= > Is there any program of computing Fibonacci numbers on a Turing Machine? > ========================================== > Alex Vinokur > http://www.simtel.net/pub/oth/19088.html > http://sourceforge.net/users/alexvn > ========================================== The IBM research people just shifted an atom over a set of stationary atoms! Making a computer from the set. And so the person is left with the proper means of programming the little atom computer. So a good project is to relate the set state to the computer state. Meaning the abstract computer as Turing's Theory appears the correct relation in this example. And the idea of Turing's infinite state requirment is then only a set's abstract necessity and not really necessary in a useful working computer. And the simple relation of the set to its cause is defined by the little atom computers design. So the shifted atom computes! Meaning a project is to copy this new atom computer in a software and learn to write the defined shift caused set, without relation to the number of places the atom jumped. That is all the darn thing is! And your special sequence may then be programmed correctly without mathematical theory. Meaning, how is your special sequence related to Turing's machine? Related independent of its mathematical definition. And so cryptography appears another subject. How that? Douglas Eagleson Gaithersburg,MD USA ==== Here is description of a Turing Machine which computes Fibonacci numbers. %%============================================ %%============== Turing Machine ============== %%======= Computing a Fibonacci number ======= %%======== Machine Definition : BEGIN ======== %%============================================ ====== Description ====== Computing Fibonacci numbers. The program computes a Fibonacci number. A number 'n' is represented by n 1-s. Sample : 5 is represented as 1 1 1 1 1 3 is represented as 1 1 1 Input : number 'n' Sample (n = 7) : 1 1 1 1 1 1 1 Output : Fibonacci#n Sample (Fibonacci#7) : 1 1 1 1 1 1 1 1 1 1 1 1 1 ====== States ====== Initial state : q0 Halting state : qf Internal states : q101 q102 q103 q104 q105 q106 q107 q108 q109 q201 q202 q203 q204 q301 q302 q303 q304 q305 q306 q307 q308 q309 q310 q311 q401 q402 q403 q404 q501 q502 q503 q601 q602 q603 q604 q701 q702 q703 q704 q801 q802 q803 q804 q805 q806 q807 q808 q809 ====== Alphabet ====== Empty symbols alphabet : b Input alphabet : 1 Internal alphabet : x * ====== Transition Table ====== Rule# 0 : q0 [ 1 ] ---> q101 [ (x, R) ] Rule# 1 : q101 [ 1 ] ---> q101 [ (1, R) ] Rule# 2 : q101 [ b ] ---> q102 [ (1, R) ] Rule# 3 : q102 [ b ] ---> q103 [ (*, R) ] Rule# 4 : q103 [ b ] ---> q104 [ (1, R) ] Rule# 5 : q104 [ b ] ---> q601 [ (*, L) ] Rule# 6 : q105 [ b ] ---> q106 [ (1, L) ] Rule# 7 : q106 [ * ] ---> q701 [ (*, L) ] Rule# 8 : q107 [ * ] ---> q108 [ (*, L) ] Rule# 9 : q107 [ 1 ] ---> q107 [ (1, L) ] Rule# 10 : q108 [ * ] ---> q109 [ (*, N) ] Rule# 11 : q108 [ 1 ] ---> q108 [ (1, L) ] Rule# 12 : q109 [ * ] ---> q109 [ (*, R) ] Rule# 13 : q109 [ 1 ] ---> q109 [ (1, R) ] Rule# 14 : q109 [ b ] ---> q201 [ (*, N) ] Rule# 15 : q201 [ * ] ---> q202 [ (*, L) ] Rule# 16 : q201 [ 1 ] ---> q201 [ (1, L) ] Rule# 17 : q202 [ * ] ---> q203 [ (*, R) ] Rule# 18 : q202 [ 1 ] ---> q202 [ (1, L) ] Rule# 19 : q202 [ b ] ---> q203 [ (b, R) ] Rule# 20 : q203 [ * ] ---> q301 [ (*, N) ] Rule# 21 : q203 [ 1 ] ---> q204 [ (b, R) ] Rule# 22 : q204 [ * ] ---> q204 [ (*, R) ] Rule# 23 : q204 [ 1 ] ---> q204 [ (1, R) ] Rule# 24 : q204 [ b ] ---> q201 [ (1, L) ] Rule# 25 : q301 [ * ] ---> q302 [ (*, L) ] Rule# 26 : q302 [ * ] ---> q303 [ (*, L) ] Rule# 27 : q302 [ b ] ---> q302 [ (b, L) ] Rule# 28 : q303 [ * ] ---> q304 [ (*, R) ] Rule# 29 : q303 [ 1 ] ---> q303 [ (1, L) ] Rule# 30 : q303 [ b ] ---> q304 [ (b, R) ] Rule# 31 : q304 [ * ] ---> q308 [ (b, N) ] Rule# 32 : q304 [ 1 ] ---> q305 [ (b, R) ] Rule# 33 : q305 [ * ] ---> q306 [ (*, R) ] Rule# 34 : q305 [ 1 ] ---> q305 [ (1, R) ] Rule# 35 : q306 [ * ] ---> q307 [ (*, L) ] Rule# 36 : q306 [ 1 ] ---> q307 [ (1, L) ] Rule# 37 : q306 [ b ] ---> q306 [ (b, R) ] Rule# 38 : q307 [ b ] ---> q302 [ (1, L) ] Rule# 39 : q308 [ 1 ] ---> q309 [ (1, L) ] Rule# 40 : q308 [ b ] ---> q308 [ (b, R) ] Rule# 41 : q309 [ b ] ---> q310 [ (*, L) ] Rule# 42 : q310 [ * ] ---> q311 [ (*, R) ] Rule# 43 : q310 [ b ] ---> q310 [ (1, L) ] Rule# 44 : q311 [ * ] ---> q501 [ (*, R) ] Rule# 45 : q311 [ 1 ] ---> q311 [ (1, R) ] Rule# 46 : q401 [ * ] ---> q402 [ (*, L) ] Rule# 47 : q401 [ 1 ] ---> q401 [ (1, L) ] Rule# 48 : q402 [ * ] ---> q403 [ (*, L) ] Rule# 49 : q402 [ 1 ] ---> q402 [ (1, L) ] Rule# 50 : q403 [ * ] ---> q403 [ (*, L) ] Rule# 51 : q403 [ 1 ] ---> q404 [ (*, L) ] Rule# 52 : q404 [ * ] ---> q404 [ (*, R) ] Rule# 53 : q404 [ 1 ] ---> q404 [ (1, R) ] Rule# 