mm-1669 === Subject: [JSH] Re: Pure math, physics, and FLT Mr. Harris, Where is the non-polynomial factors of polynomials described? I don't see it described in your latest post. Don't you agree that your post does not contain a full definition? All I see is that you have substituted a value for x. Does 5 times 4 describe the non-polynomial factors? What do you mean by a non-polynomial? > Now I've extended mathematics yet again by considering non-polynomial factors of polynomials. ... Now I've pushed with the factoring of \ polynomials into non-polynomial factors. > Factoring polynomials into polynomial factors is like factoring > integers into integer factors. > Like x^2 + 3x + 2 = (x+2)(x+1), with x=3, you have 20=5(4). > 3^2 + 3(3) + 2 = 20 > ... > James Harris === Subject: Re: ATAN2 function, Single Precision -- Any example routines >I'm trying to write ATAN2 function for a small basic language that has >IEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are >availible in the language. >I've tried a few methods I've found but the results are way off due to >low precision, rounding, etc. >Are there any repositorys of old fortran routines or algorithms that I >could use to get a good accuracy single precision routine. Speed or >space aren't important, only good accuracy. > It suffices to be able to compute arctan(x) for x in [0,1]. > Try > .0318159928972*y+.950551425796+3.86835495723/(y+8.05475522951+ > 39.4241153441/(y-2.08140771798-.277672591210/(y-8.27402153865+ > 95.3157060344/(y+10.5910515515)))) > where y = 2*x-1. The maximum error is about 8*10^(-10). > 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: Axiom of choice: is it wrong? > Well, definable isn't actually definable, formally. Informally, > a real number is definable if there is some description of that > number such that there is exactly one number meeting that description. > For example: That number whose square is equal to 2 uniquely > describes the number square-root(2). No it doesn't. That number whose square is equal to 2 equally well describes the number minus square-root(2). Mark Atherton === Subject: Re: Axiom of choice: is it wrong? >> The axiom of choice, that says it's always possible to choose one >> element of each member of a family of sets, sounds intuitively right. >> The axiom of choice doesn't really say that. It says that whenever A is >> a family of non-empty sets, then there is a choice set C, s.t. for all a \ >> in A a / C = {z} (where / tries to denote intersection). It's a set >> existence axiom, and it's not about our ability to actually choose >> anything. For example, in case of a family of alpeh_omega disjoint sets, \ >> what could possible amount to our ability to choose a single member from \ >> each of these sets? Our ability to write down a 1st order formula to >> pick one from each? > If a set can't be possibly defined in any way, what is the sense of > saying that it exists? We can't describe what's inside the event horizon of a black hole, but black holes nevertheless exist. If a set exists (such as the power set of the naturals), then we would like to say that each of its members exists, even though there may not be enough descriptions to go around. If we couldn't say that, we would lose large parts of real analysis (such as the completeness of the reals, for example). That would invalidate such things as the intermediate value theorem. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Axiom of choice: is it wrong? Mark says... >> For example: That number whose square is equal to 2 uniquely >> describes the number square-root(2). >No it doesn't. That number whose square is equal to 2 equally well >describes the number minus square-root(2). Whoops! You're right. On the other hand, for any description such that only *finitely* many reals satisfy the description, there is another description such that only one real satisfies the description: you can always say The smallest (or largest) real such that ... An undefinable real would be one such that every description of it also equally well describes infinitely many other reals. -- Daryl McCullough Ithaca, NY === Subject: Re: Axiom of choice: is it wrong? >> Well, definable isn't actually definable, formally. Informally, >> a real number is definable if there is some description of that >> number such that there is exactly one number meeting that description. >> For example: That number whose square is equal to 2 uniquely >> describes the number square-root(2). >No it doesn't. That number whose square is equal to 2 equally well >describes the number minus square-root(2). That number whose square is equal to 2, with no other conditions, describes nothing. It only has meaning if there is exactly one such number. -- 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, Deptartment of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Can you copyright an equation? > > Also, public description by the developer of the algorithm is a patent > claim. This is obvious if the developer is claiming something unique, > Is it? It could certainly be used as prior art to defeat a subsequent > patent claim, but public description is not a patent claim in and of > itself. > original, or proprietary. (This is fundamental or natural law but > structured > law adds a twenty year limit...and also recommends actual patent > registration.) > Rubbish. > To understand patent law as natural law or fundamental law just realize that > if someone in your own network is associated with a development that you are > expected within that network to respect the developer. Then formal law just > expands that situation across society... To understand patent law as an artificial creation of government with no natural standing whatsoever, consider the case where government does not grant patents (monopolies) nor enforces them with threats by police. Patents and other IP cannot exist in a naive free market. They are all about restricting the (re-)production and sales of goods. Ideas do not exist as property in places like Somalia which have no government. Property can be protected by putting it in a safe place and stationing guards to restrict access. Intellectual property is like putting your bicycle in everyone's garage and expecting the state to control what people do with it. It's a whole different thing altogether. This doesn't say that there are not valid arguements for patents and copyrights nor do I mean that they are inherently bad. They are, however, created out of whole cloth by government fiat and enforcement. [1] the word patent derives from mercantilism where the king would grant monopolies to his cronies. > subsequent future claim by showing previous claim...Well, this is obvious \ if > the public description was by a developer or inventor rather than by a > technical or historical editor. > Actual patent registration is a formality. Not so as another poster points out. > This can protect the independent > developer who may not have legal resources. But in practice it seems to enforce the status quo and is often a tool of large corporations to keep out competition by smaller entities. Only once in a while does a small company upset a big one's applecart and it usually makes the news while the reverse is rather commonplace. -- === Subject: Re: Complex Numbers >Is there a special name for functions which are like analytic functions >except that they reverse the orientation (such as complex conjugation)? >> >>I might call them conjugate-holomorphic. >> >> >Such >functions and analytic functions seem to be about equally fundamental. > >> >>You think the function z (the identity function) and zbar (the >>conjugation operation) are equally fundamental? > Hard to say, but I think the group consisting of rotations and reflections > of a one-dimensional real vector space is about as fundamental as the group [*] > consisting of rotations only. I think a mapping f: V->V from a complex > vector space to itself which is such that f(v1+v2)=f(v1)+f(v2) and > f(z.v)=(z^bar)f(v) (that is, an anti-linear transformation) is about as > fundamental as a transformation g:V->V such that g(v1+v2)=g(v1)+g(v2) and > g(z.v)=z.g(v) (that is, a linear transformation). I think a mapping from an > analytic manifold to itself which is locally an anti-linear transformation > is about as fundamental as a mapping which is locally a linear > tranformation. [...] > Mattias, > Perhaps you intended two-dimensional real vector space in line [*]? > I went looking for a difference between the group of > (R-linear) rotations and reflections that act on the euclidean plane > R^2, on the one part, and the group consisting solely of > (R-linear) rotations acting on R^2. > What I found is this: the rotations only form a commutative > group, whereas the rotations and reflections combined > form a group which is not commutative. I will have to think more about this. > For instance, consider the reflection R_x about > the x-axis. > Then, R_x(x,y) = (x, -y). and then another reflection, > this time about the line x=y, which I denote R_xy: > R_xy(x, y) = (y,x). > An easy case to check is using the vector (1,1): > R_x(R_xy((1,1))) = (1, -1) and on the other hand > R_xy(R_x((1,1))) = (-1, 1) . > David > P.S. z |-> z^{bar}, is not an analytic map; z |-> z is. > If there is a basic difference between +i and -i in > complex analysis, I'd be interested to know... Children have to learn to distinguish between left and right, that is, to impose the same orientation on any surface they see (when they look at the surface from its backside, it is a new surface). Perhaps mathematicians have to learn to not automatically to impose that orientation on any orientable two-dimensional surface they consider. Mattias === 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.electrochemistrySoftware.com === Subject: Re: Factorial/Exponential Identity, Infinity > lim_n->oo n!