mm-1029 === Subject: Re: ellipticity condition for fractional laplacian > Does an ellipticity condition for fractional Laplacian exist? If not, whats > the substitute? What is an ellipticity condition? (Here are my guesses - maybe the symbol is positive, but this is obviously true. Maybe it is some kind of maximum principal - this is true if the power is between 0 and 1. I must admit to not really knowing the PDE terminology well, and I would like to learn what ellipticity means in this context.) -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Re: ellipticity condition for fractional laplacian > Does an ellipticity condition for fractional Laplacian exist? If not, whats > the substitute? > What is an ellipticity condition? > (Here are my guesses - maybe the symbol is positive, but this is obviously > true. Maybe it is some kind of maximum principal - this is true if the power is > between 0 and 1. I must admit to not really knowing the PDE terminology well, > and I would like to learn what ellipticity means in this context.) In the usual (classic) sense, ellipticity depends on the matrix of coefficients of the operator, the matrix being positive definite in this case. So for the fractional laplacian might the question, very vaguely, be one involving geometry? A second question is, these fractional operators are usually defined through the Fourier characterisation of Sobolev spaces, and also turns up as pseudodifferential operators. Are there other equally successful ways of defining fractional ops? Perhaps linking to the question I ask about geometry, and thereby giving conditions akin to that of an ellipticity condition? === Subject: Re: ellipticity condition for fractional laplacian >Does an ellipticity condition for fractional Laplacian exist? If not, > whats >the substitute? >>What is an ellipticity condition? >>(Here are my guesses - maybe the symbol is positive, but this is > obviously >>true. Maybe it is some kind of maximum principal - this is true if the > power is >>between 0 and 1. I must admit to not really knowing the PDE terminology > well, >>and I would like to learn what ellipticity means in this context.) > In the usual (classic) sense, ellipticity depends on the matrix of > coefficients of the operator, the matrix being positive definite in this > case. So for the fractional laplacian might the question, very vaguely, be > one involving geometry? Yes, I remember that a differential operator of the form - sum_{i,j} a_{i,j}(x) d^2 / dx_i dx_j is called elliptic if the matrix a_{i,j} is (uniformly) postive definite. In terms of pseudodifferential operators, this has symbol sum_{i,j} a_{i,j} xi_i xi_j (xi is the Greek letter) so the questions becomes whether this is positive. In these terms, the fractional laplacian operator is |xi|^(2a) which is obviously positive and hence elliptic in this sense. I like to think of the operator in terms of the differential equation: du/dt = -(lap)^a u u(x,0) = f(x). where lap = -del^2 as usual. So the solution is u(. ,t) = exp(-t lap^a) f. The ellipticity condition might become the notion that the operator exp(-t lap^a) is norm decreasing on L_p, or smoothing, or something like that. If a=1, then this operator is convolving against the Gaussian kernel (or the pdf of the normal distribution). If a=1/2 this is convolving against the Poisson kernel (or the pdf of the Cauchy distribution). If 0 A second question is, these fractional operators are usually defined through > the Fourier characterisation of Sobolev spaces, and also turns up as > pseudodifferential operators. Are there other equally successful ways of > defining fractional ops? Perhaps linking to the question I ask about > geometry, and thereby giving conditions akin to that of an ellipticity > condition? As for other ways to define fractional powers - any reasonable definition of the laplacian will make it be a positive self-adjoint operator. (For example on Riemannian manifolds you would use the d*d*+*d*d construction, where d is the exterior derivative and * is the Hodge dual - I hope I got my terminology right here.) Thus the fractional powers of the laplacian would simply be the usual fractional powers of a positive self-adjoint operator on a Hilbert space, i.e. via spectral mapping. For the Laplacian on R^n, this becomes the Fourier characterization. What exactly are you trying to get at? What is the problem you are interested in? -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Re: ellipticity condition for fractional laplacian > In the usual (classic) sense, ellipticity depends on the matrix of > coefficients of the operator, the matrix being positive definite in this > case. So for the fractional laplacian might the question, very vaguely, be > one involving geometry? > Yes, I remember that a differential operator of the form > - sum_{i,j} a_{i,j}(x) d^2 / dx_i dx_j > is called elliptic if the matrix a_{i,j} is (uniformly) postive definite. In > terms of pseudodifferential operators, this has symbol > sum_{i,j} a_{i,j} xi_i xi_j > (xi is the Greek letter) so the questions becomes whether this is positive. In > these terms, the fractional laplacian operator is |xi|^(2a) which is obviously > positive and hence elliptic in this sense. Yes of-course that is correct, it could be one way to state ellipticity. > I like to think of the operator in terms of the differential equation: > du/dt = -(lap)^a u > u(x,0) = f(x). > where lap = -del^2 as usual. So the solution is u(. ,t) = exp(-t lap^a) f. The > ellipticity condition might become the notion that the operator exp(-t lap^a) is > norm decreasing on L_p, or smoothing, or something like that. Yes indeed, thats also possible as the infinitesimal generator of a heat type kernel. > If a=1, then this operator is convolving against the Gaussian kernel (or the pdf > of the normal distribution). If a=1/2 this is convolving against the Poisson > kernel (or the pdf of the Cauchy distribution). If 0 against a positive kernel (the pdf of the so called 2a-stable random variables). > I dont know if other people would think of this as ellipticity - it is just my > own private notion of what it might be. See this is part of the problem, I feel. Its probably even more fundamental than the question I ask about having a universally accepted ellipticity condition for fractional laplacian say. That basically being, how to properly define fractional ops. in the first place. > A second question is, these fractional operators are usually defined through > the Fourier characterisation of Sobolev spaces, and also turns up as > pseudodifferential operators. Are there other equally successful ways of > defining fractional ops? Perhaps linking to the question I ask about > geometry, and thereby giving conditions akin to that of an ellipticity > condition? > As for other ways to define fractional powers - any reasonable definition of the > laplacian will make it be a positive self-adjoint operator. (For example on > Riemannian manifolds you would use the d*d*+*d*d construction, where d is the > exterior derivative and * is the Hodge dual - I hope I got my terminology right > here.) Thus the fractional powers of the laplacian would simply be the usual > fractional powers of a positive self-adjoint operator on a Hilbert space, i.e. > via spectral mapping. For the Laplacian on R^n, this becomes the Fourier > characterization. Yes of-course, there is a similar op., the d-bar laplacian: DD* + D*D : L^2 -> L^2 which appears in several complex variables, D = bar @. > What exactly are you trying to get at? What is the problem you are interested in? There is a problem like -(lap)u = u^p which involves critical exponents etc. and has been somewhat well studied in the context of usual PDEs involving maximum/comparison principles and sub/super solns. A modification of this problem is to investigate something like -(lap)^a = u^p. But before meddling around with this problem it would be good to know how viable this latter problem really is, especially given the situation with fractional laplacian and a compatible ellipticity condition. Just speculating for the moment.. === Subject: Re: ellipticity condition for fractional laplacian > There is a problem like -(lap)u = u^p which involves critical exponents etc. > and has been somewhat well studied in the context of usual PDEs involving > maximum/comparison principles and sub/super solns. A modification of this > problem is to investigate something like -(lap)^a = u^p. But before meddling > around with this problem it would be good to know how viable this latter > problem really is, especially given the situation with fractional laplacian > and a compatible ellipticity condition. Just speculating for the moment.. Yes, I have a friend who studies problems like this - well maybe the problem is more like -(lap)u = u^p + v. It seems to me that the first thing they do is to rewrite the problem as u = G(u^p) + w where w=Gv, and G is convolution against the Greens function for -lap (e.g. Newtonian potential on R^n). Thus one way to define ellipticity would be in terms of the existence of such a positive Greens function. If you contact my personally, I can tell you who my friend is, who is one of the experts in this field. I myself am merely a spectator in this stuff. -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Re: ellipticity condition for fractional laplacian > Yes, I have a friend who studies problems like this - well maybe the > problem is more like -(lap)u = u^p + v. It seems to me that the > first thing they do is to rewrite the problem as > u = G(u^p) + w > where w=Gv, and G is convolution against the Greens function for > -lap (e.g. Newtonian potential on R^n). Thus one way to define > ellipticity would be in terms of the existence of such a positive > Greens function. > If you contact my personally, I can tell you who my friend is, who is > one of the experts in this field. I myself am merely a spectator in > this stuff. OK thats great, have just seen you a personal message. === Subject: Re: ellipse equation > Hi! > I have to find the equation of an ellipse: i know the foci coordinates, the > lenght of the major axis and the length of the minor axis. > > (Note: the axis are not parallel to the x and y axis of th coordinates > system) > > Can anyone help me? > > > Francesco Musicman There are a number of ways, for example: the major and minor vertices, 4 points on the ellipse. The equation of an ellipse with center (h,k) must be in the form A*(x-h)^2 + B*(x-h)*(y-k) + C*(y-k)^2 = 1. That it passes through the 4 vertices gives you 4 linear equations (but not independent ones) from which to determine the three unknown coefficients, A, B, and C. === Subject: Re: [JSH] Re: OOPS! False alarm on twin primes >No apology is required, as history knows many theorems being valid for a >short period of time. Its the US: for a success you dont have to be >correct, you just have to be loud. >>Some people have pointed out that this statement goes a bit far. On the >>other hand, JSH does seem to belong to a category that is more populated >>in the US than anywhere else. Something about the blend of ignorance >>and arrogance. > Its so nice that you corrected his stupid stereotype so that you > might state your own ever-so-slightly less offensive stupid > claim. > Youre a regular enlightened liberal, aintcha? world, but arrogance (a side-effect of power) is rather concentrated in the U.S.A., for obvious reasons. Gib === Subject: Re: [JSH] Re: OOPS! False alarm on twin primes <87isogmzwj.fsf@phiwumbda.org> the U.S.A., for obvious reasons. I suppose, by virtue of your .nz address, this opinion of yours is thankfully devoid of arrogance, however chock-full of ignorance it may be. -- Im talking about mathematics--hard, brutal, extreme ... pushing your mind beyond the limits to understand what no one else can because theyre afraid to risk it all, to lose their freaking worthless minds in the push to know. --James Harris, for the Nike Derivator === Subject: Re: [JSH] Re: OOPS! False alarm on twin primes > world, but arrogance (a side-effect of power) is rather concentrated in > the U.S.A., for obvious reasons. I found that arrogance is also pretty evenly distributed among people who live comfortably. I found as many arrogant/ignorant people (the two coming often together, though it would at first seem paradoxal) in all countries from which I know people. I think it has more to do with somehow successful people (mostly wealth-related success) than citizenship or origin. To put it short, it is a side-effect of power, but of individual power rather than state power. Sam -- So if you meet me, have some courtesy, have some sympathy, and some taste Use all your well-learned politesse, or Ill lay your soul to waste - The Rolling Stones, Sympathy for the Devil === Subject: Re: New letter from Ramanujan discovered >>Nathan Deeth >>Age 11 > Is this the same Nathan who has been carrying the title Age 11 for four > Helmut Richter They say you are only as old as you feel. Also, I read somewhere that mathematicians often have an emotional age that is less than their chronological age. Von Neumann for example was about 12 all his life. Gib === Subject: Re: Factorials question Peter Webb a dit : >> Peter Webb a dit : >> Excellent reply! >> Probably worth mentioning that if you (the OP) are only after an >> approximation to n! (say you are using it for some probability >> calculations), then there are good approximations for large n. >> You can read about them here: >> http://mathworld.wolfram.com/StirlingsApproximation.html >> I have been told not to use stirling as the computing of a power and >> a square root takes more time than an approximate computing of n! by >> the na.95ve way. >> Stirling is for theoretical purposes. > Thats not correct. > The original poster was asking about 100 digit numbers. > You can calculate an approximation to (10^100)! - at least the > approximate digit count - in less than a minute using a basic > scientific calculator and Stirlings approximation. (Somebody in this > thread has already done it). > Actually doing 10^100 multiplications, on a 1000 Gigahertz Pentium 5, > would take far longer than the age of the Universe. I think I answered too fast ... I would have calculated ln(gamma) with a series expansion to avoid stirling But I was tortured with stirling at school ... sure it would help here. -- Alexandre Charitopoulos mailto:a.charito@wanadoo.fr Em6 / Eb7(5b) / Dm7 / Db7(5b, 9b) / Cmaj7 === Subject: Erdos-Mordell Theorem, Oppenheim Variant Summary: Help with Proof of Oppenheim Variant Keywords: Erdos-Mordell Theorem Can anyone point me toward a proof of this variant (or corollary)? The theorem and variant are in Coxeters Intoduction to Geometry, p.9. The main theorem (which Ive solved with Prof. Coxeters hints) is: If O is any point inside a triangle ABC and P, Q, R are the feet of the perpendiculars from O upon the respective sides BC, CA, AB, then OA + OB + OC >= 2(OP + OQ + OR). The Oppenheim variant, which I cant prove, is OA x OB x OC >= (OQ + OR)(OR + OP)(OP + OQ). TIA, Jonathan === Subject: Re: 3x+1 - Not so interesting >systems, >(Z*C)/(X-Y) >the conjecture is false when that ratio is an integer. To be an factors >of X-Y. The conjecture can be true for C=1 without being true for C>1 >as is the case when C=5. >Now it may be that a 3 is NEVER amongst the factors of X-Y, in which >case I would say that 3x+3 is true iff 3x+1 is true. > Ive been thinking about this and have come to the realization that since X is > a power of 2 and Y is a power of 3, X-Y cannot have 3 as a factor.So when C is factors in the denominator > of the Crossover Point fraction. So those systems with C a power of 3 will have > the same loops as 3n+1. I dont think I know what the goal is here. Can you explain? Cross over is ?? a relationship of what to what? Id search back but a huge block of posts seem to be missing from Ernst === Subject: Re: 3x+1 - Not so interesting === >Subject: Re: 3x+1 - Not so interesting >>systems, >>(Z*C)/(X-Y) >>the conjecture is false when that ratio is an integer. To be an the factors >>of X-Y. The conjecture can be true for C=1 without being true for C>1 >>as is the case when C=5. >>Now it may be that a 3 is NEVER amongst the factors of X-Y, in which >>case I would say that 3x+3 is true iff 3x+1 is true. >> Ive been thinking about this and have come to the realization that since X >> a power of 2 and Y is a power of 3, X-Y cannot have 3 as a factor.So when C factors in the >denominator >> of the Crossover Point fraction. So those systems with C a power of 3 will >have >> the same loops as 3n+1. > I dont think I know what the goal is here. Can you explain? > Cross over is ?? a relationship of what to what? > Id search back but a huge block of posts seem to be missing from >Ernst Ok, heres a summary. Its all about finding loops. Every 3n+C system has a trivial loop at C. Some systems, such as 3n+5 have more than one loop in the positive integers and thus, the Collatz Conjecture is false for those systems. The unsolved question is: does 3n+1 have only its trivial loop in the positive integers? One way to answer the question is to simply do a brute force search looking for loops. Im trying to develop a smarter approach. To that end, Ive come up with Hailstone Functions and Crossover Points, which Ive explained before but will consolidate and repeat here. ------------------------------------------------------------- --------------- -- Observations about loops in 3n+C systems: Take your numbers from any 3n+C sequence and create a sequence vector (where [1]=n/2 and [2]=3n+C). For the (3n+1) loop 1 4 2 1, the sequence vector is [2][1][1] Now graph this symbolically using down for [1] and right for [2]. k_l m n ----------------------------------------------------- The Hailstone Function is k as a function of n and uses Inverse Rules. ----------------------------------------------------- m = 2*n l = 2*2*n k = (2*2*n - C)/3 This simplifies to k = (4*n - C)/3 ----------------------------------------------------- Every Hailstone Function simplifies to the form (X*n - Z*C)/Y In the above case X=4 Z=1 C=1 and Y=3. ----------------------------------------------------- Note that the Hailstone Function can be seen as the equation of a straight line with slope X/Y. The slope cannot be 1 so there exists a point at which it intersect the line f(x)=x, whose slope is 1. This intersection is called the Crossover Point. ----------------------------------------------------- (Z*C)/(X-Y) IF THE CROSSOVER POINT IS AN INTEGER, THEN THE SEQUENCE VECTOR IS A LOOP IN THAT 3n+C SYSTEM. ----------------------------------------------------- The [2][1][1] sequence vectors Crossover Point is (Z*C)/(X-Y) = (1*C)/(4-3) = C/1 = C Thus, 3n+1 has a loop at 1, 3n+3 has a loop at 3, 3n+5 has a loop at 5, etc. The sequence vector [2][1] has a Hailstone Function (2*n - C)/3 and a Crossover Point of C/(2-3) = C/(-1) = -C Thus, 3n+1 has a loop at -1, 3n+3 has a loop at -3, 3n+5 has a loop at -5, etc. The sequence vector [2][1][2][1][1] has a Hailstone Function (8*n - 5*C)/9 and a Crossover Point of (5*C)/(8-9) = 5*C/(-1) = -5*C Thus, 3n+1 has a loop at -5, 3n+3 has a loop at -15, 3n+5 has a loop at -25, etc. The sequence vector [2][1][2][1][2][1][2][1][1][2][1][2][1][2][1][1][1][1] has a Hailstone Function (2048*n - 2363*C)/2187 and a Crossover Point of (2363*C)/(2048-2187) = 2363*C/(-139) = -17*C Thus, 3n+1 has a loop at -17, 3n+3 has a loop at -51, 3n+5 has a loop at -85, etc. ----------------------------------------------------- EVERY 3n+C system has a loop at C, -C, -5C and -17C. Note that only one of these is in the positive integers. ----------------------------------------------------- Now heres where it gets interesting. The sequence vector [2][1][1][1] has a Hailstone Function (8*n - C)/3 and a Crossover Point of (C)/(8-3) = C/5 This sequence vector is only a loop in systems where C is a multiple of 5. Thus, it is a loop in 3n+5, with root 1, but it is NOT a loop in 3n+1 or 3n+3. The Collatz Conjecture is therefore false in 3n+5 because there are two loops in the positive integers. In the general Hailstone function (X*n - Z*C)/Y X is always a power of 2, Y is always a power of 3, and Z is a mixture based on how the sequence zig-zags. (Z*C)/(X-Y) If the Crossover Point is an integer, then there is a loop. ------------------------------------------------------------- ------ For EVERY possible sequence vector, a 3x+C system exists in which that sequence is a loop. ------------------------------------------------------------- ------ Take an extreme example ag_af ae ad ac_ab aa z y x w v u_t s r q p o n m_l k_j i h g f_e d c_b a the Crossover Point works out to be (35343985*C)/33552245 which factors to (5*23*307339*C)/(5*6710449) which reduces to (7068797*C)/6710449 So the system 3x+6710449 has a loop at 7068797 (in addition to the trivial loop at 6710449) which is easily verified 7068797 27916840 13958420 6979210 3489605 17179264 8589632 4294816 2147408 1073704 536852 268426 134213 7113088 3556544 1778272 889136 444568 222284 111142 55571 6877162 3438581 17026192 8513096 4256548 2128274 1064137 9902860 4951430 2475715 14137594 7068797 Whats special about 3x+1 is that with C=1, the Crossover function is Z/(X-Y) which means the only way to get an integer is if the factors of Z daunting task. There are many candidate sequences (where Z > (X-Y)), but finding one with even one common factor, let alone all, is difficult. Very often X-Y is prime. A survey of all sequences of length 4 and depth 8 ended up with 28 prime and 41 composite X-Y values. And often the composites have large prime factors. ------------------------------------------------------------- -------------- Re-iterating what you replied to, the denominator of a Crossover Point cannot have a factor of 3. Thus, when C is a power of 3, it will NOT factors in the denominator. That is exactly the same situation as when C=1. Thus 3n+3, 3n+9, 3n+27, 3n+81... will have the exact same loops (if any) as 3n+1. Which is why WB stated that there is no reason to study 3n+3. At the time of our discussion, I had not yet figured out that have a factor of 3 and my Crossover Point formula left open a possibility that 3n+1 and 3n+3 could be different. That possibility has been closed. -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: 3x+1 - Not so interesting >Re-iterating what you replied to, the denominator of a Crossover Point >cannot have a factor of 3. Thus, when C is a power of 3, it will NOT >factors in the denominator. That is exactly the same situation as when C=1. >Thus 3n+3, 3n+9, 3n+27, 3n+81... will have the exact same loops (if any) >as 3n+1. Which is why WB stated that there is no reason to study 3n+3. >At the time of our discussion, I had not yet figured out that >have a factor of 3 and my Crossover Point formula left open a possibility >that 3n+1 and 3n+3 could be different. That possibility has been closed. Of cours I never used (or understood) the crossover function, but just the simple observation that loops with the 3n+3 system will have all numbers divisible by 3, and that there is a very simple 1 to 1 correspondence between those loops and the loops of the 3n+1 system -- Wim Benthem === Subject: Re: 3x+1 - Not so interesting === >Subject: Re: 3x+1 - Not so interesting >Message-id: Re-iterating what you replied to, the denominator of a Crossover Point >>cannot have a factor of 3. Thus, when C is a power of 3, it will NOT >>factors in the denominator. That is exactly the same situation as when C=1. >>Thus 3n+3, 3n+9, 3n+27, 3n+81... will have the exact same loops (if any) >>as 3n+1. Which is why WB stated that there is no reason to study 3n+3. >>At the time of our discussion, I had not yet figured out that X-Y cannot >>have a factor of 3 and my Crossover Point formula left open a possibility >>that 3n+1 and 3n+3 could be different. That possibility has been closed. >Of cours I never used (or understood) the crossover function, but just >the simple observation that loops with the 3n+3 system will have all numbers >divisible by 3, and that there is a very simple 1 to 1 correspondence >between those loops and the loops of the 3n+1 system And if I cant explain this in terms of my Crossover function, then Ive got a problem. There is only one Truth. If my Crossover calculations are correct, they have to reach the same conclusion. When Ernst Berg mentioned that 3x+3 and 3x+9 didnt have any attractors, my first question was is that true for powers of 3 or multiples of 3? The Crossover Point function tells you that it is powers, not multiples, and explains why. And at this point, Im satisfied understanding _why_ things work. >-- >Wim Benthem -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: 3x+1 - Not so interesting [snip nice expo] _This_ expo should be on your web-pages, imnsho. Did you change them recently? I could not locate anything of the sort except a binary nova explosion. > -- > Mensanator > 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm -- Ioannis http://users.forthnet.gr/ath/jgal/ ___________________________________________ Eventually, _everything_ is understandable. === Subject: Re: 3x+1 - Not so interesting === >Subject: Re: 3x+1 - Not so interesting >Message-id: <3F51C3F5.424E@ath.forthnet.gr[snip nice expo] >_This_ expo should be on your web-pages, imnsho. Did you change them >recently? I could not locate anything of the sort except a binary nova >explosion. Yeah, I should make a new web page so I dont have to keep repeating it.Most of this stuff Ive just worked out recently. Ive been using Hailstone Functions for a long time and have always had to find Z by laboriously multiplying out the algebraic equations by hand. But Ive just developed an algorithm to calculate Z from any arbitrarily long sequence vector. That in itself is kind of neat, so Ill make a page on that also. >> -- >> Mensanator >> 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm >-- >Ioannis >http://users.forthnet.gr/ath/jgal/ >___________________________________________ >Eventually, _everything_ is understandable. -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: x7 + 29x6 + 29x5 - 5887x4 - 44573x3 + 219501x2 + 3219348x + 8511761 >Am I wrong here, but does this solve the above equation? > >29*2 = 58 +1= 59*2 =118 >I am not at all sure what you mean in the line above. A standard >you have done, that a = d, or 29*2 = 118, which is false. >Do you mean > 29*2 = 58, then > 58+1 = 59, then > 59*2 = 118 ? >This would make more sense, and is much more in the standard usage >of equal signs. You are right. I should have entered it the way you did above in three different steps. >Note that the standard usage requires every addition, >multiplication, etc., be performed BEFORE looking at the truth of >ANY equal sign, but then allows multiple equal signs on in one >expression, e.g., 2 + 2 = 3 + 1 = 8/2 = sqrt(16). > >1/(118/(29*2)) * -1 = x = -.491525423728813559322033898305.. > >Dan > >Virgil Wrote: >If Ôx8 - x7 + 29x2 + 29 = 0 were to have a rational root, it would >have to be be an integer factor of 29, meaning the set ofpossible >values is {1,-1, 29, -29}, none of which work. >If you plug in this value for x= -.4915254.. in the equation >x8 - x7 + 29x2 + 29 = 0 where x appears 3 times, x satisfies the >equation. >Then again, because of my poor math ability I could be interpreting >the equation wrong. >Also where x = -.4915254.., x is a rational because 1/(118/58)* -1 > is a rational. >Dan >If your equation means the same as > Ôx^8 - x^7 + 29*x^2 + 29 = 0, >where Ô* indicates multiplication and Ô^ means Ôraised to the >power of, then it is a monic polynomial (x^8 has coefficient 1) >with integer coefficients. >It is well known that any rational solutions must have numerators >which are integer factors of the constant term, 29, and denominators >which are integer factors of the coefficient of x^8, namely 1. >Substituting -1/(118/58) = -29/59 for x I get > x^8 - x^7 + 29*x^2 + 29 = 2*3^3*5^2*7*29*73*264342409/59^8 > or approximately 36.0166584048, nowhere near zero. The above 11 lines are not what I intended it to mean. ^^^^^^^ No matter how I calculated this value for x, and as you can see it is a closed form, the $64,000 question is, is it the correct x value for this equation, Ôx8 - x7 + 29x2 + 29 = 0? I believe it is. BTW: Being a rational, x = -.491525423728813559322033898305.. has a repeating decimal expansion with a period of 58. What a coincidence huh! (29*2) Dan === Subject: Re: x7 + 29x6 + 29x5 - 5887x4 - 44573x3 + 219501x2 + 3219348x + 8511761 >Am I wrong here, but does this solve the above equation? > >29*2 = 58 +1= 59*2 =118 >I am not at all sure what you mean in the line above. A standard >you have done, that a = d, or 29*2 = 118, which is false. >Do you mean > 29*2 = 58, then > 58+1 = 59, then > 59*2 = 118 ? >This would make more sense, and is much more in the standard usage >of equal signs. > You are right. I should have entered it the way you did above in three > different steps. >Note that the standard usage requires every addition, >multiplication, etc., be performed BEFORE looking at the truth of >ANY equal sign, but then allows multiple equal signs on in one >expression, e.g., 2 + 2 = 3 + 1 = 8/2 = sqrt(16). > >1/(118/(29*2)) * -1 = x = -.491525423728813559322033898305.. > >Dan > > > >Virgil Wrote: >If Ôx8 - x7 + 29x2 + 29 = 0 were to have a rational root, it would >have to be be an integer factor of 29, meaning the set ofpossible >values is {1,-1, 29, -29}, none of which work. > >If you plug in this value for x= -.4915254.. in the equation >x8 - x7 + 29x2 + 29 = 0 where x appears 3 times, x satisfies the >equation. >Then again, because of my poor math ability I could be interpreting >the equation wrong. >Also where x = -.4915254.., x is a rational because 1/(118/58)* -1 > is a rational. > > >Dan >If your equation means the same as > Ôx^8 - x^7 + 29*x^2 + 29 = 0, >where Ô* indicates multiplication and Ô^ means Ôraised to the >power of, then it is a monic polynomial (x^8 has coefficient 1) >with integer coefficients. >It is well known that any rational solutions must have numerators >which are integer factors of the constant term, 29, and denominators >which are integer factors of the coefficient of x^8, namely 1. >Substituting -1/(118/58) = -29/59 for x I get > x^8 - x^7 + 29*x^2 + 29 = 2*3^3*5^2*7*29*73*264342409/59^8 > or approximately 36.0166584048, nowhere near zero. > The above 11 lines are not what I intended it to mean. > ^^^^^^^ > No matter how I calculated this value for x, and as you can see it is > a closed form, the $64,000 question is, is it the correct x value for > this equation, > Ôx8 - x7 + 29x2 + 29 = 0? > I believe it is. > BTW: Being a rational, x = -.491525423728813559322033898305.. has a > repeating decimal expansion with a period of 58. > What a coincidence huh! (29*2) > Dan It depends entirely on what you mean by Ôx8 - x7 + 29x2 + 29 Can you give me some idea of the sequence of operations you are performing to evaluate Ôx8 - x7 + 29x2 + 29 for a given value of x, say when x = 5 or x = 2 to make calculations easy. And what has the x7 + 29x6 + 29x5 - 5887x4 - 44573x3 + 219501x2 + 3219348x + 8511761 from the subject line have to do with the problem? === Subject: Re: x7 + 29x6 + 29x5 - 5887x4 - 44573x3 + 219501x2 + 3219348x + 8511761 >Am I wrong here, but does this solve the above equation? > >29*2 = 58 +1= 59*2 =118 > >I am not at all sure what you mean in the line above. A standard >you have done, that a = d, or 29*2 = 118, which is false. >Do you mean > 29*2 = 58, then > 58+1 = 59, then > 59*2 = 118 ? >This would make more sense, and is much more in the standard usage >of equal signs. > > You are right. I should have entered it the way you did above in three > different steps. > >Note that the standard usage requires every addition, >multiplication, etc., be performed BEFORE looking at the truth of >ANY equal sign, but then allows multiple equal signs on in one >expression, e.g., 2 + 2 = 3 + 1 = 8/2 = sqrt(16). > >1/(118/(29*2)) * -1 = x = -.491525423728813559322033898305.. > >Dan > > > >Virgil Wrote: >If Ôx8 - x7 + 29x2 + 29 = 0 were to have a rational root, it would >have to be be an integer factor of 29, meaning the set ofpossible >values is {1,-1, 29, -29}, none of which work. > >If you plug in this value for x= -.4915254.. in the equation >x8 - x7 + 29x2 + 29 = 0 where x appears 3 times, x satisfies the >equation. >Then again, because of my poor math ability I could be interpreting >the equation wrong. >Also where x = -.4915254.., x is a rational because 1/(118/58)* -1 > is a rational. > > >Dan >If your equation means the same as > Ôx^8 - x^7 + 29*x^2 + 29 = 0, >where Ô* indicates multiplication and Ô^ means Ôraised to the >power of, then it is a monic polynomial (x^8 has coefficient 1) >with integer coefficients. >It is well known that any rational solutions must have numerators >which are integer factors of the constant term, 29, and denominators >which are integer factors of the coefficient of x^8, namely 1. >Substituting -1/(118/58) = -29/59 for x I get > x^8 - x^7 + 29*x^2 + 29 = 2*3^3*5^2*7*29*73*264342409/59^8 > or approximately 36.0166584048, nowhere near zero. > > The above 11 lines are not what I intended it to mean. > ^^^^^^^ > > No matter how I calculated this value for x, and as you can see it is > a closed form, the $64,000 question is, is it the correct x value for > this equation, > Ôx8 - x7 + 29x2 + 29 = 0? > I believe it is. > > BTW: Being a rational, x = -.491525423728813559322033898305.. has a > repeating decimal expansion with a period of 58. > What a coincidence huh! (29*2) > > > Dan > It depends entirely on what you mean by Ôx8 - x7 + 29x2 + 29 > Can you give me some idea of the sequence of operations you are > performing to evaluate Ôx8 - x7 + 29x2 + 29 for a given value of > x, say when x = 5 or x = 2 to make calculations easy. > And what has the x7 + 29x6 + 29x5 - 5887x4 - 44573x3 + 219501x2 + > 3219348x + 8511761 from the subject line have to do with the > problem? You are right, it is not in the subject line, but is the first equation listed in the OP. Ok, this is more than likely wrong but here goes --- ^^^^ ^^^^ ^^^^^^ ^^^^^ Given the equation Ôx8 - x7 + 29x2 + 29 = 0 All I am doing is solving for x. where x = (-1/(59/29)) = -.491525423728813559322033898305.. (a repeating decimal with a period of 58) x8 = -3.93220339.. x7 = -3.440677966.. Then subtract (x8 - x7) x8 - x7 = -.4915254237.. 29*x*2 = -28.508474576.. Then add the left side of the equation (x8 - x7) to (29x2) -.4915254237.. + -28.508474576.. = -29 Then add the final right side value 29 to -29. 29 + -29 = 0 Naturally the longer the decimal length of this rational x, the greater the accuracy. Dan === Subject: Re: x7 + 29x6 + 29x5 - 5887x4 - 44573x3 + 219501x2 + 3219348x + 8511761 > It depends entirely on what you mean by Ôx8 - x7 + 29x2 + 29 > > Ok, this is more than likely wrong but here goes --- > ^^^^ ^^^^ ^^^^^^ ^^^^^ > > Given the equation Ôx8 - x7 + 29x2 + 29 = 0 > All I am doing is solving for x. > where x = (-1/(59/29)) = -.491525423728813559322033898305.. > (a repeating decimal with a period of 58) > x8 = -3.93220339.. I get 8 times x as your value, whereas in a polynomial you should be calculating x to the 8th power, that is, x*x*x*x*x*x*x*x, which is .0034069667102. Apparently we are interpreting polynomials differently! Where x^8 means 8 xs multiplied together, as in x*x*x*x*x*x*x*x, and x*8 means 8 times x, to me the polynomial is Ôx^8 - x^7 + 29*(x^2)+ 29, with powers and products preceding all addition/subtraction, or in another form, but equivalently, Ô((x -1)*(x*x*x*x*x) + 29)*(x*x) + 29. Dooing things your way, the equation simplifies to 59*x + 29 = 0, which is a linear equation, and unrelated to Ôx^8 - x^7 + 29*(x^2)+ 29 = 0 When you do the calculations my way, you get much diferent results! === Subject: Re: Norm of Legendre Polynomials Using Rodrigues Formula > In Erwin Kreyszigs book, Advanced Engineering Mathematics, he states without proof that (Pn, Pn) = 2/(2n+1) can be shown using Rodrigues formula >for Legendre Polynomials and integrating by parts n times. I have sat down >and tried doing this, without success. ........... >...I dont really know very much about this subject at all, and the book I > have on it is not with me, but is it possible that it would be easier to >compute (x^n,Pn)? Then using orthogonality relations for Pn it should be >easy to relate this to (Pn,Pn). My opinion is that the computation of (P_n,P_n) or (x^n,P_n) have the same level of difficulty.When we try to prove orthogonality we find (P_n,P_n). Namely, let us denote INT =Integral_{x=-1 to x=1} {E(x)}_(a,b)=E(b)-E(a) . f^{(p)} = the p-th derivative of f . Consider that f,W:[-1,1]-->R have continuous derivatives of order p on [-1,1]. Then integrating p-times by parts one finds INT W(x)f^{(p)}(x) dx = (1) = (-1)^p *INT W^{(p)}(x)f(x) dx + + SUM_{k=0 to k=p-1}(-1)^{p-1-k}{W^{(p-1-k)}(x)f^{(k)}(x)}_{(-1,1)} . It may be said that (1) is ,,Green-Lagrange identitiy. Further denote (*) g(x) = (1/(n! 2^n))*(x^2 -1)^n . Then P_n(x)=g^{(n)}(x)= Legendre polynomial of degree n . Moreover observe that g(-1)=g(-1)=...