54 : q404 [ b ] ---> q201 [ (*, N) ] Rule# 55 : q404 [ x ] ---> q801 [ (x, N) ] Rule# 56 : q501 [ * ] ---> q502 [ (1, N) ] Rule# 57 : q501 [ 1 ] ---> q501 [ (1, R) ] Rule# 58 : q502 [ 1 ] ---> q502 [ (1, R) ] Rule# 59 : q502 [ b ] ---> q503 [ (b, L) ] Rule# 60 : q503 [ 1 ] ---> q401 [ (b, L) ] Rule# 61 : q601 [ * ] ---> q602 [ (*, L) ] Rule# 62 : q601 [ 1 ] ---> q601 [ (1, L) ] Rule# 63 : q602 [ 1 ] ---> q603 [ (*, L) ] Rule# 64 : q603 [ 1 ] ---> q604 [ (1, R) ] Rule# 65 : q603 [ x ] ---> q801 [ (x, N) ] Rule# 66 : q604 [ * ] ---> q604 [ (*, R) ] Rule# 67 : q604 [ 1 ] ---> q604 [ (1, R) ] Rule# 68 : q604 [ b ] ---> q105 [ (b, N) ] Rule# 69 : q701 [ * ] ---> q702 [ (*, L) ] Rule# 70 : q701 [ 1 ] ---> q701 [ (1, L) ] Rule# 71 : q702 [ * ] ---> q702 [ (*, L) ] Rule# 72 : q702 [ 1 ] ---> q703 [ (*, L) ] Rule# 73 : q703 [ 1 ] ---> q704 [ (1, R) ] Rule# 74 : q703 [ x ] ---> q801 [ (x, N) ] Rule# 75 : q704 [ * ] ---> q704 [ (*, R) ] Rule# 76 : q704 [ 1 ] ---> q704 [ (1, R) ] Rule# 77 : q704 [ b ] ---> q107 [ (b, L) ] Rule# 78 : q801 [ * ] ---> q801 [ (*, R) ] Rule# 79 : q801 [ 1 ] ---> q801 [ (1, R) ] Rule# 80 : q801 [ b ] ---> q802 [ (b, L) ] Rule# 81 : q801 [ x ] ---> q801 [ (x, R) ] Rule# 82 : q802 [ * ] ---> q808 [ (b, L) ] Rule# 83 : q802 [ 1 ] ---> q808 [ (1, L) ] Rule# 84 : q803 [ * ] ---> q803 [ (*, L) ] Rule# 85 : q803 [ 1 ] ---> q803 [ (*, L) ] Rule# 86 : q803 [ x ] ---> q804 [ (x, R) ] Rule# 87 : q804 [ * ] ---> q804 [ (*, R) ] Rule# 88 : q804 [ 1 ] ---> q805 [ (*, L) ] Rule# 89 : q804 [ b ] ---> q809 [ (b, N) ] Rule# 90 : q805 [ * ] ---> q805 [ (*, L) ] Rule# 91 : q805 [ 1 ] ---> q806 [ (1, R) ] Rule# 92 : q805 [ x ] ---> q806 [ (*, N) ] Rule# 93 : q806 [ * ] ---> q807 [ (1, R) ] Rule# 94 : q807 [ * ] ---> q804 [ (*, R) ] Rule# 95 : q808 [ * ] ---> q803 [ (*, L) ] Rule# 96 : q808 [ 1 ] ---> q808 [ (1, L) ] Rule# 97 : q809 [ * ] ---> q809 [ (b, L) ] Rule# 98 : q809 [ 1 ] ---> qf [ (1, N) ] Rule# 99 : q809 [ b ] ---> q809 [ (b, L) ] %%============================================ %%======== Machine Definition : BEGIN ======== %%======= Computing a Fibonacci number ======= %%============== Turing Machine ============== %%============================================ C++ Simulator of a Turing Machine * http://sourceforge.net/projects/turing-machine * http://alexvn.freeservers.com/s1/turing.html has been used to compute several Fibonacci numbers. Raw Logs : * http://groups.google.com/groups?th=1e653c4ef60faa44 ===================================== Alex Vinokur http://mathforum.org/library/view/10978.html ===================================== ==== Suppose function f(x)=x^2 is given and we have to find out half range fourier series, we will carry out in Maple like this a0 := 2/(2*2)*int(x^2, x = 0..2); an := 2/2*int(x^2*cos(n*Pi*x/2), x = 0..2); S1 := a0 + sum(an*cos(n*Pi*x/2), n = 1..10); plot(S1, x = -2..2, scaling = constrained); bn := 2/2*int(x^2*sin(n*Pi*x/2), x = 0..2); S2 := sum(bn*sin(n*Pi*x/2), n = 1..10); plot(S2, x = -2..2, scaling = constrained); So my question is what inference can be drawn if plot sum of the two half range series like this plot((S1+S2), x = -2..2); or it is just meaningless. ==== > Suppose function f(x)=x^2 is given and we have to find out half range > fourier series, we will carry out in Maple like this a0 := 2/(2*2)*int(x^2, x = 0..2); > an := 2/2*int(x^2*cos(n*Pi*x/2), x = 0..2); > S1 := a0 + sum(an*cos(n*Pi*x/2), n = 1..10); > plot(S1, x = -2..2, scaling = constrained); > bn := 2/2*int(x^2*sin(n*Pi*x/2), x = 0..2); > S2 := sum(bn*sin(n*Pi*x/2), n = 1..10); > plot(S2, x = -2..2, scaling = constrained); So my question is what inference can be drawn if plot sum of the two > half range series like this plot((S1+S2), x = -2..2); or it is just meaningless. > S1 is the Fourier series of the even extension of x^2. S2 is the Fourier series of the odd extension of x^2. Add the two together - it is the Fourier series of f(x) = 2x^2 if 0 R such that (1) p(f) is positive or zero for all f in [0,B] (2) int_0^B p(f) df = P (P and B are some fixed positive real numbers) (3) p(f) is continuous on [0,B] Let Q: S-> R such that Q(p(f)) = int_0^B log( p(f) + u(f) ) df where u(f) is a fixed continuous function which is positive for all f in [0,B] It is easy to show that Q is bounded on S, so there's a least upper bound. However, S is not (in general) compact. I can therefore not prove the following conjecture: Conjecture: that there exists p(f) in S such that Q(p(f)) = sup_S Q. Note: just an existence proof is needed! ==== Let S = the set of all functions p(f): [0,B] -> R such that (1) p(f) is positive or zero for all f in [0,B] (2) int_0^B p(f) df = P (P and B are some fixed positive real numbers) (3) p(f) is continuous on [0,B] Let Q: S-> R such that Q(p(f)) = int_0^B