/(2(n/2)!) = lim_n->oo 2^n > > There's only one variable, n. If it was x on the left and y on the > right then it wouldn't be true uness x=y. > Wrong! The variable on each side is a bound (or dummy) variable > which has n meaning outside of the limit statement, so changing the > variable within either limit statement has no effect on the meaning > of that limit statement. That's fair. I concede that and say that what you can do is use those terms as an equality and then take the limit after moving the variables to one side of the equation. I went through my proof about the identity again and found a mistake. It's trivial, now my opinion and belief is that: lim_n->oo n! ( (n/2)! 2^n) = 1 Notice there is no longer the lone 2 in the denominator. Let me describe my rationale for this statement and derivation of this identity. The binomial probability distribution describes how many of a given sequence of coin tosses will be heads and how many tails. The coin toss is an event with random results that has equal probability of achieving either of two possible results. There is a function that determines with some exactitude the probability of a given number of successes for a given number of trials. The function is often denoted or symbolized as the number of trials over the number of successful results, contained in parentheses. This notation doesn't easily transfer to ASCII or other text that doesn't contain mathematical symbols. The function thus describes for a sequence of true/false, yes/no results the probablity that a given fraction of the results will be one or the other. The derivation considers all of the infinite sequences of binary values, zero and one, 0 and 1. The first thing to notice is among all of the sequences A, at a given index i in the sequence half of the sequences have a value of zero and the other half have a value of 1. Here it is convenient to consider the infinite bit sequences as being representative of the expansions of the reals between zero and one inclusive, yet the same should hold true for all reals. Consider the first element of the sequence, i=1, it is so that for all of the values less than one half that the value for each A_n of element i is equal to zero, and for the other half of the values it is equal to one. .000... = 0 ... .100... = .011... = 1/2 ... .111... = 1 Another example is to consider for any finite and even integer the enumeration of all the sequences of that length to see that half of the values of A_n_i are zero and the other half one, to show it true for all finite cases. 00 01 10 11 Notice that in the first column there are two zeros and two ones, and in the second column two zeros and two ones. So what can be understood from this is that for any given sequence A_n of the sequences A, that for any index i that the probability of the value of A_n at i being either value is equal and is equal to one half. The binomial probability function is as follows: the probability P of k successes in n trials with p probability of success and q probability of failure is: P = (n! / k!(n-k)!) p^k q^(n-k) We have that p = q = 1/2. In the case of the infinite bit sequences, n->oo. The value k = n/2. This is because we are concerned with the sequences of length n where half of the elements are successes. Thus k = n-k = n/2. Thusly, k! = (n-k)! = (n/2)!. In multiplying p^k and q^(n-k), p = q = 1/2 and k = n-k = n/2 thus that (p^k)(q^(n-k)) = (1/2)^(n/2) * (1/2)^(n/2) = (1/2)^n. P = lim n->oo (n!/ (n/2)! (n/2)!) (1/2)^n = n! / (2 (n/2)! 2^n) Here's the major assumption: half of the sequences of A contain equal numbers of zeros and ones. Thus the probability of an infinite sequence having equal numbers of zeros and ones is 1/2, that is to say, P = 1/2. 1/2 = lim n->oo (n!/ (n/2)! (n/2)!) (1/2)^n = n! / (2 (n/2)! 2^n) At this point I am thinking that I made an error in that I forgot to so I correct that to get: lim n->oo n! / ((n/2)! 2^n)) = 1 As the expression evaluates to equal to one, so does its reciprocal. Notice that I make a huge assumption: that half of the sequences contain equal numbers of zeros and ones. This is not only that half the values at a given index are zero and the other half one, it is saying that of all the sequences that half of the sequences have equal numbers of ones and zeros, and that the entire population describes the distribution. This is true for all finite even values of n. Ross === Subject: Re: Factorial/Exponential Identity, Infinity ........... > I went through my proof about the identity again and found a mistake. > It's trivial, now my opinion and belief is that: > lim_n->oo n! ( (n/2)! 2^n) = 1 > Notice there is no longer the lone 2 in the denominator. ................. > Ross Not quite right. If you apply Stirling's theorem in the form n! ~ sqrt(2*pi*n)*(n/e)^n, you get (2n)!/(4^n*(n!)^2) ~ 1/sqrt(pi*n) Martin Cohen === Subject: Re: Gleichungen > 5x - 8 = 16x - 63 | - 16 x > - 11x - 8 = - 63 | + 8 > - 11x = - 55 | : 11 > x = - 5 ; L = (- 5) Aber Falsch! Wenn jede Seite der Gleichung mit - 1 multipliziert, kriegt man die selbe L.9asung, d.h. x = +5 ! === Subject: Re: help needed with infinite series > I want to sum following infinite series to obtain a closed form expression. > sum { ( x ^ ( 2 ^ 2n ) ) * 2^n } , n=-infinity to infinity , 0 ******************************************************** > If f(x) is the value of this series at x, then f satisfies > the functional equation f(x^4) = f(x)/2. >>************************************************ >>The function g(x) = k/sqrt(|ln(x)|) satisfies this >>functional equation, but that is not sufficient to say that >>f = g for some k. The best I can say is that there are >>values for k for which the function f-g satisfies this >>functional equation and has an infinite number of zeros, the >>set of which has 0 and 1 as limit points. >Numerically it appears that f(x) sqrt(|ln(x)|) does oscillate, >in a range between about 1.27563 and 1.28148. One can get a better handle on this by using the Poisson summation formula. Let x = 2^{-y^2}, and consider h(x) = yf(x). Write z(n) = 2^{n + log_2(y)}; then h(x)=sum z(n)*2^{-z(n)^2}. Then h(x) = sum c_k exp(2k*i*pi*log_2(y)), where c_j = int 2^{t-2^{2t}}*exp(-2jitpi) dt, which can be expressed in terms of Gamma functions of complex arguments. -- 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, Deptartment of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: help needed with infinite series > Did you test for convergence and plotted the function for n-> > infinity ? For negative n , convergence is obvious becuase of 2^n term. Let's consider the case of positive n. Let a_n be nth term of the series. then , for all real k as n->infinity , ( (n^k) * a_n ) can be shown to tend to 0. => for sufficiently large n , a_n < 1/(n^k) since series sum{ 1/(n^k) } converges for k>1 , sum {a_n } also converges > Perhaps not every convergent series has a closed form, i.e., > expressible in terms of transcendental/special functions, but sure can > be always named. If the the series sum establishes some mathematical > relationships, identities or links to significant scientific > application, it can be named, gradually gains currency and then later > it becomes closed form! I agree. But, the form of series looked so standard , I presumed it must have been already encountered in some other applications. === Subject: Re: help needed with infinite series > > > Hi , > I want to sum following infinite series to obtain a closed form expression. > > sum { ( x ^ ( 2 ^ 2n ) ) * 2^n } , n=-infinity to infinity , 0 > ******************************************************** > If f(x) is the value of this series at x, then f satisfies > the functional equation f(x^4) = f(x)/2. > > ************************************************ > The function g(x) = k/sqrt(|ln(x)|) satisfies this > functional equation, but that is not sufficient to say that > f = g for some k. The best I can say is that there are > values for k for which the function f-g satisfies this > functional equation and has an infinite number of zeros, the > set of which has 0 and 1 as limit points. Earlier, I had posted another form of this functional equation , related by substitution x = exp(- y^2) , but the solution I got from follow-ups was too general. I include that ( along with one follow-up ) here . hope it helps ----------------------------------------------------------------------- >I want general soln. of following recursive relation, >f ( (2^n) x ) = 2^(-n) f(x) >one obvious form for soln. is f(x)=K/x , but I am not sure whether any >other solution exist. I presume n is supposed to be an integer here. If we define g(t) = exp(t) f(exp(t)), your equation says g(t + ln(2)) = g(t). Thus you can take any function g that is periodic with period ln(2), and a solution for x > 0 will be f(x) = g(ln(x))/x. Similarly for x < 0 you can take f(x) = h(ln(-x))/x where h is any periodic function with period ln(2). For x = 0 your equation says f(0) = f(0)/2, so f(0) = 0. ------------------------------------------------------------------- Sanjeev === Subject: Re: Homo High > OK, if you want to nit-pick, change children to high-school students > in my original reply and it still applies. > If they're so sexually active, why aren't they being taught in a > sexually active way? Teachers often do inject sexually provocative comments to interest bored high school pupils. > 2. Homosexual high school students need the extra encouragement to develop > their intellect in topics that often don't interest homosexuals, such as > math. > And what about the topics tht often don't interest heterosexuals? Yes, there are math and science programs for girls, for example. > There's bound to be some. Would you recommend sexual provocation to make > them more