=g^{(n-1)}(-1)=0 (2) g(1)=g(1)=...=g^{(n-1)}(1)=0 Therefore for an arbitrary polynomial h (2.1) {h(x)*g^{(k)}(x)}_{(-1,1)}=0 for k=0,1,...,n-1 . for any polynomial of degree < n .) g(x)=x^{2n}/(n! 2^n) + polynomial of degree =< 2n-1 (3) P_n(x)=Coef*x^n+... , Coef :=(2n)!/((n!)^2 2^n) . Further take in (1) the following (parameters) : p=n , W(x)=P_n(x) , f(x)=g(x) . Using (2)-(2.1) we find INT P_n(x)*P_n(x) dx = ||P_n||^2 = (1.1) = (-1)^n *INT P_n^{(n)}(x)g(x) dx =...(see (*) and (3))...= =(-1)^n * n!*Coef/(2^n n!)*INT (x^2-1)^n dx =B*INT (1-x^2)^n dx . where B= (2n)!/(4^n n!^2) .If G denote Gamma function , then for a>-1 ,b>-1 (4) INT (1-x)^a(1+x)^b dx = 2^{a+b+1}G(a+1)G(b+1)/G(a+b+2) (For proof use Beta Function).Using (4) with a=b=n , from (1.1) we find ||P_n||^2 =B*2^{2n+1}(n!)^2/(2n+1)! = 2/(2n+1) === Subject: Re: Strong Bridge Hands > * Paul Sperry > In a standard bridge deal - 13 cards to each of four players - I would > like a reference or solution to these three problems: > > What is the probability that > (a) at least one player has at least one six card suit but no more than > six cards of a suit - i.e. exactly six cards of the same suit but > perhaps two six card suits; > (b) at least one player has a 7 or more cards of the same suit; > (c) at least one player has at least one void - no cards of at least > one suit. > Questions like that are easiest solved by simulations -- or by > googling... Searching for bridge hands probabilities I found the > following site: > http://www.wunderland.com/WTS/Jake/Bridge_Stats.html > Of course it is more challenging to calculate your questions exactly, combinatorial setup. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Strong Bridge Hands > In a standard bridge deal - 13 cards to each of four players - I would > like a reference or solution to these three problems: > What is the probability that > (a) at least one player has at least one six card suit but no more than > six cards of a suit - i.e. exactly six cards of the same suit but > perhaps two six card suits; > (b) at least one player has a 7 or more cards of the same suit; > (c) at least one player has at least one void - no cards of at least > one suit. ----- One can crudely estimate the odds thus: (a) the probability of a 6 card suit but no longer (in one hand) is about p=.165; so the odds that any hand has that feature is about 1-(1-p)^4 = .514. (b) the probability of a 7+ card suit (i.o.h.) is ~.040; any hand = ~.151; (c) the probability of a void (i.o.h) is ~.051; any hand = ~.189. But the probabilities of the different hands are dependent, and the absence of a feature in one hand tends to lower the odds of that feature in other hands. So those are overestimates. ----- Those questions are best solved by combinatoric counting. (If I may contradict a previous respondant.) It only takes a few minutes of run-time, with a modern computer. There are 53644737765488792839237440000 ways to deal out the four hands. a) In 26734857955043414221247634432 of them, at least one hand has at least one six-card suit, but no longer suit; b) in 7847870163258230433568719360 of them, at least one hand has a seven-card or longer suit; and c) in 9858096432031419875988460800 of them, at least one hand has at least one void. The odds are therefore (a) .49836..., (b) .14629..., and (c) .18376.... -- Don Reble djr@nk.ca === Subject: Re: Spews A new status symbol for me: an imposter at yahoo. LH === Subject: Probability - Help! This is simple probability. Whats wrong with my work below? Heres the problem: Three soldiers are engaged in battle. Soldier X hits his target with probability 1/3, Y with 1/4, Z with 1/5. If they all randomly aim at one of the others and fire at the same time, what is the probability that no soldiers survive? Only X survives? X and Y survive? Etc. Ive broken this down into: P(X hits Y) = P(X aims at Y) * P(X hits his target) = 1/2 * 1/3 = 1/6 P(X hits Z) = 1/6 P(Y hits X) = P(Y hits Z) = 1/2 * 1/4 = 1/8 P(Z hits X) = P(Z hits Y) = 1/2 * 1/5 = 1/10 Now whats wrong with this argument?: P(X gets hit) = P(Y hits X OR Z hits X) = P(Y hits X) + P(Z hits X) - P(Y hits X AND Z hits X) = 1/8 + 1/10 - (1/8 * 1/10) = 17/80 Similarly: P(Y gets hit) = 1/4 P(Z gets hit) = 13/48 Then I could say: P(no soldiers survive) = P(X gets hit AND Y gets hit AND Z gets hit) = 17/80 * 1/4 * 13/48 But this doesnt give the expected 1/240, which I have confirmed using a more tedious method. The problem is this more tedious method is nearly impossible to calculate P(only X survives). It would be much easier to say: P(only X survives) = [1 - P(X gets hit)] * P(Y gets hit) * P(Z gets hit). -jk === Subject: Re: Zenos paradoxes (was Re: reciting N and other sets) > The Zenos Paradoxes are not logical or mathematical, > but empirical and experiential. I disagree. Were not watching a magician do card tricks here. To those who succumb to the paradox, it is the application of a mental construct, namely the infinite subdivision of time and distance, to a common experience, namely the motion of a runner, that causes all the distress. Therefore the paradoxes, whatever theyre supposed to be (Im still waiting for a cogent explanation of their content), are not just empirical and experiential. === Subject: Number Theory Problem-CAUTION: HARD lol Hey! My friend gave me this math problem which I solved..however he said that there was a stronger result which he found but then *forgot* take a look at it though.. however, he and I cannot seem to be able to come up with it again (how lame, yes lol)...I am dying to figure out but I have no clue hm..anyways here is the problem: Prove that, for any integers a, b, c, there exists a positive integer n such that sqrt(n^3+an^2+bn+c) is not an integer. Here is the solution I came up with. let F(n)=n^3+an^2+bn+c. suppose F(n) is a square for n=1, 2, 3, 4. Because F(2) and F(4) are squares of the same parity..their difference 12a+2b+56 is a multiple of 4. which implies that b must be even. F(1) and F(3) are squares of the same parity..their difference 8a+2b+26 is a multiple of 4..so b odd in this case. this is a contradiction so therefore sqrt(n^3+an^2+bn+c) is not an integer. ok so theres my proof/solution thingy..can you help me by finding another proof? hmmm.. === Subject: Re: Number Theory Problem-CAUTION: HARD lol >Hey! My friend gave me this math problem which I solved..however he >said that there was a stronger result which he found but then *forgot* >take a look at it though.. however, he and I cannot seem to be able to >come up with it again (how lame, yes lol)...I am dying to figure out >but I have no clue hm..anyways here is the problem: Prove that, for >any integers a, b, c, there exists a positive integer n such that >sqrt(n^3+an^2+bn+c) is not an integer. >Here is the solution I came up with. >let F(n)=n^3+an^2+bn+c. suppose F(n) is a square for n=1, 2, 3, 4. >Because F(2) and F(4) are squares of the same parity..their difference >12a+2b+56 is a multiple of 4. which implies that b must be even. >F(1) and F(3) are squares of the same parity..their difference >8a+2b+26 is a multiple of 4..so b odd in this case. >this is a contradiction so therefore sqrt(n^3+an^2+bn+c) is not an >integer. >ok so theres my proof/solution thingy..can you help me by finding >another proof? hmmm.. If f(x) = sqrt(x^3 + a*x^2 + b*x + c), argue f(x) behaves like x^(3/2) for large x. Therefore f(x + 1) - f(x) behaves like (x + 1)^(3/2) - x^(3/2) or (3/2)x^(1/2). And f(x + 2) - 2f(x+1) + f(x) behaves like (3/4)x^(-1/2). This cannot be an integer for large x. Many details have been omitted in this argument. Your proof is much simpler, and gives the stronger result F(n) cannot be square for four consecutive values of n if F is a monic cubic polynomial. On the other hand this asymptotic argument will extend to cubics such as F(n) = 4*(n-1)*(n-2)*(n-3) + 1 (a square for n = 1, 2, 3, 4). -- Wanted: Experts at choosing the best of 100+ applicants for a position. Register as a California voter by September 22, and vote on October 7. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Re: Number Theory Problem-CAUTION: HARD lol Sam Rosenberg > Hey! My friend gave me this math problem which I solved..however he > said that there was a stronger result which he found but then *forgot* Did it use a volume argument? Write f(n) = n^3+an^2+bn+c Now if e.g. f(k)f(l)f(m) is a nonsquare, then one of the three factors is a nonsquare. So, it will be enough to show that g(l,m,n)=f(k)f(l)f(m) takes more than 1+sqrt(x) values in [0,x], for some x. But Im not sure this can be pushed through. > Prove that, for > any integers a, b, c, there exists a positive integer n such that > sqrt(n^3+an^2+bn+c) is not an integer. LH === Subject: Re: Number Theory Problem-CAUTION: HARD lol > for any integers a, b, c, there exists a positive integer n such that > f(n) = sqrt(n^3+an^2+bn+c) is not an integer. > Here is the solution I came up with. > let F(n)=n^3+an^2+bn+c. suppose F(n) is a square for n=1, 2, 3, 4. > Because F(2) and F(4) are squares of the same parity..their difference > 12a+2b+56 is a multiple of 4. which implies that b must be even. > F(1) and F(3) are squares of the same parity..their difference > 8a+2b+26 is a multiple of 4..so b odd in this case. > this is a contradiction so therefore sqrt(n^3+an^2+bn+c) is not an > integer. > ok so theres my proof/solution thingy..can you help me by finding > another proof? hmmm.. f(1)^2 = 1 + a + b + c (mod 4) f(2)^2 = 2b + c (mod 4) f(3)^2 = a - b + c - 1 (mod 4) f(4)^2 = c (mod 4) f(4)^2 - f(2)^2 = 2b (mod 4) f(3)^2 - f(1)^2 = 2b + 2 (mod 4) makes the arithmetic easier. Now prove lemma, if x^2 - y^2 is even, then x^2 - y^2 = 0 (mod 4) Thus 2b = 0 (mod 4) 2b + 2 = 0 (mod 4) 2 = 0 (mod 4) Nothing different, just slicker and quicker. ;-) === Subject: Re: Sanford PhD pencil > The quite possibly wise Carl Devore was heard saying: > In March, in this newsgroup, in a thread entitled Non math question: > What type of earser do you use? I gave high praise to the Sanford Phd > mechanical pencil for mathematical writing. Very unfortunately, this > pencil is now being marketed as the Paper:Mate Phd^Ultra. It has been > redesigned, and the new model is crap. > Terribly sorry to hear about that. Ive been looking for a fairly > permanent replacement for my Bic mechanical pencils recently. Is there > another pencil you would recommend? > Colin I recommend the Alvin Draft/Matic series. Very sharp looking pencil. The grip is a finely textured stainless steel and the pencil is built to last quite a few years. I use the .3 mm model, but they come in all sizes. Ive had mine for over a year. Price: A little over $10. Alex Solla Junior Reed College === Subject: Superset of Salem numbers Suppose we consider the set of algebraic integers each of which is such that all of its conjugates are smaller in absolute value, and such that each is real and positive. Is there a name for this type of number? They have the property that the associated linear recurrences have successive ratios which converge to them. === Subject: Re: 27 straight lines on a cubic surface >If your ground field isnt algebraically complete--e.g., if >its the real numbers instead of the complex numbers--I do believe >that some of these 27 lines may not be defined over that field >itself, but only over its algebraic closure (the complex numbers, >if the ground field is the real numbers). You get a Galois extension which has nice properties, but its actually easier to get this by using the fact that it is a reßection group. About 15 years ago I computed the generic polynomial for the 27 line group, but never published. === Subject: Question about infinitely differentiable functions with certain known values... Heh, one of the problems with upper level advanced topics is its difficult to give an adequate subject line for a post where youre asking a convoluted question. :P OK, suppose I have a function f such that: 1. f is infinitely differentiable 2. f(0) = 0 3. for all x = n/(2^m), where m,n are nonnegative integers, f(x) is known With the 3 things above, is it possible to somehow interpolate f(r) when r is some positive real NOT of the form n/(2^m)? === Subject: Re: Question about infinitely differentiable functions with certain known values... > Heh, one of the problems with upper level advanced topics is its > difficult to give an adequate subject line for a post where youre > asking a convoluted question. :P Just prefix the subject line with [JSH]. Thatll ensure everybody reads it. Jack Rudd === Subject: Re: Question about infinitely differentiable functions with certain known values... > OK, suppose I have a function f such that: > 1. f is infinitely differentiable > 2. f(0) = 0 > 3. for all x = n/(2^m), where m,n are nonnegative integers, f(x) is > known > With the 3 things above, is it possible to somehow interpolate f(r) > when r is some positive real NOT of the form n/(2^m)? Unless Im missing something, it is possible to do this -- and we dont even need infinite differentiability, since continuity is enough. The set of numbers on which f is specified (by conditions 2 and 3) is dense in the non-negative reals. Hence f is uniquely defined on the non-negative reals by the specified conditions. (For a more constructive approach: Given positive real r, there is a sequence of values x_0, x_1,... such that a) x_i is of the form n/(2^i) for some nonnegative integer n, and b) | r - x_i | <= 1 / 2^(i+1). Then x_i -> r, and f is continuous, so f(x_i) -> f(r).) -- Terry Boon, London, UK terry@counterfactual.org === Subject: Re: Question about infinitely differentiable functions with certain known values... > OK, suppose I have a function f such that: > 1. f is infinitely differentiable > 2. f(0) = 0 > 3. for all x = n/(2^m), where m,n are nonnegative integers, f(x) is > known > With the 3 things above, is it possible to somehow interpolate f(r) > when r is some positive real NOT of the form n/(2^m)? Indeed, { n/2^m | n,m in N_0 } is dense subset of [0,oo) and f is continuous over [0,oo). Now apply theorem, if two continuous functions are identical over a dense set, then they are the same. === Subject: Re: Question about infinitely differentiable functions with certain known values... > OK, suppose I have a function f such that: > 1. f is infinitely differentiable > 2. f(0) = 0 > 3. for all x = n/(2^m), where m,n are nonnegative integers, f(x) is > known > With the 3 things above, is it possible to somehow interpolate f(r) > when r is some positive real NOT of the form n/(2^m)? > Indeed, { n/2^m | n,m in N_0 } is dense subset of [0,oo) and f is > continuous over [0,oo). Now apply theorem, if two continuous functions > are identical over a dense set, then they are the same. What branch of math talks about these types of things? Is any consideration given to questions of determining whether a given function is continuous, given its values over a dense set? it sounds like something i need to learn === Subject: Re: Question about infinitely differentiable functions with certain known values... > Heh, one of the problems with upper level advanced topics is its > difficult to give an adequate subject line for a post where youre > asking a convoluted question. :P > OK, suppose I have a function f such that: > 1. f is infinitely differentiable > 2. f(0) = 0 > 3. for all x = n/(2^m), where m,n are nonnegative integers, f(x) is > known > With the 3 things above, is it possible to somehow interpolate f(r) > when r is some positive real NOT of the form n/(2^m)? One only needs that f is continuous. If a > 0 there is a sequence (a_k) where a_k -> a and eaxh a_k has the form n/2^m with m, n as in (3). Then f(a) = lim_{k->infinity} f(a_k) by continuity so f(a) is determined. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Factorial/Exponential Identity, Infinity I read some more about integer partitions. The integer partitions are the sequences A_m that sum to n, with A_i >= A_(i+1). Equivalently they are the sets the elements of which sum to n. Another consideration is a superset of the integer partitions which are each permutation of the sets of which the elements sum to n. What Im interested in are the subset of the permutations of the integer partitions which are having the cyclic characteristic of being distinct on rotation, cyclic permutation. For example, the sequence (2, 1, 2, 1) is indistinct from (1, 2, 1, 2), but distinct from (1, 1, 2, 2). http://mathworld.wolfram.com/Permutation.html (That reference lists permutations as sets, I think they should be sequences instead, i.e. {1, 2, 3} is the same set as {3, 2, 1} but (1, 2, 3) is a different sequence than (3, 2, 1), a notation quibble. ) The reason is that I require algorithms to generate a variety of integer partitions. I want to generate the rotationally distinct unordered partition sequences of a given number of partitions of for a number A_k_p, and then generate the unordered partition sequences with the same number of partitions for another larger number, B_(n-k)_p. The idea then is that for A_k_p x B_(n-k)_p that there is a sequence (a_1_p, b_1_p, ..., ..., a_i_p, b_i_p). This sequence then can represent a sequence with a_1_p many ones, then b_1_p many zeros, etcetera, to a_i_p many ones and b_i_p many zeros. The resulting sequence represents a k-subset of an n-set, and can be rotated to generate n or a fraction of n many other k-subsets of the n-set. The idea is that it is an algorithm to enumerate k-subsets, sets of a given size, of an n-set. The idea behind generating k-subsets is then to multiply their elements together to sum those proucts and get the result of the sum of the product of each k-subset of an n-set, {1, ..., n}. There are simpler methods to enumerate the k-subsets. Well this is pretty simple, the sum of the products of the unordered pairs of {1, ..., n} are the Stirling numbers of the first kind s(n+1, n-1). http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?An um=A000914 http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html Theres an error in the Encyclopedia listing, it says Sum of product of unordered pairs of numbers from {1..n+1}, where it is actually Sum of product of unordered pairs of numbers from {1..n}. I search the Encyclopedia of Integer Sequences for Sum of product and get some search results, but none of them is the Sum of product of unordered trios of {1, ..., n}. Would that be s(n+2, n-1)? Gee, wouldnt that be great. Perhaps not, as theyre all negative, but -s(n+1, n-2) might be closer to the mark. Lets see, going back to (n+1)...(n+n)/n^n, the first coefficient of the polynomial is 1, the second is then n(n+1)/2, the third is here being s(n+1, n-1). Yet, I want an equation in terms of n that is this value, the Stirling number of the first kind does not really help here, unless I find a way to derive a simple form. It is so that ((sum n)^2 - sum (n^2))/2 = s(n+1, n-1), for n. It does appear that the sum of the products of each unordered trio of elements from {1 ,..., n}, for n>=3, is the Stirling number of the first kind s(n+1, n-2). It isnt so that ((sum n)^3) - sum (n^3))/6 = -s(n+1, n-2). It appears to be some other function. Here are the first values of -s(n+1, n-2) from a table: n=3, -s(4, 1)=6 n=4, -s(5, 2)=50 n=5, -s(6, 3)=225 n=6, -s(7, 4)=735 n=7, -s(8, 5)=1960 n=8, -s(9, 6)=4536 n=9, -s(10, 7)=9450 n=10, -s(11, 8)=18150 The sum of i for i=1 to n is the nth pyramidal number, n(n+1)/2. Its cube is (sum n)^3. n=3,(sum n)^3=6^3=216 n=4,(sum n)^3=10^3=1000 n=5,(sum n)^3=15^3=3375 n=6,(sum n)^3=21^3=9261 n=7,(sum n)^3=28^3=21952 n=8,(sum n)^3=36^3=46656 n=9,(sum n)^3=45^3=91125 n=10,(sum n)^3=55^3=166375 The sum of i^3 for i=1 to n is perhaps a numerate number. It equals (sum n)^2. n=3, sum (n^3)=1+8+27=36 n=4, sum (n^3)=1+8+27+64=100 n=5, sum (n^3)=1+8+27+64+125=225 n=6, sum (n^3)=1+8+27+64+125+216=441 n=7, sum (n^3)=1+8+27+64+125+216+343=784 n=8, sum (n^3)=1+8+27+64+125+216+343+512=1296 n=9, sum (n^3)=1+8+27+64+125+216+343+512+729=2025 n=10, sum (n^3)=1+8+27+64+125+216+343+512+729+1000=3025 The difference of (sum n)^3 and sum (n^3) is going to be a multiple of -s(n+1, n-2), the sum of the products of each trio of {1, ..., n}, t was for small values and maybe for these yet small values it will show a pattern based upon n. n=3, ((sum n)^3-sum(n^3))=216-36=180 n=4, ((sum n)^3-sum(n^3))=1000-100=900 n=5, ((sum n)^3-sum(n^3))=3375-225=3150 n=6, ((sum n)^3-sum(n^3))=9261-441=8820 n=7, ((sum n)^3-sum(n^3))=21952-784=21168 n=8, ((sum n)^3-sum(n^3))=46656-1296=45630 n=9, ((sum n)^3-sum(n^3))=91125-2025=89100 n=10, ((sum n)^3-sum(n^3))=166375-3025=163350 Then, I will divide ((sum n)^3-sum(n^3)) by -s(n+1, n2), to see what is the ratio of the difference of the two constructions. n=3, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=180/6 = 30 n=4, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=900/50 = 18 n=5, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=3150/225 = 14 n=6, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=8820/735 = 12 n=7, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=21168/1960 = 10.8 n=8, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=45630/4536 = ~10.05952 n=9, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=89100/9450 = ~9.428751 n=10, ((sum n)^3-sum(n^3))/(-s(n+1, n-2))=163350/18150 = 9 So then Im looking for a function thus that f(3)= 1/30 f(4)= 1/18 f(5)= 1/14 f(6)= 1/12 f(7)= 1/10.8 f(8)= 1/ ~10.05952 f(9)= 1/ ~9.428751 f(10)= 1/9 ... where the function f(n) gives -s(n+1, n-2) for ((sum n)^3 - sum(n^3)) * f(n), in a similar way to how s(n+1, n-1) = ((sum n)^2 -sum (n^2)) * 1/2. The table of Stirling numbers has an entry for n=24, -s(25, 22) = 3795000. Lets see, sum n for n=24 is 24(25)/2= 300, (sum n)^3 is thus 27000000, sum (n^3) is (sum n)^2 is 90000, their difference is 27000000-90000= 26910000, twenty six million nine hundred and ten thousand. Dividing that by -s(25, 22) yields around 7.090909, 7 1/11. Is there a function that works for all values of n that maps ((sum n)^3 - sum (n^3)) to -s(n+1, n-2)? What is it? Where its known, it might be simple to calculate s(n+1, n-2). Its pretty simple to calculate s(n+1, n-1). How about for ((sum n)^4 - sum (n^4)) and s(n+1, n-3), and so on for ((sum n)^x - sum (n^x)) and s(n+1, n-(x-1))? The Stirling number of the first kind, s(n, m), is (-1)^(n-m) times the number of permutations of n symbols which have exactly m cycles. So, the sum of the products of trios of a set {1, ..., n} is the number of permutations of n+1 symbols which have exactly n-2 cycles. Well, thats about that. This is about something different, whats the sum of the reciprocals of n!? That is, what is 1/(1!) + 1/(2!) + 1/(3!) ... + 1/(n!) + ...? What are the coefficients of the polynomial (n+1)(n+2)...(n+n)? Ross === Subject: Re: existence proof > A similar situation... > It is easy to prove that at least one of pi+e and pi*e is > transcendental. But no proof is known that pi+e is transcendental, and > no proof is known that pi*e is transcendental. Is it easy enough to sketch the proof for a general audience like sci.math? -- [R]eality has a fascinating ability to check us when we get a little too big for our britches... Make no mistake. There isnt a mathematician alive today that I cant now touch, and not a mathematical career on the planet that I cant now affect. --James Harris, render of worlds === Subject: Re: existence proof >> A similar situation... >> It is easy to prove that at least one of pi+e and pi*e is >> transcendental. But no proof is known that pi+e is transcendental, and >> no proof is known that pi*e is transcendental. >Is it easy enough to sketch the proof for a general audience like >sci.math? Yes. ************************ David C. Ullrich === Subject: Re: existence proof <87n0dqxftp.fsf@phiwumbda.org> > A similar situation... > It is easy to prove that at least one of pi+e and pi*e is > transcendental. But no proof is known that pi+e is transcendental, and > no proof is known that pi*e is transcendental. >>Is it easy enough to sketch the proof for a general audience like >>sci.math? > Yes. See, now, Lee should learn from the precision with which David answers simple yes/no questions. Now *thats* a helpful response, and not a byte wasted (which cannot be said of this critique). -- Jesse Hughes By definition m is a variable. By definition all then (sic) numbers represented by letters are variables--thats algebra[,] Magidin. -- James Harris shows deep understanding of algebra === Subject: Re: existence proof >> A similar situation... >> It is easy to prove that at least one of pi+e and pi*e is >> transcendental. But no proof is known that pi+e is transcendental, and >> no proof is known that pi*e is transcendental. >Is it easy enough to sketch the proof for a general audience like >sci.math? Lessee. The *definition* of algebraic number is root of a polynomial with rational coefficients, but then those clever algebraists prove--by some maneuver which never sticks in my mind, no matter how often I see it, but which always turns out to be really simple (maybe it has to do with companion matrices?)-- that every root of a polynomial with coefficients that are algebraic numbers is an algebraic number. So if both pi+e and pi*e are algebraic numbers, then both roots of x^2-(pi+e)x+pi*e are algebraic numbers. But theyre pi and e, which we know (because we have been told so so many times) are transcendental; so at least one of pi+e and pi*e is transcendental. Was that easy enough? Was it a proof? At least it was sketchy. Is one out of three good or bad? Lee Rudolph === Subject: Re: existence proof > A similar situation... > It is easy to prove that at least one of pi+e and pi*e is > transcendental. But no proof is known that pi+e is transcendental, and > no proof is known that pi*e is transcendental. >>Is it easy enough to sketch the proof for a general audience like >>sci.math? >Lessee. The *definition* of algebraic number is root of a >polynomial with rational coefficients, but then those clever >algebraists prove--by some maneuver which never sticks in my mind, >no matter how often I see it, but which always turns out to be >really simple (maybe it has to do with companion matrices?)-- >that every root of a polynomial with coefficients that are algebraic >numbers is an algebraic number. Well, if I explain this in terms so simple that even _I_ can understand whats going on maybe itll stick. If a is algebraic of degree (order? whatever) n then the set of all Q-linear combination of a^j, 0 <= j < n, is closed under multiplication. So if a and b are both algebraic then the Q-linear span of a^j*b^k (0 <= j < n, 0 <= k < m) is closed under multiplication. Similarly for finitely many algebraic numbers. So say x is a root of an equation with algebraic coefficients. Let F = Q[the coefficients]. Then F is a field and F has finite dimension when regarded as a Q-vector space. Let G = F[x]. The fact that x satisfies an equation with coefficients in F shows that the set of F-linear combinations of x^j, 0 <= j < n, is closed under multiplication. That is, G is a field and G has finite dimension as an F-vector space; since F has finite dimension as a Q-vector space it follows that G has finite dimension as a Q-vector space, hence x is algebraic. > So if both pi+e and pi*e are >algebraic numbers, then both roots of x^2-(pi+e)x+pi*e are algebraic >numbers. But theyre pi and e, which we know (because we have been >told so so many times) are transcendental; so at least one of pi+e >and pi*e is transcendental. >Was that easy enough? Was it a proof? At least it was sketchy. >Is one out of three good or bad? >Lee Rudolph ************************ David C. Ullrich === Subject: Re: existence proof <87n0dqxftp.fsf@phiwumbda.org> > A similar situation... > It is easy to prove that at least one of pi+e and pi*e is > transcendental. But no proof is known that pi+e is transcendental, and > no proof is known that pi*e is transcendental. >>Is it easy enough to sketch the proof for a general audience like >>sci.math? > Lessee. The *definition* of algebraic number is root of a > polynomial with rational coefficients, but then those clever > algebraists prove--by some maneuver which never sticks in my mind, > no matter how often I see it, but which always turns out to be > really simple (maybe it has to do with companion matrices?)-- > that every root of a polynomial with coefficients that are algebraic > numbers is an algebraic number. So if both pi+e and pi*e are > algebraic numbers, then both roots of x^2-(pi+e)x+pi*e are algebraic > numbers. But theyre pi and e, which we know (because we have been > told so so many times) are transcendental; so at least one of pi+e > and pi*e is transcendental. > Was that easy enough? Was it a proof? At least it was sketchy. > Is one out of three good or bad? I asked a simple yes/no question. I dont know why you cant just give me a one word answer. Damn, youre unhelpful. -- Jesse F. Hughes I think the burden is on those people who think he didnt have weapons of mass destruction to tell the world where they are. -- White House spokesman Ari Fleischer === Subject: Newton-Raphson method Does anyone know if there are any equations where the Newton-Raphson method fails no matter the value of x0 (the starting value)? I know that equations such y=x^2 fail when the starting value is 0 because f(x) = 2x = 0 so the tangent never touches the succeeds when the starting value is anything but 0. Any help greatly appreciated. Allan Lewis. === Subject: Re: Newton-Raphson method > Does anyone know if there are any equations where the Newton-Raphson > method fails no matter the value of x0 (the starting value)? Thomas points out that the method will oscillate for x |-> sgn(x-r) sqrt(|x-r|). Seems to me that this is true for any odd S-shaped curve with f(0) = 0. Of course, in the first example, the derivative does not exist at the zero. > I know that equations such y=x2 fail when the starting value is 0 > because f(x) = 2x = 0 so the tangent never touches the > this succeeds when the starting value is anything but 0. Huh? The method stops successfully immediately if f(x0) = 0. Perhaps you mean something like f(x) = x^3 + 1. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Differential Equations System in matrix form I need to solve the following differential system in matrix form: dN(x)/dx=M(x)*N(x) N(a)=I where I, N(x) and M(x) are matrices (n x n). I is the identity matrix, a is any initial point. Since M is not constant, the solution cannot be taken as matrix exponential. Nevertheless, M(x) can be put in the following form: M(x)=M0 + (1/x)*M1 + (1/x^2)*M2 where M0, M1, M2 are numerical matrices (not functions of x). The question is: does exist any analytical solution for this kind of matrix equation? In particular, I need to evaluate the matrix N(x) in x= 0, where the matrix M is divergent (as you can easily notice). I tried with a sixth-order implicit Runge-Kutta method, but I would like to know if any analitycal solution is possibile to reduce the working time and to improve the results. Please, help me as soon as possible. I would like to receive your Roberto Dott. Ing. Roberto Diana Politecnico di Bari Dipartimento di Elettrotecnica ed Elettronica Via Re David, 200 70125 Bari email: rdiana@libero.it === Subject: Re: Differential Equations System in matrix form > I need to solve the following differential system in matrix form: > dN(x)/dx=M(x)*N(x) > N(a)=I > where I, N(x) and M(x) are matrices (n x n). I is the identity matrix, a > is any initial point. > Since M is not constant, the solution cannot be taken as matrix > exponential. > Nevertheless, M(x) can be put in the following form: > M(x)=M0 + (1/x)*M1 + (1/x^2)*M2 > where M0, M1, M2 are numerical matrices (not functions of x). > The question is: does exist any analytical solution for this kind of > matrix equation? In particular, I need to evaluate the matrix N(x) in x= > 0, where the matrix M is divergent (as you can easily notice). I tried > with a sixth-order implicit Runge-Kutta method, but I would like to know > if any analitycal solution is possibile to reduce the working time and > to improve the results. > Please, help me as soon as possible. I would like to receive your > Roberto > Dott. Ing. Roberto Diana > Politecnico di Bari > Dipartimento di Elettrotecnica ed Elettronica > Via Re David, 200 > 70125 Bari > email: rdiana@libero.it Take a look in the mathematical methods book by Goertzel & Tralli. IIRC they discussed cases where this could be done explicitly. For general constant matrices M_0, M_1 and M_2 I do not think a closed-form solution is possible. -- Julian V. Noble Professor Emeritus of Physics ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ Science knows only one commandment: contribute to science. -- Bertolt Brecht, Galileo. === Subject: Re: Differential Equations System in matrix form >I need to solve the following differential system in matrix form: >dN(x)/dx=M(x)*N(x) >N(a)=I >where I, N(x) and M(x) are matrices (n x n). I is the identity matrix, a >is any initial point. >Since M is not constant, the solution cannot be taken as matrix >exponential. >Nevertheless, M(x) can be put in the following form: >M(x)=M0 + (1/x)*M1 + (1/x^2)*M2 >where M0, M1, M2 are numerical matrices (not functions of x). >The question is: does exist any analytical solution for this kind of >matrix equation? In particular, I need to evaluate the matrix N(x) in x= >0, where the matrix M is divergent (as you can easily notice). There probably isnt a closed-form solution. Time-ordered exponentials are sometimes used for this sort of thing. You can write the solution (for x > a) as a series N(x) = N(a) + sum_{k=1}^infinity int M(s_k) M(s_{k-1}) ... M(s_1) N(a) ds_1 ... ds_k where the integral is over the k-simplex a < s_1 < ... < s_k < x or for x < a, N(x) = N(a) + sum_{k=1}^infinity (-1)^k int M(s_k) ... M(s_1) N(a) ds_1 ... ds_k where x < s_k < ... < s_1 < a. In your case, its not at all clear that N(0) exists at all. For the k=1 term you have a divergent integral. In the n=1 case, where you have a closed-form solution N(x) = c exp(M0 x + M1 ln(x) - M2/x) this will diverge as x -> 0 if M2 < 0 (or if M2 = 0 and M1 < 0). Similarly for any n, if M0, M1 and M2 commute and M2 has a positive eigenvalue the solution will diverge as x -> 0. 