mm-1031 === Subject: Re: Dark Matter Puzzle > arent you the shrinking earth guy > from Auz?... did you go to a Public School? > Gordon, poor lad. You think youve got a handle on how the universe > --les ducs dEnron! > http://members.tripod.com/~american_almanac No, you moron. Im not that guy. Ive never advocated a Ôshrinking Earth and Ive never been to Australia. And Ive been to both public and private colleges and universities. CC === Subject: differentiability question If I have a function, the derivative f of which is say, f(x) = x^2 Is f differentiable as x->infty? The reason I ask is because as I understand the definition of differentiability at a point, it is that the limit that defines the derivative must both exist and be finite at this point. Does f meet the criteria of the definition as x->infty. Bruce. === Subject: Re: differentiability question Bruce Percy > If I have a function, the derivative f of which is say, > f(x) = x^2 > Is f differentiable as x->infty? Do you mean f instead of f here? > The reason I ask is because as I understand the definition of > differentiability at a point, it is that the limit that defines the > derivative must both exist and be finite at this point. Does f meet the > criteria of the definition as x->infty. f doesnt have a limit at infinity, but even if it did, there is no meaning to the statement that f is differentiable at infinity or differentiable in the limit. If we _adjoin_ a point at infinity, and f behaves suitably in a neighbourhood of the new point (this f doesnt) then the notion of differentiable at infinity can be validated; but that expression is not in conventional use, as far as I know. LH === Subject: Re: differentiability question > Bruce Percy >> If I have a function, the derivative f of which is say, >> f(x) = x^2 >> Is f differentiable as x->infty? > Do you mean f instead of f here? I mean df/dx = x^2. Sorry about the confusion.. f in this case would be x^3/3. >> The reason I ask is because as I understand the definition of >> differentiability at a point, it is that the limit that defines the >> derivative must both exist and be finite at this point. Does f meet the >> criteria of the definition as x->infty. > f doesnt have a limit at infinity, but even if it did, there is no meaning > to the statement that f is differentiable at infinity or differentiable > in the limit. If we _adjoin_ a point at infinity, and f behaves suitably in > a neighbourhood of the new point (this f doesnt) then the notion of > differentiable at infinity can be validated; but that expression is not in > conventional use, as far as I know. > LH === Subject: Re: differentiability question I dont think I comprehend your question. Are you asking if the function x^2 is differentiable on the interval (-oo, oo)? Or, are you asking if x^2 is differentiable where x = {x : x If I have a function, the derivative f of which is say, > f(x) = x^2 > Is f differentiable as x->infty? > The reason I ask is because as I understand the definition of > differentiability at a point, it is that the limit that defines the > derivative must both exist and be finite at this point. Does f meet the > criteria of the definition as x->infty. > Bruce. === Subject: Re: differentiability question > I dont think I comprehend your question. Are you asking if the function > x^2 is differentiable on the interval (-oo, oo)? Or, are you asking if x^2 > is differentiable where x = {x : x it is differentiable because it is a constant. > Lurch >> If I have a function, the derivative f of which is say, >> f(x) = x^2 >> Is f differentiable as x->infty? >> The reason I ask is because as I understand the definition of >> differentiability at a point, it is that the limit that defines the >> derivative must both exist and be finite at this point. Does f meet the >> criteria of the definition as x->infty. >> Bruce. Maybe I should have used the notation /frac{df}{dx}=x^2.... My question is to do with the definition of finite (I think??). As approaches the vertical. Does the definition of the derivative hold for all x in R? Bruce. === Subject: Re: differentiability question ... stuff deleted ... > Maybe I should have used the notation /frac{df}{dx}=x^2.... > My question is to do with the definition of finite (I think??). As > x->pminfty then f(x)->infty, so the tangent line to the curve > approaches the vertical. Does the definition of the derivative hold for > all x in R? > Bruce. If the definition doesnt hold for all x in R, then it fails for some x in R. That is, there must be a finite value of x (since all reals are finite), for which the derivative isnt as advertised. However, if f(x) = x^2, youll have x = x^3/3 + constant, and the derivation of the formula for f holds with x as a variable. It then specializes to a correct formula for any particular value of x, right? Im sure I dont understand that query, then. Dale. === Subject: Re: differentiability question > ... stuff deleted ... >> Maybe I should have used the notation /frac{df}{dx}=x^2.... >> My question is to do with the definition of finite (I think??). As >> x->pminfty then f(x)->infty, so the tangent line to the curve >> approaches the vertical. Does the definition of the derivative hold for >> all x in R? >> Bruce. > If the definition doesnt hold for all x in R, then it fails for some > x in R. That is, there must be a finite value of x (since all reals > are finite), for which the derivative isnt as advertised. > However, if f(x) = x^2, youll have x = x^3/3 + constant, and the > derivation of the formula for f holds with x as a variable. It then > specializes to a correct formula for any particular value of x, right? > Im sure I dont understand that query, then. > Dale. Bruce. === Subject: Re: Machs Principle > >> The following is with reference to a couple of your recent mail, >> only some of what I can understand. > > Why do you pretend that people are asking you questions? > Its a busy life in la-la land. So many people to talk to. > I think it was Mike Varney who got to see Sarfatti at a conference. How > very, very embarassing for Sarfatti to be surrounded by real scientists > who thought hes a crank. Mike Varney is an evil mouthed dickhead himself as well as being a pseudoscientist so those two could really get along if it werent for the fact that they both have such enormous egos. Theres a Varney-Sarfatti exclusion principle that says that no two such dickheads can occupy the same room without either of them changing state to an even bigger dickhead. :-). CC. === Subject: Re: Machs Principle > The following is with reference to a couple of your recent mail, > only some of what I can understand. > Why do you pretend that people are asking you questions? Jack is now and always has been a loon. He managed to get a Ph.D. in the days when they were handing them out to anyone and everyone that supplied the faculty with some of their own stash. Jacks existence depends upon the existence of other people which is to say that he loves to be a name dropper thinking somehow that the fact that he met some famous person or other conveys a level of immortality onto himself. The poor son of a bitch is scared less of dying and so hes trying to intellectually masturbate his way into immortality by endorsing schemes like Applegates where hell be recognized as a great intellect when the aliens finally arrive. Hes in for a big disappointment when they dig a hole for him (or put the torch to his petered out frame and scatter him off of the Golden Gate Bridge) and theres no aliens but the demons that have been haunting the poor jackass all of his life. I think that it is likely that some deluded nitwits actually think Jack is a brilliant physicist. After all he endorses their desire to have UFOs and aliens to be real so they figure that if this guy has a PhD then he must know what hes talking about and it makes them feel all warm and fuzzy to think that hes supporting their ideas. In reality he couldnt care less about them except those who send him money. But it isnt that some people ask him questions but that his ego is so starved for attention that he has to publish his private coorespondences. CC. === Subject: PGL Graphics Questions Has anyone had any experience with the PGL graphics package for doing simple 2-D scientific visualization under Visual C++? If so, how did you like it, and which other products are its prime competitors in the $50 price range? === Subject: Re: abstract algebra questions >how to solve this question? >What is the greatest integer that divides p^4-1 for >every prime number p greater than 5? >(A) 12 (B) 30 (C) 48 (D) 120 (E) 240 Im a little rusty on this stuff - in fact, others have pointed out that my answer to the first question below is incorrect; the correct answer is C, since the difference of two polynomials of even degree may have odd degree. I would approach your question above this way, but there may be a faster way: p^4 - 1 = (p-1) (p^3 + p^2 + p +1). p-1 is even. By doing the arithmetic mod 4, one can determine that the second factor is divisible by 4. Hence, 8 always divides the number, ruling out answers (A) and (B). 7^4 -1 = 6 x 500, which is not divisible by 16, hence (E) is ruled out. If p = 1 mod 5, then 5 | (p-1); else, you may check that 5 divides the second factor. Hence, I get (D) to be the answer. Is this correct? I have posted this to sci.math, in case others have a faster way. >--- Stephen J. Herschkorn >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. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: abstract algebra questions >how to solve this question? >What is the greatest integer that divides p^4-1 for >every prime number p greater than 5? >(A) 12 (B) 30 (C) 48 (D) 120 (E) 240 > > Im a little rusty on this stuff - in fact, others have pointed out that > my answer to the first question below is incorrect; the correct answer > is C, since the difference of two polynomials of even degree may have > odd degree. > I would approach your question above this way, but there may be a faster > way: > p^4 - 1 = (p-1) (p^3 + p^2 + p +1). p-1 is even. By doing the > arithmetic mod 4, one can determine that the second factor is divisible > by 4. Hence, 8 always divides the number, ruling out answers (A) and (B). > 7^4 -1 = 6 x 500, which is not divisible by 16, hence (E) is ruled > out. If p = 1 mod 5, then 5 | (p-1); else, you may check that 5 > divides the second factor. Hence, I get (D) to be the answer. Is this > correct? I compute 7^4-1 to be 2400. Id factor p^4-1 as (p^2-1)(p^2+1) = (p-1)(p+1)(p^2+1) Then you can see that all 3 factors are divisible by 2, plus one of (p-1) or (p+1) must be divisible by 4, so p^4-1 must be divisible by 16. Also, one of (p-1) or (p+1) must be divisible by 3. By letting p = 1,2,3,4 (mod 5) you can see that one of the 3 factors must be divisible by 5. I believe the answer is 16 * 3 * 5 = 240 which is (E). David > I have posted this to sci.math, in case others have a faster way. >--- Stephen J. Herschkorn > > > >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. === Subject: Re: abstract algebra questions >kazu3273@yahoo.com (kazu) >> What is the greatest integer that divides p^4-1 for >> every prime number p greater than 5? >> (A) 12 (B) 30 (C) 48 (D) 120 (E) 240 > Id factor p^4-1 as (p^2-1)(p^2+1) = (p-1)(p+1)(p^2+1) > Then you can see that all 3 factors are divisible by 2, > plus one of (p-1) or (p+1) must be divisible by 4, so > p^4-1 must be divisible by 16. > Also, one of (p-1) or (p+1) must be divisible by 3. > By letting p = 1,2,3,4 (mod 5) you can see that one of > the 3 factors must be divisible by 5. > I believe the answer is 16 * 3 * 5 = 240 which is (E). More generally use the Carmichael lambda function (see below). y(240) = y(2^4 3 5) = lcm(2^2,2,4) = 4 => x^4 = 1 (mod 240) for x coprime to 2,3,5 ------------ : : In my number theory homework I showed that if n is an integer : not divisible by 2 or 3 then n^2 + 23 must be divisible by 24. : Is there a general theory that relates to this? You showed that a^2 = 1 (mod 24) if (a,24) = 1. More generally, k = y(m) => a^k = 1 (mod m) if (a, m) = 1 where y(m) is the Carmichael lambda function, defined as is the Euler phi function on prime powers, but combined via lcm vs. times: y(1) = y(2) = 1, y(4) = 2, y(2^e) = 2^(e-2) for e > 2 y(p^e) = p^(e-1) (p-1) for odd prime p y(p^i q^j ...) = lcm( y(p^i), y(q^j), ...) In your case y(24) = y(2^3*3) = lcm(2,2) = 2, as you showed, and its easy to see that 24 is the largest m with y(m) = 2. In group theoretic language one says the y(m) is the exponent of the unit group of Z/(m), i.e. the smallest k such that x^k = 1 for all elts x of the group. -Bill Dubuque === 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 the >>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 Pardon but, I assume that all 3x+3^q {q|q>0} share one result. 3^q. I am not so sure what attractor is here. Ernst === Subject: Re: 3x+1 - Not so interesting === >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 the >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 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 >Pardon but, I assume that all 3x+3^q {q|q>0} share one result. 3^q. > I am not so sure what attractor is here. How can you not know what attractor is? YOU came up with that word, meaning the smallest number in a loop. Dont you recall asking if anyone had data on attractor sets? I only used that word so you would know what Im talking about. >Ernst -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: 3x+1 - Not so interesting === >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 the >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 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 >Pardon but, I assume that all 3x+3^q {q|q>0} share one result. 3^q. > I am not so sure what attractor is here. > How can you not know what attractor is? YOU came up with that word, meaning > the smallest number in a loop. Dont you recall asking if anyone had data on > attractor sets? I only used that word so you would know what Im talking > about. >Ernst Yes I know what attractor is.. I was thinking one thing and typing I figured out what I didnt get as well. So you have connected 3x+1 to 3x+3 ? Wow I was thinking they are different. I saw some data that suggested to me 3x+1 and 3x+3^q {q|q>0} were different. Hardly a rigid study I add. Also I posted that data I wanted to share as a reply to your explanation of your latest work which I enjoyed reading. It gave me some ideas. I hope you find some ideas in what I have shared. 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 >>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 > 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 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. Okay! That is how I can understand things. I actually understand what you are saying. This moment seems to be a good time to share a point of view. This may contrast some of the things you are doing so it may be useful. Perhaps we are seeing not a complete path to any C ( using your variable name ) but each instance of 3x+C and x/2^q {q|q>0} is the complete functioning. Allow me to outline : Seperating 3x+C into its own, it is an arithmetic series ( I hope I am labeling that correctly ) Start-- Enter the value for Y of 3x+y:5 Formula is 3x+ 5 Enter a value:33552245 Arithmetic series report for 3(33552245) + 5 The q = 0 : The result > 3728025 The q = 1 : The result > 11184080 The q = 2 : The result > 33552245 < Calling value The q = 3 : The result > 100656740 The q = 4 : The result > 301970225 The q = 5 : The result > 905910680 The q = 6 : The result > 2717732045 The q = 7 : The result > 3858228844 The q = 8 : The result > 2984751945 The q = 9 : The result > 364321248 Primary Diff is 7456055 End --- This shows that 33552245 is a member of the progression that is a power of 3 times 7456055. Such that the difference between 3728025 and 11184080 is 3^0 times 7456055 and the difference between 33552245 and 11184080 is 3^1 times 7456055.and so on.. I think this holds for all 3x+Cs so it is a growth. Let me know if you want a copy of that short program. Then the Odd value we have called x before is not a result of 3x+C only the first even after it. So only the first even in in the 3x+C set. The rest is a member of the x/2 set including the odd value because it is a result of one or the other. If this is a bad way to look at it someone tell me. So I was thinking there are two sets of data values. Ones that are the result of 3x+C and values that are the result of x/2^q {q|q>0} To make this short x/2 apporaches a limit of zero if allowed to run forever and 3x+c goes the other way by 3x+C but has a common structure of those differences. So We cause the pattern. We stop one and start the other. With that in mind then each instance of 3x+C and its partner x/2^q must be expressing the same functioning. That removing powers of two and steping up values in a progression allows for a defined path to exist. That it runs until some value points back to the path to x/2^q * itself( Joke > exit vector? ). I havent come up with a term to discribe a value for 3x+C that creates this effect . Attractor may not be correct. I havent seen anyone saying this before but they could have: please someone point out the link if I have repeated some work(s), It will save me time. It is hard to see how to predict the formation of factors other than two (2^1); however, there is a relationship with the first evens after 3x+C. There was a trend I saw where the growth reduced over distances until many x/2s happened in a row. Example : Again done by hand so error is possible. Start -- 3x+7 First evens. 57: 178 274 418 634 958 1444 1090 1642 2470 3712 94 148 118 184 76 64 10 22 40 178 274 418 634 958 1444 1090 1642 2470 3712 94 148 118 184 76 64 10 22 274/178 = 1.539325843 418/274 = 1.525547445 Is less than previous. 634/418 = 1.516746411 is less 958/634 = 1.511041009 is less 1444/958 = 1.507306889 is less 1090/1444 = 0.754847645 less than one 1642/1090 = 1.506422018 is less 2470/1642 = 1.504263094 is less 3712/2470 = 1.502834008 is less 94/3712 = 0.025323276 less than one 148/94 = 1.574468085 is larger 118/148 = 0.797297297 less than one 184/118 = 1559322034 is less 76/184 = 0.413043478 64/76 = 0.42105263 10/64 = 0.15625 22/10 = 2.2 End -- It is tempting to suggest that there is a factor of 6 in the differences of the first evens; however, that fails for some Cs and some attractors. Only a factor of 3 looks to be valid. Maybe 3^q is true {q|q>0} Over all I think that it can be shown that any 3x+C,x/2^q approaches zero by using the first even after 3x+C and the x/2^q data in a ratio since the amount of decay is greater than growth. Can someone who knows how to do statistics share how please. Also I found interesting structure in holding the value for X and processing C to some point. Example using -1 for x ( I did this by hand so there may be errors ) Y 1 3 5 7 9 11 13 15 17 X -1 -------------------------------------------------------------- -2 0 2 4 6 8 10 12 14 -1 3 1 2 3 4 5 6 7 -2 12 8 1 18 2 28 3 38 -1 6 4 10 9 1 14 24 19 -2 3 2 5 36 14 7 12 56 This suggests to me that its all related. Also it would be good to study this as it may open a door on factor formation as well as other things. I plan to. Thats basically all.. I know there is structure to the Attractors but I havent done the work needed to present it. Perhaps you may find this data interesting. Like you I see relationships between instances of 3x+C and x/2^q. Depending on the q. I was thinking a private group might help us all. I know enough to know I might not know if I know the answer :) Ernst === Subject: Re: [JSH] Mathematical Certainty James Harris > 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. ... > The math says Im right. You lie. prime numbers are there less than 19,591,119? If you had anything to say about that one, it would spread around the world like wildfire, requiring no help from usenet or from the scientific press. For newbies: www.crank.net/harris.html LH === Subject: Re: Fraud in Computer Science Publishing > 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? You may call it a programming language. What Im telling you is that some languages, including C, do not define the results of all situations on purpose. Therefore, you statement was not true. Maybe the standard you are preparing for your language does not do so, but just dont generalize to all programming languages. > Sam, for you its free. Free means that *everyone* has he following freedoms: 0. freedom to run the program, for any purpose 1. freedom to study how the program works, and adapt it to your needs (access to the source code is a precondition) 2. freedom to redistribute copies 3. freedom to improve the program, and release your improvements to the public, so that the whole community benefits (access to the source code is a precondition) I will not describe the reasons I only use free software here, but they are pretty much summarized at http://www.gnu.org/philosphy/ > 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. Im afraid it is not really topical for sci.math Personally Im going back to my normal very busy life in a couple of days, so I wont be able to help much, but Im sure you will find people interested on other newsgroups. i hope you do manage to get a working system. Our disagreements here were more on terminology and use of Lisp as an alternative that on your system per se. Sam -- [...] but the delight and pride of Aule is in the deed of making, and in the thing made, and neither in possession nor in his own mastery; wherefore he gives and hoards not, and is free from care, passing ever on to some new work. - J.R.R. Tolkien, Ainulindale (Silmarillion) === 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 It should be added that the sum is, in a way, also the continuous definition of the harmonic numbers (1 +1/2 +1/3 +...+1/m), but as defined only for the reals x, where -1 < x < 1. So, if, for example we have H(x) = sum{k=2 to oo} (-1)^k zeta(k) x^(k-1), which converges for -1 < x < 1; then, for example: H(x) - H(x-1) = 1/x if 0 < x < 1. The original sum can be rearranged to as to form a sum-form for H(x) which converges for all values of x where H(x) is defined. (But I am NOT claiming that this rearrangement is rigorously justified for all those x where H(x) is defined, even if it is.) sum{k=2 to oo} (-1)^k zeta(k) x^(k-1) = sum{k=2 to oo} sum{j=1 to oo} (-1)^k x^(k-1) /j^k = x sum{j=1 to oo} 1/(j(j+x)) = sum{j=1 to oo} (1/j - 1/(x+j)). (Each of the individual terms in the last sum form divergent series. But summing each term together, as they are within the parentheses and as in the next-to-last sum, forms a telescoping series for x = positive integer.) (So H(x) has, according to this, infinities at x = negative integers.) Leroy Quet === 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 > It should be added that the sum is, in a way, also the continuous > definition of the harmonic numbers (1 +1/2 +1/3 +...+1/m), but as > defined only for the reals x, where -1 < x < 1. > So, if, for example we have H(x) = > sum{k=2 to oo} (-1)^k zeta(k) x^(k-1), > which converges for -1 < x < 1; then, for example: > H(x) - H(x-1) = 1/x > if 0 < x < 1. > The original sum can be rearranged to as to form a sum-form for H(x) > which converges for all values of x where H(x) is defined. > (But I am NOT claiming that this rearrangement is rigorously justified > for all those x where H(x) is defined, even if it is.) > sum{k=2 to oo} (-1)^k zeta(k) x^(k-1) = > sum{k=2 to oo} sum{j=1 to oo} (-1)^k x^(k-1) /j^k > = x sum{j=1 to oo} 1/(j(j+x)) = > sum{j=1 to oo} (1/j - 1/(x+j)). > (Each of the individual terms in the last sum form divergent series. > But summing each term together, as they are within the parentheses and > as in the next-to-last sum, forms a telescoping series for x = > positive integer.) > (So H(x) has, according to this, infinities at x = negative integers.) > Leroy Quet when i was playing around with harmonic numbers. I found that the nth harmonic number can be written as H(n) = int(x=0 to x=1) (1-x^n)/(1-x) dx, but later i noticed that this integral still converges for every real number n between 0 and 1. I also tried to expand it into its maclaurin series, and thats how i found the series. But still i would like to know more on continuation of the harmonic === Subject: Re: ellipse equation > Martin Cohen ha scritto nel messaggio >>Semi-automatic response: >>Let the foci be P1=(x1,y1) and P2=(x2,y2). >>Then, if P=(x,y) is a general point on the ellipse, >>use dist(P,P1) + dist(P,P2) = const >>to get a quadratic in x and y for the ellipse with coefficients >>depending on x1, y1, x2, y2, and const. > ok to this point, now I have: > sqrt[(x-x1)^2+(y-y1)^2]+sqrt[(x-x2)^2+(y-y2)^2]=const At this point, do some algebraic manipulations to eliminate the sqrts. E.g., move one sqrt to the other side of the =, square both sides, isolate the remaining sqrt, square both sides again, and see what happens. (Hint: you should get a quadratic in x and y with coefficients depending on x1, y1, x2, y2, and const.) === Subject: Re: ellipse equation > .... > I have to find the equation of an ellipse: I know the foci coordinates, the > length of the major axis and the length of the minor axis. > .... In that case you know too much. :-) As Ignacio Larrosa Ca.96estro has explained, you can calculate the whole thing from the two foci and the major axis 2a. The length of the minor axis can then be found, and if it disagrees with your given value then your data are inconsistent. I shant prove the consistency condition, but in fact it is (x - x1)^2 + (y - y1)^2 = (2a)^2 - (2b)^2 = (major axis)^2 - (minor axis)^2. Do your given numbers satisfy that? If not, then youre looking for an ellipse which doesnt exist. Ken Pledger. === 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. Why? There are a lot of those numbers (but still countably infinite, so not all transcedental numbers are amongst them). But first of all, *how do you define a^b with a and b algebraic integers*? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Polynomials and 2^sqrt{3} ... >>However, take something like 2^sqrt{3}, which I havent figured out >>how to associate with a polynomial with integer coefficients. ... >I remind readers that Gauss began the process by considering numbers >a+bi, where Ôa and Ôb are integers, and those numbers are now called >gaussian integers in his honor. > Non sequitur. O, it is not entirely, because it gives me some ideas. Apparently James thinks that in some ring of integer-like things the operation a^b should be defined. This comes from a number of mistakes. 1. James thinks that taking powers is a well-defined operation for arbitrary numbers. It is not. It is only well-defined if the exponent is a (rational) integer. 2^(1/2) is *not* well-defined, it is only defined by convention. 2. James thinks that a ring should be closed under taking powers. This is not true, even if they were well-defined. Also the Gaussian integers are not closed under taking powers. For instance, under the customary definition of powering, the value of 2^i (both Gaussian integers) is [cos(ln(2)) + i.sin(ln(2))], which (apart from the fact that in such context the logarithm is really multi-valued), is certainly *not* a Gaussian integer. 