log( p(f) + u(f) ) df where u(f) is a fixed continuous function which is positive for all f in >[0,B] It is easy to show that Q is bounded on S, so there's a least upper bound. >However, S is not (in general) compact. I can therefore not prove the >following conjecture: Conjecture: that there exists p(f) in S such that Q(p(f)) = sup_S Q. Taking B = 1 for simplicity, and letting U = int u: If P >= U then there is a unique u which makes u + p constant, and this is the u that gives the inequality shows that for any other u' we have int log(u' + p) <= log(int (u' + p)) = log(U + P) = int log(u + p). So it could be that when P < U the extremal u is one which makes u + p as close as possible to constant, in some sense (and then depending on what sense that is it should be clear whether there is such a u...) >Note: just an existence proof is needed! > ************************ David C. Ullrich ==== >Let >S = the set of all functions p(f): [0,B] -> R such that >(1) p(f) is positive or zero for all f in [0,B] >(2) int_0^B p(f) df = P (P and B are some fixed positive real numbers) >(3) p(f) is continuous on [0,B] >Let >Q: S-> R such that >Q(p(f)) = int_0^B log( p(f) + u(f) ) df >where u(f) is a fixed continuous function which is positive for all f in >[0,B] >It is easy to show that Q is bounded on S, so there's a least upper bound. >However, S is not (in general) compact. I can therefore not prove the >following conjecture: >Conjecture: that there exists p(f) in S such that Q(p(f)) = sup_S Q. Note: just an existence proof is needed! It seems to me fairly clear that the maximum is obtained with p(f) = { c - u(f) if u(f) < c { 0 otherwise where the constant c is chosen so that int_0^B p(f) df = P. The simplest existence proof may be just to show that this is the maximum. A main ingredient is the concavity of ln. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >Let > >S = the set of all functions p(f): [0,B] -> R such that > >(1) p(f) is positive or zero for all f in [0,B] > >(2) int_0^B p(f) df = P (P and B are some fixed positive real numbers) > >(3) p(f) is continuous on [0,B] > >Let > >Q: S-> R such that > >Q(p(f)) = int_0^B log( p(f) + u(f) ) df > >where u(f) is a fixed continuous function which is positive for all f in >[0,B] > >It is easy to show that Q is bounded on S, so there's a least upper bound. >However, S is not (in general) compact. I can therefore not prove the >following conjecture: > >Conjecture: that there exists p(f) in S such that Q(p(f)) = sup_S Q. Note: just an existence proof is needed! > It seems to me fairly clear that the maximum is obtained with > p(f) = { c - u(f) if u(f) < c > { 0 otherwise where the constant c is chosen so that int_0^B p(f) df = P. > The simplest existence proof may be just to show that this > is the maximum. A main ingredient is the concavity of ln. Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 Call your function p_0(x) (I can't bring myself to use f for the independent variable!). For any other fixed admissible function p(x), define p_t(x) = p_0(x) + t*dp(x) where dp(x) = p(x) - p_0(x) and 0 <= t <= 1. Note that int_0^B dp(x) dx = 0. Compute the derivatives of G(t) = Q(p_t) with respect to t: G' = int_0^B dp(x) / (p_t(x) + u(x)) dx Note that p_0(x) + u(x) >= c for all x in [0, B]. This implies that G'(0) <= int_0^B dp(x) / c = 0 Furthermore, G''(t) = int_0^B {-(dp(x))^2 / (p_t(x) + u(x))^2} dx < 0 for all t in [0, 1]. This shows that G(1) > G(0), or in other words, Q(p_0) > Q(p), so p_0 is the unique maximum of Q. John Mitchell ==== > ... >Conjecture: that there exists p(f) in S such that >Q(p(f)) = sup_S Q. >Note: just an existence proof is needed! > > It seems to me fairly clear that the maximum is obtained with > p(f) = { c - u(f) if u(f) < c > { 0 otherwise > ... > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 ... > This shows that G(1) > G(0), or in other words, Q(p_0) > Q(p), so p_0 ^^^^^^^^^^^ This should have been G(0) > G(1). John Mitchell ==== This argument looks correct, but won't it work for any candidate p_0(x) which is bounded above by some constant c? What is it that makes p_0(x) so special? > Call your function p_0(x) (I can't bring myself to use f for the > independent variable!). For any other fixed admissible function p(x), > define p_t(x) = p_0(x) + t*dp(x) where dp(x) = p(x) - p_0(x) and 0 <= t <= 1. Note that int_0^B dp(x) dx = 0. Compute the derivatives of G(t) = > Q(p_t) with respect to t: G' = int_0^B dp(x) / (p_t(x) + u(x)) dx Note that p_0(x) + u(x) >= c for all x in [0, B]. This implies that > G'(0) <= int_0^B dp(x) / c = 0 Furthermore, > G''(t) = int_0^B {-(dp(x))^2 / (p_t(x) + u(x))^2} dx < 0 for all t in > [0, 1]. This shows that G(1) > G(0), or in other words, Q(p_0) > Q(p), so p_0 > is the unique maximum of Q. John