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: Differential Equations System in matrix form Sorry, I messed the previous post up > I need to solve the following differential system in matrix form: > dN(x)/dx=M(x)*N(x) > N(a)=I > where I, N(x) and M(x) are matrices (n x n). I is the identity matrix, a > is any initial point. > Since M is not constant, the solution cannot be taken as matrix > exponential. > Nevertheless, M(x) can be put in the following form: > M(x)=M0 + (1/x)*M1 + (1/x^2)*M2 Which other information do you have? Do the generators M0,M1,M2 commute (this would faciliate the solution)? Do you know anything about the spectrum of M2 and M1 > where M0, M1, M2 are numerical matrices (not functions of x). > The question is: does exist any analytical solution for this kind of > matrix equation? In particular, I need to evaluate the matrix N(x) in x= > 0, where the matrix M is divergent (as you can easily notice). You must expect some singularity at x=0. If you consider the 1-d case dy/dx =y/x^2 you get a solution like exp(-1/x) dz/dx =alpha*z/x you get a solution like x^alpha which is not diofferentiable at x=0 if alpha<1. I would expect the first behaviour (y) for small x and z-behaviour for larger x. With matrices there need the singular behaviour may be oscillatory hth klaus > I tried > with a sixth-order implicit Runge-Kutta method, but I would like to know > if any analitycal solution is possibile to reduce the working time and > to improve the results. > Please, help me as soon as possible. I would like to receive your > Roberto > Dott. Ing. Roberto Diana > Politecnico di Bari > Dipartimento di Elettrotecnica ed Elettronica > Via Re David, 200 > 70125 Bari > email: rdiana@libero.it === Subject: Re: Differential Equations System in matrix form > I need to solve the following differential system in matrix form: > dN(x)/dx=M(x)*N(x) > N(a)=I > where I, N(x) and M(x) are matrices (n x n). I is the identity matrix, a > is any initial point. > Since M is not constant, the solution cannot be taken as matrix > exponential. > Nevertheless, M(x) can be put in the following form: > M(x)=M0 + (1/x)*M1 + (1/x^2)*M2 Which other information do you have? Do the generators M0,M1,M2 commute (this would faciliate the solution)? Do you know anything about the spectrum of M2 and M1 > where M0, M1, M2 are numerical matrices (not functions of x). > The question is: does exist any analytical solution for this kind of > matrix equation? In particular, I need to evaluate the matrix N(x) in x= > 0, where the matrix M is divergent (as you can easily notice). You must expect some singularity at x=0. If you consider the 1-d case dy/dx =y/x^2 you get a solution like exp(1/x^2) dz/dx =alpha*z/x you get a solution like x^alpha which is not diofferentiable at x=0 if alpha<1. I would expect the first behaviour (y) for small x and z-behaviour for larger x. With matrices there need the singular behaviour may be oscillatory hth klaus >I tried > with a sixth-order implicit Runge-Kutta method, but I would like to know > if any analitycal solution is possibile to reduce the working time and > to improve the results. > Please, help me as soon as possible. I would like to receive your > Roberto > Dott. Ing. Roberto Diana > Politecnico di Bari > Dipartimento di Elettrotecnica ed Elettronica > Via Re David, 200 > 70125 Bari > email: rdiana@libero.it === Subject: JSH: Dont miss this one Sorry to add to the noise, but there are various sci.math readers who collect Harris quotes, and theres one current on alt.writing that they dont want to miss. A post in the thread A sperm and egg analogy to my situation contains the sentence She could be the first intellectual sperm woman. Ok, you can all go back to work now. ************************ David C. Ullrich === Subject: Attn: Jim Heckman - follow up question/clarification getting it (I have been out of school too long). Here is what you posted a while back: > The key to this is the fact, which can be proved by induction, > that the angle subtended at the center of an n-d simplex by any > two vertices is acos(-1/n). One convenient Cartesian coordinate > system is to choose the first coordinate axis through an > arbitrary vertex, the second through the projection of another > arbitrarily chosen vertex onto the hyperplane perpendicular to > the first axis, and so on. In this system, the coordinates of > the vertices v_i are: > v_1 = (1,0,0,...) > v_2 = (-1/n,sqrt(1 - (1/n^2)),0,...) > v_3 = (-1/n,sqrt(1 - (1/n^2))*(-1/(n-1)),...) > ... Could you expand on this generic case a bit further? I am trying to extend this to the 4d example you gave (below), and I cant quite make the connection. In particular, how is the 3rd element of v_5 equal to -sqrt(5/24)? I dont doubt that it is, I just dont see how to calculate it. Craig > For example, in 2d its: > v_1 = (1,0) > v_2 = (-1/2,sqrt(3/4)) > v_3 = (-1/2,-sqrt(3/4)) > In 3d its: > v_1 = (1,0,0) > v_2 = (-1/3,sqrt(8/9),0) > v_3 = (-1/3,-sqrt(2/9),sqrt(2/3)) > v_4 = (-1/3,-sqrt(2/9),-sqrt(2/3)) > In 4d: > v_1 = (1,0,0,0) > v_2 = (-1/4,sqrt(15/16),0,0) > v_3 = (-1/4,-sqrt(5/48),sqrt(5/6),0) > v_4 = (-1/4,-sqrt(5/48),-sqrt(5/24),sqrt(5/8)) > v_5 = (-1/4,-sqrt(5/48),-sqrt(5/24),-sqrt(5/8)) === Subject: Re: My prime research, focus on a feature REVISED OOPS! Forgot to include the program in my last post. Notice the dx in it as it is setup so that a small change, which is to take away the (int) cast will shift it to the continuous field. Then its just a matter of letting your dx approach 0. Thing is the number of calculations for small dx get HUGE rather quickly. Im VERY interested in some experts on numerical methods for evaluating partial differential equations stepping forward, but my attempts so far to get an answer in that area from the math world have failed. My guess is that it would be a slamdunk in my favor, so they refuse to talk about it. James Harris ------------------------------------ #include #include double S(double x, double yin); double pi(double xin, double yin) { return ((int)xin-S(xin,yin)-1); } double min(double x, double y) { return x>y?y:x; } double S(double x, double yin) { double sum=0, i, sum1, sum2, dx=1; for(int i=2; i<=yin; i++) { sum1 = ( pi(x/i,min(i-dx,sqrt(x/i))) - pi(i-dx,sqrt(i-dx))) ; sum2 = ( pi(i,sqrt(i)) - pi(i-dx,sqrt(i-dx))); sum+=(sum1 * sum2); } return sum; } int main() { int input; cout <>input; cout << pi(input,sqrt(input)) << endl; return 0; } === Subject: Re: My prime research, focus on a feature REVISED In sci.physics, James Harris Notice the dx in it as it is setup so that a small change, which is > to take away the (int) cast will shift it to the continuous field. > Then its just a matter of letting your dx approach 0. Thing is the > number of calculations for small dx get HUGE rather quickly. And you wonder why no one takes your algorithm seriously. It has nice space complexity, Ill give it that. But the time complexity is atrocious. [rest snipped] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: My prime research, focus on a feature REVISED > OOPS! Forgot to include the program in my last post. > Notice the dx in it as it is setup so that a small change, which is > to take away the (int) cast will shift it to the continuous field. Also notice that treating dx in this manner (as a finite difference: dx = 1), renders the equation a simple Ôdifference equation, not a Ôpartial differential equation as you have consistently claimed. Partial differentials refer to ratios, not infinitesimals. > Then its just a matter of letting your dx approach 0. Thing is the > number of calculations for small dx get HUGE rather quickly. Note, however, that dy/dx does not necessarily get HUGE and can often be approximated accurately for Ôdx approaching zero. Thats what limiting processes are all about. Note also that partial differential equations are routinely solved numerically without that problem you suggest. See any text on the numerical solution of partial differential equations. You have apparently tried to sidestep an alleged problem with calculations involved Ôsmall dx by the assignment Ôdx = 1. This is a finite difference, not a partial differential. Although finite differences are often used in the solution of partial differential equations, they are used to approximate ratios. > Im VERY interested in some experts on numerical methods for > evaluating partial differential equations stepping forward, but my > attempts so far to get an answer in that area from the math world have > failed. Well, I dont claim to be an expert, in fact I am a fry cook with a national burger chain, but I have solved many differential and partial differential equations numerically. The equation you are solving in your program is NOT a partial differential equation. It is a Ôdifference equation, i.e. you are using finite differences as substitutes for infinitesimals rather than for partial differentials to perform the computations. > My guess is that it would be a slamdunk in my favor, so they refuse to > talk about it. Your guess is wrong. They dont talk about it because the label Ôpartial differential equation does not apply to your programmed solution. Your program computes its values using finite differences, not partial differentials. Just how stupid are you, James Harris? > James Harris > ------------------------------------ > #include double pi(double xin, double yin) > return ((int)xin-S(xin,yin)-1); > double min(double x, double y) > return x>y?y:x; > double S(double x, double yin) > double sum=0, i, sum1, sum2, dx=1; > for(int i=2; i<=yin; i++) > { > sum1 = ( pi(x/i,min(i-dx,sqrt(x/i))) - pi(i-dx,sqrt(i-dx))) ; > sum2 = ( pi(i,sqrt(i)) - pi(i-dx,sqrt(i-dx))); > sum+=(sum1 * sum2); > } > return sum; > int main() > int input; > cout < cin>>input; > cout << pi(input,sqrt(input)) << endl; > return 0; -- 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: My prime research, focus on a feature REVISED > Ill show you a difference equation. > Take y=x^2, and use y+dy = (x+dx)^2, so multiplying out you get > y+dy = x^2 + 2x dx + dx^2, so > dy = 2x dx + dx^2, so dy/dx = 2x + dx, > so 2x + dx is a difference equation, > 2x + dx is *not* an equation. It is an expression. dy/dx = 2x + dx > and in discrete math, the minimum > difference is 1, so with dx=1, you have 2x + 1. Thats a difference equation. > 2x + 1 is *not* an equation. It is an expression. dy = 2x + 1 > Its the difference equation for y=x^2. > The corresponding differential equation for y=x^2, is, of course, 2x, > as in the continuous field you let dx approach 0. > Ô2x is *not* an equation. It is an expression. y = 2x > But now that you have demonstrated your willingness to post a difference equation and (if we can read between the > lines) a differential equation, why not go ahead and post the Ô*partial* differential equation you claim your > difference equation leads to, and show us how it supports counting primes? Your silence on this point had become > obvious. My silence? Im surprised at anyone claiming Ive been silent, and Ive posted the partial differential many times. Here it is again. J_y(x,y) = [J(y, sqrt(y)) - J(x/y, y)] J_x(y, sqrt(y)) My numerical calculations with the program that Ill include here show that it is closer to the prime distribution than Li(x) but further than R(x), the Riemann function, which makes it a great candidate for the *reason* behind something that has intrigued mathematicians since Gauss. Besides that, its just really cool. James Harris === Subject: Re: My prime research, focus on a feature REVISED >> Ill show you a difference equation. >> Take y=x^2, and use y+dy = (x+dx)^2, so multiplying out you get >> y+dy = x^2 + 2x dx + dx^2, so >> dy = 2x dx + dx^2, so dy/dx = 2x + dx, >> so 2x + dx is a difference equation, >> 2x + dx is *not* an equation. It is an expression. > dy/dx = 2x + dx >> and in discrete math, the minimum >> difference is 1, so with dx=1, you have 2x + 1. Thats a difference equation. >> 2x + 1 is *not* an equation. It is an expression. > dy = 2x + 1 > >> Its the difference equation for y=x^2. >> The corresponding differential equation for y=x^2, is, of course, 2x, >> as in the continuous field you let dx approach 0. >> Ô2x is *not* an equation. It is an expression. > y = 2x >> But now that you have demonstrated your willingness to post a difference equation and (if we can read between the >> lines) a differential equation, why not go ahead and post the Ô*partial* differential equation you claim your >> difference equation leads to, and show us how it supports counting primes? Your silence on this point had become >> obvious. >My silence? Im surprised at anyone claiming Ive been silent, and >Ive posted the partial differential many times. >Here it is again. > J_y(x,y) = [J(y, sqrt(y)) - J(x/y, y)] J_x(y, sqrt(y)) Not that its all that important, but that is _not_ a pde. (A pde is a relation between a function and its partial derivatives. Whats a above is not that, its a relation between a function and the _composition_ of the partial derivatives with various other functions.) >My numerical calculations with the program that Ill include here show >that it is closer to the prime distribution than Li(x) but further >than R(x), the Riemann function, which makes it a great candidate for >the *reason* behind something that has intrigued mathematicians since >Gauss. Define here - I dont see any calculations or program included. >Besides that, its just really cool. >James Harris ************************ David C. Ullrich === Subject: Re: My prime research, focus on a feature REVISED > Ill show you a difference equation. > > Take y=x^2, and use y+dy = (x+dx)^2, so multiplying out you get > > y+dy = x^2 + 2x dx + dx^2, so > > dy = 2x dx + dx^2, so dy/dx = 2x + dx, > > so 2x + dx is a difference equation, > 2x + dx is *not* an equation. It is an expression. > dy/dx = 2x + dx > and in discrete math, the minimum > difference is 1, so with dx=1, you have 2x + 1. Thats a difference equation. > 2x + 1 is *not* an equation. It is an expression. > dy = 2x + 1 > Its the difference equation for y=x^2. > > The corresponding differential equation for y=x^2, is, of course, 2x, > as in the continuous field you let dx approach 0. > Ô2x is *not* an equation. It is an expression. > y = 2x > But now that you have demonstrated your willingness to post a difference equation and (if we can read between the > lines) a differential equation, why not go ahead and post the Ô*partial* differential equation you claim your > difference equation leads to, and show us how it supports counting primes? Your silence on this point had become > obvious. > My silence? Im surprised at anyone claiming Ive been silent, and > Ive posted the partial differential many times. > Here it is again. > J_y(x,y) = [J(y, sqrt(y)) - J(x/y, y)] J_x(y, sqrt(y)) Ha-ha. You have now fallen into your own trap! > My numerical calculations with the program that Ill include here show > that it is closer to the prime distribution than Li(x) but further > than R(x), the Riemann function, which makes it a great candidate for > the *reason* behind something that has intrigued mathematicians since > Gauss. See my comments following your post in which you actually include the program you claim is included here. > Besides that, its just really cool. Give me a break. That so-called partial differential equation you posted above is completely ignored in your program. > James Harris -- 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: My prime research, focus on a feature REVISED <3F513626.771F2246@ix.netcom.com> so 2x + dx is a difference equation, >> 2x + dx is *not* an equation. It is an expression. > dy/dx = 2x + dx dy = 2x dx + dx^2. -- Jesse F. Hughes Thats whats annoying about Usenet as some loser will state a case, get their ass kicked, but STILL keep coming back as if nothing happened. -- James Harris explains his strategy. === Subject: Re: My prime research, focus on a feature REVISED > > But do the research. They missed it. > > How would a guy who is proud never to have read a mathematics > textbook or paper know what mathematicians did or didnt miss? >>Ive never made any such claims, and in fact Ive read many math >>textbooks, and admittedly few math papers, but I have looked over >>some. >> > But mathematicians found that method ever harder to work out as you > expanded as you have to keep adding and subtracting these > combinations. > > Over a hundred years they found various tricks to speed things up, and > actually got close to what I discovered but apparently never made the > leap to considering a pi(x,y) function versus a pi(x) function. > > in square brackets []): > > We first describe the basic structure of algorithms of > Meissel-Lehmer type, following the treatment of Lehmer > [1959 reference]... > > let phi(x,a)... denote the partial sieve function which > counts numbers <=x with no prime factor less than or equal > to p_a [the a-th prime]. [Does this sound familiar yet, > James?] >>Yes, as I pointed out, mathematicians got *close* to the actual prime >>counting function, but failed--possibly because they couldnt accept >>the possibility--to realize that a pi(x,y) functional definition >>worked better than looking just for a pi(x) function. >>Here notice that it is phi not pi that is used. >> Excellent. This may be a new one: The reason yours is better >> is that you use the letter pi instead of phi. >Are you sure youre really a math professor? Well they have you on >their website at Oklahoma State University last time I checked so I >guess you are, at least in name. >I use pi(x,y) because pi(x,sqrt(x)) = pi(x) the traditional math >function as it is a prime counting function. >Their phi is a different function. It doesnt give a prime count. Are _you_ sure youre not a blithering idiot? >>The next step for the mathematicians would have been to use pi(x,y), >>and then realize that pi(y,sqrt(y)) - pi(y-1,sqrt(y-1)) equals 0 when >>y is prime, which would have given them my definition. >>Those who wish a better *look* at what mathematicians have, since >>Randy Poes copy into a text area drops formatting can consider >> http://mathworld.wolfram.com/MeisselsFormula.html >>and I want to emphasize the amount of detail theyll find there. >>Since mathematicians never quite made it to the full definition, what >>they have is clunkier. >> > and let Pk(x,a) denote the k-th partial sieve function, > which counts numbers <=x with exactly k prime factors, > none smaller than or equal to p_a.... >>Oh yeah, thats why you see p_a which just means the a_th prime. >>Like if a=1, its p_1 = 2. >>My prime counting function doesnt have p in it as it figures out >>the prime numbers itself, which when you think about it should be a >>minimum requirement for any prime counting function! >> You really must not realize how idiotic youre being here. >> Hint: The Meissel algorithm figures out the prime numbers itself. >> Honest. >No it doesnt. >Thats why it needs prime lists. >And yes people David Ullrich has given an outright lie. >But you see, from what Ive seen, mathematicians are used to lying to >society, which is why I contacted the FBI, as mathematicians are key >to the defense of this nation. >If a math professor like David Ullrich will lie so casually then maybe >lots of other mathematicians are used to casually lying. >And maybe theyre telling the government theyre working for us, when >theyre really working for other nations. >You see, if mathematicians cant be trusted, they cant be trusted. >>My guess is that feeling frustrated at years of work, mathematicians >>just called their prime counting *algorithms* functions to make >>themselves feel better. >>After all, remember it was Fermats Last Theorem many years before >>anyone thought it had been proven, when theorem means that something >>has been proven. >>And it bears mentioning that mathematicians make some wacky choices >>for their terminology as they use pi(x), where the symbol pi is used >>or pi for the count of prime numbers. Why? I dont know. >>Though Ive seen some people get confused as they think prime counting >>for some reason involves the number pi!!! >>The math world is not nearly as orderly or sensible as some people >>might believe, but you see, mathematicians are people too. > phi(x,a) = P0(x,a) + P1(x,a) + P2(x,a) + ... > where the sum on the right has only finitely many nonzero > terms. Now P1(x,a) = Pi(x) - a so that if one can compute > phi(x,a), P2(x,a), P3(x,a), etc, one can obtain Pi(x). > [Does this sound familiar yet, James? Going to keep > claiming there are no two-argument prime counting > functions that have ever appeared in mathematics before?] > > We now consider the computation of the sum phi(x,a). > Meissel-Lehmer methods [recall that LMO are still > describing a 1959 algorithm, not their 1996 refinement] > accomplish this by repeated use of the recurrence > > phi(x,a) = phi(x,a-1) - phi(x/pa, a-1) > > [Oh, gee. A recursive difference equation on the > two-argument prime-counting function. Guess nobody ever >>Well, its debatable whether or not you can call it a difference >>equation because its actually a partial sieve function as the >>authors themselves say above. >> Another hint: An equation is not a function. The equation >> above is a difference equation satisfied by that partial >> sieve function. (The same partial sieve function as the >> one you use, btw.) >Thats another lie. The authors *themselves* call it a partial sieve >function and not a difference equation as it uses prime lists, which >is given away by the pa above. Cuz you sure sound like an idiot sometimes. Yes, the authors refer to that partial sieve function as a partial sieve function, because thats what it is, just as your pi(x,y) is a partial sieve function. And that difference equation is a difference equation. >The p stands for prime, and Randy Poe should have put that in as >p_a since he lost formatting in copying to text. And p_a is the >a_th prime. For instance, p_5 = 11. >First David Ullrich lied earlier claiming the method found primes on >its own when it uses a prime list, And the algorithm gets that prime list from God? No, it constructs it, all by itself. Duh. >now he lies here claiming its a >difference equation. I didnt claim the _method_ was a difference equation. I said the difference equation was a difference equation. Heres a hint: A typical English sentence has both a subject and a predicate. Whether the sentence is true depends not only on the predicate, it depends on the subject. In all of your comments below you seem to be missing this - I say A is B and you say no thats a lie, C is not B. >Ill show you a difference equation. >Take y=x^2, and use y+dy = (x+dx)^2, so multiplying out you get > y+dy = x^2 + 2x dx + dx^2, so > dy = 2x dx + dx^2, so dy/dx = 2x + dx, >so 2x + dx is a difference equation, No, 2x + dx is not an equation at all. You can tell because theres no equals sign - an equation includes this symbol: =. Truly fascinating - you dont even know what the word equation means... >and in discrete math, the minimum >difference is 1, so with dx=1, you have 2x + 1. >Thats a difference equation. Its the difference equation for y=x^2. >The corresponding differential equation for y=x^2, is, of course, 2x, >as in the continuous field you let dx approach 0. >>The difference between a partial sieve function and what I have goes >>back to my prime counting function knowing numbers are prime, while >>above you see pa which should be p_a, though I guess Randy Poe got >>lazy in transcribing, so he lost formatting from the page. >>That is the a_th prime like before, so again, for a=1, p_a = 2, or for >>a=3, p_3 = 5. >>The equation needs a list of primes, which is why Id say its not a >>difference equation, as the only difference between a difference >>equation and a regular one is that it only uses integers. >>Arguing otherwise, is like saying that y=p_a^2 where you use only >>primes is a difference equation like y=x^2, where you can just put in >>integers for x. > [Heres part of their refinement] > In the Extended Meissel-Lehmer method, we will take > a = pi(x^1/3) [That is, theyll work out the terms using > primes up the cube root, rather than the square root > of x], in which case P3(x,a) = P4(x,a) = ... = 0, so > we need only consider the computation of P2(x,a) here. > [Clever combinatoric computation of P2(x,pi(x^(1/3)) > follows]. > > [The sieving up to the cube root rather than the square > root is the main reason this method is so fast and so compact > in storage requirements. It is O(x^2/3 + epsilon) in > computation, O(x^1/3 + epsilon) in storage. Consider > x = 10^15, x^1/2 ~ 3*10^7, x^1/3 = 10^5. The difference > between x^1/2 and x^1/3 in storage is a factor of 300. > The difference between x and x^2/3 in computation is > a factor of 100,000. > > What follows in the next 40 pages is not a bunch of claims > of how nobody has done anything like this before. Note that > they call this computation the Meissel-Lehmer method, > not the Lagarias-Miller-Odlyzko method. What > follows is a detailed description of their refinement, > claims of the complexity in terms of run-time and storage, > theoretical PROOFS of those claims, and then some actual > run-time experiments in the last couple of pages. The > mathematical description of the algorithm in equation > form suitable for rigorous proof of correctness (along > with those proofs) is what takes up the bulk of the paper. > > - Randy >>And that right there should be the final clue. >>To Randy Poe, a person from math culture with an emphasis on >>celebrity, it might seem like impressive evidence to mention 40 >>pages with multiple proofs, as he slipped up and revealed more about >>math culture than he might have realized. >>For mathematicians length and complexity might be great, but the rest >>of the world can appreciate short and simple. >>James Harris >> ************************ >> David C. Ullrich >Make no mistake, the lies David Ullrich told here are lies that most >mathematicians *know* are lies if they know about so-called prime >counting functions. >But will they help you the public by telling you that David Ullrich is >lying? >In my experience, no they will not. >Now if you all believe that mathematicians confident in lying to you, >where other mathematicians refuse to tell you when youre being lied >to, is ok, then dont be surprised if later we find out that the FBI >*should* have investigated, only to find that a math culture, which >lies so easily, acted against the people of the United States. >After all, if mathematicians have reached a point where they place >their society above world society, then how do you know exactly where >that might lead? >Society has values for a reason. Mathematicians are *supposed* to >follow rules like reporting important math results. Math professors >like David Ullrich are not supposed to make bold public lies to the >public. >Now if he thought he were going to get caught, would David Ullrich lie >so boldly? >What do you think that tells you about what he thinks of you? >James Harris ************************ David C. Ullrich === Subject: Re: My prime research, focus on a feature REVISED > For mathematicians length and complexity might be great, but the rest > of the world can appreciate short and simple. But what do your physical and mental characteristics have to do with math? -- Wayne Brown | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: My prime research, focus on a feature REVISED > Mathematicians are *supposed* to > follow rules like reporting important math results. Who is suppose to report Z[1/2] = R result and how? === Subject: Re: Reply from another German editor > ÔWinkel-Tri-Sektierer rarely get a rude reply. > This is a polite answer. But you should try to > understand what the last sentence really says. Oh, you mean contacting a local mathematician? I tried. As I live in Atlanta metro there are several universities. Phone calls and emails didnt get me more than one math professor referring me to another university. She said hers lacked the expertise. Beyond that an earlier version of the paper that the editor is referring to, was commented on by no less than Barry Mazur. Another earlier version went by Andrew Granville. So basically Ive already had comments on the paper from a mathematician at a level beyond what even some of those who would consider themselves excellent mathematicians get in their entire careers despite dozens if not hundreds of *published* papers. If they just did their jobs and reported then I wouldnt be having these problems, but apparently some mathematicians believe they can sit and wait. The trouble with that is they must be hoping that the truth wont gain traction, which is that their math discipline has this wacky, esoteric ßaw in whats called the ring of algebraic integers. They can run and hide, but after all, I found a short proof of Fermats Last Theorem, so tracking things down is what I do. Yeeehaa!!! James Harris === Subject: Re: Reply from another German editor > ÔWinkel-Tri-Sektierer rarely get a rude reply. > This is a polite answer. But you should try to > understand what the last sentence really says. > Oh, you mean contacting a local mathematician? > I tried. As I live in Atlanta metro there are several universities. > Phone calls and emails didnt get me more than one math professor > referring me to another university. She said hers lacked the > expertise. > Beyond that an earlier version of the paper that the editor is > referring to, was commented on by no less than Barry Mazur. And what, exactly, did Barry Mazur say? -- Bob Day === Subject: Re: Reply from another German editor > ÔWinkel-Tri-Sektierer rarely get a rude reply. > This is a polite answer. But you should try to > understand what the last sentence really says. >> Oh, you mean contacting a local mathematician? >> I tried. As I live in Atlanta metro there are several universities. >> Phone calls and emails didnt get me more than one math professor >> referring me to another university. She said hers lacked the >> expertise. >> Beyond that an earlier version of the paper that the editor is >> referring to, was commented on by no less than Barry Mazur. > And what, exactly, did Barry Mazur say? Damn, I was gonna ask that one. Well, James? Were really curious. === Subject: Re: Reply from another German editor >>Winkel-Tri-Sektierer rarely get a rude reply. >>This is a polite answer. But you should try to >>understand what the last sentence really says. >Oh, you mean contacting a local mathematician? >I tried. As I live in Atlanta metro there are several universities. >Phone calls and emails didnt get me more than one math professor >referring me to another university. She said hers lacked the >expertise. >Beyond that an earlier version of the paper that the editor is >referring to, was commented on by no less than Barry Mazur. >>And what, exactly, did Barry Mazur say? > Damn, I was gonna ask that one. > Well, James? > Were really curious. Heres my guess. Its only a guess, but I believe it cant be too far off the mark: >BEGIN GUESS>> I am sitting in the smallest room of my house, with your paper before me. Presently, it will be behind me. Yours Truly, et cetera... <<<<<<<<@ > I only submit my paper Advanced Polynomial Factorization to a single >@ > journal at a time; >[cut] >@ > But heres what the German editor said: >@ @ > I had a short look at your paper. My impression is that it cannot be >@ > given to a referee in the current form. There are no citations. Any >@ > reference to exsting work is missing, such that it cannot be seen if >@ > you have checked the originality of your investigations. You also may >@ > try to contact an expert in number theory at a university near to your >@ > place. >@ @ @> Bernd Wegner >@ > The gist of it is that the editor decided that it wasnt ok because I >@ > didnt reference other mathematicians!!! >@ No. Thats not it at all. Read his response again >@ and get a clue. He is telling you that you did not put >@ in your paper what portions (if any) of your work you >@ are claiming to be original and what portions (if any) >@are derived from the work of others and from where >@ they were so derived. If thats what the reviewer wants, >@ then salute and give it to him. >I agree with James Harris that the editors response is poor. >Since James Harris did not include any reference to existing >mathematical works, he is saying that everything in his >submission is original. In fact, this is true and a short >look at the paper would show this (although I havent seen the >paper, but I am going by what has appeared in this newsgroup). >I understand that this is most unusual for a submitted paper, >but I could see where some paper might not have any references, >yet still be of mathematical worth. I have a (half-page) paper appearing soon that contains no references. (I considered adding some but decided not to; any information contained in the references would already be very well-known to anyone who was going to be reading it.) >I think the editor should have given a better response than >what he did. >-- Bill Hale ************************ David C. Ullrich === Subject: Re: Reply from another German editor >I only submit my paper Advanced Polynomial Factorization to a single >journal at a time; however, heres a case where an editor apparently >decided to sit on my submission to the journal Algebra und Geometrie, >so I finally get a reply long after Id moved on, and yes, its >another German paper. The Germans have been a bit more interesting >than others in terms of how theyve behaved and I find it kind of >humorous. >The paper is currently at an American journal which has told me that >it has been given to referees, so it is under whats called peer >review. >But heres what the German editor said: >given to a referee in the current form. There are no citations. Any >reference to exsting work is missing, such that it cannot be seen if >you have checked the originality of your investigations. You also may >try to contact an expert in number theory at a university near to your >place. >Bernd Wegner >didnt reference other mathematicians!!! Not quite... >So he didnt even bother to send it forward for peer review. Just like you didnt bother to read the letter and try to understand the point to it before posting it here... >So the gist of it is that math society members ONLY need apply as far >as that editor is concerned, which is the kind of thing that Ive been >telling you about when it comes to the current celebrity math culture. Uh, no. The gist of it is hes not interested in publishing work that has nothing original about it. Hes being very polite, not mentioning the fact that his journal also requires that papers be comprehensibly written and contain _correct_ results. >Then again, hopefully Im wrong, and the American journal that >currently has the paper will demonstrate that I am. >Still Im not holding my breath, as I think that math society has gone >rogue. Consider that my paper highlights an error with what >mathematicians call core mathematics, so theres little chance that >Im going over old ground!!! >But then consider what that editor said anyway. >The behavior is bizarre to the level that I must say that clearly >mathematicians are behaving in a way that is not normal. >James Harris ************************ David C. Ullrich === Subject: Re: Reply from another German editor ..not mentioning the fact that his journal >also requires that papers be comprehensibly written and contain >_correct_ results. I particularly liked the editors hint to go to school. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Reply from another German editor > @ > I only submit my paper Advanced Polynomial Factorization to a > single @ > journal at a time; > [cut] > @ > But heres what the German editor said: > @ @ > I had a short look at your paper. My impression is that it cannot > be @ > given to a referee in the current form. There are no > citations. Any @ > reference to exsting work is missing, such that it > cannot be seen if @ > you have checked the originality of your > investigations. You also may @ > try to contact an expert in number > theory at a university near to your @ > place. > @ @ @> Bernd Wegner > @ > The gist of it is that the editor decided that it wasnt ok > because I @ > didnt reference other mathematicians!!! > @ No. Thats not it at all. Read his response again > @ and get a clue. He is telling you that you did not put > @ in your paper what portions (if any) of your work you > @ are claiming to be original and what portions (if any) > @are derived from the work of others and from where > @ they were so derived. If thats what the reviewer wants, > @ then salute and give it to him. > I agree with James Harris that the editors response is poor. > Since James Harris did not include any reference to existing > mathematical works, he is saying that everything in his > submission is original. In fact, this is true and a short > look at the paper would show this (although I havent seen the > paper, but I am going by what has appeared in this newsgroup). If everything is original, it would be a new math. Wouldnt it be nice if it was consistent with existing math? === Subject: Re: Reply from another German editor JSH, Why havent you posted to sci.chem? That seems to be the only NG that I visit and you do not. Why dont you start experimenting with ammonium nitrate and diesel fuel? I am quite certain that you could make some earth shaking discoveries. Lurch > I only submit my paper Advanced Polynomial Factorization to a single > journal at a time; however, heres a case where an editor apparently > decided to sit on my submission to the journal Algebra und Geometrie, > so I finally get a reply long after Id moved on, and yes, its > another German paper. The Germans have been a bit more interesting > than others in terms of how theyve behaved and I find it kind of > humorous. > The paper is currently at an American journal which has told me that > it has been given to referees, so it is under whats called peer > review. > But heres what the German editor said: > given to a referee in the current form. There are no citations. Any > reference to exsting work is missing, such that it cannot be seen if > you have checked the originality of your investigations. You also may > try to contact an expert in number theory at a university near to your > place. > Bernd Wegner > didnt reference other mathematicians!!! > So he didnt even bother to send it forward for peer review. > So the gist of it is that math society members ONLY need apply as far > as that editor is concerned, which is the kind of thing that Ive been > telling you about when it comes to the current celebrity math culture. > Then again, hopefully Im wrong, and the American journal that > currently has the paper will demonstrate that I am. > Still Im not holding my breath, as I think that math society has gone > rogue. Consider that my paper highlights an error with what > mathematicians call core mathematics, so theres little chance that > Im going over old ground!!! > But then consider what that editor said anyway. > The behavior is bizarre to the level that I must say that clearly > mathematicians are behaving in a way that is not normal. > James Harris === Subject: Re: Reply from another German editor In sci.physics, C. Bond <3F50DA42.8841869F@ix.netcom.com>: >> Still Im not holding my breath, > Please reconsider! The problem with holding ones breath is that one ultimately falls down unconscious and then starts breathing again. This obviously isnt quite the solution desired in some cases. Of course, I think in Mr. Harris case, he doesnt need to have oxygen shut off to his brain; he needs *more* oxygen in his brain in order to have a thought approaching coherency. Evidently he thinks theres a worldwide conspiracy afoot to prove to the world that his math is bunk. (Im not sure about his physics; Ive yet to see any. His writing is reasonably good although Im not sure thats the idea behind alt.writing.) [rest snipped] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: defining the factorial of a positive real... I know about Gamma of course but Ive never liked how Gamma, while indeed taking on factorial values at N, is completely unlike the factorial curve elsewhere (plot the factorials and draw a smooth line, similar-looking to y=e^x or y=x^x, to see what i mean). The way the gamma function is often presented- ie, people making a big deal about how it equals the factorials- i was under an impression that the problem of finding a continuous increasing-everywhere function that passes through the factorials. i undertook that problem on a boring night to pass some time, and was surprised to find quite the opposite. indeed, simply define r! = product (i in S) i where S = { r - n | n a nonnegative integer and r - n > 1 } Considering I have never seen this before, and yet it is clearly continuous and increasing everywhere... whats going on?? === Subject: Re: defining the factorial of a positive real... >I know about Gamma of course but Ive never liked how Gamma, while >indeed taking on factorial values at N, is completely unlike the >factorial curve elsewhere (plot the factorials and draw a smooth >line, similar-looking to y=e^x or y=x^x, to see what i mean). >The way the gamma function is often presented- ie, people making a big >deal about how it equals the factorials- i was under an impression >that the problem of finding a continuous increasing-everywhere >function that passes through the factorials. i undertook that problem >on a boring night to pass some time, and was surprised to find quite >the opposite. indeed, simply define >r! = product (i in S) i >where S = { r - n | n a nonnegative integer and r - n > 1 } >Considering I have never seen this before, and yet it is clearly >continuous and increasing everywhere... whats going on?? It is not differentiable at the natural numbers, though. There is a theorem by Harald Bohr (according to my copy of Courant) that every convex solution of the difference equation u(x+1)-u(x) = log x is of the form u(x) = log Gamma(x)+C, where C is an arbitrary constant. It follows that there is only one function u : (0,infinity) -> R such that u(x+1)-u(x) = log x for all x > 0 and u(1) = 0, and that solution is u(x) = log Gamma(x). Courant also supplies a proof. I think that this result makes the ordinarily defined Gamma function a very natural choice for the function. David McAnally -------------- === Subject: Re: defining the factorial of a positive real... > I know about Gamma of course but Ive never liked how Gamma, while > indeed taking on factorial values at N, is completely unlike the > factorial curve elsewhere (plot the factorials and draw a smooth > line, similar-looking to y=e^x or y=x^x, to see what i mean). > The way the gamma function is often presented- ie, people making a big > deal about how it equals the factorials- i was under an impression > that the problem of finding a continuous increasing-everywhere > function that passes through the factorials. i undertook that problem > on a boring night to pass some time, and was surprised to find quite > the opposite. indeed, simply define > r! = product (i in S) i > where S = { r - n | n a nonnegative integer and r - n > 1 } In this definition r is a bound variable, so how does r! = product (i in S) i define something that varies with r ? > Considering I have never seen this before, and yet it is clearly > continuous and increasing everywhere... whats going on?? -- G.C. === Subject: Re: defining the factorial of a positive real... > I know about Gamma of course but Ive never liked how Gamma, while > indeed taking on factorial values at N, is completely unlike the > factorial curve elsewhere (plot the factorials and draw a smooth > line, similar-looking to y=e^x or y=x^x, to see what i mean). > The way the gamma function is often presented- ie, people making a big > deal about how it equals the factorials- i was under an impression > that the problem of finding a continuous increasing-everywhere > function that passes through the factorials. i undertook that problem > on a boring night to pass some time, and was surprised to find quite > the opposite. indeed, simply define > r! = product (i in S) i > where S = { r - n | n a nonnegative integer and r - n > 1 } > Considering I have never seen this before, and yet it is clearly > continuous and increasing everywhere... whats going on?? Is r real? If so S is uncountable. How do you define the product in that case? -- G.C. === Subject: Re: defining the factorial of a positive real... > I know about Gamma of course but Ive never liked how Gamma, while > indeed taking on factorial values at N, is completely unlike the > factorial curve elsewhere (plot the factorials and draw a smooth > line, similar-looking to y=e^x or y=x^x, to see what i mean). > The way the gamma function is often presented- ie, people making a big > deal about how it equals the factorials- i was under an impression > that the problem of finding a continuous increasing-everywhere > function that passes through the factorials. i undertook that problem > on a boring night to pass some time, and was surprised to find quite > the opposite. indeed, simply define > r! = product (i in S) i > where S = { r - n | n a nonnegative integer and r - n > 1 } > Considering I have never seen this before, and yet it is clearly > continuous and increasing everywhere... whats going on?? > Is r real? If so S is uncountable. How do you define the product > in that case? r is real, but S is finite. Her definition is equivalent to saying that r! is the product of (r - i) as the index i ranges from 0 to ßoor(r)-1. Of course, the trouble, as William pointed out, is that the function so obtained is not smooth. (It is not differentiable when r is an integer.) To answer one of Williams questions, BTW, the empty product is 1. So at least that aspect of her definition isnt problematic. David === Subject: Re: defining the factorial of a positive real... > The way the gamma function is often presented- ie, people making a big > deal about how it equals the factorials- i was under an impression > that the problem of finding a continuous increasing-everywhere > function that passes through the factorials. Would you complete your sentence, hence your thought? > simply define > r! = product (i in S) i > where S = { r - n | n a nonnegative integer and r - n > 1 } If r <= 1, then S = nulset. How is product of no numbers defined? If 1 < r <= 2: S = { r }; r! = r; 2! = 2 If 2 < r <= 3: S = { r, r-1 }; r! = r^2 - r; 3! = 6 If 3 < r <= 4: S = { r, r-1, r-2 }; r! = r^3 - 3r^2 + 2r; 4! = 24 > Considering I have never seen this before, and yet it is clearly > continuous and increasing everywhere... whats going on?? Whats d(r!)/dr when r an integer? Compare dGamma(r)/dr. === Subject: Re: My prime research, focus on a feature Hello there. Im new here, and already found out that James Harris is working hard in the field of number theory. Well, so am I, although not from a mathematical point of view, but from a computer scientists view. The dS function J.H. gave - be it correct or not - is sure an interesting one, but from the point of complexity it is of no practical use whatsoever, because when analyzing bigger numbers, the complexity skyrockets and calculations would take way too long to be of any use. Just for the sake of it I give you another example my research work has yielded: The limit of the function a(t) = ßoor(X/ceil(X/a(t-1)) with t going to infinity, yields a factor of (the integer) X. Might sound remarkable, but when you express it in terms of a programming language, it prooves to be the same as the trivial division algorithm - one of the worst known, in terms of complexity. Im not trying to discourage anyone, just encouraging you to speed things up ;-) CM Wintersteiger (Excuse my bad english, Im german.) >Repeatedly Ive given out a definition for counting prime numbers, but >Im afraid that something has not been communicated. So first Im >going to give an example, counting the number of prime numbers up to >100, and afterwards Ill give the function. In that way I hope to >point clearly to a key feature of my research. >What my function does is count the number of composites for each >integer up to the square root of your base number. So for counting >primes up to 100 the base number is 100 and its square root is 10. >Here are the composite counts: > dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) = 6; dS(100,7) = 3; dS(100,8) = 0; > dS(100,9) = 0; dS(100,10)= 0; >and summing the dS values gives you 74. Notice when dS equals 0. >Now you add 1 for the number 1, as its not considered prime, and >subtract from 100 to get 25, which is the count of primes up to 100. >And those prime numbers are >2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, >67, 71, 73, 79, 83, 89, and 97. >Now a key feature of dS that I want to emphasize is that for dS(x,y) >if y is not prime dS equals 0. I want to repeat that for emphasis, >for dS(x,y) if y is not prime dS equals 0. >Amazingly enough, mathematicians never discovered such a function as >my dS throughout their entire history. >Some posters have tried to convince you otherwise. While some >apparently have tried to convince you that it doesnt matter if thats >true. Basically I think many of them have just worked to confuse you. >Now heres that definition Ive given so many times before: >dS(x,y) = [pi(x/y, y-1) - pi(y-1, sqrt(y-1))] > [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))], >S(x,1) = 0. >And pi(x, y) = ßoor(x) - S(x, y) - 1, and you get S as the sum of dS >from dS(x,1) to dS(x,y). >Note: pi(x,sqrt(x)) here gives the same value as the traditional >pi(x). >For faster calculations you need to use > dS(x,y) = [pi(x/y, sqrt(x/y)) - pi(y-1, sqrt(y-1)] > [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))] >when sqrt(x/y) < y-1. >See http://groups.msn.com/AmateurMath/primecountingfunction.msnw >If your eyes glazed over, or you found yourself wishing to quickly run >away from what looks very complicated then you might understand how >some posters could so easily manipulate the discussion away from the >truth. >The truth is that the dS function does the amazing, as you see a >complete definition above, where you dont see any prime numbers >except 2 but *somehow* as you noticed above dS(x,y) equals 0 when y is >a prime number. >You see, the function *knows* when a number is prime, making it >unique. >Those values from above were > dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) = 6; dS(100,7) = 3; dS(100,8) = 0; > dS(100,9) = 0; dS(100,10)= 0; >and you can see the definition I gave above at work, as it picks its >way through the numbers dropping to 0, whenever y is not prime. >Thats so huge that I cant emphasize it enough. Its also something >that should give you a certain sadness because what should have >happened is that mathematicians should have excitedly embraced such a >result. >And I think for some of you, disbelieving *me* is easier than >considering even the possibility that a discipline like mathematics >could have enough corruption that mathematicians would fight what they >should be celebrating. >But focusing on me is a mistake. >If you really wish to consider human achievement as some personal >issue, where you wish to make me that powerful as a single human being >that the accomplishments of thousands of years of effort can be >reduced to just being about the individuals involved, then you shatter >the foundations of human civilization, and make it just some fashion >show. >The truth is that the discoveries are more important than the >discoverers, and in letting it turn into some celebrity issue, you >demean human society and its accomplishments. >For some of you I fear that the idea that I might become a celebrity >is all that matters to you. And I fear you think thats all that ever >mattered through all of human history, as if the truth were just a >word, and life has always just been a big popularity contest. >The truth is not just a word and emphasizing truth over social issues >has helped humanity to make the technological achievments which make >my communication through this medium possible. >Where would we be if engineers had paused to figure out who they liked >and disliked before they bothered to use the information discovered? >Mathematicians arent doing you any favors. I suggest that all of you >focus less on me than on the information, and forget about what >benefits telling the truth will give me, as you consider that the >value to society is far, far greater. >So how can mathematicians behave as they have? >Thats an important question, and I suggest we find out why popularity >has so much importance in their world. >James Harris === Subject: Re: quaternions-- whats the point? >> The real simple classical Lie groups fall into 10 categories, up to >> local isomporphism: >You both seem to me to love complication. Not complication. Precision. >There are 4 families of simple Lie algebras over C, >and each such algebra has a unique compact real form. So why do you want to restrict your attention to compact not real simple compact classical Lie groups. If I had wanted to restrict attention to compact groups, I would have specified that the groups of interest are compact. Noncompact groups are also of importance, for example, in physics, Sp(2n,R), SO(3,1) and SO(4,2) are of importance. SO(2,1) is the symmetry group of the hyperbolic plane. Why should I have restricted the real groups that I discussed to compact groups, when I did not mention any such restriction? >Apparently (from the above) you are only interested in groups >up to local isomorphism, ie you are interested in Lie algebras. I did not feel like specifying ALL the real Lie groups (e.g. Spin(p,q) and its quotient groups, and what do I do about the universal covering group for a group like SO(4,2), which is known to be infinitely connected, so that the covering group covers SO(4,2) infinitely many times, with the result that there is infinitely many groups locally isomorphic to SO(4,2), which are mutually non-isomorphic), so I gave a generalization with a representative from each class. I knew enough to be able to comment that SU(2) and Sp(2) are isomorphic, rather than just locally isomorphic. The following real simple classical Lie groups are infinitely connected (and so there are infinitely many groups locally isomorphic to the group, which are mutually non-isomorphic): SL(2,R); SU(p,q) for p > 0 and q > 0; SO(p,2) for p = 1 and also for p > 2 (SO(2,2) is infinitely connected, but it is not simple); SO^*(2n) for n > 0; Sp(2n,R) for n > 0. SO(2), which is also isomorphic to U(1), is, strictly speaking, a simple Lie group, but it is also abelian. >In that case my statement above seems to me much more likely >to be relevant to any issue actually likely to arise, >as well as being far simpler. No, it doesnt, because the restriction to compact real forms is an artificial restriction. A discussion on all real forms of these groups is exactly that: a discussion on ALL real forms, and not just on compact forms. And as I noted above, there are noncompact groups of significant importance, including Sp(2n,R), SO(3,1) and SO(4,2) in physics. >Nothing you have said has changed my view >that in virtually all cases (of any interest) >where the symplectic group plays a role >the structure arising can be described in terms of quaternions. Is this true in the case of phase space in Lagrangian Mechanics, particularly if the number of conjugate pairs is odd? David McAnally -------------- === Subject: Re: Memory requirement, prime counting clincher > >> For those who dont know Christian Bau is one of the posters who has >> spent time trying to call my prime counting function Legendres >> Method. >I am not calling it Legendres method, it _is_ Legendres method in an >obfuscated form. Readers should not that is Christian Baus claim which Richard Tobin questions in what follows. > The Mathworld page that James mentioned > (http://mathworld.wolfram.com/PrimeCountingFunction.html) lists the > space complexity of Legendres method as O(N^1/2), whereas Jamess > program appears to be O(log(N)) while still being O(N) in time. > > Maybe its obvious that Legendres method can be modified to have that > property (perhaps by greatly increasing the constant for the O(N) > time); I dont know. > That is Harris ingenious invention: In the variant of his algorithm > that he published first (the slow one) he doesnt calculate a table of > prime numbers but checks each number individually to see whether it is > prime or not; the faster variant of his algorithm uses a table of - > guess what - size O (N^(1/2)) containing all the primes. Back to what Richard Tobin said, the answer is that my prime counting function is NOT Legendres Method, and Christian Bau in claiming it is has been directly refuted by the memory usage of a program that implements my prime counting function. And in fact as revealed at http://mathworld.wolfram.com/PrimeCountingFunction.html all the practical methods for counting primes known in mainstream mathematics need prime lists. So a key difference is that my prime counting function figures out the numbers which are prime that it needs versus the other methods that need to be given a list of primes, so a direct implementation doesnt use nearly as much memory. James Harris === Subject: Re: Memory requirement, prime counting clincher >A poster mentioned something that I thought it worth emphasizing which >is the memory requirement for my prime counting *function* which >clearly distinguishes it from what mathematicians discovered before >me. > The program uses the stack as its working space. Each recursive call > makes its own copy of the local variables in a stack frame. Being an > accomplished programmer, you knew that, right? Yup. And I mentioned space for the variables used by the program. My reference on storage complexity was http://mathworld.wolfram.com/PrimeCountingFunction.html and a table on the page with the headers method, time complexity, and storage complexity. My interpretation for the values given is that they represent a rough estimate of the space used for the prime lists that each of those methods require. Since my prime counting function, an integration of a partial difference equation, doesnt need prime lists, based on that interpretation, its storage complexity I figured would be 0, within that context. So the context of that table is important. Ive had various disagreements with my assessment, and I dont mind shifting to whats generally agreed upon. > However, the stack is allocated once at program startup time, and > experiments show that the stack depth only grows as about log10(N). So > its fair to say that your program is O(1) in storage complexity. > It depends what your compiler and OS does. If your stack is static and > non-extendable, then James algorithm will crash at high enough numbers to > blow the stack (typically 128k). If the stack is extendable (ARM > architecture, for example), then you have to consider the stack depth as > memory requirements. > Either way, the algorithm will only work with log(N) memory, assuming your > log10(N) observation was specific to this program, rather than general. (and > I must say, Im sceptical about that!) And even with that assessment a comparison with the other methods on the page http://mathworld.wolfram.com/PrimeCountingFunction.html shows that my prime counting function is not any of them, which directly refutes claims that various posters have made that it is. Theyll say its just Legendres Formula or Meissels Formula, but readers can see that NONE of the methods shown, including those, are even close on the storage complexity. > Your conclusion is too kind. James algorithm is clearly not O(1) in storage > complexity, unless you limit yourself to N < M, a silly limitation. > Roland What can be lost in the arguments about just how negligible is the memory use is the fact that it is negligible in comparison to the other methods listed at http://mathworld.wolfram.com/PrimeCountingFunction.html which directly refutes those posters who have claimed that its just one of those methods and is nothing new. James Harris === Subject: Re: Memory requirement, prime counting clincher > That is, indications are that it is proof to them that one man cant >> make a discovery so huge as to be bigger than the mathematicians who >> society has put in place to judge such finds. >> While they block me, they prove to themselves that theyre more >> important than I am, no matter what I discovered. >> They cant block you from publishing on usenet, obviously. You have >> succeeded in putting the word out. >> Now you have to be very vigilant. They will probably start using your >> results in their own work without crediting you. You should start reading >> all the number theory journals right now to guard your rights. You dont >> want to be cheated out of your proper glory. >I wish thatd happen. Instead areas that I work on are probably, >unfortunately, blocked off, as mathematicians cant start using my >research finds without realizing that eventually itll lead to my >proper recognition. >So as I work on my research and talk about it, areas of mathematics >get continually blocked off. Damm, we didnt think youd catch on to that so fast. Heres how you can bring the entire world of mathematics to its knees without involving Congress or the FBI: Branch out. Start working on problems in _every_ area of mathematics. Before too long all of mathematics will be blocked off. Imagine how silly us mathematicians will feel when when that happens. >I figure though that eventually *some* mathematician might be driven >to break with their society. >It is an odd situation, but my prime counting work simply doesnt >leave any other explanation as who can imagine physicists spending so >much energy trying to totally dismiss a simple, but profound result >that they know is correct? >Sure physicists get bothered by people claiming something is correct, >when they say its not, but here the mathematicians admit my work is >correct. >Anyone who actually *looks* at my prime counting function, and then >looks at what mathematicians are touting as so much greater than it, >can see that theyve just gone batty. >Oh, so I *wish* theyd start using my work, even if they try to use it >secretly. But you see, they probably know that once papers got >published, thatd take away their ability to deny. Maybe no one would >notice, but if anyone did and talked, theyd lose all. >Theyve painted themselves into a corner. >So I figure they wont allow any mathematicians to publish research >that might reveal the truth about my work. >James Harris ************************ David C. Ullrich === Subject: Re: Memory requirement, prime counting clincher [...] > What tools are you using to track memory usage? [...] Im letting my computer run Jamess PrimeCountH.java program for n=10^16, so its working on pi(10^16). This is what Ive got so far: In July, I computed pi(10^15) using Jamess PrimeCountH.java program. Here is a link to my post: I monitored the memory used by the program using the Windows Task Manager in Windows 2000. What I noticed was that the memory usage increased as time went by. The computation took about 5 hours. Once it completed, I think all the memory used was freed. So I dont know how much memory was being used 1 second before the computation was done... Now for the pi(10^16) case, if I allow a maximum amount of memory of 500 megabytes, the program runs for a few minutes and runs out of memory it is allowed to use before printing Sieve Time: [...]. If I allow up to 600 megabytes, it completes the sieving and prints: Sieve Time: 593680 m_max=100000001 Then, looking at the Task Manager, java.exe is using about 46 megs soon after printing the Sieve Time message. A bit later, java.exe is using only 1072 kbytes. After 3 hours CPU time, java.exe is using 1944 kbytes according to the Task Manager. I guess Ill let the pi(10^16) computation continue to run and see what happens... David === Subject: Re: Memory requirement, prime counting clincher > That is, indications are that it is proof to them that one man cant > make a discovery so huge as to be bigger than the mathematicians who > society has put in place to judge such finds. > > While they block me, they prove to themselves that theyre more > important than I am, no matter what I discovered. > > They cant block you from publishing on usenet, obviously. You have > succeeded in putting the word out. > > Now you have to be very vigilant. They will probably start using your > results in their own work without crediting you. You should start reading > all the number theory journals right now to guard your rights. You dont > want to be cheated out of your proper glory. > I wish thatd happen. Instead areas that I work on are probably, > unfortunately, blocked off, as mathematicians cant start using my > research finds without realizing that eventually itll lead to my > proper recognition. > So as I work on my research and talk about it, areas of mathematics > get continually blocked off. > I figure though that eventually *some* mathematician might be driven > to break with their society. > It is an odd situation, but my prime counting work simply doesnt > leave any other explanation as who can imagine physicists spending so > much energy trying to totally dismiss a simple, but profound result > that they know is correct? > Sure physicists get bothered by people claiming something is correct, > when they say its not, but here the mathematicians admit my work is > correct. > Anyone who actually *looks* at my prime counting function, and then > looks at what mathematicians are touting as so much greater than it, > can see that theyve just gone batty. > Oh, so I *wish* theyd start using my work, even if they try to use it > secretly. But you see, they probably know that once papers got > published, thatd take away their ability to deny. Maybe no one would > notice, but if anyone did and talked, theyd lose all. > Theyve painted themselves into a corner. > So I figure they wont allow any mathematicians to publish research > that might reveal the truth about my work. > James Harris Are you saying you found a pattern in the occurrences of primes in a linear count? Websters Revised Unabridged Dictionary (1913) Occurrence Oc*currence, n. Cf. F. occurrence. See Occur. 1. A coming or happening; as, the occurence of a railway collision. Voyages detain the mind by the perpetual occurrence and expectation of something new. --I. Watts. 2. Any incident or event; esp., one which happens without being designed or expected; as, an unusual occurrence, or the ordinary occurrences of life. All the occurrence of my fortune. --Shak. Websters Revised Unabridged Dictionary (1913) Linear Line*ar (--e*~er), a. L. linearis, linearius, fr. linea line: cf. F. lineaire. See 3d Line. 1. Of or pertaining to a line; consisting of lines; in a straight direction; lineal. 2. (Bot.) Like a line; narrow; of the same breadth throughout, except at the extremities; as, a linear leaf. Linear differential equation (Math.), an equation which is of the first degree, when the expression which is equated to zero is regarded as a function of the dependent variable and its differential coefficients. I am curious. Websters Revised Unabridged Dictionary (1913) Curious Curi*ous (k?r?-?s), a. OF. curios, curius, F. curieux, L. curiosus careful, inquisitive, fr. cura care. See Cure. 1. Difficult to please or satisfy; solicitous to be correct; careful; scrupulous; nice; exact. Obs. Little curious in her clothes. --Fuller. How shall we, If he be curious, work upon his faith? --Beau. & Fl. Ernst === Subject: Re: Memory requirement, prime counting clincher ... > |Ackermans function is not primitive recursive. > The primitive recursive functions are the ones that can be computed by > programs having only bounded looping: repeating a block of code a number > of times fixed at the start of the loop by the value of a variable, as well > as the usual if-then-else conditional constructs, and basic operations such > as incrementing indefinite-size integers. Thats quite a big class. Right. > A non-primitive-recursive function seems to me to have a curious open-ended > quality to it. All upper bounds on the number of steps required to compute > a non-primitive-recursive function also arent primitive recursive. If you > have a primitive recursive way of computing an upper bound on the number > of steps, then you can perform the computation itself in a primitive > recursive manner by simulating the original algorithm for N steps, where N > is the upper bound previously computed. ... > Ackermanns function illustrates how recursion can put you outside of that > class. If we fix one of the parameters, then the function of the other > argument is primitive recursive, computable with k nested loops. The > general function can be computed in a sense by setting up a variable number > of nested loops. Right also. The Ackermann function essentially diagonalizes over the primitive recursive functions. The PR functions can be put into nested computational classes (depending on the depth of nesting of the bounded loops) and equivalently using the dependencies of equations describing them. The PR functions correspond to a well-ordering of type omega (call it w); and Ackermann is the w for this. You can continue to play this game, with types of order w+1, w+2, ... 