3. James thinks that when we consider a ring of integers that that is something special, different from any other ring. Also wrong. A ring of integers is always a subring of another ring, where the elements have a particular feature, otherwise it makes almost no sense. So in Q(sqrt(3)) we can take a ring of integers to be all numbers of the form a+b.sqrt(3), with a and b integer. This is a ring (which is easily verified). However for some reasons it is more common to use those numbers where a and b are both either odd or even multiples of 1/2. An introductiory book on number theory (the chapter about quadratic fields) will show why. Within the definitions of a ring there is *no* power operation. It is only introduced as a shorthand for a (finite) repeated multiplication. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Polynomials and 2^sqrt{3} [cut] > So in Q(sqrt(3)) we can take a ring of integers to be all > numbers of the form a+b.sqrt(3), with a and b integer. This is a > ring (which is easily verified). However for some reasons it is > more common to use those numbers where a and b are both either odd > or even multiples of 1/2. The reason is that you want to take *all* numbers of Q(sqrt(3)) that are integral over Q to be your ring of integers so that it will be a Dedekind domain and thus you will have unique factorizations of ideals as a product of prime ideals. Other nice properties also follow and are present. Rings like the numbers of the form a+b.sqrt(3), with a and b integer, are called orders, I believe. I am not familiar with their properties or their importance. -- Bill Hale === Subject: stupid cal II question I have a stupid cal II question. Im trying to find the area enclosed by 3 lines. Heres the problem Find A when y=2, y=1/sqrt(x),y=x I can do a nice simple A under the curve ...and A b/t two curves, but this is 3 lines. Im not sure where to go on this one. I did manage to find the x and y intercepts of the 3 lines cause I thought theyd be helpful and i didnt want to come here empty handed. They are A. 1/4,2 B.2,2 C. 1,1. I was thinking of a divide and conquer approach but I got lost during conquest. See, I told you it was a stupid question. But then again There are no ... thx === Subject: Re: stupid cal II question In sci.math, sibersmith by 3 lines. > Heres the problem > Find A when y=2, y=1/sqrt(x),y=x > I can do a nice simple A under the curve ...and A b/t two curves, but > this is 3 lines. Im not sure where to go on this one. I did manage > to find the x and y intercepts of the 3 lines cause I thought theyd > be helpful and i didnt want to come here empty handed. They are A. > 1/4,2 B.2,2 C. 1,1. > I was thinking of a divide and conquer approach but I got lost during > conquest. > See, I told you it was a stupid question. But then again There are > no ... > thx Well, y=2 is a horizontal cut line, which cuts y=1/sqrt(x) at the point (1/4,2). y=2 cuts y=x at (2,2), of course. y=x cuts y=sqrt(1/sqrt(x)) at (1,1). Add three vertical drop lines x=1/4, x=1, and x=2, and you get an integral and a triangle; the problem should then be fairly trivial. Visualization using a plotting package such as gnuplot, or just doing it the old-fashioned way with pencil and graph paper, helps as well. -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Basic Maple Question - Solving a DE |>Im in a physics class where I have to churn out solutions |>periodically with either MathLAB or Maple. Im choosing to use Maple |>because this is what I have on my computer. Incidentally, Im not very |>familar or confident with it, so Im having the following trouble. |>Basically, I have a function, say F(t) = F0 - ct. And of course, |>Newtons law, F(t) = m*x. So my differential equation is F0 - ct = |>m*x. A very basic brute-force integration to solve by hand, but Im |>having a difficult time with Maple. |>Ive made two different attempts. This is what Im doing now -- |>I have my with(DEtools); with(plots); with(plottools); You dont need these here. |>DEFunction := F - c*t^2 = m * D(D(x))(t); Why are you squaring the t? |>init_con := x(0) = 0, D(x)(0) = 0; |>dsolve( { DEFunction, init_con } , { x(t) } ); |>The solution it gives is |>x(t) = -(1/2)((-F + ct)t^2) / m No, I dont think so. Without the extra ^2 I get > DEFunction := F - c*t = m * D(D(x))(t); init_con := x(0) = 0, D(x)(0) = 0; dsolve( { DEFunction, init_con } , { x(t) } ); 2 3 F t c t x(t) = ---- - ---- 2 m 6 m 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: Finding extrema with Maple >Ive been trying to find the following extrema with Maple: >extrema(arctan(x*y),{x,y}); Dont use extrema. It has some very bad bugs and weaknesses. For example, it doesnt properly handle complex roots. Your best bet is to find the partial derivatives and look for extreme points directly, as well as considering limits at infinity. 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: 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 Maybe he was born on a leap year. adam === Subject: Re: 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 Your problem is the underlined passage above. The events {X gets hit},{Y gets hit} {Z gets hit} are not independent. For instance, X getting hit and Y getting hit both depend on who Z chooses to target. I dont think its difficult to get 1/240. That none of them survive would require that each is targeted by someone: a probability of 1/4, and that each hits their target: probability of 1/60. These events are independent so you get 1/240. Rob === Subject: statistics software? I was wondering if anyone can suggest any good statistical software, freeware or shareware or trial limit. I need a software package for regression, least squares, etc..... ....... any help is appreciated === Subject: Re: statistics software? R, no doubt, its a free version of S-Plus > I was wondering if anyone can suggest any good statistical software, > freeware or shareware or trial limit. > I need a software package for regression, least squares, etc..... ....... > any help is appreciated === Subject: Re: statistics software? > I was wondering if anyone can suggest any good statistical software, > freeware or shareware or trial limit. > I need a software package for regression, least squares, etc..... ....... > any help is appreciated As the saying goes, Google is your friend. Just search for free statistics software and you get lots. Look for octave, scilab, r-project, ... You dont pay your money and you get good software. === 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?? Im not all that frequent a user of the Gamma function, so I just did some plotting, courtesy of Matlab. I plotted Gamma(x) from x=0 to x = 1000 (with a resolution of 0.001 along x), and seeing nothing very much of detail in the area of its minimum value, I looked at the interval [1.46,1.464] surrounding the minimum, using 1.e6 points (a resolution of about 1.e-9). Heres what I found out: On the positive reals (I assume you dont really care about non-positive reals in this exercise), Gamma(x) is only decreasing on the interval (0, 1.461632]. Since this interval appears to fall outside your domain of interest, Im led to ask: Whats the problem with Gamma(x)? I dont get your complaint. After all, 0! = 1! = 1, so one must inquire as to what *you* expect Gamma to act like between two points (x=1,2) where it takes the same value. Analyticity (from the somewhat parochial perspective of a mathematician) seems to be a more useful characteristic than simple monotonicity. Given that, Gamma(x) cant really do other than to dip below the line y=1 over that interval. Well, it *could* have risen above, only to return, but you have to admit that monotonicity is pretty much out the window, once you have two locations where the function takes the same value. If my reliance on the graphs Ive plotted is causing me to miss some egregious failure of Gamma(x) to behave properly, please educate me. Dale. === Subject: Re: Complex variable >> If f is an holomorphic function on U (an open set of C) >> Suppose, moreover, that f(x) <> 0 and f(x) <> 1 for all x in U. >> Can we say that there exists h, an holomorphic function on U such that f(z) >> = >> exp(2*i*Pi*cos(h(z))). >No, take U = {0 < |z| < 1} and f(z) = z. Then f is never 0 or 1 in U. If we >had f = exp(g) for some g holomorphic in U, then f would have a holomorphic >square root in U, namely exp(g/2), contradiction. However, the assertion is true if U is simply connected. Proof: Since f is never 0 in U, we can write f = exp(2 pi i g) for some g analytic in U. Since f is never 1, g does not take any integer values in U. Now 1 - g^2 is an analytic function on U that doesnt take the value 0, therefore has an analytic square root in U, say v(z). Thus (g(z) + i v(z))(g(z) - i v(z)) = g(z)^2 + v(z)^2 = 1 so g(z) + i v(z) is never 0 on U. So g(z) + i v(z) = exp(i h(z)) for some analytic function h(z) on U, and g(z) - i v(z) = 1/exp(i h(z) = exp(-i h(z)) so g(z) = 1/2 (exp(i h(z)) + exp(-i h(z))) = cos(h(z)) and f(z) = exp(2 pi i cos(h(z))) 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: [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. Maybe, but in the case of the US individual power and state power go hand in hand, since besides great state power it has a large number of wealthy individuals. Gib === 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 suppose, by virtue of your .nz address, this opinion of yours is > thankfully devoid of arrogance, however chock-full of ignorance it may > be. You might suppose that. But in fact I did live in the US for 13 years. Gib === Subject: Re: Factorial/Exponential Identity, Infinity I figured out a function for f(n) of ( (sum n)^3 - sum (n^3) ) / f(n) = | s(n+1, n-2) |. It appears that f(n)= 6(n+2)/(n-2) works. With this then there is a way to calculate the Stirling number of the first kind s(n+1, n-2) without calculating intermediate results of a recurrence relation. The value of (sum n) is n(n+1)/2, (n^2+n)/2 the value of sum (n^3) is (sum n)^2, or (n(n+1)/2)^2. The expression ((n^2+n)/2)^2 is (n^4+2n^3+n^2)/4, ((n^2+n)/2)^3 is (n^6+2n^5+n^4 + n^5 + 2n^4 + n^3)/8 = (n^6+3n^5+3n^4+n^3)/8. So, there is: ( (sum n)^3 - (sum n)^2 ) (n-2)/ 6(n+2) ( ((n^2+n)/2)^3 - ((n^2+n)/2)^2 ) (n-2) / 6 (n+2) ( ((n^6+3n^5+3n^4+n^3)/8) - ((n^4+2n^3+n^2)/4) ) (n-2) / 6(n+2) ( ((n^6+3n^5+3n^4+n^3)/8) - ((2n^4+4n^3+2n^2)/8) ) (n-2) / 6(n+2) ( n^6+3n^5+3n^4+n^3 - (2n^4+4n^3+2n^2) ) (n-2) / 48(n+2) ( n^6+3n^5+n^4-3n^3-2n^2) (n-2) / 48(n+2) ( n^7 + 3n^6 +n^5-3n^4-2n^3 -2n^6 -6n^5 -2n^4 +6n^3 +4n^2) / 48(n+2) ( n^7 +n^6 -5n^5 -5n^4 +4n^3 +4n^2 ) / 48(n+2) Evaluating for n=3: ( 2187 + 729 -1215 - 405 + 108 + 36 ) / (48 * 5) 1440 / (48*5) = 6 Yep, that there is -s(4, 1). So, it appears that the absolute value of s(n+1, n-2), for n >= 3, is: ( n^7 +n^6 -5n^5 -5n^4 +4n^3 +4n^2 ) / 48(n+2) I wonder if there is a similar method to get a formula representing Stirling numbers s(n+1, n-3). One thing that would not be as simple to represent is that where sum (n^3) = (sum n)^2, sum (n^4) is a recurrence relation instead of a function. Whats the most simple way to represent sum (n^4) as a function? For what other values of x, if any, is sum (n^x) = (sum n)^f(x)? Ross === Subject: Re: Factorial/Exponential Identity, Infinity Is this useful for determining an arbitrary unsigned Stirling number of the first kind | s(n, m) | ? It can be. I search plaintively and come across a form for s(n, n-2). http://mathforum.org/library/drmath/view/61259.html s(n, n-2) = n(n-1)(n-2)(3n-1)/24 s(n, n-2) = (n^2-n)(3n^2-7n+2)/24 s(n, n-2) = (3n^4 -7n^3 +2n^2 -3n^3 +7n^2-2n)/ 24 s(n, n-2) = (3n^4 -10n^3 +9n^2 -2n) / 24 Is the sum of products of pairs of {1, 2, 3} equal to |s(3+1, 3-1)|? (3*4^4 -10*4^3 +9*4^2-2*4) / 24 (768 -640 +144 -8 ) / 24 = 11 ((sum n)^2 - sum (n^2)) / 2 (n(n+1)/2)^2 - sum (n^2)) / 2 (3(4)/2)^2 - (1+9+16) / 2 (36-24) / 2 11 Yes, it is. So now there are these equations: s(n, n-2) = (3n^4 -10n^3 +9n^2 -2n) / 24 s(n+1, n-2) = ( n^7 +n^6 -5n^5 -5n^4 +4n^3 +4n^2 ) / 48(n+2) Also something to consider is that: (3n^4 -10n^3 +9n^2 -2n) / 24 = ((sum (n-1))^2 - sum ((n-1)^2)) / 2 (3n^4 -10n^3 +9n^2 -2n) / 12 = ( (n(n-1)/2)^2 - sum ((n-1)^2)) (3n^4 -10n^3 +9n^2 -2n) / 12 = ((n^2-n)/2)^2 - sum ((n-1)^2)) (3n^4 -10n^3 +9n^2 -2n) / 12 = (n^4-2n^3+n^2)/4 - sum ((n-1)^2)) (3n^4 -10n^3 +9n^2 -2n) / 12 = (3n^4-6n^3+3n^2)/12 - sum ((n-1)^2)) (3n^4-6n^3+3n^2)/12 - (3n^4-10n^3+9n^2-2n)/12 = sum ((n-1)^2)) (4n^3-6n^2+2n)/12 = sum ((n-1)^2)) Huh. Ill try that for n=2, (4*8-6*4+4)/12 = 1. how about for n=3, (4*27-6*9+2*3)/12 = 5 = 1^2+2^2, excellent! So now besides a function for n for sum (n^3) I have a function for sum ((n-1)^2). Lets see, in terms of sum (n^2) then the expression would be: (4(n+1)^3 - 6(n+1)^2 +2(n+1)) / 12 (4(n^2+2n+1)(n+1) - 6(n^2+2n+1) + 2(n+1) ) / 12 ( 4(n^3+n^2+2n^2+2n+n+1) - 6(n^2+2n+1) + 2(n+1) ) / 12 ( 4n^3 +12n^n +12n +4 -6n^2 -12n -6 +2n +2 ) / 12 ( 4n^3 +6n^2 +2n ) / 12 Is it so that 1+2^2 = 5 = (4*8 + 6*4 + 2*2)/12? It is! So I have something along the lines of sum_(i=1^n) i^2 = ( 4n^3 +6n^2 +2n ) / 12 sum_(i=1^n) i^3 = ( n^4 +2n^3 +n^2) / 4 The sum of the products of the pairs of {1, ..., n}, | s(n+1, n-1) |, is: (n^4 +2n^3 -3n^2 -2n) / 24 The sum of the products of the trios of {1, ..., n}, | s(n+1, n-2) |, is: ( n^7 +n^6 -5n^5 -5n^4 +4n^3 +4n^2 ) / 48(n+2) So now I have the forms for the first four coefficients of (n+1)(n+2)...(n+n). Heres something I want you to write if you know: what are forms for sum(n^x)? That is, what is a function the result of which is the sum of fourth, fifth, etcetera powers of one through n? Those might enable the evaluation of Stirling numbers of the first kind non-recursively. I notice Leroy Quet had mentioned an integral form of the Stirling number of the first kind but I cant make it out. Heres something, on http://mathworld.wolfram.com/ StirlingNumberoftheFirstKind.html, in the little permutation diagrams, doesnt it look like the third diagram of S_1(5, 4) is missing a dot? About evaluating more Stirling numbers, heres one function that finds ready application here: s(n+2, n-1) = (n+1) s(n+1, n-1) + s(n+1, n-2) Theres already a form for s(n+1, n-2). For what other values of x, if any, is sum (n^x) = f(x)? Ross === Subject: least squares question How do I derive a least squares formula for y = mx + b, and y = Ae^-bx? ive searched numerous books, but have yet to find one that actually goes through the process. Philip === Subject: Re: least squares question >How do I derive a least squares formula for y = mx + b, and y = Ae^-bx? >ive searched numerous books, but have yet to find one that actually goes >through the process. >Philip Look in an introductory or elementary statistics textbook. G C === Subject: Re: least squares question let Y be the actual data reading so the error on each point is Y-y then calculate S=sum over all data points of (Y-y)^2 so S= um (Y-mx-b)^2 minimise S with respect to m and b by differentiation wrt to m and b and solving the two equations. for the other, take logs first so you have log(y) = log(A) - bx, so just do as above with (sum (log(Y)-log(y))^2 rod > How do I derive a least squares formula for y = mx + b, and y = Ae^-bx? > ive searched numerous books, but have yet to find one that actually goes > through the process. > Philip === 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. Heres a page that answered most of my topological questions: http://www.geom.uiuc.edu/java/IFSoft/IFSs/ 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. Ah!! According to the definition at the given URL, your T_i arent just any mappings; they are contraction mappings. This means that any two points x, y in S always get closer together after application of any T_i; and to make sense of get closer, S must be a metric space - as it is (implictly) in the case of R^n. A single contraction mapping T maps every point in S to a single fixed point at the limit of multiple compositions; thus as you note, sets of {T_i} of cardinality 1 are not very interesting :). Limits require some notion of convergence; which is in this case supplied by the metric on the space S. In R^n, T(z) = z/2 is contractive, with fixed point z = 0. A similar action of taking each point or element of S towards some fixed point occurs whenever T is contractive in a metric space S. The URL above then extends this observation to a _different_ metric space: the space H(S) whose points, or elements, are non-empty, compact subsets of S; and with a metric (method of determining distance between two such subsets), making H(S) into a metric space. The function acting by s :-> union(s.T_i) for s a compact subset of S then acts as a contraction mapping in this new space, so by the preceding logic, there is some unique fixed point A in H(S). In the limit, the function acts to take any non-empty compact subset of S (including the subset consisting of any single point {x}) arbitrarily close to the same point of H(S) - a particular non-empty compact subset of the original space S. Of course, any finite iterative application of s :-> union(s.T_i) to the singleton set {x} will only have a finite number of points in it; but were only concerned with the idea that the resulting (finite) subset of S gets closer and closer to A as we continue to iterate. That gives the uniqueness and non-triviality that I was looking for, by adding a few caveats to your original definition: Given a _metric_ space S, and a set of _contraction_ mappings T_i, the attractor A is the (unique) _non-empty compact_ subset of S such that union(A.T_i) = A. One remaining question I have - how do we know in general that, given s is a compact subset of S, then union(s.T_i) is compact as well? If the T_i are always considered to be continuous (in the topological sense), then this follows automatically; but that isnt explicitly stated in the URL. This might have an effect (granted, rather abstract) on the original question you asked - if X is not a continuous function, then although A would be compact, A.X might not be; so it might be the case that A* != A.X. > The other definition is the limit of one of several iterative > algorithms, including the random, recursive and xerox algorithms. This limit process is surely going to invoke some metric on S; so Im guessing that these approaches in some way always translate to some contractive function of H(S), the space of compact subsets of S, yielding a unique result. > 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 Ooof! But at what cost? It seems like youd want A to be infinite, possibly even uncountable (e.g., the Sierpinski Gasket, etc.) > 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? I think compact deals with the previous paragraphs concerns, and other problems. Compact in R^n is the same as closed and bounded. This is good for us! First, If T(z) = z/2, then the line L = {z : z.x = 0} has the property that L.T = L; but L is not compact (its unbounded), and so it is not a member of H(R^2). A more annoying example thusly excluded: the set U = {z : |z| = 2^n, n an integer} also has U.T = U; again, fortunately U is unbounded, and so not compact. Another excluded set: U = {z : |z|<1}; U.T = U; but U is open, and so not compact. Second, suppose that A is the attractor for some set {T_i} in R^2. A is compact; now let A = {x in A : x in Q^2} (the rational points of A). A might be compact in Q^2; but it is not compact in R^2 (A is dense in A, so the closure of A is not A, so A is not closed, so its not compact). Union(A.T_i) = A might be true for T_is with rational coefficients; in which case, it would be the attractor in Q^2 for {T_i}. However, A is not a compact subset of R^2, and so doesnt meet the definition of attractor in R^2. The same logic follows for the algebraics. Conversely, if A is the attractor of {T_i} in S, and S is a subspace of S (i.e., inherits the same metric as S), then if each T_i is closed w.r.t. S, we have that A = A intersect S is the attractor of {T_i}, restricted to S. Thus, if the T_i all have algebraic parameters, then if A is the attractor in R^n, A intersect (algebraics^n) is the attractor in (algebraics^n). >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. ..., but as defined above, (i) z*|z| is not contractive on R^2, and (ii) A_2 and A_3 are not compact. Even restricted to S = A_0 u A_1, T is still not contractive on S; since the contractivity factor is 1. Interesting stuff. === Subject: Re: Iterated Function Systems <+SijSCAO4cU$Ewj8@meden.demon.co.uk> You might like a look at http://www.meden.demon.co.uk/Images/Fractals/ before I reclaim the space (Im running short, but I had forgotten that I had these images online.) >> 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. >Heres a page that answered most of my topological questions: >http://www.geom.uiuc.edu/java/IFSoft/IFSs/ That page imposes three restrictions that I dont think are necessary: that the transforms are locally contractive; that they individually converge on a point; and that they map the metric space to itself. (Note that this page talks about hyperbolic IFSs, implying that there is a more general IFS concept.) For more comments on contractivity see below. generate the attractor of the transforms of Sierpinski Triangle combined with projection onto the exterior or interior circle. Although this projection doesnt map R^2 to itself, but to a subspace of dimension 1, it does produce interesting attractors; the first, IIRC, generates fractally nested circles, and the second a Sierpinski Triangle with circles inscribed in each void. Other transforms that dont map R^2 to itself include projection onto a (straight) line, contraction towards a circle (contraction along the vectors perpendicular to a circle), and piecewise linear contraction towards or projection onto the edges of a polygon. Another is a piecewise linear transform including x -> |x|/2. Note that some of these are not contractive over the whole of R^2, and none of them converge on a point. > 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. >Ah!! According to the definition at the given URL, your T_i arent >just any mappings; they are contraction mappings. This means that >any two points x, y in S always get closer together after >application of any T_i; and to make sense of get closer, S must be a >metric space - as it is (implictly) in the case of R^n. >A single contraction mapping T maps every point in S to a single fixed >point at the limit of multiple compositions; thus as you note, sets of >{T_i} of cardinality 1 are not very interesting :). Limits require >some notion of convergence; which is in this case supplied by the >metric on the space S. >In R^n, T(z) = z/2 is contractive, with fixed point z = 0. A similar >action of taking each point or element of S towards some fixed point >occurs whenever T is contractive in a metric space S. Barnsley, in Peitgen and Saupe, says averagely contractive, but I dont follow what he means by this. Ive been trying to avoid mentioning contraction mappings because Im fairly sure, subject to matters of nomenclature, that the transforms of an IFS dont have in general to be contraction mappings. For linear transformations over R^n the rule appears to no transforms are expansive parallel to any vector; at least one transform is contractive parallel to any vector; each transform is contractive parallel to at least one vector, i.e. the transforms are contraction mappings. However ... z -> z/(|z|^0.5), is expansive within the circle, and contractive without, and maps R^2 to R^2. Ive generalised my inversion in a circle class from z -> z/|z*z| to z -> z*(|z|^w), and tried a few IFSs. Taking a Sierpinski triangle, and adding z -> z/(|z|^0.5), I find that if the Sierpinski triangle lies wholly within the circle the attractor is the enclosed disc - no doubt with other choices of transforms we could get a disc with voids. If the Sierpinski triangle extends beyond the disc the result is more interesting, but still not very interesting. Also some of the transforms mentioned above transforming R^2 to a subspace of R^2 are expansive for some pairs of points in R^2. In other spaces contractivity becomes even less clearly defined; on a torus x -> x/2, y -> 3*y converges on a line, though it is everywhere locally expansive. It seems to me that some concept of global contractivity is required, and this may be difficult to define in a way that includes all spaces. Therefore I was hoping that it would be possible to define an attractor in such a way that the contractivity requirements were implicit in the definition. >The URL above then extends this observation to a _different_ metric >space: the space H(S) whose points, or elements, are non-empty, >compact subsets of S; and with a metric (method of determining >distance between two such subsets), making H(S) into a metric space. >The function acting by s :-> union(s.T_i) for s a compact subset of S >then acts as a contraction mapping in this new space, so by the >preceding logic, there is some unique fixed point A in H(S). In the >limit, the function acts to take any non-empty compact subset of S >(including the subset consisting of any single point {x}) arbitrarily >close to the same point of H(S) - a particular non-empty compact >subset of the original space S. >Of course, any finite iterative application of s :-> union(s.T_i) to >the singleton set {x} will only have a finite number of points in it; >but were only concerned with the idea that the resulting (finite) >subset of S gets closer and closer to A as we continue to iterate. >That gives the uniqueness and non-triviality that I was looking for, >by adding a few caveats to your original definition: >Given a _metric_ space S, and a set of _contraction_ mappings T_i, the >attractor A is the (unique) _non-empty compact_ subset of S such that >union(A.T_i) = A. >One remaining question I have - how do we know in general that, given >s is a compact subset of S, then union(s.T_i) is compact as well? If >the T_i are always considered to be continuous (in the topological >sense), then this follows automatically; but that isnt explicitly >stated in the URL. I have considered (but never investigated nor coded) piecewise linear functions, which are need not be continuous at all points. Thinking on the matter, one could define a set of piecewise linear functions, such that there is one attractor in the subspace x<0, and another in the subspace x>=0. These individually and jointly satisfy A.T{i} = A, and are compact and non-empty. So I seem to have :-( outlined a proof that an attractor, as defined above, is not necessarily unique. >This might have an effect (granted, rather abstract) on the original >question you asked - if X is not a continuous function, then although >A would be compact, A.X might not be; so it might be the case that A* >!