interesting to heterosexual students? My math problems were not necessarily about sex. > 3. Math is a political instrument, despite its pretenses to an objectivity > devoid of context. > I disagree completely. Math is the most objective science there is. Its > applications might be political instruments but math itself is another > thing. Then it is better to have a mathematically literate population to raise the level of political discourse. In today's world, the math used in politics is full of basic errors, like equating correlation with causality. > You seem to still think that all there is to homosexuals is the fact > that they're homosexual. Didn't say they are. > Have you ever met any? If you have, have they > ever done anything except have sex? For people who live in, say, rural environments where nobody is open about their homosexuality, their only encounter with a known homosexual might be a man who approached them in the bathroom during a visit to a big city. For such people, the notion would be true. Residents of big cities, who have been carefully indoctrinated to tolerate homosexuality, might be inclined to tell rural people that they deserved the incident. Homosexuals still like to have their gay pride parades in which they have public displays of nudity or even outdoor sex, which do not help their image. === Subject: Re: How to start with this problem? > Let A,B and C be sets such that AUB=AUC and > A intersection B = A intersection C, show that > B=C. > Well, it seems obious to me that B=C, but mathematically, > where do I start from? Sure its a homework question, > but can any one just please give me the first one or > two steps to begin? > msk Some hints: Two sets B and C are equal if, for every element b in B, b is also in C, AND for every element c in C, c is also in B. If b is in A intersection B, b is in B of course; and it's given that then b is in A intersection C. Since every element in A intersection C is in C (why?), then... If b is in B, and b is NOT in A intersection B, then b is not in A. In that case, since b is surely in AUB, it's given that b is in AUC - how did it get there? That will let you show that b in B implies that b is in C. Then turn the above argument around, starting with c in C. --------------------------------------------------- C Brown Systems Designs Multimedia Environments for Museums and Theme Parks --------------------------------------------------- === Subject: Re: Interesting question. > In the ring 2Z / 18Z, whose elements can be given as > {0, 2, 4, 6, 8, 10, 12, 14, 16}, > the additive identity is 0 and the multiplicative identity > is 10. > What are the rules for a ring (versus a field)? > I noticed that 6x6 = 6x12 = 12x12 = 0, so quotients are > not unique within this ring. A ring has two operations, which let's call addition and multiplication. The addition is associative and commutative, has an identity element (which let's call 0), and each element has an inverse. The multiplication is associative and distributes over the addition. To get from here to a field, you need the multiplication to be commutative; you need an identity element for the multiplication (which let's call 1); and you need each non-zero element to have a multiplicative inverse. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Johnny Can't Add But Suresh Venktasubramanian Can > Here is a pattern I've noticed in countless organizations > at the high end of the research spectrum. In the > personnel lists, certain groups are phenomenally over- > represented with respect to their appearance in the > general American population: Chinese, Koreans, Indians, > and, though it doesn't show in the above lists, Jews. Surely many of the followups have missed an obvious point. Chinese and Indians may make up a relatively small percentage of the American population but they make up an enormous percentage of the world population. It just so happens that many of the top research institutions in the world are located in America, so it's no surprise that top researchers from around the world would gravitate there. The take-home message is not that Americans are lazy or stupid or are falling behind non-Americans. It's simply that America still happens to have some of the best academic and industry opportunities on the planet, and---hardly surprisingly---those opportunities attract a diverse group whose ethnicity is going to be little more world class than the ethnicity of a random sample of Americans. In a similar vein, would anybody seriously find it surprising that most of the people in, say, Cornell's math Ph.D. program aren't originally *from* upstate New York and then conclude that upstate New Yorkers aren't trying hard enough or that upstate New York educators are drowning our children in self-indulgent social engineering, political correctness, and feel-good substitutes for learning? I mean, get a ing grip. As for the mention of Jews, that, too, is easily explained. Whenever pin-shaped head, Jews top the list as an obvious target group. As noted jackass Fred Reed admits, his random clicking on websites turned up no hard evidence that Jews had taken over American research activities, but that's because Jews typically disguise themselves using false names to better infiltrate and destroy American society. Heck, points out Fred, even a guy named Miller could be a secret Jew! Fred probably had a few dozen paragraphs on the Protocols of the tragically cut to make room for more whenever you burn a flag, a kitten dies public service advertisements. -- Kevin === Subject: Re: Mapping of integers to reals ~ Cantor's disproof >>I'm using the fact that H(n) is noncomputable to prove that H(555) cannot >>be equal to 1 *and* that H(555) cannot be equal to 0. >> >> Well, that doesn't make any sense. The fact that H is a noncomputable >> function doesn't imply that H(555) is noncomputable. >But H(555) can be noncomputable, > No, it can't. H(555) is some natural number, and every natural > number is computable. *some* virtually means noncomputable! BusyBeaver(555) is some natural number, and every natural number is computable. Are you saying BusyBeaver(555) is computable? Herc === Subject: Re: Mathematicians, helping each other? >You don't necessarily need a correct paper or a proper argument. >You just need your group. >James Harris How not correct or not proper could the paper and the argument be? I almost hesitate but suggest the quoted stuff above should be clarified. G C === Subject: Re: Mathematicians, helping each other? yes, but we would-be, professional, and avowedly NOT mathematicians should all in deed do have a big hug! > You don't necessarily need a correct paper or a proper argument. > > You just need your group. > Eventually, a consensus was reached. Others here have done the same. > You have not. > It isn't about mathematicians having a group hug. It's about everyone http://members.tripod.com/~american_almanac/newdark.htm === Subject: Re: Mathematicians, helping each other? > Some of you don't realize that having rather basic algebra in my work > means that I now know what mathematicians are capable of, when it > comes to lying about even basic mathematics. > So I can look over the entire field recognizing just what is possible > from mathematicians themselves, with information from my own > experiences to see exactly how it can be done. > Basically in mathematics the community aspect allows mathematicians to > help each other. > You don't necessarily need a correct paper or a proper argument. > You just need your group. > James Harris Why not just stick with your recent efforts to haunt the literary newsgroups and leave math alone? Your 'right brain' probably works better than your 'left brain' -- it certainly couldn't be worse. Failing at science and reality won't prevent you from succeeding at fiction and fantasy, but stay away from science fiction. We don't need bad science, even in a fictionalized platform. HTH -- 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 === Subject: Re: Math in my favor, short FLT proof found > TWO of the a's have a factor of 2 that is in fact sqrt(2) I thought you were going to clean up this confusing terminology? V. -- mail me at lastname at cs utk edu === Subject: Re: Math in my favor, short FLT proof found < snip > Let us know when whatever it is your talking about gets published in a reputable journal. -- Bob Day === Subject: Re: Math in my favor, short FLT proof found > What is amazing is how many of you seem to be taken in by bad > arguments for bogus mathematics as I've given > 2x^3 - 3x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) > where only TWO of the a's have a factor of 2 that is in fact sqrt(2). > 2x^3 - 3x + 1 = (x-1)(2x^2 + 2x - 1) > That's a reducible equation so you can check. It comes from > (v^3+1)x^3 - 3vx + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) > where v=-1 + mf^2, so with the example above I have v=1, so m=1, > f=sqrt(2). > Now I *prove* mathematically that ONLY two of the a's will ever have a > non unit factor of f, when m and 3 are coprime to f, and that factor > will be sqrt(f). And I can show that with an example as I do above, > but people arguing against rather basic algebra, claim that > reducibility is what matters, but they can't show you an example, like > I just did. Here's an example: P(x) = 2*x^3 + 1. IRREDUCIBLE, and if it is factored in the form (a1*x + 1)*(a2*x + 1)*(a3*x + 1), then each of a1, a2, and a3 is divisible by 2^(1/3). Explicitly, a1 = -2^(1/3), a2 = 2^(1/3) * (1 + sqrt(-3))/2, and a3 = 2^(1/3) * (1 - sqrt(-3))/2. This should make it obvious that irreducibility is a key requirement. Your example, being reducible, is irrelevant. > http://groups.msn.com/AmateurMath > and if you don't want to join the group to get the pdf, you can find > links to the