2*w, and onward through the countable ordinals. All the functions are manifestly computable in the theoretical sense, because the intermediate computations are all bounded by a function lower in the hierarchy. Of course they are grotesquely impractical to actually evaluate.. Dennis === Subject: Re: Iterated Function Systems > The attractor, A, of a set of transforms T{i} of a space S is the subset > of that space such that the union of A.T{i} is A. > >>Since you use the term the attractor, I assume that this set is >>maximal; i.e., if there is another A subset of S with the above >>property, then A is a subset of A. >> Yes. In most cases the attractor is unique, i.e. has no proper subsets >> with the same property. Exceptions include the cases in R^2 where y->y >> for all transforms, but one tends to forget these because neither random >> nor recursive iteration generates the attractor in this case. >Wed also want to watch out for the empty set: {}.X.T.X = {} is >always true regardless of X or T. I picked up the definition at the top from a paper on the web somewhere, as I thought a non-algorithmic definition of the attractor of an IFS would be useful, but it was part of the definition of self-affine tiles, and it would appear that I havent abstracted it perfectly. The other definition is the limit of one of several iterative algorithms, including the random, recursive and xerox algorithms. Trying again, the attractor, A, of a set of transforms T{i} of a space S is the finite, non-empty union of all subsets S* of that space such that the union of S*.T{i} is S*. The finiteness requirement eliminates systems like x -> x/2 + by + 1, y -> y x -> x/2 - by - 1, y -> y Im in two minds as to whether maximality needs to be explicitly stated (as above), or whether then attractor is unique. However, if the transforms are linear and all coeffecients are rational, then it seems to me that the intersection, B, of A and Q^2, is such that B.T{I} = C. Similarily, if all coeffecients are algebraic. Are any of the terms closed, compact and dense relevant? >The problem to me is with functions like T(z) = z*|z|; this separates >R^2 into four disjoint sets, A_0 {z : z=0}, A_1 = {z : |z|= 1}, A_2 = >{z : |z|>1}, and A_3 = {z : |z|<1}; with A_i.T = A_i for each i. I >think wed _want_ that the attractor be A_0 u A_1 = {z | z=0 or >|z|=1}, but I dont quite see the right definition to yield that. We are talking about the attractors of sets of functions, these sets having a cardinality strictly greater than one. The function T(z) = z*|z| cannot be a function within an IFS over R^2, as if one starts with, or ever reaches, a point z, |z| > 1, the iterations can diverge to infinity. It can be a function with an IFS over the unit circle. >Probably something topological is required; and since R^n has an >implicit topology, its typically not directly referred to...? -- Stewart Robert Hinsley === Subject: Re: Mathematical Paranoia [...] > So anyway, Im curious as to the level of security you guys put > on your private research. You see, I hate to admit it, but Im > extremely paranoid when it comes to sharing my in-progress ventures in > mathematics. I havent had anything published, so Im worried if > people can just look at my work and then submit it to journals. Do you > guys think this is not warrented, that perhaps I should be slightly > precautious, or have any other opinions? Id be interested into the > input anyone has over their own experiences over math paranoia and > what-not. much greater concern for an author is that he or she not be scooped; scooping means that someone else, working independently, gets results at least as good as your own. Generally, the first to publish gets almost all the credit; a simpler proof of an important result, or a proof by elementary methods can earn some credit. Its natural not to discuss work in progress except with collaborators. If some researchers use encryption or a bank safe to protect their manuscripts, that would be news to me. David Bernier === Subject: Re: Mathematical Paranoia > So anyway, Im curious as to the level of security you guys put > on your private research. You see, I hate to admit it, but Im > extremely paranoid when it comes to sharing my in-progress ventures in > mathematics. I havent had anything published, so Im worried if > people can just look at my work and then submit it to journal. I think it all depends on how you look at your work. I personally dont do (yet) any research in mathematics, but I do something that share some aspects, which is writing computer programs. Now, I dont want to start an ßame war about that, but my opinion is that ideas do not belong to individuals, but to everyone, and that everyone should be free to use them, as long as they dont try to deprive people from this freedom. Therefore all the programs I write are free software (free as in freedom; see http://www.gnu.org for a definition). In the same way, if I start doing research in math, I will welcome anyone to work with me, help me, improve my works or base his/her works upon mine (just as I do now with my software). I do know there is a risk of people using my works and claiming they are theirs, but I still rely on the integrity of the scientific community. Furthermore, you should be aware that the cases of researchers claiming ideas that were not theirs in the first place are quite rare. What happens sometimes though, is that the people who did most of the work on a discovery or result do not get as much credit as they deserved, the ones getting most of the credit being those who did the last step. But this has much more to do with being famous in the community (which implies publishing a lot, so you definitely should) than ideas. Sam -- Fear is the path to the dark side. Fear leads to anger, anger leads to hatred, hatred leads to suffering. I sense much fear in you. === Subject: Re: Mathematical Paranoia > private research... Actually, this being my first post, I would like > to just introduce myself quickly. > You see, Ive been visiting the sci.math discussion group for > quite a while over the past month or so, but I have rarely posted, if > at all. You see, I like to read to try and learn things from you guys, > and I doubt anything I could post would be that insightful anyhow. > After all, Im only in highschool, but Ill try not to sound > sophmoric, and my apologies if I do. > So anyway, Im curious as to the level of security you guys put > on your private research. You see, I hate to admit it, but Im > extremely paranoid when it comes to sharing my in-progress ventures in > mathematics. I havent had anything published, so Im worried if > people can just look at my work and then submit it to journals. Do you > guys think this is not warrented, that perhaps I should be slightly > precautious, or have any other opinions? Id be interested into the > input anyone has over their own experiences over math paranoia and > what-not. Are you related to James Harris? Gib === Subject: Re: Mathematical Paranoia === >Subject: Re: Mathematical Paranoia >Message-id: private research... Actually, this being my first post, I would like >> to just introduce myself quickly. >> You see, Ive been visiting the sci.math discussion group for >> quite a while over the past month or so, but I have rarely posted, if >> at all. You see, I like to read to try and learn things from you guys, >> and I doubt anything I could post would be that insightful anyhow. >> After all, Im only in highschool, but Ill try not to sound >> sophmoric, and my apologies if I do. >> So anyway, Im curious as to the level of security you guys put >> on your private research. You see, I hate to admit it, but Im >> extremely paranoid when it comes to sharing my in-progress ventures in >> mathematics. I havent had anything published, so Im worried if >> people can just look at my work and then submit it to journals. Do you >> guys think this is not warrented, that perhaps I should be slightly >> precautious, or have any other opinions? Id be interested into the >> input anyone has over their own experiences over math paranoia and >> what-not. >Are you related to James Harris? Harris isnt secretive about his work, much to the annoyance of everyone else. >Gib -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: [JSH] Re: Mathematical Paranoia Note subject change. In sci.math, Mensanator === >>Subject: Re: Mathematical Paranoia >>Message-id: > private research... Actually, this being my first post, I would like > to just introduce myself quickly. > You see, Ive been visiting the sci.math discussion group for > quite a while over the past month or so, but I have rarely posted, if > at all. You see, I like to read to try and learn things from you guys, > and I doubt anything I could post would be that insightful anyhow. > After all, Im only in highschool, but Ill try not to sound > sophmoric, and my apologies if I do. > So anyway, Im curious as to the level of security you guys put > on your private research. You see, I hate to admit it, but Im > extremely paranoid when it comes to sharing my in-progress ventures in > mathematics. I havent had anything published, so Im worried if > people can just look at my work and then submit it to journals. Do you > guys think this is not warrented, that perhaps I should be slightly > precautious, or have any other opinions? Id be interested into the > input anyone has over their own experiences over math paranoia and > what-not. >>Are you related to James Harris? > Harris isnt secretive about his work, much to the > annoyance of everyone else. Yeah, but look on the bright side, if anyone else ever manages to go along the same route, Harris will have blazed a, erm, trail for him so that the other poster knows what to avoid (like notifying the FBI, mathematician conspiracy theories, interesting allegations regarding irrational numbers in Z[1/2], unique factorization of 3 in the ring of algebraic integers, etc.) :-) [.sigsnip] -- #191, ewill3@earthlink.net -- and those are just the ones *Ive* seen Its still legal to go .sigless. === Subject: Mathematical Certainty So how do I know that my work is correct, that Ive found a short proof of Fermats Last Theorem, that I have THE prime counting function, and that I have found a problem with the ring of algebraic integers? Good question. Ive been looking for neat math since 1995 when I started out basically with a prayer and strange confidence, given its history, that I could find a short and simple proof of Fermats Last Theorem. I also decided rather quickly to talk about my work rather than try to hide it, like with that worry of people stealing your results. To me thats just plain silly as a worry. Well, in any event, I made a lot of mistakes, said a lot of dumb things, made a lot of dumb mistakes, made stupid mistakes, and said many things that were stupid and dumb, and basically, yeah, I made a LOT of mistakes. One thing I learned is that when youre publicizing your work, get really confident that youve made a great discovery, congratulate yourself, and begin partying, only to have it pointed out that youre wrong--and not in a nice way mind you--resist that, only to realize that you are wrong, it sucks. Several years back I found myself considering situations where I was just wrong, after being extremely confident that I was right, only to find out that the mathematics was against me. I learned that if the math is against you, youre wrong. Now sure you may say that it was silly of me to take years to learn such a thing everyone *should* just know, but then look at the situation now. The math says Im right. How I know is that I can trace through each and every step in my work, which thankfully is short enough, and basic enough to allow that process, and make certain that each is a logical step which follows from the previous step. Thats mathematics. In looking over past mistakes, I found that there were always hidden assumptions or missing steps where Id made a leap for an answer that I wanted. When faced with social humiliation after publicizing that I was right, only to see something wrong, Id proceed to put a lot of mental effort into making myself right, and learned that the truth does not change. Ultimately I realized that its better to search for mathematical truth, than to try and make mathematical truth, because it cannot be made. Now I see a puzzling situation where some of you seem intent on learning the lessons I learned the hard way, in an even harder situation in terms of the possible outcomes. For some of you, refusing to accept mathematics, as if you can *make* mathematical truth versus accepting it, involves betraying the public trust, like by giving young people false information. And you are being paid to accept mathematics and report truthfully about it. Well I wasnt being paid. I financed my math research since 1995, personally. And squeezed in thinking about mathematics while holding down full-time jobs. My mistakes were my mistakes. Your mistakes can have terrible consequences. My challenge to you is to *believe* in mathematics. If you do so, then I can take you step-by-step into a wonderful mathematical world of greater power than you ever might have expected to see in your lifetimes. After all, the methodology announces itself with a short proof of Fermats Last Theorem. Id think thats something of a statement. My proofs are out there. http://groups.msn.com/AmateurMath/explainingtheproof.msnw James Harris === Subject: Re: Mathematical Certainty >So how do I know that my work is correct, that Ive found a short >proof of Fermats Last Theorem, that I have THE prime counting >function, and that I have found a problem with the ring of algebraic >integers? Lemme guess. You know youre right because every competent person whos commented, online, journal editors, famous mathematicians youve emailed, says its wrong? >Good question. >Ive been looking for neat math since 1995 when I started out >basically with a prayer and strange confidence, given its history, >that I could find a short and simple proof of Fermats Last Theorem. >I also decided rather quickly to talk about my work rather than try to >hide it, like with that worry of people stealing your results. To me >thats just plain silly as a worry. >Well, in any event, I made a lot of mistakes, said a lot of dumb >things, made a lot of dumb mistakes, made stupid mistakes, and said >many things that were stupid and dumb, and basically, yeah, I made a >LOT of mistakes. >One thing I learned is that when youre publicizing your work, get >really confident that youve made a great discovery, congratulate >yourself, and begin partying, only to have it pointed out that youre >wrong--and not in a nice way mind you--resist that, only to realize >that you are wrong, it sucks. >Several years back I found myself considering situations where I was >just wrong, after being extremely confident that I was right, only to >find out that the mathematics was against me. >I learned that if the math is against you, youre wrong. >Now sure you may say that it was silly of me to take years to learn >such a thing everyone *should* just know, but then look at the >situation now. >The math says Im right. >How I know is that I can trace through each and every step in my work, >which thankfully is short enough, and basic enough to allow that >process, and make certain that each is a logical step which follows >from the previous step. >Thats mathematics. No, thats a fantasy of yours. Of the steps that are comprehensible, many of them do _not_ follow from previous steps, except in your imagination. >In looking over past mistakes, I found that there were always hidden >assumptions or missing steps where Id made a leap for an answer that >I wanted. >When faced with social humiliation after publicizing that I was right, >only to see something wrong, Id proceed to put a lot of mental effort >into making myself right, and learned that the truth does not change. >Ultimately I realized that its better to search for mathematical >truth, than to try and make mathematical truth, because it cannot be >made. >Now I see a puzzling situation where some of you seem intent on >learning the lessons I learned the hard way, in an even harder >situation in terms of the possible outcomes. >For some of you, refusing to accept mathematics, as if you can *make* >mathematical truth versus accepting it, involves betraying the public >trust, like by giving young people false information. >And you are being paid to accept mathematics and report truthfully >about it. >Well I wasnt being paid. I financed my math research since 1995, >personally. And squeezed in thinking about mathematics while holding >down full-time jobs. >My mistakes were my mistakes. >Your mistakes can have terrible consequences. >My challenge to you is to *believe* in mathematics. If you do so, >then I can take you step-by-step into a wonderful mathematical world >of greater power than you ever might have expected to see in your >lifetimes. >After all, the methodology announces itself with a short proof of >Fermats Last Theorem. >Id think thats something of a statement. >My proofs are out there. > http://groups.msn.com/AmateurMath/explainingtheproof.msnw >James Harris ************************ David C. Ullrich === Subject: Re: Mathematical Certainty > So how do I know that my work is correct, that Ive found a short > proof of Fermats Last Theorem, that I have THE prime counting > function, and that I have found a problem with the ring of algebraic > integers? > Good question. [...] > Well, in any event, I made a lot of mistakes, said a lot of dumb > things, made a lot of dumb mistakes, made stupid mistakes, and said > many things that were stupid and dumb, and basically, yeah, I made a > LOT of mistakes. So, the probability that youre mistaken again is pretty high isnt it? Why should it be different this time? If you overlooked mistakes in the past because you jumped too fast to conclusions, how would you know you didnt do that again? > How I know is that I can trace through each and every step in my work, > which thankfully is short enough, and basic enough to allow that > process, and make certain that each is a logical step which follows > from the previous step. Look, in the past you probably also traced each and every step in your work, with no luck. That leads to the logical conclusion you cant find mistakes reliably. Am I wrong about that? -- Edgar === Subject: Re: Mathematical Certainty > Ultimately I realized that its better to search for mathematical > truth, than to try and make mathematical truth, because it cannot be > made. Did you discovered the objects one day? No, youre trying to *make* them. It took a lot of time, and you changed the definition so many times, its obvious that youre simply trying to *make* a ring that fits your purposes. But how do you know that the currect definition is correct? How do you know that such ring exist? === Subject: Re: Mathematical Certainty > So how do I know that my work is correct, that Ive found a short > proof of Fermats Last Theorem, that I have THE prime counting > function, and that I have found a problem with the ring of algebraic > integers? > Good question. Nice of you to ask. > Ive been looking for neat math since 1995 when I started out > basically with a prayer and strange confidence, given its history, > that I could find a short and simple proof of Fermats Last Theorem. > I also decided rather quickly to talk about my work rather than try to > hide it, like with that worry of people stealing your results. To me > thats just plain silly as a worry. > Well, in any event, I made a lot of mistakes, said a lot of dumb > things, made a lot of dumb mistakes, made stupid mistakes, and said > many things that were stupid and dumb, and basically, yeah, I made a > LOT of mistakes. Like Nixon said of his advisors, They arent always right, but theyre always sure. > One thing I learned is that when youre publicizing your work, get > really confident that youve made a great discovery, congratulate > yourself, and begin partying, only to have it pointed out that youre > wrong--and not in a nice way mind you--resist that, only to realize > that you are wrong, it sucks. Unfortunately, that learning experience was not accompanied by learning anything else. You learned that Ôit sucks to be wrong. You did not, however, learn to rigorouslyidentify and correct your errors -- only to defend them with more passion than ever. > Several years back I found myself considering situations where I was > just wrong, after being extremely confident that I was right, only to > find out that the mathematics was against me. > I learned that if the math is against you, youre wrong. You should have also learned to suspect your own judgment. Any rational person would have realized that he had cognitive difficulties. > Now sure you may say that it was silly of me to take years to learn > such a thing everyone *should* just know, but then look at the > situation now. Youre still wrong and you still believe you are right. What has changed? > The math says Im right. No, it doesnt. Your work has been completely and irrevocably refuted. > How I know is that I can trace through each and every step in my work, > which thankfully is short enough, and basic enough to allow that > process, and make certain that each is a logical step which follows > from the previous step. Unfortunately, if you missed an error the first time, and fail to learn anything, you will continue to miss them. > Thats mathematics. > In looking over past mistakes, I found that there were always hidden > assumptions or missing steps where Id made a leap for an answer that > I wanted. Add to that list logic errors, non-sequitors ambiguous terminology, and contradictions. > When faced with social humiliation after publicizing that I was right, > only to see something wrong, Id proceed to put a lot of mental effort > into making myself right, and learned that the truth does not change. > Ultimately I realized that its better to search for mathematical > truth, than to try and make mathematical truth, because it cannot be > made. > Now I see a puzzling situation where some of you seem intent on > learning the lessons I learned the hard way, in an even harder > situation in terms of the possible outcomes. > For some of you, refusing to accept mathematics, as if you can *make* > mathematical truth versus accepting it, involves betraying the public > trust, like by giving young people false information. > And you are being paid to accept mathematics and report truthfully > about it. > Well I wasnt being paid. I financed my math research since 1995, > personally. You should demand your money back. > And squeezed in thinking about mathematics while holding > down full-time jobs. > My mistakes were my mistakes. > Your mistakes can have terrible consequences. > My challenge to you is to *believe* in mathematics. If you do so, > then I can take you step-by-step into a wonderful mathematical world > of greater power than you ever might have expected to see in your > lifetimes. > After all, the methodology announces itself with a short proof of > Fermats Last Theorem. > Id think thats something of a statement. > My proofs are out there. Yes. But they are wrong, and the errors have been clearly identified by others. -- 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: Teaching ßawed information > It is vital that the field be held to a very high standard, which > includes a very high *moral* standard. >>How could a thief, liar and charlatan like you know *anything* about >>moral standards? > How else who he avoid them so completely and so successfully for such > a long period of time? Well, you can learn a little about the shape of a building by walking around the outside without entering. I suppose he could have learned a bit about moral standards by slithering around them long enough to learn a little of their shape from the outside... -- Wayne Brown | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Teaching ßawed information >> Mathematicians are important members of the security apparatus of the >> United States, and it is VERY important if there is a question of the >> integrity of the group. > Im a member of NO apparatus. By the way, which apparatus are YOU a > member of? The humor apparatus? The bigotry apparatus? The punk > apparatus? The Hip-Hop apparatus? The drunken fool apparatus? How about the all-of-the-above apparatus? -- Wayne Brown | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Teaching ßawed information >> >> >> You claim that if you have a polynomial of the form >> >> >> P(x) = (v^3 + 1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3, >> >> >> where v = -1 + m*f^2, and m, u, and f are integers, >> with f prime and m coprime to f, then P(x)/f^2 >> can be factored in the form >> >> >> P(x)/f^2 = (b1*x + u)*(b2*x + u)*(b3*x + u*f) [1] >> >> >> where b1, b2, and b3 are algebraic integers. >> >> No I dont. >> >> >> >> I downloaded Advanced Polynomial Factorization >> from your website last night. You said, assuming >> >> >> P(x) = (a1*x + uf)*(a2*x + uf)*(a3*x + uf), >> >> where a1, a2, and a3 are algebraic integers, then >> letting g1 = a1*x + uf, g2 and g3 defined similarly, >> >> two of the gs should have a factor of f which >> would force two of the as to have a factor >> that is f. >> >> Assume without loss of generality that a1 >> and a2 have a factor that is f. >Yup, two of the gs *should* have a factor of f. > In other words: all your complaints about the algebraic integers being > broken or ßawed really mean the algebraic integers do not have > the property of magically making my argument correct. No. In the ring of objects the gs *do* have a factor that is f. The problem in the ring of algebraic integers is that you can have algebraic integers Ôa, Ôb, Ôc, and a prime p, such that abc = p but neither Ôa, Ôb, nor Ôc shares any non-unit factors with p. Intriguingly enough, by the definition of coprime used by Arturo Magidin, only one of them is coprime to p. So there are multiple ßaws within what mathematicians currently believe and teach. James Harris === Subject: Re: Teaching ßawed information Visiting Assistant Professor at the University of Montana. >>> > > You claim that if you have a polynomial of the form > > > P(x) = (v^3 + 1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3, > > > where v = -1 + m*f^2, and m, u, and f are integers, > with f prime and m coprime to f, then P(x)/f^2 > can be factored in the form > > > P(x)/f^2 = (b1*x + u)*(b2*x + u)*(b3*x + u*f) [1] > > > where b1, b2, and b3 are algebraic integers. > > No I dont. > > > > I downloaded Advanced Polynomial Factorization > from your website last night. You said, assuming > > > P(x) = (a1*x + uf)*(a2*x + uf)*(a3*x + uf), > > where a1, a2, and a3 are algebraic integers, then > letting g1 = a1*x + uf, g2 and g3 defined similarly, > > two of the gs should have a factor of f which > would force two of the as to have a factor > that is f. > > Assume without loss of generality that a1 > and a2 have a factor that is f. >>Yup, two of the gs *should* have a factor of f. >> In other words: all your complaints about the algebraic integers being >> broken or ßawed really mean the algebraic integers do not have >> the property of magically making my argument correct. >No. In the ring of objects the gs *do* have a factor that is f. Which is fine. But the reason you claim the algebraic integers are ßawed is because they do not, even though you really, really, really, really ->WANT<- them to. In other words, because they do not magically make your argument correct. >The problem in the ring of algebraic integers is that you can have >algebraic integers Ôa, Ôb, Ôc, and a prime p, such that > abc = p >but neither Ôa, Ôb, nor Ôc shares any non-unit factors with p. Nonsense. a is a factor of both a and p; b is a factor of both b and p; c is a factor of both c and p. If each of a, b, and c are units, then p would be a unit, which is impossible. Therefore, at least one of a, b, and c are not units. If a is not a unit, then a shares a non-unit factor with p, namely, a itself. If b is not a unit, then b shares a non-unit factor with p, namely b itself. If c is not a unit, then c shares a non-unit factor with p, namely c itself. In summary, at least one of a, b, and c must share a non-unit factor with p in the situation above. Is it not amazing how every time you try to make a SPECIFIC claim about how the ring of algebraic integers is ßawed, you end up saying something which is CLEARLY WRONG? >Intriguingly enough, by the definition of coprime used by Arturo >Magidin, only one of them is coprime to p. Nonsense. >So there are multiple ßaws within what mathematicians currently >believe and teach. Nonsense. [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee fixe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn awy from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a ßash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do not find it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Teaching ßawed information > The problem in the ring of algebraic integers is that you can have > algebraic integers Ôa, Ôb, Ôc, and a prime p, such that > abc = p > but neither Ôa, Ôb, nor Ôc shares any non-unit factors with p. Ôa is a factor of Ôa. Ôb is a factor of Ôb. Ôc is a factor of Ôc. Thus you wrong. === Subject: Re: Teaching ßawed information > The problem in the ring of algebraic integers is that you can have > algebraic integers Ôa, Ôb, Ôc, and a prime p, such that > abc = p > but neither Ôa, Ôb, nor Ôc shares any non-unit factors with p. Still wrong, James Harris. If Ôa, Ôb and Ôc are (non-unit) algebraic integers and Ôp is an algebraic integer of any description, including ordinary integers and prime integers, then Ôbc is an algebraic integer and Ôp/a is therefore an algebraic integer. Hence Ôp shares a common factor with Ôa, namely Ôa. And just as you have accepted that Ô2 and Ô4 share a common, non-unit factor of Ô2, so Ôp shares a non-unit factor of Ôa. The same argument applies to Ôb and Ôc. Therefore, and by simple algebra, *each* of the algebraic integers, Ôa, Ôb, and Ôc, share non-unit factors with Ôp. QED Your claim is false and a simple proof has been given (along with more complex proofs by other posters.) If you persist in repeating this refuted claim, you are obliged to provide support by producing a single number (just one will do) which should be in the ring of algebraic integers, but which is not. A single value of Ôa, for example, which is a non-unit algebraic integer but which does not divide Ôp in the ring of algebraic integers. Your refusal to post such a number will be taken as conclusive proof that you *cannot* do so, and hence that you *know* you are a liar. -- 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: Teaching ßawed information > Are math professors willing to teach their students information they > know is wrong? > > > I glanced at these threads of yours a few times. I was immediately > suspicious of your claims but I never managed to summon the energy to > check the claims myself (supposing I first managed to make sense of > them). > Well given your low energy, Ill help you out by giving them to you. No thank you. I have seen your arguments often enough. I used to study mathematics at a high level but that is a very long time ago. Now I program computers. My maths is occasionally useful in this but disappointingly rarely. To me, maths is mostly a hobby and a mental equivalent of going to the gym. I look in this newsgroup since quite often I find things that are interesting enough that I want to spend a few hours of my spare time looking at them. However nothing that you have posted has inspired me in this way though occasionally the rebuttals of others have. I have resisted for a long time posting any response since as I said, I never had the energy to check your tedious polynomial factorisations. The prime counting stuff seemed insufficiently different from existing stuff to have any interest. I finally posted since this 1/3 in Z[1/2] nonsense was so obviously wrong that no energy was needed. Any tiniest doubt that you were posting complete nonsense faded away immediately. But your recent thread on 1/3 being in the ring Z[1/2] is so > ridiculous that I did not need to expend any energy checking or trust > the opinions of others. It was obviously completely wrong. As many > others have pointed out, a ring does need to be closed under infinite > sums. However my immediate reaction, which was hinted at by others > but not as far as noticed explicitly mentioned, was that a ring is not > required to have a topology or metric and an infinite sum cannot in > general be given a meaning. > Well that sounds like you have a sense of decidability and your > intuition is that infinite sums arent always decidable in a ring. How do you get that impression? Like many others on this group, I would reserve decidability for a completely different meaning. The relevant concept for infinite sums is convergence which (as far as I know) requires a topology or metric. Rings have no requirement to have either. So infinite sums have no meaning at all in rings unless they also possess some other mathematical structure (e.g. topology). My impression of you is that you are applying some naive intuition to infinite sums. If you feel that you can associate some answer to the sum you say it is decidable. My feeling is that if there is a structure which permits infinite sums to be defined then I will test the sum using that structure. If not, the infinite sum is nonsense. Your 1/3 sum converges in the reals (the question can be asked since the reals have a topology). It does not converge in Z[1/2]. There are two possible explanations for the lack of convergence. If you consider Z[1/2] as only a ring then infinite sums are not definable. If you are studying it as a ring with a topology then the sum can be defined but does not converge. Naively it would seem to converge since the partial sums increase ever more slowly and are bounded. This would be enough to deduce convergence in the reals but not in an arbitrary ring with topology (actually the notion of increase still does not apply unless you add the further condiiton that the structure has an ordering). > That is true. For instance, in integers 1+1+1+... is not decidable, > as infinity is not a number. > However, 1/3 = 1/2 + 1/2^2 +1/2^3 +... is decidable, and in fact > theres no way to specifically exclude it by the definition of a ring. I have the same opinion as others. We do not need to find a rule which excludes 1/3 from the ring. You need to find a proof from the rules which shows that it is in the ring. Your intuition that the sum converges (regardless of the structure you are studying) is not sufficient. I believe that my door mat is a member of Z[1/2]. Can you show me the ring axiom that forbids a door mat from being a member of the ring? I admit that I cannot prove from the axioms that my door mat is in the ring but if you fail to find the rule that excludes it then surely it is a member together with your 1/3. > What Ive noticed is that posters cheat, as they *want* to exclude the > result, so they start talking but never actually make any sense, as > you begin to realize that really they just want to exclude the > result. the matter. I dont want to exclude 1/3, I just have noticed that it just is not in the ring. I want my door mat to be in the ring but sadly my wish is probably not enough to get it in. > That is, mathematically objections Ive seen to including 1/3 in > Z[1/2] are empty, but make sense if you realize that the posters just > WANT to exclude 1/3 so badly that really theyve just decided its > excluded and dont give a damn what the math actually says. > You seem to only interested in rings which are sub-rings of the > complex numbers. So all of your rings have an obvious topology. But > there are many other rings and many of them do not have any standard > or interesting topology. For example the integers modulo N (for some > positive natural number N) cannot be given an interesting topology. > Hence infinite sums are not possible or interesting in these rings. I > am quite sure that nonetheless they are rings. > Well you just said that infinite sums are not possible in finite > rings. Actually I was trying to say that infinite sums are not definable in arbitrary rings whether finite or not. Say that you are talking abut rings with a topology and then you can define infinite sums. You could then ask the question whether your sum converges in Z[1/2] but the answer would still be no. You could try to define a notion of extending a ring so that it was closed under certain well behaved infinite sums. The likely result would be that the closure of Z[1/2] in this way would be isomorphic to the reals. This is possibly an interesting field of research (maybe it already has a name) but it is not the study of rings in general since they are not required to have a topology. > Duh. > Again, what is clear from your post is that you do NOT want 1/3 > included in Z[1/2] and the math be damned. Quite the opposite. Want has nothing to do with my opinion that 1/3 is not in Z[1/2]. On the contrary you seem to want it in yet will not show how this can be proved from the axioms of a ring alone. > I now know, just from my own readings, that there is no significant > chance that anything you post is likely to be valid or interesting > mathematics. > > You have not damaged my faith in mathematics or its teaching. Heck, > even if you were right, how many maths courses even mention algebraic > integers, so how many courses and teachers would be invalidated. I > already knew that bad maths teachers existed, some making far worse > errors. But fortunately I have also met some very good ones. > I was a physics undergrad, and for me and my fellow physics > undergrads, it was extremely important that what we were told was the > truth to the limits of the professors ability to give us the truth. Of course that is desirable but I live in the real world and accept that people are human and sometimes make mistakes. As long as the mistakes are not too frequent and they accept them as mistakes when revealed, I have no problem. If someone makes mistakes frequently or refuses to acknowledge them when they are pointed out, then my respect for them diminishes. > But you see in physics, its understood that learning is a continual > process which involves constant checking at all levels. > Mathematicians have instead opted for a coral process, where what is > built upon is assumed to be perfect, when math students should be > taught to challenge. I have not got the impression that mathematicians believe that they are perfect. Certainly I have met some well respected mathematicians who have made mistakes. But they do acknowledge their mistakes. > The FBI nonsense is absolutely incredible. Fortunately I am outside > their jurisdiction so I can continue to believe that the algebraic > integers are a ring and 1/3 is not in the ring Z[1/2] without fear of > arrest. > > J > Mathematicians are important members of the security apparatus of the > United States, and it is VERY important if there is a question of the > integrity of the group. > After all, mathematicians may start willing to lie about important > math results, and having proven to be corrupt, later move to more > directly acting against the interests of the United States of America. > It is vital that the field be held to a very high standard, which > includes a very high *moral* standard. > James Harris Anyway, enough of this stuff. I got tried of your personal attacks on me and others long ago. I now know only from my own observations that your mathematics posting are highly unlikely to be correct or interesting. As I said I do not expect people to be perfect but your mistakes are frequent and rarely acknowledged. J === Subject: Re: Teaching ßawed information > Are math professors willing to teach their students information they > know is wrong? > > > I glanced at these threads of yours a few times. I was immediately > suspicious of your claims but I never managed to summon the energy to > check the claims myself (supposing I first managed to make sense of > them). > > Well given your low energy, Ill help you out by giving them to you. > No thank you. I have seen your arguments often enough. I used to > study mathematics at a high level but that is a very long time ago. > Now I program computers. My maths is occasionally useful in this but > disappointingly rarely. To me, maths is mostly a hobby and a mental > equivalent of going to the gym. Ok. James Harris === Subject: Re: Teaching ßawed information > James Harris > >> It is a test for my own entertainment as the issue has been settled by >> humanitys representatives--mathematicians. > > Shit! Who voted for mathematicians to be the representatives of > humanity. If anything mathematicians represent the abscense of what it > means to be human. > Absence. > And thats a particularly verminous comment by the way. Indeed. >139--141. ... Robin Chapman: An involution on derangements. >Discrete Math. ... --John === Subject: Re: Teaching ßawed information I downloaded Advanced Polynomial Factorization > from your website last night. You said, assuming > > > P(x) = (a1*x + uf)*(a2*x + uf)*(a3*x + uf), > > where a1, a2, and a3 are algebraic integers, then > letting g1 = a1*x + uf, g2 and g3 defined similarly, > > two of the gs should have a factor of f which > would force two of the as to have a factor > that is f. > > Assume without loss of generality that a1 > and a2 have a factor that is f. > > Yup, two of the gs *should* have a factor of f. > Yes, that is the wording in APF. So what you > must now be saying is that they *should* have a > factor that is f, but they do not. Right? Actually they *do* have a factor that is f in the object ring. However, the ring of algebraic integers is ßawed, which cannot be seen from that position in the paper. So it follows that they *should* have a factor of f in that ring, from what is in the paper. The Object Ring is a commutative ring that includes all numbers such that -1 and 1 are the only members that are both a unit and an integer, where no non-unit member is a factor of any two integers that are coprime. Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw > See also below. > > To equate this to what I said above, let > > b1 = a1/f, and b2 = a2/f. > > Since you say essentially that a1 and a2 > have a factor that is f, this means that > b1 and b2 are algebraic integers. > > They should be algebraic integers. > > Aha. Should again. Actually they *do* have a factor that is f in the object ring. However, the ring of algebraic integers is ßawed, which cannot be seen from that position in the paper. So it follows that they *should* have a factor of f in that ring, from what is in the paper. The Object Ring is a commutative ring that includes all numbers such that -1 and 1 are the only members that are both a unit and an integer, where no non-unit member is a factor of any two integers that are coprime. Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw > The question here is: why are you denying the > results in your own paper? > > Nope. > > The problem is that the ring of algebraic integers doesnt include > them. > > Which goes to why I use the word should as I do. > > And should yet again. Actually they *do* have a factor that is f in the object ring. However, the ring of algebraic integers is ßawed, which cannot be seen from that position in the paper. So it follows that they *should* have a factor of f in that ring, from what is in the paper. The Object Ring is a commutative ring that includes all numbers such that -1 and 1 are the only members that are both a unit and an integer, where no non-unit member is a factor of any two integers that are coprime. Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw > > The problem is that the bs arent included in the ring of algebraic > integers, which Ive pointed out more than once. > > > See above. When you say two of the as have a > factor that is f, and you are working in the ring > of algebraic integers, that can mean only one thing: > b1 = a1/f and b2 = a2/f are algebraic integers. > > Or are you now saying that your results in APF > are incorrect ??? > > Nope. > > Right. Not incorrect because you said should > rather than, e.g., must. > I think what we have here is the following. You > are saying should because you now know that in > fact b1 and b2 cannot be algebraic integers. In > other words, you have somehow arrived at the same > point as the rest of us. Theres no somehow about it as Ive explained repeatedly and in detail. As pointed out by several posters, an algebraic integer cannot be the root of a non-monic primitive polynomial. It *is* true that for an infinity of values the bs are provably not algebraic integers, though as Ive demonstrated with f=sqrt(2), m=1, there is at least one value where they are. And in fact that one value can be used to refute any attempts at challenging the result, as Ive pointed out before to you Nora Baron, as I noted that you can just use m=1(mod 2). You then acted as if you didnt understand exactly what I meant by m=1(mod 2). I think you asked something like if m=3 would work, and I answered, yes. > But we do not agree on what one must conclude > from this. I conclude that your original claim, > in the original version of APF, was incorrect and > that your proof was in error. You conclude now, > apparently, that you have arrived at a basic > contradiction: elementary algebraic number theory > shows that b1 and b2 cannot be algebraic integers, > but you think they *must* be, therefore algebraic > number theory is wrong somehow. You have phrased > this as, the ring of algebraic integers is > incomplete. But that is just dressing it up. > If you were right, you would have discovered a > fundamental inconsistency in mathematics. Actually, I found a mistake with what mathematicians believe as before they never were able to test the definition of algebraic integers as I have. I mean, consider that definition: An algebraic integer is the root of a monic polynomial with integer coefficients. Theres nothing there to prove that it wouldnt have a problem like this, and mathematicians worked for over a hundred years believing it didnt. Now Ive proven it does. You may think you have something with my use of should but in the object ring, two of the as *do* have a factor that is f. Its just that algebraic integers are wacky. The Object Ring is a commutative ring that includes all numbers such that -1 and 1 are the only members that are both a unit and an integer, where no non-unit member is a factor of any two integers that are coprime. Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw > Evidently you are not willing to bite that > particular bullet. > It still gets back to the word should. In > order to justify should, you have to return to > your argument about m = 0 (the degenerate case) Actually I use the constant term of the polynomial P(m), which, of course, is given by P(0). Now your thinking that something basic is proof Im wrong just tells a lot about how much you really understand mathematics, or more likely, how willing you are to try and pervert others understanding. Besides, as Ive pointed out to you, I can use m=1 with f=sqrt(2), so wheres your degenerate case then Nora Baron? > and f = 3 (the case where in general the associated > polynomials are not irreducible). We have > discussed ßaws in your reasoning for both of these Is that the royal we? > cases in detail and I do not agree with your > conclusions. *Even if I did*, I do not see how > showing a pattern for m = 0 and for f = 3 > proves anything about f = prime > 3 and m coprime > to f. For example, how would a statement for > m = 0 and f = 3 tell me anything about m = 1 > and f = 5? I think even you must admit that that part > of your argument is missing. All you have at > best is circumstantial evidence for two > peripheral cases. That evidence, even if one > believed it, would not constitute a proof. What I can show with f=3 is that your objections, like claiming m=0 is just a degenerate case, are cow dung. But when you see you cant make that claim when f=3, you bounce around it as if people are supposed to believe that something checks to see if f is coprime to 3. What you write in your posts does not make sense Nora Baron. And using f=3, is just an attempt at helping people that I came up after people like you managed to sow so much confusion about things like my very basic and simple lemma: ANY factor g of a polynomial is g=r+c, where c is a factor of the constant term, and r varies with the polynomials main variable, and is 0, when that variable is 0. When I saw how posters like yourself could get people agreeing with you when you were questioning something that basic I realized that math people werent about math. > But this is your basis for saying should. The ring of algebraic integers is wacky. The as *should* have f as a factor in that ring, as the paper notes. > > So how is such an error a moral test? > > > Here the poster Nora Baron has done something done before which is > to delete out context. > > I left in the *math* context. I deleted out the > pontificating and social commentary. Mathematicians are NOT to teach ßawed information to students. If they try then they will be stopped. > The test is how much society cares for its students, with the queston > of whether or not it will allow mathematicians to teach ßawed > information. > > Im curious about that question. > > I disagree with your conclusion. I have good, logical > mathematical grounds for that disagreement. If you want > to discuss the math and the logic I am happy to do so. > I am not interested in the moralizing. After all, if you > have not in fact proved what you claim mathematically, then > all your talk of mathematicians teaching ßawed information > is just hot air. You must first rigorously prove that we > are wrong before you can get on a soapbox and claim we > teach the wrong math. > You have no such rigorous proof. Well when people like you come along and manage to question basic things like my lemma, or that m=3 means that m=1(mod 2) then what can I do? The proof is there, but then someone like you comes along, says its not, I refute you point by point and you just lie or avoid. Then you come back later and act like I never refuted your objections! Ive seen that behavior from many posters over *years* on sci.math and repeatedly posters get away with it. Then they come back and *say* that theyre using math, when in fact theyre not, and simply are relying on social forces. > > Yes, perhaps it is. > > Nora B. > > It is a moral question which goes to the heart of academia. > > Not until you have a proof that you are right. > And at the moment, as at all previous moments, > you are empty-handed. > Nora B. Yet Ive broken you time and time again. When youre broken by the math you just act like it didnt happen, and then come back and make claims like you did here. I can break you now, point-by-point, with the math, and youll probably just come back later and claim otherwise. Usenet makes that possible. You can *say* whatever makes you feel happy. > > Mathematicians might believe they can sit back with my prime counting > function or my proof of Fermats Last Theorem, but ignoring an error > requires allowing it to be taught to students. > > My assessment is that mathematicians lack the morals to make the right > decision, but more interesting to me at this point is whether or not > society will allow mathematicians to proceed with teaching ßawed > information. > > That is, will humanity protect its children in this area? > > The answer is, yes. And the arena is outside of Usenet, so while some of you may feel rather good about your ability to be refuted by me, but still keep coming back on sci.math where readers dutifully act like youve been winning, if youre dragged into court or before a congressional committee, then that will be gone. And Nora Baron if the United States Government is convinced that youre a proper target for prosecution for committing fraud against the people of the United States, then the world will find out if that is a pseudonym or not. James Harris === Subject: Re: Teaching ßawed information > I downloaded Advanced Polynomial Factorization > from your website last night. You said, assuming > > > P(x) = (a1*x + uf)*(a2*x + uf)*(a3*x + uf), > > where a1, a2, and a3 are algebraic integers, then > letting g1 = a1*x + uf, g2 and g3 defined similarly, > > two of the gs should have a factor of f which > would force two of the as to have a factor > that is f. > > Assume without loss of generality that a1 > and a2 have a factor that is f. > > Yup, two of the gs *should* have a factor of f. > > > > Yes, that is the wording in APF. So what you > must now be saying is that they *should* have a > factor that is f, but they do not. Right? > Actually they *do* have a factor that is f in the object ring. Which means that b1 and b2 are objects which are not algebraic integers, evidently. They are quotients of algebraic integers by f, so they must be algebraic *numbers* which are not algebraic *integers*. > However, the ring of algebraic integers is ßawed, which cannot be > seen from that position in the paper. So it follows that they > *should* have a factor of f in that ring, from what is in the paper. But when you say *should* have a factor of f, does that mean that you actually have a *proof* that they have a factor of f, or that you just desperately *want* them to have a factor of f? > The Object Ring is a commutative ring that includes all numbers such > that -1 and 1 are the only members that are both a unit and an > integer, where no non-unit member is a factor of any two integers that > are coprime. > Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw > See also below. > > > To equate this to what I said above, let > > b1 = a1/f, and b2 = a2/f. > > Since you say essentially that a1 and a2 > have a factor that is f, this means that > b1 and b2 are algebraic integers. > > They should be algebraic integers. > > > Aha. Should again. > Actually they *do* have a factor that is f in the object ring. > However, the ring of algebraic integers is ßawed, which cannot be > seen from that position in the paper. So it follows that they > *should* have a factor of f in that ring, from what is in the paper. > The Object Ring is a commutative ring that includes all numbers such > that -1 and 1 are the only members that are both a unit and an > integer, where no non-unit member is a factor of any two integers that > are coprime. > Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw Whoa. Deja vu here. > The question here is: why are you denying the > results in your own paper? > > Nope. > > The problem is that the ring of algebraic integers doesnt include > them. > > Which goes to why I use the word should as I do. > > > And should yet again. > Actually they *do* have a factor that is f in the object ring. > However, the ring of algebraic integers is ßawed, which cannot be > seen from that position in the paper. So it follows that they > *should* have a factor of f in that ring, from what is in the paper. > The Object Ring is a commutative ring that includes all numbers such > that -1 and 1 are the only members that are both a unit and an > integer, where no non-unit member is a factor of any two integers that > are coprime. > Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw Gee, this sounds familiar. Proof by repetition? > > The problem is that the bs arent included in the ring of algebraic > integers, which Ive pointed out more than once. > > > See above. When you say two of the as have a > factor that is f, and you are working in the ring > of algebraic integers, that can mean only one thing: > b1 = a1/f and b2 = a2/f are algebraic integers. > > Or are you now saying that your results in APF > are incorrect ??? > > Nope. > > > Right. Not incorrect because you said should > rather than, e.g., must. > > I think what we have here is the following. You > are saying should because you now know that in > fact b1 and b2 cannot be algebraic integers. In > other words, you have somehow arrived at the same > point as the rest of us. > Theres no somehow about it as Ive explained repeatedly and in > detail. > As pointed out by several posters, an algebraic integer cannot be the > root of a non-monic primitive polynomial. You need irreducible in there also. For example, 2*x^3 - x^2 - 10*x + 5 is non-monic and primitive but has sqrt(5) as a root. But its also reducible. > It *is* true that for an infinity of values the bs are provably not > algebraic integers, though as Ive demonstrated with f=sqrt(2), m=1, > there is at least one value where they are. Doesnt it seem odd to you that the cases where you think you can show what you want are all irrelevant to your main goal? I mean, for your main proofs you need f to be a prime integer > 3, and m to be relatively prime to f. So the cases you keep citing are: 1. m = 0 2. f = 3 3. f = sqrt(2) In the latter case, the polynomial in question doesnt even have rational coefficients, so it is not irreducible over the rationals! > And in fact that one value can be used to refute any attempts at > challenging the result, as Ive pointed out before to you Nora > Baron, as I noted that you can just use m=1(mod 2). See just above. Its not relevant to what you want. > You then acted as if you didnt understand exactly what I meant by > m=1(mod 2). > I think you asked something like if m=3 would work, and I answered, > yes. Yes. But again, it doesnt matter. The polynomial when f = sqrt(2) does not have rational coefficients, let alone integer coefficients. Its an irrelevant case. You need f to be a prime integer > 3. > But we do not agree on what one must conclude > from this. I conclude that your original claim, > in the original version of APF, was incorrect and > that your proof was in error. You conclude now, > apparently, that you have arrived at a basic > contradiction: elementary algebraic number theory > shows that b1 and b2 cannot be algebraic integers, > but you think they *must* be, therefore algebraic > number theory is wrong somehow. You have phrased > this as, the ring of algebraic integers is > incomplete. But that is just dressing it up. > If you were right, you would have discovered a > fundamental inconsistency in mathematics. > Actually, I found a mistake with what mathematicians believe as before > they never were able to test the definition of algebraic integers as I > have. > I mean, consider that definition: > An algebraic integer is the root of a monic polynomial with integer > coefficients. > Theres nothing there to prove that it wouldnt have a problem like > this, and mathematicians worked for over a hundred years believing it > didnt. > Now Ive proven it does. > You may think you have something with my use of should but in the > object ring, two of the as *do* have a factor that is f. > Its just that algebraic integers are wacky. If you are right, you have shown both that a1/f is an algebraic integer and that it is not. That does not represent a problem with the algebraic integers. That represent an inconsistency in mathematics. If you were correct, the whole edifice of math collapses. You can prove anything (a false premise implies anything). There is no point in dealing with objects. The game is completely over. Mathematicians must re-examine the basics, the Peano postulates of arithmetic. I think you will object that this is not what you have concluded. But look at it closely. If you do NOT have a proof that b1 = a1/f is an algebraic integer, you are simply agreeing with the rest of us. Your proof in APF is simply an incorrect argument. If you DO have a proof that b1 = a1/f is an algebraic integer, then you run up against the argument I have given and others have given which says it cannot be. You have proven an inconsistency in mathematics. You either do NOT have a proof or you DO have a proof, right? If you do NOT, then your whole structure of claims, etc., collapses. If you DO, then either we have a proof that you have made a mistake, OR there is an inconsistency in mathematics, and there is no reason to pursue object math or anything else. You have wrecked the whole thing. You know my view on this. I have pointed out specific places where your proof has gaps. You have not shown how you go from m = 0 and f = 3 to more general m and f. You just claim it is true. You think you see patterns with m = 0 and f = 3 that, you then say, must be true in general. But really you have no proof of that must be true in general part. > The Object Ring is a commutative ring that includes all numbers such > that -1 and 1 are the only members that are both a unit and an > integer, where no non-unit member is a factor of any two integers that > are coprime. > Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw Others have recently pointed out that if the O.R. contains sqrt(2), then by your arguments regarding infinite series, it must contain 1/2; therefore, like Z[1/2], it must contain all real numbers, and you have infinitely many units that are not integers, a contradiction. And evidently you have agreed that it must contain sqrt(2). Conclusion: your claims about the O.R. also lead to an inconsistency. No such ring can exist. > Evidently you are not willing to bite that > particular bullet. > > It still gets back to the word should. In > order to justify should, you have to return to > your argument about m = 0 (the degenerate case) > Actually I use the constant term of the polynomial P(m), which, of > course, is given by P(0). Now your thinking that something basic is > proof Im wrong just tells a lot about how much you really understand > mathematics, or more likely, how willing you are to try and pervert > others understanding. > Besides, as Ive pointed out to you, I can use m=1 with f=sqrt(2), so > wheres your degenerate case then Nora Baron? f = sqrt(2) is irrational. The polynomial in that case cannot have rational, let alone integer, coefficients. Further, of course, there is no reason to bother with f = sqrt(2). The cases that are relevant to what you want to prove are f = prime > 3. > and f = 3 (the case where in general the associated > polynomials are not irreducible). We have > discussed ßaws in your reasoning for both of these > Is that the royal we? No. Others have discussed these ßaws also. > cases in detail and I do not agree with your > conclusions. *Even if I did*, I do not see how > showing a pattern for m = 0 and for f = 3 > proves anything about f = prime > 3 and m coprime > to f. For example, how would a statement for > m = 0 and f = 3 tell me anything about m = 1 > and f = 5? I think even you must admit that that part > of your argument is missing. All you have at > best is circumstantial evidence for two > peripheral cases. That evidence, even if one > believed it, would not constitute a proof. > What I can show with f=3 is that your objections, like claiming m=0 is > just a degenerate case, are cow dung. Oh. Thats a new argument at least. Your polynomial is ordinarily of degree 3. When m = 0, it is of degree 1. I think that qualifies as degenerate. Also two of your coefficients, say a1 and a2, are 0. They are divisible not simply by f, but by anything. That too seems to qualify as degenerate. But I see your point. What you are really talking about in that case is cow dung. > But when you see you cant make that claim when f=3, you bounce around > it as if people are supposed to believe that something checks to see > if f is coprime to 3. The old anthropomorphic argument. If I am right, the ws must be cognizant beings. That too is a new one in mathematics. > What you write in your posts does not make sense Nora Baron. > And using f=3, is just an attempt at helping people that I came up > after people like you managed to sow so much confusion about things > like my very basic and simple lemma: > ANY factor g of a polynomial is g=r+c, where c is a factor of the > constant term, and r varies with the polynomials main variable, and > is 0, when that variable is 0. Your Ôlemma is an utter triviality. In itself it gets you nowhere. > When I saw how posters like yourself could get people agreeing with > you when you were questioning something that basic I realized that > math people werent about math. > But this is your basis for saying should. > The ring of algebraic integers is wacky. The as *should* have f as a > factor in that ring, as the paper notes. If should means you have a proof, then math is inconsistent. Forget about Ôobjects and math. Take up air hockey or something. > > So how is such an error a moral test? > > > Here the poster Nora Baron has done something done before which is > to delete out context. > > > I left in the *math* context. I deleted out the > pontificating and social commentary. Actually I made a mistake here. I had not deleted out anything. The context was all there. > Mathematicians are NOT to teach ßawed information to students. > If they try then they will be stopped. Go for it. Swing the hammer. > The test is how much society cares for its students, with the queston > of whether or not it will allow mathematicians to teach ßawed > information. > > Im curious about that question. > > > I disagree with your conclusion. I have good, logical > mathematical grounds for that disagreement. If you want > to discuss the math and the logic I am happy to do so. > I am not interested in the moralizing. After all, if you > have not in fact proved what you claim mathematically, then > all your talk of mathematicians teaching ßawed information > is just hot air. You must first rigorously prove that we > are wrong before you can get on a soapbox and claim we > teach the wrong math. > You have no such rigorous proof. > Well when people like you come along and manage to question basic > things like my lemma, or that m=3 means that m=1(mod 2) then what can > I do? You were saying 3 = 1(mod sqrt(2)). I dont disagree with that. I dont disagree with your lemma either. Generalizing from m = 0 and f = 3 to the values of m and f that are actually of interest to you, that I disagree with. > The proof is there, but then someone like you comes along, says its > not, I refute you point by point and you just lie or avoid. You are going to have to get specific: Where did I lie? Where did I avoid ? > Then you come back later and act like I never refuted your objections! > Ive seen that behavior from many posters over *years* on sci.math and > repeatedly posters get away with it. You had many incorrect arguments over those years. Posters claimed you were in error. You said no at great length. Ultimately you had to say yes. Were those posters lying ? What did they get away with ? Proving you wrong? > Then they come back and *say* that theyre using math, when in fact > theyre not, and simply are relying on social forces. Again, in the innumerable previous cases where you were forced to admit you had made a mistake: were the people who had been claiming you were wrong relying on Ôsocial forces? Or were they mathematically correct ? Have you forgotten what happened in all those cases ? > > Yes, perhaps it is. > > Nora B. > > It is a moral question which goes to the heart of academia. > > > Not until you have a proof that you are right. > And at the moment, as at all previous moments, > you are empty-handed. > > Nora B. > Yet Ive broken you time and time again. Like when ? > When youre broken by the math you just act like it didnt happen, and > then come back and make claims like you did here. > I can break you now, point-by-point, with the math, and youll > probably just come back later and claim otherwise. Do it and see what happens. > Usenet makes that possible. You can *say* whatever makes you feel > happy. Not my practice. Yours? > > Mathematicians might believe they can sit back with my prime counting > function or my proof of Fermats Last Theorem, but ignoring an error > requires allowing it to be taught to students. > > My assessment is that mathematicians lack the morals to make the right > decision, but more interesting to me at this point is whether or not > society will allow mathematicians to proceed with teaching ßawed > information. > > That is, will humanity protect its children in this area? > > The answer is, yes. > And the arena is outside of Usenet, so while some of you may feel > rather good about your ability to be refuted by me, but still keep > coming back on sci.math where readers dutifully act like youve been > winning, if youre dragged into court or before a congressional > committee, then that will be gone. > And Nora Baron if the United States Government is convinced that > youre a proper target for prosecution for committing fraud against > the people of the United States, then the world will find out if that > is a pseudonym or not. Absolutely. Why are you holding back? > James Harris One last point here. If Advanced Polynomial Factorization now says that the as should have a factor that is f, and doesnt explain that further, it seems like a rather useless paper. There is no mention of the object ring in the current draft***. A reviewer would simply conclude that you have arrived at a contradiction which shows that your argument is wrong. Have you actually submitted it to a journal in that form ? If so, it is unlikely to go anywhere. If you submitted to a journal in its previous form, where you said two of the as actually DO have a factor that is f, it is simply wrong as you concede here, and should be withdrawn. So what is its status as submitted ? Wrong or useless? Nora B. *** That is, the current draft as of 2 days ago. I find that I am now excluded again from the Amateur Math group. An interesting tactic: keep anyone who disagrees with your arguments from seeing them. You can claim all sorts of things without worrying about criticism. Not the usual practice among real mathematicians, who are generally willing to put their ideas on the line. But whatever works, right? === Subject: need help about this series Does anyone have some valuable information on the series: Sum [n=2 to inf] (-1)^n * Zeta(n) * x^(n-1) ? === Subject: Re: need help about this series > Does anyone have some valuable information on the series: > Sum [n=2 to inf] (-1)^n * Zeta(n) * x^(n-1) ? Yes Psi(z+1)= -gamma -sum_{n=1}^oo zeta(n+1)*(-x)^n with gamma the Euler constant 0.5772156649.. and Psi(x)= (ln(Gamma(x)) (and Gamma(n+1)=n! ...) your series is Psi(z+1)+gamma Try a search on polygamma for more information or go here http://functions.wolfram.com/GammaBetaErf/PolyGamma/ Raymond === Subject: trying to find 5 equi-angular 4d vectors Five vectors are: A = [0,0,0,1] B = [a,e,0,-1/4] C = [b,f,i,-1/4] D = [c,g,j,-1/4] E = [d,h,k,-1/4] where (a..k) are to be determined. The dot product of any 2 vectors is to be -1/4. This is satisfied by A with any of the others. A+B+C+D+E = 0, the resultant of all 5 is 0. magnitude(A) = magnitude(B) = magnitude(C) = magnitude(D) = magnitude(E) = 1 or all 5 vectors are unit vectors. Basically the five vectors are equal angular, the same angle exists between any two. These 5 vectors are the axial lines from the origin {0,0,0,0] of a hyperregular tetrahedron? (not sure of name of this figure) to the vertices of this 4d figure. Can someone determine the values for a,b,c,d,e,f,g,h,i,k? Randall p.s. reply to sci.math group === Subject: Re: trying to find 5 equi-angular 4d vectors One answer: The figure is called a simplex and the coordinates of the vertices are given. k=sqrt(5)/4 A = [ 0, 0, 0, 1 ] B = [ -k, k, k,-1/4 ] C = [ k,-k, k,-1/4 ] D = [ k, k,-k,-1/4 ] E = [ -k,-k,-k,-1/4 ] I just have to rotate the matrix so that B has a 0 in the 3rd position. Randall p.s. Found answer on http://astronomy.swin.edu.au/~pbourke/polyhedra/platonic4d/ === Subject: abstract algebra questions Can you solve the following questions? they are from Math GRE subject test 1)first question Let R be the field of real numbers and R[x] the ring of polynomials in x with coefficients in R. Which of the following subsets of R[x] is a subring of R[x]? I. ALl polynomials whose coefficient of x is zero II. All polynomials whose degree is an even integer, together with the zero polynomial III. all polynomials whose coefficients are rational numbers (A) I only (B) II only (C) I and III only (D)II and III only (E)I,II, and III 2)second question A cyclic group of order 15 has an element x such that the set {x^3,x^5,x^9} has exactly two elements. The number of elements in the set {x^(13n): n is a positive integer} is (A) 3 (B) 5 (C) 8 (D) 15 (E) infinite === Subject: Re: abstract algebra questions >Can you solve the following questions? they are from Math GRE subject >test >1)first question >Let R be the field of real numbers and R[x] the ring of polynomials in >x with coefficients in R. Which of the following subsets of R[x] is a >subring of R[x]? >I. ALl polynomials whose coefficient of x is zero >II. All polynomials whose degree is an even integer, together with the >zero polynomial >III. all polynomials whose coefficients are rational numbers >(A) I only >(B) II only >(C) I and III only >(D)II and III only >(E)I,II, and III All of these contain both 0 and 1. Ask yourself which of these are closed under addition and multiplication. In I, for example, if you add the polynomials 3*x^3 + 7*x^2 - pi and sqrt(3)*x^4 - 6.456*x^2 + 1 (both of which has a zero coefficient of x^1), will the sum have a zero coefficient of x^1? What about the product? >2)second question >A cyclic group of order 15 has an element x such that the set >{x^3,x^5,x^9} has exactly two elements. The number of elements in the >set {x^(13n): n is a positive integer} is >(A) 3 >(B) 5 >(C) 8 >(D) 15 >(E) infinite Since the set has exactly two elements, either x^3 = x^5 or x^3 = x^9 or x^5 = x^9 holds, but not all three are equal. In each case figure out the order of x and the order of x^13. Or you can start by noting that the order of x must divide 15, hence be 1, 3, 5, or 15. Which of these orders is/are consistent with the hypothesis? Ask yourself how many different values {x^13, x^26, x^39, ...} will have. For these test questions, where there is no answer (F) Cannot be determined from the hypothesis, once you find one order for x consistent with the hypothesis you can count the corresponding {x^(13n)} and select your answer, not worrying about other values for the order of x. -- Wanted: Experts at choosing the best of 100+ applicants for a position. Register as a California voter by September 22, and vote on October 7. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Re: abstract algebra questions >1)first question >Let R be the field of real numbers and R[x] the ring of polynomials in >x with coefficients in R. Which of the following subsets of R[x] is a >subring of R[x]? >I. ALl polynomials whose coefficient of x is zero >II. All polynomials whose degree is an even integer, together with the >zero polynomial >III. all polynomials whose coefficients are rational numbers >(A) I only >(B) II only >(C) I and III only >(D)II and III only >(E)I,II, and III (E). Think about the definition of a ring. >2)second question >A cyclic group of order 15 has an element x such that the set >{x3,x5,x9} has exactly two elements. The number of elements in the >set {x^(13n): n is a positive integer} is >(A) 3 >(B) 5 >(C) 8 >(D) 15 >(E) infinite (A). Just work in Z_15. Unless I made a mistake, x = 5 there. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: abstract algebra questions Stephen J. Herschkorn escribi.97 en el >> 1)first question >> Let R be the field of real numbers and R[x] the ring of polynomials >> in x with coefficients in R. Which of the following subsets of R[x] >> is a subring of R[x]? >> I. ALl polynomials whose coefficient of x is zero >> II. All polynomials whose degree is an even integer, together with >> the zero polynomial >> III. all polynomials whose coefficients are rational numbers >> (A) I only >> (B) II only >> (C) I and III only >> (D)II and III only >> (E)I,II, and III > (E). Think about the definition of a ring. The difference of two polynomials of degree k evenwith the same doefficient of x^k an distinct coeeficients of x^(k-1) is a polynomial of odd degree, among other cases. The correct answer is (C). -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Questions for James Harris I asked: >>4. a. Is the Object Ring a subring of the reals? Is it a subring of >> the complex numbers? and you replied: > No, as its too big. > There is an intersection between the object ring and both of those > sets. Did you really want to say that? Im okay with the fact that the object ring contains some numbers that arent real--that was implied by your answer to one of my previous questions, but are there actually elements of the object ring that arent complex numbers? If thats what you intended to write, I must confess that Im puzzled. Is the object ring something like C[[x]], the ring of formal power series in the indeterminate x with complex coefficients? You could help me get a handle on this if youd give an example of an element of the object ring thats not a complex number. Rick === Subject: Re: Questions for James Harris >> [.snip.] > I was paying attention; it just wasnt enough attention evidently. I > Magidin for explaining what you mean. You could use more than a few > helpful tips on clarity. Presumably you want to be heard and > understood by people who can recognize the validity of your work. > Learning how to speak and present ideas clearly would help you advance > your ideas, if indeed they are valid and if you truly believe them to > be valid. Or if you dont believe in the validity of your work, you > can be a coward and duck behind your conspiracy theories. > > ---- David >>You werent paying attention. I gave the definition of the object >>ring, which follows. >>The Object Ring is a commutative ring that includes all numbers such >>that -1 and 1 are the only members that are both a unit and an >>integer, where no non-unit member is a factor of any two integers that >>are coprime. >> Point the zeroth: So everyone knows the quality of your work and >> definitions, be sure to include your definition of commutative ring >> from that same page: >> A commutative ring is a set of numbers where for any members a and >> b, a+b=b+a and ab=ba. >> (from http://www.msnusers.com/AmateurMath/objectmathematic.msnw) >Well that is true, though I guess you think I should have said more. >Unfortunately, your full intent is unknown as you werent specific. His full intent was made clear with the words So everyone knows the quality of your work and definitions. Lemme spell it out for you: Anyone who (i) knows what the definition of Ôcommuttative ring is and (ii) sees you give that definition will immediately realize that youre utterly clueless when it comes to mathematics. This is why you should include the definition - it saves time for the reader. (Hint: Assuming that by number you mean complex number, according to the definition you give _every_ set of numbers is a commuttative ring.) >[...] >I focus on how people actually think, where 1/2 means 1 of 2; >whereas, trying to make it into a *thing* without reference to a set >of 2 makes it harder, and isnt the way people think. >Now that gives me a far more powerful methodology, which is >demonstrably true as with it a short proof of Fermats Last Theorem is >available. >However, those of you trained in other, less effective methodologies, >may hate my work just because having learned many things that are >unnecessary, or that you didnt really understand, you feel despair. Tee-hee. >Like, how many of you knew that convergence never was about anything >other than decidability, that is that you can get a single number >answer? Just like we all despaired when we realized we never knew that integers were irrational. (You current blather really is _just_ like that incident. Honest.) >James Harris ************************ David C. Ullrich === Subject: Re: Fraud in Computer Science Publishing > 1. I use quantum computing, which is more powerful than Turing > Machines. > I dont believe you; > to this day, only a few (less than 10 I believe) quantum > computers were made, and the things you are talking about have never yet been > even come close to. And nobody had ever come up with a Program Synthesis system or a Quine Atom before, either (or a generator of Paradoxes and Incompleteness Theorems, going beyond Godel). But that never stops me! > AFAIK, the most advanced computations made on quantum > computers are factorizations in prime terms (15 = 3 * 5). FAC(x,I)^PRIME(x) > Please dont talk about things you have no clue about. You already showed part > of your cluelessness when you stated that a program always produces the same > result regardless of the compiler used. Sam, what am I gonna do with you? No offense (just between you and me), but it doesnt matter what anybody has ever defined to be a programming language (or how they define programming etc.) Once again, I am simply talking about something and using what I think is a correct term for it, and I am told that other things are included in that term. Well, Im not talking about those other things. I am talking about deterministic machines (as I said.) Any base of computing could do, although my examples use a normal programming language that a person might design for effective communication with computers. You can use any deterministic base of computing that you find defined in modern theory of computation texts (typically miniature, though of course Turing complete, programming languages.) Is there a better term than programming language that I should use instead? > 2. I use metamathematics with which one can decide questions that > proposition undecidable in the system PM is thus decided by > metamathematical arguments. > Should I assume that you were born before 1931? Godels results are timeless and as valid today as they were then. Besides, my system includes PA and ZFC etc. as a special case: Define N#P(x) [Q(a,b)] to mean P(a) iff Q(N,a), i.e., 2-place relation Q(a,b) contains one place relation P(a) at N, where substituting N for the 1st component of Q yields Q(N,a) which is equivalent to relation P(a). P(I) mean P(x) and ~P(x). Define: YES(a,b) : Turing Machine a halts yes on input tape b. PR(a,b) : Wff a with its free variable replace by b is provable. TW(a,b) : Wff a with its free variable replaced by b is true. SE(a,b) : Set a contains element b. CE(a,b) : Class a contains b. TS(a,b) : English sentence a with noun b substituted for its pronoun is true. DEF: P,Q mean wffs P and Q are defined to be equivalent. P, Q, . . . are relation variables. DEF: P(a),P(a)^TRUE(a) : TRUE(a) is the universal set. Then TRUE(x)[PR] is Peanos 5 Axioms. That is, Peanos 5 axioms are true iff the universal set is recursively enumerable. All Peano is doing is giving a little program [X=0 ; write X ; add 1 to X ; go to the 2nd step] to list the natural numbers. If that is a valid program, then the 0 in step 1 must exist (Peanoi axiom 1: There exists a zero.), the add 1 in step 3 must work (Peano axiom 2: Every number has a successor.) etc. And conversely, if the 5 Peano axioms are true, then this program is valid and the universal set is recursively enumerable. Thus TRUE(x)[PR] is equivalent to Peanos 5 axioms. But see how much simpler this is than formalizing Peanos axioms as 5 complex wffs in Logic (that is done in published papers)? And its a lot more powerful, too. For example, ZFC is easily formalized in APC (A Program Calculus): 1. Extensionality: Is the use of DEF in APC. ZF: allX ( X e A = X e B ) => A = B APC: DEF: P(a),Q(a) => we can substitute P for Q and vice versa 2. Pairing: Is a theorem in APC. ZF: existsX allY ( Y e X = (Y=A v Y=B ) ) APC: Theorem EQ(I,x)vEQ(J,x)[SE] DEF: P(a) , (eA)P(A)^EQ(A,a) [Definition of Equality] 3. Subsets (Separation, Comprehension): Is a rule in APC. ZF: existsX allY ( Y e X = Y e S ^ A(y) ) APC: Rule P(x)[SE] + Q(x)[TW] => P(x)^Q(x)[SE] na.95ve: P(x)[TW] => P(x)[SE] NBG: P(x)[CE] + Q(x)[SE] => P(x)^Q(x)[SE] 4. Sum Set (Union): Is rule QUIT in APC. ZF: existsX allY ( Y e X = existsZ e S (Y e Z) ) APC: Rule QUIT: P(x,y) => (eA)P(A,x) or P(x) => yes(P(x)) 5. Power Set: Is an axiom in APC. ZF: allX existsY allZ Z e Y = Z c X APC: Axiom (aA)~SE(x,A) v SE(I,A)[SE] 6. Infinity: Is an axiom in APC. ZF: existsX(0 e X ^ allY e X(Y e X)) APC: DEF:P(a),P(a)^TRUE(a) + TRUE(x)[SE] 7. Replacement: Is a Theorem used as a Rule in APC. ZF: existsX allY e S ( existsZ A(Y,Z) .88 existsZ e X A(Y,Z)) APC: P(x)[SE] + Q(I,x)[TW] => (eA)P(A)^Q(A,x)[SE] ( or P(I)^Q(I,x)I[TW] or Q(x,y)[TW] ) 8. Empty Set: Is an axiom in APC. ZF: existsX allY ~ (Y e X) APC: ~TRUE(x)[SE] Note very importantly: ZFC set notation is free-ßowing and ill-defined. APC syntax is shorter, simpler, completely defined, and easily formally manipulated (my Program Synthesis system does so.) It is much easier to work with sets using my formal wffs than using the bulky, cryptic syntax of ZFC Set Theory. APC is also working at a higher level of abstraction than Set Theory, Theory of Computation, Incompleteness in Logic, etc. APC theorems apply to all of these bases (where variables occur in the wffs.) For example, that is how I discovered the worlds first actual Quine Atom (search for QUINE ATOM in Google and click on the 1st find) and variations beyond Quine. I simply applied 3 theorems from Recursion Theory to Set Theory. Here are 4: 1. N # N There is a program that outputs itself. 2. f(I) => N#f(N) Substitution Theorem. 3. f(I,J) => N#f(N,I) Recursion Theorem. 4. f(I) => N ~ f(N) Fixed Point Theorem universal rules of inference, I formalize (automate): 1. Program Synthesis - Mathematical Programs (Number Theory - Primes) 2. Program Synthesis - DataBase Query Processing (overpaid employees) 3. Theory of Computation (Turing) 4. Recursion Theory (Kleene) 5. Set Theory (ZFC) 6. Incompleteness in Logic (Godel, Rosser, Smullyan) 7. Paradoxes (Liar, Russell, God) Note that there has been very little overlap published in these research areas. Some in 6=7 but none in 1=2 or 1=3 or 3=5 etc. Researchers still dont even realize that these are all the same problem (and can all be solved in one system - APC.) Some (e.g. Minsky and Rogers) have even explicitly stated that there is no relationship between 1 and 3, when in fact they are the same principles and results simply in different domains. > 3. Using AI self-reference techniques, the system is always expanding > beyond what you prove it cannot do - always one step ahead of you! > > You talk the talk, but do you walk the walk? That is, show us the > implementation of your system youve supposedly written. You are on my distribution list. I have implemented abour 6 of the rules of inference. I can post the current problems that I am facing if anyone is interested in helping. > Sam Charlie Volkstorf Cambridge, MA http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/ 20021008.1/1 http://www.arxiv.org/html/cs.lo/0003071 === Subject: Re: Fraud in Computer Science Publishing >> 1. I use quantum computing, which is more powerful than Turing >> Machines. >> I dont believe you; I have; but I still do not believe that are actually using your system on a quantum computer, since the computations necessary to implement such a sytem have not yet been achieved on a quantum computer. If you are saying that you intend on using the *theoretical* model of quantum computers, fine, but until you have one to use, yuor system will not be tested. >> to this day, only a few (less than 10 I believe) quantum >> computers were made, and the things you are talking about have never yet >> been even come close to. > And nobody had ever come up with a Program Synthesis system or a Quine > Atom before, either (or a generator of Paradoxes and Incompleteness > Theorems, going beyond Godel). But that never stops me! Im not talking about ideas, or code. Im talking about a physical system that allows computations based on quantum computing theory. AFAIK, the few implementations that exists are based either on NMR, or on ion traps, or on Cooper boxes, and researchers are not yet able to perform complex operations. >> AFAIK, the most advanced computations made on quantum >> computers are factorizations in prime terms (15 = 3 * 5). > FAC(x,I)^PRIME(x) Again, Im not talking about programming here, Im talking of just being able to have a physical machine that does the computations. > Sam, what am I gonna do with you? No offense (just between you and > me), but it doesnt matter what anybody has ever defined to be a > programming language (or how they define programming etc.) Indeed, the definition doesnt matter here; if you acknowledge that C is a programming language (and I doubt you could claim it is not), then your statement about results independant of the compiler are false. >> Should I assume that you were born before 1931? > Godels results are timeless and as valid today as they were then. You missed the humor (I should have put a smiley). The way your sentence is phrased implies that *your* system existed in 1931, thus that you were born before then. I dont know if it was done or purpose or not, but I really doubt Godel talked about *your* system. He may have talked about a system, and it happens that yours is similar, but he certainly did not talk about *yours*. >> You talk the talk, but do you walk the walk? That is, show us the >> implementation of your system youve supposedly written. > You are on my distribution list. I have implemented abour 6 of the > rules of inference. I can post the current problems that I am facing > if anyone is interested in helping. programming language (you stated you are writing your implmentation in it) However, if the implementation is free software (http://www.gnu.org for a definition), Ill be more than happy to try it and give feedback. Otherwise, Im afraid I wont, since I only use free software. Sam -- If sharing a thing in no way diminishes it, it is not rightly owned if it is not shared. - St Augustine === Subject: Re: Fraud in Computer Science Publishing > Indeed, the definition doesnt matter here; if you acknowledge that C is a > programming language (and I doubt you could claim it is not), then your > statement about results independant of the compiler are false. What do you (shall we) call something that allows you to represent and calculate every recursive function and which is deterministic? Programming Language? Base of Computing? Deterministic Machine? Turing Machine? > I really > doubt Godel talked about *your* system. He may have talked about a system, > and it happens that yours is similar, but he certainly did not talk about > *yours*. Let me put it this way: Do you remember when Gregory Chaitin called Kurt Godel on the phone and tried to show Godel his improvement over Godels Theorems by using a differnt paradox than the Liar, and Godel told him, No, it doesnt matter which paradox you use. and over time Chaitin discovered that using a different paradox made the theorems a lot better (even though Godel was not smart enough to realize that but Chaitin was)? Well, if you believe that, then believe me when I tell you that GC handed the phone to me and I said, Mr. Godel, if I may. and told him about APC and he said, Im really interested. I dont like Chaitins stuff because he only talks about my watered down version based on soundness and never even mentions Rossers improvement, so how can he even talk about improving something that was already improved? I wanted to ask him why anyone would try to convey a mathematical finding over the phone rather than with a letter, but KG was more interested in my findings. I also wonder when Chaitin learned German? (The only German that I know is BEW means provable, from which I get my English nick-name.) >> You talk the talk, but do you walk the walk? That is, show us the >> implementation of your system youve supposedly written. > > You are on my distribution list. I have implemented abour 6 of the > rules of inference. I can post the current problems that I am facing > if anyone is interested in helping. > > programming language (you stated you are writing your implmentation in it) > However, if the implementation is free software (http://www.gnu.org for a > definition), Ill be more than happy to try it and give feedback. Otherwise, > Im afraid I wont, since I only use free software. Sam, for you its free. I am talking about posting the technical issues. Right now I am struggling a little with the issue of when to apply variable substitutions: with each rule, as a separate rule (as in my papers), hidden by normalizing everything, etc. The problem is getting each solution to come out exactly once. I will post details if anyone might wish to help tackle the problems. Charlie Volkstorf Cambridge, MA > Sam === Subject: Re: Polynomials and 2^sqrt{3} that something has been left out or we failed to continue the work boils down, at least to me, to this: 1) We could solve equations of the form 2x + 1 = 3 with Z. 2) We could solve 2x = 3 with Q. 3) We could solve x^2 = 2 with R. 4) We could solve x^2 = -1 with C. 5) We took a step back and specified two classes of numbers, two subsets of C - the algebraic numbers, which are roots of polynomials with integral coefficients, and a subset, the algebraic integers, which are roots of polynomials with integral coefficients with leading coefficients of 1. 6) Is there where mathematicians supposedly failed? Are we supposed to relate a larger class of numbers to polynomials? Are you suggesting that we look at roots of infinite series (what complex number wouldnt that include? - it is important that it be not all of C if this new classification is to be new at all) or perhaps that we should look at still finite polynomials, but with different coefficients (again, which coefficients and, given that choice, what parts of C doesnt it include?)? What exactly is the next step again? Justin Davis === Subject: Re: Polynomials and 2^sqrt{3} >>Now to you Arturo Magidin it might make sense to try and continually >>dodge the point, and then toss out transcendental as if that changes >>anything, but the only rational conclusion is that mathematicians >>stopped, after algebraic integers, as otherwise 2^sqrt{3} would have >>some other label besides transcendental, that includes integer. > Bwahahah! > So, why this stupid insistence on full names James Harris? Because you > James Harris are once again picturing yourself in a courtroom > battering the poor witness? > LOL! Ill bet its because his mamma used his full name when she scolded him, and Ill also bet she used to scold him a LOT. (She probably still does; it wouldnt surprise me to find that he lives in her basement. He must be a horribly embarrassing disappointment to his family.) -- Wayne Brown | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Polynomials and 2^sqrt{3} > Mathematicians are being warned yet again NOT to teach ßawed > information to their students. James is being laughed at yet again as he attempts to give orders as if his words have any more authority than the waste emissions of slime mold. -- Wayne Brown | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Polynomials and 2^sqrt{3} > Did you bother to look up the link I provided for you? >Nah, and still havent. What do you think it shows? > You are FLOUNTING your ignorance? How refreshingly honest of you. > I even gave you a link to mathworld. > The link above explains Gelfonds Theorem, which shows that > 2^{sqrt(3)} is a transcendental number. By an easy link, you would > have also found out just what transcendental number means. And you > would have learned ->something<-. I know what transcendental number means. Now what Ive pointed out is the fact that mathematicians failed to follow up on Dedekind so that a number like 2^sqrt{3} would have been covered as numbers like 1+i and sqrt(2) were covered by Gauss and Dedekind respectively. You know, maybe something like transcendental integer for 2^sqrt{3}. Given the context, your link wasnt worth my time, and still isnt. > Which is obviously something you are terrified of, and avoid at all > costs. Youre dodging the real issue and trying to make it seem like its about me. The issue is that 2^sqrt{3} wasnt included, and instead mathematicians left it to older terms like transcendental numbers and the catch-all of reals. Mathematicians dropped the ball. I picked it up, over a hundred years after Dedekind made his step after Gauss, and people like you Arturo Magidin, refuse to be mathematicians, and instead act like upset kids. The number 2^sqrt{3} is an object. The Object Ring is a commutative ring that includes all numbers such that -1 and 1 are the only members that are both a unit and an integer, where no non-unit member is a factor of any two integers that are coprime. Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw >>Dedekind followed up by considering the roots of monic polynomials >>with integer coefficients, and those are called algebraic integers. >> >>His work brought in numbers like sqrt(2) as well as 1+i. >> >>What youre learning now is that mathematicians failed to follow-up on >>Dedekind, and thus dropped the ball over a hundred years ago, so that >>2^sqrt{3} got left out. >> >> Its not a question of being left out. You continually miss the >> point, and get things perfectly backwards. >> >> The definition of rational preceded the definition of field by close >> to two thousand years. The definition of algebraic and >> transcendental preceded the definition of algebraic integer. >Your point? > That you are being stupid in thinking it went the other way > around. That all the conclusions you are deriving about the > mathematical community are based on (a) your ignorance of facts: > and (b) your ignorance of history. And therefore, they are, at the > very least, suspect. Yup, try and act like its all about me. Well hey, Im going to go out on a limb here, and suggest that *maybe* its about the MATH. And maybe on a math newsgroup like sci.math there might be people who would be interested in the math, and not in your need to say Im being stupid. Its like that freaking show. Damn, whats it called? You know, the one where people argue and holler at each other? Oh yeah, The Jerry Springer Show. But you see Arturo Magidin, math is not only not a fashion show, its not the freaking Jerry Springer Show. >And thats why Im posting so that you can check with mathematicians >throughout the land to see how it fits into your system. > > In the standard system, where rationals and reals are not the same > thing, where polynomials are finite expressions, and where subrings of > C are not required to be closed under the norm topology, it is a > transcendental number. > > [.snip.] >>> >>Well sqrt(2) is a radical, but its also an algebraic integer. >> >> Non sequitur again. Nobody is distinguishing between radicals and >> algebraic integers, or even algebraic numbers. You are missing the >> point. >> >> A transcendental number is, by definition, a real number which is >> not algebraic; hence, not the solution of any polynomial with rational >> coefficients. >Well the point is that 2^sqrt{3} is part of a progression started by >Gauss with numbers like 1+i, followed up on by Dedekind who considered >algebraic integers, which should have been followed up by some other >discoverer over a hundred years ago. > And this shows that you do not know history. No, 2^sqrtr(3) is not > part of a progression started by Gauss with [the gaussian integers], > followed up on by Dedekind who considered algebraic integers. > 2^sqrt(3) is a number that was well known before then: at least as a > consequence of Napiers work on logarithms and exponential functions, > which allows the definition of any real exponent. So was sqrt(2). You know, sqrt(2) is 2^{1/2}, right? However, Dedekind had the wonderful idea to associate it with integers by calling it an algebraic integer. Similarly 2^sqrt{3} can be associated with integers. If someone else had picked up where Dedekind left off, it might be a transcendental integer, but it waited until now, so I get to name. I have named it and other numbers like it, objects. Objects are members of the object ring. The Object Ring is a commutative ring that includes all numbers such that -1 and 1 are the only members that are both a unit and an integer, where no non-unit member is a factor of any two integers that are coprime. Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw >Now you may fixate on a number being algebraic, but thats losing the >basic principle here which goes to that question of numbers than can >be associated with integers, like gaussian integers and algebraic >integers. > You are babbling. Worse, you are vomiting incorrect history, badly > digested. You are losing the debate and turning instead to tricks. The *point* is that 2^sqrt{3} should have been next in a progression that saw numbers like 1+i caught by gaussian integers, and numbers like sqrt(2) caught by algebraic integers. But over a hundred years passed before its true name was declared: object. >That was a poweful and surprising concept which allowed Gauss to first >associate numbers like i and 1+2i with the set of integers, then >Dedekind could associate numbers like sqrt(2) and sqrt(3) with >integers. > Again, you are vomiting incorrect and badly digested history. sqrt(2) > was known to the Pythagoreans, and they understood PERFECTLY WELL how > to associate to integers. In fact, Euclid knew perfectly well how to > associate the square root of any constructible number to the number > itself, and thus to integers in the end. Are you now claiming that Euclid should have the credit for algebraic integers. What about gaussian integers? Anyone else you want to give those to instead of Gauss, like youre trying to rip Dedekind? How about Archimedes? > Why dont you ->learn<- something before trying to lecture? Still trying to focus on me. Well it seemed to work for you for years, now didnt it Arturo Magidin. But you see, Im a *discoverer* so I was busy discovering. Now Im done discovering for a while so I can get to other business. And now Im a pissed off discoverer, perfectly capable of taking down the entire current math establishment worldwide, including having some big name mathematicians ruined, and in some cases, put in prison. >Here the point is that after Dedekind the ball was dropped, and Im >showing that by focusing on 2^sqrt{3}. > No, the point is that you have no idea what you are talking about. But > keep talking: it only makes it even more painfully clear. Then explain again where 2^sqrt{3} goes in your current system. >>And 1+i is a complex number, but its also a gaussian integer, as well >>as an algebraic integer. >> >> Non sequitur yet again. >Only if youre trying to skew the discussion from the real point. >Now to you Arturo Magidin it might make sense to try and continually >dodge the point, and then toss out transcendental as if that changes >anything, but the only rational conclusion is that mathematicians >stopped, after algebraic integers, as otherwise 2^sqrt{3} would have >some other label besides transcendental, that includes integer. > Bwahahah! Then explain again where 2^sqrt{3} goes in your current system. > So, why this stupid insistence on full names James Harris? Because you > James Harris are once again picturing yourself in a courtroom > battering the poor witness? > LOL! Then explain again where 2^sqrt{3} goes in your current system. > I dont toss out Ôtranscendental as if that changes anything. I > pointed out that the number IS transcendental, and that means > something to those who are not willfully ignorant, like you. In > particular, it means that there is no polynomial with rational > coefficients which has it as a root. Yet Gauss found gaussian integers, and Dedekind found algebraic integers, though those numbers *had* other labels before. Are you questioning their wisdom, after all, listening to you, why did they bother? After all, Dedekind already had radical, so why use algebraic integer for numbers like sqrt(2)? > Now, of course, since YOU claim that rationals are the same as > reals, I do have to wonder. Why have you NOT associated the > polynomial x-2^sqrt(3) to it? It doesnt have integer coefficients. >Maybe, transcendental integer? > What useful properties do transcendental integers have James Harris? > Why are they interesting James Harris, as opposed to the rest of the > real numbers? What can you do with them James Harris that requires the > label? What is a precise, complete, accurate, definition of > transcendental integer James Harris? The proper name is object. Objects are members of the object ring. The Object Ring is a commutative ring that includes all numbers such that -1 and 1 are the only members that are both a unit and an integer, where no non-unit member is a factor of any two integers that are coprime. Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw The object ring is an extension from Dedekind which includes Dedekinds algebraic integers just as Dedekind included gaussian integers. > (See how stupid it sounds when you insist on using full names where > they do not belong, James Harris?) > Unless you can answer all those questions, it becomes a moot point. Why put in so much emotion Arturo Magidin? The mathematics will sit there. Its not a moving target, so you dont have to get excited as though it might vanish like a fog under a hot sun. The mathematics is that 2^sqrt{3} got left out when the revolution that Gauss started, which Dedekind extended, stopped for over a hundred years, until now. >>Tossing out labels doesnt change the situation Arturo Magidin. >> >> Im not tossing out labels. I am describing, accurately and >> precisely, exactly what sort of number 2^{sqrt(3)} is: it is a >> transcendental number: a real number which is not the root of any >> polynomial with rational coefficients. >> >> Surely even you can figure out by a simple counting argument that >> there must be quite a few of these? There are only countably many >> roots of polynomials with rational coefficients, after all, but >> uncountably many real numbers. >And if Gauss hadnt been followed by Dedekind, and I were now talking >about sqrt(2) as being integer-like, you could be calling it a >radical, and pointing out that it couldnt result from a finite sum of >integers. > The algebraic integers appeared in the work of Eisenstein, who proved > that the sum and product of any two algebraic integers is again an > algebraic integer, that the integers are the only rationals which are > algebraic integers. Others had also stated these basic properties. It > was Dedekind who realized that the concept was important and WHY it > was important. What is it you have against Dedekind? First you were tossing his work off to Euclid (see above readers) and now its about Eisenstein. Whats with you Arturo Magidin? Whats going on inside your head? >Youre like a mathematical version of a Luddite Arturo Magidin, > You are calling ->me<- a luddite? Thats rich, coming from someone who > is so scared of knowledge that he refuses to look up a link. I value my time. Right now Im having to spend it arguing with you because I have to break a math society that has gone rogue. I could be doing other things, like hanging out on the beach. Theres no point in checking your link when my central point about 2^sqrt{3} remains. >fighting against progress using the terms you know, rather than >working to understand the mathematical principles involved. > Whatever. What is going on in your head Arturo Magidin??!!! >Someone like you might have been at Dedekind or even Gausss heels, >constantly attacking their innovations, but today, you wouldnt dare >as theyre part of the history that you fight to hold intact, while >still assaulting the spirit of the discipline. > Whatever. And you know what Gauss would have done to you? Hed have ripped you to shreds. You know that Gauss wasnt a nice guy, dont you? He didnt have to be nice. >>You need to learn more mathematics in-depth. Understanding is not >>just being able to toss out an answer or a label. You need to look at >>deeper principles, as did Gauss and Dedekind, if youre able. >> >> LOL! >Here the mathematics, as usual, proves you wrong as I need only give >2^sqrt{3} and remind of the progression which Gauss and Dedekind >followed, which should have lead to *someone* including that number. > You can give it and you can remind me of all you want. It only > proves you have no idea of what you are talking about, and that you do > not know any history. > It confirms you as a blowhard. > [.rest of willful ignorance, nonsense, > arrogance removed.] Then explain again where 2^sqrt{3} goes in your current system. >Mathematicians are being warned yet again NOT to teach ßawed >information to their students. > Arrogance and stupidity, all in the same package. How efficient of > you. And *again* I warn mathematicians NOT to teach ßawed information to their students. I give warnings partly so that if you dont heed them you can be punished that much more severely, since society will know how clearly you were warned. James Harris === Subject: Re: Polynomials and 2^sqrt{3} [cut] > The number 2^sqrt{3} is an object. > The Object Ring is a commutative ring that includes all numbers such > that -1 and 1 are the only members that are both a unit and an > integer, where no non-unit member is a factor of any two integers that > are coprime. > Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw If 2^sqrt{3} is an object, then 1/2 + 2^sqrt{3} is not an object. But, why couldnt 1/2 + 2^sqrt{3} be an object while 2^sqrt{3) is not. -- Bill Hale === Subject: Re: Polynomials and 2^sqrt{3} > [cut] > The number 2^sqrt{3} is an object. Clarification: Taking the positive result of the square root. With the negative result its not an object. > > The Object Ring is a commutative ring that includes all numbers such > that -1 and 1 are the only members that are both a unit and an > integer, where no non-unit member is a factor of any two integers that > are coprime. > > Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw > If 2^sqrt{3} is an object, then 1/2 + 2^sqrt{3} is not an object. > But, why couldnt 1/2 + 2^sqrt{3} be an object while 2^sqrt{3) is not. > -- Bill Hale Because then youd have another integer besides -1 or 1 that would have to be a unit, or some non-unit member would be a factor of two integers that are coprime. James Harris === Subject: Re: Polynomials and 2^sqrt{3} > [cut] > The number 2^sqrt{3} is an object. [cut] > > The Object Ring is a commutative ring that includes all numbers such > that -1 and 1 are the only members that are both a unit and an > integer, where no non-unit member is a factor of any two integers that > are coprime. > > Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw > > If 2^sqrt{3} is an object, then 1/2 + 2^sqrt{3} is not an