= A.X. It seems to me the discontinuous piecewise linear functions arent bad enough to break this, but a mapping z -> z/2 (z algebraic), z -> z/3 (z transcendental) strikes me as doing something nasty. >> The other definition is the limit of one of several iterative >> algorithms, including the random, recursive and xerox algorithms. >This limit process is surely going to invoke some metric on S; so Im >guessing that these approaches in some way always translate to some >contractive function of H(S), the space of compact subsets of S, >yielding a unique result. >> 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 >Ooof! But at what cost? It seems like youd want A to be infinite, >possibly even uncountable (e.g., the Sierpinski Gasket, etc.) Clash of terminologies: by infinite I meant unbounded (I said I didnt know much mathematics). I think Ill by taking up your use of compact. >> 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? >I think compact deals with the previous paragraphs concerns, and >other problems. >Compact in R^n is the same as closed and bounded. This is good for >us! As you say that compact is R^n is closed and bounded I presume that that is not the case for other spaces, and our results may not apply in those other spaces. Ill have to wander over the MathWorld and refresh my memory as to what all these terms mean. >First, If T(z) = z/2, then the line L = {z : z.x = 0} has the property >that L.T = L; but L is not compact (its unbounded), and so it is not >a member of H(R^2). >A more annoying example thusly excluded: the set U = {z : |z| = 2^n, n >an integer} also has U.T = U; again, fortunately U is unbounded, and >so not compact. >Another excluded set: U = {z : |z|<1}; U.T = U; but U is open, and so >not compact. I dont follow which transform youre defining here. >Second, suppose that A is the attractor for some set {T_i} in R^2. A >is compact; now let A = {x in A : x in Q^2} (the rational points of >A). >A might be compact in Q^2; but it is not compact in R^2 (A is dense >in A, so the closure of A is not A, so A is not closed, so its not >compact). >Union(A.T_i) = A might be true for T_is with rational coefficients; >in which case, it would be the attractor in Q^2 for {T_i}. However, A >is not a compact subset of R^2, and so doesnt meet the definition of >attractor in R^2. >The same logic follows for the algebraics. >Conversely, if A is the attractor of {T_i} in S, and S is a subspace >of S (i.e., inherits the same metric as S), then if each T_i is closed >w.r.t. S, we have that A = A intersect S is the attractor of {T_i}, >restricted to S. >Thus, if the T_i all have algebraic parameters, then if A is the >attractor in R^n, A intersect (algebraics^n) is the attractor in >(algebraics^n). Whats the conventional abbreviation for the set of algebraics? >>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. >..., but as defined above, (i) z*|z| is not contractive on R^2, and >(ii) A_2 and A_3 are not compact. Even restricted to S = A_0 u A_1, T >is still not contractive on S; since the contractivity factor is 1. > Interesting stuff. Indeedly interesting; its a bit of a novelty for me to get this deep into the maths, rather than exploiting various heuristics. I never did get round to writing the What Is An IFS page; perhaps its going to be rather different from what I expected. -- Stewart Robert Hinsley === Subject: Re: Reply from another German editor Maybe, I am missing something, but I couldnt quite see how JSH got from Given: x^2 + y^2 = z ^2 ==> x^2 + y^2 + vz^2 == 0 (mod x^2 + y^2 + vz^2) is it just me? Lurch >>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. > This is such a cop-out its incredible. There are three universities in > the Atlanta area that Ive looked at: Emory, Georgia State, and Georgia > Tech. I found several names of professors (at varying ranks, from > assistant to full professor) who would doubtless have the background > to provide guidance. Whether they would be willing is an entirely > different matter. Envision, if you will, the following two scenarios: > ONE: > SMITH: Yes, can I help you? > JSH: Please pretend youre not a lying sack of for a > minute, and read this paper that shows that everything > youve been teaching is a full-blown lie, and by the > way you dont really know enough to be allowed to wipe > my furry little ass with that lying tongue of yours, > so please bow down and worship me, before I sic the > generals on you! > SMITH: No, thank you, but have a nice life! > TWO: > SMITH: Yes, can I help you? > JSH: Why, yes, I hope. Ive written this paper for publication > and I cant seem to get anywhere with it. Its about > algebraic integers and factoring polynomials, which should > be basic material, but all I get are complaints about how > either everything is trivial and I have no proofs here, > or how Im claiming that something is true that just isnt, > and so dont have any proofs here. When *I* read it it all > looks so obvious, yet when people who *claim* to be > mathematicians read it, there are these complaints. > Further, I just got this back from the prestigious German > journal Zeitschrift f.9fr der Vaterland und ihrer > Grenzgebiete und Wienerschnitzel .9fber alles, and the > editor sent this letter back (hands letter over) ... > SMITH: Im a little busy now, but if you could allow me > to Xerox your papers there, I would be glad to give > it a quick look-over and we could discuss it in a > week or so. OK? > JSH: Well, I really wanted to get on with taking down the > corrupt mathematical establishment, if its all the same > to you. After all, ßawed ideas are just wriggling all > over the corpus of mathematical thought, like little > lying maggots lying all over this mathematical corpse, > and it really gives me the Jimmies, er, Smithies, er, > heebie-jeebies. How soon was that? > SMITH: A week or so; um, look, give me a call during my > office hours next Thursday, and Ill see whether > Ive made any headway. Until then, dont you think > that this corrupt mathematical establishment will > pretty much remain in a status quo? I mean, its > been a few hundred years already, I dont see where > a week more or less can make much difference. > JSH: But theres the FBI, and the NSA and the NSC, also.... > SMITH: OK, well Ill do my best. Thursday, then? Between, > oh, one-thirty and four in the afternoon? Heres > my card. It has my office numbers and e-mail on > it. Ill talk to you soon. Gotta go. > JSH: OK, Professor Smith. Thursday it is, then! Bye!! > [sotto voce]: running lapdog of mathematical imperialism. > See, in approach ONE, our mock JSH was too up-front with his bigotry > and hatred of the mathematical establishment, and it caused a classical > defensive-troll response. > In approach TWO, the mock JSH held back, and the mathematician SMITH > was tricked into responding as a helpful teacher-troll. Even when the > bigotry was unveiled (by referring to taking down the mathematical > establishment), the teacher-troll adopted a more accommodation-troll > type or troll-response. He gave the mock JSH a troll-card to help > him find the teacher-accommodation-trolls office and phone number, > and even made the extra effort to push the review ahead to accommodate > JSHs urgency in his plans for demolition of the corrupt mathematical > establishment. > The approach can make a big difference, thats all Im saying. > 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. > No Less Than Barry Mazur. > 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. > Yes, Mazur is well-respected. His set of works is as significant as > anyones. So what? > 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. > And Mazurs job is to help *you* in some fashion? Do you truly believe > that he has a contract that even has *your* name in it? Or, is it that > you think his contractual agreements include some text such as this: > ... shall assist every person with a mathematical paper, > irrespective of the clarity of ideas expressed in said > paper, or of the correctness of the results stated therein, > to publish and claim the rights and privileges heretofore > accorded to the mathematical priesthood. These rights shall > include, but not necessarily be limited to: > 1. Fame > 2. Vast Riches > 3. The Babes. Dont forget the Babes. > 4. Power. Unlimited Power. Plus the Babes! > 5. The right to ßout all convention, especially > anything to do with farting in elevators, dropping > big pieces of poo into punchbowls, and threatening > others lives and livelihoods. > 6. Did I mention the Babes? Well, the Babes, then. > Any person who shows up with any scribbling that is purported > to be profound mathematical proof must be the number one > priority; all other work must cease because it could be > REALLY REALLY BIG!!!! and since society pays us [note how > your paycheck is signed Very Truly Yours. The Society], we > have a legal, moral, and fiduciary obligation to act as the > servants, nay, slaves, of everyone who waltzes in the door > with a slip of paper with any sort of marking on it. > 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. > Its WACKY!!!! > Its ESOTERIC!!!!! > Its A FLAW!!!!!!!!!!!!!!! > But > JSH CANT SHOW IT!!!!!!! > ...thats JUST how *wacky* and *esoteric* that ßaw is! > 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. > This would be slightly *less* pathetic if one other human could be > convinced of the truth of this sentence. > However, there is not another human being who accepts JSHs version > of mathematics. Shouldnt he have a friend that he discusses his > greatness with, to whom he could preach the word, and who would at > least stand up and say Amen to brother JSHs gospel? > Doesnt his Mom even believe him? > His Sweetheart? > His pet dog? > Yeeehaa!!! > Once again, JSH imagines hes king of the world. Time for him > to get that Burger King crown out and prance in front of the > mirror. > James Harris > Dale === Subject: Re: Reply from another German editor > Maybe, I am missing something, but I couldnt quite see how JSH got from > Given: > x^2 + y^2 = z ^2 > ==> x^2 + y^2 + vz^2 == 0 (mod x^2 + y^2 + vz^2) > is it just me? > Lurch The basic idea is that, in any ring in which A is defined, one immediately has A == 0 (mod A). This follows from the definition of the notion of equivalence modulo an integer (in the ring Z of integers) or modulo an ideal in the more general case. Since JSH is oblivious to what ideals are (in probably both senses, but Ill restrict discussion to the mathematical interpretation), what must be meant is p == q (mod r) <==> p-q = rs for some integer s. The condition x^2 + y^2 = z^2 is irrelevant in that regard. It does appear that some consideration be paid to what ring all these congruences take place, and it is certainly the case that JSH actively avoids specifying anything he doesnt personally identify as being critical to his argument. That said, the congruence must hold wherever it makes adequate sense. JSH imagines this as being a stroke of genius. I interpret all JSHs strokes as being of a somewhat cruder nature, and best performed in the privacy of ones own room. Dale === Subject: Re: Reply from another German editor Hi W. Dale, Yes, I understood the x^2 + y^2 + vz^2 == 0 (mod x^2 + y^2 + vz^2). I just dont get how he gets this result from the Pythagorean theorem. And like you said, he doesnt take the time to show you how he suppossedly got there. I think JSH needs to study more than just ideals, but also math writing. I could only tolerate reading the first few sentences because of his style and illogical steps. The above was just the first ßaw to catch my eye. Lurch > Maybe, I am missing something, but I couldnt quite see how JSH got from > Given: > x^2 + y^2 = z ^2 > ==> x^2 + y^2 + vz^2 == 0 (mod x^2 + y^2 + vz^2) > is it just me? > Lurch > The basic idea is that, in any ring in which A is defined, one > immediately has > A == 0 (mod A). > This follows from the definition of the notion of equivalence > modulo an integer (in the ring Z of integers) or modulo an ideal > in the more general case. Since JSH is oblivious to what ideals > are (in probably both senses, but Ill restrict discussion to the > mathematical interpretation), what must be meant is > p == q (mod r) <==> p-q = rs for some integer s. > The condition x^2 + y^2 = z^2 is irrelevant in that regard. It does > appear that some consideration be paid to what ring all these > congruences take place, and it is certainly the case that JSH > actively avoids specifying anything he doesnt personally identify > as being critical to his argument. That said, the congruence must > hold wherever it makes adequate sense. > JSH imagines this as being a stroke of genius. I interpret all JSHs > strokes as being of a somewhat cruder nature, and best performed in the > privacy of ones own room. > Dale === Subject: Re: [JSH] Reply from another German editor > Hi W. Dale, > Yes, I understood the x^2 + y^2 + vz^2 == 0 (mod x^2 + y^2 + vz^2). I just > dont get how he gets this result from the Pythagorean theorem. And like > you said, he doesnt take the time to show you how he suppossedly got there. > I think JSH needs to study more than just ideals, but also math writing. I > could only tolerate reading the first few sentences because of his style and > illogical steps. The above was just the first ßaw to catch my eye. > Lurch You might want to take a look at http://www.cs.hamilton.edu/~rdecker/FLT/ where Ive unpacked some of what James is saying. This page refers to an older version of his paper (before he discovered algebraic integers, rejected them, and invented his Object Ring), so only sections 1 and 2 apply to his latest paper. In spite of this, it might help get you up to speed on the first part of what hes saying. Rick === 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. Go ask him. 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? >> Damn, I was gonna ask that one. >> Well, James? >> Were really curious. >Go ask him. Strategic error here, James. You should really tell us what he said. I mean surely Mazur must have had something positive to say, because if his comments were simply negative like everyone elses then thered be no point to your proudly telling us that hed commented on it, right? But when you dont tell us what he said people are going to come to the conclusion that his comments _were_ the same as everyone elses... >James Harris ************************ David C. Ullrich === 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. >Go ask him. > Strategic error here, James. You should really tell us what he said. > I mean surely Mazur must have had something positive to say, because > if his comments were simply negative like everyone elses then thered > be no point to your proudly telling us that hed commented on it, > right? But when you dont tell us what he said people are going to > come to the conclusion that his comments _were_ the same as > everyone elses... Go ask him. James Harris === Subject: Re: Reply from another German editor >> And what, exactly, did Barry Mazur say? > Damn, I was gonna ask that one. >> Well, James? >> Were really curious. > Go ask him. >> Strategic error here, James. You should really tell us what he said. >> I mean surely Mazur must have had something positive to say, because >> if his comments were simply negative like everyone elses then >> thered >> be no point to your proudly telling us that hed commented on it, >> right? But when you dont tell us what he said people are going to >> come to the conclusion that his comments _were_ the same as >> everyone elses... > Go ask him. No, James, thats what 5 year olds say (and repeat). Better say I dont want to talk about it. You dont wanna look like a kid, do you? === Subject: Re: 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. An algebraic integer is called a Perron number if all of its conjugates are smaller in absolute value. This isnt exactly what you want but you still might find the literature on these numbers useful. Suggest you search Math Reviews. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Attn: Jim Heckman - follow up question/clarification in message <2dm4b.55905$0u4.33846@news1.central.cox.net>: > 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. Well, in 4d the 3rd element of v_5 is the same as the 3rd element of v_4. Its obtained by dividing the 3rd element of v_3, namely sqrt(5/6), by -1/2. In general, the i-th coordinate of v_i is whatever positive number it takes to make v_i a unit vector, and the i-th coordinate of all subsequent vertices v_{i+1} through v_{n+1} is the i-th coordinate of v_i divided by -1/(n-i+1). Another way to think of it is that the vector sum of the n+1 unit-vector vertices must be 0. Youre welcome. > 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)) -- Jim Heckman === Subject: Re: Attn: Jim Heckman - follow up question/clarification > in message <2dm4b.55905$0u4.33846@news1.central.cox.net>: > 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. > Well, in 4d the 3rd element of v_5 is the same as the 3rd > element of v_4. Its obtained by dividing the 3rd element of > v_3, namely sqrt(5/6), by -1/2. > In general, the i-th coordinate of v_i is whatever positive > number it takes to make v_i a unit vector, and the i-th > coordinate of all subsequent vertices v_{i+1} through v_{n+1} is > the i-th coordinate of v_i divided by -1/(n-i+1). Sorry for the multiple posts, but is that last line is the i-th coordinate of v_i divided by -1/(n-i+1) supposed to say divided or multiplied? Craig === Subject: Re: Attn: Jim Heckman - follow up question/clarification > in message <2dm4b.55905$0u4.33846@news1.central.cox.net>: > 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. > Well, in 4d the 3rd element of v_5 is the same as the 3rd > element of v_4. Its obtained by dividing the 3rd element of > v_3, namely sqrt(5/6), by -1/2. Why is it divided by -1/2? Is that a constant that I missed, or is that calculated somehow? Craig === Subject: Re: Attn: Jim Heckman - follow up question/clarification > in message <2dm4b.55905$0u4.33846@news1.central.cox.net>: > 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. > Well, in 4d the 3rd element of v_5 is the same as the 3rd > element of v_4. Its obtained by dividing the 3rd element of > v_3, namely sqrt(5/6), by -1/2. > In general, the i-th coordinate of v_i is whatever positive > number it takes to make v_i a unit vector, and the i-th > coordinate of all subsequent vertices v_{i+1} through v_{n+1} is > the i-th coordinate of v_i divided by -1/(n-i+1). > Another way to think of it is that the vector sum of the n+1 > unit-vector vertices must be 0. > Youre welcome. > 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)) > -- > Jim Heckman Could you explain a simpler case, the 3d one: 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)) Based on the formula you gave before, I understand how to compute the 2nd elements of v_2 through v_4, but how is sqrt (2/3) obtained as the 3rd element for v_3? I apologize, I am just really confused. Craig === Subject: Re: Not Math but Should Be |It is perfectly correct mathematics to state that a result holds |provided certain conjecture holds or fails to hold. Probably a better example is the way physicists use informal calculations involving Feynman integrals and the like to arrive at mathematical results. Progress has been made in turning this kind of calculation into rigorous mathematics, but as far as I know they still have some work to do before their quantum field theory calculations are entirely math. Keith Ramsay === Subject: Re: Not Math but Should Be > If I am not mistaken, for instance, the Weil Conjectures were proven > by first showing that If the Riemann Hypothesis is true, then the > Weil Conjectures are true, and then proving that If the Riemann > Hypothesis is false, then the Weil Conjectures are true. I think the result you have in mind is the one that says that the class number of an imaginary quadratic number field goes to infinity with the discriminant. That one was first proved by proving it was a consequence of both the Riemann Hypothesis (or maybe it was the GRH) and of the negation of the Riemann Hypothesis. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Not Math but Should Be | |> If I am not mistaken, for instance, the Weil Conjectures were proven |> by first showing that If the Riemann Hypothesis is true, then the |> Weil Conjectures are true, and then proving that If the Riemann |> Hypothesis is false, then the Weil Conjectures are true. | |I think the result you have in mind is the one that says |that the class number of an imaginary quadratic number field |goes to infinity with the discriminant. That one was first |proved by proving it was a consequence of both the Riemann |Hypothesis (or maybe it was the GRH) and of the negation of |the Riemann Hypothesis. changes sign infinitely many times was first proven by proving it from RH and from the falsity of RH. Here pi(x) is the number of primes <=x, and li(x) is the logarithmic integral. Keith Ramsay === Subject: Re: Maximum Number of Smaller Cicles in Larger Circles? Here is very simple pro |Eckard Specht also sent me an asymptotic formula valid for large N | | D/d = 1.198765 * N^0.483330 | |with D, d = diameters of large and small circle, respectively. It turns out this is a curve-fit of part of his data for large N. So I would say the correct asymptotic formula is what I deduced before, D/d ~ (2*sqrt(3)*N/pi)^(1/2) or 1.050075... * N^0.5, corresponding to the density of the hexagonal packing of disks in the plane. In his data, the density has an upward trend, naturally, and the formula above fit it well for awhile, but of course it cant be extrapolated validly to where it implies a density of > 1. Keith Ramsay === Subject: more abstract algebra questions (from GRE subject) Please explain how to solve the following questions, 1) Let p and q be distinct primes. THere is a proper subgroup J of the additive group of integers which contains exactly three elements of the set {p,p+q,pq,p^q,q^p}. Which three elements are in J? (A) pq, p^q,q^p (B) p+q, pq, p^q (C) p,p+q,pq (D) p,p^q,q^p (E) p,pq, p^q 2) An automorphism phi of a field F is a one-to-one mapping of F onto itself such that phi(a + b) = phi(a) + phi(b) and phi(ab) = phi(a)phi(b) for all a, b in F. If F is the field of rational numbers, then the number of distinct automorphisms of F is (A) 0 (B) 1 (C) 2 (D) 4 (E) infinite === Subject: Re: more abstract algebra questions (from GRE subject) [...] > 2) An automorphism phi of a field F is a one-to-one mapping of F onto > itself such that phi(a + b) = phi(a) + phi(b) and phi(ab) = > phi(a)phi(b) for all a, b in F. If F is the field of rational numbers, > then the number of distinct automorphisms of F is > (A) 0 > (B) 1 > (C) 2 > (D) 4 > (E) infinite Hint 1: phi(a + b) = phi(a) + phi(b) -> phi(0) = 0 Hint 2: phi(ab) = phi(a)phi(b) -> phi(1) = 1 -- Jim Heckman === Subject: Re: Teaching ßawed information > And you can use math to show that Et = h (plancks constant h which is > the limit of human certaintiy is measured in energy * time). Does math mean mathematics? An you cannot do this, since energy and time are physical, not mathematical, concepts. -- 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: Teaching ßawed information |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. You might be entertained to figure out which algebraic numbers u have the property that u is contained in a subring R of the complex numbers, with the property that 1 and -1 are the only units in R, and that any integers m and n having no common factor greater than 1 in the integers also have no common factors in R other than units of R. (This is an exercise suggested by James definition of object ring. I dont see that he necessarily means to restrict only to algebraic numbers, though, and its looking like he might want number to be broader than complex number.) You might also enjoy determining whether the set of all such us is a ring. Keith Ramsay === 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? Analysis -- 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: 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. Bad advise, I dont JSH nor FLT. For ill humors, newsgroup alt.politics is more fun. === 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 I used a topological theorem. === Subject: Re: partial differential equations >> The first thing you should realize is that your f is not a >> function but an operator. If it were just a function you would >> have an ODE with a parameter x. So, you dont actually have a PDE >> at all. > Ok. >> There are _lots_ of methods that may be applicable to your problem. >> Without further information it is all but impossible to recommend >> one. > Well, everything I know is that u(x,t) is a probability distribution, > and the equation describes its time evolution. The only concrete > (in other cases I *will* have real PDEs...) or integral equations... --snip-- >> Another idea would be to transfer your problem to a true PDE by >> rearranging terms and differentiation. > I forgot to say that the integral is definite, between 0 and 1, so > unfortunately this would not work either... But this may be good news. It can make the treatment of your equations easier. Here is how to do it with the example you gave: d/dt u(x,t) = u(x,t) (int(x u(x,t),x) - x) with u(x,0) = u0(x). Now, int(x u(x,t),x) is a function depending only on t. So, for now I give it a new name: v(t) = int( x u(x,t), x=0..1) (*) and solve d/dt u(x,t) = u(x,t) (v(t) - x) u(x,0) = u0(x) This yields u(x,t) = u0(x) exp( - x t + V(t) ) (**) where V(t) = int( v(s), s=0..t) (i.e. V is the antiderivative of v with V(0) = 0). The function u given above will be a solution of your equation if it verifies the condition (*), which translates to v(t) = int( x u0(x) exp(- x t), x=0..1) exp(V(t)) which can be written as - d/dt ( exp( - V(t))) = int( x u0(x) exp(- x t), x=0..1) and so exp(- V(t)) = 1 + int( u0(x) exp( - x t), x =0..1). Putting this into (**) gives the solution: u(x,t) = u0(x) exp(-x t)/(1 + int(u0(x) exp(- x t), x=0..1)) HTH, Michael. -- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&& Dr. Michael Ulm FB Mathematik, Universitaet Rostock michael.ulm@mathematik.uni-rostock.de === Subject: Re: Memory requirement, prime counting clincher > [...] >> What tools are you using to track memory usage? > [...] [...] > 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... Currently, java.exe has 623,644 kbytes of virtual memory, or address space, committed to it. It is using 10,156 kbytes of physical memory. David === Subject: Re: Memory requirement, prime counting clincher > |No. Ackermans function is computable. Among > |the computable or recursive functions, there is a proper > |subset known as the primitive recursive functions. > | > |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. Yes, it comes very close to programming theory. > A non-primitive-recursive function seems to me to have a curious open-ended > quality to it. Indeed. Any function that is not primitive recursive is the other. Nothing new here. But the major point is that Ackermans function is computable. That means that the value is computed in finite time without being primitive recursive, and that for each possible input. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Memory requirement, prime counting clincher ... > I thought one *couldnt* program Ackermans function. Isnt its > significance due to the fact that its not computable? > I have a little Ackerman program that also counts > the recursions (the number of times A(x,y) is called). > Recursions for A(4,1) = 65533: 2,862,984,010 More interesting is recursion depth. It is the depth rather than the number of recursions that limit the function memory-wise. Knuth used this function as a man or boy test back in the sixties to determine whether a computer was a man or a boy. If I remember right, A(3,1) was boy, A(4,1) was man. Our computer (an Electrologica X-8 with 32,768 27-bit words) was a man. The recursion level depends on how you define it. But my memory must be faulty. I just checked, with tail recursion removed the function recursion depth is at most 65,531 for A(4,1). This would exceed the capabilities of the X8. My memory tells me that the recursion depth is proportional to the value calculated; and at least that seems to be right. (rec-depth(A(n,1) = A(n,1) - 2.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Telecommuting Math Jobs SchoolAnswers.com needs math experts for part-time telecommuting jobs. Part-time, up to $2,000 per month. Work from home and get paid via PayPal for solving math problems. To apply, send resume to: http://schoolanswers.com/become_tutor.jsp === Subject: Re: Factorials question > 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. It depends on what youre really doing. If you are using infiinite precision math then I agree, just doing the eact computation is probably just as fast (or even faster) than Stirlings approximation. But if all you care about is a handful of the most significant digits and the number of digits (i.e. using ßoating point calculations), then Stirlings takes just the constant number of clock cycles for those few operations (exponentiation, square root, division). (assuming of course that all the n you care about have n! less than the ßoating pt representation can handle. So if youre trying to work out probabilities on the ßy (with your pocket calculator), I think Stirling is probably a good idea. -- Mitch Harris Lehrstuhl fuer Automatentheorie, Fakultaet Informatik Technische Universitaet Dresden, Deutschland http://tcs.inf.tu-dresden.de/~harris === Subject: Jacobian If z = z(x,y) is a solution to p(x,y) z_x + q(x,y) z_y = 0 then for any differentiable function f u = f(z(x,y)) is a solution since u_x = f(z(x,y)) z_x(x,y) u_y = f(z(x,y)) z_y(x,y) p(x,y) u_x + q(x,y) v_y = f(z(x,y)) [p(x,y) z_x + q(x,y) z_y] = 0 Likewise: if z1,.. zk are solutions and f(x1,..xk) is a partially differentiable function, then another solution is f(z1(x,y),.. zk(x,y)). -- Conversely If u = u(x,y), v = v(x,y) are solutions, then p(x,y) u_x + q(x,y) u_y = 0 p(x,y) v_x + q(x,y) v_y = 0 u_x/u_y = -q(x,y)/p(x,y) = v_x/v_y and the Jacobian J = J(u,v/x,y) = u_x v_y - v_x u_y = 0. Were J /= 0, then u(x,y), v(x,y) would be invertible x = x(u,v), y = y(u,v) But as J = 0, can I conclude as u,v arent invertible, that theyre functionally dependent? That theres a differentiable function f with u(x,y) = f(v(x,y)) or v(x,y) = f(u(x,y)) ? -- valid conclusion ? If z = z(x,y) is a non-constant solution to p(x,y) z_x + q(x,y) z_y = 0 then for any other solution u = u(x,y) p(x,y) u_x + q(x,y) v_y = 0 theres differentiable function f with u = f(z(x,y)) ? ---- === 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... Its historical development. The modern slick version is all hindsight. === Subject: my question is~~~~ if a_n < 0 , then lim(root a_n) = root(lim a_n) ------------------------ it is possible?? i know that if a_n >= 0 , then lim(root a_n) = root(lim a_n) please~ advice for me, my doctor~ === Subject: Re: my question is~~~~ > if a_n < 0 , then lim(root a_n) = root(lim a_n) > ------------------------ > it is possible?? > i know that if a_n >= 0 , then lim(root a_n) = root(lim a_n) Yes, it is possible if root is an odd root such as the cube root, for example. It isnt possible if root is the square root (or any other even root) , unless you want to consider complex valued functions. _________________________________________________________ Eric J. Wingler (wingler@math.ysu.edu) Dept. of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, OH 44555-0001 330-941-1817 === Subject: Re: Mongolian BBQ Challenge http://mygate.mailgate.org/mynews/sci/sci.math/ 087dbe670e049818536ed925f3f61f 7f.48257%40mygate.mailgate.org > 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? Well, it isnt calculus, just algebra. I think most answerers have missed the real world thought that the combinations arent a matter of just choosing sauce sets and entre sets, but that each entre can be _separately_ spiced with different combinations of sauces, which probably makes the number of possible combinations a lot higher than the claim. The answers as given assume you are making soup, with everything stirred together, but that is not necessarily the case in the real world. xanthian. -- === 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? > Well, it isnt calculus, just algebra. > I think most answerers have missed the real world thought that the > combinations arent a matter of just choosing sauce sets and entre sets, > but that each entre can be _separately_ spiced with different > combinations of sauces, which probably makes the number of possible > combinations a lot higher than the claim. > The answers as given assume you are making soup, with everything stirred > together, but that is not necessarily the case in the real world. > xanthian. For the sake of reaching a total number of possible combinations, the assumption is that only 1 sauce may be included in a unique combination. For example, mongo + garlic sauce = mongo + garlic + garlic sauce. My friend thought the answer was reached by using derivatives. Given these parameters, what total to do you reach? === Subject: Re: Mongolian BBQ Challenge http://mygate.mailgate.org/mynews/sci/sci.math/ 9da1deea5f01b7c0228382b210422b 10.48257%40mygate.mailgate.org > For the sake of reaching a total number of possible combinations, the > assumption is that only 1 sauce may be included in a unique > combination. For example, mongo + garlic sauce = mongo + garlic + > garlic sauce. I have no clue what that first sentence means, and the example, with garlic in it twice, just doubles the confusion. > My friend thought the answer was reached by using derivatives. Your friend was trying to make an impression by using words he didnt understand. > Given these parameters, what total to do you reach? So far the problem isnt well enough defined to allow any total to be reached. xanthian. -- === 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? ... > I think most answerers have missed the real world thought that the > combinations arent a matter of just choosing sauce sets and entre sets, > but that each entre can be _separately_ spiced with different > combinations of sauces, which probably makes the number of possible > combinations a lot higher than the claim. > The answers as given assume you are making soup, with everything stirred > together, but that is not necessarily the case in the real world. As there are 512 possible sauce sets, separate saucing would lead to not less than C(512,1)+C(512,2)+...+C(512,17) possibilities which is about 2.5 x 10^31. This is about 10^22 times larger than the advertised count, so thats probably not it. -jiw === 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? Theres no URL for the restaurant, its printed on the front of their menu. I can give you the address if youd like to start a class action lawsuit... === 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? I made a number of grammatical errors and typos in my first reply, so I am reposting with corrections. This is how I figure it, which is not what I think the others replying meant. (controversy!....) ;) If I understand, you need to choose 1 main ingredient. There are 17 of these. So we have, to start, 17 first-to-be-picked main ingredients. Then we have 16 remaining ingredients, which each may occur or not in the combination. So we have 17* 2^16 possible combinations of main ingredients. And, we have 2^9 possible combinations of sauces, since any sauce may either occur or not in the combination. So, I get the total as: 17 *2^(16+9) = 570425344, which is NOT near 4.2 billion. Now, 4.2 billion IS a little bit under 2^32. So, if someone has not made a mistake in the math, perhaps someone (including me) has made a mistake in the interpretation of the problem. Leroy Quet === Subject: Re: Mongolian BBQ Challenge >Now, 4.2 billion IS a little bit under 2^32. They dont by any chance have six optional spices you can add? -- Richard -- Spam filter: to mail me from a .com/.net site, put my surname in the headers. FreeBSD rules! === 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? > I made a number of grammatical errors and typos in my first reply, so > I am reposting with corrections. > This is how I figure it, which is not what I think the others replying > meant. > (controversy!....) > ;) > If I understand, you need to choose 1 main ingredient. > There are 17 of these. > So we have, to start, 17 first-to-be-picked main ingredients. > Then we have 16 remaining ingredients, which each may occur or not in > the combination. > So we have 17* 2^16 possible combinations of main ingredients. > And, we have 2^9 possible combinations of sauces, since any sauce may > either occur or not in the combination. > So, I get the total as: > 17 *2^(16+9) = 570425344, which is NOT near 4.2 billion. > Now, 4.2 billion IS a little bit under 2^32. > So, if someone has not made a mistake in the math, perhaps someone > (including me) has made a mistake in the interpretation of the > problem. > Leroy Quet It appears that you violated the Prime Directive of Mongolian BBQ Command, which Ill repeat for your edification: >>Duplicate combinations in different orders do not >>count. For example,(chicken + noodes) is the same >>as (noodes + chicken). I think this will explain any discrepancy between your results and the others Dale. === Subject: Re: Mongolian BBQ Challenge <3F52AE2A.1050805@pacbell.net >>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? > This is how I figure it, which is not what I think the others replying > meant. > If I understand, you need to choose 1 main ingredient. > There are 17 of these. > So we have, to start, 17 first-to-be-picked main ingredients. > Then we have 16 remaining ingredients, which each may occur or not in > the combination. > So we have 17* 2^16 possible combinations of main ingredients. > And, we have 2^9 possible combinations of sauces, since any sauce may > either occur or not in the combination. > So, I get the total as: > 17 *2^(16+9) = 570425344, which is NOT near 4.2 billion. > It appears that you violated the Prime Directive of Mongolian BBQ > Command, which Ill repeat for your edification: > >>Duplicate combinations in different orders do not > >>count. For example,(chicken + noodes) is the same (Homer Simpson voice: Mmmm... noodes...) > >>as (noodes + chicken). Well, I can *almost* see Leroys answer as fitting the rules. Hes thinking something like this: Carnivore: I want chicken with broccoli. Apostate vegetarian: I want broccoli with chicken. These could be very different dishes! (I know I always want to order cashew nuts with chicken, myself...) But I think the stated problem is on the soup model: I want chicken and broccoli. I want broccoli and chicken. These two statements mean the same thing. However, I bet the real answer is that some main ingredients can be individually sauced. For example: General Tsos chicken with sweet sauce *plus* beef with lemon sauce would be different from chicken with lemon sauce *plus* beef with sweet sauce. If thats done indiscriminately, then we get way more than 4.2 billion combinations, though, so there must be more rules involved, that we dont know about yet. Rules like Nobody orders just plain rice with lemon sauce *and* soy sauce, because thats just nasty. Rules like that. -Arthur === 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? > > > I made a number of grammatical errors and typos in my first reply, so > I am reposting with corrections. > > > This is how I figure it, which is not what I think the others replying > meant. > (controversy!....) > ;) > > If I understand, you need to choose 1 main ingredient. > There are 17 of these. > > So we have, to start, 17 first-to-be-picked main ingredients. > Then we have 16 remaining ingredients, which each may occur or not in > the combination. > So we have 17* 2^16 possible combinations of main ingredients. > And, we have 2^9 possible combinations of sauces, since any sauce may > either occur or not in the combination. > > So, I get the total as: > > 17 *2^(16+9) = 570425344, which is NOT near 4.2 billion. > > Now, 4.2 billion IS a little bit under 2^32. > So, if someone has not made a mistake in the math, perhaps someone > (including me) has made a mistake in the interpretation of the > problem. > > Leroy Quet > It appears that you violated the Prime Directive of Mongolian BBQ > Command, which Ill repeat for your edification: > >>Duplicate combinations in different orders do not > >>count. For example,(chicken + noodes) is the same > >>as (noodes + chicken). > I think this will explain any discrepancy between your results > and the others > Dale. I appreciate the responses. Let me try and further clarify, esp. for those of you who dont frequent Mongolian BBQs. When it comes to the 17 main ingredients, you can select 1, or 17, or any number in between. (For example, you can pick chicken --or-- chicken, mushrooms, brocolli, and onion --or-- all 17 items. In addition to the main ingredients, you can add any combination of 0-9 sauces. The sauces alone dont constitute a meal, but expand the total number of combinations possible when added to one or more main ingredients. For example, you could select all the main indgredients and all the sauces for a total of 26 items. Again, duplicates must be excluded in that different sequences of the same items dont represent unique combinations. That is, chicken + beef is the same as beef + chicken. My colleague, an electrical engineering graduate, is quite sure that the 4.2 billion number is accurate. I dont have a strong math background, but just dont see Carl Sagans eloquent articulation of the concept of a billion in a lowly Mongolian BBQ... Jon === Subject: Re: Mongolian BBQ Challenge >I appreciate the responses. Let me try and further clarify, esp. for >those of you who dont frequent Mongolian BBQs. When it comes to the >17 main ingredients, you can select 1, or 17, or any number in >between. (For example, you can pick chicken --or-- chicken, >mushrooms, brocolli, and onion --or-- all 17 items. >In addition to the main ingredients, you can add any combination of >0-9 sauces. The sauces alone dont constitute a meal, but expand the >total number of combinations possible when added to one or more main >ingredients. For example, you could select all the main indgredients >and all the sauces for a total of 26 items. >Again, duplicates must be excluded in that different sequences of the >same items dont represent unique combinations. That is, chicken + >beef is the same as beef + chicken. >My colleague, an electrical engineering graduate, is quite sure that >the 4.2 billion number is accurate. I dont have a strong math >background, but just dont see Carl Sagans eloquent articulation of >the concept of a billion in a lowly Mongolian BBQ... Your friend is quite wrong. As others have pointed out, there are 2^17 -1 possible combinations of main ingredients and 2^9 possible combination of sauces. This makes (2^17 -1) 2^9 = 67,108,352 possible dishes. This analysis follows from the fact that there are 2^n distinct subsets (including the empty set) of a set with n elements. Are you leaving out details about side dishes? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Mongolian BBQ Challenge >>My colleague, an electrical engineering graduate, is quite sure that >>the 4.2 billion number is accurate. I dont have a strong math >>background, but just dont see Carl Sagans eloquent articulation of >>the concept of a billion in a lowly Mongolian BBQ... > Your friend is quite wrong. As others have pointed out, there are 2^17 > -1 possible combinations of main ingredients and 2^9 possible > combination of sauces. This makes (2^17 -1) 2^9 = 67,108,352 > possible dishes. > This analysis follows from the fact that there are 2^n distinct > subsets (including the empty set) of a set with n elements. Are you > leaving out details about side dishes? At the Mongolian BBQ I go to, after one adds sauces, then there are sesame seeds, peanuts (at least two types), and those red things that look like nuts but arent and maybe some other things one can add. Bart === 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? This is how I figure it, which is not what I think the others replying meant. (controversy!....) ;) If I understand, you need to choose 1 main ingredient. There are 17 of these. So we have, to start, 17 first-to-be-picked main ingredients. Then we have 16 remaining ingredients, which may each all occur or not in the combination. So we have 17* 2^16 possible combinations of main ingredients. And, we have 2^9 possible combinations of sauces, since sauce may either occur or not in the combination. So, I get the total as: 17 *2(16+9) = 570425344, which is NOT near 4.2 billion. Now, 4.2 billion IS a little bit under 2^32. So, if someone has not made a mistake in the math, perhaps someone (including me) has made a mistake in the interpretation of the problem. Leroy Quet === Subject: Re: Mongolian BBQ Challenge >So we have, to start, 17 first-to-be-picked main ingredients. >Then we have 16 remaining ingredients, which may each all occur or not >in the combination. >So we have 17* 2^16 possible combinations of main ingredients. Thats not right - youve overcounted combinations by a lot. Doug === Subject: Re: Mongolian BBQ Challenge >> A single combination must consist of at least 1 main ingredient (e.g. >> chicken), up to 17 plus 0 to 9 sauces. >Youve been cheated. There are 2^17 ways of choosing the main >ingredients, and 2^10 ways of choosing the sauces, Surely 2^17-1, and 2^9. -- Richard -- Spam filter: to mail me from a .com/.net site, put my surname in the headers. FreeBSD rules! === 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? (2**17) different main-ingredient combos, but one of them (the null combo) isnt allowed. So (2**17)-1 main ingredient combos. (2**9) different sauce combos. ((2**17)-1) * (2**9) total ingredient-sauce combos, for a grand total of 67108352, or just a little over 67 million dishes. Nowhere near a billion. Maybe theyre including side dishes, too? :-) -Arthur === 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? > (2**17) different main-ingredient combos, but one of them (the null > combo) isnt allowed. So (2**17)-1 main ingredient combos. > (2**9) different sauce combos. > ((2**17)-1) * (2**9) total ingredient-sauce combos, for a grand total > of 67108352, or just a little over 67 million dishes. Nowhere near > a billion. > Maybe theyre including side dishes, too? :-) > -Arthur Arthur, If they were to include side dishes, and I added those to the list of main ingredients, the number would increase from 17 to 21. In that case, whats the possible number of combinations with 21 main ingredients and 9 sauce options.... === 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. The other thing is, when you restrict your calculations to dx = 1 youre not testing _anything_ about the behavior of the actual pde - in particular youre not obtaining _any_ evidence for your stupid claims about how the solution to the pde is a better approximation to pi(x) than various other things. Remember the example? Consider the recurrence [1] a[0] = 1, a[n+1] - a[n] = a[n] versus the corresponding differential equation [2] f(0) = 1, f = f. Weve noted many times that the solution to [2] has nothing to do with the solution to [1]. But suppose you attempt to numerically calculate the solution to [2] using, say, Eulers method with a step size of dx = 1. The approximation you get to the solution to [2] is exactly the solution to [1] (and hence this approximation is totally worthless in terms of evidence regarding the behavior of the _actual_ solution to [1].) Its fascinating. Incredible genius finds pde describing pi(x). How does he verify that the solution is actually a good approximation to pi(x)? By _ignoring_ the pde and going back to the difference equation... >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 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; ************************ David C. Ullrich === Subject: Re: My prime research, focus on a feature REVISED > In sci.physics, James Harris > 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. > 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] Actually my prime counting function is faster than Legendres Method, which shows you how badly posters lied. Still Legendres Method is quite slow, which is why mathematicians worked for so many years on various improvements. Yet from my prime counting function I found fast algorithms in only a few months. My efficiency blows the entire math world away, and Im just one man. If math professors werent so damn sensitive theyd acknowledge my work, so I could move on to other things instead of trying to convince you people. James Harris === Subject: Re: My prime research, focus on a feature REVISED In sci.physics, James Harris In sci.physics, James Harris >> 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. >> 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] > Actually my prime counting function is faster than Legendres Method, > which shows you how badly posters lied. It probably is, since Legendres Formula (according to http://mathworld.wolfram.com/LegendresFormula.html anyway) requires p_i to drive it (p_1 = 2, p_2 = 3, etc.). The recurrence relation mentioned: phi(x,n) = phi(x, n-1) - phi(x / p_n, n - 1) may be of some use to somebody, with the initial conditions phi(x, 0) = ßoor(x) for any x >= 0 and phi(x, n) = 0 for any x < 2 and n >= 0. Im estimating at least x^(1/2) space complexity, mostly because of the prime table; however, initial runs suggest far more than that. Meissels or Lehmers formulas might yield better results but would still require the prime table. The author asserts phi(x,x) = pi(x); I dont know if phi(x,sqrt(x)) would work better or not. It will be interesting to see how well my implementation works; it basically simply maintains a small vector of primes and uses the obvious recursion; if it needs a new of it yours beats it handily, but I had low expectations anyway. :-) Hell, the bloody *precalc* of all primes up to 1,000,000, when optimized, took less time than my unoptimized sieve implementation, making Legendres method of largely theoretical interest anyway. (Yes, it got the right answer.) For very large n pi(n) ~ li(n), where li(n) = integ(t = mu to n)(dt/ln t), where mu is Soldners constant, equal to approximately 1.4513692346... . Presumably this could be expanded as an infinite series in 1/n, with the usual attendant ugliness. > Still Legendres Method is quite slow, which is why mathematicians > worked for so many years on various improvements. If one wants pi(n) the last thing one wants to compute is p_n, which is much bigger than n. > Yet from my prime counting function I found fast algorithms in only a > few months. > My efficiency blows the entire math world away, and Im just one man. Your time efficiency still needs work; my sieve blows your system out of the water. Granted, I have more space complexity but, given a gigabyte of memory I can theoretically compute 24 billion primes. [*] And a gig is cheap. A sieve is at most of order O^(N^(1.5)) in time complexity. (The mesh is of size N and has to be run through at most sqrt(N) times. In fact, the mesh is run through pi(sqrt(N)) times -- plus 1 for the actual printout.) What is the time complexity of your algorithm? > If math professors werent so damn sensitive theyd acknowledge my > work, so I could move on to other things instead of trying to convince > you people. Im convinced your algorithm *works*. It just doesnt work *well*. > James Harris [*] by prefiltering out multiples of 2 and 3; all subsequent primes are of the form 6k-1 or 6k+1 for some k >= 1. -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: My prime research, focus on a feature REVISED > It probably is, since Legendres Formula (according to > http://mathworld.wolfram.com/LegendresFormula.html > anyway) requires p_i to drive it (p_1 = 2, p_2 = > 3, etc.). The recurrence relation mentioned: > phi(x,n) = phi(x, n-1) - phi(x / p_n, n - 1) > may be of some use to somebody, with the initial conditions > phi(x, 0) = ßoor(x) for any x >= 0 and phi(x, n) = > 0 for any x < 2 and n >= 0. Im estimating at least > x^(1/2) space complexity, mostly because of the prime > table; however, initial runs suggest far more than that. > Meissels or Lehmers formulas might yield better results > but would still require the prime table. The author > asserts phi(x,x) = pi(x); I dont know if phi(x,sqrt(x)) > would work better or not. The recurrence relation is _exactly_ what you should use in an implementation, it very nicely adds up all the terms in Legendres formula and conveniently leaves those out that evaluate to 0. Note that the definition of Legendres formula on the webpage you quote seems completely messed up: First, phi (x, a) is correctly defined as the number of integers <= x that are not divisible by the first primes; the summation formula then doesnt write how far the sums are supposed to go. In the second formula, you dont choose a = x, that is nonsense; you choose a = pi (ßoor (sqrt (x))). Once you fix the obvious bugs on the webpage, you get Legendres formula = Harris formula. As to space requirements: Who says that you should use ONE sieve to find all the primes <= sqrt (x) ? Use a sieve to find all primes <= x ^ (1/3), then observe that all primes > x^(1/3) are used only once if you use the recursion formula, so there is no need to store them. Use several sieves of size x^(1/3) to find those primes, and your space requirement is O (N^(1/3)). === Subject: Re: My prime research, focus on a feature REVISED > It probably is, since Legendres Formula (according to > > http://mathworld.wolfram.com/LegendresFormula.html > > anyway) requires p_i to drive it (p_1 = 2, p_2 = > 3, etc.). The recurrence relation mentioned: > > phi(x,n) = phi(x, n-1) - phi(x / p_n, n - 1) > > may be of some use to somebody, with the initial conditions > phi(x, 0) = ßoor(x) for any x >= 0 and phi(x, n) = > 0 for any x < 2 and n >= 0. Im estimating at least > x^(1/2) space complexity, mostly because of the prime > table; however, initial runs suggest far more than that. > Meissels or Lehmers formulas might yield better results > but would still require the prime table. The author > asserts phi(x,x) = pi(x); I dont know if phi(x,sqrt(x)) > would work better or not. > The recurrence relation is _exactly_ what you should use in an > implementation, it very nicely adds up all the terms in Legendres > formula and conveniently leaves those out that evaluate to 0. > Note