mm-401 === Subject: ((log x)^2) + ((log x)^-1) -12 = 0 ((log x)^2) + ((log x)^-1) -12 = 0 ...could someone please point me in the direction to solving this equation. I would request that you do not reveal the answer; I just need a little help. === Subject: Re: ((log x)^2) + ((log x)^-1) -12 = 0 Content-type: text/plain; charset=US-ASCII Content-transfer-encoding: 7bit 94b9ce07.0403061834.4f9414d5@posting.google.com, jussy > ((log x)^2) + ((log x)^-1) -12 = 0 > ...could someone please point me in the direction to solving this > equation. I would request that you do not reveal the answer; I just > need a little help. Think of log x as just x. Solve the cubic. Answer below... I assume that log x is the natural log -> lnx. Substitute log x with x. x^2 + 1/x - 12 = 0. Multiplying both sides by x gives the cubic x^3 - 12x + 1 = 0 Solving gives 3 real roots, which are (approx.) -3.5050397246... .0833816425608... 3.421658082059... Now, there is no real number (but there _is_ an imaginary one) whose natural log is -3.505... So, the two real roots are approx. 1.08695655828 And 30.62014366415 Barnaby === Subject: Re: ((log x)^2) + ((log x)^-1) -12 = 0 ETAtAhUAhWozBhcFdeflCEzQfpgGQ7cJkioCFGwcFnGUNajliYQDo05cFM35r6H A Sure there is a real number whose natural log is -3.5. It's exp(-3.5) = 0.03 or something like that. === Subject: Re: ((log x)^2) + ((log x)^-1) -12 = 0 OK, so im trying to solve this equation algebraically. I just know my teacher put this in here for me specifically. Unfortunatly, we have not been _enlightened_ as to systematicly solving cubic functions. Its obviously in the nice form of __x^3 -12x +1 = 0__. ******** first of all ----- x^3 -12x +1 =0 x^3 +px +q =0 x^3 +(-12)x +(1) =0 p=-12 q=1 ******** So, using what I could find on this process, I created the dummy variables u and t as follows: u-t=1 -> u=t+1 -> t=u-1 tu=(p/3)^3 -> u=(-64)/t -> t=(-64)/u when I substitute to find the variables they are not real. > 94b9ce07.0403061834.4f9414d5@posting.google.com, jussy > ((log x)^2) + ((log x)^-1) -12 = 0 > ...could someone please point me in the direction to solving this > equation. I would request that you do not reveal the answer; I just > need a little help. > Think of log x as just x. Solve the cubic. > Answer below... > I assume that log x is the natural log -> lnx. Substitute log x with x. > x^2 + 1/x - 12 = 0. > Multiplying both sides by x gives the cubic > x^3 - 12x + 1 = 0 > Solving gives 3 real roots, which are (approx.) > -3.5050397246... > .0833816425608... > 3.421658082059... > Now, there is no real number (but there _is_ an imaginary one) whose natural > log is -3.505... > So, the two real roots are approx. > 1.08695655828 > And > 30.62014366415 > Barnaby === Subject: Re: ((log x)^2) + ((log x)^-1) -12 = 0 Content-type: text/plain; charset=US-ASCII Content-transfer-encoding: 7bit 94b9ce07.0403071410.6a2bf629@posting.google.com, jussy > OK, so im trying to solve this equation algebraically. I just know my > teacher put this in here for me specifically. Unfortunatly, we have > not been _enlightened_ as to systematicly solving cubic functions. > Its obviously in the nice form of __x^3 -12x +1 = 0__. > ******** > first of all ----- > x^3 -12x +1 =0 > x^3 +px +q =0 > x^3 +(-12)x +(1) =0 > p=-12 > q=1 > ******** > So, using what I could find on this process, I created the dummy > variables u and t as follows: > u-t=1 -> u=t+1 -> t=u-1 > tu=(p/3)^3 -> u=(-64)/t -> t=(-64)/u > when I substitute to find the variables they are not real. I think you'll find that when you use the formula for solving cubics, that even if there are three real roots, the expressions you'll generate will all have imaginary numbers in them, numbers that will cancel out and disappear with algebraic manipulation. Barnaby === Subject: Re: ((log x)^2) + ((log x)^-1) -12 = 0 > Now, there is no real number (but there _is_ an imaginary one) whose natural > log is -3.505... What a strange thing to say. === Subject: Re: ((log x)^2) + ((log x)^-1) -12 = 0 Content-type: text/plain; charset=US-ASCII Content-transfer-encoding: 7bit >Now, there is no real number (but there _is_ an imaginary one) whose > natural >log is -3.505... > What a strange thing to say. Sorry, Clive. I was asleep at the wheel. I was thinking there was no real number for ln(-3.505...), but of course, that doesn't apply here. I was backwards. The third root is .03004558... Barnaby === Subject: Re: ((log x)^2) + ((log x)^-1) -12 = 0 > ((log x)^2) + ((log x)^-1) -12 = 0 > ...could someone please point me in the direction to solving this > equation. I would request that you do not reveal the answer; I just > need a little help. Let y = log(x), and you get a cubic equation in y. There are 3 real but irrational roots for the equation in y, which makes exact solution for y difficult. Try numerical methods. === Subject: Dividing Into Numerator of Harmonic-Number Difference If p = an odd prime, and m = positive integer, m < p, then: p divides into the numerator of |H(p-m-1) - H(m)| which either is 1/(p-m-1) + 1/(p-m-2) +...+1/(m+1) or is 1/m +1/(m-1) +...+1/(p-m). (H(m) = 1 + 1/2 +1/3 +...+1/m.) (So, we have a semi-generalization of Wolstenholme's theorem, where Wolstenholme's theorem deals with m=0, and says instead that *p^2* divides into the numerator, not just p.) So, I wonder, what is the highest power of p that divides the numerator for any given m? If p^a(m) is this highest power of p which divides into the numerator of |H(p-m-1) - H(m)|, I believe: if a(m-1) >= 2, then a(m) = 1; if a(m-1) = 1, then p^(a(m)-1)|(num(m-1)-den(m-1)), where num(m)/den(m) = |H(p-m-1) - H(m)|, and num(m) and den(m) are integers where GCD(num(m),den(m))=1. - Bonus, an reposted related theorem: If p = odd prime, then p divides (2*numerator(H(p-3)) -3*denominator(H(p-3))) Leroy Quet === Subject: Re: Dividing Into Numerator of Harmonic-Number Difference > If p = an odd prime, > and m = positive integer, m < p, > then: > p divides into the numerator of > |H(p-m-1) - H(m)| > which either > is 1/(p-m-1) + 1/(p-m-2) +...+1/(m+1) > or is > 1/m +1/(m-1) +...+1/(p-m). > (H(m) = 1 + 1/2 +1/3 +...+1/m.) > (So, we have a semi-generalization of Wolstenholme's theorem, where > Wolstenholme's theorem deals with m=0, and says instead that *p^2* > divides into the numerator, not just p.) > So, I wonder, what is the highest power of p that divides the > numerator for any given m? > If p^a(m) is this highest power of p which divides into the numerator > of > |H(p-m-1) - H(m)|, > I believe: > if a(m-1) >= 2, then a(m) = 1; > if a(m-1) = 1, then p^(a(m)-1)|(num(m-1)-den(m-1)), > where num(m)/den(m) = > |H(p-m-1) - H(m)|, > and num(m) and den(m) are integers where > GCD(num(m),den(m))=1. > - No progress on the above. But as for the result below, I have bottom-posted some things. > Bonus, an reposted related theorem: > If p = odd prime, then > p divides (2*numerator(H(p-3)) -3*denominator(H(p-3))) > Leroy Quet I had forgotten my original proof. But here is an alternative proof I have written-up earlier today: First, we have: n(k)/d(k) = H(k), where n and d are integers and GCD(n(k),d(k))=1, And: p^2|n(p-1) by Wolstenholme's theorem. (All we need here is p^1|n(p-1).) And H(p-3) = H(p-1) -1/(p-2) -1/(p-1) = ((p-1)(p-2)n(p-1) -d(p-1)(2p -3))/(d(p-1)(p-1)(p-2)). So, n(p-3)*m = (p-1)(p-2)n(p-1)-d(p-1)(2p-3), and d(p-3)*m = d(p-1)(p-1)(p-2), (*) for some m. So, 2*n(p-3) -3*d(p-3) = (1/m)*(2(p-1)(p-2) n(p-1)-2d(p-1) (2p-3) -3 d(p-1)(p-1)(p-2)) = (1/m) (k*p +6 d(p-1) -6 d(p-1)), for some integer k, which is a multiple of p if m is coprime with p. But since GCD(d(p-1),p) = GCD(p-1,p) = GCD(p-2,p) = 1 (GCD(d(p-1),p)=1 because p|n(p-1)), then, from (*), GCD(d(p-3)*m,p) = 1, and so GCD(m,p) =1. Qed. Leroy Quet === Subject: Re: upper bound for a sum? >Ignore my other reply to this thread! I did not see that the OP wanted >the sum over the PRIMES <= n-1. >Still I wonder if I was right anyway, assuming the sum is over all >positive integers <= n. >Leroy Quet >> I'm looking for a good upper bound of > n-1 >> ----- >> n (mod p) >> f(n) = ) ---------- >> / p >> ----- >> p = 2 > where the sum is over the primes (less than n). >> I think I might be able to get something like f(n) <= n, >> but I was hoping for something better (asymptotic bounds >> are good enough). > -Tyler >>limit{m -> oo} >>(1/m) sum{k=2 to m} (m(mod k))/k = >>1 -c, >>where 0 <= m(mod k) <= k-1, >>and where c = Euler's constant (which is sometimes referred to by gamma). >So, I do not know about upper/lower bounds from this limit, but it is >interesting that >f(n) is therefore asymptotical to n*(1-c), question works. Instead of letting I(n,k) be the primes in the interval [n/(k+1),n/k), let it be all of that interval. Then we get the number of elements of I(n,k) is n/(k(k+1)). Then asymptotically the sum is 1 n ( 1-gamma + O( - ) ) [5] n Rob Johnson 0 and all integer n 1. find int f(z) dz , C is any simple closed curve on complex plane C 2. suppose that there is exist positive number K such that |f(z)| <= K let f(i) = 3i find f(2+i) -------------------------------------------- um.....i listened that this problem is contained error. but i can't find error. if this problem is right, how to solve it? let me advice.....please.....thank you === Subject: Re: complex problem.... > let complex function f(z) is possible to expand taylor series about z = z_0 > let |f^(n)(z_0)| <= M for all integer M >0 and all integer n oh...i am sorry...i miss let |f^(n)(z_0)| <= M for som integer M >0 and all integer n very sorry....... > 1. find > int f(z) dz , C is any simple closed curve on complex plane > C > 2. suppose that there is exist positive number K such that |f(z)| <= K > let f(i) = 3i > find f(2+i) > -------------------------------------------- > um.....i listened that this problem is contained error. > but i can't find error. > if this problem is right, how to solve it? > let me advice.....please.....thank you === Subject: Re: complex problem.... > oh...i am sorry...i miss > let |f^(n)(z_0)| <= M for som integer M >0 and all integer n > very sorry....... Where is f defined? Is it analytic? You're asking for help from us, so help us by being as clear as you can. === Subject: Re: complex problem.... hot-girl > let complex function f(z) is possible to expand taylor series about z = z_0 > let |f^(n)(z_0)| <= M for all integer M >0 and all integer n Meaning for some constant M>0? > 1. find int f(z) dz , C is any simple closed curve on complex plane See below about bounded functions. > 2. suppose that there is exist positive number K such that |f(z)| <= K > let f(i) = 3i > find f(2+i) If a meromorphic function is bounded on the whole complex plane, it is constant, so we get f(2+i) = 3i. This gives part of the answer, but I don't think the problem is accurately stated. === Subject: Quadratic Fields, Class Numbers and Fundamental Units. Piqued a bit by John Robertson's reply, I hacked up a quick-and-dirty bignum package to calculate fundamental units of real quadratic fields. (And, the fundamental unit of Q(9601) contains 84 digit integers.) I further did look at class numbers. For complex quadratic fields with William Hale's applet and for real quadratic fields with some stuff by Keith Matthews (and of course, Henri Cohen's book would be more comprehensive). I found two surprising things: 1. Class numbers of complex quadratic fields increase much faster than those for real quadratic fields. The largest class number for complex quadratic fields with -10000 < D < 0 is 139 (for D = -9239), for real quadratic fields with 0 < D < 10000 it is 27 (D = 8761). Gauss already had a conjecture about the minimal value of the class number of a complex quadratic field. Somesuch has later been proven. my data it appears that there is no minimal class number for large discriminants. There are still plenty UFD's with 9000 < D < 10000. 2. More surprising, to me at least. It appears that the magnitude of the integers found in fundamental units in real quadratic fields is related to the class number. A high class number appears to imply small integers. I have set up two charts, one showing records in the integers in fundamental units, the second being records in class numbers. The first shows only UFD's, the second only fundamental units with integers < 1000. Is there someone who has an explanation? 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: Quadratic Fields, Class Numbers and Fundamental Units. >I found two surprising things: A good summary of much of what you have found, or are asking about, is in Paulo Ribenboim's, My Numbers, My Friends, Springer, 2000, Chapter 6. This is an expansion of a popular lecture and gives lots of history, formulas, conjectures, you name it. It is well worth tracking down or buying. A few things you might know, and along the lines of what Robin discussed: 1. It is known that for every n >=1, there are only finitely many discriminants D < 0 with class number n. out every negative discriminant with class number n for n up to 100. 3. (Further to Robin's comment) For positive discriminants D, in 1936 Siegel showed that log(h(D) log epsilon_D) ~ log sqrt(D) asymptotically as D goes to infinity, where epsilon_D is the fundamental unit. 4. It is conjectured that there are arbitrarily large positive discriminants with class number 1. 5. Below I give a way to compute log epsilon_D when epsilon_D is too big to compute even approximately. Apologies for not having time to un-Tex it. 6. As you noted Henri Cohen's Course in Computational Algebraic Number Theory is a good source. Others are Harvey Cohen's A Classical Introduction to Algebraic Numbers and Class Fields and Richard Mollin's Quadratics. Also, the GP calculator in PARI computes class numbers easily. John Robertson Computing the log of the fundamental unit: For a fundamental discriminant Delta_0, here's a method to compute $ln(epsilon)$ when $epsilon$ (the fundamental unit) is too large to compute conveniently (I think this works just as well for non-fundamental disciminants generalized Pell equation,'' at http://hometown.aol.com/jpr2718/ for this algorithm and the notation used in what follows. If $Delta_0 equiv 1 pmod{4}$ then for the PQa algorithm take $P_0=1$, $Q_0=2$ and $D=Delta_0$. If $Delta_0 equiv 0 pmod{4}$ then take $P_0=0$, $Q_0=1$ and $D=Delta_0/4$. Don't bother computing $A_i$, $B_i$, or $G_i$. Compute a value we will denote $L_i$ as follows: [ L_0 = ln(sqrt{D} + Q_0 a_0 - P_0) - P_0 ln(2), ] [ L_i = L_{i-1} + ln( frac{P_i + sqrt{D}}{Q_i} ). ] If the length of the period of the continued fraction expansion of $frac{P_0 + sqrt{D}}{Q_0}$ is $ell$, then $ln(epsilon) = L_{ell - 1}$. === Subject: Re: Quadratic Fields, Class Numbers and Fundamental Units. Possibly also worth mentioning: For d > 0, squarefree, let t, u be the least positive solution to x^2 - dy^2 = 4. Let e=(t + u*sqrt(d))/2. Then Schur, Gottingen Nachrichter, 1918, pp. vol 48, 1942, p 731, shows for d == 0 or 1 (mod 4) that log e < (d^(1/2))*((1/2) log d + 1). John Robertson === Subject: Re: Quadratic Fields, Class Numbers and Fundamental Units. > Possibly also worth mentioning: > For d > 0, squarefree, let t, u be the least positive solution to x^2 - dy^2 = > 4. Let e=(t + u*sqrt(d))/2. Then Schur, Gottingen Nachrichter, 1918, pp. > vol 48, 1942, p 731, shows for d == 0 or 1 (mod 4) that log e < > (d^(1/2))*((1/2) log d + 1). 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: Quadratic Fields, Class Numbers and Fundamental Units. > Piqued a bit by John Robertson's reply, I hacked up a quick-and-dirty > bignum package to calculate fundamental units of real quadratic fields. > (And, the fundamental unit of Q(9601) contains 84 digit integers.) I > further did look at class numbers. For complex quadratic fields with > William Hale's applet and for real quadratic fields with some stuff > by Keith Matthews (and of course, Henri Cohen's book would be more > comprehensive). I found two surprising things: > 1. Class numbers of complex quadratic fields increase much faster than > those for real quadratic fields. The largest class number for > complex quadratic fields with -10000 < D < 0 is 139 (for D = -9239), > for real quadratic fields with 0 < D < 10000 it is 27 (D = 8761). > Gauss already had a conjecture about the minimal value of the class > number of a complex quadratic field. Somesuch has later been proven. > my data it appears that there is no minimal class number for large > discriminants. There are still plenty UFD's with 9000 < D < 10000. > 2. More surprising, to me at least. It appears that the magnitude of > the integers found in fundamental units in real quadratic fields is > related to the class number. A high class number appears to imply > small integers. I have set up two charts, one showing records in > the integers in fundamental units, the second being records in class > numbers. The first shows only UFD's, the second only fundamental > units with integers < 1000. Is there someone who has an explanation? Look up the Brauer-Siegel theorem. This is an asymtpotic formula in terms of the discriminant for h_K R_K where K ranges through fields of a given degree with a given number of real embeddings, h_K is the class number and R_K is the regulator. For imaginary quadratic fields the regulator is 1. For real quadratic fields it is log epsilon, where epsilon is the fundamental unit. We thus have a good estimate for h log epsilon in 2. This means that for small fundamental unit we expect large classnumber and vice versa. Of course h_K R_K is what the analytic class number formula spits out. Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: easy algebra problem.... let order of finite group G is 155 if G is not abelian , the number of Sylow 5-subgroup is 31 ------------------------------------- um......i know |G| = 155 = 5*31 the number of sylow 5-subgroup is form 5k+1 thus 1,6,11,16,21,26,31,36,41....... thus the number of sylow 5-subgroup is 1 or 31 if the number of sylow 5-subgroup is 1 , what's happen contradiction? help....me......please...... === Subject: Re: easy algebra problem.... > let order of finite group G is 155 > if G is not abelian , > the number of Sylow 5-subgroup is 31 > ------------------------------------- > [...] > the number of sylow 5-subgroup is 1 or 31 > if the number of sylow 5-subgroup is 1 , > what's happen contradiction? Observe that G already has exactly 1 Sylow-31 subgroup. === Subject: Logic question #7 Hello again guys, I've seen the following question appear several times on the scholarship past papers but I'm not too sure I understand them. Anyway, I'm putting one on here to see how it is approached, and then once I understand, hopefully I'll be able to work the others. 3 days to go till the exam! -------------------------------------------------------------- -------------- -- Several points, labelled P, Q, ..., etc., are drawn on a piece of paper in such a way that no three of the points lie on the same straight line. (I) If there are 4 points, in how many different ways can pairs of these points be connected by straight lines in such a way that each point is connected to two others? (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 (II) If there are 4 points, in how many different ways can pairs of these points be connected by straight lines in such a way that each point is connected to three others? (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 (III) If there are 5 points, in how many different ways can pairs of these points be connected by straight lines in such a way that each point is connected to two others? (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 (IV) If there are 5 points, in how many different ways can pairs of these points be connected by straight lines in such a way that each point is connected three others? (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 -------------------------------------------------------------- -------------- -- === Subject: Re: Logic question #7 Here is some other way to explain the reasoning: * heidiannjames@yahoo.com > (I) If there are 4 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to two others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 This is equivalent to place the points in a loop, each point connected to its predecessor and antedecessor. How many ways are thre? Well, point A may have 3 predecessors which leavs 2 choices for the follower point, i.e. 6. But then we have counted all configurations twice, because a forward loop is the same as a backward loop. > (II) If there are 4 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to three others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 Well, obviously just one. > (III) If there are 5 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to two others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 This is the difficult one. But we can use the same reasoning as in (I). We also here can observe that we must count the number of different loops. Why? Because once first point, A, is connected to two points, say B and E, both B and E needs another point. If they are connected to each other, then C and D cannot be connected to any of these three, and it is impossible to fulfill the requirements. Thus B must be connected to a new point, say C, again we cannot connect C to E (leaving out D all alone), so we need to put D into the loop. Therefore any legal configuration is a loop. How many loops are there? Well, there are 5!=120 ways to order five objects. But in a loop there are no start point, so we have to divide by 5, leaving 24 different loops. And again we cannot regard a forward loop (say ABCDE) different from the reversed loop (EDCBA), so we have to divide by two giving 12 as the correct answer. > (IV) If there are 5 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected three others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 This is the fun question. If there are three lines leaving each of the five points we can count line end points: 5 times 3 = 15. But we cannot have an odd number of line end points. Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92 === Subject: Re: Logic question #7 > Hello again guys, > I've seen the following question appear several times on the > scholarship past papers but I'm not too sure I understand them. > Anyway, I'm putting one on here to see how it is approached, and then > once I understand, hopefully I'll be able to work the others. 3 days > to go till the exam! > -------------------------------------------------------------- ------------ ---- > Several points, labelled P, Q, ..., etc., are drawn on a piece of > paper in such a way that no three of the points lie on the same > straight line. > (I) If there are 4 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to two others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > (II) If there are 4 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to three others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > (III) If there are 5 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to two others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > (IV) If there are 5 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected three others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > -------------------------------------------------------------- ------------ ---- I. C - 3. ab, bc, cd, da; ab, bd, dc, ca; ac, cb, bd, da. II. B - 1. III. E - 12. ab, bc, cd, de, ea; ac, ce, eb, bd, da; ab, bc, ce, ed, da (plus four similar rotations); ac, cb, be, ed, da (plus four similar rotations). IV. A - 0. === Subject: Re: Logic question #7 > Hello again guys, > I've seen the following question appear several times on the > scholarship past papers but I'm not too sure I understand them. > Anyway, I'm putting one on here to see how it is approached, and then > once I understand, hopefully I'll be able to work the others. 3 days > to go till the exam! > -------------------------------------------------------------- ------------ > ---- > Several points, labelled P, Q, ..., etc., are drawn on a piece of > paper in such a way that no three of the points lie on the same > straight line. > (I) If there are 4 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to two others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > (II) If there are 4 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to three others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > (III) If there are 5 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to two others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > (IV) If there are 5 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected three others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > -------------------------------------------------------------- ------------ > ---- > I. C - 3. ab, bc, cd, da; ab, bd, dc, ca; ac, cb, bd, da. > II. B - 1. > III. E - 12. ab, bc, cd, de, ea; ac, ce, eb, bd, da; ab, bc, ce, ed, da > (plus four similar rotations); ac, cb, be, ed, da (plus four similar > rotations). > IV. A - 0. Is there anyway to work out the answers without having to list the possible combinations? A formula perhaps (Wishful thinking? :D) === Subject: Re: Logic question #7 > Hello again guys, > I've seen the following question appear several times on the > scholarship past papers but I'm not too sure I understand them. > Anyway, I'm putting one on here to see how it is approached, and then > once I understand, hopefully I'll be able to work the others. 3 days > to go till the exam! > -------------------------------------------------------------- ------------ > ---- > Several points, labelled P, Q, ..., etc., are drawn on a piece of > paper in such a way that no three of the points lie on the same > straight line. > (I) If there are 4 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to two others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > (II) If there are 4 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to three others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > (III) If there are 5 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to two others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > (IV) If there are 5 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected three others? > (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > -------------------------------------------------------------- ------------ > ---- > I. C - 3. ab, bc, cd, da; ab, bd, dc, ca; ac, cb, bd, da. > II. B - 1. > III. E - 12. ab, bc, cd, de, ea; ac, ce, eb, bd, da; ab, bc, ce, ed, da > (plus four similar rotations); ac, cb, be, ed, da (plus four similar > rotations). > IV. A - 0. > Is there anyway to work out the answers without having to list the > possible combinations? A formula perhaps (Wishful thinking? :D) I started with a simple diagram of 4 dots arranged roughly in a square and drew in all the line pairs. It was easy then to see the answers to I and II. With 5 points, I started with the fully-connected sketch again, then drew a few trial sketches, recognizing the equivalence of rotating some of the figures, === Subject: Re: Logic question #7 > Hello again guys, I've seen the following question appear several times on the > scholarship past papers but I'm not too sure I understand them. > Anyway, I'm putting one on here to see how it is approached, and then > once I understand, hopefully I'll be able to work the others. 3 days > to go till the exam! -------------------------------------------------------------- ------------ > ---- Several points, labelled P, Q, ..., etc., are drawn on a piece of > paper in such a way that no three of the points lie on the same > straight line. (I) If there are 4 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to two others? (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > (II) If there are 4 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to three others? (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 > (III) If there are 5 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected to two others? (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 (IV) If there are 5 points, in how many different ways can pairs of > these points be connected by straight lines in such a way that each > point is connected three others? (A) 0 (B) 1 (C) 3 (D) 6 (E) 12 -------------------------------------------------------------- ------------ > ---- > I. C - 3. ab, bc, cd, da; ab, bd, dc, ca; ac, cb, bd, da. > II. B - 1. > III. E - 12. ab, bc, cd, de, ea; ac, ce, eb, bd, da; ab, bc, ce, ed, > da > (plus four similar rotations); ac, cb, be, ed, da (plus four similar > rotations). > IV. A - 0. > Is there anyway to work out the answers without having to list the > possible combinations? A formula perhaps (Wishful thinking? :D) > I started with a simple diagram of 4 dots arranged roughly in a square and > drew in all the line pairs. It was easy then to see the answers to I and > II. > With 5 points, I started with the fully-connected sketch again, then drew a > few trial sketches, recognizing the equivalence of rotating some of the > figures, Oh ok...thanks Richard. Forgive me if I sound a bit (dumb?), I left high school years ago (5) and never went on to college so my brain's a bit slow right now. Wish me luck on my exam, I have it on Tuesday. Heidi === Subject: Re: Group >If H,K are group then HK is also a group ? Well, if H,K are *not* considered as subsets of ->one<- set with a group or other algebraic structure including an operation written as multiplication, HK doesn't mean anything, as far as I know ... (Do you mean the cartesian product HxK ? BTW, that ->has<- in fact a standard group structure defined.) If H,K *are* subgroups of a group G, then HK = set of all products hk for all pairs (h,k) in HxK, which ->may be<- not a subgoup, when G is non-abelian. If G *is* abelian, then (h,k) |-> hk is a homomorphism from the group HxK into G, so that its image, which is HK, *is* a subgroup of G. Counter-example with G non-abelian: take G=the group of permutations of a set with 3 elements, h and k two different elements of order 2 in G, so that H={1,h} and K={1,k} are subgroups of G. The elements 1,h,k,hk of HK are all different, so HK has 4 elements, but G has no subgroup of order 4 (the order of G is 6 and the order of a subgroup of G must therefore be a divisor of 6) ! In fact, kh does not belong to HK. >When H,K are groups, >If H or K are finite group then HK is also a group ? Not necessarily, all I say above still applies. Z_5 = {5k, 5k+1, 5k+2, 5k+3, 5k+4|k in Z} and > Z_6={6k, 6k+1, ... , 6k+5 | k in Z} These definitions are not standard writing. But I guessed what you mean, my guess seeming confirmed by what you write next ... That would be the set of classes of integers modulo 5 respectively 6. (This set is *not* made of integers, as the notation above might suggest, only the *members* of this set - the mentioned classes - are, and one of them for instance only contains 5k+2, but for all k in Z) > are finite subgroup of addictive group Z It is called additive (coming from the verb 'to add') - groups (of math.) usually do not drink alcohol or take drugs or loose money in casinos etc. ;-) (just your 'typo', I suppose ...) And: these groups you call Z_5 and Z_6 are *not* subgroups of Z (which has only the singleton {0} as a finite subgroup), they are so-called quotient groups of Z: in fact, they are usually named Z/5Z and Z/6Z (5Z and 6Z meaning the - infinite - subgroups of Z made of all multiples of 5 respectively 6) >AND order of Z_5 = 5 , order of Z_6 =6 True - if I guessed right about the intention of your definitions. >Are these 2 statements TRUE ? >(...) === Subject: Re: F(x) = 0 for all x in [0,1] ? Content-Length: 672 Originator: rusin@vesuvius > Let F:[0,1]-->R be continuous on [0,1] and such that Integral_{t=0 to t=1}F(2xt/(x+1)) dt = 0 for all x in [0,1]. > It's true that F=0 on [0,1] ? Suppose F in C[0,1]. For which functions k:[0,1]x[0,1]--->[0,1] from [1] Integral_{t=0 to t=1}F(k(x,t)) dt = 0 for all x in [0,1], it follows that [2] F=0 on [0,1] ? For instance, if k(x,t)=x*g(t) , g:[0,1]-->R being increasing with g(0)=0 , g(1)=1 , then [1] implies [2] ? === Subject: Re: F(x) = 0 for all x in [0,1] ? Content-Length: 1040 Originator: rusin@vesuvius >Let F:[0,1]-->R be continuous on [0,1] and such that Integral_{t=0 to t=1}F(2xt/(x+1)) dt = 0 for all x in [0,1]. >It's true that F=0 on [0,1] ? Will a proof do? Note that a(x) = 2x/(x+1) maps [0,1] to [0,1] nicely. Therefore, let's rephrase your hypothesis to |1 | F(at) dt = 0 [1] | 0 for all a in [0,1]. Now, changing variables, we get 1 |a - | F(t) dt = 0 [2] a | 0 Multiplying [2] by a then differentiating with respect to a, we get F(a) = 0 [3] When a = 0 there is the difficulty of dividing by 0, but we can easily handle the case a = 0 with [1] which says directly that F(0) = 0. Rob Johnson Prove that the sum of tan^2(2x+1) , where x is from 0 to 44, = 4005. >tan^2(1) + tan^2(3) + tan^2(5) + ... + tan^2(89) = 4005 >Are you sure that you've got the problem formulated correctly? A quick >Mathematica calculation gives me a sum of approximately 59,934. > It's 4005. I was given a hint to use a polynomial of degree 45 but just can't think of it. === Subject: Re: Satellite Dish by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i22DUBU01730; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i223Wli03493 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i223WlX20381; >
