicians rather than go for the novel and new idea. That's human nature. They're using human nature in the hope that they can manage the lie. Many of you are probably intimidated by mathematicians, so they can lie to you, know that you're terrified of looking stupid by challenging them, and get away with lying to your face. It's a frustrating situation for me, as I search for the brave among you. James Harris === Subject: Re: Simple principle, core error proven > Arturo Magidin is a mathematician who was educated at Berkeley > University, which I verified by going to the Berkeley website. It's not Berkeley University! IDIOT IDIOT IDIOT... Gee, I get a bit frustrated when someone can't get a simple fact straight after being told over and over and over... -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Simple principle, core error proven > Luckily for me the mathematical argument that proves that I've been > correct all along can be further simplified by the use of *numbers* > instead of variables, as while algebra as an idea is well-founded, so > letter symbols should do, when mathematicians are lying about the > mathematics, you need to use what you can, and pray. 1. First the problematic definition: Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. > No, the definition just doesn't include enough to make things work the > way you should. If the definition had an error of the type you suggest, > the contradiction would lie in one of the theorems *about* algebraic > integers, not in the definition. For example, the theorem that the > algebraic integers form a ring. > I'm merely stating the problem and giving my assertion, so there's no > reason to get into a debate here. > 2. The important tool I use is a polynomial: P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) > Oh yeah, that's > P(m) = 14706125 m^3 - 900375 m^2 + 899640 m + 22 > and I think it's easier to work with as I give it originally. > Notice that what I do is creatively break the polynomial up, so that I > can factor it in a special way. > But it's still just a polynomial. where m varies in the ring of algebraic integers. Some may find it looks odd. But the entire point of that form is to > do something a little different than anyone else apparently has done > before, which is to get non-polynomial factors. And the factorization with those factors is P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) where the a's are roots of the following cubic: a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m). So you can see that the a's are functions of m, and if you really want > headaches you can go ahead and solve the cubic to get a view of them. However, I can move on without needing their explicit form. > g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7) and I can find those terms that are free of m by setting m=0, > or by inspection you could simply rewrite a_1 as a_1(m) etc, and realize > that the 7's are the constant terms, and therefor *independent* of m. > That's wrong. Posters in the past have made a similar claim, but > readers should note that the reality is that the term independent of m > with ONLY TWO of the factors is 7, while for the third it is 3(5) + 7, > that is 22. > Now here's a good place for me to emphasize to readers to read > *carefully* as the problem I'm facing is that mathematicians appear to > be trying to protect themselves from an over one hundred year old > problem, so they will just say wrong things to you, hoping you'll just > trust. > just > like with any polynomial, you know like with S(m) = 2(m^2 + 2m + 1), > S(0) = 2, which you can just look at and see here, but with my P(m) > it's just a little harder, so I set m=0, which gives me P(0) = 49(3(5) + 7), which fits with the cubic as at m=0 it gives a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3, to show that at m=0, the three factors are g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22. Now dividing P(m) by 49 gives P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7 and the question is what happens to the g's, but look now at P(0)/49 > as that is P(0)/49 = 3(5) + 7 as two factors of 7, each 7, have beeen divided off, which is easy to > see. > So far, so good. > The math is rather basic, and like I said, it's easy to see. The only way that can happen is if P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) where only two of the a's have 7 as a factor, > Wrong. Factors of 7 will be distributed as determined by the values of > each of the a's for each value of m. The factorization depends on m > because the a's depend on m. This has been demonstrated with other > polynomials through explicit computations. I don't recall if Arturo has > done the computations for this particular one or not. > That's not how it works, but the problem is that for many a_1, a_2 > and a_3 are mysteries, so I'll help out by asking some what if > questions. > What if, a_1 has sqrt(m) as a factor? It has some factor of m, since > it equals 0, when m is a factor. So what then? > What if that factor could be m, would you believe Will Twentyman > then? > Some of you may be confused by a_3, but what if it were m+3? > Let's look at that possibility with x, y and z algebraic integers > introduced for any remaining factors: > Then you'd have > P'(m) = (5xm + 7)(5ym + 7)(5z(m+3) + 7) > and how many of you NOW would accept that 7 divides off *dependent* on > the value of m? > I hope none. > What works with the what if scenario is > P'(m) = (5(7)m + 7)(5(7)m + 7)(5(m+3) + 7) > and hopefully some of you can now see how simple it really is, and how > diabolical mathematicians in arguing against my result have been. > They've convinced many of you that independent, constant terms can > vary based on the value of m. > Here you can see how it can happen that only two of the factor have 7 > as a factor while for the last it's blocked. Now the a's aren't quite > that simple in one sense, but they are in another as you can see at > m=0, that > P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) gives > P(0) = (7)(7)( 5(3) + 7). > Now finally, can any of you come up with a *rational* way for 49 to > divide off P(m), such that the independent terms vary based on the > value of m? > But, they're independent terms, so they can't--rationally. > I'm telling you that mathematicians are trying to hide an error in > their discipline. As they are mathematicians they KNOW HOW TO LIE TO > YOU, and you can see that they're quite willing to do it. > Their lies are messing up my life, and can't be helping world society. > Meanwhile we, the public, can't know just how bad the problem is, > though the longer mathematicians lie about the over hundred year old > error, the more likely it seems that it is a meltdown. > James Harris How are they messing your life up? They're not making you post. If you post to Usenet, you should expect replies and people to disagree with you. You just have to know how to counter their arguments. Once again, you should factor terms out and not dividing them off. Doing the latter changes the polynomial, unless you only divide off constants. David Moran === Subject: Re: Simple principle, core error proven > Luckily for me the mathematical argument that proves that I've been > correct all along can be further simplified by the use of *numbers* > instead of variables, as while algebra as an idea is well-founded, so > letter symbols should do, when mathematicians are lying about the > mathematics, you need to use what you can, and pray. 1. First the problematic definition: Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. No, the definition just doesn't include enough to make things work the > way you should. If the definition had an error of the type you suggest, > the contradiction would lie in one of the theorems *about* algebraic > integers, not in the definition. For example, the theorem that the > algebraic integers form a ring. > I'm merely stating the problem and giving my assertion, so there's no > reason to get into a debate here. > 2. The important tool I use is a polynomial: P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) > Oh yeah, that's > P(m) = 14706125 m^3 - 900375 m^2 + 899640 m + 22 > and I think it's easier to work with as I give it originally. > Notice that what I do is creatively break the polynomial up, so that I > can factor it in a special way. > But it's still just a polynomial. where m varies in the ring of algebraic integers. Some may find it looks odd. But the entire point of that form is to > do something a little different than anyone else apparently has done > before, which is to get non-polynomial factors. And the factorization with those factors is P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) where the a's are roots of the following cubic: a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m). So you can see that the a's are functions of m, and if you really want > headaches you can go ahead and solve the cubic to get a view of them. However, I can move on without needing their explicit form. > g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7) and I can find those terms that are free of m by setting m=0, or by inspection you could simply rewrite a_1 as a_1(m) etc, and realize > that the 7's are the constant terms, and therefor *independent* of m. > That's wrong. Posters in the past have made a similar claim, but > readers should note that the reality is that the term independent of m > with ONLY TWO of the factors is 7, while for the third it is 3(5) + 7, > that is 22. > Now here's a good place for me to emphasize to readers to read > *carefully* as the problem I'm facing is that mathematicians appear to > be trying to protect themselves from an over one hundred year old > problem, so they will just say wrong things to you, hoping you'll just > trust. > just > like with any polynomial, you know like with S(m) = 2(m^2 + 2m + 1), > S(0) = 2, which you can just look at and see here, but with my P(m) > it's just a little harder, so I set m=0, which gives me P(0) = 49(3(5) + 7), which fits with the cubic as at m=0 it gives a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3, to show that at m=0, the three factors are g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22. Now dividing P(m) by 49 gives P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7 and the question is what happens to the g's, but look now at P(0)/49 > as that is P(0)/49 = 3(5) + 7 as two factors of 7, each 7, have beeen divided off, which is easy to > see. So far, so good. > The math is rather basic, and like I said, it's easy to see. The only way that can happen is if P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) where only two of the a's have 7 as a factor, Wrong. Factors of 7 will be distributed as determined by the values of > each of the a's for each value of m. The factorization depends on m > because the a's depend on m. This has been demonstrated with other > polynomials through explicit computations. I don't recall if Arturo has > done the computations for this particular one or not. > That's not how it works, but the problem is that for many a_1, a_2 > and a_3 are mysteries, so I'll help out by asking some what if > questions. > What if, a_1 has sqrt(m) as a factor? It has some factor of m, since > it equals 0, when m is a factor. So what then? > What if that factor could be m, would you believe Will Twentyman > then? > Some of you may be confused by a_3, but what if it were m+3? > Let's look at that possibility with x, y and z algebraic integers > introduced for any remaining factors: > Then you'd have > P'(m) = (5xm + 7)(5ym + 7)(5z(m+3) + 7) > and how many of you NOW would accept that 7 divides off *dependent* on > the value of m? > I hope none. > What works with the what if scenario is > P'(m) = (5(7)m + 7)(5(7)m + 7)(5(m+3) + 7) > and hopefully some of you can now see how simple it really is, and how > diabolical mathematicians in arguing against my result have been. > They've convinced many of you that independent, constant terms can > vary based on the value of m. > Here you can see