that the definition of Legendres formula on the webpage you quote > seems completely messed up: First, phi (x, a) is correctly defined as > the number of integers <= x that are not divisible by the first primes; > the summation formula then doesnt write how far the sums are supposed > to go. In the second formula, you dont choose a = x, that is nonsense; > you choose a = pi (ßoor (sqrt (x))). > Once you fix the obvious bugs on the webpage, you get Legendres formula > = Harris formula. And notice that Christian Bau is back to his lie. The facts havent changed as readers can *still* look at http://mathworld.wolfram.com/PrimeCountingFunction.html and go down to that table with the headers method, time complexity and storage complexity and readers can still see that my prime counting function is off the charts in terms of memory usage in terms of being *low*. But Usenet allows a dedicated liar like Christian Bau to just keep coming back as hes done now for months. > As to space requirements: Who says that you should use ONE sieve to find > all the primes <= sqrt (x) ? Use a sieve to find all primes <= x ^ > (1/3), then observe that all primes > x^(1/3) are used only once if you > use the recursion formula, so there is no need to store them. Use > several sieves of size x^(1/3) to find those primes, and your space > requirement is O (N^(1/3)). And clearly he *knows* hes lying as he tries to dance around the memory issue. The fact remains that my prime counting function is unique in the math world. A poster like Christian Bau who hasnt a major achievement to his name can come on Usenet and tell lies, and I guess get some satisfaction that way. While mathematicians behave oddly, Im posting on Usenet, so readers can have the fun of posts like this one. Usenet makes it possible. James Harris === Subject: Re: My prime research, focus on a feature REVISED can come on Usenet and tell lies, and I guess get some satisfaction >that way. >While mathematicians behave oddly, Im posting on Usenet, so readers >can have the fun of posts like this one. >Usenet makes it possible. >James Harris Sorry. But why do you post your research work on usenet, if you dont want people to comment on it? As much as I have followed this here, you have never prooven your work right, while he has not proven it wrong either. Criticism is a good thing, no matter if its against you, or for you, as it helps as all to develop. You could just as well publish your work in scientific journals, but there will be no less fighting against your ideas. Just my two cent, and Im sorry for them. CMW === Subject: Re: My prime research, focus on a feature REVISED Followups reset. In sci.physics, Christian Bau It probably is, since Legendres Formula (according to >> http://mathworld.wolfram.com/LegendresFormula.html >> anyway) requires p_i to drive it (p_1 = 2, p_2 = >> 3, etc.). The recurrence relation mentioned: >> phi(x,n) = phi(x, n-1) - phi(x / p_n, n - 1) >> may be of some use to somebody, with the initial conditions >> phi(x, 0) = ßoor(x) for any x >= 0 and phi(x, n) = >> 0 for any x < 2 and n >= 0. Im estimating at least >> x^(1/2) space complexity, mostly because of the prime >> table; however, initial runs suggest far more than that. >> Meissels or Lehmers formulas might yield better results >> but would still require the prime table. The author >> asserts phi(x,x) = pi(x); I dont know if phi(x,sqrt(x)) >> would work better or not. > The recurrence relation is _exactly_ what you should use in an > implementation, it very nicely adds up all the terms in Legendres > formula and conveniently leaves those out that evaluate to 0. > Note that the definition of Legendres formula on the webpage you quote > seems completely messed up: First, phi (x, a) is correctly defined as > the number of integers <= x that are not divisible by the first primes; > the summation formula then doesnt write how far the sums are supposed > to go. In the second formula, you dont choose a = x, that is nonsense; > you choose a = pi (ßoor (sqrt (x))). > Once you fix the obvious bugs on the webpage, you get Legendres formula >= Harris formula. > As to space requirements: Who says that you should use ONE sieve to find > all the primes <= sqrt (x) ? Use a sieve to find all primes <= x ^ > (1/3), then observe that all primes > x^(1/3) are used only once if you > use the recursion formula, so there is no need to store them. Use > several sieves of size x^(1/3) to find those primes, and your space > requirement is O (N^(1/3)). Well, somethings still messed up. I feel like a schoolkid again... :-) Heres *my* program, which is fairly fast but gives out the wrong answer. (The source should be fairly obvious even though Im using STL.) ----------8< >8---------- #include #include #include std::vector primeArray; // Routine to compute the ixth prime. static unsigned prime(unsigned ix) { while(ix >= primeArray.size()) { // glory be, weve run out of primes! unsigned last = primeArray[primeArray.size()-1]; bool foundPrime = false; while(!foundPrime) { if(last % 6 == 1) last += 4; else last += 2; // dont bother testing 2 and 3 for(unsigned j = 3; j < primeArray.size(); j++) { if(primeArray[j] * primeArray[j] > last) { // weve run the gauntlet and know Ôlast is prime primeArray.push_back(last); foundPrime = true; break; } if(last % primeArray[j] == 0) { break; // not prime, sorry } } } } return primeArray[ix]; } // Routine to compute the number of primes less than or equal to lim // This routine has a number of latent bugs which are not elicited // by the rest of the implementation; a more general routine would // use a binary search if the last prime is greater than the limit. static unsigned stupidpi(unsigned lim) { while(primeArray[primeArray.size()-1] < lim) prime(primeArray.size()); return primeArray.size() - 2; } // Legendres function. static double phi(double x, unsigned n) { if(x < 2) return 0; if(n == 0) return ßoor(x); // Since the primes are strictly monotonic, if we know // the next iteration is going to be phi(y,something) where // y < 2, dont bother explicitly computing prime(n). if(n > primeArray.size() && primeArray[primeArray.size() - 1] > x/2) return phi(x, n-1); return phi(x, n-1) - phi(x / prime(n), n-1); } // Main program. int main(int argc, char **argv) { primeArray.push_back(1); // p_0 is not defined so it doesnt matter anyway primeArray.push_back(2); primeArray.push_back(3); primeArray.push_back(5); int a = stupidpi(1000); std::cout << Precalculation done; a = << a << std::endl; std::cout << phi(1000000,a) << std::endl; std::cout << primeArray.size() << << primeArray[primeArray.size()-1] << std::endl; int b = stupidpi(1000000); std::cout << b << std::endl; return 0; } ----------8< >8---------- It gives me the wrong answers, so Ive obviously missed something. Sigh. $ legendrephi Precalculation done; a = 168 81142 170 1009 78498 $ sieve 1000000 1000000 78498 $ -- #191, ewill3@earthlink.net -- but at least its fast :-) Its still legal to go .sigless. === Subject: Re: My prime research, focus on a feature REVISED > static double phi(double x, unsigned n) > if(x < 2) return 0; This is not correct. phi (x, n) is the number of integers 1 <= i <= x which are not divisible by the first n primes. 1 is not divisible by any prime, so if x = 1 then phi (x, n) = 1. === Subject: Re: My prime research, focus on a feature REVISED >Followups reset. >In sci.physics, Christian Bau >> It probably is, since Legendres Formula (according to > > http://mathworld.wolfram.com/LegendresFormula.html > > anyway) requires p_i to drive it (p_1 = 2, p_2 = > 3, etc.). The recurrence relation mentioned: > > phi(x,n) = phi(x, n-1) - phi(x / p_n, n - 1) > > may be of some use to somebody, with the initial conditions > phi(x, 0) = ßoor(x) for any x >= 0 and phi(x, n) = > 0 for any x < 2 and n >= 0. Im estimating at least > x^(1/2) space complexity, mostly because of the prime > table; however, initial runs suggest far more than that. > Meissels or Lehmers formulas might yield better results > but would still require the prime table. The author > asserts phi(x,x) = pi(x); I dont know if phi(x,sqrt(x)) > would work better or not. >> The recurrence relation is _exactly_ what you should use in an >> implementation, it very nicely adds up all the terms in Legendres >> formula and conveniently leaves those out that evaluate to 0. >> Note that the definition of Legendres formula on the webpage you quote >> seems completely messed up: First, phi (x, a) is correctly defined as >> the number of integers <= x that are not divisible by the first primes; >> the summation formula then doesnt write how far the sums are supposed >> to go. In the second formula, you dont choose a = x, that is nonsense; >> you choose a = pi (ßoor (sqrt (x))). >> Once you fix the obvious bugs on the webpage, you get Legendres formula >>= Harris formula. >> As to space requirements: Who says that you should use ONE sieve to find >> all the primes <= sqrt (x) ? Use a sieve to find all primes <= x ^ >> (1/3), then observe that all primes > x^(1/3) are used only once if you >> use the recursion formula, so there is no need to store them. Use >> several sieves of size x^(1/3) to find those primes, and your space >> requirement is O (N^(1/3)). >Well, somethings still messed up. I feel like a schoolkid again... :-) >Heres *my* program, which is fairly fast but gives out the >wrong answer. (The source should be fairly obvious even >though Im using STL.) Note that phi(x,pi(sqrt(x)) = pi(x) - pi(sqrtx) + 1 To calculate pi(x) from phi (x) you can use the recursion pi(x) = phi(x,pi(sqrt(x)) + pi(sqrt(x)) - 1 The only difference between what James does and the recursion on the webpage, is that he got rid of this extra recursion. This isnt good for execution time as now EVERY invocation of pi(x,y) is more complicated, wheras with doing it with legendres phi, calculating pi(sqrt(x)) takes only a tiny fraction of the time of pi(x) -- Wim Benthem === Subject: Re: Catalan sequence problem > The number of ways to partition a set of n elements into clusters > is given by the Bell number B(n). > > Now suppose that these n elements are cities lying on a circle, > and that roads are built inside the circle. > > Any placement of roads partitions the n cities into clusters > of inter-driveable cities. How many such partitions are there? > > (Overpasses are not allowed: if they were, the answer would just > be B(n).) > > For n = 1, 2, 3, 4, and 5 there are 1, 2, 5, 14, and 42 partitions, > respectively. Highly suggestive, but I couldnt come up with a > mapping between this problem and one of the standard Catalan number > problems. >> Take a regular n-polygon and dissect it fully (that is put in as many >> non crossing segments as possible). (I think those are the appropriate >> formally defined technical terms for what you are asking about). > This is a different problem. Oh. reading comprehension. fitting the answer to the request. Sorry, lets try again. > For example, one of the partitions has > no line segments (namely, {{1},{2},{3},...,{n}}). Let me give examples > of what I mean. > n C_n B(n) > 1 1 1 > 2 2 2 > 3 5 5 > 4 14 15 > 5 42 52 > For n = 4, the only partition of {1,2,3,4} disallowed is {{1,3},{2,4}}. > There is no way to connect 1 to 3 and 2 to 4 without connecting 1-3 to > 2-4. For n = 5 the disallowed partitions are {{1,3},{2,5}} + 4 symmetric > cases and {{1,3,4},{2,5}} + 4 symmetric cases, giving 52 - 10 = 42 > allowable partitions. > Im not convinced that this is, in fact, a Catalan sequence, but so > many things are. OK, this sounds close to noncrossing partitions. Which are yet another Catalan counted structure (with all sorts of bijections with other Catalan structures) problem pp from Stanley (with only references, no explanation)). However, the cyclic nature of your description might make a difference. But then it should show up with n = 5. Er... no it doesnt (rethinking the definitions), because the partitions can enclose another without crossing: e.g. {{1,2,5},{3,4}} is not crossing. Im thinking about what the bijection really should be (I have seen it before but cant remember, and Im having trouble with obvious possibilities). Maybe DFS labelling on an ordered tree? > Usually Id write a Mathematica program to get an unambiguously > long sequence and then Sloane it, but our license server is down. Mathematica is a crutch (speaking form experience). Anyway. look through Sloanes page for the Catalan #s at the EIS for references and possibly clues. >> If you ever have a structure that looks like it might be catalan like, >> look in here for the correspondence: >> http://www-math.mit.edu/~rstan/ec/catalan.ps.gz >> http://www-math.mit.edu/~rstan/ec/catsol.ps.gz > Wow. Presumably no one has ever been sadistic enough to assign > problem 19 in its full glory: prove these 66 sets have the same > number of elements by exhibiting bijections between each pair. Or even better come up with a new one. -- Match Harris Lehrstuhl fuer Automatentheorie, Fakultaet Informatik Technische Universitaet Dresden, Deutschland http://tcs.inf.tu-dresden.de/~harris === Subject: Re: quaternions-- whats the point? >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. >You make my point. So how would YOU describe the universal covering group for SO(4,2), which covers SO(4,2) infinitely many times? >I am not arguing with your mathematics, >just whether it is a wise approach to the subject. >As far as I can see, you use the term classical group >to mean any Lie group with a simple Lie algebra. No, I dont mean that. I use the term classical in the same way that the word has been used for decades (see, for instance, Hermann Weyls book about the Classical Groups). The simple complex classical groups are the groups in the four infinite sequences (types A, B, C, D). There are five complex simple Lie groups (up to local isomorphism) which are not classical (seven complex simple Lie groups up to isomorphism: E_6, E_6/Z_3, E_7, E_7/Z_2, E_8, F_4 and G_2). These Lie groups are called exceptional. Also, why do you think that authors like Jacobson, Gilmore and Barut take so much trouble to describe the real forms of simple classical groups, noncompact as well as compact? >As your remarks above indicate, this leads one into a morass. The study of real forms of the complex simple Lie groups is important. >Since in practice one is nearly always interested >in representations of these groups over C, >rather than the groups themselves, But one is typically interested in representations of the REAL groups. As I previously noted, Sp(2n,R), SO(3,1) and SO(4,2) are important for physics, and the unitary representations of these groups are important. So it is not enough just to have representations for Sp(2n,C), SO(4,C) and SO(6,C), but these representations must also be unitary representations of Sp(2n,R), SO(3,1) and SO(4,2) (and therefore infinite-dimensional). Further, why should a unitary representation for any of these real groups extend to a representation of the complex forms of the groups? Particularly, one has to consider multi-valued unitary representations of these real Lie groups, and the question of whether these representations can be extended to multi-valued representations of the complex forms. >it is simpler and more appropriate to consider >the four families of simple Lie algebras over C. Not for practical applications like physics, where the appearance of groups like Sp(2n,R), SO(3,1) and SO(4,2) is more important than the fact that their complex forms are Sp(2n,C), SO(4,C) and SO(6,C). The symmetry group for the isotropic simple harmonic oscillator is isomorphic to U(n), not to GL(n,C). The symmetry group of the bound states of the nonrelativistic hydrogen atom is SO(4), not SO(4,C), so that the representation theory uses unitary representations of SO(4). The symmetry group of the ionized states of the hydrogen atom is SO(3,1), not SO(4,C), so that the representation theory uses unitary representations of SO(3,1). The bound states of the hydrogen atom form a unitary representation of SO(4,2). The ionized states of the hydrogen atom form a unitary representation of SO(4,2). The symmetry group for the hyperbolic plane is isomorphic to SO(2,1), not to SO(3,C). It is not more appropriate to restrict attention to the groups and algebras over C and their compact real forms (and yes, I do know that U(n) and SO(4) are compact). I should also mention here that I first mentioned these groups in response to a suggestion that quaternions are no longer of use, i.e. by pointing out that ceratin real simple Lie groups are most easily described using quaternions. >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. >There is nothing artificial about compactness. I didnt say that compactness is artificial. I said that the restriction of consideration to the compact real forms (especially to the exclusion of the noncompact real forms) is artificial. Please dont put words into my mouth. David McAnally -------------- === Subject: Re: quaternions-- whats the point? > it seems to me that a quaternion is nothing more than a simple > ordered quadruplet. And that multiplication/addition is simply > extended in such a way that it is defined over such ordered > quadruplets. Its as if I were to write a geometry book where I > called points binarinions and write them as a#{(b}) instead of > (a,b).. the point being that it would be arbitrary and accomplish > nothing and, in fact, greatly obfuscate something for no reason. > This, it seems, is the entire nature of quaternions. > I assume that when you say quaternions you are referring to elements of > the form a+bi+cj+dk. Well, the set of complex numbers is basically > nothing more than R^2 with a specific multiplication- namely that > (a,b)*(c,d)=(ac-bd,ad+bc). So in saying that the quaternions are nothing > more than ordered quadruplets with a special multiplication you are > right. However, it seems to me that the i,j,k notation is much simpler > to deal with. If you didnt know that (a,b)*(c,d)=(ac-bd,ad+bc) in the > complex numbers off hand, you could easily arrive at that by multiplying > (a+bi)*(c+di) and knowing that i^2=-1. Similarly for the quaternions. > SC Because, 203 = 4 + 9 + 16 + 25 + 36 + 49 + 64 = 2^2 + 3^3 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 and -1 = ( i^2 + j^2 + k^2 + E^2 + I^2 + J^2 + K^2 ) / 7 and 203 = 9 + 25 + 169 = 3^2 + 5^2 + 13^2 and -1 = ( i^2 + j^2 + k^2 ) / 3 The quaternions are a normed division algebra, one of only four such number systems, reals, complex, quaternion, and octonion, which have 1, 2, 4, 8 dimensions with 0, 1, 3, 7, imaginary number bases, and someone long ago knew that there would come a time when human imagination and inventiveness would be generating all sorts of alternative possible theories to explain the universe, but there were only these four systems that were needed to develop physical theory. So they built a big monument of stone and stuck it in the desert in Egypt and built it out of square platforms of stone added upon other square platforms to signify the importance adding the squares of numbers. And then, they made the height exacty 203 steps tall, which number happens to be the sum of three squares and also the sum of seven squares, to signify the imaginary units of the quaternions and octonions, and the relationship to time which is the vertical verses space which is horizontal. And they even broke off the capstone, which is a similar shape to the whole pyramid again, and made the capstone linear size one part in 16 of the pyramids inorder to tell us that when you get to the 16 dimension numbers the algebra structure is broken, and the shape of the pyramid tellus about inversions and rotations which are performed by multiplications, so we know that the breakdown in the algebra of issue is with that multiplicative structure to signify the importance of the idea, because its hard to see while you are on the path of discovery and dont already know all the facts and results that come out of the applications of the theory. With that compas to guide us we can follow its suggestion, or go our own way and try to see if we can find something better ourselves. pmj http://www.hypercomplex.com === Subject: Re: quaternions-- whats the point? >>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. You make my point. I am not arguing with your mathematics, just whether it is a wise approach to the subject. As far as I can see, you use the term classical group to mean any Lie group with a simple Lie algebra. As your remarks above indicate, this leads one into a morass. Since in practice one is nearly always interested in representations of these groups over C, rather than the groups themselves, it is simpler and more appropriate to consider the four families of simple Lie algebras over C. >>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. There is nothing artificial about compactness. -- Timothy Murphy e-mail: tim@birdsnest.maths.tcd.ie tel: +353-86-233 6090 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Data analysis software > It plots and analyses any x-y data for peak location, peak height, peak > width, semi-derivative, derivative, integral, semi-integral, > convolution, > deconvolution, curve fitting, and separating overlapped peaks and > background. > > www.chemSoftware.com === Subject: Re: [JSH] Polynomials and 2^sqrt{3} > Arturo Magidin >>Nonsense: surely it is reasonable to associate 2^{sqrt(3)} with the >>polynomial x-2^{sqrt(3)}! > The word rational seems to have gone missing from my post :) This whole time, I was reading 2^{sqrt{3}} as the cube root of 3. and wondering about others sanity (because if I doubt my own, that would certainly be rational). Note to self: learn to read. Mitch === Subject: Counting complex roots in a region? Using Sturm sequences, one can count exactly the number of real roots of a polynomial in a given region. Is there a related method for counting complex roots in a given complex region? Mitch === Subject: Re: Counting complex roots in a region? > Using Sturm sequences, one can count exactly the number of real roots of > a polynomial in a given region. > Is there a related method for counting complex roots in a given complex > region? You can evaluate the contour integral p(x)/p(x)/(2pi) numerically. I think it is recommended to use a circle (region=disc) and take the trapez-rule. Of course you may stop as soon as the integer result is manifest hth klaus > Mitch === Subject: Re: Counting complex roots in a region? >>Using Sturm sequences, one can count exactly the number of real roots of >>a polynomial in a given region. >>Is there a related method for counting complex roots in a given complex >>region? > You can evaluate the contour integral p(x)/p(x)/(2pi) numerically. I > think it is recommended to use a circle (region=disc) and take the > trapez-rule. Of course you may stop as soon as the integer result is manifest I have trouble seeing how this is similar to the Sturm sequence method. I would think instead of an integral, youd be taking residues (remainders) of that division, or rather of p(x)/p(x). Basically, Im looking for something that doesnt use numerical approx. (note I just want a procedure to -count- the roots, Ill use some sort of 2-d binary search to isolate the roots). Also note that I have little idea how the contour integral is working in your idea, and so I could very well be missing the obvious entirely. Mitch === Subject: Re: Counting complex roots in a region? >Using Sturm sequences, one can count exactly the number of real roots of >a polynomial in a given region. >Is there a related method for counting complex roots in a given complex >region? If theres not an algorithm for doing that in Henricis _Applied and Computational Complex Analysis_, Id be very surprised (but its been a long time since Ive looked at the book). A lot depends on how your given complex region is given, I suppose. Lee Rudolph === Subject: Re: Counting complex roots in a region? >>Using Sturm sequences, one can count exactly the number of real roots of >>a polynomial in a given region. >>Is there a related method for counting complex roots in a given complex >>region? > If theres not an algorithm for doing that in Henricis > _Applied and Computational Complex Analysis_, Id be > very surprised (but its been a long time since Ive looked > at the book). A lot depends on how your given complex > region is given, I suppose. I wasnt thinking of some fancy region, just a simple 2 intervals: one for the real part one for the imaginary, or 1 for the abs and one for the arg. If some generalization works, so much the better. Mitch === Subject: Re: symbolic engine > May I suggest you try SymbMath.com? > 17.5. Interface with Other Software > You can run SymbMath from another software as a engine. Another software > sends a text file to SymbMath, then run SymbMath in background, get result > back from SymbMath. > Please read its document for details. > www.SymbMath.com === Subject: Re: SymbMath.com: web-based computer algebra system > SymbMath For Java is web-based symbolic math and computer algebra > system, > which runs in any computer with Java. You can play it online. > > www.SymbMath.com === Subject: out-of-kilter algorithm Hello! I was just wondering if there is someone who could help me in finding some usefull site with explanation of out-of-kilter algorithm and examples of problems solved with this algorithm. Mimmy === Subject: dual sum integral identites Given the coincidence: e^x = Sum[x^t / t!, {t, 0, oo}] x! = Integral[t^x / e^t, {t, 0, oo}], are there any other similar sum/integral correspondences? (Im not really talking about discrete/continuous correspondences. Or am I?) -- Mitch Harris Lehrstuhl fuer Automatentheorie, Fakultaet Informatik Technische Universitaet Dresden, Deutschland http://tcs.inf.tu-dresden.de/~harris === Subject: Re: dual sum integral identites >Given the coincidence: > e^x = Sum[x^t / t!, {t, 0, oo}] > x! = Integral[t^x / e^t, {t, 0, oo}], >are there any other similar sum/integral correspondences? >(Im not really talking about discrete/continuous correspondences. Or am >I?) How about int_0^1 1/x^x dx = sum_{n=1}^infinity 1/n^n 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 > 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. M(x)* M(x)=M(x)*M(x) => N(x)=exp(Int(M(y),{y, a ,x}))*a, -- extrabyte === Subject: Re: Differential Equations System in matrix form > M(x)* M(x)=M(x)*M(x) => N(x)=exp(Int(M(y),{y, a ,x}))*a, ops, N(x)=exp(Int(M(y),{y, a ,x}))*a) -- extrabyte === Subject: Re: Differential Equations System in matrix form > M(x)* M(x)=M(x)*M(x) => N(x)=exp(Int(M(y),{y, a ,x}))*a, > ops, > N(x)=exp(Int(M(y),{y, a ,x}))*a) This is only true if M(x) and M(y) commute,[M(x),M(y)]=0 > -- > extrabyte === 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 >M(x)=M0 + (1/x)*M1 + (1/x^2)*M2 > 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). ^^^^^^ I think M2>=0. N(0) can be either 0 or infty. If we integrate from x=-1 to x=-0 we need the singular eigenvalues Re(M2) < 0 (or if Re(M2) = 0 then Re(M1)/x < 0) to get N(0)=0. If we integrate backward from +a to +0 the signs of Re(M2) and Re(M1)/x must be reversed. I think the OP could take the analytical solution N0(x) for M0=M1=0 and integrate N(x)/N0(x) numerically. If the real parts of the eigenvalues of M2 are positive N0 will disappear for a quite large x, where the singularity of M1/x doesnt matter sincerely Klaus > 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 > this will diverge as x -> 0 if M2 < 0 (or if M2 = 0 and M1 < 0). > ^^^^^^ > I think M2>=0. I somehow mixed up the sign and the meaning of diverge, sorry klaus > Klaus > 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: Polynomial algebra (2 questions) Consider a field B, p(x)= b_0 + b_1 x +...