object. > > But, why couldnt 1/2 + 2^sqrt{3} be an object while 2^sqrt{3) is not. > > -- Bill Hale > Because then youd have another integer besides -1 or 1 that would > have to be a unit, or some non-unit member would be a factor of two > integers that are coprime. I dont think you are understanding my question. I am asking how do you prove that 2^sqrt{3} is an object? It seems like the same proof would prove that 1/2 + 2^sqrt{3} is an object. I understand that both numbers cannot simultaneously be objects. Can you give the proof that 2^sqrt{3} is an object? -- Bill Hale === Subject: Re: Polynomials and 2^sqrt{3} Visiting Assistant Professor at the University of Montana. [.snip.] >> And this shows that you do not know history. No, 2^sqrtr(3) is not >> part of a progression started by Gauss with [the gaussian integers], >> followed up on by Dedekind who considered algebraic integers. >> 2^sqrt(3) is a number that was well known before then: at least as a >> consequence of Napiers work on logarithms and exponential functions, >> which allows the definition of any real exponent. [.snip.] >>That was a poweful and surprising concept which allowed Gauss to first >>associate numbers like i and 1+2i with the set of integers, then >>Dedekind could associate numbers like sqrt(2) and sqrt(3) with >>integers. >> Again, you are vomiting incorrect and badly digested history. sqrt(2) >> was known to the Pythagoreans, and they understood PERFECTLY WELL how >> to associate to integers. In fact, Euclid knew perfectly well how to >> associate the square root of any constructible number to the number >> itself, and thus to integers in the end. [.snip.] > Surely even you can figure out by a simple counting argument that > there must be quite a few of these? There are only countably many > roots of polynomials with rational coefficients, after all, but > uncountably many real numbers. >>And if Gauss hadnt been followed by Dedekind, and I were now talking >>about sqrt(2) as being integer-like, you could be calling it a >>radical, and pointing out that it couldnt result from a finite sum of >>integers. >> The algebraic integers appeared in the work of Eisenstein, who proved >> that the sum and product of any two algebraic integers is again an >> algebraic integer, that the integers are the only rationals which are >> algebraic integers. Others had also stated these basic properties. It >> was Dedekind who realized that the concept was important and WHY it >> was important. >What is it you have against Dedekind? Have you stopped beating your wife? What do you have against truth? >First you were tossing his work off to Euclid (see above readers) Yes, please do, and tell me how I was tossing his work off to Euclid. > and now its about Eisenstein. YOU claimed we owe sqrt(2) to Dedekind: Thats false. You insinuate that nobody had even thought about algebraic integers before Dedekind; thats also patently false. Hell, Dedekind HIMSELF makes a reference to the work of Eisenstein. If you are going to lecture about history, then it is incumbent upon you to be correct on that history. You are not. As to what do I have against Dedekind? The answer is nothing. You, on the other hand, seem to think that his work is so beneath contempt that you have to bolster it with things he did not do. I happen to think his accomplishments are more than enough without having to ignore the contributions of others. And, I also have an obligation to correct your false statements, in case someone who does not know better gets the wrong impression. Now, are you trying to claim that the concept of algebraic integer DID NOT appear in the work of Eisenstein? Are you claiming that Eisenstein did NOT prove that the sum of two algebraic integers and the product of two algebraic integers was an algebraic integer? Are you claiming that Dedekind was wrong in saying Eisenstein had proven these things? If not, then why do you complain that I have made a correct statement? Because it contradicts the image you have created in your head about how the history happened? Do you even know who Eisenstein was? Did you know that when asked who the three greatest mathematicians in history had been, Gauss replied Archimedes, Newton, and Eisenstein? [.snip.] >>Youre like a mathematical version of a Luddite Arturo Magidin, >> You are calling ->me<- a luddite? Thats rich, coming from someone who >> is so scared of knowledge that he refuses to look up a link. >I value my time. Right now Im having to spend it arguing with you >because I have to break a math society that has gone rogue. The point is: you accuse me of being a luddite. Your basis is that I You, on the other hand, consistently refuse to review anybody elses work. You refuse to look up basic texts on algebra and Galois theory, while pontificating about both; you refuse to look up a link that explains the situation you are asking, lest you learn something new. You claim that your new discoveries will destroy the mathematical establishment, that the old lies will be swept away in the light of your truth. Luddites do not reject after examining. Luddites are scared to examine and prefer to destroy sight unseen. Just like you. [.snip.] >>fighting against progress using the terms you know, rather than >>working to understand the mathematical principles involved. >> Whatever. >What is going on in your head Arturo Magidin??!!! I am wondering how you manage to hang on to your illusions in the face of reality. I find it perplexing. [.snip.] >>Someone like you might have been at Dedekind or even Gausss heels, >>constantly attacking their innovations, but today, you wouldnt dare >>as theyre part of the history that you fight to hold intact, while >>still assaulting the spirit of the discipline. >> Whatever. >And you know what Gauss would have done to you? >Hed have ripped you to shreds. Why so much anger James Harris? Why the focus on me James Harris? What are you scared of James Harris that you have to threaten me with physical violence a the hands of historical figures whose accomplishments you do not even know or understand? [.snip.] >>Mathematicians are being warned yet again NOT to teach ßawed >>information to their students. >> Arrogance and stupidity, all in the same package. How efficient of >> you. >And *again* I warn mathematicians NOT to teach ßawed information to >their students. >I give warnings partly so that if you dont heed them you can be >punished that much more severely, since society will know how clearly >you were warned. Such arrogance and such stupidity, all in the same package. Truly, a marvel of efficiency. [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee fixe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn awy from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a ßash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do not find it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Polynomials and 2^sqrt{3} Visiting Assistant Professor at the University of Montana. >> Did you bother to look up the link I provided for you? >>Nah, and still havent. What do you think it shows? >> You are FLOUNTING your ignorance? How refreshingly honest of you. >> I even gave you a link to mathworld. >> The link above explains Gelfonds Theorem, which shows that >> 2^{sqrt(3)} is a transcendental number. By an easy link, you would >> have also found out just what transcendental number means. And you >> would have learned ->something<-. >I know what transcendental number means. But you did not, apparently, realize that 2^sqrt(3) is one. Hence the link to Gelfonds Theorem. >Now what Ive pointed out is the fact that mathematicians failed to >follow up on Dedekind so that a number like 2^sqrt{3} would have been >covered as numbers like 1+i and sqrt(2) were covered by Gauss and >Dedekind respectively. Tell me: how do you know what mathematicians may or may not have done? mathematics? Or just from what youve heard here in sci.math? The number 2^sqrt(3) is perfectly well covered. Just because it doesnt have an impressive sounding name that YOU like doesnt mean its been ignored. >You know, maybe something like transcendental integer for 2^sqrt{3}. I asked you very specific questions about this. And you ignored them. Just tossing out labels right and left is stupid. And it is ->exactly<- what you are doing. You are trying to invent a name for something like 2^sqrt(3) for no good reason, except perhaps that you imagine it is important to come up with a new name. What is it about 2^sqrt(3) that makes it special? Why is the concept important? What use does it have, other than for you to say heres something no one ever thought to give a name before? >Given the context, your link wasnt worth my time, and still isnt. In short: you are scared of learning something. You always have been. You accuse others of being luddites, but everyone here has bothered to look at your arguments. The only one who willfully closes his eyes is you. The only one scared of new knowledge is you. Not worth your time. Thats just what you tell yourself to pretend you are not scared. >> Which is obviously something you are terrified of, and avoid at all >> costs. >Youre dodging the real issue and trying to make it seem like its >about me. >The issue is that 2^sqrt{3} wasnt included, and instead >mathematicians left it to older terms like transcendental numbers and >the catch-all of reals. Wasnt included in ->what<-? Its not in the rationals because it is not a rational. It is not in the algebraic integers because it is not an algebraic integer. It is not in the algebraic numbers because it is not an algebraic number. What special property does 2^sqrt(3) has that demands that it be singled out? Gauss did not single out the gaussian integers simply because they had been left out. He singled them out because he had a SPECIFIC and SPECIAL use for them. Dedekind did not single out the algebraic integers because they had not been singled out before (they had). He singled them out because he had a specific use for them. Thats why you need names for things: when you need to use them. Not just because it is nice to have names. [.personal insults removed.] [.mindless repetition of the ßawed attempt at defining object ring removed. See the multiple posts in which its problems have been pointed out in extenso.] >>And thats why Im posting so that you can check with mathematicians >>throughout the land to see how it fits into your system. >> >> In the standard system, where rationals and reals are not the same >> thing, where polynomials are finite expressions, and where subrings of >> C are not required to be closed under the norm topology, it is a >> transcendental number. >> >> [.snip.] >Well sqrt(2) is a radical, but its also an algebraic integer. > > Non sequitur again. Nobody is distinguishing between radicals and > algebraic integers, or even algebraic numbers. You are missing the > point. > > A transcendental number is, by definition, a real number which is > not algebraic; hence, not the solution of any polynomial with rational > coefficients. >>Well the point is that 2^sqrt{3} is part of a progression started by >>Gauss with numbers like 1+i, followed up on by Dedekind who considered >>algebraic integers, which should have been followed up by some other >>discoverer over a hundred years ago. >> And this shows that you do not know history. No, 2^sqrtr(3) is not >> part of a progression started by Gauss with [the gaussian integers], >> followed up on by Dedekind who considered algebraic integers. >> 2^sqrt(3) is a number that was well known before then: at least as a >> consequence of Napiers work on logarithms and exponential functions, >> which allows the definition of any real exponent. >So was sqrt(2). Sqrt(2) was known to the PYTHAGOREANS. Rational exponents were well understood by the greeks. You do realize that a nonrational exponent is a different thing, though, right? >You know, sqrt(2) is 2^{1/2}, right? Non sequitur. Do try to stay on topic, James Harris. >However, Dedekind had the wonderful idea to associate it with integers >by calling it an algebraic integer. Please learn some history before spouting nonsense. >Similarly 2^sqrt{3} can be associated with integers. How? Why? You have failed to answer both questions. You just go on and on about it being overlooked. [.mindless repetition of ßawed definition removed.] >>Now you may fixate on a number being algebraic, but thats losing the >>basic principle here which goes to that question of numbers than can >>be associated with integers, like gaussian integers and algebraic >>integers. >> You are babbling. Worse, you are vomiting incorrect history, badly >> digested. >You are losing the debate and turning instead to tricks. Whatever. I seem to always be losing the debate, and yet you keep trying to answer. Given so much energy expended by you in showing I am some kind of madman, one has to wonder how I could possibly be losing the debate so much. >The *point* is that 2^sqrt{3} should have been next in a progression >that saw numbers like 1+i caught by gaussian integers, and numbers >like sqrt(2) caught by algebraic integers. Why? Because you say so? What is so important, or useful, about 2^sqrt(3) that requires that it be part of a progression? There are at least uncountably many rings sitting between the rationals and the complex numbers. It is logically impossible to give names to all. Only those which are somehow USEFUL are singled out. What, exactly and explicitly, is useful about 2^sqrt(3)? >But over a hundred years passed before its true name was declared: >object. You are declaring it an object by fiat; unfortunately for you, your definition is non-operative, so you cannot prove that it is one other than by saying it is and getting all angry and annoyed if anyone asks why it is one. All you do is repeat your ßawed definition over and over, mindlessly. >>That was a poweful and surprising concept which allowed Gauss to first >>associate numbers like i and 1+2i with the set of integers, then >>Dedekind could associate numbers like sqrt(2) and sqrt(3) with >>integers. >> Again, you are vomiting incorrect and badly digested history. sqrt(2) >> was known to the Pythagoreans, and they understood PERFECTLY WELL how >> to associate to integers. In fact, Euclid knew perfectly well how to >> associate the square root of any constructible number to the number >> itself, and thus to integers in the end. >Are you now claiming that Euclid should have the credit for algebraic >integers. Are you having trouble comprehending written English? How is square root of any constructible number the same as algebraic integers? cuberoot(2) is an algebraic integer, but it is not a constructible number. >What about gaussian integers? >Anyone else you want to give those to instead of Gauss, like youre >trying to rip Dedekind? I am pointing out the historical fact that the square roots of the integers were known to the Pythagoreans and that how to construct square roots of constructible numbers was known to Euclid. That when you try to insinuate that sqrt(2) is somehow to be laid at Dedekinds feet, you are demonstrating a complete ignorance of the subject you are pretending to lecture on. I am also pointing out the historical fact that the basic properties of algebraic integers (that they are closed under addition and multiplication) was known to Eisenstein. Dedekind ->himself<- makes that attribution. So when you imagine history as if Dedekind created them out of whole cloth, you are misrepresenting and misunderstanding history. >How about Archimedes? >> Why dont you ->learn<- something before trying to lecture? >Still trying to focus on me. You are trying to lecture on the history of mathematics. You are demonstrating an appalling ignorance of that subject. If that is not worth mentioning, then what is? [.snip inane and insults and libel.] >>Here the point is that after Dedekind the ball was dropped, and Im >>showing that by focusing on 2^sqrt{3}. >> No, the point is that you have no idea what you are talking about. But >> keep talking: it only makes it even more painfully clear. >Then explain again where 2^sqrt{3} goes in your current system. It is a transcendental number. It is a member of an uncountable number of rings lying between the integers and the complex numbers. Unless you can explain WHY it should be given a special name, then there is no more reason to give it one than there is to explain where pi^sqrt(e) fits. [.snip inane repetition.] >> I dont toss out Ôtranscendental as if that changes anything. I >> pointed out that the number IS transcendental, and that means >> something to those who are not willfully ignorant, like you. In >> particular, it means that there is no polynomial with rational >> coefficients which has it as a root. >Yet Gauss found gaussian integers, and Dedekind found algebraic >integers, though those numbers *had* other labels before. Non sequitur, yet again. >Are you questioning their wisdom, after all, listening to you, why did >they bother? Have you stopped beating your wife? The reason Gauss singled out the gaussian integers is because he had a specific use for them. The reason Dedekind singled out the algebraic integers is because he had a specific use for them. Not just because it is neat to give new names to new things. >After all, Dedekind already had radical, so why use algebraic >integer for numbers like sqrt(2)? Because radical is something else; because not every algebraic integer can be expressed through radicals; because he wanted an evocative name to refer to a specific well-defined, well-understood subset of the algebraic numbers, because he was going to USE them for something. So he did not want to say root of a monic polynomial with integer coefficients every time he wanted to refer to one of these algebraic numbers. That would be too long-winded, and it would obfuscate the presentation. In the hopes of ->clarity<-, he gave them a special, short name. (And, for the record, Gauss did not call them gaussian integers or integers. He refered to them as numbers of the form a+bi with a and b integers. How does that fit in your theory?) >> Now, of course, since YOU claim that rationals are the same as >> reals, I do have to wonder. Why have you NOT associated the >> polynomial x-2^sqrt(3) to it? >It doesnt have integer coefficients. Ah, well. But is 2^qrt(3) is a rational, as you claim, then surely you can just clear denominators! >>Maybe, transcendental integer? >> What useful properties do transcendental integers have James Harris? >> Why are they interesting James Harris, as opposed to the rest of the >> real numbers? What can you do with them James Harris that requires the >> label? What is a precise, complete, accurate, definition of >> transcendental integer James Harris? >The proper name is object. >Objects are members of the object ring. >The Object Ring is a commutative ring that includes all numbers such >that -1 and 1 are the only members that are both a unit and an >integer, where no non-unit member is a factor of any two integers that >are coprime. >Source: http://groups.msn.com/AmateurMath/objectmathematic.msnw >The object ring is an extension from Dedekind which includes >Dedekinds algebraic integers just as Dedekind included gaussian >integers. Your definition has no referent. There is no set that satisfies the conditions you require of your ring of objects. You cannot define something into existence. You are talking literally, about nothing. >> (See how stupid it sounds when you insist on using full names where >> they do not belong, James Harris?) >> Unless you can answer all those questions, it becomes a moot point. >Why put in so much emotion Arturo Magidin? Im not, you are. You are the one who imagines himself in a Court of Law trying the evil mathematicians. You are the one who calls the FBI and the CIA to bring them on top of me. You are the one who threatens me with legal action, with physical harm at the hand of your pals the generals of the army, and with jail. Im the one pointing out you sound ridiculous. Who is putting so much emotion James Harris? >The mathematics will sit there. Its not a moving target, so you >dont have to get excited as though it might vanish like a fog under a >hot sun. The mathematics will sti there. But your lack of ability is no measure for the amount of harm an ignoramus can do if he is taken seriously. It is the professional obligation of those who know better to correct what is wrong when they can. Thats why I correct you: because I have a much higher regard for truth than you will ever do. [.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 mans 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 critize. A great many people are staggered to this extend, 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: Polynomials and 2^sqrt{3} [cut] > Gauss did not single out the gaussian integers simply because they had > been left out. He singled them out because he had a SPECIFIC and > SPECIAL use for them. David A. Cox claims ... the whole reason Gauss introduced the Gaussian integers was so that he could state biquadratic reciprocity. This claim of his appears in his excellent book Primes of the Form x^2 + ny^2. Biquadratic reciprocity is a generalization of quadratic reciprocity. For cubic reciprocity, the ring Z[w] is used, where w is a non-trivial cube root of 1 (say, w = (-1 + sqrt(-3))/2). Of course, just because Cox claims this does not mean that he is right. But, it sounds reasonable. -- Bill Hale === Subject: Re: Polynomials and 2^sqrt{3} Visiting Assistant Professor at the University of Montana. >[cut] >> Gauss did not single out the gaussian integers simply because they had >> been left out. He singled them out because he had a SPECIFIC and >> SPECIAL use for them. >David A. Cox claims ... the whole reason Gauss introduced the Gaussian >integers was so that he could state biquadratic reciprocity. >This claim of his appears in his excellent book Primes of the >Form x^2 + ny^2. >Biquadratic reciprocity is a generalization of quadratic reciprocity. >For cubic reciprocity, the ring Z[w] is used, where w is a non-trivial >cube root of 1 (say, w = (-1 + sqrt(-3))/2). >Of course, just because Cox claims this does not mean that he is right. >But, it sounds reasonable. I believe this is indeed the case, that he was interested in biquadratic reciprocity. The biquadratic reciprocity symbol takes values 0, 1, -1, i, and -i (the biquadratic roots of unity), just like the quadratic reciprocity symbol takes values 0, 1, and -1 (the square roots of unity). Eisenstein developed the theory of Z[w] to state cubic reciprocity, and Kummer worked in the cyclotomic fields Z[zeta_p] to develop reciprocity of p-th powers (p a prime, zeta_p a primitive p-th root of unity). 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 mans 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 critize. A great many people are staggered to this extend, 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: Polynomials and 2^sqrt{3} > [cut] > Gauss did not single out the gaussian integers simply because they had > been left out. He singled them out because he had a SPECIFIC and > SPECIAL use for them. > David A. Cox claims ... the whole reason Gauss introduced the Gaussian > integers was so that he could state biquadratic reciprocity. > This claim of his appears in his excellent book Primes of the > Form x^2 + ny^2. > Biquadratic reciprocity is a generalization of quadratic reciprocity. > For cubic reciprocity, the ring Z[w] is used, where w is a non-trivial > cube root of 1 (say, w = (-1 + sqrt(-3))/2). > Of course, just because Cox claims this does not mean that he is right. > But, it sounds reasonable. > -- Bill Hale Its also irrelevant to the subject which is how 2^sqrt{3} got left out of the revolution started by Gauss. It might seem academic to some but math is about rules. The rules that apply when you associate numbers like 1+i and sqrt(2) with integers, also lead to the necessity of an association for numbers like 2^sqrt{3}. And while Arturo Magidin and others may dislike mathematical rules that does not take them away. Worse, the ring of algebraic integers is built by a provably ßawed rule, which is that they are the roots of a monic polynomial with integer coefficients. Now people may *wish* that the definition removed the possibility Ive proven mathematically, but wishes are not math rules. That definition offers no protection from the problem as it cant given that the problem exists. It IS mathematics after all. Im puzzled by people who supposedly are mathematicians who spend so much time fighting math rules as if they were arbitrary and changeable. Its just really damn odd. James Harris === Subject: Re: Polynomials and 2^sqrt{3} >> [cut] >> Gauss did not single out the gaussian integers simply because they had >> been left out. He singled them out because he had a SPECIFIC and >> SPECIAL use for them. >> David A. Cox claims ... the whole reason Gauss introduced the Gaussian >> integers was so that he could state biquadratic reciprocity. >> This claim of his appears in his excellent book Primes of the >> Form x^2 + ny^2. >> Biquadratic reciprocity is a generalization of quadratic reciprocity. >> For cubic reciprocity, the ring Z[w] is used, where w is a non-trivial >> cube root of 1 (say, w = (-1 + sqrt(-3))/2). >> Of course, just because Cox claims this does not mean that he is right. >> But, it sounds reasonable. >> -- Bill Hale >Its also irrelevant to the subject which is how 2^sqrt{3} got left >out of the revolution started by Gauss. >It might seem academic to some but math is about rules. >The rules that apply when you associate numbers like 1+i and sqrt(2) >with integers, also lead to the necessity of an association for >numbers like 2^sqrt{3}. Exactly _which_ rule is that? >And while Arturo Magidin and others may dislike mathematical rules >that does not take them away. And the fact that you feel something should be so does not mean it follows from the rules. >Worse, the ring of algebraic integers is built by a provably ßawed >rule, which is that they are the roots of a monic polynomial with >integer coefficients. >Now people may *wish* that the definition removed the possibility Ive >proven mathematically, but wishes are not math rules. >That definition offers no protection from the problem as it cant >given that the problem exists. >It IS mathematics after all. >Im puzzled by people who supposedly are mathematicians who spend so >much time fighting math rules as if they were arbitrary and >changeable. >Its just really damn odd. >James Harris ************************ David C. Ullrich === Subject: Re: Polynomials and 2^sqrt{3} > You are FLOUNTING your ignorance? How refreshingly honest of you. I think you meant FLAUNTING. (The rest of your post was spot on, though.) Gib === Subject: Re: Definition of Complete Measure Space > I have seen at least 2 definitions. Given measure space (O,F,v), it > is a complete measure space if and only if: > > 1.-Any subset of a null set in F is a null set (this is the one in > Chungs book) > > 2.-Any subset of a null set in F is also an element of F. > > I can see how 2.- would imply 1.-, but does 2.- follow from 1.- as > well? In practice, is either preferable to the other? > > Definition 2 is better. Definition 1 can be used (and is equivalent) > provided the definition of null set includes element of F. === Subject: Transfer functions- Control systems Can someone explain how to complete this question. Im totally stumped: The forward path of a unity feedback control system is represented by: G(s) = K s(4s +200) gain K is set at 50000, what is the natural frequency and the damping ratio? Do i need to multiply the bottom out? Help please, Ive got an exam tomorrow and am totally stuck at the moment. Ben J === Subject: Re: Transfer functions- Control systems > Can someone explain how to complete this question. Im totally stumped: > The forward path of a unity feedback control system is represented by: > G(s) = K > s(4s +200) > gain K is set at 50000, what is the natural frequency and the damping ratio? > Do i need to multiply the bottom out? > Help please, Ive got an exam tomorrow and am totally stuck at the moment. You want to look at 1/(1+G(s)). The denominator will be a degree two polynomial in s. The damping ratio is half the coefficient of s IIRC. The natural frequency comes from the roots of the polynomial. Hint: the feedback is stabilizing (G(s) is only conditionally stable). Chuck -- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net === Subject: Re: Twin Primes Formula > My previous post on Twin Primes had a mistake. Figuring out that > mistake (it was just a goof, but oh well) is crucial to getting a > piece of the glory of proving the Twin Primes Conjecture. > > Heres the Twin Primes Formula: > > Given X a prime, then iff > > ßoor(X/3) - ßoor(X/6) = ßoor((X+2)/3) - ßoor((X+2)/6) > > then X+2 is a prime as well. > Well thats not necessarily so, as I had one of those ideas that > ßashed through but didnt pass muster. > Oh well, have a lot of ideas; toss a lot of ideas. My two cents and a bit of hand waving... [~ means asymptotically approaches, ~= means approximately equal to] Given pi(x) ~= x/ln(x) for large x, Let F(x) = x/ln(x), then F(x) = f(x) = 1/ln(x) - 1/ln(x)^2 What does f(x) represent? In reality, x is either prime or it is not, but if one knew nothing about x beforehand, then one could consider f(x) to be the probability that x is prime. As an empirical example: For small k, f(8*10^9 + k) ~ 0.042, using -1000 <= k <= 0, we would expect about 42 of those 1000 to be prime, and sure enough pi(8*10^9) - pi(8*10^9 - 1000) = 41 (i.e. about 42). Consider the pair x and x+2, The probability that both x and x+2 are prime is the product f(x)f(x+2). For large x, f(x) ~= f(x+2), or f(x)f(x+2) ~= f(x)^2 Let g(x) = f(x)^2, then G(x) approximates the number of twin primes to x Continuing, G(x) = li(x)/6 - x/6ln(x) + 5x/6ln(x)^2 - x/3ln(x)^3 For large x, li(x)/6 ~ x/6ln(x), so G(x) ~ 5x/6ln(x)^2 - x/3ln(x)^3 The largest term 5x/6ln(x)^2 is positive, hence G(x) has no upper limit, and from that one can conclude there are infinitely many twin primes. Empirically, for x to 3*10^15, pi2(x) / G(x) ~ k, with k < 1.70086492. Russell === Subject: Why ÔEarly Transcendentals? Why do so many authors now offer early transcendentals versions of their texts? Is it just a ploy to sell more books? I really cant understand the benefit of introducing transcendental functions, specifically the exponential and natural log, before they can be properly understood in terms of the area under the hyperbola, and so on... John === Subject: Re: Why ÔEarly Transcendentals? > Why do so many authors now offer early transcendentals versions of > their texts? Is it just a ploy to sell more books? > I really cant understand the benefit of introducing transcendental > functions, specifically the exponential and natural log, before they > can be properly understood in terms of the area under the hyperbola, > and so on... > John Freshmen (freshpersons?) now take physics starting in their first semester. So math courses have to adapt and squeeze things in for their use. Back in the Good Old Days, physics was a sophomore course, so it could assume a full year of calculus, and calculus could be done in a logical order. Just my guess... === Subject: Re: Homeomorphy > This is a problem I have trouble solving: > Assume that M is compact,, non-empty, perfect, and homeomorphic to > its Cartesian square, M cong M times M. Must M be homeomorphic to > the Cantor set, the Hilbert cube, or some combination of them? -- > Charles Pugh You will also need to rule out things too small (a single point or the emptyset) and too large (not metrizable). -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: When does Newton-Raphson fail?? Is there an equation, f(x), such that the Newton-Raphson mathod fails no matter the value of x0?? (excluding equations with no real roots, e.g. x^2 + 1) If not, are there equations such that Newton-Raphson fails when a suitable value for x0 (i.e. very close to the root) is chosen (e.g. if the root is 1.55, a suitable value would be 2)?? Allan Lewis. Newbie-extraordinaire ;) === Subject: Re: When does Newton-Raphson fail?? >Is there an equation, f(x), such that the Newton-Raphson mathod fails no >matter the value of x0?? (excluding equations with no real roots, e.g. >x^2 + 1) >If not, are there equations such that Newton-Raphson fails when a >suitable value for x0 (i.e. very close to the root) is chosen (e.g. if >the root is 1.55, a suitable value would be 2)?? If x0 is the initial value, then the next approximation is x1 where x1 = x0 - f(x0)/f(x0) Try to set up a differential equation for f (with initial condition f(0) = 0) do that the sequence (x0, x1, x2, ...) will diverge when x0 <> 0. For example, perhaps try forcing x1 = x0 + 1, getting the differential equation x + 1 = x - f(x)/f(x), or f(x) = -f(x), whose solution is f(x) = c*exp(-x). Alas, the initial condition f(0) = 0 forces c = 0 and f(x) = 0 everywhere, which doesnt give us a solution to the original problem. Try other differential equations such as x1 = -x0. -- Wanted: Experts at choosing the best of 100+ applicants for a position. Register as a California voter by September 22, and vote on October 7. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Mongolian BBQ Challenge A friend and I recently visited a Mongolian BBQ that claims on their menu that you can create over 4.2 billion combinations from the ingredients available when filling your bowl. Can someone do the calculus to determine if this is right? Here are the simple variables: 17 main ingredients 9 kinds of sauces A single combination must consist of at least 1 main ingredient (e.g. chicken), up to 17 plus 0 to 9 sauces. A single sauce or combination of sauces doesnt consist of a meal (must include 1 main ingredient). Duplicate combinations in different orders do not count. For example, (chicken + noodes) is the same as (noodes + chicken). Given these rules, whats the total number of possible combinations? === Subject: Re: Mongolian BBQ Challenge > A friend and I recently visited a Mongolian BBQ that claims on their > menu that you can create over 4.2 billion combinations from the > ingredients available when filling your bowl. Can someone do the > calculus to determine if this is right? > Here are the simple variables: > 17 main ingredients > 9 kinds of sauces > A single combination must consist of at least 1 main ingredient (e.g. > chicken), up to 17 plus 0 to 9 sauces. A single sauce or combination > of sauces doesnt consist of a meal (must include 1 main ingredient). > Duplicate combinations in different orders do not count. For example, > (chicken + noodes) is the same as (noodes + chicken). > Given these rules, whats the total number of possible combinations? Youve been cheated. There are 2^17 ways of choosing the main ingredients, and 2^10 ways of choosing the sauces, for a total of 2^27 = 134,217,728 possible meals. Demand your money back. (Or your stomach.) --Ron Bruck === Subject: Re: Mongolian BBQ Challenge Jon escribi.97 en el > A friend and I recently visited a Mongolian BBQ that claims on their > menu that you can create over 4.2 billion combinations from the > ingredients available when filling your bowl. Can someone do the > calculus to determine if this is right? > Here are the simple variables: > 17 main ingredients > 9 kinds of sauces > A single combination must consist of at least 1 main ingredient (e.g. > chicken), up to 17 plus 0 to 9 sauces. A single sauce or combination > of sauces doesnt consist of a meal (must include 1 main ingredient). > Duplicate combinations in different orders do not count. For example, > (chicken + noodes) is the same as (noodes + chicken). > Given these rules, whats the total number of possible combinations? Combinations of 1 up 17 main ingretients times combinations of 0 to 9 sauces ... Can you give us the URL? -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com