how it can happen that only two of the factor have 7 > as a factor while for the last it's blocked. Now the a's aren't quite > that simple in one sense, but they are in another as you can see at > m=0, that > P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) gives > P(0) = (7)(7)( 5(3) + 7). > Now finally, can any of you come up with a *rational* way for 49 to > divide off P(m), such that the independent terms vary based on the > value of m? > But, they're independent terms, so they can't--rationally. > I'm telling you that mathematicians are trying to hide an error in > their discipline. As they are mathematicians they KNOW HOW TO LIE TO > YOU, and you can see that they're quite willing to do it. > Their lies are messing up my life, and can't be helping world society. > Meanwhile we, the public, can't know just how bad the problem is, > though the longer mathematicians lie about the over hundred year old > error, the more likely it seems that it is a meltdown. > James Harris > How are they messing your life up? They're not making you post. If you post > to Usenet, you should expect replies and people to disagree with you. You > just have to know how to counter their arguments. This is not true. Although there are differences between sci.math/sci.logic posters and the bot described by Herbert Simon in the cite following my sig, these are not differences that lend themselves to the flowering of new mathematics. For sci.math/sci.logic posters (unlike bots) fancy nothing more than the spoor of blood and the joy of inflicting pain. (If this sounds like your math prof, the resemblance is not coincidental.) Bots are indifferent, moreover, to *unauthorized* originality, whereas sci.math/sci.logic posters do their utmost to stay its course. To be continued. --John A performance system is designed to work in a defined task domain, accepting particular goals and seeking to reach them by some kind of highly selective search. The system must be told what goal is to be reached and must be given a description of the structure and characteristics of the task domain in which it is to operate: its problem space. [...] In contrast, a learning system, is capable of acquiring a problem space, in whole or part, by interacting with the external environment and without being instructed about it directly. Herbert Simon http://www.isi.edu/~shen/book-foreword.html === Subject: Re: Simple principle, core error proven > Luckily for me the mathematical argument that proves that I've been > correct all along can be further simplified by the use of *numbers* > instead of variables, as while algebra as an idea is well-founded, so > letter symbols should do, when mathematicians are lying about the > mathematics, you need to use what you can, and pray. > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. > 2. The important tool I use is a polynomial: > P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) > where m varies in the ring of algebraic integers. > Some may find it looks odd. But the entire point of that form is to > do something a little different than anyone else apparently has done > before, which is to get non-polynomial factors. > And the factorization with those factors is > P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) > where the a's are roots of the following cubic: > a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m). > So you can see that the a's are functions of m, and if you really want > headaches you can go ahead and solve the cubic to get a view of them. > However, I can move on without needing their explicit form. > g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7) > and I can find those terms that are free of m by setting m=0, just > like with any polynomial, you know like with S(m) = 2(m^2 + 2m + 1), > S(0) = 2, which you can just look at and see here, but with my P(m) > it's just a little harder, so I set m=0, which gives me > P(0) = 49(3(5) + 7), > which fits with the cubic as at m=0 it gives > a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3, > to show that at m=0, the three factors are > g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22. > Now dividing P(m) by 49 gives > P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7 > and the question is what happens to the g's, but look now at P(0)/49 > as that is > P(0)/49 = 3(5) + 7 > as two factors of 7, each 7, have beeen divided off, which is easy to > see. > The only way that can happen is if > P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) > where only two of the a's have 7 as a factor, where the idea is almost > trivially simple as consider a polynomial like > S(m) = 7(m^2 + 2m + 1) = (b_1 m + 7)(b_2 m + 1) > and notice that S(0) = 7, while S(m)/7, gives you > S(m)/7 = m^2 + 2m + 1, > which means that > S(m)/7 = (b_1 m/7 + 1)(b_2 m + 1) = m^2 + 2m + 1. The point being that b_1 must be divisible by 7, while b2 = 1 is coprime to 7. Yes, this is a good example. I like this example. It does reveal how you are thinking about this. Here is a closely related example: S(m) = 7*(m^2 + m + 7) = (b_1 m + 7)*(b_2 m + 7). In fact, I think this example is even closer to the polynomial P(m) that you have talked about above. Notice that S(0) = 7 * 7. This is consistent with b1 = 0 (which is divisible by 7) and b2 = 1. You can find the roots of S(m) explicitly, and you can use those to compute b_1 and b_2 explicitly. You find that: b_1 = (1 + sqrt(-27))/2 and b_2 = (1 - sqrt(-27))/2. Now, paralleling what you found for your polynomial S(m), I would expect to find that b_1 or b_2 is divisible by 7. Let's say I thought b_1 is divisible by 7. Let u = b_1/7 = (1 + sqrt(-27))/14. This implies 14*u - 1 = sqrt(-27), or 196*u^2 - 28*u + 28 = 0, or 7*u^2 - u + 1 = 0. This is an irreducible, primitive, non-monic polynomial in u. Therefore the roots cannot be algebraic integers. I started with u = b_1/7. Therefore b_1/7 is not an algebraic integer. Therefore b_1 is not divisible by 7. Similarly neither is b2. So this differs strikingly from YOUR example. Note, however, that b_1 * b_2 = 28/4 = 7. Therefore both b_1 and b_2 are *divisors* of 7. That is, neither is *coprime* to 7 either. The key difference between your S(m) and my S(m): IRREDUCIBILITY. Yours is too simple. Mine is more like your polynomial P(m): irreducibility is the rule; reducibility is the exception. I think if you look at this example very carefully and objectively, you will see where your reasoning goes off the track. > It's the same basic idea while what I have is more complicated as I'm > showing you an over hundred year old flaw, and it turns out that it > takes a somewhat complicated expression to show it which > P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7 > is, but the same principle works as just like with S(m), with P(m), > dividing out 49, affects the independent or constant terms, revealing > factors of the a's. > That is, from the distributive property, factors that are 7 must > divide through. That's where you are wrong. 7*7 divides out, but not in the way you think. > So now I know that the correct factorization is > P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) > which is like > S(m)/7 = (b_1 m/7 + 1)(b_2 m + 1). No - the parallel is much closer with the S(m) that I specified, not the one you have given. > 3. So *two* of the a's *should* have 7 as a factor, and in fact they > do, in a proper ring, but the ring of algebraic integers has problems, > so that for certain values of m, they won't. No - in the case of P(m), for most values of m, *none* of the a's is coprime to 7. > It's that *inconsistency* which shows you there's a problem because > mathematics isn't about being wishy-washy, where sometimes something > works and then other times it doesn't. There is no inconsistency. Your reasoning is based on an overly simplistic [reducible] example. Why is that so hard to understand ? > That error has sat in mathematics--the body of discoveries commonly > called mathematics--for over a *hundred* years. The error has sat in your head for *quite a few* months. > Since I found the error I should probably get rich and famous from it, > but so far mathematicians I've contacted seem more interested in > denying or hiding the error than in telling the truth. However, that > means there is this error, which may sink lots of proofs over the > past hundred years in an area of mathematics called algebraic number > theory. Drivel. > And in fact, mathematicians may be engaging in a bigger fraud because > they may *know* by now just how big the problem is, and it may be a > bigger embarrassment than even I am aware of at this point. Hogwash. > They may see themselves as having no choice but to hide it to keep > their money, their prestige, and their history as it is known to most > as of now. Nonsense. > Some of them may be trying to hide it partly out of envy or jealousy > of my discovery as well. Mathematicians can be VERY vicious for petty > and childish reasons I've found. Ridiculous. > If you are a math student, you probably will want to stay out of the > area of algebraic number theory, or consider carefully things your > professors supposedly prove in the area. If you are a math student, use your own judgement. Don't take anyone's word for anything. You don't have to. Math should always be all out in the open. Algebraic number theory is a wonderful, beautiful subset of mathematics. You might want to start with Ian Stewart's book, Algebraic Number Theory and Fermat's Last Theorem (with David Tall). Not an easy book, but it is well worth reading. Nora B. > While mathematicians behave this way, you have to wonder now about > what they teach you. > James Harris > ------------------------- > Intellectual laziness is about deciding > ahead of time what you wish to believe, > and daring God to be different. > http://lostincomment.blogspot.com/ === Subject: Re: Simple principle, core error proven [...] >2. The important tool I use is a polynomial: >P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) [...] >And the factorization with those factors is >P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) >where the a's are roots of the following cubic: >a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m). >So you can see that the a's are functions of m, [...] >g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7) >and I can find those terms that are free of m by setting m=0, just >like with any polynomial, you know like with S(m) = 2(m^2 + 2m + 1), ^^^ It may seem irrelevant, but you probably mean a here. The formulation used suggests that g_1(m), g_2(m), g_3(m) are given by polynomial expressions in m. I'll explain a bit later why this is important. [...] >to show that at m=0, the three factors are >g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22. >Now dividing P(m) by 49 gives >P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7 >and the question is what happens to the g's, but look now at P(0)/49 >as that is >P(0)/49 = 3(5) + 7 >as two factors of 7, each 7, have beeen divided off, which is easy to >see. >The only way that can happen is if >P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) >where only two of the a's have 7 as a factor, where the idea is almost >trivially simple as consider a polynomial like There are two mistakes here. First of all, 7 is not a prime (in the ring of algebraic integers). So even for a specific algebraic number m, from the fact that 49 divides P(m) = (5 a_1(m) + 7) (5 a_2(m) + 7) (5 a_3(m) + 7) you cannot automatically conclude that any of those three factors is divisible by 7. You might be able to get around this problem by looking at some factorization of 7 in the ring of algebraic integers. But I think you are making another mistake as well here. I think you are trying to use something like: Let q(M), r(M) be polynomials (over some ring). Let p be a prime (in the same ring). Assume that p divides q(M) r(M), that p divides q(0) and that p does not divide r(0). Then p divides q(M). I guess you are trying to apply this with some ring being the algebraic integers, q(M) = (5 a_1(M) + 