methods provided that are used in the paper. Those > methods are also key in a short proof of Fermat's Last Theorem, which > is also there. > So all you have to do is look at > 2x^3 - 3x + 1 = (x-1)(2x^2 + 2x - 1) > and see an example where my methods work. Those methods are > extraordinarily basic, but they are radical as I factor polynomials > into non-polynomial factors. See above. Your methods do not work in the situation in which you need to apply them. > That's it. Think carefully, what I do that's so extraordinary is > factor polynomials into non-polynomial factors. Why haven't > mathematicians done that already? Why would they fight correct > mathematics which uses such a fascinating approach? It's not correct. Several proofs that your Advanced Polynomial Factorization claims are incorrect have been given. You have not answered them. > The paper in near current form went by the New York Journal of > Mathematics without claim of error, as the chief editor said it didn't > fit their format. It's currently at another journal and I'm waiting > to hear more from them. > So here you have my claim that I've found a *short* proof of Fermat's > Last Theorem, where the methods used involve factoring polynomials > into non-polynomial factors, which you can't find in all of > mathematics outside of my work, meanwhile I face a lot of hostility > over my work, from people who can't show an error within the work > itself. People have identified exactly where the error is and what it is (m = 0, etc.). You refuse to listen. Andrzej > James Harris === Subject: Re: Nonprimes & Primes (All of Them) > I have reworked my original paper ( > http://www.tln.net/~reriker/prime.html ) based on feedback from other > sci.math posters. > I am not sure that I have demonstrated it in a rigorous enough manner, > but I have attempted to demonstrate that it is possible to generate > all of the nonprime numbers. Once you have generated these, any gaps > left are the prime numbers. > Perhaps this is too unwieldy to be of any general use. However, > perhaps it is of some interest (assuming, of course, that I have not > made any blatant errors ; of course, if I have, I am sure that someone > will point them out :-;) I'm getting a blank page for your link... is it me? === Subject: Re: n-prime sets ##### This is what the thread is about: # Following David Conlon we define: # A multi-set of n positive integers {x_1,...,x_n} is # called n-prime if for each j with 1 < j <= n and # k from 1 to n there is a subset S of {1,...,n} # with |S|=j, k in S, and sum_{i in S} x_i prime. > (Explaining the not found for n=6:) > I was looking for examples with just two values. Looking especially for 6-prime multi-sets with only 1's and 2's as members, I was able to give a proof for the non-existence. There are only two prime-adding families of size 6 and consisting of 1's and 2's alone: A = {1,1,1,1,1,2} and B = {1,2,2,2,2,2} For size 3 there are only these two families: C = {1,1,1} and D = {1,2,2}. For size 5 there are only these two: E = {1,1,1,1,1} and F = {1,1,1,2,2} A is not 6-prime: Family C is the only sub-family with size j=3. And element x_6 = 2 is not a member of E. B is not 6-prime: There is no prime-adding sub-family with size j=5, since neither of E and F is contained in B. This was a small exercise in putting the words correctly. I hope it worked. Corrections are very welcome. I still like the puzzle. Rainer Rosenthal r.rosenthal@web.de === Subject: Re: Please recommend some entertainment math books > > Abbot EA. Flatland. (Stewart also has an annotated version, which I > have not yet read.) >> >> >No, Stewart's book is a sequel, not an annotation. It's called >Flatterland. You needn't have read Flatland first. It belongs >on the list. > Incorrect, Sir. In addition to Flatterland, there is another book > entitled The Annotated Flatland. Perseus Publishing: Cambridge, MA, > 2002. It is sitting right here on my shelf. I sit corrected. Stewart has indeed published an annotated Flatland, as well as the sequel, Flatterland. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Pure math, physics, and FLT Today many mathematicians engage in what they call pure math. However, if pure math had been the rule for mathematics, would humanity even know of irrational numbers like sqrt(2)? Irrational square roots come up in construction. Construction drove a lot of early mathematical research, like working out pi. You know, for building things with circles. Now I've extended mathematics yet again by considering non-polynomial factors of polynomials. Go ahead, take a look in all of established mathematics where you see polynomials, and notice all there is about polynomial factors, but what about non-polynomial factors? It's like if pure math mathematicians had refused to consider irrational numbers, like sqrt(2), and were busily occupying themselves with integer factors. Mathematicians were pushed not only into using irrational numbers--notice the name--but into using imaginary numbers, now also called complex numbers. Intriguingly, factoring polynomials of degree 3 and higher pushed mathematicians there, but those of you in physics also know that complex numbers have their role in helping humanity understand the real world. Throughout history mathematicians have been pushed and they push back. Numbers that weren't perfect integers were called irrational. When the square root of negative numbers were needed mathematicians fought back calling them imaginary. Um, mathematicians don't have a great record here people. Now I've pushed with the factoring of polynomials into non-polynomial factors. Factoring polynomials into polynomial factors is like factoring integers into integer factors. Like x^2 + 3x + 2 = (x+2)(x+1), with x=3, you have 20=5(4). 3^2 + 3(3) + 2 = 20 Why am I pushing this issue? Because mathematicians, pushing back against a new concept, are fighting a short proof of Fermat's Last Theorem that uses factoring of polynomials into non-polynomial factors. That's like mathematicians fighting use of sqrt(2), but construction workers disagreed. Or like mathematicians fighting sqrt(-1), saying it was imaginary, but physics now disagrees. Now they fight non-polynomial factorization, and I'm stuck. What gives? The problem is pure math which means that mathematicians don't have to go down paths they don't like. If previous mathematicians were into pure math then humanity might not even have sqrt(2), oh, but those construction workers wouldn't have gone along. So now I need people *outside* of mathematics to push. Make them follow mathematical logic instead of doing what they like, and show that not only construction workers can make mathematicians follow the math. James Harris === Subject: Re: Pure math, physics, and FLT > Mathematicians were pushed not only into using irrational > numbers--notice the name--but into using imaginary numbers, now also > called complex numbers. > Intriguingly, factoring polynomials of degree 3 and higher pushed > mathematicians there, but those of you in physics also know that > complex numbers have their role in helping humanity understand the > real world. Actually, this demostrates just how little JSH knows about math, either pure or applied. Both irrationals, like the square root of 2, and complex numbers, like the square root of -1, first arise from polynomials of degree 2, not 3. === Subject: Re: Pure math, physics, and FLT > Today many mathematicians engage in what they call pure math. > However, if pure math had been the rule for mathematics, would > humanity even know of irrational numbers like sqrt(2)? Irrational > square roots come up in construction. Construction drove a lot of > early mathematical research, like working out pi. You know, for > building things with circles. Surely not. One might need an approximate value of pi for building things with circles. One can't need any irrational values for constructing anything, else nothing would get constructed. GC === Subject: Re: Pure math, physics, and FLT A8EwTYfhf*u~,Eu,tf6$HN*MY&)u0G =N' x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|'1{w70ZY=ih,=)kMY_}?{%)x0)];K~@J6m5.EN?>Zh Xh;Y V|',x(js'Jfq02joVpj|#x >> Mathematicians were pushed not only into using irrational >> numbers--notice the name--but into using imaginary numbers, now also >> called complex numbers. >> Intriguingly, factoring polynomials of degree 3 and higher pushed >> mathematicians there, but those of you in physics also know that >> complex numbers have their role in helping humanity understand the >> real world. > Actually, this demostrates just how little JSH knows about math, > either pure or applied. > Both irrationals, like the square root of 2, and complex numbers, > like the square root of -1, first arise from polynomials of degree > 2, not 3. Stunningly, James persists in believing that irrational is a pejorative applied to certain numbers. He was explicitly told the origin of irrational (from the root word ratio, of course) at least as early as August 12, 2000. Well, perhaps it isn't *really* stunning. -- It's an exercise in game theory[...] I've been brutally logical in my analysis on this point.