+ b_n x^n is an irreducible polynomial of B[x]. Let I=(p(x)) be the ideal generated by p(x), then B[x]/I is a field. Consider the subfield B= {I+b | b in B}. We denote with X a variable in B. (1) In terms of formal series, is X= (0, I+1, 0,...,0,...) ??? With the same premises, consider the polynomial p(X) of B[X]: p(X) = (I+b_0) + (I+b_1)X +...+ (I+b_n)X^n. (2) Why is p(X) irreducible in B[X] ??? TIA === Subject: 1+1= ? donÇÇt know it. Can anyone help me ? === Subject: Re: 1+1= ? > don??t know it. > Can anyone help me ? Very often in applications 1+1=1 Zdenek Hurak === Subject: Re: 1+1= ? > donÇÇt know it. > Can anyone help me ? Sometimes 1+1=3. They call it synergy. === Subject: Re: 1+1= ? > donÇÇt know it. > Can anyone help me ? I dont know it either. Look at these: 1+.=! -+|=+ a+O=@ S+|=$ -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Re: 1+1= ? > donÇÇt know it. > Can anyone help me ? Yes, its 0 (sometimes). :-) Nijmegen, Netherlands === Subject: Re: Three-signed arithmetic : T Space I have now found that products exactly match complex arithmetic. By choosing the star branch to be the real axis the product arithmetic works out. The transform from three-signed Y to a complex C is: s(y) - (1/2)(m(y) + p(y)) + i(sqrt(3)/2)(m(y) - p(y)) Where y is in Y and m(),p(),s() return the minus,plus and star components of their argument. These equations come from pure graphical analysis of the three-signed branches being at angles 2pi/3. The components resolve via a 30-60-90 triangle. - minus pole - - - - - - . . . . . . . p1 = * 7 - 4 - . - . . - . . 0 * * * * * * * * * * star pole . + origin . p3 . + . . + . + . + . + . . . . . . . . . p2 = * 9 + 6 + + + + plus pole It can be shown that (y1)(y2) = y3 is equivalent to the product of the transformed values (a1+b1i)(a2+b2i). Proof of Equivalent Product Between Y and C ------------------------------------------- The following transform allows the conversion of a value in Y to a value in C: z = s(y)-(1/2)(m(y)+p(y)) + i(sqrt(3)/2)(m(y)-p(y)). This is a geometrical analysis of the three-signed plane suprposed with the complex plane with the star pole aligning with the positive real axis. The star component matches the real and the minus and plus components derive from right triangles. To simplify the notation I will simply use s,m,and p to represent the three components. I will also use q3 to represent the square root of three. All of the math for this proof is in the real domain. Now prove: y1y2 => z1z2 We have z1 = s1-(1/2)(m1+p1) + i(q3/2)(m1-p1) and z2 = s2-(1/2)(m2-p2) + i(q3/2)(m2-p2) Their product is then: (s1-m1/2-p1/2 + iq3m1/2-iq3p1/2)(s2-m2/2-p2/2 + iq3m2/2-iq3p2/2) = s1s2 -s1m2/2 -s1p2/2 +s1m2iq3/2-s1p2iq3/2 + -m1s2/2 +m1m2/4 +m1p2/4 -m1m2iq3/4 +m1p2iq3/4 + -p1s2/2 +p1m2/4 +p1p2/4 -p1m2iq3/4 +p1p2iq3/4 + +m1s2iq3/2 -m1m2iq3/4 -m1p2iq3/4 -m1m2(3/4) +m1p2(3/4) + -p1s2iq3/2 +p1m2iq3/4 +p1p2iq3/4 +p1m2(3/4) -p1p2(3/4). Gathering terms with precedence in the order of s,m,p: = s1s2 -s1m2/2 +s1m2iq3/2 -s1p2/2 -s1p2iq3/2 + -m1s2/2 +m1s2iq3/2 +m1m2/4 -m1m2(3/4) -m1m2iq3/4 -m1m2iq3/4 +m1p2/4 +m1p2iq3/4 -m1p2iq3/4 +m1p2(3/4) + -p1s2/2 -p1s2iq3/2 +p1m2/4 +p1m2(3/4) +p1m2iq3/4 -p1m2iq3/4 +p1p2/4 -p1p2(3/4) +p1p2iq3/4 +p1p2iq3/4 = s1s2 -s1m2/2 +s1m2iq3/2 -s1p2/2 -s1p2iq3/2 + -m1s2/2 +m1s2iq3/2 -m1m2/2 -m1m2iq3/2 +m1p2 + -p1s2/2 -p1s2iq3/2 +p1m2 -p1p2/2 +p1p2iq3/2 The product y1y2 has the following properties based on its definition: m = m1s2 + p1p2 + s1m2. p = m1m2 + p1s2 + s1p2. s = m1p2 + p1m2 + s1s2. Converted to a complex value it is: +m1p2 +p1m2 +s1s2 -(1/2){m1s2 +p1p2 +s1m2 +m1m2 +p1s2 +s1p2 } +i(q3/2){m1s2 +p1p2 +s1m2 -(m1m2 + p1s2 + s1p2)} = m1p2 +p1m2 +s1s2 -m1s2/2 -p1p2/2 -s1m2/2 -m1m2/2 -p1s2/2 -s1p2/2 +m1s2iq3/2 +p1p2iq3/2 +s1m2iq3/2 -m1m2iq3/2 -p1s2iq3/2 - s1p2iq3/2 Now gathering terms from the above to better match the complex product further up: = +s1s2 -s1m2/2 +s1m2iq3/2 -s1p2/2 - s1p2iq3/2 -m1s2/2 +m1s2iq3/2 -m1m2/2 -m1m2iq3/2 +m1p2 -p1s2/2 -p1s2iq3/2 +p1m2 -p1p2/2 +p1p2iq3/2 This exactly matches the reduced form of the complex product and represents the equation of the product of two three-signed values y1y2 (where s1,m1,p1 represents y1 and s2,m2,p2 represents y2 converted to the complex representation. Sums are also equivalent. That is: y1 * y2 => z1 + z2. The proof is trivial. === Subject: Math rules, algebraic integers What puzzles me is what Im facing now when math is about rules. There are certain rules that when followed give you conclusions which can be trusted. Ive shown the math rules that lead to conclusions about various things, only to have posters make fun of me, or deny the conclusions, even when they are mathematicians. But mathematicians are supposed to follow math rules. The math rules here are so easy I can give them quickly in this post. For instance, for a factor g of a polynomial P(m), g=r+c exists, where c=g when m=0, and r = g-c. How much simpler can you get? Lets say you were one of the people who accepted that, and yes I know some of you wish to have the specific ring, but if you know much about polynomials, then youll realize that P(x) being a polynomial is whats important. Since I know many of you dont understand the details of many math rules, Ill leave that as a place where you can reply and Ill explain those rules in more detail, while for others Ill continue. So now Ill go to the following expression: P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) If youre a purists you may have problems with me calling that P(m) as you have all those variables running around. Ill try and help out a bit by letting f=5, so that you have P(m) = 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) and you still may be bothered by a lot of variables, but I need the remaining ones for placement, and you may also wonder what the ring is. Well consider the ring to be algebraic integers, so all of the variables are algebraic integers. While Im looking at the expression as a polynomial P(m), the remaining variables are there as placemarkers so that I can factor it into non-polynomial factors. That is, so that I have 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) = (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u). That may seem awkward but its purpose provides the need for it. Still by math rules what Ive done is quite correct. Now though you can see the purpose of generalizing a factor g, as here I finally have more than a trivial use for g=r+c. Letting g_1 = a_1 x + 5u, I want to know what g_1 equals when m=0, because g_1 is a factor of P(m). Letting m=0, with P(m), I get P(0) = 25(3xu^2 + 5u^3) = 25u^2(3x + 5u), and now you can see another reason for why I need to keep the other variables as placemarkers. Here that reveals that when m=0, (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) = 25u^2(3x + 5u), which means that exactly two of the as go to 0, when m=0. Well that makes it easy, as my indices are arbitrary, so I can pick a_1 and a_2 to go to 0 when m=0, and now figure out that g_1 = 5u when m=0, which gives c_1 = 5u, and then I have r_1 = g_1 - c_1 = a_1 x + 5u - 5u = a_1 x. You may be wondering whats the point, since r and c here map to what would be the r and c if x were the key variable. Well Im working towards the point as that 25 factor of 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) has just been sitting there, when normally youd separate that off when factoring, as *normally* itd make the factorization ambiguous. Ill do that now and consider P(m)/25, so that I have P(m)/25 = (625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3. But what now? Well before I had 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) = (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) so I can wonder what happens to (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) when the 25 is divided off. It turns out that focusing on the constant term of P(m) provides an answer as from before I have P(0) = 25u^2(3x + 5u), where intriguingly the 25 is again visible as a factor. So dividing both sides by 25 gives me P(0)/25 = u^2(3x + 5u), and there are no more visible factors of 5, but you might imagine that maybe Ôu or Ôx could have a factor that is 5. Well possibly for specific cases, but definitely not in general, right? After all, I might pick x=7 and u=11, as theres been no constraint on those variables, and Ive left them in as placemarkers. So whats the significance of that result? Well focusing now on g_1 as I go from the general to the specific, as a variable of P(m), its also affected by the dividing off of 25. Ill imagine a factor w_1 divides off of g_1 when 25 is divided off of P(m). That gives me g_1/w_1 = r_1/w_1 + c_1/w_1 = a_1 x/w_1 + 5u/w_1 but as g_1/w_1 is a factor of P(m)/25, I also have an r and c for it. So using g_1/w_1 = r + c, I can set use P(0)/25, which is P(0)/25 = u^2(3x + 5u), to see that c = u. That forces w_1 = 5, so that I have g_1/w_1 = a_1 x/5 + u which *should* mean that a_1 has a factor that is 5 in the ring. You might know that there has been a lot of debate about that result where several posters have claimed that the result is only true if m=0. Well, remember c is a factor of the *constant* term of the polynomial, so theyre trying to make a new nonsensical math rule. For instance, by such an argument maybe with 2(x^2 + 2x+1) = (x+1)(2x+2) the 2 could in fact somehow, someway be a variable dependent on x. Obviously the posters know that no one would believe such a thing with a *polynomial* factor, so they rely on the complexity of my using non-polynomial factors to try and sneak in this nonsensical rule. And it has worked for them, as the sci.math newsgroup has let them go on for months pushing this new, nonsensical math rule, which Ive called voodoo math. Why would posters try to make up such a new rule? Why dont you ask them? James Harris === Subject: Re: Math rules, algebraic integers > You might know that there has been a lot of debate about that result > where several posters have claimed that the result is only true if > m=0. Actually, what I remember them saying is that being true at m=0 does not *guarantee* that its true at m<>0. You have to *prove* , not just *assume* that the result holds for other values of m. the result is only true if m=0 being true at m=0 does not *guarantee* that its true at m<>0 See the difference? Probably not. -- 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: Is this subset complete? Hi all, suppose we are working in the space of bounded functions on the natural numbers, equipped with the supremum norm. Now, this set A:={f: f(n)->0 when n->oo} is closed and complete. This set B:={f: f(1),f(2),... is summable} is a subset of A, and is not closed. But is it complete under the supnorm? Nijmegen, Netherlands === Subject: Re: Is this subset complete? >Hi all, >suppose we are working in the space of bounded functions on the natural >numbers, equipped with the supremum norm. Now, this set A:={f: f(n)->0 >when n->oo} is closed and complete. This set B:={f: f(1),f(2),... is >summable} is a subset of A, and is not closed. But is it complete under >the supnorm? No. Its easy to show that if X is a complete metric space and A is a subset of X then A is closed if and only if A is complete (in the metric it inherits from X.) >Nijmegen, Netherlands ************************ David C. Ullrich === Subject: degrees of freedom Does anyone have a good reference for mathematical explanation of the degrees of freedom for the sum of squares equation ? === Subject: difficult (?) problem Ive been trying to solve the following problem with no success. Find all functions f: N>=0 -> N>=0 such that f(n^2+m^2)=f(n)^2+f(m)^2 for all n,m in N>=0. (where N>=0 means the natural numbers plus the zero). Any ideas? nojb. === Subject: Re: difficult (?) problem >Ive been trying to solve the following problem with no success. >Find all functions f: N>=0 -> N>=0 such that >f(n^2+m^2)=f(n)^2+f(m)^2 for all n,m in N>=0. >(where N>=0 means the natural numbers plus the zero). >Any ideas? >nojb. Show us what progress youve made. One obvious solution is the identity f(n) = n for all n -- have you found any others, or do you suspect this is the only one? Substituting m = n = 0 into the identity gives f(0) = 2*f(0)^2, whence f(0) = 0 since f(0) is an integer (but it might be instructive to pursue f(0) = 1/2 if you want to look for solutions over the real numbers rather than over the nonnegative integers). Can you determine f(n) for other values of n, especially 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: Estimating the logarithmic integral Ive read in numerous places that the leading term of li(x) is x/ln(x). Is there an approximation for the leading term in the following? li(x) - x/ln(x) = ? In particular, Im interested in whether it can be shown that: li(x) - x/ln(x) < 5x/[ln(x)]^2 for some sufficiently large x Russell === Subject: Re: Estimating the logarithmic integral > Ive read in numerous places that the leading term of li(x) is > x/ln(x). Is there an approximation for the leading term in the > following? > li(x) - x/ln(x) = ? > In particular, Im interested in whether it can be shown that: > li(x) - x/ln(x) < 5x/[ln(x)]^2 for some sufficiently large x Integration by parts gives li(x) = x/log(x) + x/log(x)^2 + O(x/log(x)^3) etc. so I suppose the answer is yes. -- 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: Estimating the logarithmic integral Abramowitz & Stegun gives asymptotic expansions for li(x). === Subject: Math rules REVISED What puzzles me is what Im facing now when math is about rules. There are certain rules that when followed give you conclusions which can be trusted. Ive shown the math rules that lead to conclusions about various things, only to have posters make fun of me, or deny the conclusions, even when they are mathematicians. But mathematicians are supposed to follow math rules. The math rules here are so easy I can give them quickly in this post. For instance, for a factor g of a polynomial P(m), g=r+c exists, where c=g when m=0, and r = g-c. How much simpler can you get? Ok, so you need a ring, and while my point is it has problems, Ill use them here so that I can say that the other condition is that g is in the ring of algebraic integers for any algebraic integer m. So now Ill go to the following expression: P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) If youre a purists you may have problems with me calling that P(m) as you have all those variables running around. Ill try and help out a bit by letting f=5, so that you have P(m) = 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) and you still may be bothered by a lot of variables, but I need the remaining ones for placement, and you may also wonder what the ring is. Well consider the ring to be algebraic integers, so all of the variables are algebraic integers. While Im looking at the expression as a polynomial P(m), the remaining variables are there as *placemarkers* so that I can factor it into non-polynomial factors. That is, so that I have 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) = (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u). That may seem awkward but its purpose provides the need for it. Still by math rules what Ive done is quite correct. Now though you can see the purpose of generalizing a factor g, as here I finally have more than a trivial use for g=r+c. Letting g_1 = a_1 x + 5u, I want to know what g_1 equals when m=0, because g_1 is a factor of P(m). Letting m=0, with P(m), I get P(0) = 25(3xu^2 + 5u^3) = 25u^2(3x + 5u), and now you can see another reason for why I need to keep the other variables as placemarkers. Here that reveals that when m=0, (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) = 25u^2(3x + 5u), which means that exactly two of the as go to 0, when m=0. Well that makes it easy, as my indices are arbitrary, so I can pick a_1 and a_2 to go to 0 when m=0, and now figure out that g_1 = 5u when m=0, which gives c_1 = 5u, and then I have r_1 = g_1 - c_1 = a_1 x + 5u - 5u = a_1 x. You may be wondering whats the point, since r and c here map to what would be the r and c if x were the key variable. Well Im working towards the point as that 25 factor of 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) has just been sitting there, when normally youd separate that off when factoring, as *normally* itd make the factorization ambiguous. Ill do that now and consider P(m)/25, so that I have P(m)/25 = (625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3. But what now? Well before I had 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) = (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) so I can wonder what happens to (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) when the 25 is divided off. It turns out that focusing on the constant term of P(m) provides an answer as from before I have P(0) = 25u^2(3x + 5u), where intriguingly the 25 is again visible as a factor. So dividing both sides by 25 gives me P(0)/25 = u^2(3x + 5u), and there are no more visible factors of 5, but you might imagine that maybe Ôu or Ôx could have a factor that is 5. Well possibly for specific cases, but definitely not in general, right? After all, I might pick x=7 and u=11, as theres been no constraint on those variables, and Ive left them in as placemarkers. So whats the significance of that result? Well focusing now on g_1 as I go from the general to the specific, as a variable of P(m), its also affected by the dividing off of 25. Ill imagine a factor w_1 divides off of g_1 when 25 is divided off of P(m). That gives me g_1/w_1 = r_1/w_1 + c_1/w_1 = a_1 x/w_1 + 5u/w_1 but as g_1/w_1 is a factor of P(m)/25, I also have an r and c for it. So using g_1/w_1 = r + c, I can set use P(0)/25, which is P(0)/25 = u^2(3x + 5u), to see that c = u. That forces w_1 = 5, so that I have g_1/w_1 = a_1 x/5 + u which *should* mean that a_1 has a factor that is 5 in the ring. You might know that there has been a lot of debate about that result where several posters have claimed that the result is only true if m=0. Well, remember c is a factor of the *constant* term of the polynomial, so theyre trying to make a new nonsensical math rule. For instance, by such an argument maybe with 2(x^2 + 2x+1) = (x+1)(2x+2) the 2 could in fact somehow, someway be a variable dependent on x. Obviously the posters know that no one would believe such a thing with a *polynomial* factor, so they rely on the complexity of my using non-polynomial factors to try and sneak in this nonsensical rule. And it has worked for them, as the sci.math newsgroup has let them go on for months pushing this new, nonsensical math rule, which Ive called voodoo math. Why would posters try to make up such a new rule? Why dont you ask them? James Harris === Subject: Re: Math rules REVISED What did Barry Mazur say about your paper? === 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? Hallucinogens? > 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. Yes, it would be silly for you to worry that people might steal your results. [rest snipped] === 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? No. >My proofs are out there. > http://groups.msn.com/AmateurMath/explainingtheproof.msnw James Harris === Subject: Help - How to write in factorial form? Given two bi-variable functions depending on variabels x and y : poly1=(x-a1)...(x-an).(y-b1)...(y-bn) poly2=(x-c1)...(x-cn).(y-d1)...(y-dn) How can i write (poly1+poly2) in factorial form? If not possible theoretically, is there software available to do it? (ai,bi, ci and di for i=1:n are real numbers) J. Miller === Subject: Q: On 3x+1 problem I have been delving into the 3x+1 problem and have been wondering if there is a table showing the number of starting numbers (seeds) at each level of the Collatz tree. For instance there is 18 (seeds) @ the 15th level of the Collatz tree. Where each of these seeds starts a unique path back to level 3,2 and 1 where these seeds are 4,2,1 respectively. When I refer to a unique path I mean each sequence could have only 1 or more terms that are different from another sequence starting @ any particular level of the Collatz tree. Starting with level 1 of the Collatz tree--- Level # of Paths (seeds) 1 1 2 1 3 1 4 1 5 1 6 2 7 2 8 4 9 4 10 6 11 6 12 8 13 10 14 14 15 18 16 24 17 29 18 36 19 44 20 58 Etc. Is there a table like the one above showing many more levels and # of paths? I have searched the Web but to no avail. Explaining it further... Level The starting seeds # of seeds on this (paths) level are -- 10 12,13,80,84,85,512 6 9 6,40,42,256 4 8 3,20,21,128 4 7 10,64 2 6 5,32 2 5 16 1 4 8 1 3 4 1 2 2 1 1 1 1 Dan === Subject: Re: Q: On 3x+1 problem >> I have been delving into the 3x+1 problem and have been wondering if >> there is a table showing the number of starting numbers (seeds) at >> each level of the Collatz tree. For instance there is 18 (seeds) @ the >> 15th level of the Collatz tree. Where each of these seeds starts a >> unique path back to level 3,2 and 1 where these seeds are 4,2,1 >> respectively. When I refer to a unique path I mean each sequence could >> have only 1 or more terms that are different from another sequence >> starting @ any particular level of the Collatz tree. >> Starting with level 1 of the Collatz tree--- >> Level # of Paths (seeds) >> 1 1 >> 2 1 >> 3 1 >> 4 1 >> 5 1 >> 6 2 >> 7 2 >> 8 4 >> 9 4 >> 10 6 >> 11 6 >> 12 8 >> 13 10 >> 14 14 >> 15 18 >> 16 24 >> 17 29 >> 18 36 >> 19 44 >> 20 58 >> Etc. >> Is there a table like the one above showing many more levels and # of >> paths? >> I have searched the Web but to no avail. >> >> Explaining it further... >> Level The starting seeds # of seeds >> on this (paths) >> level are -- >> 10 12,13,80,84,85,512 6 >> 9 6,40,42,256 4 >> 8 3,20,21,128 4 >> 7 10,64 2 >> 6 5,32 2 >> 5 16 1 >> 4 8 1 >> 3 4 1 >> 2 2 1 >> 1 1 1 >> Dan >Whenever I play with the Collatz conjecture, I find it easier to use a >slightly different (but equivalent) set of rules: > if x is odd....: x -> (3x+1)/2 > if x is even...: x -> x/2 >This makes sense since for x odd, 3x+1 is always even, so the next step >in the traditional procedure is always the even step. Arranging the >rules this alternative way eliminates a pointless extra use of the >even rule. >With this Ôreduced ruleset a tree may be drawn from the number showing >the paths to/from other integers (except 1, 2, 4 and any counterexamples >to the collatz conjecture). >If we draw this tree from the root, placing the inverses of the reduced >rules [ x -> (2x-1)/3 and/or x -> 2x ] applied to the parent as >children, we obtain the following (monospaced font recommended): > .------ 8 ------. > .-------- `--------. > .--- 5 --. 16 > .-- `--. | > 3 10 .-- 32 --. > | | .- `-. > 6 .- 20 --. 21 64 > | .- `--. | | > 12 13 40 42 .- 128 -. > | | | | . `. > 24 26 80 84 85 256 > | / / | | | > 48 17 52 53 160 168 170 512 >To prove the Collatz conjecture, all we need to do(TM) is prove that >the tree listed contains all integers except 1, 2 and 4 (which fall >above 8). >I notice that there seems to be a pattern in that the tree from 5 is >very similar to the tree from 32 - whether this means anything is up to >someone smarter and less prone to mistakes than I. :) >Carl I beliveve you have this tree wrong! It should start with 16 as the first branch connecting directly to 32 and 5. Also many more like 20 and 13, 85 and 128, 53 and 80, all of these should not be directly connected! Dan === Subject: Re: Q: On 3x+1 problem > I have been delving into the 3x+1 problem and have been wondering if > there is a table showing the number of starting numbers (seeds) at > each level of the Collatz tree. For instance there is 18 (seeds) @ the > 15th level of the Collatz tree. Where each of these seeds starts a > unique path back to level 3,2 and 1 where these seeds are 4,2,1 > respectively. When I refer to a unique path I mean each sequence could > have only 1 or more terms that are different from another sequence > starting @ any particular level of the Collatz tree. > Starting with level 1 of the Collatz tree--- Whenever I play with the Collatz conjecture, I find it easier to use a slightly different (but equivalent) set of rules: if x is odd....: x -> (3x+1)/2 if x is even...: x -> x/2 This makes sense since for x odd, 3x+1 is always even, so the next step in the traditional procedure is always the even step. Arranging the rules this alternative way eliminates a pointless extra use of the Ôeven rule. With this Ôreduced ruleset a tree may be drawn from the number showing the paths to/from other integers (except 1, 2, 4 and any counterexamples to the collatz conjecture). If we draw this tree from the root, placing the inverses of the reduced rules [ x -> (2x-1)/3 and/or x -> 2x ] applied to the parent as children, we obtain the following (monospaced font recommended): .------ 8 ------. .-------- `--------. .--- 5 --. 16 .-- `--. | 3 10 .-- 32 --. | | .- `-. 6 .- 20 --. 21 64 | .- `--. | | 12 13 40 42 .- 128 -. | | | | . `. 