I was wondering, what exactly is the shape of a satellite
dish? Could
>you graph the shape on a three-dimensional graph?
>Hamed
>
I was wondering, what exactly is the shape of a satellite dish? Could >you graph the shape on a three-dimensional graph? When a parallel microwave train travelling parallel to axis hits an antenna surface, they bounce off the dish to become spherical waves and arrive at a point focus in phase. A parabolic reflector does this function due its property that sum of distances from (aperture) plane and focus is constant for any point of the reflector dish. A simple Mathematica program for unit focal length fl: fl=1 ; dish[u_,v_]:={u*Cos[v], u*Sin[v], u^2 /(4 fl)} ; ParametricPlot3D[dish[u,v]//Evaluate, {u,0,2 },{v,0,2 Pi }] ; === Subject: Re: cantor's theorem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i242CrI13587; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i241vki11702 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i241vkZ13812; >Well, I'm glad that there is an answer and we don't have to reinvent >set theory. Thank you. Isn't it ironic that in Cantor's time, he was >considered the renegade mathematician by many and now his stuff is >well accpepted and anyone who opposes it is considered renegade? >Although those sci.maths posters who oppose Cantor may consider >themselves renegades and outlaws, the rest of us consider them >idiots and imbeciles. It's true that the arguments of the anti-Cantor posters are worthless, but it is interesting that amongst real mathematicians there is a small minority who do not accept the Cantorian realist position, and that this minority of intuitionists and contructivists contains some some very distinguished names. My own position is a Platonist/Goedelian realist one, so like nearly all mathematicians I don't accept the sophisticated arguments for constructivism either. The sophisticated arguments are of course nothing like the anti-Cantorian nonsense typically posted on this site though. www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 >Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: cantor's theorem >> Your world of computation is mystical. > I think you would have tremendous difficulty supporting > that assertion. The world of computation, as I have > defined it before, is something that is entirely objective. > Your world of computation arbitrarily excludes the activity > of constructing proofs. That's entirely false. I have tried often to make it clear that proofs do lie within the world of computation. The interpretation of the proofs can be another story. === Subject: Re: cantor's theorem Discussion, linux) >> Well, I'm glad that there is an answer and we don't have to reinvent >> set theory. Thank you. Isn't it ironic that in Cantor's time, he was >> considered the renegade mathematician by many and now his stuff is >> well accpepted and anyone who opposes it is considered renegade? >Although those sci.maths posters who oppose Cantor may consider >themselves renegades and outlaws, the rest of us consider them >idiots and imbeciles. > It's true that the arguments of the anti-Cantor posters are worthless, > but it is interesting that amongst real mathematicians there is a > small minority who do not accept the Cantorian realist position, and > that this minority of intuitionists and contructivists contains some > some very distinguished names. My own position is a > Platonist/Goedelian realist one, so like nearly all mathematicians > I don't accept the sophisticated arguments for constructivism > either. The sophisticated arguments are of course nothing like the > anti-Cantorian nonsense typically posted on this site though. In particular, those arguments are properly philosophical, not mathematical, arguments. Many argue that its programmers have turned out shoddy programs, but [their] objective is to make profit, not superlative programs per se. By the profit criterion, Microsoft has been one of the greatest companies in the history of this country. -- ADTI defends Microsoft === Subject: Re: simple question for cluster point. > I haven't been reading this group for a while: I assume > people have already commented on the advantages of using > a monicker like hot-girl when posting basic questions > about point-set topology? I'm surprised it doesn't catch > on. It seems you're reading the group often now ;-) I wondered about that handle also, but I've come to the conclusion that there's a culture clash here. An example of Engrish, so to speak. (Engrish is usually used in regard to the Japanese, but sometimes the same kind of funniness occurs with Koreans.) And no, I don't think people have commented on the advantages of that kind of handle until now. I'm writing a kind of unofficial FAQ for this newsgroup, and I'll be sure to put in the tip to pick this kind of handle (to get more responses to homework questions). === Subject: Re: simple question for cluster point. >I haven't been reading this group for a while: I assume >people have already commented on the advantages of using >a monicker like hot-girl when posting basic questions >about point-set topology? I'm surprised it doesn't catch >on. >It seems you're reading the group often now ;-) >I wondered about that handle also, but I've come to the conclusion that >there's a culture clash here. An example of Engrish, so to speak. >(Engrish is usually used in regard to the Japanese, but sometimes the >same kind of funniness occurs with Koreans.) >And no, I don't think people have commented on the advantages of that >kind of handle until now. I kept quiet because I was watching how well it worked and trying to predict when the males would catch on. >I'm writing a kind of unofficial FAQ for this newsgroup, and I'll be >sure to put in the tip to pick this kind of handle (to get more >responses to homework questions). I didn't mind guys swallowing that bait; what I do mind is that the poster doesn't seem to learn a damned thing. /BAH Subtract a hundred and four for e-mail. === Subject: No such thing as a random number? OK, let talk about a sequence of random numbers. Use base 2. e.g. 0,1,0,1,1,1,0,1,1,0 On the wolfram website it says such a (finite) sequence can never be proved to be random. Cab anyone give any references to a proof for this? === Subject: Re: No such thing as a random number? > OK, let talk about a sequence of random numbers. > Use base 2. > e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can never be proved > to be random. Cab anyone give any references to a proof for this? > The wolfram website is wrong. A sequence of integers can be truely random given two conditions: 1. The cardinality of the sequence is infinite. 2. Their is total entropy in the sequence. Condition 2 occurs when the following propositions are true: 1. The probability of drawing x from R where n <= x <= m is 1/[(m-n)+1]. 2. The probability of drawing x after any finite sequence S is 1/[(m-n)+1]^(|S| + 1) where S : some finite sequence where all elements are within [n,m] (sequence is not necessarily bound by [n,m]) Property 1 forces all elements in R to be equally probable. Property 2 forces total entropy. JS === Subject: Re: No such thing as a random number? > OK, let talk about a sequence of random numbers. > Use base 2. > e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can never be proved > to be random. Cab anyone give any references to a proof for this? The wolfram website is wrong. > A sequence of integers can be truely random given two conditions: > 1. The cardinality of the sequence is infinite. > 2. Their is total entropy in the sequence. > Condition 2 occurs when the following propositions are true: > 1. The probability of drawing x from R where n <= x <= m is > 1/[(m-n)+1]. > 2. The probability of drawing x after any finite sequence S is > 1/[(m-n)+1]^(|S| + 1) > where > S : some finite sequence where all elements are within [n,m] > (sequence is not necessarily bound by [n,m]) > Property 1 forces all elements in R to be equally probable. Property 2 > forces total entropy. > JS So I guess the infinite decimal expansion of pi is random then? === Subject: Re: No such thing as a random number? > OK, let talk about a sequence of random numbers. > Use base 2. > e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can never be > proved > to be random. Cab anyone give any references to a proof for this? The wolfram website is wrong. > A sequence of integers can be truely random given two conditions: > 1. The cardinality of the sequence is infinite. > 2. Their is total entropy in the sequence. > Condition 2 occurs when the following propositions are true: > 1. The probability of drawing x from R where n <= x <= m is > 1/[(m-n)+1]. > 2. The probability of drawing x after any finite sequence S is > 1/[(m-n)+1]^(|S| + 1) > where > S : some finite sequence where all elements are within [n,m] > (sequence is not necessarily bound by [n,m]) > Property 1 forces all elements in R to be equally probable. Property 2 > forces total entropy. > JS > So I guess the infinite decimal expansion of pi is random then? No it is not since arbitrary finite sequences of digits are not all equally probable. You might have better luck with sqrt(2) - 1. === Subject: Re: No such thing as a random number? > OK, let talk about a sequence of random numbers. > Use base 2. e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can never be > proved > to be random. Cab anyone give any references to a proof for this? The wolfram website is wrong. > A sequence of integers can be truely random given two conditions: > 1. The cardinality of the sequence is infinite. > 2. Their is total entropy in the sequence. > Condition 2 occurs when the following propositions are true: > 1. The probability of drawing x from R where n <= x <= m is > 1/[(m-n)+1]. > 2. The probability of drawing x after any finite sequence S is > 1/[(m-n)+1]^(|S| + 1) > where > S : some finite sequence where all elements are within [n,m] > (sequence is not necessarily bound by [n,m]) > Property 1 forces all elements in R to be equally probable. Property 2 > forces total entropy. > JS > So I guess the infinite decimal expansion of pi is random then? > No it is not since arbitrary finite sequences of digits are not all > equally probable. You might have better luck with sqrt(2) - 1. You have a proof that not all arbitratry digit sequences are equally probable in the decimal expansion of pi? I was under the impression that pi had not been proved to be normal (but everybody believes it to be normal) in any number base. If what you say is correct, pi can be demonstrated to be not normal in some base. URL please? === Subject: Re: No such thing as a random number? >You have a proof that not all arbitratry digit sequences are equally >probable in the decimal expansion of pi? He means the Schoenfeld Pi. It's much nicer but you can't use it without paying him royalties. I'm not interested in mathematics that might have anything to do with reality. -- Easterly, in sci.math === Subject: Re: No such thing as a random number? > OK, let talk about a sequence of random numbers. > Use base 2. > e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can never be proved > to be random. Cab anyone give any references to a proof for this? The wolfram website is wrong. Doubtful. === Subject: Re: No such thing as a random number? > OK, let talk about a sequence of random numbers. > Use base 2. > e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can never be > proved > to be random. Cab anyone give any references to a proof for this? The wolfram website is wrong. > Doubtful. I gave you an equation. === Subject: Re: No such thing as a random number? > OK, let talk about a sequence of random numbers. > Use base 2. e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can never be > proved > to be random. Cab anyone give any references to a proof for this? The wolfram website is wrong. > Doubtful. > I gave you an equation. So? === Subject: Re: No such thing as a random number? > OK, let talk about a sequence of random numbers. > Use base 2. e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can never be > proved > to be random. Cab anyone give any references to a proof for this? > The wolfram website is wrong. > Doubtful. > I gave you an equation. > So? You're right. Arguments such as doubtful and so? are much more convincing. === Subject: Re: No such thing as a random number? > OK, let talk about a sequence of random numbers. > Use base 2. > e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can > never be > proved > to be random. Cab anyone give any references to a proof for > this? The wolfram website is wrong. Doubtful. > I gave you an equation. > So? > You're right. Arguments such as doubtful and so? are much more > convincing. === Subject: Re: No such thing as a random number? > On the wolfram website it says such a (finite) sequence can never be > proved to be random. Cab anyone give any references to a proof for this? Without more information, such as what you mean by random, your question has no meaning. === Subject: Re: No such thing as a random number? > OK, let talk about a sequence of random numbers. > Use base 2. > e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can never be > proved > to be random. Cab anyone give any references to a proof for this? I'm a little surprised with the qualifier finite being there. Consider such a sequence of length n. If it were truly random, you would expect (1a) the ratio of the number of ones in the sequence to the length of the sequence to approach 1/2 as the length of the sequence increases without bound. (1b) Some sort of rule that whether the (n + 1)st bit is 0 or 1 cannot be determined as a function of the ratios of zeros and ones in the first n terms of the series. Here's a sequence that meets that condition: 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, ... That is, converting successive pairs of bits to base 10: 0, 1, 2, 3, 0, 1, 2, 3, ... Clearly that sequence isn't random. So condition (1) isn't enough to specify what one means by a sequence to be random. One would want some additional sort of condition like: (2a) the ratios of number of subsequences (0, 0), (0, 1), (1, 0), and (1, 1) in the sequence to the length of the sequence approaches 1/4 as the length of the sequence approaches without bound. (2b) the (n/2) + 1 and (n/2) + 2 terms cannot be determined as a function of the ratios of these subsequences in the first n/2 terms of the series. But then consider a sequence like: 0,0,0,0 0,0,0,1, 0,0,1,1, 0,1,0,0, 0,1,0,1, ..., 1,1,1,1, repeat... Clearly such a sequence is not random. So we'd have to extend conditions (1) and (2) to conditions (1), (2), and (3). You may see that the number of conditions needed to define an infinite binary sequence as random would be countably infinite. But the number of sequences satisfying these conditions is zero. So there is no such thing as an infinite sequence of random bits. Some such argument, no doubt better expressed and without any misunderstandings I may have introduced, is in Knuth's three volume Art of Computer Programming. Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: No such thing as a random number? Well, no. You try and create your own definition of random, and then try and show that no infinite string satisfies it. All that proves is that you have picked a lousy definition of random, as it is (according to you) not satisfied by any finite or infinite string. You may as well have defined the set of random sequences as being the empty set, for all the good your definition does. The most useful definition of random is the one that I previously gave - basically a sequence is random if it cannot be defined by a Turing machine of shorter length than the sequence itself. By this definition, no computable infinite sequence is random. To define a random infinite sequence, we have to be able to specify a string that cannot be generated by a Turing Machine. One example is the sequence of decimal digits in the following number: http://en.wikipedia.org/wiki/Chaitin's_constant These would appear to be random according to any reasonable definition of the term; the only problem is we have no way of generating these terms (and indeed if we could, they wouldn't be random). > OK, let talk about a sequence of random numbers. > Use base 2. > e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can never be > proved > to be random. Cab anyone give any references to a proof for this? > I'm a little surprised with the qualifier finite being there. > Consider such a sequence of length n. If it were truly random, you > would expect > (1a) the ratio of the number of ones in the sequence to the > length of the sequence to approach 1/2 as the length of the sequence > increases without bound. > (1b) Some sort of rule that whether the (n + 1)st bit is 0 or 1 cannot > be determined as a function of the ratios of zeros and ones in the > first n terms of the series. > Here's a sequence that meets that condition: > 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, ... > That is, converting successive pairs of bits to base 10: > 0, 1, 2, 3, 0, 1, 2, 3, ... > Clearly that sequence isn't random. So condition (1) isn't enough > to specify what one means by a sequence to be random. One would > want some additional sort of condition like: > (2a) the ratios of number of subsequences (0, 0), (0, 1), (1, 0), > and (1, 1) in the sequence to the length of the sequence approaches > 1/4 as the length of the sequence approaches without bound. > (2b) the (n/2) + 1 and (n/2) + 2 terms cannot be determined as > a function of the ratios of these subsequences in the first n/2 > terms of the series. > But then consider a sequence like: > 0,0,0,0 0,0,0,1, 0,0,1,1, 0,1,0,0, 0,1,0,1, ..., 1,1,1,1, repeat... > Clearly such a sequence is not random. So we'd have to extend > conditions (1) and (2) to conditions (1), (2), and (3). You may > see that the number of conditions needed to define an infinite > binary sequence as random would be countably infinite. > But the number of sequences satisfying these conditions is zero. > So there is no such thing as an infinite sequence of random bits. > Some such argument, no doubt better expressed and without any > misunderstandings I may have introduced, is in Knuth's three > volume Art of Computer Programming. > -- > Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html > To solve Linear Programs: .../LPSolver.html > r c A game: .../Keynes.html > v s a Whether strength of body or of mind, or wisdom, or > i m p virtue, are found in proportion to the power or wealth > e a e of a man is a question fit perhaps to be discussed by > n e . slaves in the hearing of their masters, but highly > @ r c m unbecoming to reasonable and free men in search of > d o the truth. -- Rousseau === Subject: Re: No such thing as a random number? > But then consider a sequence like: > 0,0,0,0 0,0,0,1, 0,0,1,1, 0,1,0,0, 0,1,0,1, ..., 1,1,1,1, repeat... Did you mean 0,0,0,0 0,0,0,1, 0,0,1,0, 0,0,1,1, 0,1,0,0, 0,1,0,1, ..., 1,1,1,1, repeat...? === Subject: Re: No such thing as a random number? > But then consider a sequence like: > 0,0,0,0 0,0,0,1, 0,0,1,1, 0,1,0,0, 0,1,0,1, ..., 1,1,1,1, repeat... > Did you mean 0,0,0,0 0,0,0,1, 0,0,1,0, 0,0,1,1, 0,1,0,0, 0,1,0,1, ..., > 1,1,1,1, repeat...? Yes. Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: No such thing as a random number? > On the wolfram website it says such a (finite) sequence can never be proved > to be random. Cab anyone give any references to a proof for this? Of course, there's a big difference between (1) not being able to prove that something is random and (2) saying that no random numbers exist. === Subject: Re: No such thing as a random number? >OK, let talk about a sequence of random numbers. >Use base 2. >e.g. 0,1,0,1,1,1,0,1,1,0 >On the wolfram website it says such a (finite) sequence can never be proved >to be random. Cab anyone give any references to a proof for this? One can never observe random numbers. Once they are observed, they are no longer random. One can describe unknown phenomena (e.g., the outcome of the next roll of the dice, tomorrow's temperature at noon, you mother's blood pressure right now) via random variables, but not known quantities. Mathematically, a random variable (or number) is a *function* on a sample space, not a number. Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: No such thing as a random number? > OK, let talk about a sequence of random numbers. > Use base 2. > e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can never be proved > to be random. Cab anyone give any references to a proof for this? > It would have been helpful if you provided a link to where it says this, so that we could see exactly what it says. However, any sequence of numbers a_1, a_2, a_3 ... a_n can (for example) be generated as the f(1), f(2) ... f(n) where f is an nth degree polynomial in x. There are many other ways of producing an equation which has a_1, a_2, a_3 ... (0,1,0,1,1,1,0,1,1,0 in your example) as its first n values. Does this make the string random? There is another definition of random which says (basically) that there is no shorter way of defining the sequence than simply listing it. I would be surprised if it was impossible to prove a sequence was random using this definition as you say Wolfram states, as it could be accomplished through a finite search. === Subject: Re: No such thing as a random number? wok > OK, let talk about a sequence of random numbers. > Use base 2. > e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can never be proved > to be random. Cab anyone give any references to a proof for this? The data might just be dumped from a compressed file, as this was (base 16): ED32A498064990DE 50A432F3E3E9CFC8 in which case it is not random, whatever they mean by that word. It's a badly posed statement. I think the website should be more careful in how it uses the word prove, as well. === Subject: Re: No such thing as a random number? > OK, let talk about a sequence of random numbers. > Use base 2. > e.g. 0,1,0,1,1,1,0,1,1,0 > On the wolfram website it says such a (finite) sequence can never be proved > to be random. Cab anyone give any references to a proof for this? > Any finite sequence of integers, such as the above, can be produced by successive integer substitutions into a polynomial function with rational coefficients. === Subject: ring I have this equation x(a/(p^k)) + y = 1/(p^k) where p is a prime, k is an integer >= 0 and a is an integer. Now I need to find integers x and y so the equation holds. Can anyone help me with that? Thank you. -Greg R. === Subject: Re: ring > x(a/(p^k)) + y = 1/(p^k) where p is a prime, k is an integer >= 0 and a > is an integer. Now I need to find integers x and y so the equation holds. > Can anyone help me with that? ax + y p^k = 1 ax = 1 (mod p^k) So find for x multiplicative inverses of a modulus p^k, if any. Then y = (1 - ax)/p^k is an integer === Subject: Re: ring > I have this equation > x(a/(p^k)) + y = 1/(p^k) where p is a prime, k is an integer >= 0 and a > is an integer. Now I need to find integers x and y so the equation holds. > Can anyone help me with that? > Thank you. > -Greg R. if y = (1 - a*x)/p^k is to be an integer, the 1-a*x = b*p^k, for some integer b or x = (1-b*p^k)/a is an integer. Then b*p^k == 1 mod a. Then b is congruent to a multiplicative inverse, mod a , of p^k, provided that such inverse exists, and it will exist whenever a is not a multiple of p. === Subject: Re: Inverse Error Function >I am looking for an algorithm for computing the Inverse Error Function. >This is the the number of standard deviations as a function of >probability. I am hoping to find it in a continued division equation, >such at the one Laplace derived for the Error Function. A power series >would be second best. According to Maple, if F(x) is the cdf of the standard normal distribution, its inverse has the power series about p=1/2: 2^(1/2)*Pi^(1/2)*(p-1/2)+1/3*Pi^(3/2)*2^(1/2)*(p-1/2)^3 +7/30*Pi^(5/2)*2^(1/2)*(p-1/2)^5+127/630*Pi^(7/2)*2^(1/2)*(p-1 /2)^7 +4369/22680*Pi^(9/2)*2^(1/2)*(p-1/2)^9+O((p-1/2)^10) and the continued fraction (p-1/2)/(1/2*2^(1/2)/Pi^(1/2)+(p-1/2)^2/ (-3*2^(1/2)/Pi^(1/2) +(p-1/2)^2/ (5/11*2^(1/2)/Pi^(1/2)+(p-1/2)^2/ (-847/241*2^(1/2)/Pi^(1/2)+(p-1/2)^2/...)))) I don't know about convergence properties. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Inverse Error Function >I am looking for an algorithm for computing the Inverse Error Function. >This is the the number of standard deviations as a function of >probability. I am hoping to find it in a continued division equation, >such at the one Laplace derived for the Error Function. A power series >would be second best. > According to Maple, if F(x) is the cdf of the standard normal > distribution, its inverse has the power series about p=1/2: > 2^(1/2)*Pi^(1/2)*(p-1/2)+1/3*Pi^(3/2)*2^(1/2)*(p-1/2)^3 > +7/30*Pi^(5/2)*2^(1/2)*(p-1/2)^5+127/630*Pi^(7/2)*2^(1/2)*(p-1 /2)^7 > +4369/22680*Pi^(9/2)*2^(1/2)*(p-1/2)^9+O((p-1/2)^10) Then, to get a series for the inverse error function itself, just replace p by (x+1)/2 in the above, and divide that result by Sqrt(2). Doing so, and letting u denote Sqrt(pi)/2*x for convenience, we get inverf(x) = u + u^3/3 +7*u^5/30 + 127*u^7/630 + ... That series can be found, for example, at . Convergence of the series is terribly slow when |x| is close to 1. Due to that, let me suggest that the inverse error function can be approximated fairly well, when |x| is close to 1, by # inverf(x) ~ sgn(x)*Sqrt(1/2*(log(v) - log(log(v)))) where v = 2/(pi*(1 - |x|)^2). For 0.9 < |x| < 1, |relative error| < 0.005 . Relative error approaches 0 as |x| approaches 1. [BTW, those familiar with the Lambert W function may already have guessed, correctly, that it was involved in obtaining #.] I was just about to ask Has anyone seen # before? when I found that it is already mentioned at . David Cantrell === Subject: Re: Inverse Error Function >I am looking for an algorithm for computing the Inverse Error Function. >This is the the number of standard deviations as a function of >probability. I am hoping to find it in a continued division equation, >such at the one Laplace derived for the Error Function. A power series >would be second best. >According to Maple, if F(x) is the cdf of the standard normal >distribution, its inverse has the power series about p=1/2: >2^(1/2)*Pi^(1/2)*(p-1/2)+1/3*Pi^(3/2)*2^(1/2)*(p-1/2)^3 >+7/30*Pi^(5/2)*2^(1/2)*(p-1/2)^5+127/630*Pi^(7/2)*2^(1/2)*(p- 1/2)^7 >+4369/22680*Pi^(9/2)*2^(1/2)*(p-1/2)^9+O((p-1/2)^10) >and the continued fraction >(p-1/2)/(1/2*2^(1/2)/Pi^(1/2)+(p-1/2)^2/ > (-3*2^(1/2)/Pi^(1/2) +(p-1/2)^2/ > (5/11*2^(1/2)/Pi^(1/2)+(p-1/2)^2/ > (-847/241*2^(1/2)/Pi^(1/2)+(p-1/2)^2/...)))) >I don't know about convergence properties. The power series can easily be seen to converge for |p - 1/2| < 1/2, and approach the appropriate infinity as p approaches 0 or 1. Look instead at the solution for y of the equation x = int_0^y exp(-.5*t*2) dt; all the coefficients are positive, so the nearest singularity is when y becomes infinite. This gets rid of the powers of sqrt(2pi) which make it hard to read. An easy way to compute the coefficients is to let z = exp(-.5*y^2). Then z' = -y, and z'' = -1/z. Clearly z is even and z(0) = 1, so z*z'' = -1 gives a simple recursion for the power series. This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: How can this be correct? Consider the following 2nd order non-homogenous diff'l equation: mu'' + cu' + ku = FoCos(wt). if the damping constanct c = 0, then, mu'' + ku = FoCos(wt). Now, for the case where wo is not equal to w ( w = frequency), then y = C1 Cos(wot) + C2 Sin(wot) + (FoCos(wt))/m(wo^2-w^2), [1] is a solution. Now, if we treat wo equal to w, then the above solution is the same with the exeption of the last term. The last term is FotSin(wt)/2mw, [2]. This is known as Resonance. Anyway, what is interesting is that you can apply L'Hopital's rule to the last term of [1] and get [2]. Try it! Then take the Limit as wo approaches w. Any ideas as to why this works when in fact the last term in [1] is NOT indeterminate? Gord Bramfield. === Subject: Re: quadratic inequalities >You mean you're given the matrix P and you want to find some X such >that X^T P X > 0? I assume the entries are all real. >yes. And more generally find all the solutions X of this inequality, >i.e. find a domain D such that all x in D satisfy this inequality (may >be something like a cone). Well, yes, it is like a cone in that if X satisfies the inequality so does any positive scalar multiple of X. Also, of course it's an open set. But it's not convex in general. I doubt that there's a nice characterisation of all solutions. >If P + P^T has a positive eigenvalue lambda, an eigenvector X for >that eigenvalue will do. So will any nonzero linear combination of >eigenvectors for the positive eigenvalues of P + P^T. >On the other hand, if P + P^T has no positive eigenvalues, such an >X can't exist. >Can you give me a reference ? No, but it's not hard to prove. Note first that X^T P^T X = X^T P X = (1/2) X^T Q X where Q = P + P^T. If X is an eigenvector of Q for eigenvalue lambda, X^T Q X = lambda X^T X > 0. For a linear combination of eigenvectors, use the fact that eigenvectors of a real symmetric matrix for distinct eigenvalues are orthogonal. And if Q has no positive eigenvalues, then it is negative semidefinite. >And what about the case where X is a matrix, i.e. find X such that X^T >P X is positive definite ? (should we just write [v_1, v_2,...v_r] >with r the dimension of X and v_i are eigenvectors of P+P^T ?) Assuming it is symmetric, X^T P X is positive definite iff (X x)^T P X x > 0 for all nonzero vectors x, which says Ker(X) = {0} and Ran(X){0} is contained in your set D. In particular it will be true if the columns of X are linearly independent and each is a linear combination of eigenvectors for positive eigenvalues. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: quadratic inequalities >> You mean you're given the matrix P and you want to find some X such >> that X^T P X > 0? I assume the entries are all real. >yes. And more generally find all the solutions X of this inequality, >i.e. find a domain D such that all x in D satisfy this inequality (may >be something like a cone). > Well, yes, it is like a cone in that if X satisfies the inequality so does > any positive scalar multiple of X. Also, of course it's an open set. > But it's not convex in general. I doubt that there's a nice > characterisation of all solutions. >> If P + P^T has a positive eigenvalue lambda, an eigenvector X for >> that eigenvalue will do. So will any nonzero linear combination of >> eigenvectors for the positive eigenvalues of P + P^T. > On the other hand, if P + P^T has no positive eigenvalues, such an >> X can't exist. >Can you give me a reference ? > No, but it's not hard to prove. Note first that X^T P^T X = X^T P X > = (1/2) X^T Q X where Q = P + P^T. If X is an eigenvector of Q for > eigenvalue lambda, X^T Q X = lambda X^T X > 0. For a linear combination > of eigenvectors, use the fact that eigenvectors of a real symmetric matrix > for distinct eigenvalues are orthogonal. And if Q has no positive > eigenvalues, then it is negative semidefinite. >And what about the case where X is a matrix, i.e. find X such that X^T >P X is positive definite ? (should we just write [v_1, v_2,...v_r] >with r the dimension of X and v_i are eigenvectors of P+P^T ?) > Assuming it is symmetric, X^T P X is positive definite iff > (X x)^T P X x > 0 for all nonzero vectors x, which says Ker(X) = {0} > and Ran(X){0} is contained in your set D. In particular it will be > true if the columns of X are linearly independent and each is a linear > combination of eigenvectors for positive eigenvalues. thanks a lot. Please ignore my precedent message, as you message was not displayed by then. thanks again. === Subject: Re: quadratic inequalities good evening, > If P + P^T has a positive eigenvalue lambda, an eigenvector X for > that eigenvalue will do. So will any nonzero linear combination of > eigenvectors for the positive eigenvalues of P + P^T. > On the other hand, if P + P^T has no positive eigenvalues, such an > X can't exist. or at least show me how to prove it. thank you. === Subject: Re: Help! Need a book recommendation... Well I won't be posting here any more, there's too many dickheads who think they're clever. It's obvious you aren't, as a gifted person just wouldn't act in the way you have. Why do you feel the need to ridicule someone who has less maths knowledge than yourself? Why do you feel the need to answer straightforward questions in such a cryptic manner as to only be meaningful to those who already know the answer? I'll tell you why - you're average. You are outstanding in NO way at all. But you want to believe you are not. === Subject: Re: Motivate Originator: baez@math-cl-n03.math.ucr.edu (John Baez) Karl M. Bunday Can someone please explain to me the usage of the word motivate, as in a >recent posting Motivation for e? >The context implies that the word is being used as a synonym for explain, >but I have never seen this usage in a dictionary or thesaurus. >The particular meaning of motivate in this kind of context appears to be provide background for understanding why this notation/problem/conjecture >would be interesting to mathematicians. In other words, to explain to a >callow undergraduate, who hasn't developed enough mathematical maturity to >know why the issue is important to mathematicians, what got mathematicians >interested in the issue in the first place. Right. To motivate a concept is not to explain it, but to explain why someone should give a damn about the explanation. So, apparently motivating a concept arose as some sort of jargonistic shorthand for motivating the reader to learn the concept. People who use this jargon tend to be bad at explaining things. They think okay, I need to explain X, so first I'll motivate it and then [breathing a sigh of relief that this chore is done] I'll explain it. You can feel it when someone is thinking this, so by the time they're done motivating the concept you're already bored. It's as if a musician decided that the first 2 minutes of a piece should get you interested enough to put up with the rest. === Subject: Re: Motivate |Karl M. Bunday > Can someone please explain to me the usage of the word motivate, |>> as in a recent posting Motivation for e? | |>> The context implies that the word is being used as a synonym for |>explain, but I have never seen this usage in a dictionary or |>> thesaurus. | |>The particular meaning of motivate in this kind of context appears |>to be provide background for understanding why this |>notation/problem/conjecture would be interesting to mathematicians. |>In other words, to explain to a callow undergraduate, who hasn't |>developed enough mathematical maturity to know why the issue is |>important to mathematicians, what got mathematicians interested in |>the issue in the first place. | |Right. To motivate a concept is not to explain it, but to explain |why someone should give a damn about the explanation. | |So, apparently motivating a concept arose as some sort of |jargonistic shorthand for motivating the reader to learn the |concepti'm not sure how serious you're being here. it's more likely that the math jargon motivate arose (as karl bunday indicated and you seemed to agree with) in connection with trying to understand what got mathematicians interested in the issue in the first place. that is, it was about the motivations of the original creators, not of the students, though with some hope that the students may adopt those motivations for themselves. you and herman seem to be trying to repeat the old joke about the method actor who says what's my motivation, man? (meaning something like what's my character's motivation, so that i can adopt it and thus become, like, that character) and the director who replies your motivation is that i'm going to beat you up if you don't do what i say, except that herman at least seems to be serious. |People who use this jargon tend to be bad at explaining things. |They think okay, I need to explain X, so first I'll motivate it and then |[breathing a sigh of relief that this chore is done] I'll explain it. | |You can feel it when someone is thinking this, so by the time they're |done motivating the concept you're already bored. | |It's as if a musician decided that the first 2 minutes of a piece |should get you interested enough to put up with the rest. there are lots of different ways to be bad at explaining things and i doubt that use of the math jargon motivate correlates much one way or the other with explanatory ability. heavy use of motivate may sometimes be associated with thinking of motivation as a formulaic chore to be gotten done with, but sometimes it's just descriptive terminology alluding to the phenomenon of how if you only hear the last 2 minutes of a piece and don't know what came before it then you may be bewildered by the applause. [e-mail address jdolan@math.ucr.edu] === Subject: Re: Motivate > Karl M. Bunday > Can someone please explain to me the usage of the word motivate, as in a >> recent posting Motivation for e? >> The context implies that the word is being used as a synonym for explain, >> but I have never seen this usage in a dictionary or thesaurus. >The particular meaning of motivate in this kind of context appears to be provide background for understanding why this notation/problem/conjecture >would be interesting to mathematicians. In other words, to explain to a >callow undergraduate, who hasn't developed enough mathematical maturity to >know why the issue is important to mathematicians, what got mathematicians >interested in the issue in the first place. > Right. To motivate a concept is not to explain it, but to explain > why someone should give a damn about the explanation. > So, apparently motivating a concept arose as some sort of jargonistic > shorthand for motivating the reader to learn the concept> People who use this jargon tend to be bad at explaining things. > They think okay, I need to explain X, so first I'll motivate it and then > [breathing a sigh of relief that this chore is done] I'll explain it> You can feel it when someone is thinking this, so by the time they're > done motivating the concept you're already bored. > It's as if a musician decided that the first 2 minutes of a piece > should get you interested enough to put up with the rest. More nonsense from John Baez. When I was an undergraduate at the University of Chicago I had a mathematics professor explain the reason for using jokes in lecture. Undergraduate students tend to fall asleep every fifteen minutes. Graduate students tend to fall asleep every ten minutes. A lecturer is lucky if they can keep the attention of professors for even five minutes. Motivating a subject is often motivated by the know-it-all attitude of the audience. :-) === Subject: Re: Motivate >Karl M. Bunday Can someone please explain to me the usage of the word motivate, as in a >>recent posting Motivation for e? >>The context implies that the word is being used as a synonym for explain, >>but I have never seen this usage in a dictionary or thesaurus. >The particular meaning of motivate in this kind of context appears to be provide background for understanding why this notation/problem/conjecture >would be interesting to mathematicians. In other words, to explain to a >callow undergraduate, who hasn't developed enough mathematical maturity to >know why the issue is important to mathematicians, what got mathematicians >interested in the issue in the first place. Right. To motivate a concept is not to explain it, but to explain >why someone should give a damn about the explanation. >So, apparently motivating a concept arose as some sort of jargonistic >shorthand for motivating the reader to learn the concept. >People who use this jargon tend to be bad at explaining things. >They think okay, I need to explain X, so first I'll motivate it and then >[breathing a sigh of relief that this chore is done] I'll explain it. >You can feel it when someone is thinking this, so by the time they're >done motivating the concept you're already bored. >It's as if a musician decided that the first 2 minutes of a piece >should get you interested enough to put up with the rest. I cannot believe the amount of responses along this line I have seen here. I cannot imagine the introduction of a concept without an explanation of its reason for being: applications of the matter either in mathmeatics or elsewhere. Would you just start talking about integration without talking about area, volume, work, etc.? Differential equations without any discussion of why one would even consider them? Topological spaces, groups or differentiable manifolds without examples of how these object appear enough to merit abstract study? Hypothesis testing without an example of the kind of issues it addresses? Math programming without any sample models? I guess there are a lot of lousy teachers out there. (What a surprise!) This carries over into presentations as well. Personally, regardless of the content, I find the report of Wiles' presentation, as described by Aczel, horrifying. Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Motivate >Karl M. Bunday Can someone please explain to me the usage of the word motivate, as in a >recent posting Motivation for e? >The context implies that the word is being used as a synonym for explain, >but I have never seen this usage in a dictionary or thesaurus. >>The particular meaning of motivate in this kind of context appears to be provide background for understanding why this notation/problem/conjecture >>would be interesting to mathematicians. In other words, to explain to a >>callow undergraduate, who hasn't developed enough mathematical maturity to >>know why the issue is important to mathematicians, what got mathematicians >>interested in the issue in the first place. >Right. To motivate a concept is not to explain it, but to explain >why someone should give a damn about the explanation. >So, apparently motivating a concept arose as some sort of jargonistic >shorthand for motivating the reader to learn the concept. >People who use this jargon tend to be bad at explaining things. >They think okay, I need to explain X, so first I'll motivate it and then >[breathing a sigh of relief that this chore is done] I'll explain it. >You can feel it when someone is thinking this, so by the time they're >done motivating the concept you're already bored. >It's as if a musician decided that the first 2 minutes of a piece >should get you interested enough to put up with the rest. >I cannot believe the amount of responses along this line I have seen here. >I cannot imagine the introduction of a concept without an explanation of >its reason for being: applications of the matter either in mathmeatics >or elsewhere. Would you just start talking about integration without >talking about area, volume, work, etc.? In fact, I would not use ANY of these. They use unnecessary geometry or physics to confuse the student. An integral is a sum of products of a function by a measure, or the limit of such. What was the first computation of an integral? It was not any of the ones cited; it was the computation of a merchant's , shortly after the idea of a unit of exchange was developed. Measures (and integrals) are discrete or limits of discrete. They belong early, not after years of computation. Differential equations without >any discussion of why one would even consider them? How many of the students taking differential equations now could understand such a discussion? One needs to know about limits, and what a derivative is, not how to compute them. Topological spaces, >groups or differentiable manifolds without examples of how these object >appear enough to merit abstract study? One should give examples, but only AFTER the general concept has been introduced. It is only by looking at nasty spaces that the need for the additional good assumptions can be understood. Abstract concepts are much easier if they are directly introduced; building up to them requires unlearning, which is quite difficult. Hypothesis testing without an >example of the kind of issues it addresses? Frankly, hypothesis testing without lots of probability, and a discussion of decision theory, which is how it is taught, leads to the use of statistics as religious dogma, unsound. Math programming without >any sample models? You might have a point here. >I guess there are a lot of lousy teachers out there. (What a >surprise!) This carries over into presentations as well. Personally, >regardless of the content, I find the report of Wiles' presentation, as >described by Aczel, horrifying. Very little presentation now, even with motivation, explains the concepts. This is not at all surprising; few have seen a clear use of the concepts, and most mathematicians only have a working knowledge of them. One can prove theorems formally without understanding the underlying concepts. A good example is that of the integers. There are several quite different concepts here, all satisfied by the same objects. It is only those who look at the question who seem to notice this, and teaching arithmetic the usual way makes all of them harder. At least two different concepts are needed for early treatment of the counting numbers This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Motivate > Very little presentation now, even with motivation, explains > the concepts. This is not at all surprising; few have seen a > clear use of the concepts, and most mathematicians only have a > working knowledge of them. One can prove theorems formally > without understanding the underlying concepts. To be honest, this is a sad indictment. How then does a curriculum committee make its decisions? Certainly not on the basis of understanding. Unfortunately, I have a complaint in my mailbox that pretty much illustrates the problem: a case in point---are impenetrable because you make one vague statement using the jargon from one field of math, and then without any clarification you wildly jump to a vague statement using jargon from a totally different field of math. Experts in pseudosemilattices and double quasiorders, whatever those are, are not going to be experts in the other fields whose terminology you are borrowing from. They therefore will understand at most one or two understand those sentences, because you don't go into enough detail to explain what you're talking about. Do you really think that the communication problems that you've generated by throwing together vague thoughts from fifteen different fields will be solved by throwing in some more vague thoughts from a sixteenth field? If you met someone who spoke an incomprehensible medley of English, Basque, Finnish, Hebrew, Japanese, Greek, Hausa, and Telugu, and who complained that nobody understood him, would you solve his problem by recommending that he inject some Navajo into the mix since Navajo has the perfect turn of phrase for one of the thirty-two things he's trying to say? You have to pick an audience---let's say, the experts in pseudosemilattices---and then phrase your question in terms that they can understand. If you want to say something that is most naturally expressed using some other language, resist the temptation to switch languages, but take the time to develop the relevant concept in the chosen language. This is from a person *who never once asked* me to clarify any statement. The actual history behind these remarks are not relevant to this post. But, I did pick the correct audience, and, I did try to develop the concept in the given area. concepts. In all fairness, I will also provide a defense for the Tower of Babble that John Wilkins posted to alt.philosophy: Scientific subdisciplines and disciplines have a basic constraint on them that necessitates a degree of specialisation and internal languages: the time it takes to become an expert in it. To be sure, disciplines would benefit if you could cross the disciplinary boundaries, but how could you? It takes on average 25 years to become a front rank professional in a given discipline, from birth. Suppose that the specialisation begins at 16 - that means you must add 9 years or so to each new discipline added. Let us assume that if you sample widely, you can do three lots of specialisations and cover most of modern knowledge to the extent that you won't make gross mistakes in cross-disciplinary work. We are asking professionals to wait until they get to their 40s before they can contribute. Now note that most people in such fields are most productive at 28 or so... The Learning Curve is the basic limiter. But there's another - to whom would you communicate? If your language requires you to spend 43 years getting up to speed, the chances anyone else will even understand you is minimal, exponentially relative to the number of subdisciplinary specialisations that there are. The real problem I have with these statements, however, comes from my background as a tradesman. They do not seem to emphasize quality. That is, they sound like excuses for poor work. :-) === Subject: Re: Motivate > (... ) What was the first computation of an > integral? It was not any of the ones cited; it was the > computation of a merchant's , shortly after the idea of > a unit of exchange was developed. That's interesting. Can you give a reference for this? Herman Jurjus === Subject: Re: Motivate >(... ) What was the first computation of an >integral? It was not any of the ones cited; it was the >computation of a merchant's , shortly after the idea of >a unit of exchange was developed. >That's interesting. Can you give a reference for this? I do not have any reference for this, but most of the clay tablets are merchants' accounts. They certainly did compute amounts of s. This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Sum where some terms are added, others subtracted Here is one of those results which seem like they should be common-knowledge, yet I have never seen them before. Let a(k,m) = 1 if k <= sqrt(m); Let a(k,m) = -1 if k > sqrt(m). If m = any positive integer, then: sum{k=1 to m} floor(m/k) a(k,m) always equals floor(sqrt(m))^2, which is simply m if m is a square, of course. (unless I am wrong.) (this result reminds me of the result I posted a couple of weeks ago: sum{k=1 to m,k=squarefree} floor(sqrt(m/k)) always equals m, I believe was the result.) Leroy Quet === Subject: Re: Cosets in Q...index... > Hi all > I am staring at the following problem: > Prove that Q (rational numbers with +) has no proper subgroup of finite > index. I'm trying to find a constructive way to show it, in the sense > of explicitly constructing an infinite list of distinct cosets. I > would rather have such a proof than a suppose x1,...,xn are are finite > set of coset reps, here is a contradiction... proof, which I have > already. A hint. If K were such a subgroup, then Q/K would be a nontrivial finite abelian group and so there'd be a surjective homomorphism from Q to suck a group. So, can you show that all homomorphisms from Q to a finite group are trivial? Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Cosets in Q...index... Robin Unfortunately the problem in the book lies in the chapter when all the students have to work with is the definition of a coset and Lagrange's Theorem. Factor groups have yet to exist, so since LT doesn't apply, I need some really basic approach. Thank you, Justin : A hint. If K were such a subgroup, then Q/K would be a nontrivial finite : abelian group and so there'd be a surjective homomorphism from Q : to suck a group. So, can you show that all homomorphisms from Q : to a finite group are trivial? === Subject: Re: Cosets in Q...index... Justin top-posted: > Robin > Unfortunately the problem in the book lies in the chapter when all the > students have to work with is the definition of a coset and Lagrange's > Theorem. Factor groups have yet to exist, so since LT doesn't apply, I > need some really basic approach. > Thank you, > Justin > : A hint. If K were such a subgroup, then Q/K would be a nontrivial finite > : abelian group and so there'd be a surjective homomorphism from Q > : to suck a group. So, can you show that all homomorphisms from Q > : to a finite group are trivial? In which case I shall revise my hint .... follow my original hint and then translate the solution back into naive terms :-) OK a different hint. Suppose there are n cosets. For a in Q consider the cosets K, a + K, 2a + K, ..., na + K. Deduce that ma is in K for some m with 1 <= m <= n, and hence that n! a is in K .... etc. Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Cosets in Q...index... Content-transfer-encoding: 8bit > Hi all > I am staring at the following problem: > Prove that Q (rational numbers with +) has no proper subgroup of finite > index. I'm trying to find a constructive way to show it, in the sense > of explicitly constructing an infinite list of distinct cosets. I > would rather have such a proof than a suppose x1,...,xn are are finite > set of coset reps, here is a contradiction... proof, which I have > already. > Any suggestions? > Justin > ps. As elementary as possible would be nice! You are asking whether Q/G can ever be finite for G <> Q. The quick way is to note that Q/G <> {0} is divisible and hence cannot be finite. I don't think that is what you are looking for. I would despair of coming up with an enumeration of the elements of Q/G considering the number of possibilities for its structure. Paul Sperry Columbia, SC (USA) === Subject: Re: topolgy problem... === Subject: topolgy problem... >let Y in X Y is a memeber of X? No, you mean Y subset X. >let X and Y be connected. >show that if A and B form a separation of X-Y , You mean if A and B are a disconnection of X-Y. >then Y U A and Y U B are connected. Are you using the theorem that for E subset S, the subspace S|E is disconnected iff some separated nonnul A,B with A / B = E ? A,B separated when (cl A) / B = nulset = A / cl B >i will show that only Y U A is connected. >let Y U A is disconnected >separation C,D is exist such that >C != empty, D != empty, cls(C) ^ D = C ^ cls(D) = empty >define E = C ^ Y, F = D ^ Y >if E != empty and F != empty , then E U F = Y and >cls(E) ^ F = cls(C ^ Y) ^ F in cls(C) ^ cls(Y) ^ F = empty (cd E) / F = (cl C/Y) / F subset cl C / cl Y / D / Y = nulset >similary, E ^ cls(F) = empty >thus E, F is separation of Y >this is a contradiction. because Y is connected. >thus either E or F is must empty set. >without loss of generality, if E = empty then C in A C / Y = nulset; D / Y = Y; Y subset D Why is C subset A ? C / D = A / Y C / (C / D) = C / (A / Y) C = (C / A) / (C / Y) = C / A >now, C, D U B will show that X is separation. >C U (D U B) = X, C / D / B = Y / A / B = Y / X-Y = X >cls(C) ^ (D U B) = cls(C) ^ B in cls(A) ^ B = empty (cl C) / B subset (cl A) / B provided you show C subset A >and >C ^ cls(D U B) = {C ^ (cls(D))} U {C ^ cls(B)} in A U cls(B) = >empty >thus X is disconnected. it's contradiction. Because C and D are nonnul. Interesting approach to a natty problem. ---- === Subject: Re: algebraic -> reverse polish > I'm searching a tool (Win/Linux) that turns the algebraic notation of > formulas into reverse polish notation. Can you help me? If you can program, then there are a couple ways to do it. One way is to write a parser for your algebraic expressions that converts them to an expression tree, then output that tree in RPN. If you aren't up to writing complete parsers, there is a simple procedure you can do to handle simple expressions. You'll need a stack, which can hold operators. Here's how to handle expressions that have no parenthesis. I'll tell you afterward how to handle parenthesis: Scan through the expression. When you come to a number, output it. When you come to an operator, check the operator stack: 1. If the operator stack is empty, push the operator. 2. Otherwise, compare to the operator on the top of the stack. If the operator on the top of the stack is not lower precedence than the new operator, pop the operator off the stack and output it, and go back to step #1. If the operator on the stack is lower precedence, push the new operator. When you reach the end of the expression, while the operator stack is not empty, pop and output the operators on it. For example, consider this (I'll use = to mean end of expression): 2 + 3 * 4 + 5 = This would be handled like this: 2: output 2 +: push + 3: output 3 *: + is lower than *, so push * 4: output 4 +: * is not lower than +, so pop and output * + is not lower than +, so pop and output + push + 5: output 5 =: pop and output + So the output is: 2 3 4 * + 5 + Whereas for 2 * 3 + 4 * 5 = we would have this: 2: output 2 *: push * 3: output 3 +: * is not lower than +, so pop and output * push + 4: output 4 *: + is lower than *, so push * 5: output 5 =: pop and output * pop and output + giving 2 3 * 4 5 * + How about parenthesis? Simply keep a count of the nesting level, and treat an operator at nesting level N as higher priority than any operator of lower nesting level. I'll use the notation +N or *N to mean a + or * at nesting level N. So consider this: ( 2 + 3 ) * ( 4 + 5 ) = (: nesting level is 1 2: output 2 +: push +1 3: output 3 ): nesting level is 0 *: +1 is not lower than *0, so pop and output + push *0 (: nesting level is 1 4: output 4 +: *0 is lower than +1, so push +1 5: output 5 ): nesting level is 0 =: pop and output + pop and output * giving 2 3 + 4 5 + * As an alternative to tracking nesting levels explicitly, and having to have a way to record on the operator stack which nesting level an operator was at, you can also treat ( and ) as operators, and push them on the operator stack. When a ( is on the top of the stack, treat the stack as if it is empty, and when you encounter ), you pop and output everything until you find a (, and then pop that. That would make ( 2 + 3 ) * ( 4 + 5 ) = work out like this: (: push ( 2: output 2 +: push + 3: output 3 ): pop and output + pop ( *: push * (: push ( 4: output 4 +: push + 5: output 5 ): pop and output + pop ) =: pop and output * Result: 2 3 + 4 5 + * Note that this is easy to change from a converter to an evaluator. Just change the places that output numbers to push those onto a number stack, and change the places that output operators to instead perform those operations using the top two items on the number stack, pushing the result onto the number stack. === Subject: Re: algebraic -> reverse polish < snip and will keep it for future reference. -- Bob Day reverse polish Tim Smith schrieb: > If you can program, then there are a couple ways to do it. I can't :-) > [program sample] Oh, thats to difficult for me. I'm only searching a ready-to-run tool :-) But thanks a lot. Markus === Subject: Re: Pi and Po > Given Pi, is there a formula for calculating the n'th digit of Pi that does > NOT use floors and ceilings? If we allow floors, then it's easy to just let > k = Pi*10^(n-1), and j = k - Floor[k], then finally take Floor[10j] to get > the n'th digit. But I wanted a formula that didn't use this. > Anyway, I the ultimate goal was to try to find a formula in terms of Pi for > the number Po=33.114411559922..., I got one obviously, using an infinite > series and the above method for extracting digits, but I was trying to see > if there was something more elegant. > By the way, I realize this number is not useful for anything, I was just > curious. I wonder if the continued fraction expansion of such a number > would have interesting properties like those of the original number. > Zach Given Pi? Assuming you are doing this with a computer and you already have a stored representation of PI to the m-th digit: What is the use of calculating the n-th digit? Why don't you use the representation as a string, build po as a string and then proceed to convert the po string into a number and do whatever it is you want to do? No calculations needed, just basic string manipulations. regards. === Subject: Re: integer quadratic program ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i27D2KX04578; I need help with this equation, 3(w+2)=2(w-1). How do u slove it? What are the steps? === Subject: Re: integer quadratic program ? > I need help with this equation, 3(w+2)=2(w-1). How do u slove it? What are the steps? This is how u slove it: 3w + 6 = 2w - 2 3w + 8 = 2w w + 8 = 0 w = -8 -24 + 6 = -16 -2 -18 = -18 === Subject: Re: integer quadratic program ? > I need help with this equation, 3(w+2)=2(w-1). How do u slove it? What are the steps? What makes you think it is a quadratic equation? To e-mail me get rid of the cats and dogs. === Subject: Re: integer quadratic program ? > I need help with this equation, 3(w+2)=2(w-1). How do u slove it? What > are the steps? Apply the distributive law to the left hand side: 3 w + 6 = 2(w - 1) Apply the distributive law to the right hand side: 3 w + 6 = 2 w - 2 Can you go on from there? Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Documents about multiplicative order in general and mersenne numbers > Let a be relatively prime to m. > The order of a (mod m) is the smallest positive integer r such that > a^r = 1 (mod m). This is identical to saying it's the smallest > positive integer r such that m divides a^r - 1. The values of s > such that m divides a^s - 1 are precisely the multiples of this r. > Unless I've misunderstood (which is quite possible), everything > you've said about order and mersenne numbers and primes is > contained in the paragraph above or is easily deduced from it. > Surely the information in my paragrpah is also in lots of intro > number theory texts. In any event, I think the paragraph brings out > the link between numbers of the form a^s - 1 and order of a (mod m). Thank you for your answer. than I thought at the beginning. Maybe I was just surprised when I recognized the real power of this statement :o) It is realy very powerfull. I would like to show it, but I need some more time, because I need to check a few things more. But please don't expect something new. I guess you will already know all that. > Because I couldn't find it in my books I was wondering what else is > already known about multiplicative order besides what is covered in my > book. I would really like to learn more about this topic :o) > And I'm still wondering what else you want to know about multiplicative > order. Your books surely talk about quadratic reciprocity; that can > be interpreted as being about order, since the quadratic residues > modulo a prime number p are precisely the residues of order dividing > (p - 1) / 2. Quadratic reciprocity is for sure worth learning about. I would really like to learn all :o) Maybe that sound stupid, but thats the way it is. I recognized how powerfull just this basic statements are and I wonder what else could be based on that. Thank you for your suggestion. I'll look at that. To me this link was not obvious before. Maybe I should read my books more systhematically. === Subject: Re: Documents about multiplicative order in general and mersenne numbers >... > Thank you for your answer. > than I thought at the beginning. Maybe I was just surprised when I > recognized the real power of this statement :o) > It is realy very powerfull. I would like to show it, but I need some > more time, because I need to check a few things more. But please don't > expect something new. I guess you will already know all that. I just recognized that I did a terrible mistake :o( I asumed that there is at least one prime p for each x and a in x=order a (mod p). But I recognized that this assumption was unfortunately wrong. For example, there is no prime p which has Ord 2 (mod p) = 6 If that had been true, it would have been possible to tell some interesting things about the distribution of primitive roots :o( === Subject: Tree Diagram - Help I am stumped with one problem and need help in constructing a tree diagram and its probabilities for this problem: The credit department of a local hardware store reported that 30% of sales are cash, 30% are paid by cheque and 40% are charged at time of purchase. 