[...] My analysis indicates that the optimal strategy for mathematicians is to acknowledge the result today. --JSH gives practical reasons to accept his FLT proof === Subject: Re: Pure math, physics, and FLT > Mathematicians were pushed not only into using irrational > numbers--notice the name--but into using imaginary numbers, now also > called complex numbers. > Intriguingly, factoring polynomials of degree 3 and higher pushed > mathematicians there, but those of you in physics also know that > complex numbers have their role in helping humanity understand the > real world. > Actually, this demostrates just how little JSH knows about math, > either pure or applied. > Both irrationals, like the square root of 2, and complex numbers, > like the square root of -1, first arise from polynomials of degree > 2, not 3. The cubics come into the story because Cardano (or was it Bombelli?) found that even to solve a cubic with real roots he had to use complex numbers. The quadratic with real roots can be solved without an excursion through the complex numbers. In modern terms, it is impossible to solve x^3 + px + q = 0 by real radicals unless the equation is reducible in the base field. This is the casus irreducibilis of Cardano. GC === Subject: Re: Pure math, physics, and FLT > Today many mathematicians engage in what they call pure math. > However, if pure math had been the rule for mathematics, would > humanity even know of irrational numbers like sqrt(2)? Just after the first unit triangle. Learn something about the Pythagoreans and their beans, moron. > Now I've extended mathematics yet again http://w0rli.home.att.net/youare.swf http://b5.sdvc.uwyo.edu/bab5/snds/argcstpd.wav > So now I need people *outside* of mathematics to push. Solicit vegetarians. May you obtain the largest turd you can imagine. -- Uncle Al http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: Pure math, physics, and FLT > Today many mathematicians engage in what they call pure math. > However, if pure math had been the rule for mathematics, would > humanity even know of irrational numbers like sqrt(2)? Irrational > square roots come up in construction. Construction drove a lot of > early mathematical research, like working out pi. You know, for > building things with circles. Math does not contain physics. Physics contains only small portions of math, here and there, but it contains the complete scientific method. S = 4*pi*r^2 doesn't always work in physics. [Old Man] > James Harris === Subject: related rates problems I am stuck on the following three problems. Any help would be greatly appreciated. 1. A point P(x,y) is moving along the line y=.5x. If x is decreasing at the rate of 3 units/s, at what rate is the distance between P and the fixed point Q(0,11) changing at the instant when P is at (6,3)? I've tried various stuff, ie. d=((x-x)^2+(y-y)^2), but nothing seems to be working. 2. A camera is situated at a point 100m from where a rocket is launched. If the rocket is launched at an angle of 15 degrees to the vertical, in a direction away from the camera, find the rate of change of the distance between them when the rocket has travelled 150m and is travelling 70m/s. I tried using cosine law on this one, but since we haven't been taught how to find the derivative of sin/cos,etc, I think that there must be another way that i can't figure out. 3. The cross section of a trough for holding water is an inverted equilateral triangle. The trough is 6m long and 50cm deep. If water is flowing in at a rate of 0.6 m^3/min, find the rate at which the height is changing when the water is 40cm deep. I am having trouble relating the rates with this one. Scott Eliason -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Reminder: Wages, Employment Not Determined By Supply, Demand > More like a complete disregard for neo-classical > economics and its assumptions. Robert's model has no reference to the > treatment he is supposedly attacking. The neo-classical assumption is > that the firm maximizes *economic* profits. > > Robert wishes us to think of an integrated firm, which produces corn > while maintaining it's inventories of inputs. As such, he wishes us > to view corn and iron in inventory as durable goods. > In my example, corn and iron are totally used up in production > in each period. But you force the firm to replicate the original capital stock. I agree my interpretation below is not the natural one. However, there is no other way to reconcile your integrated firm with standard notion. > Fine. What is > the appropriate expression for economic profits? > First, we can write the production technology as > > y=F(L,I,C) > > where y is net output of corn, I is iron in inventory and C is corn in > inventory. > Apparently JimT has no idea how to develop an argument or conduct a > rational discussion. I have registered my doubts before about whether > one can express the firm in my example as facing a single production > function. I have asked JimT to outline how to construct such a > function for my example. He just refuses to. Above he dishonestly > pretends he hasn't refused to present an argument. It won't be a function. But we can just as easily replace it with a choice of techniques. You've already outlined two that produce a unit of corn from maintained inventories of corn and iron. The problem is then to calculate profits using each technique and pick the technique that yields the highest profits. > What Robert has failed to do (and is likely incapable of doing) is > showing that his so-called firm equilibrium is consistent with the > central hypothesis of maximization of economic profits. > JimT contradicts himself: > This is the right problem for a firm that behaves competitively > on all markets. > -- JimT commenting on my formulation of the primal LP, 7 July Book V., Chapter V., Section 8. > It is an implication of the theory of the long period, when > all inputs are approriately adjusted, that the firm in > competitive equilibrium does not earn economic profits. I > showed in the dual LP in my post kicking off this thread > that for a firm to be willing to continue production at > unchanged levels, the following equation must be satisfied: > ([pi pc] A + a0 w)(1 + r) = [pi pc] > where a0 and A comprise the Leontief input-output matrix for > the cost-minimizing technique. This equation implies that > no pure economic profits are being earned when the > accounting rate of profits is r. You forget that entry and exit are included in the notion of the long-run. So your rate of accounting profits must the same as rate of profits in all other sectors of the economy. The Sraffarian Long-Period General Equilibrium is derived around an economy-wide profit rate. (This is roughly equivalent to Marshall's idea of zero profits.) This is motivated around the hypothesis that capitalists seek the highest return, and all commodities will be produced in the long-period only if the rate of return is the same. You claim to have a model of a single firm (or industry). As a result, there is no reason that the accounting profits of your solution are equal to such a profit rate. As I've repeatly said before, your problem is overdetermined...one the profit rate is taken as a given, production in both sectors requires allowing the price of corn to adjust. > So I laugh at JimT's ignorant suggestion that I have not > showed the solution to my example is consistent with the > central hypothesis of maximization of economic profits. The joke's on you, laughing boy.... I finally had a glance at some of Duncan Foley's lecture notes. He's coherent and seems to know what he's talking about. You should just post references to his writing...they're worthy introductions to Sraffa's ideas. === Subject: Re: Search algorithm needed for specially structured data Distribution: inet James Waldby You have a rectangle with 4 borders, left, right, top and bottom. Using binary search you can efficiently search each border to find adjacent elements such that your target number is between them. If you actually find the target you are done. If a number on the top boundary is larger than the target, that column and all greater columns can be eliminated. If a number on the left boundary is larger than the target, that row and all greater rows can be eliminated. If a number on the right boundary is smaller than the target, that row and all lesser rows can be eliminated. If a number on the bottom boundary is smaller than the target, that column and all lesser columns can be eliminated. repeat this process. It must terminate with a single square because the top right element is either larger or smaller, so a row or column is eliminated. Same for the lower left corner. If you start with (n x m) elements, worst case is (n x log(m))+ (m x log(n)) comparisons, or something like that. For 1000x1500 thats about 11*1000+10*1500, or 26000 comparisons for a 1.5 million element array. Best case is much better. -- Gene Wagenbreth ISI (310)448-8213 genew@isi.edu === Subject: Re: Search algorithm needed for specially structured data > I have a table containing numbers: > > a1,a2,a3,a4,a5,a6,a7,... > b1,b2,b3,b4,b5,b6,b7,... > c1,c2,c3,c4,c5,c6,c6,... > ... > Typically, these tables have 1000 columns and 1500 rows. > ... > a1 < a2 < a3 < a4 ... < an > ... that is, row cells are in increasing order. > > Also, the column cells are in increasing order, so that > > a1 < b1 < c1 ... > ... > How can I efficiently search for a given number? > You haven't provided sufficient information for conclusive > answers. How many tables? How many searches per table? > Distinct integer numbers in limited range? Is answer > just Present/Not Present, or nearest number, or coordinates? > Source of data? I see. First of all, 1000 by 1500 was to low, it's about 10 million columns and 15 million rows. An integer may occur more than once and I need only present answer and it's coordinates (row column) if it is indeed present. I'm going to search the table only once and then throw it away. Can we do better if the data has even more structure? I now know that the difference a[n] - a[n-1] is constant, the same for b[n] - b[n-1] and so on. In a similar fasion, the difference between subsequent cells in a given row is constant. === Subject: Re: Search algorithm needed for specially structured data Originator: daw@mozart.cs.berkeley.edu (David Wagner) Let me restate your problem. You're given a list of m rows, with each row a sequence of integers, like this: x_1, x_1 + y_1, x_1 + 2 y_1, ..., x_1 + n y_1 x_2, x_2 + y_2, x_2 + 2 y_2, ..., x_2 + n y_2 ... x_m, x_m + y_m, x_m + 2 y_m, ..., x_m + n y_m Also, we're promised that x_i + j y_i < x_{i+1} + j y_{i+1} for all i,j. Given z, we want to check whether it appears in the above table, and if so, where. Did I get that right? If we can't assume anything else, the best I can think of is an O(n) time algorithm. For each i, reduce z mod y_i and check whether it is equal to x_i; if so, check whether 0 <= (z - z_i)/y_i <= n. Do you know anything else about how the x_i's and y_i's are distributed? === Subject: Re: Search algorithm needed for specially structured data ... > You have a rectangle with 4 borders, left, right, top and bottom. > Using binary search you can efficiently search each border to > find adjacent elements such that your target number is between > them. If you actually find the target you are done. > If a number on the top boundary is larger than the target, that > column and all greater columns can be eliminated. [snip other 3 rules] > repeat this process. It must terminate with a single square because > the top right element is either larger or smaller, so a row or > column is eliminated. Same for the lower left corner. > If you start with (n x m) elements, worst case is (n x log(m))+ > (m x log(n)) comparisons, or something like that. ... > Best case is much better I agree with your O(n log(m)+ m log(n)) bound, which is slightly different than the O(m log(m)+ n log(n)) bound I gave for my method (b). Anyhow, the method you give probably has better usual-case performance than my method (b), and indeed probably could usually outperform my method (c), which has better worst- case performance, giving O(m+n+lg(m)) by following an isoline. In the new situation that hantheman mentions, where the elements of each row and each column are in arithmetic progression clearly the problem can be solved in O(n) time. Denote the elements by aij and target by t. Let di = ai2 - ai1. For i = 1 to n see if di evenly divides t-ai1. I imagine that there is an O(1) solution for this new problem, because no matter how large the matrix is, it is completely determined by the 4 values a11, a12, a21, a22, so one can write aij = f(i, j, a11, a12, a21, a22) and perhaps solve for t = aij. -jiw === Subject: Solving equation (Correct again) (Jim, parantheses below is now balanced ;-) Sorry!) I posted the attached message earlier today, but as Ashlie pointed out, the equation was wrong. Here is the corrected version: N = (A * (x-1)) + ((B + 4*(x-1)) * y-1) As before everything is integers and upper bounds for x is 8 and 16 for y. Lower bound for both x and y are 1. Let N=552, A=66 and B=34. The solution in this case is unique: x=3 and y=11 How do I solve this in general? Do I need to minimize using linear programming? I only need one solution if that make things easier. Can I solve this in a computer program efficiently, even if A,B and N are huge (hundreds of digits) and x and y are less than, say, 5000? (I know C and Java, not Mathematica, etc) -han > Hi all, > I've been pondering about the following equation, but cannot solve it > in general: > N = (A + 4*x) + (B + 4*y) > where N is known and A and B are known constants and the unknowns x > and y have known upper bounds. Everything is positive integers. How do > I find x and y in general? > For instance, A=34 and B=66 and N=1215, upper bounds are 8 for x and > 16 for y. > Solution is x=3 and y=3. === Subject: Re: Solving equation (Correct again) >(Jim, parantheses below is now balanced ;-) Sorry!) The parentheses are now balanced, but not correctly. >Here is the corrected version: >N = (A * (x-1)) + ((B + 4*(x-1)) * y-1) The numbers you supplied as an example suggest that you REALLY meant N = (A * (x-1)) + ((B + 4*(x-1)) * (y-1) ) Solving that equation for x and y is easy. Multiply by 4 and add A*B to both sides; you get 4*N + A*B = (4*x + B-4) * (4*y + A-4) So all you have to do is to find ways to factor 4N+AB into pairs of integers, one of the factors being congruent to A mod 4, and the other being congruent to B mod 4. Any upper and lower bounds you have for x and y are equivalent to upper and lower bounds for the factors. Let's try it: >Let N=552, A=66 and B=34. ... so 4N+AB = 2^2 * 3 * 7 * 53. You want both factors to be congruent to 2 mod 4, i.e., each is to be twice an odd number, so the possible pairs are (2,2226), (6,742), (14,318), (42,106), or the reversals of these. >As before everything is integers and upper bounds for x is 8 and 16 >for y. Lower bound for both x and y are 1. In that case the first factor 4x+B-4 lies between 34 and 62, and the second factor 4y+A-4 lies between 66 and 126. The first constraint already requires that the two factors be 42 and 106 respectively, and fortunately that's consistent with the second constraint. Then 4x+B-4 = 42 requires x=3 and 4y+A-4 = 106 requires y=11. >The solution in this case is unique: x=3 and y=11 Bingo! >How do I solve this in general? Do I need to minimize using linear >programming? What would you minimize, and what makes you think to use _linear_ programming for a nonlinear problem? >Can I solve this in a computer program efficiently, even if A,B and N >are huge (hundreds of digits) and x and y are less than, say, 5000? (I >know C and Java, not Mathematica, etc) Ooh, that's harder. I have assumed above that you could factor 4N+AB. That's not completely unreasonable when A, B, and N have dozens of digits; hundreds makes this less likely to be efficient. But if on the other hand you really know that x,y are to be positive integers less than 5000, why not just check them all? C'mon - what's 25M cases on a modern machine? As it turns out you can combine these two ideas: just check each x up to 50000, compute 4x+B-4, and if it divides 4N+AB evenly, just solve for y and see whether it's an integer in the right range. In pseudo-code: for x from 1 to 5000 do { let y = ( (4N+AB)/(4x+B-4) + 4 - A )/4 if (y is an integer) and (y>lowerbound) and (y I posted the attached message earlier today, but as Ashlie pointed > out, the equation was wrong. > Here is the corrected version: > N = (A * (x-1)) + ((B + (4*(x-1)) * y-1) Looks significantly different from the first version! Actually, the parentheses in this expression are not balanced, so there is some ambiguity... Expect to get varied answers unless you correct this. Jim === Subject: Re: stuck proving simple(?) tautology > (for any T1(x,y)) (for any T2(x,y)) (for any G(x,y)) (for any I(x,y)) > ((I(x,y)<=>x = y) & T1 = I / G / G / T1 & T2 = I / G / T2 / T2 => T1 > = T2 ) And this one can be cleaned even more (for any T1(x,y)) (for any T2(x,y)) (for any G(x,y)) (T1 <=> (x=y) / G / G / T1 & T2 <=> (x=y) / G / T2 / T2 |-- T1 = T2 ) Sorry for the noise. === Subject: Re: Taking Calculus without taking precalcu >I could claim that most students who take calculus without taking >pre-calculus >probably do better than those that do take pre-caculus. This ignores the >reason why the students who took pre-calculus choose to do so. >Then again, I remember a math major who fogot how to derive the quadratic >equation using the completing the square method. So many students of math are quite young in the community college or the university; they will take what the prerequites are recommended to be. If Calculus 1 has College Algebra and Trigonometry as a prerequisite, many students will take the prerequisite. Some of the students truely need the rigorous review. On the other hand, some students may recall intermed algebra and trigonom very well and do not need the College Algebra and Trigonometry before starting Calculus 1. Schools also have mathematics assessment tests to indicate what the student should study. G C === Subject: Re: Tell me your paradoxes! > I have put two sealed envelopes in a bowl. Each of them has money in > it; one has exactly twice the amount as the other. You may choose one > and keep the money. Your choice is not final until you open the > envelope. You cannot tell how much money is in the envelope by feeling it. > You choose. Then you realize that, if the envelope you have chosen has > x dollars, the expected amount in the other envelope is 1/2 * x/2 + > 1/2 * 2x = 5x/4 dollars. Therefore, you decide to switch. > However, when you choose the first envelope, you had the same odds. > One of the envelopes has x dollars. No matter which one you choose, > the probably outcomes are 1/2 * x/2 + 1/2 * 2x = 5x/4. So there's > no point in switching, as the first envelope has the same expected > return as the second. Let R (resp. L) be the amount in the envelope on the right (resp. left). Then E(R|L) = 1.5 L E(L|R) = 1.5 R E(R) = E( E(R|L) ) = E( 1.5 L ) = 1.5 E(L) E(L) = E( E(L|R) ) = E( 1.5 R ) = 1.5 E(R) E(R) = 2.25 E(R) E(L) = 2.25 E(L) Therefore, since R>0 and L>0, neither R nor L has a real expected value; i.e., E(R) = oo and E(L) = oo. === Subject: Re: Tell me your paradoxes! [oops ... typo correction: please ignore my previous post] > I have put two sealed envelopes in a bowl. Each of them has money in > it; one has exactly twice the amount as the other. You may choose one > and keep the money. Your choice is not final until you open the > envelope. You cannot tell how much money is in the envelope by feeling > it. > You choose. Then you realize that, if the envelope you have chosen has > x dollars, the expected amount in the other envelope is 1/2 * x/2 + > 1/2 * 2x = 5x/4 dollars. Therefore, you decide to switch. > However, when you choose the first envelope, you had the same odds. > One of the envelopes has x dollars. No matter which one you choose, > the probably outcomes are 1/2 * x/2 + 1/2 * 2x = 5x/4. So there's > no point in switching, as the first envelope has the same expected > return as the second. Let R (resp. L) be the amount in the envelope on the right (resp. left). Then E(R|L) = 1.25 L E(L|R) = 1.25 R E(R) = E( E(R|L) ) = E( 1.25 L ) = 1.25 E(L) E(L) = E( E(L|R) ) = E( 1.25 R ) = 1.25 E(R) E(R) = 1.25^2 E(R) E(L) = 1.25^2 E(L) Therefore, since R>0 and L>0, neither R nor L has a real expected value; i.e., E(R) = oo and E(L) = oo. === Subject: Re: The basic idea behind my great forthcoming proof <87znj4l6p9.fsf@phiwumbda.localnet> <87n0f38zl9.fsf@phiwumbda.localnet> <87fzku5vx8.fsf@phiwumbda.localnet> <874r17donm.fsf@phiwumbda.localnet> <87ptju5esn.fsf@phiwumbda.localnet> A8EwTYfhf*u~,Eu,tf6$HN*MY&)u0G =N' x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|'1{w70ZY=ih,=)kMY_}?