7) (5 a_2(M) + 7), r(M) = 5 a_3(M) + 7 and p = 7 (which is not a prime, as I said, but I'll ignore it for a minute). You know that p divides q(0) and that p does not divide r(0). Also p does divide q(M) r(M), since that is the polynomial you started with, and hence p must divide q(M). Now you can see why the difference between any and a above is important! In your case q(M) and r(M) are not polynomials. -- Peter van Rossum, | Universal law of linearity: for all Dept. of Mathematics, New Mexico | f : R -> R and for all x, y in R: State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y) === Subject: Re: Simple principle, core error proven > [...] >2. The important tool I use is a polynomial: >P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) > [...] >And the factorization with those factors is >P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) >where the a's are roots of the following cubic: >a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m). >So you can see that the a's are functions of m, > [...] >g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7) >and I can find those terms that are free of m by setting m=0, just >like with any polynomial, you know like with S(m) = 2(m^2 + 2m + 1), > ^^^ > It may seem irrelevant, but you probably mean a here. The formulation > used suggests that g_1(m), g_2(m), g_3(m) are given by polynomial expressions > in m. I'll explain a bit later why this is important. That's dumb. If you have P(x), setting x=0, that is, looking at P(0) works. That works with ANY polynomial. > [...] >to show that at m=0, the three factors are >g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22. >Now dividing P(m) by 49 gives >P(m)/49 = (2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7 >and the question is what happens to the g's, but look now at P(0)/49 >as that is >P(0)/49 = 3(5) + 7 >as two factors of 7, each 7, have beeen divided off, which is easy to >see. >The only way that can happen is if >P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) >where only two of the a's have 7 as a factor, where the idea is almost >trivially simple as consider a polynomial like > There are two mistakes here. First of all, 7 is not a prime > (in the ring of algebraic integers). So even for a specific > algebraic number m, from the fact that 49 divides > P(m) = (5 a_1(m) + 7) (5 a_2(m) + 7) (5 a_3(m) + 7) > you cannot automatically conclude that any of those three factors > is divisible by 7. You might be able to get around this > problem by looking at some factorization of 7 in the ring > of algebraic integers. I don't assume anything. The independent term for P(m) is 7(7)(22). The independent term for P(m)/49 is 22. Therefore, the factors in the factorization of P(m)/49 must NOT have 7 as a factor of their independent terms. It's BASIC ALGEBRA. If you expect the newsgroup to believe that your competence is so low that you can't handle such basic algebra then why should anyone listen to you on a math newsgroup at all? The other option is that you're LYING and lying rather badly as well. You incompetent lying boob, why can't you just tell the ing truth, you sick, twisted ??!!! James Harris === Subject: Re: Simple principle, core error proven > [...] >g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7) >and I can find those terms that are free of m by setting m=0, just >like with any polynomial, you know like with S(m) = 2(m^2 + 2m + 1), > ^^^ > It may seem irrelevant, but you probably mean a here. >That's dumb. If you have P(x), setting x=0, that is, looking at P(0) >works. >That works with ANY polynomial. ^^^ You're doing it again. This formulation suggests that the g_i's are polynomials in m, which they aren't. I've explained why this is important: if q(m) and r(m) are polynomials and p is a prime such that p divides q(m)r(m) and p does not divide r(0), then p divides q(m). You seem to be applying this in a case where you neither have polynomials, not a prime. >You incompetent lying boob, why can't you just tell the ing truth, >you sick, twisted ??!!! Interesting... -- Peter van Rossum, | Universal law of linearity: for all Dept. of Mathematics, New Mexico | f : R -> R and for all x, y in R: State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y) === Subject: Re: Simple principle, core error proven Adjunct Assistant Professor at the University of Montana. > [...] >g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7) and I can find those terms that are free of m by setting m=0, just >like with any polynomial, you know like with S(m) = 2(m^2 + 2m + 1), > ^^^ > It may seem irrelevant, but you probably mean a here. >That's dumb. If you have P(x), setting x=0, that is, looking at P(0) >works. >That works with ANY polynomial. > ^^^ >You're doing it again. This formulation suggests that the g_i's are >polynomials in m, which they aren't. I've explained why this >is important: if q(m) and r(m) are polynomials and p is a prime >such that p divides q(m)r(m) and p does not divide r(0), then >p divides q(m). If by p divides q(m)r(m) you mean the constant polynomial p divides the polynomial q(m)r(m) in A[x] or in Z[x]. If you mean, as James has claimed he does, just that p divides every VALUE of q(m)r(m), and you restrict m to the integers (as James usually does), then it is not true. q(m) = x+2 r(m) = x+1 p = 2 p divides q(m)r(m); p does not divide r(0); but p does not divide q(m). If you extend m to all algebraic integers, then Dot gave a proof that an integer w divides all values of a polynomial f(m) for arbitrary m in the algebraic integers if and only if it divides all coefficients of f(m). Then one can apply Dedekind's Prague Theorem to deduce that, if q(m) and r(r) have integer coefficients, then under the conditions you give p divides q(m) (every coefficient of q(m)): DEDEKIND's PRAGUE THEOREM. If the coefficients a0,...,a_n, b_0,...,b_m of two polynomials in one variable F(x) = a_0 +...+a_n*x^n G(x) = b_0 +...