Mitchell ==== In my algebra textbook, the product of two ideals I,J is defined as { sum_{i=1..n} a_i b_i | n >= 1 , a_i in I and b_i in J } Now it is rather easy to prove that IJ is an ideal in R. The last question of the exercise is: Is A = { ab | a in I, b in J } an ideal of R. Now the preceding questions strongly suggest that the answer in general is no, but I can't find a counterexample. Clearly, (since it is understood that R is commutative), if one of I or J is a principal ideal, the set A is an ideal, so a counterexample has to consist of a non-PID and to ideals generated by at least two elements each. My strategy was to construct I and J so that I could find ab and a'b' in A with ab + a'b' = 1, and then argue that 1 could not be written as a product of elements from I and J (because then each would contain a unit, so I = J = R), but so far my search has failed. Does anybody have a hint (maybe I'm looking the wrong places, or even after the wrong thing; I am not 100% certain that the answer _is_ no, but then again, I see no reason why it should be yes). Rasmus -- ==== > In my algebra textbook, the product of two ideals I,J is defined as { sum_{i=1..n} a_i b_i | n >= 1 , a_i in I and b_i in J } Now it is rather easy to prove that IJ is an ideal in R. The last > question of the exercise is: Is A = { ab | a in I, b in J } an ideal of R. Now the preceding questions strongly suggest that the answer in > general is no, but I can't find a counterexample. Clearly, (since it > is understood that R is commutative), if one of I or J is a principal > ideal, the set A is an ideal, so a counterexample has to consist of a > non-PID and to ideals generated by at least two elements each. My strategy was to construct I and J so that I could find ab and a'b' > in A with ab + a'b' = 1, and then argue that 1 could not be written as > a product of elements from I and J (because then each would contain a > unit, so I = J = R), but so far my search has failed. It's bound to. If a, a' in I and ab + a' b' = 1, then 1 in I as I is an ideal. Hence I = R is principal (and also J = R).0 You need to look at nonprincipal ideals. Two rings where it's easy to find nonprincipal ideals are the polynomial rings Z[X] and Q[X,Y]. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== > In my algebra textbook, the product of two ideals I,J is defined as { sum_{i=1..n} a_i b_i | n >= 1 , a_i in I and b_i in J } Now it is rather easy to prove that IJ is an ideal in R. The last > question of the exercise is: Is A = { ab | a in I, b in J } an ideal of R. Now the preceding questions strongly suggest that the answer in > general is no, but I can't find a counterexample. Clearly, (since it > is understood that R is commutative), Why is this an assumption? > if one of I or J is a principal > ideal, the set A is an ideal, so a counterexample has to consist of a > non-PID and to ideals generated by at least two elements each. How did you establish that the product was an ideal? Go through a parallel proof for A, and see whether the proof is valid. Don't try to guess whether the answer is yes or no just by the pattern of yes-es and no-s that precedes it. (BTW, though I'm no expert on this stuff, I think the answer is yes by an elementary argument. That's why I suggest going through your more complex proof for IJ.) - Randy ==== >In my algebra textbook, the product of two ideals I,J is defined as { sum_{i=1..n} a_i b_i | n >= 1 , a_i in I and b_i in J } Now it is rather easy to prove that IJ is an ideal in R. The last >question of the exercise is: Is A = { ab | a in I, b in J } an ideal of R. Now the preceding questions strongly suggest that the answer in >general is no, but I can't find a counterexample. Clearly, (since it >is understood that R is commutative), Good thing to mention in general when talking about rings and ideals... > if one of I or J is a principal >ideal, the set A is an ideal, so a counterexample has to consist of a >non-PID and to ideals generated by at least two elements each. Indeed. >My strategy was to construct I and J so that I could find ab and a'b' >in A with ab + a'b' = 1, and then argue that 1 could not be written as >a product of elements from I and J (because then each would contain a >unit, so I = J = R), but so far my search has failed. Wouldn't you have a rather hard time doing that? Note that the set A is a subset of IJ, and also note that IJ is contained in the intersection of I and J; if you could find I and J with a,a' in I, b,b' in J, and with ab + a'b' = 1, then 1 would be an element of BOTH I and J, and hence you would not be able to have a contradiction. > Does anybody >have a hint (maybe I'm looking the wrong places, or even after the >wrong thing; I am not 100% certain that the answer _is_ no, but