{%)x0)];K~@J6m5.EN?>Zh Xh;Y V|',x(js'Jfq02joVpj|#x >> Cantor introduced that idea that the sizes of >> infinite sets can be compared. >> He gave a definition of cardinality, one which apparently seemed >> natural to Frege, Dedekind and other contemporaries. >> Surely you're not suggesting that his *definition* is the source of >> all problems? > You could certainly say that making the definition > magnified the problem. Before he made his definition, > all uses of infinity in mathematics were consistent > with the idea that infinity had a potential existence, > rather than an actual existence. So your beef is with the axiom of infinity? If so, this is a rather longwinded way to say so. Finitism is not a new philosophical position (though, I don't think that most finitists hold such a grudge against Cantor's role). There must be better ways to express this position than the vague references to computation/observation/whatever. -- Even if [...] a communistic regime should come [to China], the old tradition [...] will break Communism and change it beyond recognition, rather than Communism [...] break the old tradition. It must be so. -- Lin Yutang on Socialism with Chinese characteristics in 1935 === Subject: Re: The basic idea behind my great forthcoming proof david_lawrence_petry@yahoo.com (David Petry) says... >> As long as it is logically consistent, it is valid >> mathematics. >That is a very dangerous and very wrong belief. >Logical consistency alone doesn't give us a >way to distinguish reality from fantasy, or >science from mysticism. How is it dangerous? How is it wrong? What's an example of something dangerous or evil that follows from studying set theory? The worst that can be said about it is that it may be a waste of time. Since when is wasting time considered dangerous? -- Daryl McCullough Ithaca, NY === Subject: Re: The basic idea behind my great forthcoming proof >> As long as it is logically consistent, it is valid mathematics. > That is a very dangerous and very wrong belief. Logical consistency alone > doesn't give us a way to distinguish reality from fantasy, or science from > mysticism. First off, you eliminated the context of my sentence, resulting in what sounds like an endorsement of said philosophy. I have reinserted the original text below: Acceptance of Cantor's work came about *because* mathematicians realized that arguing about the metaphysics of mathematical work was silly and unfruitful. As long as it is logically consistent, it is valid mathematics. So, in a sense, Cantor's revolution is really about the ridding of the quasi-religious mythology in mathematics. ----- This dangerous and very wrong belief is the mainstream view today. Yours is the minority view. Without casting insults on either view, let me point out that the mainstream view lets you, David, and others who believe as you do, to continue doing mathematics. Nobody really cares what you believe as long as you keep producing mathematics. Interestingly, it doesn't work the other way. If you had your way, no doubt the dangerous practitioners of the evil Cantorian religion would be stopped one way or another. Is your point to distinguish science from mysticism? That's interesting, since science is not the same as mathematics, so I don't see the relevance. Not to mention that everything you've said thus far indicates that you are far more prone to mysticism than anyone else in the thread. \ === Subject: Re: The basic idea behind my great forthcoming proof >> First, Dedekind's definition of the reals was formulated around the same >> time as Cantor's definition, and yes, Cantor did formulate a definition >> of the reals, which *was* new at the time: it's called (unjustly) the >> Cauchy completion of the rationals. Dedekind was, in fact, >> corresponding with Cantor heavily at the time, and so was inspired by >> Cantor to some degree. The most important point here, I think, is that >> there was no existing definition of R before Cantor. Some >> mathematicians, like Cauchy, had made some attempts, but their >> definitions were circular. > I appreciate the corrections. You're right that I was fuzzy on the > history, and didn't have the appropriate texts at hand. I see that > Dedekind published his work on irrationals and continuity in 1872. When > did Dedekind and Cantor begin correspondence? I have a book here (by > Boyer) that says they first met in 1874, but it is silent on whether they > had previously corresponded. In any case, Boyer is writing a broad book > for a wide audience, and perhaps should not be taken as a reliable source > on these matters. Boyer is inaccurate here. Right now I'm working on my recollections of the Dauben book, according to which Cantor discovered that the reals had cardinality greater than the Also, I recall they had been corresponding before then. Just as a reality check, I looked at http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Cantor.html and found: Cantor was promoted to Extraordinary Professor at Halle in 1872 and in that year he began a friendship with Dedekind who he had met while on holiday in Switzerland. Cantor published a paper on trigonometric series in 1872 in which he defined irrational numbers in terms of convergent sequences of rational numbers. Dedekind published his definition of the real numbers by Dedekind cuts also in 1872 and in this paper Dedekind refers to Cantor's 1872 paper which Cantor had sent him. In 1873 Cantor proved the rational numbers countable, i.e. they may be placed in one-one correspondence with the natural numbers. He also showed that the algebraic numbers, i.e. the numbers which are roots of polynomial equations with integer coefficients, were countable. However his attempts to decide whether the real numbers were countable proved harder. He had proved that the real numbers were not countable by December 1873 and published this in a paper in 1874. It is in this paper that the idea of a one-one correspondence appears for the first time, but it is only implicit in this work. The references for this webpage list Dauben's book, but also I Grattan-Guinness, The rediscovery of the Cantor-Dedekind correspondence, Jahresberichte der Deutschen Mathematiker book, he did not have access to this correspondence. I'm not sure what the story behind that is. [snipped] > David claims Cantor introduced the setting in which different sizes of > infinity coexist. On the contrary, I claim that the proof |N| < |R| is > not due to any feature of analysis introduced by Cantor, at least not any > feature solely introduced by him. Instead, all one needs for the diagonal > proof is the usual decimal representation of R (and that N and R are sets, > of course). So, it is misleading in the extreme to claim that Cantor is > to blame for the mathematics in which |N| < |R| is provable. I agree. It is clear that even before Cantor, mathematicians freely used the view, the diagonal argument forced mathematicians of the time to accept that their presupposed ideas about the reals, led to something their intuitions regarded as counter-intuitive. So in this way Cantor only forced mathematicians to accept new counter-intuitive ideas in light of already accepted notions. I think your point is at least valid to that degree: Cantor did not (and did not have to) force mathematicians to accept his religious beliefs in order to buy his most striking results. After all, Cantor was himself working in the same tradition as his fellow mathematicians. Only of his later work, I think, can there really be any charges of a radical change in the existing mathematical setting. > This claim is partly historical (as far as whether the decimal > representation was taken for granted prior to Cantor), but largely logical > (as to what assumptions are sufficient for the proof that |N| < |R|). I > am not so concerned with who proved it and when, but whether it *could > have been* proven prior to Cantor. I acknowledge that this question is > not so well-defined, since the mathematics of the day was not explicit in > its foundations, but this is my aim. I'm still confused what the point of this debate is. It seems that you are saying: Cantor proved some cool theorems (like uncountability of the reals) and did some foundational work, but it certainly wasn't earth-shattering. Far from it, he only formalized much of what people already thought. David Petry is saying: 1)Cantor completely changed the way mathematicians thought. 