+b_m*x^m are algebraic integers, and the coefficients c_0,...,c_{n+m} of the product F(x)G(x) are all divisible by the integer w, then each of the numbers a_0b_0, a_0b_1,....,a_0b_m,a_1b_0,...,a_nb_m is also divisibly by w. So, if we know that every coefficient of q(m)r(m) is divisible by f, f is a prime, and f does not divide r(0), then it must divide r_0*b_i, where the b_i are the coefficients of q(m), and if r_0 and the b_i are integers, then f must divide all b_i, giving the result. But that requires that you assume not just that p divides every value q(m)r(m), but that it divides it as polynomials. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Simple principle, core error proven >If by p divides q(m)r(m) you mean the constant polynomial p divides >the polynomial q(m)r(m) in A[x] or in Z[x]. I meant the constant polynomial p divides q(m) r(m) in A[m], which I think is equivalent to p divides q(m) r(m) for all m in A. >If you mean, as James >has claimed he does, just that p divides every VALUE of q(m)r(m), and >you restrict m to the integers (as James usually does), then it is not >true. Ok, over the integers this doesn't work. >If you extend m to all algebraic integers, then Dot gave a proof that >an integer w divides all values of a polynomial f(m) for arbitrary m >in the algebraic integers if and only if it divides all coefficients >of f(m). So it indeed is equivalent :-) I must admit I don't know the proof right away. I guess you prove this for a prime algebraic integer w first. (Choose distinct elements m_0, ..., m_n of A/w where n is the degree of f. This is possible because this ring is infinite. Then det(m_i^j) <> 0. Write f = a_0 + a_1 m + ... + a_n m^n. Then the matrix product (m_i^j) (a_j) is zero (in A/w) since by assumption all f(m_i) are 0. Hence the a_j are 0 (in A/w). Extending this to products of distinct algebraic integer primes should be easy; the case with duplicate factors might be more complicated. Do you know if the statement is true for all algebraic integers w and not just for integers?) >Then one can apply Dedekind's Prague Theorem to deduce that, >if q(m) and r(r) have integer coefficients, then under the conditions >you give p divides q(m) (every coefficient of q(m)): Ah yes, this is more difficult if p is a prime integer instead of a prime algebraic integers. Good point. -- Peter van Rossum, | Universal law of linearity: for all Dept. of Mathematics, New Mexico | f : R -> R and for all x, y in R: State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y) === Subject: Re: Simple principle, core error proven Adjunct Assistant Professor at the University of Montana. >If by p divides q(m)r(m) you mean the constant polynomial p divides >the polynomial q(m)r(m) in A[x] or in Z[x]. >I meant the constant polynomial p divides q(m) r(m) in A[m], which >I think is equivalent to p divides q(m) r(m) for all m in A. They are equivalent for the ring of all algebraic integers. In general, the former implies the latter, but not conversely (as the example over Z shows). >If you extend m to all algebraic integers, then Dot gave a proof that >an integer w divides all values of a polynomial f(m) for arbitrary m >in the algebraic integers if and only if it divides all coefficients >of f(m). >So it indeed is equivalent :-) I must admit I don't know the proof right >away. I certainly did not either. Her proof is at It looked right at the time, though I confess I did not check it very carefully... > I guess you prove this for a prime algebraic integer w first. >(Choose distinct elements m_0, ..., m_n of A/w where n is >the degree of f. This is possible because this ring is infinite. Not entirely sure what you mean by A/w, but I am guessing you mean A/(w), moding out by the principal ideal.... >Then one can apply Dedekind's Prague Theorem to deduce that, >if q(m) and r(r) have integer coefficients, then under the conditions >you give p divides q(m) (every coefficient of q(m)): >Ah yes, this is more difficult if p is a prime integer instead >of a prime algebraic integers. Good point. Ehr; if p is a rational prime, then follows. Again, don't know what you mean by prime algebraic integer. My first impulse would be p is an element of the algebraic integers such that if p|ab, with a and b algebraic integers, then p|a or p|b. In that case, there are no such things, so I'm not sure what you could mean. My second impulse would be that p is a prime algebraic integer means that p is a prime element of the ring of integers of Q(p). In that case, I believe that Dedekind's Prague Theorem will extend to polynomials over the number field Q(p), and so you get the same result then. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Simple principle, core error proven >So it indeed is equivalent :-) I must admit I don't know the proof right >away. >I certainly did not either. >Her proof is at Ah nice. I didn't realize that you were referring to a > I guess you prove this for a prime algebraic integer w first. >(Choose distinct elements m_0, ..., m_n of A/w where n is >the degree of f. This is possible because this ring is infinite. >Not entirely sure what you mean by A/w, but I am guessing you mean >A/(w), moding out by the principal ideal.... Correct. >if q(m) and r(r) have integer coefficients, then under the conditions >you give p divides q(m) (every coefficient of q(m)): >Ah yes, this is more difficult if p is a prime integer instead >of a prime algebraic integers. Good point. >Ehr; if p is a rational prime, then follows. Again, don't know what >you mean by prime algebraic integer. My first impulse would be p is >an element of the algebraic integers such that if p|ab, with a and b >algebraic integers, then p|a or p|b. Hmm, I meant a prime element of the ring of algebraic integers, but those thing do not exist, indeed. That makes the statement true, though :-) -- Peter van Rossum, | Universal law of linearity: for all Dept. of Mathematics, New Mexico | f : R -> R and for all x, y in R: State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y) === Subject: Re: Simple principle, core error proven > You incompetent lying boob, why can't you just tell the ******* truth, > you sick, twisted ****??!!! And so the real James Gutter-Trash Harris slips off the mask and makes another appearance... -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Simple principle, core error proven > You might be able to get around this > problem by looking at some factorization of 7 in the ring > of algebraic integers. > I don't assume anything. The independent term for P(m) is 7(7)(22). > The independent term for P(m)/49 is 22. > Therefore, the factors in the factorization of P(m)/49 must NOT have 7 > as a factor of their independent terms. > It's BASIC ALGEBRA. If you expect the newsgroup to believe that your > competence is so low that you can't handle such basic algebra then why > should anyone listen to you on a math newsgroup at all? > The other option is that you're LYING and lying rather badly as well. > You incompetent lying boob, why can't you just tell the ing truth, > you sick, twisted ??!!! Ah... James... The sad thing is... He probably thinks he IS telling the truth!! I know... it hard to believe that... but it is my honest opinion. Magidin probably thinks he is telling the truth!! These mathematicians... they have been brain-washed!! You don't use mathematical terms in quite the right way... your definitions are slightly non-existent... and they use these little errors to proclaim that you are wrong!! The ONLY way you will ever overcome them... and convince them of The Truth... is to play the game on their terms. Once you start defining things according to THEIR rules... using technical terms the way THEY do... the scales will fall from their eyes! They will beat their breasts and wail!! You will be KING!!! -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Simple principle, core error proven ... > That's dumb. If you have P(x), setting x=0, that is, looking at P(0) > works. > That works with ANY polynomial. shown that with another polynomial your reasoning leads to sqrt(7) divisible by 7. What about that? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Game of pieces on 2 linear-boards I might as well post this (Backgammon-esque, kind-of) game, which could very fun,...perhaps.. *2 Players (one associated with black, the other with white). *Each player has k black and k white pieces (similar to Go pieces, but flat so as to be stackable). *2 boards, one per player, which are each a strip of (2k) positions where the pieces can be placed in stacks. *Each player places his/her pieces (secretly, so as not to be influenced by other player), in some arrangement of black and white, onto his/her board, exactly one piece per position. The boards are then revealed. At each player's move, a group of positions (each selected from 1 to 2k) is randomly chosen somehow. (In one game-variation, the number of positions chosen varies {and is 1 to 2k}; in the other game-variation, there is a fixed number of chosen positions.) At each move, a player does one of the following at EVERY picked position: A) If both stacks in same position on the boards are the player's color, then player moves one piece from one of the stacks to the other corresponding stack (in either direction). If the colors do not match in the two corresponding stacks at a picked position, these stacks are unaltered. or B) Switch all stacks (of either color) on the player's own board (only) at the picked positions with the stacks immediately to the left. (If far left stack is picked, switch with far right stack.) The game's goals vary with each game, and are predetermined by the players. Examples: 1) After fixed number of moves, player wins who has most number of his own-color pieces on left side of his board. 2) First player to get an even number (or prime number, or...) of pieces of his color on each position of his board occupied by his color (in the 1st place), wins. etc etc... Sample starting-boards setup (for visualization): * * o * o o * * * o * * o * * o o o o * o o o * * o * o * o o * o o * o * o * o * * * o Any mathematical analysis that anyone has to offer? Any game-improvement analysis either?? Leroy Quet === Subject: Re: math and selfstudie why don't you take some math courses at a local university? Bye. > I wish I could, I really wish I could, but age (being almost 30) and > money prevents it. is it that expensive to go to school in holland? > In holland they make it pretty difficult to enrole in university after > a certain age how, exactly do they do that? > and if you already have done another studie, the dutch governements > figures that you have had your shot, so that pretty much cancels it out. move to america. leave the sozialists to their putridness. > You certainly won't get money from them, not in this day and age. So > that's a no go. I should have thought of it when I was twenty. well. self study is often challengeing, specially when you lack adequate backround. if you are REALLY interested in learning something, and self study is unfeasible, is up to you to get the $$$ required to get the schooling. > Laters, > Dries === Subject: Re: math and selfstudie Dries === Subject: Re: simple question about subspaces at 06:57 PM, Jesper said: >Assume that V is a vector space >Assume that S is a subspace of V >Assume that O is the zero vector of V >Is O also the zero vector of S? In other words, is the zero vector >of V inherited by S? By the definition of subspace, only one answer is possible. Every vector in S must be a vector in V, and that includes the null vector of S. That plus the fact that addition in S gives the same results as addition in V should answer your question. Is this homework? -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: simple question about subspaces at 09:41 PM, Toni Lassila said: >I was wondering about the minimal required definition. I was told the >inclusion of the zero vector must be explicit to avoid a problem >where the empty set could be a subspace of V, Not with the standard definition of subspace. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: simple question about subspaces >The empty set is /not/ closed under >both theese operations. Of course it is. But if (S,F,+,*) is a vector space then (S,+) is a group and hence has an identity element. Thus if (S,F,+,*) is a subspace of (V,F,+,*), S cannot be null. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: simple question about subspaces >|If you are in a variety of algebras, then the empty algebra is an >|algebra in the variety if and only if the type has no nullary >|operations. So the empty set is not a vector space, since the type of >|vector spaces includes a nullary operation. >I considered mentioning that, plus the fact that one could eliminate >the references to 0 in the definition by having the axiom (x+(-x))+y=y >and so on. >Would that axiom not allow an underlying structure which is an >inverse monoid but not a group? Hmmm... Haven't thought it through; >perhaps the scalar multiplication takes care of that... I once undertook to rewrite the axioms for a vector space in such a way that any subset of them would make sense (which meant rewriting the one that says additive inverses exist since I wanted that to be independent of there is a zero vector); the goal was to see whether the axioms were independent. http://www.math-atlas.org/99/lin_alg The upshot is that you can describe vector spaces with seven axioms, all necessary in the sense that for any one of the seven there exist objects which fail to satisfy that axiom but satisfy the other six. dave === Subject: Is this a valid proof? I'm paraphrasing from Iain Adamson's Introduction to Field Theory. I don't think the proof given below is a valid proof, but is a trifle circular in argument. Can anyone agree/disagree/provide a clearer proof of the theorem? Is the theorem really a theorem? Notation: E is a field, F a subfield of E, S any subset of E. F(S), the subfield of E generated over F by S, is defined as the intersection of all subfields of E which include both F and all the elements of S. [In other words, F(S) is the subfield of E obtained by adjoining S to F.] Theorem: The elements of F(S) are precisely those elements of E expressible as quotients of finite linear combinations (with coefficients in F) of finite products of elements of S. [I guess this means the elements of F(S) are rational functions of elements of S with coefficients in F?] Proof: (a) Let K be the subset of E consisting of elements expressible in this way. Since the sum, difference, product and quotient of any two such elements can be expressed in the same form, it follows that K is a subfield of F. [OK, agreed] Further it is clear that K includes both F and S, and hence includes F(S). (b) On the other hand, since F(S) is a subfield of E including F and S, F(S) includes all finite products of elements of S, all finite linear combinations of such products with coefficients in F, and hence all quotients of such linear combinations. So, F(S) includes K. So, F(S) = K. End of proof. Having typed all this, I'm a bit more convinced that the argument constitutes a proof, but couldn't it be done better? Or is the Theorem hardly a theorem as such? Any comments gratefully received. Guy Corrigall === Subject: Re: Core error, FEAR is a natural response as no-one has resolved the issue of TheRealJSH@MSN, I'm afraid that Harris is dead, thus allowing someone who likes to go even more ofer-the-top to emulate him. if this is so, I offer my condolences to his wife etc.; on the other hand, maybe *she* can win the prize. > What the hell are you talking about? What is a prediction made by > math? > What James has found is an error in basic mathematics, which, if corrected > and followed through obvious algebraic derivations, will prove Fermat's Last > Theorem on one page using methods which can be understood by any 4-year > mathematics graduate. > The logical hingepoint of the proof is that you can determine factors > for a polynomial P(m), by focusing on the constant terms, as they are > independent of the value of m, and figure things out by watching what > happens to them after some simple algebraic manipulations. > As the proof is so simple, objectors have continually argued a > position that requires that the constant terms are in fact NOT > independent of the value of m, and they have refused to acknowledge > the full proof, choosing instead to pick at in pieces and ignore basic > algebra. --les ducs d'Enron! === Subject: Re: Core error, FEAR is a natural response Discussion, linux) > That's partly how mathematicians were fooled into thinking the ring is > without problems as it IS closed under those operations; however, > there is decomposition or factorization, even in the ring of algebraic > integers. > It's in decomposition that the problem arises as you can have an > algebraic integer that results from multiplying an algebraic integer > times a number from a more inclusive ring, like with my x=2a example, > where 'a' is an object but not an algebraic integer, while x and, of > course, 2 are both algebraic integers. But don't you have the same problem in your object ring? Let x=1. Then x is an object. Let a = 1/2. Then a is not an object[1], but a number in a more inclusive ring. Then x = 2a, where x and 2 are both objects. How does this differ from the flaw in the algebraic integers? [1] According to what you've written, at least. Of course, since lately you've been claiming that Goedel's work shows that human beings are objects, too, who can tell? By the way, what is sqrt(2) times James Harris? Also, what is James Harris + George W. Bush + 4? I'm a little perplexed about how the object ring's operations extend to humans. -- [I]t's good for the economy to charge for intellectual property, so open source software cannot be good, while Microsoft is the most far-thinking company around and is doing it all for the good of the public. -- Linus Torvalds paraphrases Microsoft VP Craig Mundie === Subject: Re: Magidin is too many > /~~~ > / | /~~| `=` |/~~ > /_/ | `` | | `` |/~~ > / | | | | `` | > { <| | | | | > | | > | / . . ` / > | . . / > . / > . / > / > / > | | > | | --les ducs d'Enron!