20% of cash purchases, 90% of cheque and 60% of charges are for more than $60. Mr. A purchased a new drill that cost $250. === Subject: Re: Tree Diagram - Help > I am stumped with one problem and need help in constructing a tree > diagram and its probabilities for this problem: > The credit department of a local hardware store reported that 30% of > sales are cash, 30% are paid by cheque and 40% are charged at time of > purchase. 20% of cash purchases, 90% of cheque and 60% of charges are > for more than $60. Mr. A purchased a new drill that cost $250. [Cash 30%] [<60:80%*30%=24%][>60:20%*30%=6%] [Cheque 30%] [<60:10%*30%=3%][>60:90%*30%=27%] [Charge 40%] [<60:40%*40%=16%][>60:60%*40%=24%] = 100% P [A|B] B>$60 A_1= Cash == 6/(6+27+24) A_2= Cheque == 27/(6+27+24) A_3= Charge == 24/(6+27+24) If its Monday then I am a fool but not ignorant. === Subject: Re: Tree Diagram - the answer that the new drill cost $250 dollars. You are most welcome. > I am stumped with one problem and need help in constructing a tree > diagram and its probabilities for this problem: > The credit department of a local hardware store reported that 30% of > sales are cash, 30% are paid by cheque and 40% are charged at time of > purchase. 20% of cash purchases, 90% of cheque and 60% of charges are > for more than $60. Mr. A purchased a new drill that cost $250. === Subject: Re: Tree Diagram - the answer >that the new drill cost $250 dollars. >You are most welcome. Yup. And I was able to also deduce that it was purchased by a male. --Lynn === Subject: Re: on the male alpha problem > : > : >Take, for instance, superstring theory. As with all ideas when they are > : >first introduced, it has been attacked quite a lot from many different > : >directions. With more research in the field, it has been able to > establish > : >itself as a topic legitimate to study in physics curricula, even > entering > : >undergraduate courses these days. So what was the content of the > attacks? > : >Excluding mathematical corrections to various papers (which I do not > : >consider attacks but legitimate scrutiny of method), the biggest attacks > : >have been on the freedom of predictability in the theory. Unique or > : >otherwise natural solutions are still not well established. There is > still > : too much freedom in the theory. Other attacks (though less frequent) > are > : >bothered by the large dimensionalities or see certain preferred frames in > : >the theories as problems despite the lack of observability. > : > : I'd put it a little differently. Some problems with string theory are: > : > : 1) it hasn't made any verifiable predictions yet, much less > : verified ones. > : 2) there is currently an enormous landscape of superstring > : vacua, each of which describes a different world with > : to choose among them. > : 3) none of these vacua corresponds to our world in a very > : convincing way, since until supersymmetry is broken all > : these worlds have the property that for every fermion there > : is a boson of the same mass, which is not the case in our > : world. Nobody has figured out the details of how supersymmetry > : can break spontaneously. So for now soft supersymmetry breaking > : terms are added to the theory in an ad hoc way to give the > : fermions and bosons different masses - which could be pretty > : much anything you want. > : 4) nobody knows how to reconcile superstring theory with an > : eternally accelerating expansion of the universe, which is > : what we seem to see. In other words, nobody knows a superstring > : vacuum that looks like de Sitter space. > : 5) unlike general relativity, which it seeks to supplant, nobody > : knows how to formulate string theory in a background-free manner. > : 6) despite frequent claims that perturbative superstring theory is > : better behaved than plain old perturbative quantum gravity, nobody > : has shown that scattering amplitudes in perturbative superstring > : theory are well-defined beyond 2 loops - which case was only dealt > : with by d'Hoker and Phong in 2001. > I agree* with most of this. These types of problems are in some way more > like goals of researche programmes than they are critiques of the theory. > Any field of mathematics has its early stages where people are asking > questions and saying things like we would like to answer questions x, y, > and z, but unfortunately we haven't figured out how yetIf that is indeed the state in which string theory finds itself, then its practitioners should set about their tasks in a much more sub rosa fahion until they actually produce some physics, instead of prematurely shouting the values of their wares in the bazaar. Franz === Subject: implicit function proof using fixed point theorem Using M2, Opera's revolutionary e-mail client: http://www.opera.com/m2/ How can I prove the implicit function theorem using a fixed point theorem. This problem has been bugging me for years. Any hints ? I can assume that df/dx(0,0)<>0 (1) Then given the functional equation F(x,y)=0 . Assume that y=f(x) for x in an open interval containing x_0 Then F(x,f(x))=0 Since (1) isn't zero I can construct a contraction mapping and get dF/dx(x,y)+(dF/dy)(x,y)y'=0 and for (x,y)near (0,0) y'=[-dF/dx(x,y)]/[dF/dy](x,y) Is this correct ? === Subject: Re: Question about iterated set operations. >Consider the collection {C_n} of sets where > oo oo >C_n = U / E_njk, where {E_njk} is the nth. mapping (j,k)---> field >G. > j=1 k=1 > I assume by field you mean sigma field. What, exactly, is the nth > mapping from (j,k)? G is a field, not necessarily a sigma field. We are considering all possible mappings of the set of pairs of positive integers into G. For instance, one possibility would be E_jk = A, (where A is an element of F) for all j and k. Sorry for my lack of precision, hope this clarifies it. Many thanks. === Subject: Re: A Paradox of The Central Limit Theorem > But you missed you turn to answer why? > Anyway I'll answer why: the left side is nonzero when > alpha < beta. The right side is zero. I usually say that > a nonzero number is unequal to zero :-) Well, in my view, the key point of this paradox is that lim{n->infinity}Pr{f(n)} = Pr{lim{n->infinity}f(n)} does not necessarily hold. How do you say about this? === Subject: Re: A Paradox of The Central Limit Theorem >But you missed you turn to answer why? >Anyway I'll answer why: the left side is nonzero when >alpha < beta. The right side is zero. I usually say that >a nonzero number is unequal to zero :-) > Well, in my view, the key point of this paradox is that lim{n->infinity}Pr{f(n)} = Pr{lim{n->infinity}f(n)} does > not necessarily hold. How do you say about this? What is f(n)? Is it an event? If so how do you define lim_n f(n)? Now even you must be aware that there are sequences of continuous functions f_n(x) such that lim_{n->infty} int_0^1 f_n(x) dx =/= int_0^1 lim_{n->infty} f_n(x) dx. So why should an identity like yours hold? Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: A Paradox of The Central Limit Theorem > But you missed you turn to answer why? > Anyway I'll answer why: the left side is nonzero when >> alpha < beta. The right side is zero. I usually say that >> a nonzero number is unequal to zero :-) > Well, in my view, the key point of this paradox is that lim{n->infinity}Pr{f(n)} = Pr{lim{n->infinity}f(n)} does > not necessarily hold. How do you say about this? > What is f(n)? Is it an event? Of course it is an event such as _n = mu> Now even you must be aware that there are sequences of continuous > functions f_n(x) such that > lim_{n->infty} int_0^1 f_n(x) dx =/= int_0^1 lim_{n->infty} f_n(x) dx. > So why should an identity like yours hold? Every paradox contains a kind of logical error. === Subject: Re: A Paradox of The Central Limit Theorem >> But you missed you turn to answer why? > Anyway I'll answer why: the left side is nonzero when >> alpha < beta. The right side is zero. I usually say that >> a nonzero number is unequal to zero :-) >Well, in my view, the key point of this paradox is that lim{n->infinity}Pr{f(n)} = Pr{lim{n->infinity}f(n)} does >not necessarily hold. How do you say about this? >What is f(n)? Is it an event? > Of course it is an event such as _n = muSo what is lim_{n-> infinity} ( = mu) ? Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: A Paradox of The Central Limit Theorem >> But you missed you turn to answer why? > Anyway I'll answer why: the left side is nonzero when > alpha < beta. The right side is zero. I usually say that > a nonzero number is unequal to zero :-) >> Well, in my view, the key point of this paradox is that >lim{n->infinity}Pr{f(n)} = Pr{lim{n->infinity}f(n)} does >> not necessarily hold. How do you say about this? > What is f(n)? Is it an event? > Of course it is an event such as _n = mu > So what is lim_{n-> infinity} ( = mu) ? What we are discussing is lim_{n-> infinity} Pr( = mu), not lim_{n-> infinity} ( = mu), aren't we ? === Subject: Re: A Paradox of The Central Limit Theorem But you missed you turn to answer why? > Anyway I'll answer why: the left side is nonzero when > alpha < beta. The right side is zero. I usually say that > a nonzero number is unequal to zero :-) >> Well, in my view, the key point of this paradox is that >lim{n->infinity}Pr{f(n)} = Pr{lim{n->infinity}f(n)} does >> not necessarily hold. How do you say about this? > What is f(n)? Is it an event? >Of course it is an event such as _n = mu >So what is lim_{n-> infinity} ( = mu) ? > What we are discussing is lim_{n-> infinity} Pr( = mu), > not lim_{n-> infinity} ( = mu), aren't we ? No. To quote from your message Pr{lim{n->infinity}f(n)} which appears to be the probability of the event lim{n->infinity}f(n) n'est-ce pas? Later you confirmed that f(n) could be the event _n = muHence lim_{n-> infinity}f(n) = lim_{n-> infinity} ( = mu) was exactly one of the things you were talking about. So, what is it? Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: (-2/3)^(-2/3) = (3/2)^(2/3)? > If iel Johnson will permit me to jump in here, his point was, > I believe, that while you CAN reduce (-2/3)^(-2/3) or (-8)^(1/3) to > real values, you cannot define (-2/3)^x or (-8)^x in a consistent > way for all real x and therefore the FUNCTION (-9)^x is not > defined. > f(x) = (-9)^x could be defined for x real with infinity many > discontinuities. > More specifically, for rational x of the form m/(2n+1) ... > For example the software GraphEq uses this fact to > draw visually continuous graph of y = (-9)^x. > It's name is GrafEq, actually. The graph it gives looks like the union > of the graphs of y = 9^x and y = -(9^x). Yes, I do. > Is it possible y = (-2)^(x^x) to have real root for -1 < x < 0, x > real? No. Let's think why not: For -1 < x < 0, if x^x is to be a real number, then x must be a rational of the form m/(2n+1). But then x^x would be irrational, and so (-2)^(x^x) would not be defined in the reals. > For example GrafEq plots visually continuous graphs (in 2nd and 3rd > quadrant) of the system: > |y = (-2)^u > |u = x^x > |x < 0 Right. But of course visually continuous doesn't mean defined everywhereNote that GrafEq plots no points for y = (-2)^(x^x) where -1 < x < 0. David === Subject: Re: (-2/3)^(-2/3) = (3/2)^(2/3)? > Note that GrafEq plots no points for y = (-2)^(x^x) where -1 < x < 0. Is there a mathematical difference between the systems: |y = (-2)^u |y = (-2)^(x^x) |u = x^x and |-1 < x < 0 |-1 < x < 0 According to GrafEq the graph of the first system consists of four curves and infinite vertical line (probably x = 0). The graph of the second consists of point (probably (-1;-0.5)) and infinite vertical line (probably x = 0). So everything is OK? === Subject: Re: (-2/3)^(-2/3) = (3/2)^(2/3)? > Note that GrafEq plots no points for y = (-2)^(x^x) where -1 < x < 0. > Is there a mathematical difference between the systems: > |y = (-2)^u |y = (-2)^(x^x) > |u = x^x and |-1 < x < 0 > |-1 < x < 0 > According to GrafEq the graph of the first system consists of four > curves and infinite vertical line (probably x = 0). The graph of the > second consists of point (probably (-1;-0.5)) and infinite vertical > line (probably x = 0). > So everything is OK? As far as I can tell, everything is OK with GrafEq. But your statements about what the graphs consist of according to GrafEq are subtly incorrect. (I assume that was not intentional on your part.) Here is what you could have said correctly: According to GrafEq the graph of the first system consists, _at most_, of four curves and the vertical line x = 0. The graph of the second consists, _at most_, of the point (-1,-1/2) and the vertical line x = 0. The reason for at most above is that GrafEq does not, for either system, ever reach the Graph (Finished) state. Rather, it ultimately says something like Graph (> 99.86% Proven, Subdivisions Exhausted). Although we can be sure that there are no points belonging to the desired graphs which are not shown, we cannot then be sure that all the points shown necessarily belong to the desired graphs. In fact, for these two systems, since the given inequality (-1I have a small problem here : >Given 2 polynomials of equal degree >X(z)=(z-x_1)(z-x_2)..(z-x_n) >and >Y(z)=(z-y_1)(z-y_2)..(z-y_n) >vith all x_i, y_i being known, positive integers, and >a=max{max_i{x_i},max_i{y_i}} >and given >D = X - Y >is there anything we can say about the roots of D ? >In particular, I would like to be able to say that they are lower >bounded by -a and upper bounded by a, but I dont know whether that >holds ? or why (not). Well, that's certainly false. Consider the case n=2: D(z) = (y1+y2-x1-x2) x + (x1 x2 - y1 y2) You could have e.g. x1 = x2 = a/2, y1 = a, y2 = 1, where the root of D(z) is a - a^2/4. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Roots of difference of polynomials > I have a small problem here : > Given 2 polynomials of equal degree > X(z)=(z-x_1)(z-x_2)..(z-x_n) > and > Y(z)=(z-y_1)(z-y_2)..(z-y_n) > vith all x_i, y_i being known, positive integers, and > a=max{max_i{x_i},max_i{y_i}} > and given > D = X - Y > is there anything we can say about the roots of D ? Since D will be quadratic in z, we can find exact expressions for the roots of D. === Subject: Re: Roots of difference of polynomials >I have a small problem here : >Given 2 polynomials of equal degree >X(z)=(z-x_1)(z-x_2)..(z-x_n) >and >Y(z)=(z-y_1)(z-y_2)..(z-y_n) >vith all x_i, y_i being known, positive integers, and >a=max{max_i{x_i},max_i{y_i}} >and given >D = X - Y >is there anything we can say about the roots of D ? >Since D will be quadratic in z, we can find exact expressions for the >roots of D. Quadratic? Try the case n=4 with x1+x2+x3+x4 <> y1+y2+y3+y4. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Roots of difference of polynomials >> I have a small problem here : > Given 2 polynomials of equal degree > X(z)=(z-x_1)(z-x_2)..(z-x_n) >> and >> Y(z)=(z-y_1)(z-y_2)..(z-y_n) >> vith all x_i, y_i being known, positive integers, and >> a=max{max_i{x_i},max_i{y_i}} >> and given >> D = X - Y >> is there anything we can say about the roots of D ? >Since D will be quadratic in z, we can find exact expressions for the >roots of D. > Quadratic? Try the case n=4 with x1+x2+x3+x4 <> y1+y2+y3+y4. Sorry, need to clean my glasses more often. === Subject: Re: Roots of difference of polynomials En el mensaje:c2ftub$ce9$1@nntp.itservices.ubc.ca, Robert Israel escribi.97: >> I have a small problem here : >> Given 2 polynomials of equal degree >> X(z)=(z-x_1)(z-x_2)..(z-x_n) >> and >> Y(z)=(z-y_1)(z-y_2)..(z-y_n) >> vith all x_i, y_i being known, positive integers, and >> a=max{max_i{x_i},max_i{y_i}} >> and given >> D = X - Y >> is there anything we can say about the roots of D ? >Since D will be quadratic in z, we can find exact expressions for the >roots of D. > Quadratic? Try the case n=4 with x1+x2+x3+x4 <> y1+y2+y3+y4. Virgil didn't saw the dots '..' ... === Subject: Re: Solving An Exponential Equation (Bleasdale Model) In this specific problem, B=0 is one solution, and unless B, p, q, or r have a negative value among them, you cannot have any others. In general, there are ways to solve this sort of thing numerically. One is to say, for example, B[n+1] = -1 + (1+2B[n])^(-q/p)*(1+3B[n])^(-r/p) and hope that the series of values B[n] converges. (It would usually converge very rapidly.) >I have 3 data points, where x=1,2,3, and y3>y2>y1 >I am trying to fit these to the Bleasdale Model, where: >y=(a+bx)^(1/c) >I arrive at the folllowing equation, where p, q, and r are known, and >I need to solve for B: >(1+B)^p*(1+2B)^q*(1+3B)^r=1 >How can I solve for B? If it helps, B is always between 0 and 1 for >my application. >Jim === Subject: Total differential?? Dear all, I understand that the total derivative of a given function z(x,y)=Dz=partial derivative of z with respect to x * dx + partial derive. of z with respect to y* dy But in one of the math books, the author is referring that the second total differential of a function y(x) as follows d(Dy) d(Dy) D^2y=D(Dy)=fy[ -----*Dx+ ------*Dy] dx dy such that d(dy)/dx is the partial derivative of Dy to x and so on. Which means that y is both a function of x and a function of itself?? Which implies that the differential of y(x) is not dy Dy= ---*Dx dx but becomes as follows dy dy Dy= ---*Dx+---*Dy, is this possible??. dx dx I know that according to chain rule in case of implicit functions if we assume that y(x) is a function of itself say u(y), which is a function of x dy du(y) d(y(x)) Then ----= -----*------ dx dy dx and hence if we have an implicit function like y^2=x taking the derivatives wrt x for both sides of the equation will yield to 2y dy/dx=1 therefore dy/dx=1/(2y), Can we apply the same concept to get the total derivative by assuming that y is a function of itself u(y) , which is in turn a function of x. (i.e. Total differential of y= partial derivative of y with respect to itself * dy+ partial derivative of y with respect to x* dx) And what does it mean in this case?? Thank you very much. P.S : this argument is found in Chiang book Fundamental Methods of Mathematical Economics Page 380. === Subject: Re: Prime factors of number near googolplexplex Dario, >4000 factors are on their way to that hotmail address. If it's an unread account, just dop me a note, and I can get them to you via some other means. Phil 1st bug in MS win2k source code found after 20 minutes: scanline.cpp 2nd and 3rd bug found after 10 more minutes: gethost.c Both non-exploitable. (The 2nd/3rd ones might be, depending on the CRTL) === Subject: CHAVEZ CALLS BUSH AN ASSHOLE Comments: This message did not originate from the above address. It was automatically remailed by one or more anonymous mail services. ***PLEASE REPORT ABUSE TO admin@italy-anonymous-remailer.it*** ***RIFERITE OGNI ABUSO A: admin@italy-anonymous-remailer.it*** Mail-To-News-Contact: abuse@dizum.com CHAVEZ CALLS BUSH AN ASSHOLE OH MY GOD, A BIG STORM IS COMING http://www.biblebelievers.org.au/przion1.htm http://users.one.se/~chribesk/sects/zog/newlight.html Jesus, save us from the evil ones!!! Questo messaggio e' stato inoltrato automaticamente da un paio di anonymous remailer. Il mittente originale === Subject: Re: CHAVEZ CALLS BUSH AN ASSHOLE > CHAVEZ CALLS BUSH AN ASSHOLE > OH MY GOD, A BIG STORM IS COMING What does this have to do with mathematics? Oil is fungible. > http://www.biblebelievers.org.au/przion1.htm Cf. http://www.yuppiesofzion.com > http://users.one.se/~chribesk/sects/zog/newlight.html > Jesus, save us from the evil ones!!! You? http://hertzlinger.blogspot.com === Subject: Re: The Schur Multiplier >>I have only just begun the Group Cohomology course, and my only background >>is a Group Theory course so be gentle. >>The Schur multiplier (M(G)) is defined as M(G)=A intersection X` , where >>X->G is a projective central extension with kernel A (X and G are groups, >>X->G is an epimorphism, and its kernel is a subgroup of Z(X), such that >>for every other Y->H with the aforementioned properties, and for every >>homomorphism G->H there exists a homomorphism X->Y such that the square is >>commutative). >>A very basic property of M(G) is that it is well-defined. Meaning it >>doesn`t depend on the Y->H and G->H picked. >The definition does not involve picking any particular Y or H. >It does involve picking X->G with the properties you mention, however, >So you have to prove that the definition does not depend (up to >isomorphism) on the choice of X->G with the properties you mention. >>I have to mention that X->G is a projective central extension if and only >>if X and G are groups, X->G is an epimorphism, and its kernel is a >>subgroup of Z(X), such that for every other Y->G with the aforementioned >>properties there exists a homomorphism X->Y such that the triangle is >>commutative. >That property follows from the one that you have already stated, just >by taking G = H. >>Somehow, using this property, they prove that M(G) is well-defined. I`m >>afraid I can`t seem to prove it on my own... Can you help me? >Normally, when you take a course, you do not expect to be able to prove >all of the theorems in the course on your own. If they prove that M(G) >is well-defined, wouldn't it make more sense to attempt to understand >their proof, rather than trying to do it yourself? >But here is an explanation anyway. >Suppose we have two projective central extensions A->X->G and B->Y->G >of G with the required properties. Then by the definition, there exist >homomorphisms f: X->Y and g:Y->X such that the triangles commute. >We cannot conclude from this, and it is not necessarily true, that f and >g are isomorphisms. But for x in X, by the commutativity of the diagrams, >g(f(x)) maps onto the same element of G as x, and so g(f(x)) = xa, for >some a in the kernel A of the map X->G. Since A is central in X, for any >x1, x2 in X, the commutator [f(g(x1)), f(g(x2))] is equal to [x1,x2]. >So fg induces the indentity on the commutator subgroup. Similarly gf >induces identity on [Y,Y], so f induces an isomorphism [X,X] -> [Y,Y] and >hence also on A ^ [X,X] -> B ^ [Y,Y]. So M(G) is uniquely defined up to >isomorphism. >But we have not yet proved that M(G) exists, because we have not proved >that projective central extensions A->X->G with the required properties >exist. To prove that, let 1 -> R -> F -> G -> 1 be an exact sequence >with F a free group. Then it is not hard to show by using the universal >property of the free group that 1-> R/[R,F] -> F/[R,F] -> G -> 1 >is a projective central extension of G. >Derek Holt. >skips certain proofs plus I'm taking this course earlier than I'm >supposed to... >Can you recommend any good books in the subject? It seems that every >book I try assumes some knowledge in topology, or defines Schur's >multiplier as the second homology rather than A intersection X'... There is book called The Schur Multiplier by Gregory Karpilovsky. My memory is that it provides a reasonable account of the topic, mainly from a group-theoretical viewpoint. You can get a long way with just group theory, but more advanced study of it would inevitably involve some homological algebra. Derek Holt. >P.S. The property I mentioned isn't proved just by saying that H=G. >That's only one direction. It doesn't matter, though. I proved the other >direction as well (which is the less trivial one). === Subject: abstract algebra question Question: if for the group G we have that |G| = 55, what possible values do we then have for the order of the elements in G. I dont get this. What is the relation between orders of elements in G and the order of the group(|G| =55) === Subject: Re: abstract algebra question Adjunct Assistant Professor at the University of Montana. >Question: >if for the group G we have that |G| = 55, what possible values do we then >have for the order of the elements in G. >I dont get this. What is the relation between orders of elements in G and >the order of the group(|G| =55) A corollary to Lagrange's Theorem states that the order of an element must divide the order of the group; this because if x is in G, then the subgroup has as many elements as the order of x, and by Lagrange's Theorem, the order of a subgroup divides the order of the group. So, if x is an element of G, then the order of G must divide the order of the group |G|=55. What are the possibilities for the order of x? ======== === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs In another thread, Theorist asked: > Anyway, as long as you're here, maybe you can provide the answer to > the GCD problem you posted a while ago, so that I can compare the > answer I obtained to it. This is odd, since I DID reply to theorist's post a while ago, and gave my puzzle's answer in that reply and in another message in that thread. (I *hate* when portals to Usenet fail to propagate my posts or succeed in deleting them.) But here is my reply again: :) Leroy > My initial guess is 1 - pi*sin(sqrt(6)) / sqrt(6). Based on a > heuristic argument. > Did you REALLY mean: > 1 - sin(sqrt(6)) / sqrt(6), > with no 'pi*'? > I and Rob Johnson got the no-pi solution. > (I am not saying you are wrong or dumb....I mis-type/mis-calculate too sometimes...) > (EVEN I!...) > ;) > Leroy Quet === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >...just because an event has probability 0 doesn't mean it can't >happen. > mmmm... I have a problem with this. Can you give an example of a probability > 0 > event that actually happened? > I would argue that you measured its probability incorrectly. Or it changed > after > you measured it. Or something. * Dave Seaman answered your question, but I think you did not understand him. Think about this: 1. Pick a random point on a line -- let's say the line from zero to one (on the positive x-axis if you like.) 2. Now, pick a second random point on the same line. What is the probability that you have picked the same point as in 1. above? 3. The answer is clearly zero. There is only one point that would satisfy 2. above but an infinite number of points that would not. Therefore the probability is zero. 4. And yet is is clearly possible to select the same point. In fact, you did it in the first case above! A probability of zero does not imply an impossibility. earle * === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs ) Think about this: ) ) 1. Pick a random point on a line -- let's say the line from zero to ) one (on the positive x-axis if you like.) One could argue that this is not possible, although that's not a very strong argument. ) 2. Now, pick a second random point on the same line. What is the ) probability that you have picked the same point as in 1. above? Assuming that you can, in fact, pick random points on a line, the probability is zero, obviously. I can state with absolute certainty that the two points are not the same, (assuming the 'random' way of picking is in fact random). ) 4. And yet is is clearly possible to select the same point. In fact, ) you did it in the first case above! One could argue, quite more strongly, that it is not possible to select the same point again. Unless the first point happens to be one of a finite subset. ) A probability of zero does not imply an impossibility. Given the second argument, one could say it implies it does. Think about this: I pick a random point on a line. I want you to pick the exact same point on the line. How are you going to do this ? You have to actually select it, you can't just say 'I'll take the same point that you selected. I assert that you can't do this, unless I happened to pick one of a finite subset of points. SaSW, Willem Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > ) Think about this: > ) > ) 1. Pick a random point on a line -- let's say the line from zero to > ) one (on the positive x-axis if you like.) > One could argue that this is not possible, although that's not a very > strong argument. > ) 2. Now, pick a second random point on the same line. What is the > ) probability that you have picked the same point as in 1. above? > Assuming that you can, in fact, pick random points on a line, the > probability is zero, obviously. I can state with absolute certainty > that the two points are not the same, (assuming the 'random' way of > picking is in fact random). *** No you can't. All you can say is the probability is zero that the two points are the same. > ) 4. And yet is is clearly possible to select the same point. In fact, > ) you did it in the first case above! > One could argue, quite more strongly, that it is not possible to select the > same point again. Unless the first point happens to be one of a finite > subset. > ) A probability of zero does not imply an impossibility. > Given the second argument, one could say it implies it does. * I suggest you grab a book. Manny Parzen's Introduction to Probability and Statistics would be a good place to start. I took his course at Stanford -- the book is great! earle * === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs )> Assuming that you can, in fact, pick random points on a line, the )> probability is zero, obviously. I can state with absolute certainty )> that the two points are not the same, (assuming the 'random' way of )> picking is in fact random). ) *** ) No you can't. All you can say is the probability is zero that the two ) points are the same. Just stating that something is true doesn't convince me one bit. I find it logical to say that if you pick two points randomly, it's impossible that you pick the same point. Counterargument please ? Also, I wonder why you're just adressing a side point, instead of the main point of my post. See below. )> ) 4. And yet is is clearly possible to select the same point. In fact, )> ) you did it in the first case above! ) )> One could argue, quite more strongly, that it is not possible to select the )> same point again. Unless the first point happens to be one of a finite )> subset. ) )> ) A probability of zero does not imply an impossibility. ) )> Given the second argument, one could say it implies it does. ) ) * ) I suggest you grab a book. Manny Parzen's Introduction to ) Probability and Statistics would be a good place to start. I took ) his course at Stanford -- the book is great! Why should I read a book ? Is there no simple argument you can use to convince me that it is possible to randomly pick a point on a line between 0 and 1, with uniform distribution, and then select that point again ? I assert that it is impossible to select that point again, because it is impossible to describe the point exactly, and selecting a point implies describing that point. SaSW, Willem Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >Only a finite subset of real numbers can be described in finite space >and/or time. That's certainly not true. The subset of describable reals is infinite. Countable, but infinite. Matthew T. Russotto mrussotto@speakeasy.net Extremism in defense of liberty is no vice, and moderation in pursuit of justice is no virtue. But extreme restriction of liberty in pursuit of a modicum of security is a very expensive vice. === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs )>Only a finite subset of real numbers can be described in finite space )>and/or time. ) ) That's certainly not true. The subset of describable reals is ) infinite. Countable, but infinite. I'm not going into the infinite-vs-unbounded discussion again, but I feel I should point out that I quantified my argument. SaSW, Willem Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > )>Only a finite subset of real numbers can be described in finite space > )>and/or time. > ) > ) That's certainly not true. The subset of describable reals is > ) infinite. Countable, but infinite. > I'm not going into the infinite-vs-unbounded discussion again, > but I feel I should point out that I quantified my argument. I submit that the set of rational numbers is a description of an infinite subset of the reals, and that the description occupies only a finite space and can be written down in finite time. Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ) I submit that the set of rational numbers is a description of an > ) infinite subset of the reals, and that the description occupies only a > ) finite space and can be written down in finite time. > In case you have missed the original discussion, we were talking about > describing single numbers, not entire sets. In that case, each single member of the set of rational numbers can be described in finite space and finite time. Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. In case you have missed the original discussion, we were talking about )> describing single numbers, not entire sets. ) In that case, each single member of the set of rational numbers can be ) described in finite space and finite time. True, but given finite space and finite time, only members belonging to a finite subset can be described. SaSW, Willem Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > )> In case you have missed the original discussion, we were talking about > )> describing single numbers, not entire sets. > ) In that case, each single member of the set of rational numbers can be > ) described in finite space and finite time. > True, but given finite space and finite time, only members belonging > to a finite subset can be described. I gave a counterexample to that statement in my earlier posting. Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. )> In case you have missed the original discussion, we were talking about )> )> describing single numbers, not entire sets. ) ) )> ) In that case, each single member of the set of rational numbers can be )> ) described in finite space and finite time. ) )> True, but given finite space and finite time, only members belonging )> to a finite subset can be described. ) ) I gave a counterexample to that statement in my earlier posting. You can't mean the example where you described 'all rational numbers', because you now know we're talking about describing single members. So what counterexample do you mean ? I must have missed it. SaSW, Willem Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > )> )> In case you have missed the original discussion, we were talking about > )> )> describing single numbers, not entire sets. > ) > ) > )> ) In that case, each single member of the set of rational numbers can be > )> ) described in finite space and finite time. > ) > )> True, but given finite space and finite time, only members belonging > )> to a finite subset can be described. > ) > ) I gave a counterexample to that statement in my earlier posting. > You can't mean the example where you described 'all rational numbers', > because you now know we're talking about describing single members. > So what counterexample do you mean ? I must have missed it. If you are describing single numbers, then each rational number has a finite description. If you are describing sets of numbers, then the set of rational numbers has a finite description. Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. Yes, and if you have finite time and space, there are rational numbers you >can not describe. >SaSW, Willem Please give an example of a rational number I can't describe :-) --Lynn === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >) You are saying that it is possible to select a first point at time t = >) 0, but not possible to select that point at a later time? What is it >) about that point that changes so that it cannot be selected later? >Well, actually all I'm saying is that it's not possible to select that >point later. I'm not saying anything about selecting it the first time. >In my view, if you actually select a point, it has to be one of a finite >subset (being the 'describable' points), but I am willing to concede that >it is possible to *randomly* select any point on a continuum. >But even if some method exists to select a random point from a continuous >interval, then you still can't select that point again, afterwards. >You seem to be assuming it's possible to select any given point, even if >you are unable to describe that point. I disagree with that assumption. But this is like saying that you disagree with the assumption that a group has an identity element. According to the standard mathematical theory of probability, with the uniform probability distribution on [0,1], any measurable subset of [0,1] has probability equal to its measure - so any singleton point, or any countable subset for that matter, has probability 0. There are no assumptions concerning actual selection or physical points involved here; this is a purely mathematical theory, which assigns probabilities to measurable subsets of [0,1]. This theory is used as a model for application to real-life probability problems. You can say that you do not like this model, because it does not properly correspond to the intuitive idea that probability 0 represents impossibility. That is a valid criticism, but in the end, the degree to which a model corresponds to intuition is a secondary consideration. What is more important is firstly that the mathematics of the model can be developed in a satisfactory fashion, and secondly that it should be capable of successful application, and this model performs excellently according to those criteria. Of course you are free to develop your own alternative mathematical model of probability in which probability 0 does correspond to impossibility, but developing a complete and successful mathematical theory from the ground up is no light undertakings. Typically, it takes many years of hard work by lots of first rate mathematicians to achieve such a thing. Derek Holt. === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs )>You seem to be assuming it's possible to select any given point, even if )>you are unable to describe that point. I disagree with that assumption. ) But this is like saying that you disagree with the assumption that a group ) has an identity element. Why ? ) According to the standard mathematical theory of ) probability, with the uniform probability distribution on [0,1], any ) measurable subset of [0,1] has probability equal to its measure - so any ) singleton point, or any countable subset for that matter, has probability 0. I agree with you here. ) There are no assumptions concerning actual selection or physical points ) involved here; this is a purely mathematical theory, which assigns ) probabilities to measurable subsets of [0,1]. And it assigns a probability of zero to the subset [N..N], which has measure zero. Therefore it is impossible that you will randomly select N. ) This theory is used as a model for application to real-life probability ) problems. You can say that you do not like this model, because it does not ) properly correspond to the intuitive idea that probability 0 represents ) impossibility. That is a valid criticism, but in the end, the degree to ) which a model corresponds to intuition is a secondary consideration. What ) is more important is firstly that the mathematics of the model can be ) developed in a satisfactory fashion, and secondly that it should be capable ) of successful application, and this model performs excellently according to ) those criteria. Then could you explain to me why it is undesirable to say that it is impossible to randomly select any given point in a continuum ? Is there some axiom that states and/or implies that a probability of zero does *not* imply impossibility, and if there is, why is that axiom so important to the mathematical model ? ) Of course you are free to develop your own alternative mathematical model ) of probability in which probability 0 does correspond to impossibility, but ) developing a complete and successful mathematical theory from the ground up ) is no light undertakings. Typically, it takes many years of hard work by ) lots of first rate mathematicians to achieve such a thing. In what way would the current mathematical model be compromised by saying that a probability of 0 implies an impossibility ? SaSW, Willem Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >Is there some axiom that states and/or implies that a probability of zero >does *not* imply impossibility, and if there is, why is that axiom so >important to the mathematical model ? The impossible event corresponds to the empty set, which has measure zero. >In what way would the current mathematical model be compromised by saying >that a probability of 0 implies an impossibility ? You would be claiming that all sets of zero measure are empty. I'm not interested in mathematics that might have anything to do with reality. -- Easterly, in sci.math === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >Why should I read a book ? So you can learn and understand why you are wrong. >Is there no simple argument you can use to >convince me that it is possible to randomly pick a point on a line between >0 and 1, with uniform distribution, and then select that point again ? Apparently not. You need to understand continuous probability density functions. >I assert that it is impossible to select that point again, because it is >impossible to describe the point exactly, and selecting a point implies >describing that point. >SaSW, Willem I think you are tied up in the notion of selectingOf course the notion of having some sort of physical spinner actually select numbers from [0,1] with uniform distribution is an unrealizable ideal in the first place. The real number line is an abstraction, and probability spaces are abstractions too. The set of outcomes in this case is the set of points or numbers in [0,1]. The probability assigned to any (every) particular value is 0. That doesn't imply that, for example, your ideal experiment can't produce the value 1/2. --Lynn === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs )>Why should I read a book ? ) ) So you can learn and understand why you are wrong. I'm guessing the book also simply states or assumes that it's possible, without giving an actual explanation of *why* it's possible. )>Is there no simple argument you can use to )>convince me that it is possible to randomly pick a point on a line between )>0 and 1, with uniform distribution, and then select that point again ? ) ) Apparently not. You need to understand continuous probability density ) functions. AIUI, these are defined in such a way that any given *range* has a probability for a randomly chosen value to be in it. And as the range [N..N] has size zero, the probability of a randomly chosen value to be inside that is zero. I don't see why you can't state it's impossible that a randomly chosen value will be equal to N. )>I assert that it is impossible to select that point again, because it is )>impossible to describe the point exactly, and selecting a point implies )>describing that point. )>SaSW, Willem ) ) I think you are tied up in the notion of selectingAnd I think you are tied up in the notion that it should be possible to select any value. Why should it be possible ? It's perfectly logical to say that it is not, and as far as I can see, it does not lead to any contradictions, and it has the benefit that something with probability zero is impossible, as one would intuitively assume. ) Of course the notion of having some sort of physical spinner actually ) select numbers from [0,1] with uniform distribution is an unrealizable ) ideal in the first place. The real number line is an abstraction, and ) probability spaces are abstractions too. The set of outcomes in this ) case is the set of points or numbers in [0,1]. The probability ) assigned to any (every) particular value is 0. That doesn't imply ) that, for example, your ideal experiment can't produce the value 1/2. I say it does imply that. After all, the probability is zero, isn't it ? So tell me, what is wrong with the notion that an ideal experiment that produces a number from the continuous range [0..1] cannot produce any *given* number ? SaSW, Willem Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > AIUI, these are defined in such a way that any given *range* has a > probability for a randomly chosen value to be in it. And as the range > [N..N] has size zero, the probability of a randomly chosen value to be > inside that is zero. [N,N] has infinitesimal size. [N,N) has zero size; it's obviously impossible to choose a value within that set, since that set is the empty set. If you are left with an infinitesimal *when you are done calculating*, then it collapses to zero. > And I think you are tied up in the notion that it should be possible to > select any value. Why should it be possible ? It's perfectly logical > to say that it is not, and as far as I can see, it does not lead to any > contradictions, and it has the benefit that something with probability > zero is impossible, as one would intuitively assume. Oh dear, this is some sort of Axiom of Choice issue, isn't it? I admit that the denial of the AoC has always been beyond my comprehension. > ) Of course the notion of having some sort of physical spinner actually > ) select numbers from [0,1] with uniform distribution is an unrealizable > ) ideal in the first place. The real number line is an abstraction, and > ) probability spaces are abstractions too. The set of outcomes in this > ) case is the set of points or numbers in [0,1]. The probability > ) assigned to any (every) particular value is 0. That doesn't imply > ) that, for example, your ideal experiment can't produce the value 1/2. > I say it does imply that. After all, the probability is zero, isn't it ? I think your argument just became circular. The probability is zero because you can't choose a given point. You can't choose a given point because the probability is zero=== Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs )> ) Of course the notion of having some sort of physical spinner actually )> ) select numbers from [0,1] with uniform distribution is an unrealizable )> ) ideal in the first place. The real number line is an abstraction, and )> ) probability spaces are abstractions too. The set of outcomes in this )> ) case is the set of points or numbers in [0,1]. The probability )> ) assigned to any (every) particular value is 0. That doesn't imply )> ) that, for example, your ideal experiment can't produce the value 1/2. ) )> I say it does imply that. After all, the probability is zero, isn't it ? ) I think your argument just became circular. The probability is zero ) because you can't choose a given point. You can't choose a given point ) because the probability is zeroYou got that wrong. The probability is zero because you told me that it is zero. You can't *randomly* choose a given point because the probability is zero. SaSW, Willem Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > )> Assuming that you can, in fact, pick random points on a line, the > )> probability is zero, obviously. I can state with absolute certainty > )> that the two points are not the same, (assuming the 'random' way of > )> picking is in fact random). > ) *** > ) No you can't. All you can say is the probability is zero that the two > ) points are the same. > Just stating that something is true doesn't convince me one bit. > I find it logical to say that if you pick two points randomly, it's > impossible that you pick the same point. Counterargument please ? > Also, I wonder why you're just adressing a side point, instead of the main > point of my post. See below. > )> ) 4. And yet is is clearly possible to select the same point. In fact, > )> ) you did it in the first case above! > ) > )> One could argue, quite more strongly, that it is not possible to select the > )> same point again. Unless the first point happens to be one of a finite > )> subset. > ) > )> ) A probability of zero does not imply an impossibility. > ) > )> Given the second argument, one could say it implies it does. > ) > ) * > ) I suggest you grab a book. Manny Parzen's Introduction to > ) Probability and Statistics would be a good place to start. I took > ) his course at Stanford -- the book is great! > Why should I read a book ? Is there no simple argument you can use to > convince me that it is possible to randomly pick a point on a line between > 0 and 1, with uniform distribution, and then select that point again ? * You are saying that it is possible to select a first point at time t = 0, but not possible to select that point at a later time? What is it about that point that changes so that it cannot be selected later? But Parzen's book is really a very good reference for things like this. earle * === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs ) You are saying that it is possible to select a first point at time t = ) 0, but not possible to select that point at a later time? What is it ) about that point that changes so that it cannot be selected later? Well, actually all I'm saying is that it's not possible to select that point later. I'm not saying anything about selecting it the first time. In my view, if you actually select a point, it has to be one of a finite subset (being the 'describable' points), but I am willing to concede that it is possible to *randomly* select any point on a continuum. But even if some method exists to select a random point from a continuous interval, then you still can't select that point again, afterwards. You seem to be assuming it's possible to select any given point, even if you are unable to describe that point. I disagree with that assumption. SaSW, Willem Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > ) You are saying that it is possible to select a first point at time t = > ) 0, but not possible to select that point at a later time? What is it > ) about that point that changes so that it cannot be selected later? > Well, actually all I'm saying is that it's not possible to select that > point later. I'm not saying anything about selecting it the first time. > In my view, if you actually select a point, it has to be one of a finite > subset (being the 'describable' points), but I am willing to concede > that it is possible to *randomly* select any point on a continuum. This is, on the face of it, pure madness. For starters, what definition are you using for describable? Do you consider the reciprocal of each positive integer to be describable? If so, then that's already an infinite subset right there. Some transcendentals are describable in the normal English sense: the reciprocal of pi, the reciprocal of e, the number whose decimal expansion is 0.10100100010000100000001... > But even if some method exists to select a random point from a > continuous interval, then you still can't select that point again, > afterwards. You selected the point the first time, so it must be describable, so you can select it the second time as well. > You seem to be assuming it's possible to select any given point, even if > you are unable to describe that point. I disagree with that assumption. Any geomtric point in (x >= 0, x <= 1, y = 0) is describable by a single real number (the x coordinate). Any real number in [0,1] is describable by its Cartesian digit expansion. === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs )> Well, actually all I'm saying is that it's not possible to select that )> point later. I'm not saying anything about selecting it the first time. ) )> In my view, if you actually select a point, it has to be one of a finite )> subset (being the 'describable' points), but I am willing to concede )> that it is possible to *randomly* select any point on a continuum. ) This is, on the face of it, pure madness. For starters, what definition ) are you using for describable? That which can be described. Any definition that covers that is fine with me. ) Do you consider the reciprocal of each ) positive integer to be describable? No, I consider the reciprocal of each describable integer to be describable. Duh. ) If so, then that's already an infinite subset right there. ) Some transcendentals are describable in ) the normal English sense: the reciprocal of pi, the reciprocal of e, ) the number whose decimal expansion is 0.10100100010000100000001... That's quite true. So ? It's still a finite subset. )> But even if some method exists to select a random point from a )> continuous interval, then you still can't select that point again, )> afterwards. ) ) You selected the point the first time, so it must be describable, ) so you can select it the second time as well. It was selected by a random process the first time. You can't make a random process do the same thing twice, and you can't select it directly. )> You seem to be assuming it's possible to select any given point, even if )> you are unable to describe that point. I disagree with that assumption. ) ) Any geomtric point in (x >= 0, x <= 1, y = 0) is describable by a single ) real number (the x coordinate). Any real number in [0,1] is describable ) by its Cartesian digit expansion. Only a finite subset of real numbers can be described in finite space and/or time. Some as a rational number, some as square roots, taylor series, whatever. But only a finite subset. SaSW, Willem Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > Only a finite subset of real numbers can be described in finite space > and/or time. Some as a rational number, some as square roots, taylor > series, whatever. But only a finite subset. Oh, *finitely* describable. That makes much more sense. (I still don't quite agree with you, but this dispenses with seems to be madness.) === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs Supersedes: ...just because an event has probability 0 doesn't mean it can't >happen. >mmmm... I have a problem with this. Can you give an example of a >probability 0 >event that actually happened? >I would argue that you measured its probability incorrectly. Or it >changed after >you measured it. Or something. > 1. Pick a random point on a line -- let's say the line from zero to one > (on the positive x-axis if you like.) > 2. Now, pick a second random point on the same line. What is the > probability that you have picked the same point as in 1. above? > 3. The answer is clearly zero. There is only one point that would > satisfy 2. above but an infinite number of points that would not. > Therefore the probability is zero. I argue that the probability is infinitesimally small, but not zero. > 4. And yet is is clearly possible to select the same point. In fact, > you did it in the first case above! > A probability of zero does not imply an impossibility. I argue that probability of zero should be synonymous with impossible, and infinitesimally small probability used to cover cases such as yours. === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >>...just because an event has probability 0 doesn't mean it can't >>happen. >> mmmm... I have a problem with this. Can you give an example of a >> probability 0 >> event that actually happened? > I would argue that you measured its probability incorrectly. Or it >> changed after >> you measured it. Or something. > 1. Pick a random point on a line -- let's say the line from zero to one > (on the positive x-axis if you like.) > 2. Now, pick a second random point on the same line. What is the > probability that you have picked the same point as in 1. above? > 3. The answer is clearly zero. There is only one point that would > satisfy 2. above but an infinite number of points that would not. > Therefore the probability is zero. > I argue that the probability is infinitesimally small, but not zero. * The probability is the ratio of number of successful cases to the total number of cases -- namely 1 / infinity, which in my book is zero. By the way, how do you define infinitesimally small but not zero? earle * === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >I argue that the probability is infinitesimally small, but not zero. > The probability is the ratio of number of successful cases to the > total number of cases -- namely 1 / infinity, which in my book is zero. Then I think your book is wrong. Consider the following: (a) The chance that a randomly chosen number in [0,1] equals 0.5 is infinitesimally small (1 / infinity). (b) The chance that a randomly chosen number in [0,1] equals 14 is zero (0 / infinity). Since (a) is clearly different from (b), there ought to be some different term for it. Infinitesimally small may as well be that term. > By the way, how do you define infinitesimally small but not zero? Based on the standard definition(s) of infinitesimal: http://dictionary.reference.com/search?q=infinitesimal adj. 1. Immeasurably or incalculably minute. 2. Mathematics. Capable of having values approaching zero as a limit. n. 1. An immeasurably or incalculably minute amount or quantity. 2. Mathematics. A function or variable continuously approaching zero as a limit. === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >1. Pick a random point on a line -- let's say the line from zero to one >(on the positive x-axis if you like.) >2. Now, pick a second random point on the same line. What is the >probability that you have picked the same point as in 1. above? >3. The answer is clearly zero. There is only one point that would >satisfy 2. above but an infinite number of points that would not. >Therefore the probability is zero. > I argue that the probability is infinitesimally small, but not zero. Probabilities, by definition, are real numbers. That rules out infinitesimals. Probability is based on measure theory. A set can have measure zero without being empty (a single point, for example, or any countable set). Analogously, an event can have probability zero without being impossible. >4. And yet is is clearly possible to select the same point. In fact, >you did it in the first case above! >A probability of zero does not imply an impossibility. > I argue that probability of zero should be synonymous with impossible, > and infinitesimally small probability used to cover cases such as yours. That's not an argument; it's merely an unsupported assertion. Given a uniform distribution on [0,1], exactly what probability do you assign to each point, and how do you go about adding up these probabilities to obtain a total of 1 for the whole space? Keep in mind that the interval [0,1/2] has the same number of points as [0,1]. The answer from measure theory is that the probability P(X in A) is obtained as the integral of the constant function 1 over the set A. But this implies that the measure of each single point is zero, not an infinitesimal. Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. > 1. Pick a random point on a line -- let's say the line from zero to >> one (on the positive x-axis if you like.) >> 2. Now, pick a second random point on the same line. What is the >> probability that you have picked the same point as in 1. above? >> 3. The answer is clearly zero. There is only one point that would >> satisfy 2. above but an infinite number of points that would not. >> Therefore the probability is zero. >I argue that the probability is infinitesimally small, but not zero. > Probabilities, by definition, are real numbers. That rules out > infinitesimals. Probability is based on measure theory. A set can have > measure zero without being empty (a single point, for example, or any > countable set). Analogously, an event can have probability zero without > being impossible. Infinitesimal probabilities *are* sometimes used. A Google search on (probability infinitesimal) turns up, among other things: http://www-zeus.roma1.infn.it/~agostini/cern/node52.html http://www.vims.edu/~david/MS505/exp_dist.pdf http://www.princeton.edu/~adame/papers/inf-laws/inf-laws.pdf >> 4. And yet is is clearly possible to select the same point. In fact, >> you did it in the first case above! >> A probability of zero does not imply an impossibility. >I argue that probability of zero should be synonymous with impossible, and infinitesimally small probability used to cover >cases such as yours. > That's not an argument; it's merely an unsupported assertion. Given a > uniform distribution on [0,1], exactly what probability do you assign to > each point, and how do you go about adding up these probabilities to > obtain a total of 1 for the whole space? Keep in mind that the interval > [0,1/2] has the same number of points as [0,1]. (infinitesimal) x (infinity) is undefined in the general case, due to insufficient context. In specific cases, some intuitively obvious cases can be constructed: a = (probability that a randomly chosen element of [0,1] equals C, where C is a member of [0,1]) b = (number of elements of [0,0.5]) a x b = 0.5 > The answer from measure theory is that the probability P(X in A) is > obtained as the integral of the constant function 1 over the set A. That certainly works, assuming that A is a subset of the set from which X is chosen. > But this implies that the measure of each single point is zero, not an > infinitesimal. I disagree with this part, and suggest the following expansion of your prior statement: The probability P(a randomly chosen member of [0,1] is in A) is obtained as the integral of y = { 1 if x is in [0,1] { 0 otherwise over the set A. For a concrete example, consider A = [-0.5,0.5]. This expansion implies that the measure of each single point in [0,1] is infinitesimal, and the measure of each single point outside it is zero. === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > Infinitesimal probabilities *are* sometimes used. A Google search > on (probability infinitesimal) turns up, among other things: > http://www-zeus.roma1.infn.it/~agostini/cern/node52.html > http://www.vims.edu/~david/MS505/exp_dist.pdf > http://www.princeton.edu/~adame/papers/inf-laws/inf-laws.pdf The first two do use the word 'infinitesimal', but it seems to me (I didn't study them carefully) that they do not assume that a probability function can take anything but a real number as a value. The second of the papers linked seems to be using 'infinitesimal' to mean 'arbitrarily small', as one speaks in calculus of arbitrarily small intervals. The last mentions infinitesimal probabilities but mainly in order to say that they are useless in solving the difficulty that the paper addresses. (By the way, that paper is by the person who unleashed the Sleeping Beauty problem on the world.) -Jamie === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >> I argue that the probability is infinitesimally small, but not zero. >Probabilities, by definition, are real numbers. That rules out >infinitesimals. Probability is based on measure theory. A set can have >measure zero without being empty (a single point, for example, or any >countable set). Analogously, an event can have probability zero without >being impossible. > Infinitesimal probabilities *are* sometimes used. A Google search > on (probability infinitesimal) turns up, among other things: > http://www-zeus.roma1.infn.it/~agostini/cern/node52.html Although the paper contains the phrase infinitesimal calculus, I see no attempt to make the concept of infinitesimal probabilities precise. > http://www.vims.edu/~david/MS505/exp_dist.pdf That looks like a standard measure-theoretic treatment to me. > http://www.princeton.edu/~adame/papers/inf-laws/inf-laws.pdf Although the author describes a scheme involving infinitesimal probabilities, he admits that the idea is a failure for his purposes. This paper is to appear in a journal of philosophy, not a mathematics journal. > 4. And yet is is clearly possible to select the same point. In fact, > you did it in the first case above! > A probability of zero does not imply an impossibility. >> I argue that probability of zero should be synonymous with >impossible, and infinitesimally small probability used to cover >> cases such as yours. >That's not an argument; it's merely an unsupported assertion. Given a >uniform distribution on [0,1], exactly what probability do you assign to >each point, and how do you go about adding up these probabilities to >obtain a total of 1 for the whole space? Keep in mind that the interval >[0,1/2] has the same number of points as [0,1]. > (infinitesimal) x (infinity) is undefined in the general case, due to > insufficient context. In specific cases, some intuitively obvious > cases can be constructed: Infinitesimals exist in non-archimedean ordered fields. In any such field, the product of any two elements is always well defined, even if one of them happens to be infinitesimal and the other infinite. > a = (probability that a randomly chosen element of [0,1] equals C, > where C is a member of [0,1]) > b = (number of elements of [0,0.5]) > a x b = 0.5 But you also have b = (number of elements of [0,1], whereupon a x b = 1.0? Sorry, it doesn't work. >The answer from measure theory is that the probability P(X in A) is >obtained as the integral of the constant function 1 over the set A. > That certainly works, assuming that A is a subset of the set from which > X is chosen. And assuming A is measurable. >But this implies that the measure of each single point is zero, not an >infinitesimal. > I disagree with this part, and suggest the following expansion of your > prior statement: > The probability P(a randomly chosen member of [0,1] is in A) is > obtained as the integral of > y = { 1 if x is in [0,1] > { 0 otherwise Since x is a randomly chosen member of [0,1], your if x is in [0,1] is superfluous; y is simply the constant function 1 on [0,1]. > over the set A. For a concrete example, consider A = [-0.5,0.5]. Your A is not an event, since it is not contained in the sample space. > This expansion implies that the measure of each single point in [0,1] is > infinitesimal, and the measure of each single point outside it is zero. Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. a = (probability that a randomly chosen element of [0,1] equals C, > where C is a member of [0,1]) >b = (number of elements of [0,0.5]) >a x b = 0.5 > But you also have b = (number of elements of [0,1], whereupon a x b = > 1.0? Sorry, it doesn't work. Oh dear, you're right. I suppose the limit-as-you-approach-zero method from integral calculus is as elegant as this is going to get. >> The answer from measure theory is that the probability P(X in A) is >> obtained as the integral of the constant function 1 over the set A. >That certainly works, assuming that A is a subset of the set from which >X is chosen. > And assuming A is measurable. Granted. >> But this implies that the measure of each single point is zero, not an >> infinitesimal. >I disagree with this part, and suggest the following expansion of your >prior statement: >The probability P(a randomly chosen member of [0,1] is in A) is >obtained as the integral of > y = { 1 if x is in [0,1] > { 0 otherwise > Since x is a randomly chosen member of [0,1], your if x is in [0,1] is > superfluous; y is simply the constant function 1 on [0,1]. >over the set A. For a concrete example, consider A = [-0.5,0.5]. > Your A is not an event, since it is not contained in the sample space. If you're using [0,1] as the universal set, then the difference between P(a randomly chosen value equals 0.5) and P(it equals 14) is admittedly less important, since 14 will never be used for *any* purpose. It still seems intuitively obvious, though, that P(it equals 0.5) should be an infinitesimally small but non-zero value - which, however, is treated as zero in certain situations. === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs ) If you're using [0,1] as the universal set, then the difference ) between P(a randomly chosen value equals 0.5) and P(it equals 14) ) is admittedly less important, since 14 will never be used for *any* ) purpose. It still seems intuitively obvious, though, that P(it ) equals 0.5) should be an infinitesimally small but non-zero value - ) which, however, is treated as zero in certain situations. Well, no. To me it seems intuitively obvious that P(it equals 0.5) is zero, which means that it's impossible to pick 0.5 randomly. Impossible to pick any rational number, for that matter. I still don't see what's wrong with this. SaSW, Willem Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: Cantor's Diagonal Argument > Perhaps we'd better start by defining our terms. What, > precisely, is a real number in this context? >> okay, > a real number is any point on the number line, like a complex number is >> a point in the argand plane. >> Define point. I can't do arithmetic on points in this form. > We'll refine to the region 0 to 1, the arguments should apply > to the entire number line by considering extra parameters. > A single point is referenced by every decimal expansion. *DE > You have not shown this, although it's not a big problem; for any > point, one can work out its decimal expansion in a fairly obvious > fashion, and the reverse is true; for any infinite decimal expansion > one can show there's a unique point, by using Dedekind cuts. > Basically, one slices up [0,1) into 10 half-intervals, then a > selected subsegment into 10 more half-intervals, and so on. > Two DEs are equal if : > DE1 = 0 . d_1 d_2 d_3 d_4.... > DE2 = 0 . e_1 e_2 e_3... > <- > ForAll nEN, d_n = e_n > To avoid the quantifier, a function is defined > NotEqual (DE1, DE2) { > DigitsEqual (DE1, DE2, 1) > Return TRUE > } > DigitsEqual (DE1, DE2, digit) { > if (DE1_digit = DE2_digit) { > DigitsEqual (DE1, DE2, digit+1) > } > } > Two real numbers are considered equal if it can be shown > NotEqual does not terminate. The [Return TRUE] line is only > reached when some n is found such that DE1_n and DE2_n are found unequal. > Eg 0.12345 =/= 0.12344 since DE1_5 = 5 and DE2_5 = 4, thus the > NotEqual function will terminate returning TRUE. > A recurring 9 detection could be added to the function to make a stronger > 'iff' definition, details are not significant at this point. > It's only an issue if d_n = x and e_n = x-1 (1 <= x <= 9), > and d_{n+k} = 0 and e_{n+k} = 9 for all k > 0. > This should not be a big problem to incorporate into your recursion. > This must of course be done both ways. > DigitsEqual(DE1,DE2,digit) { > If(DE1_digit = DE2_digit) { > DigitsEqual(DE1, DE2, digit+1); > } > else if (DE1_digit = DE2_digit-1) { > ZeroNine(DE1_digit, DE2_digit, digit+1); > } > else if (DE2_digit = DE1_digit-1) { > ZeroNine(DE2_digit, DE1_digit, digit+1); > } > } > ZeroNine(DE1, DE2, digit) { > if(DE1_digit = 0 and DE2_digit = 9) > ZeroNine(DE1, DE2, digit+1); > } Good, its only tail recursion so we can compile to while loops. Now I'll define the real count, then I just have to use this condition : > Two real numbers are considered equal if it can be shown > NotEqual does not terminate So I have to show NotEqual applied to Diag and some number *on* the list, does not terminate. Start with a modified UTM that recognises 11 symbols, -1..9. A multitasking harness applies the UTM to 2 parameters, n and i, seperated by -1 symbol. n i 1 2 3 4 5 6 7 ... 1 1 2 4 7 2 3 5 8 3 6 9 4 10 ... It allocates some m processing cyles to each parameter pair, then moves on to the next designated pair in this fashion : 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, .... So this partially evaluates UTM(n, i) where (n, i) = (1,1),(1,1),(1,2),(1,1),(1,2),(2,1)... When a parameter pair is finished computing (halts) it is removed from the processor queue, -1 output or multiple digits is considered void and treated as if the process has not halted. This results in an increasingly populated grid of digits. n i 1 2 3 4 5 6 7 ... 1 0 7 2 8 8 3 3 8 3 5 7 4 3 ... As a digit is calculated, its number n on the grid is tested for 'effectively computable', its first j digits must all be complete. j starts at 1 and increases by 1 for every effectively computable number that is found. c j 1 2 3 4 5 6 7 ... 1 x 2 x x 3 x x x 4 x x x x ... In this example, n = 3 qualifies 3 8 3 5 7 1 0.8357.. is indexed as the 1st computable number. This guarantees diagonal constructions can be applied to the list of computable numbers. OK so far? Herc > OK so far? > I missed negatives, I'd rather define it as a boolean +- sign at the head > of the leading d, to keep consistant with human representation, -0 is not > an issue, as long as the format is viable for diagonal analysis. > For the purposes of this diatribe I think the half-interval [0,1) > is quite adequate; one can always map it to the real line later. > Herc > In decimal representation : >> sequences of the form dddddddd.ddddddd... > You have yet to show the equivalence of your definitions. This >> may cause some problems. > Formally, you are defining a set R+, where R+ is all numbers >> j + sum(i=1,+oo) (d_i * 10^(-i)), where d_i is your digit, and >> j is a positive integer. > (If you want to include negative numbers of this form, that's fine too. >> I'll just change j to be an integer.) >> where >> d -> 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 >> Ready for next step? > Go for it. >> Herc > I'm surprised, does *anyone* here follow my argument that *any* > sequence of digits appears on the list of computable numbers to > *any* number of specified decimal places. > Essentially you can't give a lower bound to how many digits belong to > a digit sequence missing from the list. And no a digit sequence does not > imply rational. > Can *someone* examine my inductive proof? > Given a certain definition of the real numbers, the number of > real numbers is in fact finite. > Herc > -- > #191, ewill3@earthlink.net > It's still legal to go .sigless. >> -- >> #191, ewill3@earthlink.net >> It's still legal to go .sigless. > #191, ewill3@earthlink.net > It's still legal to go .sigless. === Subject: Re: Cantor's Diagonal Argument In sci.logic, |-|erc : > Perhaps we'd better start by defining our terms. What, > precisely, is a real number in this context? >> okay, >> a real number is any point on the number line, like a complex number is >> a point in the argand plane. >> Define point. I can't do arithmetic on points in this form. >We'll refine to the region 0 to 1, the arguments should apply >to the entire number line by considering extra parameters. >A single point is referenced by every decimal expansion. *DE >You have not shown this, although it's not a big problem; for any >point, one can work out its decimal expansion in a fairly obvious >fashion, and the reverse is true; for any infinite decimal expansion >one can show there's a unique point, by using Dedekind cuts. >Basically, one slices up [0,1) into 10 half-intervals, then a >selected subsegment into 10 more half-intervals, and so on. >Two DEs are equal if : >DE1 = 0 . d_1 d_2 d_3 d_4.... >DE2 = 0 . e_1 e_2 e_3... ><- >ForAll nEN, d_n = e_n >To avoid the quantifier, a function is defined >NotEqual (DE1, DE2) { > DigitsEqual (DE1, DE2, 1) > Return TRUE >} >DigitsEqual (DE1, DE2, digit) { > if (DE1_digit = DE2_digit) { > DigitsEqual (DE1, DE2, digit+1) > } >} >Two real numbers are considered equal if it can be shown >NotEqual does not terminate. The [Return TRUE] line is only >reached when some n is found such that DE1_n and DE2_n are found unequal. >Eg 0.12345 =/= 0.12344 since DE1_5 = 5 and DE2_5 = 4, thus the >NotEqual function will terminate returning TRUE. >A recurring 9 detection could be added to the function to make a stronger >'iff' definition, details are not significant at this point. >It's only an issue if d_n = x and e_n = x-1 (1 <= x <= 9), >and d_{n+k} = 0 and e_{n+k} = 9 for all k > 0. >This should not be a big problem to incorporate into your recursion. >This must of course be done both ways. >DigitsEqual(DE1,DE2,digit) { > If(DE1_digit = DE2_digit) { > DigitsEqual(DE1, DE2, digit+1); > } > else if (DE1_digit = DE2_digit-1) { > ZeroNine(DE1_digit, DE2_digit, digit+1); > } > else if (DE2_digit = DE1_digit-1) { > ZeroNine(DE2_digit, DE1_digit, digit+1); > } >} >ZeroNine(DE1, DE2, digit) { > if(DE1_digit = 0 and DE2_digit = 9) > ZeroNine(DE1, DE2, digit+1); >} > Good, its only tail recursion so we can compile to while loops. That was the general idea. > Now I'll define the real count, then I just have to use this condition : >Two real numbers are considered equal if it can be shown >NotEqual does not terminate > So I have to show NotEqual applied to Diag and some number *on* > the list, does not terminate. > Start with a modified UTM that recognises 11 symbols, -1..9. > A multitasking harness applies the UTM to 2 parameters, n and i, > seperated by -1 symbol. > n i 1 2 3 4 5 6 7 ... > 1 1 2 4 7 > 2 3 5 8 > 3 6 9 > 4 10 > ... > It allocates some m processing cyles to each parameter pair, then moves on to > the next designated pair in this fashion : > 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, .... > So this partially evaluates UTM(n, i) where (n, i) = (1,1),(1,1),(1,2),(1,1),(1,2),(2,1)... > When a parameter pair is finished computing (halts) it is removed from the processor > queue, -1 output or multiple digits is considered void and treated as if the process has not > halted. > This results in an increasingly populated grid of digits. > n i 1 2 3 4 5 6 7 ... > 1 0 7 > 2 8 8 3 > 3 8 3 5 7 > 4 3 > ... > As a digit is calculated, its number n on the grid is tested for > 'effectively computable', its first j digits must all be complete. > j starts at 1 and increases by 1 for every effectively > computable number that is found. > c j 1 2 3 4 5 6 7 ... > 1 x > 2 x x > 3 x x x > 4 x x x x > ... > In this example, n = 3 qualifies > 3 8 3 5 7 > 1 0.8357.. is indexed as the 1st computable number. > This guarantees diagonal constructions can be applied to the list of > computable numbers. > OK so far? This construction only governs computable numbers. I suspect these are similar to algebraic numbers in that both are countable. Are you specifying an explicit mapping (e.g., one can compute Computable(n) for all n > 0) or merely assuming one exists? > Herc >OK so far? >I missed negatives, I'd rather define it as a boolean +- sign at the head >of the leading d, to keep consistant with human representation, -0 is not >an issue, as long as the format is viable for diagonal analysis. >For the purposes of this diatribe I think the half-interval [0,1) >is quite adequate; one can always map it to the real line later. >Herc >> In decimal representation : >> sequences of the form dddddddd.ddddddd... >> You have yet to show the equivalence of your definitions. This >> may cause some problems. >> Formally, you are defining a set R+, where R+ is all numbers >> j + sum(i=1,+oo) (d_i * 10^(-i)), where d_i is your digit, and >> j is a positive integer. >> (If you want to include negative numbers of this form, that's fine too. >> I'll just change j to be an integer.) >> where >> d -> 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 >> Ready for next step? >> Go for it. >> Herc > I'm surprised, does *anyone* here follow my argument that *any* > sequence of digits appears on the list of computable numbers to > *any* number of specified decimal places. > Essentially you can't give a lower bound to how many digits belong to > a digit sequence missing from the list. And no a digit sequence does not > imply rational. > Can *someone* examine my inductive proof? > Given a certain definition of the real numbers, the number of > real numbers is in fact finite. > Herc > -- > #191, ewill3@earthlink.net > It's still legal to go .sigless. >> -- >> #191, ewill3@earthlink.net >> It's still legal to go .sigless. >-- >#191, ewill3@earthlink.net >It's still legal to go .sigless. #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Cantor's Diagonal Argument > This guarantees diagonal constructions can be applied to the list of > computable numbers. > OK so far? > This construction only governs computable numbers. I suspect these > are similar to algebraic numbers in that both are countable. > Are you specifying an explicit mapping (e.g., one can compute > Computable(n) for all n > 0) or merely assuming one exists? The mapping I outlined should be explicit except for the variable m (processing grain) and the particular UTM used. UTMs exist in real life, I could actually start producing a binary list from a UTM from a book. UTMs applied to a data input can emulate any computer function. So in theory there exists a finite x for each UTM where UTM(x,0) = 3. UTM(x,1) = 1 UTM(x,2) = 4 UTM(x,3) = 1 UTM(x,4) = 5 adding pi to the list of computables. I'm trying to show the set of computable numbers is a larger class than previously thought and includes all irrationals. We need a definition of countable!? Herc === Subject: Stepping Around A Square Grid As many of my puzzles/games, we start with an n-by-n grid. We place the integers 1 through n^2 into this grid's squares (one integer per square) so that each (m+1) is exactly m steps from m. And the 'steps' are a sequence of up/down/left/right moves (of one square per step) where the directions CAN change during the sequence of steps between any m and (m+1), and no square is visited more than once on any *particular* move, but moves can pass over squares with integers in them already or over squares moved over in previous moves. (Again, a move is the sequence of m steps between any m and (m+1).) For which n's can you find a way to fill the n-by-n grid completely? Here are a solution I have for n=3 and for n=4: 1 2 8 6 4 3 9 7 5 1 2 13 6 10 8 7 4 5 3 12 14 15 9 11 16 For example, the move which goes from 11 to 12, in n=4, is (can be): left 2, up 3, right 3, down 2, left 1, and 2 + 3 + 3 + 2 + 1 = 11. So many of you may have guessed already that, for n odd, n^2 and n^2 -1 must be in opposing corner-squares; and guessed that, for n even, n^2 and n^2 -1 must be in the corner-squares on the ends of the same side of the grid. Leroy Quet === Subject: Re: Stepping Around A Square Grid >As many of my puzzles/games, we start with an n-by-n grid. >We place the integers 1 through n^2 into this grid's squares (one >integer per square) so that >each (m+1) is exactly m steps from m. >And the 'steps' are a sequence of up/down/left/right moves (of one >square per step) where the directions CAN change during the sequence >of steps between any m and (m+1), and no square is visited more than >once on any *particular* move, but moves can pass over squares with >integers in them already or over squares moved over in previous moves. >(Again, a move is the sequence of m steps between any m and (m+1).) >For which n's can you find a way to fill the n-by-n grid completely? >Here are a solution I have for n=3 and for n=4: >1 2 8 >6 4 3 >9 7 5 >1 2 13 6 >10 8 7 4 >5 3 12 14 >15 9 11 16 >For example, the move which goes from 11 to 12, in n=4, is (can be): >left 2, up 3, right 3, down 2, left 1, >and 2 + 3 + 3 + 2 + 1 = 11. >So many of you may have guessed already that, for n odd, >n^2 and n^2 -1 must be in opposing corner-squares; >and guessed that, for n even, n^2 and n^2 -1 must be in the >corner-squares on the ends of the same side of the grid. >Leroy Quet Am I missing something? 8 6 4 7 5 3 9 2 1 meets your requirements, but 8 and 9 are not on opposite corners. -Matt === Subject: Re: Stepping Around A Square Grid >As many of my puzzles/games, we start with an n-by-n grid. >We place the integers 1 through n^2 into this grid's squares (one >integer per square) so that >each (m+1) is exactly m steps from m. >And the 'steps' are a sequence of up/down/left/right moves (of one >square per step) where the directions CAN change during the sequence >of steps between any m and (m+1), and no square is visited more than >once on any *particular* move, but moves can pass over squares with >integers in them already or over squares moved over in previous moves. >(Again, a move is the sequence of m steps between any m and (m+1).) >For which n's can you find a way to fill the n-by-n grid completely? >Here are a solution I have for n=3 and for n=4: >1 2 8 >6 4 3 >9 7 5 >1 2 13 6 >10 8 7 4 >5 3 12 14 >15 9 11 16 >For example, the move which goes from 11 to 12, in n=4, is (can be): >left 2, up 3, right 3, down 2, left 1, >and 2 + 3 + 3 + 2 + 1 = 11. >So many of you may have guessed already that, for n odd, >n^2 and n^2 -1 must be in opposing corner-squares; >and guessed that, for n even, n^2 and n^2 -1 must be in the >corner-squares on the ends of the same side of the grid. >Leroy Quet Am I missing something? > 8 6 4 > 7 5 3 > 9 2 1 > meets your requirements, but 8 and 9 are not on opposite corners. > -Matt You are not missing anything. I realized last night, long after getting off-line, that I had made an error about the placements of the n^2 and n^2-1. (For they need not necessarily even be in ANY corner.) And, also, I may have implied with my examples that the 1 must be in a corner, but it can be anywhere you wish. I found this puzzle kind of fun to do in practice, but it is actually pretty dumb mathematically... (at least when compared to some of my other grid-puzzles and grid-games I have posted here). Leroy Quet === Subject: Multiplying Dice, then adding Here is a dice game which I post more of as a question than as a game-idea. You have 2 dice and 2 players. combination, on each face next to each of the six numbers on that die. (One sign per face.) And they do this secretly before play begins. Next, the dice are cast together on each round, and then the *product* P of the 2 dice's values (including the signs) is taken. Finally, on each round, the product (which can be positive or negative) is added to a running-total. After a fixed number of rounds, player 1 wins if the running-total is positive, player 2 wins if the running-total is negative. (A total of 0 is a tie.) What would be a good strategey for either player in choosing whether to put a + or a - on any particular face of his/her die? Leroy Quet === Subject: Re: Multiplying Dice, then adding > Here is a dice game which I post more of as a question than as a > game-idea. > You have 2 dice and 2 players. > combination, on each face next to each of the six numbers on that die. > (One sign per face.) > And they do this secretly before play begins. > Next, the dice are cast together on each round, and then the *product* > P of the 2 dice's values (including the signs) is taken. > Finally, on each round, the product (which can be positive or > negative) is added to a running-total. > After a fixed number of rounds, player 1 wins if the running-total is > positive, player 2 wins if the running-total is negative. > (A total of 0 is a tie.) > What would be a good strategy for either player in choosing whether > to put a + or a - on any particular face of his/her die? > Leroy > Quet As noted already in a reply, this is definitely more of a psychological game than a mathematical game, as far as strategy. I wonder if any obvious variations on this game can make it more mathematical, however. Maybe if we simply added the products, doing away with the signs. And the winner is the player who guesses closest to the final total, after a fixed number of rounds. Or we have 3 dice and 3 players, each marking their own die with signs as before. And the winner is determined by taking the final total mod 3. Here is the list of the number, m, of products which get n, if we have two 6-sided dice. (All unlisted n's have 0 products which equal them.) n: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 m: 1, 2, 2, 3, 2, 4, 2, 1, 2, 4 n: 15, 16, 18, 20, 24, 25, 30, 36 m: 2, 1, 2, 2, 2, 1, 2, 1 Using the unequal number of ways to get to any product, there *must* be a mathematically-interesting game somewhere... Leroy Quet === Subject: Re: Multiplying Dice, then adding > Here is a dice game which I post more of as a question than as a > game-idea. > You have 2 dice and 2 players. > combination, on each face next to each of the six numbers on that die. > (One sign per face.) > And they do this secretly before play begins. > Next, the dice are cast together on each round, and then the *product* > P of the 2 dice's values (including the signs) is taken. > Finally, on each round, the product (which can be positive or > negative) is added to a running-total. > After a fixed number of rounds, player 1 wins if the running-total is > positive, player 2 wins if the running-total is negative. > (A total of 0 is a tie.) > What would be a good strategey for either player in choosing whether > to put a + or a - on any particular face of his/her die? > Leroy > Quet You have 64 possibilities. For each possibility, you can check all combinations of your opponents die and sum all possible outcomes. This sum is always zero. So, whatever be the configuration you choose, the expected result is a tie. Against a random opponent, no winning strategy exists. === Subject: Re: Multiplying Dice, then adding qqquet@mindspring.com (Leroy Quet)'s wild thoughts were fruit: >Here is a dice game which I post more of as a question than as a >game-idea. >You have 2 dice and 2 players. >combination, on each face next to each of the six numbers on that die. >(One sign per face.) >And they do this secretly before play begins. >Next, the dice are cast together on each round, and then the *product* >P of the 2 dice's values (including the signs) is taken. >Finally, on each round, the product (which can be positive or >negative) is added to a running-total. >After a fixed number of rounds, player 1 wins if the running-total is >positive, player 2 wins if the running-total is negative. >(A total of 0 is a tie.) >What would be a good strategey for either player in choosing whether >to put a + or a - on any particular face of his/her die? Player 2 puts a minus on every number, then the product will always be negative. Jan Hyde (MVP - Visual Basic) Two thieves stole a calendar . . . they each got 6 months [Abolish the TV Licence - http://www.tvlicensing.biz/] === Subject: Re: Multiplying Dice, then adding > qqquet@mindspring.com (Leroy Quet)'s wild thoughts were > fruit: >Here is a dice game which I post more of as a question than as a >game-idea. >You have 2 dice and 2 players. >combination, on each face next to each of the six numbers on that die. >(One sign per face.) >And they do this secretly before play begins. >Next, the dice are cast together on each round, and then the *product* >P of the 2 dice's values (including the signs) is taken. >Finally, on each round, the product (which can be positive or >negative) is added to a running-total. >After a fixed number of rounds, player 1 wins if the running-total is >positive, player 2 wins if the running-total is negative. >(A total of 0 is a tie.) >What would be a good strategey for either player in choosing whether >to put a + or a - on any particular face of his/her die? > Player 2 puts a minus on every number, then the product will > always be negative. Suspecting that player 2 might try this, player 1 puts a minus sign on every number on his die, so the result is always positive. === Subject: Re: Multiplying Dice, then adding > qqquet@mindspring.com (Leroy Quet)'s wild thoughts were > fruit: >Here is a dice game which I post more of as a question than as a >game-idea. >You have 2 dice and 2 players. >combination, on each face next to each of the six numbers on that die. >(One sign per face.) >And they do this secretly before play begins. >Next, the dice are cast together on each round, and then the *product* >P of the 2 dice's values (including the signs) is taken. >Finally, on each round, the product (which can be positive or >negative) is added to a running-total. >After a fixed number of rounds, player 1 wins if the running-total is >positive, player 2 wins if the running-total is negative. >(A total of 0 is a tie.) >What would be a good strategey for either player in choosing whether >to put a + or a - on any particular face of his/her die? > Player 2 puts a minus on every number, then the product will > always be negative. Not if Player 1 also puts a minus on every number, which will make every product positive. Neither player has a dominant strategy. ------------------------ Mark Jeffrey Tilford tilford@ugcs.caltech.edu === Subject: Re: Multiplying Dice, then adding Mark J. Tilford 's wild thoughts following fruit: >qqquet@mindspring.com (Leroy Quet)'s wild thoughts were >fruit: >>Here is a dice game which I post more of as a question than as a >>game-idea. >>You have 2 dice and 2 players. >>combination, on each face next to each of the six numbers on that die. >>(One sign per face.) >>And they do this secretly before play begins. >>Next, the dice are cast together on each round, and then the *product* >>P of the 2 dice's values (including the signs) is taken. >>Finally, on each round, the product (which can be positive or >>negative) is added to a running-total. >>After a fixed number of rounds, player 1 wins if the running-total is >>positive, player 2 wins if the running-total is negative. >>(A total of 0 is a tie.) >>What would be a good strategey for either player in choosing whether >>to put a + or a - on any particular face of his/her die? >Player 2 puts a minus on every number, then the product will >always be negative. >Not if Player 1 also puts a minus on every number, which will make every >product positive. Neither player has a dominant strategy. Doh! It was early.... Jan Hyde (MVP - Visual Basic) I punched him in the stomach three times, said Tom triumphantly. (Kegel Archives) [Abolish the TV Licence - http://www.tvlicensing.biz/] === Subject: Re: Multiplying Dice, then adding > Here is a dice game which I post more of as a question than as a > game-idea. > You have 2 dice and 2 players. > combination, on each face next to each of the six numbers on that die. > (One sign per face.) > And they do this secretly before play begins. > Next, the dice are cast together on each round, and then the *product* > P of the 2 dice's values (including the signs) is taken. > Finally, on each round, the product (which can be positive or > negative) is added to a running-total. > After a fixed number of rounds, player 1 wins if the running-total is > positive, player 2 wins if the running-total is negative. > (A total of 0 is a tie.) > What would be a good strategey for either player in choosing whether > to put a + or a - on any particular face of his/her die? Since each player must mark the die without knowledge of the opponent's actions, there is no sensible strategy. There are 64 ways to mark the die; these form 32 pairs of duals, where the dual of a set of markings is created by reversing all 6 signs. With no knowledge of the opponent's choices or strategy, we can only assume that each member of each pair of duals is equally likely, but they create exactly opposite results. The game is simply random. If multiple games were being played, with new markings each time, then it might be possible to attempt some strategy by watching how the opponent's markings changed and trying to anticipate them -- rather like repeatedly playing rock-paper-scissors(*) with the same opponent. (*) Also known by many other names, but you know what I mean. Mark Brader 1. remove ball from package. 