then >again, I see no reason why it should be yes). Consider the ring Z[x], a notable and easy source of nonprincipal ideals; the ideals I=(x,2) and J=(x,3), for example, are easy prove non-principal. I consists of the polynomials with integer coefficients whose constant term is a multiple of 2; J consists of the polynomials whose constant term is a multiple of 3. So what is A? It consists of all products of the form (x*p(x) + 2a)*(x*q(x)+3b) with p(x),q(x) in Z[x], a,b in Z. Since both 2x and 3x are in A (the first obtained by setting p(x)=0, a=1, q(x)=1, b=0; the second by setting p(x)=1, a=0, b=1, q(x)=0), if A were an ideal then x=3x-2x would also have to be in A. Is that possible? If so, we would have x = (x*p(x) + 2a)(x*q(x)+3b) = p(x)q(x)*x^2 + (2aq(x)+3bp(x))*x + 6ab for some choice of p(x),q(x) in Z[x], a,b in Z. So we know that either a=0 or b=0, since 6ab must be equal to 0. If a=0, then we have x = p(x)q(x)*x^2 + 3bp(x)*x; since the linear term of the right hand side is equal to 3bp(0)*x, we see that this can never be equal to x. So we cannot have a=0. But then b=0; in that case, we have x=p(x)q(x)*x^2 + 2aq(x)*x; again, the linear term on the right hand side is equal to 2aq(0)*x, and this is also never equal to x. So we cannot have b=0. In conclusion, x cannot be written as a product of an element of I=(2,x) and an element of J=(3,x), even though it lies in the smallest ideal which contains A. Therefore, A cannot be an ideal in this case. The answer to the question you have, then, is not necessarily (since sometimes A is an ideal, and sometimes it is not). ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== In my algebra textbook, the product of two ideals I,J is defined as { sum_{i=1..n} a_i b_i | n >= 1 , a_i in I and b_i in J } Now it is rather easy to prove that IJ is an ideal in R. The last > question of the exercise is: Is A = { ab | a in I, b in J } an ideal of R. Now the preceding questions strongly suggest that the answer in > general is no, but I can't find a counterexample. Clearly, (since it > is understood that R is commutative), if one of I or J is a principal > ideal, the set A is an ideal, so a counterexample has to consist of a > non-PID and two ideals generated by at least two elements each [...] HINT: Find proper ideals whose product contains an irreducible element, e.g. p in (p,a)(p,b) if (a,b) = (1) Examples abound. -Bill Dubuque ==== > In my algebra textbook, the product of two ideals I,J is defined as { sum_{i=1..n} a_i b_i | n >= 1 , a_i in I and b_i in J } Now it is rather easy to prove that IJ is an ideal in R. The last > question of the exercise is: Is A = { ab | a in I, b in J } an ideal of R. Now the preceding questions strongly suggest that the answer in > general is no, but I can't find a counterexample. Clearly, (since it > is understood that R is commutative), if one of I or J is a principal > ideal, the set A is an ideal, so a counterexample has to consist of a > non-PID and to ideals generated by at least two elements each. My strategy was to construct I and J so that I could find ab and a'b' > in A with ab + a'b' = 1, and then argue that 1 could not be written as > a product of elements from I and J (because then each would contain a > unit, so I = J = R), but so far my search has failed. Does anybody > have a hint (maybe I'm looking the wrong places, or even after the > wrong thing; I am not 100% certain that the answer _is_ no, but then > again, I see no reason why it should be yes). > Rasmus -- One counterexample could be as follows: consider polynomials over integer, Z[x], and let I = <2,x> the ideal generated by 2 and x, J = <3,x> the ideal generated by 3 and x, A as what you've defined earlier. Then 6 = 2*3 is in A, and x^2 is also in A. But x^2 + 6 is irreducible in Z[x] and hence cannot be written as a product of elements in I and J. Hence x^2+6 is not in A and A is not an ideal in Z[x]. ==== Suppose a function f: R^n to R^n is differentiable at a point x, with an invertible derivative. Is there necessarily an open neighbourhood, U, of x, such that f is injective on U, and such that the inverse of f is differentiable on f(U)? The answer is probably no, but I think counterexamples are hard to come by. Paul Epstein ==== > Suppose a function f: R^n to R^n is differentiable at a point x, with > an invertible derivative. Is there necessarily an open neighbourhood, U, of x, such that f is > injective on U, and such that the inverse of f is differentiable on > f(U)? The answer is probably no, but I think counterexamples are hard to > come by. > Paul Epstein f(x) = x + x^2*sin(1/x) on R^1 You can check that f'(0) = 1, but f'(1/2PI*n) = 0 for all positive integers n, so the inverse function is not differentiable on any neighborhood of 0. John