24 26 80 84 85 256 | / / | | | 48 17 52 53 160 168 170 512 | / | / | | / / 96 11 34 104 35 106 320 336 113 340 341 1024 | / | / / | / | / | / | 192 7 22 68 69 208 23 70 212 213 640 672 75 226 680 227 682 2048 To prove the Collatz conjecture, all we need to do(TM) is prove that the tree listed contains all integers except 1, 2 and 4 (which fall Ôabove 8). I notice that there seems to be a pattern in that the tree from 5 is very similar to the tree from 32 - whether this means anything is up to someone smarter and less prone to mistakes than I. :) Carl === Subject: Names of spectra For an operator L on a Banach space X I am looking for the names of the following subsets of the (essential) spectrum: 1) those complex numbers c such that cI-L has not-closed range or infinite-dimensional kernel (i.e. is not a Phi_+ operator) 2) those complex numbers c such that cI-L has not-closed range or range of infinite co-dimension (i.e. is not a Phi_- operator) Are there any commonly known names for these two spectra? Kai === Subject: Irrational behavior from mathematicians One thing that intrigued me over the years was how certain people could get away with posting wrong mathematics, when they argued with me. I concluded that math society didnt necessarily follow math rules. Now Ive proven it. Consider that with P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) several posters didnt like a conclusion that follows from the mathematics that if you use the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) that two of the as *should* have a factor that is f, if youre in the ring of algebraic integers, but provably you can find values where they dont, so logically theres a problem with the ring. Now the math argument which proves that is rather basic, but then again, mathematicians probably would rather hold on to the notion that theres nothing wrong with the ring of algebraic integers, and certain posters exploited your emotional needs. Basically their argument came down to claiming that w_1(m), w_2(m), and w_3(m) exist where w_1(m) w_2(m) w_3(m) = f^2 as they pushed the notion that the f^2 divided off of P(m) as a *function* of m. I found it fascinating that they were so successful, but my suspicion for a long time was that math society often did not follow math rules. I realized a while back that many of you dont actually believe in math itself, but like math society for the same reasons that some people like religion society. You know for a lot of you its like being a Buddhist, or a Christian, or a Jew, or a Muslim, or a Hindu (sorry for those left off, but I thought Id stop there). That conclusion is the only *rational* one that follows as otherwise given an argument requiring w_1(m) w_2(m) w_3(m) = f^2 youd have realized that it forces a relationship between Ôm and Ôf, while P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) does not. Basically Id found a way to actually *see* that math society was no better than regular society--no matter what it might claim--and that if you found mathematics that math society members didnt like, theyd throw out the rules of mathematics. Its been kind of fascinating watching what lengths many of you would go to, as Id test you various ways. I was curious about world society as well so Id test to see how others would react as they considered mathematicians as humanitys representatives, so hey, maybe others should care if mathematicians were wacky. Amazingly enough, consistently, over long periods math society again and again proved that if the math were something it didnt like, itd reject it, and fight math rules. I have results of contacts with Barry Mazur, Andrew Granville, major math journals like the American Journal of Mathematics, and German editors, as well as LOTS of posts. Id sit back and muse about you, wondering what exactly did any of you believe in, and realized that you believe in people. I tested that as well by writing emails to TIME magazine, so that I could baseline society itself, and noticed what they liked at TIME versus what they didnt and even managed to get a *version* of an email published. When Bushs policy in Iraq fell apart, as the weapons of mass destruction failed to appear, and later chaos began to rule as predicted, I watched how you reacted both as a math society, and as part of world society. Ultimately I realized that you did indeed believe in people, and often you didnt care about math, reality, or anything at all that contradicted things you *decided* were true. Well thats it. Ive tested mathematicians as thoroughly as I like, and besides, there are other things I want to do now. So, start obeying math rules, or if you prefer I can proceed to tear away your illusions of being rational and logical piece by piece over time, as I reduce your society using advanced psychological tools that rival the best psychological warfare techniques of world governments. I can put ideas in your heads at will. If you wish to not be mathematicians, then I will begin anew with a new program. Eventually Ill reduce the world society of mathematicians to a level that will be more amenable to logic and rationality. My estimate is that it would take less than 20 years. James Harris === Subject: Re: Irrational behavior from mathematicians James, you are missing the point. Mathematics is based on standard rules and as far as I can see, none of the rules that other posters have eluded to I have any problem with. I do have a problem with your conduct. You treat everyone with disrespect, but get pissed when someone disrespects you. How old are you by the way? I guess 5 or 6. Though I am still a mathematics student, I have no problem whatsoever with the math education Ive received. It all makes sense to me, to the point of How could it be any other way?. You know, I would bet that Magidin, Ullrich, et. al. would be more than willing to help someone who was polite to them. Theyve helped you more than I would given youve been a complete jerk to EVERYONE. Grow some skin, James. David Moran === Subject: Re: Irrational behavior from mathematicians > One thing that intrigued me over the years was how certain people > could get away with posting wrong mathematics, when they argued with > me. > I concluded that math society didnt necessarily follow math rules. > Now Ive proven it. > Consider that with > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) > several posters didnt like a conclusion that follows from the > mathematics that if you use the factorization > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > that two of the as *should* have a factor that is f, if youre in the > ring of algebraic integers, but provably you can find values where > they dont, so logically theres a problem with the ring. ...or there is a problem with your Ôproof. To rule out the possibility of an error on your part, why not post one of the numbers which you claim *should* be an algebraic integer, but which is left out of the ring of algebraic integers? THAT should put all arguments to rest, and all mathematicians to shame. Just ONE number is all it takes. > Now the math argument which proves that is rather basic, but then > again, mathematicians probably would rather hold on to the notion that > theres nothing wrong with the ring of algebraic integers, and certain > posters exploited your emotional needs. Youre in danger of practicing psychology without a license again. You are already on notice for this. [snip standard Harris irrational nonsense loaded with terms like fascinating, amazing, but, hey, Yup, etc.] > I can put ideas in your heads at will. If you can do that, why not just make them accept your Ôproof? > If you wish to not be mathematicians, then I will begin anew with a > new program. > Eventually Ill reduce the world society of mathematicians to a level > that will be more amenable to logic and rationality. THREAT ALERT! THREAT ALERT! > My estimate is that it would take less than 20 years. Personally, I hope you get 30-40 years. -- 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: Irrational behavior from mathematicians What did Barry Mazur say about your paper? === Subject: [JSH] Re: Math rules, algebraic integers >What puzzles me is what Im facing now when math is about rules. >There are certain rules that when followed give you conclusions which >can be trusted. >Ive shown the math rules that lead to conclusions about various >things, only to have posters make fun of me, or deny the conclusions, >even when they are mathematicians. But mathematicians are supposed to >follow math rules. >The math rules here are so easy I can give them quickly in this post. >For instance, for a factor g of a polynomial P(m), g=r+c exists, where >c=g when m=0, and r = g-c. >How much simpler can you get? Its simpler than that: To be more succinct, define r(x) = g(x) - g(0). That one equation is basically the content of your lemma. >Lets say you were one of the people who accepted that, Few if any of us have not accepted your lemma. After all, it says almost nothing. > and yes I know >some of you wish to have the specific ring, but if you know much about >polynomials, then youll realize that P(x) being a polynomial is >whats important. >Since I know many of you dont understand the details of many math >rules, Ill leave that as a place where you can reply and Ill explain >those rules in more detail, while for others Ill continue. >So now Ill go to the following expression: > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) >If youre a purists you may have problems with me calling that P(m) as >you have all those variables running around. It does seem odd, given that you are going to factor it as a polynomial in x, not as a polynomial in m. And in your lemma, g and r are also thought of as factors with respect to x, not m. c = g(0) refers to the value of g when x = 0, not when m = 0. But this looks like just quibbling over notation. Clearly P is a both a function of m and a polynomial in x. You might consider writing it as P(m, x) just to makes things clear. (See also below - its not just quibbling.) >Ill try and help out a >bit by letting f=5, so that you have > P(m) = 25((625 m^3 - 75 m^2 + 3m) x^3 - > 3(-1+25m )x u^2 + 5u^3) >and you still may be bothered by a lot of variables, but I need the >remaining ones for placement, and you may also wonder what the ring >is. >Well consider the ring to be algebraic integers, so all of the >variables are algebraic integers. In your applications all the variables m, f, and u are ordinary integers. >While Im looking at the expression as a polynomial P(m), the >remaining variables are there as placemarkers so that I can factor it >into non-polynomial factors. >That is, so that I have > 25((625 m^3 - 75 m^2 + 3m) x^3 - > 3(-1+25m )x u^2 + 5u^3) = > (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u). Where, presumably, a1, a2, and a3 are algebraic integers - right? >That may seem awkward but its purpose provides the need for it. >Still by math rules what Ive done is quite correct. >Now though you can see the purpose of generalizing a factor g, as here >I finally have more than a trivial use for g=r+c. >Letting g_1 = a_1 x + 5u, I want to know what g_1 equals when m=0, >because g_1 is a factor of P(m). >Letting m=0, with P(m), I get > P(0) = 25(3xu^2 + 5u^3) = 25u^2(3x + 5u), >and now you can see another reason for why I need to keep the other >variables as placemarkers. Here that reveals that when m=0, > (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) = 25u^2(3x + 5u), >which means that exactly two of the as go to 0, when m=0. >Well that makes it easy, as my indices are arbitrary, so I can pick >a_1 and a_2 to go to 0 when m=0, and now figure out that > g_1 = 5u >when m=0, which gives c_1 = 5u, and then I have > r_1 = g_1 - c_1 = a_1 x + 5u - 5u = a_1 x. Everything is OK up to this point. Of course in general, a1, a2 and a3 are functions of m, and when m <> 0, none of them are zero, in contrast to the m = 0 case, when two of them are zero and P(m) becomes a degree 1 polynomial rather than a degree 3 polynomial. >You may be wondering whats the point, since r and c here map to what >would be the r and c if x were the key variable. >Well Im working towards the point as that 25 factor of > 25((625 m^3 - 75 m^2 + 3m) x^3 - > 3(-1+25m )x u^2 + 5u^3) >has just been sitting there, when normally youd separate that off >when factoring, as *normally* itd make the factorization ambiguous. >Ill do that now and consider P(m)/25, so that I have > P(m)/25 = (625 m^3 - 75 m^2 + 3m) x^3 - > 3(-1+25m )x u^2 + 5u^3. >But what now? >Well before I had > 25((625 m^3 - 75 m^2 + 3m) x^3 - > 3(-1+25m )x u^2 + 5u^3) = > (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) >so I can wonder what happens to > (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) >when the 25 is divided off. >It turns out that focusing on the constant term of P(m) provides an >answer as from before I have > P(0) = 25u^2(3x + 5u), >where intriguingly the 25 is again visible as a factor. >So dividing both sides by 25 gives me > P(0)/25 = u^2(3x + 5u), >and there are no more visible factors of 5, but you might imagine that >maybe Ôu or Ôx could have a factor that is 5. >Well possibly for specific cases, but definitely not in general, >right? R I G H T ! What you have done so far is valid only for m = 0. >After all, I might pick x=7 and u=11, as theres been no constraint on >those variables, and Ive left them in as placemarkers. Worse yet, you might pick m = something other than 0. >So whats the significance of that result? >Well focusing now on g_1 as I go from the general to the specific, as >a variable of P(m), its also affected by the dividing off of 25. >Ill imagine a factor w_1 divides off of g_1 when 25 is divided off of >P(m). >That gives me > g_1/w_1 = r_1/w_1 + c_1/w_1 = a_1 x/w_1 + 5u/w_1 >but as g_1/w_1 is a factor of P(m)/25, I also have an r and c for it. ... noting here that w1 is a function of m, unless you have a proof to the contrary. But you have *not* provided any proof or evidence of that. >So using g_1/w_1 = r + c, I can set use P(0)/25, which is > P(0)/25 = u^2(3x + 5u), >to see that c = u. Yes. Just fine for m = 0. But c <> u in general. See below. >That forces w_1 = 5, so that I have > g_1/w_1 = a_1 x/5 + u Yes again: for m = 0, w1 = 5. But not in general. >which *should* mean that a_1 has a factor that is 5 in the ring. Since all you have considered so far is m = 0, and a1 = 0 when m = 0, a1 = 0 does indeed have a factor of 5, although for m = 0 it also has a factor of q for any other integer q. The factorization in this case is degenerate. As you have defined it above, c = c1/w1. But (unless you can prove otherwise), w1 is a function of m. Therefore also c is a function of m. This is true even though you have c = 5*u is NOT a function of m. >You might know that there has been a lot of debate about that result >where several posters have claimed that the result is only true if >m=0. >Well, remember c is a factor of the *constant* term of the polynomial, >so theyre trying to make a new nonsensical math rule. Here you are fooling yourself, and it has to do with what I said before about P(m, x). For a fixed value of m, c = 5*u/w1 is the constant *with respect to x*. But since in general, w1 must be assumed to be a function of m, c is also a function of m. Got that? Constant with respect to x, but not constant with respect to m. This is a key point which you seem to have had enormous difficulty understanding. Why? >For instance, by such an argument maybe with > 2(x^2 + 2x+1) = (x+1)(2x+2) >the 2 could in fact somehow, someway be a variable dependent on x. You have ALMOST GOT IT. The constant c here is 2, which is the constant with respect to x. You do not have an m in your oversimplified example. If you did, you would expect in general that c would be a function of m. For example: let Q(x) = x^2 + m*x + 2. Suppose Q(x) is factored in the form (x + c1)*(x + c2). Do you actually think c1 and c2 are *not* functions of m ??? >Obviously the posters know that no one would believe such a thing with >a *polynomial* factor, so they rely on the complexity of my using >non-polynomial factors to try and sneak in this nonsensical rule. You are NOT using non-polynomial factors: g(x) = a1*x + u*f and g(x) = a1*x/w1 + 5*u/w1 are quite obviously *first-degree polynomials* in x. You do NOT have the complexity of non-polynomial factors. This is NOT the problem. You have plenty of problems, but this is NOT one of them. >And it has worked for them, as the sci.math newsgroup has let them go >on for months pushing this new, nonsensical math rule, which Ive >called voodoo math. >Why would posters try to make up such a new rule? >Why dont you ask them? Absolutely! But let me ask why you keep starting new threads on this - with ideas that are now becoming *threadbare* - when you have failed utterly to answer questions like the following: You claim to have shown factorizations of the form you want in three very special cases: 1. m = 0 2. f = 3 3. f = sqrt(2) What you actually need for your proofs are: 1. f a prime integer larger than 3, AND 2. m relatively prime to f. None of the three special cases above cover this. How can you proceed from m = 0, f = 3 to, for example, m = 1, f = 5 ? Dont you think it is odd that you have not succeeded in finding any arguments which apply for such cases ? It must seem strange to your loyal fans that you claim to have a proof for exactly and *only* the cases that are irrelevant to your main argument. All you have is some circumstantial evidence (and in the case of f = 3, I dont agree that you have anything valid at all). If you had some way to show how m = 0, f = 3 can be used to show what you want for other values, that might work. But you have given no such proof. Again you just have that bit of circumstantial evidence in the degenerate case, m = 0. Why dont (or cant) you produce a proof for values like the following ? 1. m = 1, f = 5 2. m = 4, f = 7 3. m = 62, f = 107 etc. ... instead of just more irrelevant peripheral Ôevidence for m = 0, f = 3 ??? Nora B. >James Harris === Subject: Editors again, New York Journal of Mathematics Im an admirer of Einstein and Gauss who were both Germans, which may be why I was picking on the German editors a bit, as I think both were embarrassments to the rich intellectual history of Germanic people. However, there are some interesting replies from Americans as well, so here are two American editors of the New York Journal of Mathematics to an earlier version of the same paper Advanced Polynomial Factorization. I am sorry but this looks more like a submission for the chief editor, Mark Steinberger, than for me. I only accept submissions directly in number theory. Andrew Granville We generally dont publish short notes. Im not sure what journals to suggest, but NYJM probably isnt the right place for this one. Best, Mark Wow, if the *chief* editor of the New York Journal of Mathematics wouldnt know what journal to suggest, then who would? James Harris === Subject: Re: Editors again, New York Journal of Mathematics >Im an admirer of Einstein and Gauss who were both Germans, which may >be why I was picking on the German editors a bit, as I think both were >embarrassments to the rich intellectual history of Germanic people. >However, there are some interesting replies from Americans as well, so >here are two American editors of the New York Journal of Mathematics >to an earlier version of the same paper Advanced Polynomial >Factorization. >editor, > Mark Steinberger, than for me. I only accept submissions directly > in number theory. > Andrew Granville >suggest, but NYJM probably isnt the right place for this one. >Best, >Mark >wouldnt know what journal to suggest, then who would? If you ever sober up youre going to be more embarassed by this stuff than by a lot of your stuff. Really, if most people had a paper rejected by a _large_ number of journals they wouldnt be advertising it like this, like it somehow shows theres something wrong with the journals or the editors. That Im not sure what journal to suggest doesnt show ignorance, its an attempt at a polite brushoff. >James Harris ************************ David C. Ullrich === Subject: Re: Editors again, New York Journal of Mathematics > Im an admirer of Einstein and Gauss who were both Germans, which may > be why I was picking on the German editors a bit, as I think both were > embarrassments to the rich intellectual history of Germanic people. > However, there are some interesting replies from Americans as well, so > here are two American editors of the New York Journal of Mathematics > to an earlier version of the same paper Advanced Polynomial > Factorization. > editor, > Mark Steinberger, than for me. I only accept submissions directly > in number theory. > Andrew Granville > suggest, but NYJM probably isnt the right place for this one. > Best, > Mark > wouldnt know what journal to suggest, then who would? Oohh! Me! Pick me!!! Okay, how about the ÔJournal of Irreproducible Results? === Subject: Re: Editors again, New York Journal of Mathematics > Im an admirer of Einstein and Gauss who were both Germans, which may > be why I was picking on the German editors a bit, as I think both were > embarrassments to the rich intellectual history of Germanic people. > However, there are some interesting replies from Americans as well, so > here are two American editors of the New York Journal of Mathematics > to an earlier version of the same paper Advanced Polynomial > Factorization. > I am sorry but this looks more like a submission for the chief > editor, > Mark Steinberger, than for me. I only accept submissions directly > in number theory. > Andrew Granville > We generally dont publish short notes. Im not sure what journals to > suggest, but NYJM probably isnt the right place for this one. > Best, > Mark > Wow, if the *chief* editor of the New York Journal of Mathematics > wouldnt know what journal to suggest, then who would? > Oohh! Me! Pick me!!! Okay, how about the ÔJournal of Irreproducible Results? I really think he should try with the Journal of Theoretics: http://www.journaloftheoretics.com/second-index.htm He might get Ken Seto as a peer reviewer! Dirk Vdm === Subject: Re: Editors again, New York Journal of Mathematics > be why I was picking on the German editors a bit, as I think both were > embarrassments to the rich intellectual history of Germanic people. You mean they actually told you what the paper was worth, as opposed ot the American editors you quoted, who only told you to go away? V. -- mail me at lastname at cs utk edu === Subject: Re: Editors again, New York Journal of Mathematics > Im an admirer of Einstein and Gauss who were both Germans, which may > be why I was picking on the German editors a bit, as I think both were > embarrassments to the rich intellectual history of Germanic people. > However, there are some interesting replies from Americans as well, so > here are two American editors of the New York Journal of Mathematics > to an earlier version of the same paper Advanced Polynomial > Factorization. > editor, > Mark Steinberger, than for me. I only accept submissions directly > in number theory. > Andrew Granville > suggest, but NYJM probably isnt the right place for this one. > Best, > Mark > wouldnt know what journal to suggest, then who would? Journals are extremely supportive of researchers if there is any value in their submission. Had there been any value at all, the Editor would have appended a short list of objections to be remedied plus with a request to correct and resubmit, or sent a list of allied journals more suited to content. Im not sure what journals to suggest is isomorphous to go away. Since you are too ing stooopid to comprehend declarative sentences, Uncle Al will feed you your own tripe. CASEY AT THE POST-MODERNIST BAT The outlook wasnt brilliant for the theory crowd that week: Objectivism ruled, and nothing radical was chic. So when Derridas new book got panned and Beaudrillard retired The whole postmodern movement felt depressed and uninspired. They said, If only Casey had a chance to write a piece, Our paradigms imperiled life would get a second lease! But Casey was department chair at Northern Ivy U And taught a massive seminar, and summer courses, too; And what with that new cooking class and chanting every night Six years would surely pass before hed get a chance to write So in yogurt shops in Berkeley and in trains at Harvard Square, A hegemonic melancholy lingered in the air. But then there came a rumor that the profs could scarce believe: The seminar was off the books, and Caseys taken leave! Hed ordered all his students not to bug him Ôtil the fall And bought ten pounds of mocha from the Starbucks at the mall. Then cheers rang out in latt.8e bars across the eastern seaboard: For Casey, mighty Casey, was advancing to the keyboard! There was style in Caseys manner as he eased into his prose On hegemonic (dis)course tropes in global know/ledge ßows. He quoted heavy metal bands and talked of mise-en-sc.8fne. With long reßexive musings on a rash hed had at ten. He interspersed parentheses and slashes and notations That made straightforward gerunds look like chemical equations. He rounded off the essay with a trenchant aper.8du And mailed it off to Washington to breeze through peer review. He sent out copies far and wide, and found they made a hit; Deconstruct the editor! ßew the e-mails Ôcross the lands And they might have just pursued it had not Casey stayed their hands. A triße, chuckled Casey as he glanced at the reviews Then whipped off a revision that would surely make the news. He added untranslated quotes in Arabic and Hmong And one concluding sentence that was seven pages long. He closed it with a stirring plea for union rights at Yale, Then pasted in a glossy nude and popped it in the mail. And then he packed his bags to rest in Maui for a bit Fight the power! shouted angry mobs in internet cafes Until they got his valiant fax: Ill have it fixed in days! The play was gone from Caseys prose, his style was clipped and terse; Hed taken out the lit review in blank Homeric verse; Hed added charts and plots and graphs and diagrams from Venn ÔTil all agreed his piece could never get turned down again And now the printer spits it out, and now the stamp is bought, And now the world is shattered by the depth of Caseys thought! Oh, somewhere in the learned world progressive thought prevails And journals publish fresh assaults on dead Caucasian males; And somewhere deconstruction reigns, and joy is uninßected But not this week in anthro - mighty Casey got rejected. Hey Harris, http://www.apa.org/journals/psp/psp7761121.html http://w0rli.home.att.net/youare.swf http://www.mazepath.com/uncleal/sunshine.jpg -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Editors again, New York Journal of Mathematics > Hey Harris, > http://www.apa.org/journals/psp/psp7761121.html And it has references! Lots of them. === Subject: Re: Editors again, New York Journal of Mathematics > Im an admirer of Einstein and Gauss who were both Germans, which may > be why I was picking on the German editors a bit, as I think both were > embarrassments to the rich intellectual history of Germanic people. > However, there are some interesting replies from Americans as well, so > here are two American editors of the New York Journal of Mathematics > to an earlier version of the same paper Advanced Polynomial > Factorization. > editor, > Mark Steinberger, than for me. I only accept submissions directly > in number theory. > Andrew Granville > suggest, but NYJM probably isnt the right place for this one. > Best, > Mark > wouldnt know what journal to suggest, then who would? Maybe you could try asking someone on your home planet. HTH > James Harris === Subject: Re: Editors again, New York Journal of Mathematics > Im an admirer of Einstein and Gauss who were both Germans, which may > be why I was picking on the German editors a bit, as I think both were > embarrassments to the rich intellectual history of Germanic people. > However, there are some interesting replies from Americans as well, so > here are two American editors of the New York Journal of Mathematics > to an earlier version of the same paper Advanced Polynomial > Factorization. > editor, > Mark Steinberger, than for me. I only accept submissions directly > in number theory. > Andrew Granville > suggest, but NYJM probably isnt the right place for this one. > Best, > Mark > wouldnt know what journal to suggest, then who would? > James Harris The Journal of Irreproducible Results jumps to mind. I think it still exists in some form even though the last I heard, the editors were revolting. Nobel laureates have published there. Consider also Hot Air which has had a print version. The headquarters, or properly hindquarters, seem to be at MIT. Chuck -- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net <3F537F20.71565E02@earthlink.net> === Subject: Re: Editors again, New