2. place in hand. msb@vex.net 3. call dog by name. 4. throw ball. Toronto -- directions seen on rubber ball package === Subject: Re: Multiplying Dice, then adding > What would be a good strategey for either player in choosing whether > to put a + or a - on any particular face of his/her die? > Since each player must mark the die without knowledge of the opponent's > actions, there is no sensible strategy. > There are 64 ways to mark the die; these form 32 pairs of duals, where > the dual of a set of markings is created by reversing all 6 signs. > With no knowledge of the opponent's choices or strategy, we can only > assume that each member of each pair of duals is equally likely, but > they create exactly opposite results. The game is simply random. Perhaps a mixed strategy: You choose - from the total set of 64 markings - a particular combination at random. The question is now, what probabilities to assign to each combination? === Subject: Re: Factoring project seeks members > Hello everybody, > I am programmer interested in seeking members for PROOF of CONCEPT in new > approaches to Factoring Algorithms . > http://www.sourceforge.net/projects/atpg > For previous work see > http://www.lri.fr/~simon/satex/infos/benchs-generators/ VZNuri.html > and > http://www.cs.indiana.edu/cgi-pub/sabry/cnf.html > The project works follows. > 1) Build a multiplier circuit for inputs A(n/2 -bit) and B( n - bit ) and > output C (n-bit) > 2) Put a gaint AND gate D checking C such that the output of the gate is 1, > when A and B are factors of C. > For example, If the number to be factored is 5, and C is 4-bits long. > D = C_0 AND C_1' AND C_2 AND C_3' > D becomes 1, iff if C_0 = 1, C_1 = 0 , C_2 = 1, and C_3 = 0 > If the number to be factor is 8, and C is 4-bits long > D = C_0' AND C_1' AND C_2' AND C_3 > D becomes a 1, iff and only if C_0 = 0 , C_1 = 0, C_2 = 0, C_3 = 1 > 3) Find binary vectors A, B such that D = 1 > ----------- > What the project currently implements > 1) Program for simulating binary circuits > 2) multiplier generators > 3) Partly implemented PREDICT Solver > ------------- > What the goal of the project is > The project goal is to use statistical techniques, for finding A and B. > See > http://sourceforge.net/project/showfiles.php?group_id=61183& package_id=106530 > for some methods > The idea is to find a method that is able to calculate the probability of D > being a 1, as accurately as possible. If this could be accomplished most of > the time, for weightsets A , B, it would be equivalent to factoring C. Well, you're out of luck on step 1, since the circuit you described is not a new approach. It's a modified op-amp. The probability of D being a 1 is 1, since your D is a latch, not a gate. That is easily proven since the iff implies the statistics you're using are Bayesian Statistics. And the only thing that's not known about Bayesian Statistics is how much will Bayesian Statisticians pay for a real computer. > -suresh === Subject: Re: Factoring project seeks members Sorry, u need to elaborate. I know the approach isnt totally new. In fact, i know people who has studied or took similar approaches before. This is one of the reasons, why I know i have not gone completely crazy. The only thing interesting or new about the approach is calculation of probability of D. -suresh === Subject: Re: Factoring project seeks members > Hello everybody, > I am programmer interested in seeking members for PROOF of CONCEPT in new > approaches to Factoring Algorithms . > http://www.sourceforge.net/projects/atpg > For previous work see > http://www.lri.fr/~simon/satex/infos/benchs-generators/ VZNuri.html > and > http://www.cs.indiana.edu/cgi-pub/sabry/cnf.html > The project works follows. Also see http://www.theneoproject.com/ and other large scale distributed processing pages. Good luck! Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Fun with axioms -- automated theory generation >It's fun to do. The neat thing about math is one doesn't have >to include a reality check. > Well, actually mathematicians care a lot about which > axioms or definitions are more *fruitful* than others. > It's not so much a matter of whether the axioms are true > as whether they lead to useful insights, beautiful surprises, > a nice bce between complexity and simplicity, or new > connections between other things we're interested in. > So, while reality check might not be the best word for it, > good mathematicians are quite fussy about what structures defined > via axioms they are willing to bother studying. If you write down > an arbitrary list of axioms and ask a mathematician to study it's > consequences, they'll probably look at you funny. Do a Google groups search under Mathematician's AlgorithmMuch (or even all) of the process of finding those magical nexus points (like the distributivity axiom in lattice theory, the Pierce rule in Intuitionist logic, the axiom of choice in set theory, etc.) can -- in fact -- be done in a systematic way and even automated. === Subject: Re: Fun with axioms -- automated theory generation >It's fun to do. The neat thing about math is one doesn't have >to include a reality check. > Well, actually mathematicians care a lot about which > axioms or definitions are more *fruitful* than others. > It's not so much a matter of whether the axioms are true > as whether they lead to useful insights, beautiful surprises, > a nice bce between complexity and simplicity, or new > connections between other things we're interested in. > So, while reality check might not be the best word for it, > good mathematicians are quite fussy about what structures defined > via axioms they are willing to bother studying. If you write down > an arbitrary list of axioms and ask a mathematician to study it's > consequences, they'll probably look at you funny. > Do a Google groups search under Mathematician's Algorithm > Much (or even all) of the process of finding those magical nexus > points (like the distributivity axiom in lattice theory, the Pierce > rule in Intuitionist logic, the axiom of choice in set theory, etc.) > can -- in fact -- be done in a systematic way and even automated. Nobody has looked mathematicans funny concerning axioms. They've only looked at them bizzarly concerning Set Theory. Which aren't axioms, since mathematicians didn't even invented them. Born-again Einstonians and Pythagorus wannabees invented them. So since we already know that the Greeks begat the British who begat the Russians who begat the Commies, who begat the Canadians, who begat California, who begat Florida, who begat leisure suits, which begat Austrailian in-laws, it's still unknown if unknown mathema-gibberers can even run automatic dishwashers, so automatic axioms are usually left to Israelis, Egyptians and those who own over-priced real estate in Soho. === Subject: Re: Fun with axioms ..................... >But then of course there's the ultimate reality check >for mathematics: physics. >I believe you have that backwards. When physics is done ignoring the theoretical >coherence of mathematics you will find that a great many resources have been >unnecessarily directed toward an infinite regress of triangles. There is a good example from physics. Some of you may have heard about the Feynman integral. I am one of those who looked at it (I have not published on this point), and it is clear that Feynman assumed that there were mathematical objects available which do not exist. I do not believe that there is yet a reasonable fix which enables it to be used as a direct means to evaluate quantum transitions. This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Fun with axioms > But then of course there's the ultimate reality check > for mathematics: physics. > I believe you have that backwards. When physics is done ignoring the theoretical > coherence of mathematics you will find that a great many resources have been > unnecessarily directed toward an infinite regress of triangles. Galileo. Galvani Volta Joule Hooke Faraday Would you wish me to enlarge the list further? Franz === Subject: Re: Fun with axioms > But then of course there's the ultimate reality check > for mathematics: physics. > I believe you have that backwards. When physics is done ignoring the > theoretical > coherence of mathematics you will find that a great many resources have > been > unnecessarily directed toward an infinite regress of triangles. > Galileo. > Galvani > Volta > Joule > Hooke > Faraday > Would you wish me to enlarge the list further? No, you can stop at Galileo. Since he's the invented Moderne Physica. Where Gravity and Set Theory are ALL things, and Pisa is just a tiny target in quality goods: Genuine US of A Stealth Bombers. And since Faraday invented Farads, he's also one of the few physicists even worth mentioning in the last two hundred years. > Franz === Subject: Re: Fun with axioms > But then of course there's the ultimate reality check > for mathematics: physics. > I believe you have that backwards. When physics is done ignoring the > theoretical > coherence of mathematics you will find that a great many resources have > been > unnecessarily directed toward an infinite regress of triangles. > Galileo. > Galvani > Volta > Joule > Hooke > Faraday > Would you wish me to enlarge the list further? Unlike you, Franz, I am not in the habit of denying the brilliance of *great* men and women with substantive ideas. Your ignorance of the history of mathematics (as well as so many other things) is making itself plain here. What I am denying in John's remark is that physics provides any kind of reality check for mathematics whatsoever. In fact, the failure of making proper strides toward these issues leads to discussions such as, http://plato.stanford.edu/entries/mathematics-inconsistent/ That is not to say that the precious numerical calculations of physics are in any way engered. That very link includes the paragraph, Robert K. Meyer (1976) seems to have been the first to think of an inconsistent arithmetical theory. At this point, he was more interested in the fate of a consistent theory, his relevant arithmetic R#. There proved to be a whole class of inconsistent arithmetical theories; see Meyer and Mortensen (1984), for example. Meyer argued that these theories provide the basis for a revived Hilbert Program. Hilbert's program was widely held to have been seriously damaged by G.9adel's Second Incompleteness Theorem, according to which the consistency of arithmetic was unprovable within arithmetic itself. But a consequence of Meyer's construction was that within his arithmetic R# it was demonstrable by simple finitary means that whatever contradictions there might happen to be, they could not adversely affect any numerical calculations. Hence Hilbert's goal of conclusively demonstrating that mathematics is trouble-free proves largely achievable. Of course, a mathematical illiterate such as yourself probably has a difficult time reading a passage that long. Nevertheless, if you have not yet exhausted that waste of protoplasm you call a brain, it is the same kind of statement that is to be found in another paper to which I brought attention within Galathaea's threads, M. Pavicic and N. Megill, Non-Orthomodular Models for Both Standard Quantum Logic and Standard Classical Logic: Repercussions for Quantum Computers, Int. J. Theor. Phys. 39, 2349-2391 (2000), http://xxx.lanl.gov/abs/quant-ph/9906101 (1999). Hence the main aspect of our result is that the syntactical structure of classical logic corresponds to the structure of the weakly distributive lattice and not the one of the Boolean algebras. The result does not affect our usage of the models based on numerical valuation of classical logic but opens a possibility of using non-ordered lattice models which would in turn faithfully reflect the syntax of the logic. But all these statements really say is that numerical calculation is a lowest common denominator--something you have so amply demonstrated with your personality, Franz. I could try to make further explication here, Franz; but, as has often been the case, I respect your desire to remain an ignoramus. :-) === Subject: Re: Fun with axioms > But then of course there's the ultimate reality check > for mathematics: physics. > I believe you have that backwards. When physics is done ignoring the > theoretical > coherence of mathematics you will find that a great many resources have > been > unnecessarily directed toward an infinite regress of triangles. > Galileo. > Galvani > Volta > Joule > Hooke > Faraday > Would you wish me to enlarge the list further? > Unlike you, Franz, I am not in the habit of denying the brilliance of *great* > men and women with substantive ideas. Reread and try to understand your own words before spouting. Practice on the following quotation from yourself, for instance: When physics is done ignoring the theoretical coherence of mathematics you will find that a great many resources have been unnecessarily directed toward an infinite regress of triangles. > Your ignorance of the history of > mathematics (as well as so many other things) is making itself plain here. I know almost no history of mathematics, and have never laid claim to any such knowledge. You are unable to stick to a point and are raising strawmen to which I will not respond. > What I am denying in John's remark is that physics provides any kind of > reality check for mathematics whatsoever. I redirect you once again to the solitary statement of yours to which I have been responding. You are also suffering from your usual verbal diarrhoea, which I snip from here down. Franz === Subject: Re: Fun with axioms > But then of course there's the ultimate reality check > for mathematics: physics. I believe you have that backwards. When physics is done ignoring the > theoretical > coherence of mathematics you will find that a great many resources > have > been > unnecessarily directed toward an infinite regress of triangles. Galileo. > Galvani > Volta > Joule > Hooke > Faraday > Would you wish me to enlarge the list further? > Unlike you, Franz, I am not in the habit of denying the brilliance of > *great* > men and women with substantive ideas. > Reread and try to understand your own words before spouting. > Practice on the following quotation from yourself, for instance: When physics is done ignoring the theoretical coherence of mathematics you > will find that a great many resources have been unnecessarily directed > toward an infinite regress of triangles. Sure Franz. Go read Hilbert's Foundations of Geometry, especially the part about equicomplementability and equidecomposability in the sense of Cavalieri. Then go read Kant's Critique of Pure Reason (Einstein did. And it is clear that Planck was strongly influenced in that tradition.) I have no problem with people solving practical engineering problems. But, physics goes far beyond that. People cannot afford your infinity of questions. Now, if you want to build accelerators out of your own pocket... As for diarrhea, go buy yourself some Kaopectate. It's good for a little ass like you. :-) === Subject: Re: Fun with axioms > But then of course there's the ultimate reality check > for mathematics: physics. I believe you have that backwards. When physics is done ignoring the > theoretical > coherence of mathematics you will find that a great many resources > have > been > unnecessarily directed toward an infinite regress of triangles. > Galileo. > Galvani > Volta > Joule > Hooke > Faraday > Would you wish me to enlarge the list further? > Unlike you, Franz, I am not in the habit of denying the brilliance of > *great* > men and women with substantive ideas. > Reread and try to understand your own words before spouting. > Practice on the following quotation from yourself, for instance: When physics is done ignoring the theoretical coherence of mathematics you > will find that a great many resources have been unnecessarily directed > toward an infinite regress of triangles. > Sure Franz. I have erased the irrelevant straw man you raised. Franz === Subject: Re: Fun with axioms > But then of course there's the ultimate reality check > for mathematics: physics. > I believe you have that backwards. When physics is done ignoring > the > theoretical > coherence of mathematics you will find that a great many resources > have > been > unnecessarily directed toward an infinite regress of triangles. > Galileo. > Galvani > Volta > Joule > Hooke > Faraday > Would you wish me to enlarge the list further? Unlike you, Franz, I am not in the habit of denying the brilliance of > *great* > men and women with substantive ideas. > Reread and try to understand your own words before spouting. > Practice on the following quotation from yourself, for instance: When physics is done ignoring the theoretical coherence of mathematics > you > will find that a great many resources have been unnecessarily directed > toward an infinite regress of triangles. > Sure Franz. > I have erased the irrelevant straw man you raised. Ahh... relevance. Let's see, that goes back to utility and Bentham. So, following Guenther... Have you gangraped anyone lately Franz? I mean, of course, other than the mathematicians and philosophers whose ideas preceded your illustrious career? :-) === Subject: Re: Fun with axioms >Have you gangraped anyone lately Franz? All by himself? I guess he's just too much man. Experiments are the only means of knowledge at our disposal. The rest is poetry, imagination. -- Max Planck === Subject: Re: Fun with axioms > But then of course there's the ultimate reality check > for mathematics: physics. > I believe you have that backwards. When physics is done ignoring > the > theoretical > coherence of mathematics you will find that a great many resources > have > been > unnecessarily directed toward an infinite regress of triangles. > Galileo. > Galvani > Volta > Joule > Hooke > Faraday > Would you wish me to enlarge the list further? Unlike you, Franz, I am not in the habit of denying the brilliance of > *great* > men and women with substantive ideas. Reread and try to understand your own words before spouting. > Practice on the following quotation from yourself, for instance: When physics is done ignoring the theoretical coherence of mathematics > you > will find that a great many resources have been unnecessarily directed > toward an infinite regress of triangles. Sure Franz. > I have erased the irrelevant straw man you raised. > Ahh... relevance. > Let's see, that goes back to utility and Bentham. > So, following Guenther... > Have you gangraped anyone lately Franz? , That remark characterises yourself quite precisely. Franz === Subject: Re: Fun with axioms >The neat thing about building a math with axioms, is that you can >first build one, then go back and change just one axioms slightly. >Go through the exercise of building the math again, and see the >differences between the first and the second build. >Right. It's a bit like what they say about technology: >the great thing about standards is that they're so many to >choose from. That's a feature. >It's fun to do. The neat thing about math is one doesn't have >to include a reality check. >Well, actually mathematicians care a lot about which >axioms or definitions are more *fruitful* than others. You're right. I should have added the clause, As long as one doesn't need to eat,. The neat thing about doing this kind of math work is that it doesn't need a lab, can be done anywhere, and isn't dependent on another human getting a piece of work done first. >It's not so much a matter of whether the axioms are true >as whether they lead to useful insights, beautiful surprises, >a nice bce between complexity and simplicity, or new >connections between other things we're interested in. >So, while reality check might not be the best word for it, It wasn't. >good mathematicians are quite fussy about what structures defined >via axioms they are willing to bother studying. If you write down >an arbitrary list of axioms and ask a mathematician to study it's >consequences, they'll probably look at you funny. Oh, no, no, no. The whole point is to do the work yourself, NOT ask a mathematician. >Nonmathematicians have trouble grokking all the constraints >an axiom system must satisfy to be worth studying, in part >because many of these systems seem arbitrary until one has >studied lots of math, so one can naively think oh, those >guys, they'll study anything! Another reason one can be >fooled into thinking that mathematicians will study anything >is that are lots of mathematicians, so that even though >most of them are pretty fussy, tastes vary, so there will >often be *someone* willing to study even a fairly offbeat >subject - at least until it turns out to be boring. >But then of course there's the ultimate reality check >for mathematics: physics. Yea, it was nice when something we made was used by those dudes. If the stuff you've built is used by them, you know you've made something pretty good. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Fun with axioms >>The neat thing about building a math with axioms, is that you can >>first build one, then go back and change just one axioms slightly. >>Go through the exercise of building the math again, and see the >>differences between the first and the second build. Assuming the axioms are consistent. This is not automatic. .................. >good mathematicians are quite fussy about what structures defined >via axioms they are willing to bother studying. If you write down >an arbitrary list of axioms and ask a mathematician to study it's >consequences, they'll probably look at you funny. >Oh, no, no, no. The whole point is to do the work yourself, >NOT ask a mathematician. As someone who has consulted on mathematics and statistics, many of my clients could not possibly have done so. When one formulates a real world problem in mathematical terms, so that one can get further results, one must make are theorems, these are the consequences of the assumptions. If the consequences do not have acceptable real world properties, the assumptions must be changed, or it must be recognized that what the client wants is impossible. .............. >But then of course there's the ultimate reality check >for mathematics: physics. The Feynman integral is an example of contradictory assumptions. What is it? This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Fun with axioms >The neat thing about building a math with axioms, is that you can >first build one, then go back and change just one axioms slightly. >Go through the exercise of building the math again, and see the >differences between the first and the second build. >Assuming the axioms are consistent. This is not >automatic. Of course. That's a part of the learning experience. > .................. >>good mathematicians are quite fussy about what structures defined >>via axioms they are willing to bother studying. If you write down >>an arbitrary list of axioms and ask a mathematician to study it's >>consequences, they'll probably look at you funny. >Oh, no, no, no. The whole point is to do the work yourself, >NOT ask a mathematician. >As someone who has consulted on mathematics and statistics, >many of my clients could not possibly have done so. That's why you're here :-). I'm simply talking about a hobby that is self-gratifying, can be done anywhere at anytime, and doesn't need a gazillion dollars of infrastructure before getting started. A half-dozen project notebooks, pencil, and a very big eraser is all the equipment needed. A compass and straight-edge is useful for geometry. >When one formulates a real world problem in mathematical >terms, so that one can get further results, one must make >are theorems, these are the consequences of the assumptions. >If the consequences do not have acceptable real world >properties, the assumptions must be changed, or it must >be recognized that what the client wants is impossible. Yep. > .............. >>But then of course there's the ultimate reality check >>for mathematics: physics. >The Feynman integral is an example of contradictory >assumptions. What is it? I'm looking forward to this discussion. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Fun with axioms Originator: baez@math-cl-n03.math.ucr.edu (John Baez) >But then of course there's the ultimate reality check >for mathematics: physics. >Do you think physics will be able to someday resolve the truth or >falsity of the axiom of choice and the generalized continuum hypothesis? I wasn't suggesting that physics can resolve the truth or falsity of any axioms of set theory. What I was trying to say, in a jokey way, is that doing physics helps focus our attention on mathematical structures that help us model the universe we inhabit. For example, Lie groups are important; their p-adic cousins seem less so. I don't think the physical universe cares much either way for the axiom of choice and generalized continuum hypothesis: you can do physics about equally well no matter what you say about these issues. When I was talking about axioms I was talking about the axioms governing all sorts of mathematical gadgets, not just the fundamental axioms of set theory, which are less important than some people think. === Subject: Bottom-Up Karatsuba Multiplication Algorithm I have created a preliminary bottom-up Karatsuba multiplication algorithm. The premise behind the algorithm is to first do all the subtractions and base case multiplications before the interpolation. This is done by rearranging the base case multiplications in such a way that factors are guaranteed to have been calculated before they multiplied. This differs from the recursive version in that the recursive version computes the factors when they are needed. There are many improvements possible. For one, the base case multiplies can be deconstructed to prevent cache misses. The base case multiplications act on strictly positive values with a flag to denote rather the real product should be negative. That could be rewritten so that a negation could be taken out. The interpolation stage can be completely rewritten to eliminate duplicate additions. The archive is at http://www.student.gsu.edu/~zliu2/centrinia.tar.bz2 in directory centrinia/design/base/N/medium/arithmetic/multiplication/ Karatsuba__Ofman/bo ttom-up/ The current implementation depends on the GMP library. One needs to compile the files bottom_up.c, test.c, and ../rrandom.c in order to test the validity of the algorithm. As said before, this is only preliminary and there will be greatly needed work done to speed up the algorithm. I would like to thank anyone who could spare their valuable time to look at my algorithm. === Subject: Re: Bottom-Up Karatsuba Multiplication Algorithm > I have created a preliminary bottom-up Karatsuba multiplication algorithm. > The premise behind the algorithm is to first do all the subtractions and > base case multiplications before the interpolation. This is done by > rearranging the base case multiplications in such a way that factors are > guaranteed to have been calculated before they multiplied. This differs from > the recursive version in that the recursive version computes the factors > when they are needed. What's the trade off though? Normal Karatsuba [can] take little ram and is very trivial to implement. Also do you implement a cutoff point system? E.g. at X digits don't use karatsuba and just baseline it [or Comba]. > There are many improvements possible. For one, the base case multiplies can > be deconstructed to prevent cache misses. The base case multiplications act > on strictly positive values with a flag to denote rather the real product > should be negative. That could be rewritten so that a negation could be > taken out. The interpolation stage can be completely rewritten to eliminate > duplicate additions. If the technique really works you ought to consider writing a paper about it. I'd be interested in seeing a description [being the author a free MP bignum library myself ;-)] Tom === Subject: ** CHAVEZ CALLS BUSH AN A-S-S-H-O-L-E ** Comments: This message did not originate from the above address. It was automatically remailed by one or more anonymous mail services. ***PLEASE REPORT ABUSE TO admin@italy-anonymous-remailer.it*** ***RIFERITE OGNI ABUSO A: admin@italy-anonymous-remailer.it*** Mail-To-News-Contact: abuse@dizum.com CHAVEZ CALLS BUSH AN ASSHOLE --- OH MY GOD, A BIG STORM IS COMING http://www.biblebelievers.org.au/przion1.htm http://users.one.se/~chribesk/sects/zog/newlight.html Jesus, save us from the evil ones!!! Questo messaggio e' stato inoltrato automaticamente da un paio di anonymous remailer. Il mittente originale === Subject: Re: Is this proof of Hadwiger's Conjecture right? shealy > http://arxiv.org/abs/math.GM/0311475 > Its idea is very strange and simple. It said that those conjectures > are topological problems, the proof is out of graph theory. Is it > right? > He doesn't prove this critical statement on page 4: Any planar map is in C4R2, so there does not exist a solid C5. I think it is right here. The Solid Map is generated from the planar map by (plane X z). Every region in the solid map is columnar. The decrete structure of them is the same planar graph(vertice and edges). There haven't a planar C5 in the planr map, of course, a solid C5 couldn't be found in the generated solid map. > But the scheme is worth considering. In R3 take any five points w,x,y,z, and > the origin, such that no 3 of the five are collinear. The set of points u in > R3 of the form aw+bx+cy, a,b,c all positive, form a solid blob, and there > are three more such blobs made from {w,x,z}, {w,y,z}, {x,y,z} respectively. > The four are pairwise adjacent in an obvious sense. Now given a plane map, > or (more convenient) a map on the sphere, can we embed the sphere in R3 such > that its intersections with these blobs are the given countries? Yes _if_ > the graph is four-colorable! I leave it as an exercise :) > LH === Subject: ** CHAVEZ CALLS BUSH AN A-S-S-H-O-L-E ** Comments: This message did not originate from the above address. It was automatically remailed by one or more anonymous mail services. ***PLEASE REPORT ABUSE TO admin@italy-anonymous-remailer.it*** ***RIFERITE OGNI ABUSO A: admin@italy-anonymous-remailer.it*** Mail-To-News-Contact: abuse@dizum.com CHAVEZ CALLS BUSH AN ASSHOLE --- OH MY GOD, A BIG STORM IS COMING http://www.biblebelievers.org.au/przion1.htm http://users.one.se/~chribesk/sects/zog/newlight.html Jesus, save us from the evil ones!!! Questo messaggio e' stato inoltrato automaticamente da un paio di anonymous remailer. Il mittente originale === Subject: ** CHAVEZ CALLS BUSH AN A-S-S-H-O-L-E ** Comments: This message probably did not originate from the above address. It was automatically remailed by one or more anonymous mail services. You should NEVER trust ANY address on Usenet ANYWAYS: use PGP !!! Get information about complaints from the URL below X-Remailer-Contact: http://80.65.224.85/POL/ In case my abuse address is unreachable: It is because it has been flooded by , please contact , please contact I need to break the problem down into solving smaller angles ... Hint: 30 degrees turned in 5 minutes is for minute hand only. You need to consider relative angle of minute hand with respect to the slower hour hand. A clock has only 11 relative revolutions in 12 hours, or 5.5 degrees per minute. Start at 3 O clock, from the angle turned 43 X 5.5 degrees take away 90 degrees. No need of trigonometry, size of arms does not matter, only the difference angular speed. === Subject: Re: Angles of the hands of a clock - Trigonometry - How do I solve? >If you visit the link below and enter the Math Problems folder you will see >a picture of the problem I am trying to solve. I need to break the problem >down into solving smaller angles to help me out. I am having trouble >figuring out the degrees the hour hand is past the 3 hour mark. How do I go >about solving this step by step? >http://photos.yahoo.com/diatonic80 The minute hand moves at 360 degrees per hour. Current angle = 360*43/60. The hour hand moves at 1/12 that speed, or 30 degrees per hour. Current angle = 30*43/60 + 90. (90 is the 3:00 position.) --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Angles of the hands of a clock - Trigonometry - How do I solve? > If you visit the link below and enter the Math Problems folder you will see > a picture of the problem I am trying to solve. I need to break the problem > down into solving smaller angles to help me out. I am having trouble > figuring out the degrees the hour hand is past the 3 hour mark. How do I go > about solving this step by step? > http://photos.yahoo.com/diatonic80 > -- > To reply to me via email you must remove the text removethis from my email > address. I'd figure out how much the angle changes per minute assuming the hour hand doesn't move. David Moran === Subject: Re: Angles of the hands of a clock - Trigonometry - How do I solve? > If you visit the link below and enter the Math Problems folder you will > see > a picture of the problem I am trying to solve. I need to break the > problem > down into solving smaller angles to help me out. I am having trouble > figuring out the degrees the hour hand is past the 3 hour mark. How do I > go > about solving this step by step? > http://photos.yahoo.com/diatonic80 > -- > To reply to me via email you must remove the text removethis from my > email > address. > I'd figure out how much the angle changes per minute assuming the hour hand > doesn't move. The minute hand is 43/60 of the way around the circle, hence it makes an angle of 6*43 = 258 degrees with a line from the center of the clock to 12:00. The hour hand is 43/60 of the way from 3:15 to 3:20. Each five minutes is 1/12 * 360 = 30 degrees. So the hour hand is 90 degrees + 43/60 of 30 degrees, or 90 + 43/2 = 223/2 = 111.5 degrees with respect to the vertical from the center to 12:00. The difference, 258 - 111.5 = 146.5, is the angle between the hands. === Subject: Re: Angles of the hands of a clock - Trigonometry - How do I solve? > If you visit the link below and enter the Math Problems folder you will > see > a picture of the problem I am trying to solve. I need to break the > problem > down into solving smaller angles to help me out. I am having trouble > figuring out the degrees the hour hand is past the 3 hour mark. How do I > go > about solving this step by step? > http://photos.yahoo.com/diatonic80 > -- > To reply to me via email you must remove the text removethis from my > email > address. > I'd figure out how much the angle changes per minute assuming the hour hand > doesn't move. > The minute hand is 43/60 of the way around the circle, hence it makes an > angle of 6*43 = 258 degrees with a line from the center of the clock to > 12:00. The hour hand is 43/60 of the way from 3:15 to 3:20. Each five > minutes is 1/12 * 360 = 30 degrees. So the hour hand is 90 degrees + > 43/60 of 30 degrees, or 90 + 43/2 = 223/2 = 111.5 degrees with respect > to the vertical from the center to 12:00. The difference, 258 - 111.5 = > 146.5, is the angle between the hands. Since this question comes up from time to time (hey! that's a pun!) I worked it out for arbitrary time H:M. With the same reasoning as above, the angle works out to (11M - 60H)/2 Tricky though . . . at 1:00pm we have M = 0, H = 1, and the formula gives -30. That's right, because the original poster's problem was to compute the angle from the hour hand clockwise to the minute hand. At 1:00 this would be 330 degrees, or -30 degrees. This formula also solves the question of when the two hands overlap. 