Mitchell <9v03gvo0r862lt018ko75vh2m1q9r6f1hr@4ax.com> ==== > I'm just being too humble again.) I'm going to be famous. And rich: >> they'll probably give me a big fat prize, maybe a Clay prize: this >> is like P=NP ... yeah! That's it! Without the need to divide, prime >> factorization becomes simple, so P=NP, but the Riemann Hypothesis too! >> It's about primes! So that's 2 million dollars right there. Of course, >> the real money will be in the endorsements.... > > I wish I could do the accent like that. (I gave it a feeble try in my > reply to Rusin but it wasn't the same...) Did you spend a lot of time > learning to do this or is it a natural gift or what? > The key is to get in touch with your inner Harris. Many are frightened to > do this, lest they never return to the realm of logic. They pack flashlight > and compass, yet never get more than 100 feet in. You must stride boldly with > your eyes closed. Use the force. There. It's completely dark. You are alone with the void. You say, One plus one is > and your words die out without an echo. Absolute silence reigns again. Then > your shout Two! Of course it's two. But you wonder: did I say two because > it must be two, or must it be two because I said so? As your power grows within you, you begin to test your freedom. You walk about > like Socrates, spinning vast strands of reasoning into space. But then you slow > down, halt, head bent and muttering, how does it go? how is the object defined? > AH HA!!! The object IS the object, so all the zeroes have real part equal to the > spin of the fundamental object! And you find that you have taken a mighty leap > through space, and the whole universe comes into alignment, all the harmony of > the spheres resolves into one resounding chord, and you burst through into the > human realm, trailing clouds of glory. You take a moment to pity the dung beetles pushing their little equations around. > Then you begin to type. That was one of the most beautiful and frightening things I have ever read. Have you ever considered being a proffesional crank? No current crank has the sheer artistry of arrogance that you've just described, and it would be rather refreshing. -- William C. Hogg ==== > IP belongs to blg7-48.imt.net. imt.net is an ISP in Montana. Anybody else see anything else interesting in here? I live in Houston and my ISP is in Seattle (I think). Hell I don't even know where my ISP is. (I'm in San Antonio using a different ISP temporarily, in case you decide to trace my headers). It's quite common with broadband that your DSL service provider is nowhere close to you. Anyway, I'm not giving credit to the post, just pointing out that where the ISP is has nothing to do with where the person is. ==== > IP belongs to blg7-48.imt.net. imt.net is an ISP in Montana. Anybody else see anything else interesting in here? I live in Houston and my ISP is in Seattle (I think). Hell I don't even > know where my ISP is. (I'm in San Antonio using a different ISP > temporarily, in case you decide to trace my headers). It's quite common > with broadband that your DSL service provider is nowhere close to you. > Anyway, I'm not giving credit to the post, just pointing out that where the > ISP is has nothing to do with where the person is. Where's Trinity University? This year's Crypto conference is in Santa Barbara August 17-21. The early registration deadline is July 14th. Full program information is available at: It'll be great, both technically and socially! Greg. (General Chair) -- Greg Rose INTERNET: ggr@qualcomm.com Qualcomm Australia VOICE: +61-2-9817 4188 FAX: +61-2-9817 5199 Level 3, 230 Victoria Road, http://people.qualcomm.com/ggr/ Gladesville NSW 2111 232B EC8F 44C6 C853 D68F E107 E6BF CD2F 1081 A37C ==== what are they? I tried the coordinates on the webpage www.math.niu.edu/~rusin/known-math/95/polyhedr.tori3 A1=(-3,3,0) A2=(3,-3,0) A3=(-3,-3,1) A4=(3,3,1) A5=(1,2,3) A6=(-1,-2,3) A7=(0,0,5) but there seems to be a problem: (135) and (147) are faces, but they intersect. (unless i'm making some mistake). Does anybody have the coordinates? They were apparently published in Acta Sci Math Szeged 13 (1949( 140-2 ==== think i found the problem, A5 and A6 have z coordinate 2, not 3. >what are they? I tried the coordinates on the webpage >www.math.niu.edu/~rusin/known-math/95/polyhedr.tori3 >A1=(-3,3,0) >A2=(3,-3,0) >A3=(-3,-3,1) >A4=(3,3,1) >A5=(1,2,3) >A6=(-1,-2,3) >A7=(0,0,5) >but there seems to be a problem: (135) and (147) are faces, but they >intersect. (unless i'm making some mistake). ==== I am trying to prove something along the lines of: Let P and Q be polynomials, and let S be the operator which cyclically permutes the coefficients of a polynomial. Then, S(P*Q) = S(P)*S(Q). Actually I know this is generally