York Journal of Mathematics >> Im an admirer of Einstein and Gauss who were both Germans, which may >> be why I was picking on the German editors a bit, as I think both >> were embarrassments to the rich intellectual history of Germanic >> people. However, there are some interesting replies from Americans >> as well, so here are two American editors of the New York Journal of >> Mathematics to an earlier version of the same paper Advanced >> I am sorry but this looks more like a submission for the chief >> editor, >> Mark Steinberger, than for me. I only accept submissions directly >> in number theory. >> Andrew Granville >> We generally dont publish short notes. Im not sure what journals to >> suggest, but NYJM probably isnt the right place for this one. >> Best, >> Mark >> Wow, if the *chief* editor of the New York Journal of Mathematics >> wouldnt know what journal to suggest, then who would? >> James Harris > The Journal of Irreproducible Results jumps to mind. I think it > still exists in some form even though the last I heard, the editors > were revolting. Nobel laureates have published there. Consider also > Hot Air which has had a print version. The headquarters, or properly > hindquarters, seem to be at MIT. > Chuck http://www.ncbi.nlm.nih.gov:80/entrez/query.fcgi?cmd=Retrieve &db=PubMed& list_uids=12385726&dopt=Abstract I suppose if this could get published, then someone somewhere will publish the whacktards opus. === Subject: Re: Editors again, New York Journal of Mathematics > > Im an admirer of Einstein and Gauss who were both Germans, which may > be why I was picking on the German editors a bit, as I think both > were embarrassments to the rich intellectual history of Germanic > people. However, there are some interesting replies from Americans > as well, so here are two American editors of the New York Journal of > Mathematics to an earlier version of the same paper Advanced > I am sorry but this looks more like a submission for the chief > editor, > Mark Steinberger, than for me. I only accept submissions directly > in number theory. > > > Andrew Granville > > > > > We generally dont publish short notes. Im not sure what journals to > suggest, but NYJM probably isnt the right place for this one. > > Best, > > Mark > > > Wow, if the *chief* editor of the New York Journal of Mathematics > wouldnt know what journal to suggest, then who would? > > James Harris >> The Journal of Irreproducible Results jumps to mind. I think it >> still exists in some form even though the last I heard, the editors >> were revolting. Nobel laureates have published there. Consider also >> Hot Air which has had a print version. The headquarters, or properly >> hindquarters, seem to be at MIT. >> Chuck > http://www.ncbi.nlm.nih.gov:80/entrez/query.fcgi?cmd=Retrieve &db=PubMed& > list_uids=12385726&dopt=Abstract > I suppose if this could get published, then someone somewhere will > publish the whacktards opus. is some kind of serious scientfic work. Jens -- Kriegsverbrechen werden verfolgt werden, Kriegsverbrecher werden bestraft werden, und es wird keine Verteidigung sein zu sagen ÔIch habe nur Befehle === Subject: Re: Editors again, New York Journal of Mathematics >> [Clutter removed] >> Wow, if the *chief* editor of the New York Journal of Mathematics >> wouldnt know what journal to suggest, then who would? >> James Harris > The Journal of Irreproducible Results jumps to mind. I think it > still exists in some form even though the last I heard, the editors > were revolting. Nobel laureates have published there. Consider also > Hot Air which has had a print version. The headquarters, or properly > hindquarters, seem to be at MIT. > Chuck > http://www.ncbi.nlm.nih.gov:80/entrez/query.fcgi?cmd=Retrieve &db=PubMed& > list_uids=12385726&dopt=Abstract > I suppose if this could get published, then someone somewhere will > publish the whacktards opus. have the style. He does not know a thing about elevator operators and I doubt that he has any expertise in the structural engineering of strapless evening gowns. A snappy title might help. Methinks the likes of Factorization of Polynomials by Coercion would be better. Maybe Tilting at a Palindrome is more suggestive. No, not so good. Saltines and Polynomials has a certain nothing about it. Come to think, Ill take my first suggestion. I think he should add a proof that Alexander the Great did not exist and had an infinite number of limbs. This would demonstrate his prowess with mathematical logic. Chuck -- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net === Subject: Re: Editors again, New York Journal of Mathematics > Wow, if the *chief* editor of the New York Journal of Mathematics > wouldnt know what journal to suggest, then who would? He was probably being polite. If your ideas had sex appeal they might make the National Enquirer but short of that, too bad...... === Subject: Re: Editors again, New York Journal of Mathematics > Im an admirer of Einstein and Gauss who were both Germans, which may > be why I was picking on the German editors a bit, as I think both were > embarrassments to the rich intellectual history of Germanic people. > However, there are some interesting replies from Americans as well, so > here are two American editors of the New York Journal of Mathematics > to an earlier version of the same paper Advanced Polynomial > Factorization. > editor, > Mark Steinberger, than for me. I only accept submissions directly > in number theory. > Andrew Granville > suggest, but NYJM probably isnt the right place for this one. > Best, > Mark > wouldnt know what journal to suggest, then who would? > James Harris Hes probably politely saying, GET LOST. === Subject: Re: Editors again, New York Journal of Mathematics > Im an admirer of Einstein and Gauss who were both Germans, which may > be why I was picking on the German editors a bit, as I think both were > embarrassments to the rich intellectual history of Germanic people. > However, there are some interesting replies from Americans as well, so > here are two American editors of the New York Journal of Mathematics > to an earlier version of the same paper Advanced Polynomial > Factorization. > editor, > Mark Steinberger, than for me. I only accept submissions directly > in number theory. > Andrew Granville > suggest, but NYJM probably isnt the right place for this one. > Best, > Mark > wouldnt know what journal to suggest, then who would? Try submitting a paper that is full of horse , even according to *you*. Compare the replies that you will get with the replies you are getting now, and you will finally understand that you are merely getting polite rejections. Dirk Vdm === Subject: Twin prime conjecture restated without reference to primes In the mid-nineties, I stumbled across a trivial way of restating the twin prime conjecture without reference to primes. Ive sat on it since its so trivial, but found that others thought it a fun bit of trivia when I shared it. Perhaps the following has been shared many times before, but Ive not seen it elsewhere. So, thinking that some out there might find it an enjoyable bit of trivia on the side, here it is: To eliminate all positive c such that 6c-1 or 6c+1 is a composite, take the cross product of the set of all positive integers of the form (6a+-1) with itself: (6a+-1)(6b+-1) = 6(6ab+-a+-b)+-1 = 6c+-1 for all positive a,b and for all 4 combinations of + and -(only 3 are necessary, obviously). So the twin prime conjecture is true if and only if C = {c | 6ab+-a+-b = c for some a,b} leaves infinitely many positive integers uncovered. For economy, let C = {6ab+-a+-b), and so let R = {6ab-a-b}, S = S_1 = {6ab-a+b} = S_2 = {6ab+a-b}, and T = {6ab+a+b}. Although S = S_1 = S_2, I note both because of the matrical representations below, where S_1 and S_2 are the transposes of each other: C forms an infinite symmetric matrix M (that is actually just a modified subset of the infinite multiplication table on the positive integers), where each row or column is an ordered subset of the residue classes +-a modulo 6a-1 or modulo 6a+1: 4 6 9 11 14 16 . . . 6 8 13 15 20 22 . . . 9 13 20 24 31 35 . . . 11 15 24 28 37 41 . . . 14 20 31 37 48 54 . . . 16 22 35 41 54 60 . . . . . . . . . . . . . . . . . . . . . So the twin prime conjecture is true if and only if M leaves infinitely many positive integers uncovered. Notice that M can be broken up into four matrices: R forms the symmetric matrix: 4 9 14 . . . 9 20 31 . . . 14 31 48 . . . . . . . . . . . . S_1 forms: 6 11 16 . . . 13 24 35 . . . 20 37 54 . . . . . . . . . . . . S_2 forms: 6 13 20 . . . 11 24 37 . . . 16 35 54 . . . . . . . . . . . . T forms the symmetric matrix: 8 15 22 . . . 15 28 41 . . . 22 41 60 . . . . . . . . . . . . Since weve already restated the twin prime conjecture without reference to primes, well refer brießy here to a theorem on primes. Since these sets R, S, and T represent composites of the form 6c-1 or 6c+1, by Dirichlets Theorem (remember, 6c-1 = 6(c-1)+5), we know the following: R and T each leave infinitely many positive integers uncovered, and we know that R U T, the union of R and T, also leaves infinitely many positive integers uncovered. We know that S leaves infinitely positive integers uncovered. The trick now is to see whether (R U T) U S, the union of (R and S) and T, leaves infinitely many positive integers uncovered. Anyone care to try? Have fun. Im going sunset watching and classical music listening. (Some like fishing, some dont.) Paul === Subject: decidability and recursion theory questions impression that showing there is no algorithm to solve a problem (such as the word problem for groups, or Hilberts 10th problem) has thus far been done by showing that if the problem was decidable then the halting problem would also be decidable. Is there no need of recursion theory beyond the undecidability of the halting problem in these proofs? Recursion theory seems to be a relatively developed subject as Posts problem (concerning Turing degrees) has been solved and some things are known about the semilattice of recursively enumerable degrees. My question is this: What is it good for? Are there any proofs of the undecidability (or decidability) of a problem that depends on results in recursion theory. If not, are there important connections between recursion theory and other branches of logic (if so, what are they?) Is the solution to Posts problem simply a nice result, or does it have genuine connections to the rest of mathematics? Hugh === 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). === Subject: Re: simultaneous diagonalizability > can anyone suggest me some thorough reference on the simultaneous > diagonalizability of two or more matrices? > I would realy appreciate it. > NB: Two are more matrices A1,A2,... are simultaneously diagonalizable if > there is a matrix P such that P^T(Ai)P is diagonal for all i. > I know an equivalent characterization but i would need some deeper > analysis, special cases, etc. > Imre Polik If the matrices are Hermitian, commutativity is only sufficient. More possibilities crop up, but require more complicated conditions. Rao and Mitra, Generalized Inverse of Matrices and its Applications, Wiley, 1971, cover the topic exhaustively in Chapter 6. For general matrices, there is a book by Radjavi already mentioned in this thread. === Subject: Repeated exponentiation n@m It is a pholosophical rather than a concrete maths question.There appears there is no generally known answer to the following query about types of numbers from their comprehesive implied formative logic. Posting this in spite of some vagueness. Repeated addition is multiplication. Repeated multiplication is exponetiation. But the process need not end there ... what is Repeated exponetiation? Is anything the like of it defined ? When we analytically go backwards to break up the lump x^y united by exponentiation, we find by taking logarithms a product(of y and log(x)); when we keep at it again by taking logarithms, we get an added quantity , log(y)+ log(log(x)). Multiplication and division necessitated/spurred appearance of real, fractional and decimal numbers. So also, exponentiation of negative numbers to real numbers(non-integers) had unwittingly introduced complex numbers e.g., in [ (-1)^(1/3)].The Euler relation gives two extra complex roots on the Argand diagram for a comprehensive interpretation of a negative base exponentiation. n+n+n+á add m times -> n * m -> real numbers n*n*n*á multiply m times -> n ^ m -> complex numbers (...(((n^n)^n)^n)... ^n -> exponentiate m times -> n@m (say) -> ?? Qualitatively speaking, what new sort of numbers, that are supersets of complex numbers that may now be generally unknown ,could be involved/hidden in Repeated exponentiation n@m ? === Subject: Re: Repeated exponentiation n@m > It is a pholosophical rather than a concrete maths question.There > appears there is no generally known answer to the following query > about types of numbers from their comprehesive implied formative > logic. Posting this in spite of some vagueness. > Repeated addition is multiplication. > Repeated multiplication is exponetiation. > But the process need not end there ... what is Repeated exponetiation? > Is anything the like of it defined ? Its called tetration. Its usually written with the exponent to the upper left of the base. Using @, 4@2 = 2^2^2^2, etc. See Infinity and the Mind by Rucker for a reference. === Subject: Re: Newton-Raphson method > 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. No. Write out the equation for x_{n+2}=x_n and you get a simple differential equation for which that mirrored sqrt is the solution. V. -- mail me at lastname at cs utk edu === 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)? >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. As Stephen Herschkorn pointed out, Newton-Raphson succeeds immediately in this case. But how about f(x) = x^2 + 1? Or more generally, if there are no real roots, Newton-Raphson obviously cant converge. A more interesting question is what happens if there is a real root r. Note that there are cases where f(x) = 0 at infinitely many points, of which r is a limit point. Thus there are points arbitrarily close to r at which Newton-Raphson fails. Nevertheless, there will be some points where it does succeed. That is: Theorem: Suppose f is C^2 on the real line with f(r) = 0, and f(x) <> 0 for 0 < |x-r| <= epsilon. Then there is some x_0 with 0 < |x-r| < epsilon such that Newton-Raphson starting at x = x_0 converges to r (including the possibility that it hits r in finitely many iterations). Proof: WLOG take r = 0. Let N(x) = x - f(x)/f(x). If f(0) <> 0, then by well-known theorems Newton-Raphson converges to 0 if x_0 is sufficiently close to 0. So suppose f(0) = 0. Also suppose WLOG f(epsilon) > 0, and thus f(x) > 0 for 0 < x < epsilon. Let g(x) = f(x) - x f(epsilon)/epsilon. Since g(0) < 0 and g(0) = g(epsilon) = 0, g(x) has a minimum at some point b_1 in (0,epsilon). Since f(b_1) = f(epsilon)/epsilon > f(b_1)/b_1 > 0, 0 < N(b_1) < b_1. Now if 0 < N(x) < x for all x in the interval (0, b_1], its easy to see that Newton-Raphson starting at b_1 converges to 0. Otherwise there is some b_2 in (0, b_1) such that N(b_2) = 0, i.e. Newton-Raphson starting at b_2 goes directly to 0 (and terminates there). 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: Newton-Raphson method >Theorem: Suppose f is C^2 on the real line with f(r) = 0, and f(x) <> 0 >for 0 < |x-r| <= epsilon. Then there is some x_0 with 0 < |x-r| < epsilon >such that Newton-Raphson starting at x = x_0 converges to r (including >the possibility that it hits r in finitely many iterations). Isnt this false for f(x) = Arctan(x) and r = 0? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Newton-Raphson method >>Theorem: Suppose f is C^2 on the real line with f(r) = 0, and f(x) <> 0 >>for 0 < |x-r| <= epsilon. Then there is some x_0 with 0 < |x-r| < epsilon >>such that Newton-Raphson starting at x = x_0 converges to r (including >>the possibility that it hits r in finitely many iterations). >Isnt this false for f(x) = Arctan(x) and r = 0? No. If N(x) = x - f(x)/f(x), |N(x)| < |x| for 0 < |x| < 1.391745200 approximately. 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: 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)? Try y = x^(1/3); select any x0 != 0. === Subject: Re: Zenos paradoxes (was Re: reciting N and other sets) >Can anyone really claim that distances >much greater than 13.7 billion light years >or distances much smaller than those associated with elementary >have same kind of meaning as ordinary distances. Of course one can claim it. And such a claim is in accordance with common sense. Now, it may turn out that there will be a new Quantum Theory in which Spacetime is quantized. But, if so, that theory will be highly counterintuitive and, IAC, will have nothing to do with Zenos paradox. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply === Subject: ellipses equation Z=A*exp(i*t)+B*exp(-i*t) I want to show that the equation Z=A*exp(i*t)+B*exp(-i*t) describes an ellipse, where A and B are complex variable and t is real number. I tried to use the definition of ellipse to prove it, which is like |Z-Z_1|+|Z-Z_2|=2*a, where Z_1 and Z_2 are two fixed point on the complex plane and a is the axis. I found difficulties to find the point Z_1 and Z_2. Any one has ideas? I have been thinking about it for hours. :( === Subject: Re: ellipses equation Z=A*exp(i*t)+B*exp(-i*t) >I want to show that the equation Z=A*exp(i*t)+B*exp(-i*t) describes an >ellipse, where A and B are complex variable and t is real number. Expressing Z as a curve in R^2, Z(t) = M (cos t, sin t), a linear transformation of a circle, where the (constant) components of the matrix M determined by A and B in a straightforward manner. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: ellipses equation Z=A*exp(i*t)+B*exp(-i*t) > I want to show that the equation Z=A*exp(i*t)+B*exp(-i*t) describes an > ellipse, where A and B are complex variable and t is real number. > I tried to use the definition of ellipse to prove it, which is like > |Z-Z_1|+|Z-Z_2|=2*a, where Z_1 and Z_2 are two fixed point on the > complex plane and a is the axis. I found difficulties to find the > point Z_1 and Z_2. Any one has ideas? I have been thinking about it > for hours. :( I think that an easier approach would be to show that if z=x+iy, then x and y satisfy a quadratic equation ax^2 + by^2 + cxy +dx + ey + f = 0. where 4ac-b^2>0. Then show that any such equation gives an ellipse. -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Small prime factors and Christian Aebis 13 problem In http://www.oswego.edu/~baloglou/math/thirteen.html I present my answer to Trying to solve the problem: if n > 9 then n*(n+1)*(n+2)*(n+3) is divisible a question that Christian Aebi posted on sci.math about 13 months ago: as far as I recall and understood his question has so far remained unanswered. My solution is totally elementary but exhaustive (18x6 = 108 cases), not terribly imaginative but rather entertaining at times. It is clear that it would be very hard to follow my approach (unless *programmable*, that is) for the obvious generalizations of Aebis problem, so I am asking for other ideas and methods, including non-elementary ones. I suspect that we can have a Ramsey-like theory here -- for example, how many consecutive integers do guarantee a prime factor > or = 31, and so on -- presenting the same type of difficulties in finding sharp bounds. By the way, no hard feelings in case someone states that Theorem X implies that there are no counterexamples beyond Y, rest is ruled out by a computer search: I worked on Aebis problem and prepared the thirteen web page mostly for fun* and for the sake of those who would not know or understand that Theorem X; indeed I feel that my solution could find good use in the classroom, asking students to come up with alternative solutions for each case, for example. *My ultimate reward is a recurring question: all this HAD to work, but WHY? baloglouAToswego.edu === Subject: Re: Rounding to zero The round to zero dilemma is quite interesting. In the NASDAQ and NYSE and other exchanges they only trade in certain increments, where every time regulatory agencies have decreased the increments, traders profits have gone done substantially. The trader would rather have the wares be sold in only say 10 dollar increments, as opposed to say .01 dollar increments, for (with a monopoly) the buyer must buy, so every thing gets rounded UP (not to the nearest increment, not down). If the ware has value of say .01 to 10.00, with a 10 dollar increment, then all those value ranges must funnel up to 10.00 These value to price ratios will then put pressure on the 10.01 to 20.00 to go to 20.00 as opposed to 10.00, for that lower cost range is already used up. The moral of the study is the smaller the price increment the better! === 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. Very appropriate answer, considering that the whole thread is about nonconstructive existence. Those who dont like this answer should perhaps ask themselves whether they like nonconstructive existence. -- hendrik > ************************ > David C. Ullrich === Subject: Re: existence proof |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. Perhaps you have in mind multiplying together the polynomial and all of its conjugate polynomials. That gets you a polynomial with rational coefficients which has the roots of the original polynomial among them. The fact that that works kind of depends upon the facts developed in the more usual, pedestrian proof which goes like this. First you show that b is algebraic over a field F if and only if the smallest field containing F and b, F(b), is finite-dimensional over F. Then you show that a finite extension of a finite extension is also finite. (If L is finite over K, and K is finite over F, pairs consisting of an element from a basis of L over K and an element from a basis of K over F form a basis of L over F. So the dimension of L as a vector space over F is the product of the dimension of L as a vector space over K and the dimension of K as a vector space over F.) Then you show that an extension obtained by adjoining a finite number of elements algebraic over F, F(b1,...,bn), is also finite over F. Then observe that since a root r of a polynomial with coefficients b1,...,bn is algebraic over F(b1,...,bn), it is in a finite extension of F(b1,...,bn). A finite extension of a finite extension is also finite extension, so F(b1,...,bn,r) is finite over F. Then F(r) is also finite, since it is contained in F(b1,...,bn,r), which implies r is algebraic. The trick involving companion matricies which is most often mentioned on sci.math is to show that the sum and product of algebraic numbers is algebraic, I think. What I have above is enough to show that this is true already because b1+b2 and b1*b2 are elements of F(b1,b2), so they are algebraic. We want a concrete approach, however. A companion matrix has the roots of a polynomial as its eigenvalues. There are manipulations which take two matricies A and B and produce matricies whose eigenvalues are the products and sums respectively of an eigenvalue of A and an eigenvalue of B. I think I learned this from a posting of Robert Israel. The product construction is the tensor or Kronecker product of the two matricies. A tensor B is a matrix whose rows are indexed by pairs consisting of an index into the rows of A and an index into the rows of B, and whose columns are likewise indexed by pairs of indices into the columns of A and B. The entry (A tensor B)_{(i,j),(k,l)} = A_{i,k} B_{j,l}. I was having trouble remembering the sum construction, but its just A tensor I + I tensor B, where the identity matricies I are the same size as B and A respectively. He also pointed out that it shows the sum and product of algebraic integers are algebraic integers. This thread is roughly about pure existence proofs which are nonconstructive existence proofs. Ill finish by pointing out that this proof, while its not a constructive proof that one of pi+e or pi*e is transcendental (since it doesnt enable us to tell which one), is an unproblematic constructive proof that they are not both algebraic. This reßects the actual nature of the proof; saying and therefore one of them is transcendental is just a frill. Keith Ramsay === Subject: Arithm. problem (residue classes, congruence) Hi all, Here is a little problem (asked by a guy absolutely crap in arithmetics) : let p/q, p/q be two ireductible rationals in [0,1], consider the sequence (pq.i^2 + qp.i) mod (qq) when i runs over [0,qq]. I am looking for: 1. the periodicity of the sequence (for the moment, say its qq) 2. the SUM of all terms over a period. I am completely stumped by the latter question, and all I could find about residue classes and congruences did not help a lot so far. If evaluating the exact sum does not seem feasable, could it be possible to get some good bounds? Any suggestion will be deeply appreciated. Xavier PS: some examples: 1. (p,q,p,q)=(1,3,3,5) 0 14 8 12 11 5 9 8 2 6 5 14 3 2 11 0 14 8 12 11 5 9 8 2 6 5 14 3 2 11 0 14 8 12 11 5 9 8 2 6 ... periodicity 15, sum 110 2. (p,q,p,q)=(1,4,3,4) 0 0 8 8 0 0 8 8 ... periodicity 4, sum 16 3. (p,q,p,q)=(3,4,3,4) -- same as before, not a surprise 4. (p,q,p,q)=(2,5,2,5) 0 20 10 20 0 0 20 10 20 ... periodicity 5, sum 50 5. (p,q,p,q)=(2,7,3,4) 0 1 18 23 16 25 22 7 8 25 2 23 4 1 14 15 4 9 2 11 8 21 22 11 16 9 18 15 0 1 18 23 16 25 22 7 8 25 2 23 ... periodicity 28, sum 350 6. (p,q,p,q)=(3,4,3,5) 0 7 4 11 8 15 12 19 16 3 0 7 4 11 8 15 12 19 16 3 0 7 4 11 8 15 12 19 16 3 0 7 4 11 8 15 12 19 16 3 ... periodicity 10, sum 79 7. (p,q,p,q)=(3,5,3,4) 0 7 18 13 12 15 2 13 8 7 10 17 8 3 2 5 12 3 18 17 0 7 18 13 12 15 2 13 8 7 10 17 8 3 2 5 12 3 18 17 ... periodicity 20, sum 190 -- ------------------------------------------------------- Xavier HILAIRE | LORIA 34 avenue de la Garenne | 615 rue du Jardin Botanique F-54000 NANCY | F-54602 VILLERS-LES-NANCY FRANCE | FRANCE Tel: +33 3 83 41 86 46 | Tel: +33 3 83 59 20 90 E-mail: xhilaire#free!fr | E-mail: hilaire#loria!fr (#,!)=(@,.) | (#,!)=(@,.) ------------------------------------------------------- === Subject: How to factorize this polynomial? Given two bi-variate factorized functions depending on variables x and y : poly1=(x-a1)...(x-an).(y-b1)...(y-bn) poly2=(x-c1)...(x-cn).(y-d1)...(y-dn) How can i factorize (poly1+poly2)? If not possible theoretically, is there software available to do it? (ai,bi, ci and di for i=1:n are real numbers) J. Miller (PS. In a previous post, i made a mistake and said factorial form, that should be factorized form. Sorry for my bad english) === Subject: Re: derivations of the simple continued fraction for e > [re-posted] > (3) http://mathworld.wolfram.com/e.html claims the desired result > follows immediately from re-writing the usual power series as > e = 2 + (1/2)(1 + (1/3)(1 + (1/4)(1 + (1/5)(1 + ...)))). > I would appreciate a hint on how that works, as I must be missing > something obvious. > Is that a false claim by MathWorld? After 2 weeks and a re-posting, Im amazed that no one seems to have bothered to look at this. --r.e.s.