11M-60H = 0 when M = (60/11)H. Just compute that for H = 0, 1, 2, . . . 11. === Subject: Re: Angles of the hands of a clock - Trigonometry - How do I solve? > If you visit the link below and enter the Math Problems folder you will see > a picture of the problem I am trying to solve. I need to break the problem > down into solving smaller angles to help me out. I am having trouble > figuring out the degrees the hour hand is past the 3 hour mark. How do I go > about solving this step by step? > http://photos.yahoo.com/diatonic80 (Looking at your diagram, I'd say it's more like 3:41 or 3:42; but I'll take your word for it that it's supposed to be 3:43.) How about doing it as how far each hand is (moving clockwise) from 12 o'clock? SPOILER So the minute hand is (43/60)*360 past 12 o'clock; and the hour hand is ((43 + 3*60)/(12*60))*360 past. Subtract the latter from the former and simplify. rgds DVD (David Brownridge) If you visit the link below and enter the Math Problems folder you will see >a picture of the problem I am trying to solve. I need to break the problem >down into solving smaller angles to help me out. I am having trouble >figuring out the degrees the hour hand is past the 3 hour mark. How do I go >about solving this step by step? >http://photos.yahoo.com/diatonic80 This actually doesn't require any trig. (1) Calculate the angle between the hour hand and 12:00 (going clockwise). (2) Calculate the angle between the minute hand and 12:00 (going clockwise). (3) Subtract. (1a) Calculate the fraction of 12 hours that is 3 hours and 43 minutes; (1b) Multiply by 360 degrees (or 2pi if you're working in radians). (2a) Calculate the fraction of 60 minutes that is 43 minutes; (2b) Multiply by 360 degrees (or 2pi). HTH. --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- === Subject: Re: Determining the volume of barrels > IIRC, a slight variation on Simpson's rule for approximate integration > was the legal rule for taxation by volume on beer barrels in medieval > England. Let me guess... the slight variation was such as to guarantee that the numerical integration was greater than the true volume? === Subject: Re: Determining the volume of barrels > .... > By coincidence, just yesterday James Dolan was claiming there > used to be a special word for the art of determining the volume of > barrels. Do you know it? Does anyone? > And no, I don't mean stereometry> .... Gauging? Some time ago I found a second-hand book whose title-page may be worth typing out in full (but without all the variations in type size and case). Here we go. :-) A handbook of practical gauging, with instructions in the use of Sykes's hydrometer. Also a chapter on distillation, describing the process for ascertaining the strength of wines, and the original gravities of beers. Illustrated with diagrams. By James B. Keene, of H.M. Customs, London, Author of the Handbook of Hydrometry. Fourth edition, revised. London: F. Pitman Hart & Co., Ltd., 20 & 21 Paternoster Row, E.C. After all that there is, of course, no date. Perhaps I should also type out the first paragraph of the introduction (exactly as printed): Gauging is that branch of Practical Geometry which treats of the Measurement, by means of their dimensions, of vessels intended to contain liquids, in order to ascertain the quantity they are capable of holding when filled, and that which they actually contain when partially full. The former is technically called the Content, the latter the Ullage. So now you know! === Subject: Re: Determining the volume of barrels Originator: baez@math-cl-n03.math.ucr.edu (John Baez) >By coincidence, just yesterday James Dolan was claiming there >used to be a special word for the art of determining the volume of >barrels. Do you know it? Does anyone? >And no, I don't mean stereometry>Dolan wins. >http://www.faculty.fairfield.edu/jmac/ther/superlatives.htm I guess so! This translates the title of Kepler's Nova Stereometria Doliorum Vinariorum not as New Stereometry of Wine Barrels but simply as Doliometry. I.e., barrel measurement. This website is the only Google hit for the word doliometry=== Subject: Re: Determining the volume of barrels >> By coincidence, just yesterday James Dolan was claiming there >> used to be a special word for the art of determining the volume of >> barrels. Do you know it? Does anyone? >> And no, I don't mean stereometry>Dolan wins. >http://www.faculty.fairfield.edu/jmac/ther/superlatives.htm >This translates the title of Kepler's Nova Stereometria Doliorum >Vinariorum not as New Stereometry of Wine Barrels but simply >as Doliometry. >I.e., barrel measurement. >This website is the only Google hit for the word doliometryOED doesn't know of it, although it does have dolium: Rom. Antiq. A large earthenware jar or vessel, more or less spherical, for holding wine, oil, or dry commodities, etc.; hence, in mod. use, a cask. and dolioloid resembling a cask Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Calculus Stuff... does anyone here chat on efnet #math? > http://roasted.homelinux.net/~helpinghand/problem.jpg > from the second last to the last line, how did it change to that? My teacher > ask if we could find out how it is done, not homework. I think the term he > P.S. those marks and dusk are pencil. === Subject: Re: Calculus Stuff... > oh, yes, p is another variable Actually, it's the original function, p(x) = Ax^2 + Bx + C Plug in the suggested values for x and expand. === Subject: Re: Calculus Stuff... wonder why it posted a few times... > maybe i didn't put it clearly, here's the original if i'm missing > anything... > http://roasted.homelinux.net/~helpinghand/1.jpg > http://roasted.homelinux.net/~helpinghand/2.jpg > oh, yes, p is another variable > http://roasted.homelinux.net/~helpinghand/problem.jpg > from the second last to the last line, how did it change to that? My > teacher > ask if we could find out how it is done, not homework. I think the term > he > P.S. those marks and dusk are pencil. === Subject: Re: a digital game > Two players define a number x = 0. d_1 d_2 d_3 ... (base 2) as follows: > First, one player separates {1, 2, 3, ...} into two disjoint infinite > sets and defines the d_i for all i in one of the sets; the other player > then defines all the remaining d_i. By what strategy can one of the > players ensure that x is transcendental, regardless of who plays first? > --r.e.s. What does all that mean???? === Subject: Re: a digital game >Two players define a number x = 0. d_1 d_2 d_3 ... (base 2) as follows: >First, one player separates {1, 2, 3, ...} into two disjoint infinite >sets and defines the d_i for all i in one of the sets; the other player >then defines all the remaining d_i. By what strategy can one of the >players ensure that x is transcendental, regardless of who plays first? > What does all that mean???? Let's go through it one step at a time: >Two players define a number x = 0. d_1 d_2 d_3 ... (base 2) as follows: x is a real number between 0 and 1 (inclusive), defined in terms of its representation in base 2. For instance, this set of digits: d_1 = 1 d_n = 0 for all n > 1 represents 1/2 (base 10). >First, one player separates {1, 2, 3, ...} into two disjoint infinite >sets {1, 2, 3, ...} is an infinite set. The first player must pick an infinite subset (call it S), such that S' = {all elements not in S} is also infinite. >and defines the d_i for all i in one of the sets; The first player specifies all the digits corresponding to elements of S. >the other player then defines all the remaining d_i. The second player specifies all the other digits. >By what strategy can one of the players ensure that x is >transcendental, regardless of who plays first? http://dictionary.reference.com/search?q=transcendental 4. Mathematics. Of or relating to a real or complex number that is not the root of any polynomial that has positive degree and rational coefficients. The question breaks down into two parts: (a) If the first player wants to force the number to be transcendental, and the second wants to force it to be non-transcendental, then how can the first player win? (b) If the first player wants to force non-transcendental, and the second wants to force transcendental, then how can the second player win? In both cases, the first player may choose an infinite subset S and the associated digits in whatever manner supports his goal, and the second player may choose the associated digits of S' in whatever manner supports *his* goal. === Subject: Re: a digital game charset=Windows-1252 >Two players define a number x = 0. d_1 d_2 d_3 ... (base 2) as follows: >First, one player separates {1, 2, 3, ...} into two disjoint infinite >sets and defines the d_i for all i in one of the sets; the other player >then defines all the remaining d_i. By what strategy can one of the >players ensure that x is transcendental, regardless of who plays first? > This is rather reminiscent of the Banach-Mazur game, where instead of > just two turns you have an infinite sequence of turns: at each stage > the player whose turn it is chooses a finite number of digits. In that > case, either player can force x to be in any dense G_delta. However, > that's not true for r.e.s.'s game: the first player can force x to be > in any dense G_delta, but there are also nowhere dense closed sets > that the first player can force. I'll show my ignorance by asking this, but could you be more explicit as to how x being in a nowhere dense set (or in a dense G_delta) relates to whether x is transcendental, if it does? (IOW, I can't tell whether your comments above are about strategies in the present game, or concern a game with different objectives.) --r.e.s. === Subject: Re: a digital game >Two players define a number x = 0. d_1 d_2 d_3 ... (base 2) as follows: >First, one player separates {1, 2, 3, ...} into two disjoint infinite >sets and defines the d_i for all i in one of the sets; the other player >then defines all the remaining d_i. By what strategy can one of the >players ensure that x is transcendental, regardless of who plays first? >This is rather reminiscent of the Banach-Mazur game, where instead of >just two turns you have an infinite sequence of turns: at each stage >the player whose turn it is chooses a finite number of digits. In that >case, either player can force x to be in any dense G_delta. However, >that's not true for r.e.s.'s game: the first player can force x to be >in any dense G_delta, but there are also nowhere dense closed sets >that the first player can force. >I'll show my ignorance by asking this, but could you be more >explicit as to how x being in a nowhere dense set (or in a >dense G_delta) relates to whether x is transcendental, if it >does? (IOW, I can't tell whether your comments above are >about strategies in the present game, or concern a game with >different objectives.) The transcendental numbers form a dense G_delta (as is the complement of any countable set). I was thinking about the generalization of your game where the players are trying to force x to be in some given set, rather than just transcendental vs algebraic. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: a digital game charset=Windows-1252 >Two players define a number x = 0. d_1 d_2 d_3 ... (base 2) as follows: >First, one player separates {1, 2, 3, ...} into two disjoint infinite >sets and defines the d_i for all i in one of the sets; the other player >then defines all the remaining d_i. By what strategy can one of the >players ensure that x is transcendental, regardless of who plays first? > That's quite an interesting puzzle! Kudos! But I wonder whether > it can be trivially solved by the following conjectured solution > (which looks not trivial at all, but I could easily be wrong and it > could be completely trivially false): > Let whichever player is trying to make 'x' transcendental simply > list off the binary digits of pi, unless that would make the number > algebraic, in which case he should list off the digits, inverted > (0 to 1 and vice versa). > The first player might as well choose every other digit. > These both seem like really common-sense things to do, but neither > of them seems obviously correct. I wonder whether a first-player > strategy exists such that the above second-player strategy fails. It seems a 2nd-player strategy of the type you describe depends on the following proposition: In the binary expansion of an algebraic irrational, inverting the digits of any subsequence that is the expansion of a transcendental, converts the overall expansion to that of a transcendental. (The 1st player could have filled in all but such a subsequence of some algebraic irrational.) I suspect the proposition is not true, but it would be interesting to see how to prove or disprove it. --r.e.s. === Subject: Re: a digital game charset=Windows-1252 >> Two players define a number x = 0. d_1 d_2 d_3 ... (base 2) as follows: >> First, one player separates {1, 2, 3, ...} into two disjoint infinite >> sets and defines the d_i for all i in one of the sets; the other player >> then defines all the remaining d_i. By what strategy can one of the >> players ensure that x is transcendental, regardless of who plays first? >It seems to me that the 1st player can guarantee a transcendental >by going .010?00100?0000001000000?00...00100..00?.... >where she is giving long stretches of 0 broken up by single 1s >(the ? are the ones she leaves for the other player). No matter >what the other player does, a number results with rational >approximations to good to be algebraic. >The second player can also guarantee a transcendental, by cardinality, >although I'm not so sure there's a practical strategy. > If the second player is faced with, for example, > 0.0101001?1010?1?001001... > Just get a nice list of all the algebraics which match that pattern > and make like Cantor. Not sure if this counts as a practical strategyIt seems to me that diagonalizing with such a list requires the set of digit-positions for each player to be recursively enumerable -- an assumption not entailed by the game rules. In another strategy, given below, we assume that each of the two sets, rather than being r.e., contains an infinite r.e. subset. [*] (The strategy then uses the easy-to-prove fact that every infinite r.e. set of naturals has an infinite recursive subset.) With the assumption[*], the strategy is as follows ... Let S_1, S_2 be the two infinite sets that the 1st player defines to partition the naturals, and let R_1, R_2 be some two particular infinite recursive subsets of S_1, S_2 respectively. Finally, let ch() be the characteristic function of some particular infinite non-recursive subset of the naturals. Then the kth player (k=1,2) can force x to be transcendental by defining d_i = ch(j) for i = jth-smallest element in R_k (j=1,2,3,...) d_i = 0 for i in S_kR_k. Such an x must be transcendental, because ... Suppose, as hypothesis to be disproved, that x is computable. Then ch(n), n=1,2,3,... can be effectively computed as follows: for n = 1,2,3,... if n in R_k: <- this is decidable if R_k is recursive [*] compute d_n <- d_n is computable assuming x is computable output ch(n) <- ch(n) = d_n But this is a contradiction, since ch(), being the characteristic function of a non-recursive set, is not computable. Therefore, x is not computable; consequently, x is transcendental, since the algebraic reals are computable. QED Concerning [*], could anyone offer ideas on how to prove whether there exists a non-r.e. set having no infinite recursive subset? It seems to me that all 2nd-player strategies so far suggested assume that the sets of digit-positions chosen by the 1st player are recursively enumerable, or contain infinite r.e. subsets. I had assumed as much without realizing it when stating the game rules, and I'm not sure whether there's successful 2nd-player strategy without such an assumption. --r.e.s. === Subject: Re: Help needed with nonlinear fit > I need to fit a curve to a data set so that I can interpolate potential > points. Suggest you look at the chapter on interpolation in Numerical Recipes in C, which is available on-line at http://lib-www.lanl.gov/numerical/. === Subject: Has a convergent subsequence Define the sequence {Xn} by (n^2 + 20n +35) sin(n^3) / (n^2 +n +1). Prove that {Xn} has a convergent subsequence. I understand that as n--> the division of the polynomial goes to 0 and that cos(n^3) is bounded by -1 and 1 but I don't see where to go from here. Sure cos(n^3) is periodic but the === Subject: Re: Has a convergent subsequence Content-transfer-encoding: 8bit > Define the sequence {Xn} by (n^2 + 20n +35) sin(n^3) / (n^2 +n +1). Prove > that {Xn} has a convergent subsequence. I understand that as n--> the > division of the polynomial goes to 0 and that cos(n^3) is bounded by -1 and > 1 but I don't see where to go from here. Sure cos(n^3) is periodic but the ??? The quotient of the polynomials, (n^2+20n+35)/(n^2+n+1), converges to 1, not 0. And the product of this with cos(n^3) is bounded. And every bounded sequence has _ c_nv_rg_nt s_bs_q__nc_. --Ron Bruck === Subject: Elliptic Curve Parameters Hi All, I am trying to learn Elliptic Curves in an effort to some work in crypto using them. An Elliptic Curve is defined (one definition) as y^2 = x^3 + ax + b defined over Z{p} where a, b E Z{p}.Z{p}, under the condition that 4a^3 + 27b^2 != 0 (mod p). What does the b E Z{p}.Z{p} mean? (I think I know, but want to make sure). Here are some examples (how does b satisfy the above?): Curve 1: y^2 = x^3 + x + 5 (mod 17) Curve 2: y^2 = x^3 + 6x + 5 (mod 17) Curve 3: y^2 = x^3 + x + 6 (mod 11) What does it mean in the case of: y^2 + xy = x^3 + ax^2 + b where a, b E GF(2^m) with b != 0 Can you provide an example of such a curve that math these conditions? Is the condition 4a^3 + 27b^2 != 0 over GF(2^m) necessary? Thank you for any insight, Flip === Subject: Re: Elliptic Curve Parameters > An Elliptic Curve is defined (one definition) as y^2 = x^3 + ax + b > defined over Z{p} where a, b E Z{p}.Z{p}, under the condition that 4a^3 + > 27b^2 != 0 (mod p). ... > Here are some examples (how does b satisfy the above?): > Curve 1: y^2 = x^3 + x + 5 (mod 17) In this case a=1, b=5, so one has to show that 17 does not divide 4 + 27.5^2. This follows since 4 + 27.25^2 = 4 + 10.8^2 = ... mod 17 > What does it mean in the case of: y^2 + xy = x^3 + ax^2 + b > where a, b E GF(2^m) with b != 0 Basically, the non-vanishing of the discriminant is the condition for the curve to be non-singular. In characteristic 2 a curve of the form y^2 = cubic in x is always singular, so one has to consider the more general form you give. You can either look up the (rather complicated) formula for the discriminant in the general case y^2 + c_1 xy + c_3 y = x^3 + c_2 x^2 + c_4 x + c_6, which simplifies in characteristic 2 to c_1^6c_6 + c_1^5c_3c_4 + c_1^4c_2c_3^2 + c_1^4c_4^2 + c_1^3c_3^3 + c_3^4, which in your case (where c_3 = c_4 = 0 and c_1 = 1) reduces to c_6 = b, or you could determine it from first principles. Basically, this means writing the equation in homogeneous form, in your case F(X,Y,Z) = Y^2Z + XYZ + X^3 + aX^2Z + bZ^3 = 0. A singular point is given by dF/dX = YZ + X^2 = 0, dF/dY = XZ = 0, dF/dZ = Y^2 + XZ + aX^2 + bZ^2 = 0. If b=0 the point [0,0,1] is singular. Otherwise the curve is non-singular, ie elliptic. > Can you provide an example of such a curve that math these conditions? y^2 + xy = x^3 + x mod 2. > Is the condition 4a^3 + 27b^2 != 0 over GF(2^m) necessary? No. It's not relevant for the second form you give (with xy on the left). Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Elliptic Curve Parameters > I am trying to learn Elliptic Curves in an effort to some work in crypto > using them. > An Elliptic Curve is defined (one definition) as y^2 = x^3 + ax + b > defined over Z{p} where a, b E Z{p}.Z{p}, under the condition that 4a^3 + > 27b^2 != 0 (mod p). > What does the b E Z{p}.Z{p} mean? (I think I know, but want to make sure). I think this is a bad transcription of the mathematical symbols saying that a and b are elements of Z_p (the field of p elements). Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Infinity equals a divisor? Hi Leroy, I think you're suggesting that there's one and only one infinite integer for each real r thus that r times an infinite integer similar in conception to 1000..., the (a) scalar infinity, divided by that infinite integer is equal to exactly unity. Multiplying by infinity is almost the same thing as dividing by zero, except instead of zero it's an infinitesimal. In some logical/mathematical models, an infinitesimal can only be considered zero, in others they can be compared and differ from zero, qualitatively or quantitatively. In some models there are no infinite integers, in others one, or two, in others infinitely many. The result of a composition of functions can have a result that is finite and known when the individual functions have as variables divergent quantities. If you accept the use of infinite integers, infinite precision integers, then one of those can contain as much information as an irrational real number, with rational numbers being ratios of finite integers and arithmetic numbers being the solutions of the polynomials with finitely many finite integer coefficients. You suggest that infinity divides into any integer. That is so for one, one divides into any integer, each integer is a multiple of one. I've theorized that infinity is negative one and sometimes zero or even i but one is just out of the ballpark. Any integer divides into zero with an integer result, thus that every number divides into zero. In a way it is about x/x, and how lim x->0+ x/x = 1, but for x=0, x/x=0. A number theoretical result with the necessary consideration of whether infinity divides into a value would, as you suggest, help in understanding that concept. Some things infinite are quite mathematical with little real world meaning, others are quite concrete. Ross F. === Subject: Re: Infinity equals a divisor? > I will troll now: > Say we define n as a divisor of m if > m/n = integer; Yes, that's a possible definition. But another possibility is to use a definition according to which is a divisor of and is a factor of mean the same thing. Using that alternative, given integers a and b, we would say that a|b iff there is an integer c such that b = a*c. > then, for ANY real m (actually), we might be justified in calling > infinity m's divisor. But not using the alternative definition. Surely you would not wish to claim that your infinity is a factor of m. > For (in a very non-rigorous way) > m/(infinity) = 0, > and 0 is an integer. It can be non-rigorous, yes, but it need not be so. It can be made perfectly rigorous if one develops a system in which a specific object infinity is defined and division is defined. [Example: Your infinity might be the equivalence class of all rational sequences which increase without bound...] Anyway, in an extended system of some sort, it certainly might be reasonable to say that infinity|infinity (since infinity = infinity*1, for example). But I can't imagine any context in which it would seem reasonable to say that infinity is a divisor (i.e., factor) of any finite nonzero quantity. David === Subject: re:Infinity equals a divisor? Leroy, I doubt that someone would make such a mistake as to exlude the possibility of some quantity being infinite because of it being a divisor of some number in the sense you defined. Anyway, as long as you're here, maybe you can provide the answer to the GCD problem you posted a while ago, so that I can compare the answer I obtained to it. === Subject: Re: Infinity equals a divisor? In sci.math, Leroy Quet : > I will troll now: > Say we define n as a divisor of m if > m/n = integer; > then, for ANY real m (actually), we might be justified in calling > infinity > m's divisor. For > (in a very non-rigorous way) > m/(infinity) = 0, > and 0 is an integer. > But with this definition of divisor, we could also call > pi/2 > a divisor of pi. > (I am not saying pi/2 is not a divisor of pi.) > Seriously, though, have there been any famous cases of proofs being > incorrect because the mathematician forgot that, under the above > definition, infinity is to be included among m's divisors? > Leroy Quet Infinity is not a real or an integer. If one restricts the operation to all real numbers or all integers, it's not a problem. #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Infinity equals a divisor? > I will troll now: > Say we define n as a divisor of m if > m/n = integer; > then, for ANY real m (actually), we might be justified in calling > infinity > m's divisor. For > (in a very non-rigorous way) > m/(infinity) = 0, > and 0 is an integer. Not only that, zero is a multiple of every number since Nx0=0 === Subject: Re: Points on a Line, Infinity Even if where there isn't a fixed value for the number of points on a line it's concept can be used, for example as the number of parallel lines on a plane. Here I used the word parallel instead of non-intersecting, to indicate the set of parallel lines through each point of a non-parallel line on a plane. There are as many of those parallel lines, compared to one line, as there are points on the line, compared to one point. They comprise a continuous plane, where any other line on the plane has as its points one point of each of those lines. The coordinate axes of a two variable system are generally defined to be orthogonal, at right angles, with their intersection at the origin. For any non-zero magnitude of one of the variables, the distance to any point on the other axis is maximized by the lines being orthogonal, at right (90 degree) angles in less than four dimensions. A problem with considering the points on a plane to be in a grid pattern is that the distance between two points parallel to the axes is, well, it's zero, but in consideration of things like all the rays from the origin through the unit circle, then there are various mutations of the geometric logic. A really good book is John H. Conway's and N.J.A. Sloane's Sphere Packings, Lattices, and Groups. My mutation of geometric logic is that the points as spheres pack fully the continuum. In finite sizes, only squares, cubes, and hypercubes pack fully the planes, volumes, and hyperspaces. My mutation of geometric logic is that the points have qualities of spheres and cubes, they are both sphere and cube. That extends to that the hyperspace is both sphere and cube. Back to a point in a line, it's a point in a hyperspace. In considering all the lines on a plane, they are of each line through the center of a circle and each line parallel to those lines. It only takes three points, two non-collinear, to describe a plane of course, a set of lines the points of which comprise each point of the plane is each of the parallel lines through another line on that plane. What that lead to is the number of points on the unit circle. If there are as many lines on a plane as there are points on a line multiplied by the number of points on the unit circle, that leaves us with another unknown along with the number of points on a line, the number of points on the unit circle. I select unit circle with radius one instead of the circle with radius ninety-five or of the value of the point at infinity because the unit circle is much more tractable. That leads into either restriction or unusability of the concept easily, for example in considering the number of undirected chords of a circle. For each point of the unit circle, there is one chord of length two and two chords of each value between two and zero. Is the circle of radius two then supposed to have twice as many chords as the perimeter is doubled? The angle between each chord and a line through the center is unchanged. This is where I want to say that the perimeter of 2pi represents exactly 2pi times as many points as [0,1]. Its length, the distance between the endpoints along that path, has exactly that value. In terms of the lines on the plane again, we can consider how many of all the unit length line segments on a plane that intersect and overlap intersect a line L. As many unit length line segments intersect a line of all the unit length line segments as there are aninteger to all integers. This is where for each line that intersects L that it can be divided into as many unit length line segments as there are integers and only one of them intersects L. There is also the case where two segment endpoints are on L, leading to the probability of a unit length line segment intersecting a line being the probability of selecting a random integer plus that of selecting a random real. That is to say if all the unit length segments on a plane were drawn on a plane in every direction that the probability of a given one intersecting a given random line then drawn is that probability. Does it make a difference to draw the line first? There are continuous probability functions, another area in which I'm ignorant, and I'm aware that there are issues with scaling or otherwise transforming a figure and I understand some issues of the infinite sets. There are only points in the set. Please address how points on a line is equal to lines on a plane and geometry in the small. Ross F.