2)In fact, he introduced quasi-religious beliefs into mathematics that have corrupted the subject to this day. > Footnotes: > [1] I hesitate to include Frege, because I would guess that he was very > directly influenced by Cantor's work. I'm not familiar with the interaction between Frege and Cantor, but it's an interesting topic. I'll try and find out something about this. === Subject: Re: The mistical value of e > Well, it's not that mystical, how it is achieved I understand, what I don't > understand is how the limit below could have its value: > lim(x->0) (1+x)^(1/x) = e > heck I would simplify it as follows instead: > lim(x->0) (1+0)^(infinitum) = 1 > But that's wrong, or more correctly perhaps....not precise enough, but why? > Doug It's wrong for two reasons. First, the meaning of the limit of Y as X tends to A does not require X to actually have the value A, which is a good thing because... Second, if x is real then 1/x is real and there is no such real number as 1/0 (more: there's _no_ number 1/0 as far as I know) and there is no such real number infinity either (which isn't to say that there aren't any numbers infinity but none of them are found among the reals). What lim(x -> c) U = V (with x real) means is: given epsilon > 0 there is a delta (generally depending on epsilon) such that |U - V| < epsilon for all x such that 0 < |x - c| < delta. In your case c = 0 so that simplifies to given epsilon > 0 there is a delta (generally depending on epsilon) such that |U - V| < epsilon for all x such that 0 < |x| < delta. Note that, in the general case, x =/= c and in your case, x =/= 0. The reason that lim(x->0) (1+x)^(1/x) exists is that (1 + 1/n)^n increases as n increases but it is bounded above by 3. An increasing sequence of reals bounded above has a limit (at or below that upper upper bound). GC === Subject: Re: The mistical value of e Doug says... >Well, it's not that mystical, how it is achieved I understand, what I don't >understand is how the limit below could have its value: >lim(x->0) (1+x)^(1/x) = e Look at the power series expansion for (1+x)^n: (1+x)^n = 1 + nx + n(n-1)/2! x^2 + ... In the special case n = 1/x, this becomes (1+x)^{1/x} = 1 + 1 + 1(1-x)/2! + 1(1-x)(1-2x)/3! + ... In the limit as x-->0, this becomes the power series lim(x->0) (1+x)^{1/x} = 1 + 1 + 1/2! + 1/3! + ... -- Daryl McCullough Ithaca, NY === Subject: Re: The mistical value of e Doug escreveu na mensagem > Well, it's not that mystical, how it is achieved I understand, what I > don't understand is how the limit below could have its value: > lim(x->0) (1+x)^(1/x) = e > heck I would simplify it as follows instead: > lim(x->0) (1+0)^(infinitum) = 1 Let x -> 0. Without rigor, by one side, 1 + x -> 1, so the value of (1 + x)^(1 / x) (*) should decreases to 1. By another side, 1 / x -> oo, so the value of (*) should increases to infinity. This two effects balance each other in such a way that the value of (*) does not decrease to 1 and does not increase to infinity. It approach the number e, between 1 and infinity. Is like if 1 + x pulls to 1 and 1 / x pulls to infinity, but no one wins, the value of (*) goes to the number e. Jaime Gaspar ______________________________ Homepage: www.jaimegaspar.com E-mail: e-mail@jaimegaspar.com === Subject: Re: The mistical value of e > Well, it's not that mystical, how it is achieved I understand, what I don't > understand is how the limit below could have its value: > lim(x->0) (1+x)^(1/x) = e > heck I would simplify it as follows instead: > lim(x->0) (1+0)^(infinitum) = 1 > But that's wrong, or more correctly perhaps....not precise enough, but why? > Doug > It's wrong for two reasons. First, the meaning of the limit of Y as X > tends to A does not require X to actually have the value A, which is a > good thing because... Second, if x is real then 1/x is real and there is > no such real number as 1/0 (more: there's _no_ number 1/0 as far as I > know) and there is no such real number infinity either (which isn't to > say that there aren't any numbers infinity but none of them are found > among the reals). > What lim(x -> c) U = V (with x real) means is: > given epsilon > 0 there is a delta (generally depending on epsilon) such > that > |U - V| < epsilon for all x such that 0 < |x - c| < delta. > In your case c = 0 so that simplifies to > given epsilon > 0 there is a delta (generally depending on epsilon) such > that > |U - V| < epsilon for all x such that 0 < |x| < delta. > Note that, in the general case, x =/= c and in your case, x =/= 0. > The reason that lim(x->0) (1+x)^(1/x) exists is that (1 + 1/n)^n > increases as n increases but it is bounded above by 3. An increasing > sequence of reals bounded above has a limit (at or below that upper > upper bound). Soory, only one upper was intended. GC > GC === Subject: Re: The mistical value of e > > Well, it's not that mystical, how it is achieved I understand, what I don't > understand is how the limit below could have its value: > > lim(x->0) (1+x)^(1/x) = e > > heck I would simplify it as follows instead: > > lim(x->0) (1+0)^(infinitum) = 1 > > But that's wrong, or more correctly perhaps....not precise enough, but why? > > Doug > It's wrong for two reasons. First, the meaning of the limit of Y as X > tends to A does not require X to actually have the value A, which is a > good thing because... Second, if x is real then 1/x is real and there is > no such real number as 1/0 (more: there's _no_ number 1/0 as far as I > know) and there is no such real number infinity either (which isn't to > say that there aren't any numbers infinity but none of them are found > among the reals). > What lim(x -> c) U = V (with x real) means is: > given epsilon > 0 there is a delta (generally depending on epsilon) such > that > |U - V| < epsilon for all x such that 0 < |x - c| < delta. > In your case c = 0 so that simplifies to > given epsilon > 0 there is a delta (generally depending on epsilon) such > that > |U - V| < epsilon for all x such that 0 < |x| < delta. > Note that, in the general case, x =/= c and in your case, x =/= 0. > The reason that lim(x->0) (1+x)^(1/x) exists is that (1 + 1/n)^n > increases as n increases but it is bounded above by 3. An increasing > sequence of reals bounded above has a limit (at or below that upper > upper bound). > Soory, only one upper was intended. Sorry again, I meant sorry. Poltergeists probably GC > GC > GC === Subject: Re: The mistical value of e ;-) anywho, but how would you go about simplifying that statement lim(x->0) (1+x)^(1/x) to obtain e. Or more correctly, is it even possible to somehow simplify this as we do with the limit definition of a derivative? I mean, can we do anything to this equation to somehow narrow it down to e or is it simply one of the proven limits sort of like lim(x->a) (sina/a) = 1 where we simply proved it, and now we use it knowing that we could never narrow it down to 1? I hope the question is making any sense. > > > Well, it's not that mystical, how it is achieved I understand, what I don't > understand is how the limit below could have its value: > > lim(x->0) (1+x)^(1/x) = e > > heck I would simplify it as follows instead: > > lim(x->0) (1+0)^(infinitum) = 1 > > But that's wrong, or more correctly perhaps....not precise enough, but why? > > Doug > > It's wrong for two reasons. First, the meaning of the limit of Y as X > tends to A does not require X to actually have the value A, which is a > good thing because... Second, if x is real then 1/x is real and there is > no such real number as 1/0 (more: there's _no_ number 1/0 as far as I > know) and there is no such real number infinity either (which isn't to > say that there aren't any numbers infinity but none of them are found > among the reals). > > What lim(x -> c) U = V (with x real) means is: > > given epsilon > 0 there is a delta (generally depending on epsilon) such > that > > |U - V| < epsilon for all x such that 0 < |x - c| < delta. > > In your case c = 0 so that simplifies to > > given epsilon > 0 there is a delta (generally depending on epsilon) such > that > > |U - V| < epsilon for all x such that 0 < |x| < delta. > > Note that, in the general case, x =/= c and in your case, x =/= 0. > > The reason that lim(x->0) (1+x)^(1/x) exists is that (1 + 1/n)^n > increases as n increases but it is bounded above by 3. An increasing > sequence of reals bounded above has a limit (at or below that upper > upper bound). > Soory, only one upper was intended. > Sorry again, I meant sorry. Poltergeists probably > GC > GC > > GC === Subject: Re: The mistical value of e yes, but HOW if presented with this question, could you narrow the limit down to e? (heck, I'm guessing it's impossible) Doug > Doug escreveu na mensagem > Well, it's not that mystical, how it is achieved I understand, what I > don't understand is how the limit below could have its value: > lim(x->0) (1+x)^(1/x) = e > heck I would simplify it as follows instead: > lim(x->0) (1+0)^(infinitum) = 1 > Let x -> 0. Without rigor, by one side, 1 + x -> 1, so the value of > (1 + x)^(1 / x) (*) > should decreases to 1. By another side, 1 / x -> oo, so the value of (*) > should increases to infinity. This two effects balance each other in such a > way that the value of (*) does not decrease to 1 and does not increase to > infinity. It approach the number e, between 1 and infinity. > Is like if 1 + x pulls to 1 and 1 / x pulls to infinity, but no one wins, > the value of (*) goes to the number e. > Jaime Gaspar > ______________________________ > Homepage: www.jaimegaspar.com > E-mail: e-mail@jaimegaspar.com === Subject: The problem of number partition Now i have a question of the number partition. A partition is a way of writing an integer n as a sum of positive integers where the order of the addends is not significant. For example, 10 can be partitioned as 10=1+2+3+4. If we use P(N,i,j) to denote the number of partitions which satisy: N is partitioned into i parts, the largest part is j. For example, since 10=2+4+4=3+3+4. So P(10,3,4)=2. My question is: If i Now i have a question of the number partition. A partition is a way of > writing an integer n as a sum of positive integers where the order of > the addends is not significant. For example, 10 can be partitioned as > 10=1+2+3+4. > If we use P(N,i,j) to denote the number of partitions which satisy: > N is partitioned into i parts, the largest part is j. > For example, since 10=2+4+4=3+3+4. So P(10,3,4)=2. > My question is: > If i u for email) === Subject: Re: The problem of number partition > My question is: > If i