false, because I tried the following: P = x+1 Q = x+2 P*Q = x^2 + 3x + 2 S(P) = x+1 S(Q) = 2x+1 S(P*Q) = 2x^2 + x + 3 S(P)*S(Q) = 2x^2 + 3x + 1 So, my question is, what conditions do the polynomials need to satisfy in order for this to be true? It seems to be true if either of the polynomials contains only one non-zero coefficient (in other words, it's true if, for P a polynomial of degree n with coefficients a_k, there is only one k<=n such that a_k =/= 0. What other conditions would make this true? Also, it does seem to be true that if P and Q have the same degree, then S(P+Q) = S(P) + S(Q). If it's not true that would mean my proof is incorrect. ;-) Anyway, if someone can help me I'd appreciate it. Zach ==== >I am trying to prove something along the lines of: Let P and Q be polynomials, and let S be the operator which cyclically >permutes the coefficients of a polynomial. Then, S(P*Q) = S(P)*S(Q). >Actually I know this is generally false, because I tried the >following: ... >So, my question is, what conditions do the polynomials need to satisfy >in order for this to be true? It seems to be true if either of the >polynomials contains only one non-zero coefficient (in other words, >it's true if, for P a polynomial of degree n with coefficients a_k, >there is only one k<=n such that a_k =/= 0. No. Try say P=x and Q=a+bx+cx^2. The left side (if nonzero) is of degree m-1 + deg(S(P)-P), the right side (if nonzero) is of degree m+n-1. So unless both sides are 0, S(P)-P must be of degree n. Suppose P and m >= 1 are given. With b_0 = 0, any Q1 of degree m-1 would do if S(P) = P (which says all coefficients of P are the same). Otherwise, there is no Q1 unless P1 (1-x^m) is divisible by S(P)-P, in which case for each b_0 <> 0 there is exactly one. For example, with P = 1 + 2 x + 3 x^2, S(P) - P = 1 + x - 2 x^2 = (1+2x)(1-x) The factor 1+2x does not divide P1 = 2+3x or 1-x^m (whose roots are all m'th roots of unity), so there is no Q of degree >= 1 for which this will work. On the other hand, with P = 1 + 2 x + 2 x^2, S(P) - P = 1 - x^2 divides P1 (1-x^m) = 2 (1+x)(1-x^m) for all m>=1 so there will be a solution, namely Q1 = 2 b_0 (1+x+...+x^(m-1)) and Q = b_0 (1 + 2 x + 2 x^2 + ... + 2 x^m). >Also, it does seem to be true that if P and Q have the same degree, >then S(P+Q) = S(P) + S(Q). Not true in general if P+Q has lower degree, i.e. the leading terms of P and of Q cancel. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== I am doing some self study of Group Theory just because I missed out when I had the chance to study it properly and feel so ignorant about it. I am currently messing with quotient groups. I have been trying to prove some things and would love some feedback. Let Z(G) be the center of group G, show that a) Z(G) is a normal subgroup of G. b) If G/Z(G) is cyclic, then G is Abelian. I finished part a), but am struggling with part b). This is what I have tried: Let x and y be elements of G. G/Z(G) is cyclic, let aZ(G) with a an element of G be its generator. (In the following, I will use Z rather than the typographically more complicated Z(G). x, y elm's of G => xZ, yZ elm's of G/Z. Since G/Z is cyclic, xZ = (aZ)^j = (a^j)Z and yZ = (aZ)^k = (a^k)Z for some int's j and k. (xy)Z = (xZ)(yZ) (since Z is normal subgroup of G) = (a^j)Z(a^k)Z = (a^(j+k))Z = (a^k)(a^j)Z = (a^k)Z(a^j)Z = (xZ)(yZ) = (yx)Z This seems to be making some progress toward G being Abelian, but can I conclude here that xy = yx giving commutativity. I realize that (xy)Z and (yx)Z are cosets that have the same elements, but I don't think this gives me what I need. Do I have to bring in some other fact at this point, or is the whole approach incorrect. Ken ==== >I am doing some self study of Group Theory just because I missed out when I >had the chance to study it properly and feel so ignorant about it. I am currently messing >with quotient groups. I have been trying to prove some things and would love some feedback. Let Z(G) be the center of group G, show that > a) Z(G) is a normal subgroup of G. > b) If G/Z(G) is cyclic, then G is Abelian. I finished part a), but am struggling with part b). This is what I have >tried: Let x and y be elements of G. >G/Z(G) is cyclic, let aZ(G) with a an element of G be its generator. (In >the following, I will use >Z rather than the typographically more complicated Z(G). x, y elm's of G => xZ, yZ elm's of G/Z. Since G/Z is cyclic, xZ = (aZ)^j = (a^j)Z and yZ = (aZ)^k = (a^k)Z for some >int's j and k. (xy)Z = (xZ)(yZ) (since Z is normal subgroup of G) > = (a^j)Z(a^k)Z > = (a^(j+k))Z > = (a^k)(a^j)Z > = (a^k)Z(a^j)Z > = (xZ)(yZ) > = (yx)Z This seems to be making some progress toward G being Abelian, but can I >conclude here that >xy = yx giving commutativity. I realize that (xy)Z and (yx)Z are cosets >that have the same >elements, but I don't think this gives me what I need. Do I have to bring >in some other fact at >this point, or is the whole approach incorrect. You are almost there. You know that G/Z is cyclic, generated by aZ. That means that EVERY element of G can be written as a^j*z, where j is an integer and z is central, right? So write x=a^j*z, y = a^k*z'. What is xy and what is yx? Are they equal? Here's a more general result: Let G be a group, let Z(G) be the center of G. Let H be a subgroup of Z(G). (1) Prove that H is normal in G. (2) Prove that if G/H is cyclic, then G is abelian. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== if x,y in G: xH,yH in G/H; some aH in G/H generator of G/H some n,m in Z with xH = (aH)^n = a^n H, yH = (aH)^m = a^m H some h,k in H with x = a^n h, y = a^m k xy = a^n h a^m k = a^m k a^n h = yx as h,k in center ---- ==== If a group is cyclic, then it is generated by an element, say a. G = a^n, n = 1,2,3,4.... Well, is a * a *a*....*a any different than a*a*a*...*a ? Lurch > I am doing some self study of Group Theory just because I missed out when I > had the chance to study it properly and feel so ignorant about it. I am currently messing > with quotient groups. I have been trying to prove some things and would love some feedback. Let Z(G) be the center of group G, show that > a) Z(G) is a normal subgroup of G. > b) If G/Z(G) is cyclic, then G is Abelian. I finished part a), but am struggling with part b). This is what I have > tried: Let x and y be elements of G. > G/Z(G) is cyclic, let aZ(G) with a an element of G be its generator. (In > the following, I will use Z rather than the typographically more complicated Z(G). x, y elm's of G => xZ, yZ elm's of G/Z. Since G/Z is cyclic, xZ = (aZ)^j = (a^j)Z and yZ = (aZ)^k = (a^k)Z for some > int's j and k. (xy)Z = (xZ)(yZ) (since Z is normal subgroup of G) > = (a^j)Z(a^k)Z > = (a^(j+k))Z > = (a^k)(a^j)Z > = (a^k)Z(a^j)Z > = (xZ)(yZ) > = (yx)Z This seems to be making some progress toward G being Abelian, but can I > conclude here that xy = yx giving commutativity. I realize that (xy)Z and (yx)Z are cosets > that have the same elements, but I don't think this gives me what I need. Do I have to bring > in some other fact at this point, or is the whole approach incorrect. > Ken > ==== Now consider an exercise that I came across in an elementary linear > algebra book which, paraphrased (possibly erroneously) states that in > any normed linear space V whose norm satisfies the parallelogram law, > an inner product can be defined; > 2 2 > = 1/4 ||u+v|| - 1/4 ||u-v|| that satisfies: > 2 > ||v|| = That is, V can be considered an inner product space with the norm > derived from the inner product in the natural way. This exercise is for a normed linear space over the real numbers. If the book fails to say this (perhaps in its definition of normed space, or of linear space), then you have grounds for complaint. For example, the inner product in a COMPLEX linear space has some non-real values, but the one defined above has only real values. There is a corresponding formula for complex vector spaces, similar to this one, but with 4 terms instead of 2. After you prove the real version, you will probably see how to do the complex one. ==== > Now consider an exercise that I came across in an elementary linear > algebra book which, paraphrased (possibly erroneously) states that in > any normed linear space V whose norm satisfies the parallelogram law, > an inner product can be defined; > 2 2 > = 1/4 ||u+v|| - 1/4 ||u-v|| > that satisfies: > 2 > ||v|| = That is, V can be considered an inner product space with the norm > derived from the inner product in the natural way. This exercise is for a normed linear space over the real numbers. > If the book fails to say this (perhaps in its definition of normed > space, or of linear space), then you have grounds for complaint. For example, the inner product in a COMPLEX linear space has some > non-real values, but the one defined above has only real values. There is a corresponding formula for complex vector spaces, similar to > this one, but with 4 terms instead of 2. After you prove the real > version, you will probably see how to do the complex one. Ah yes, I am familiar with the formula for making a normed space on a complex vector space into an inner product space. My question is now: can an inner 2 product with the property = ||u|| always be defined on a normed space V which satisfies the parallelogram law? The answer for the case where the scalar field of V is the real or complexes. What about the general case? dan ==== > 2 > product with the property = ||u|| always be defined on a normed > space V which satisfies the parallelogram law? The answer for the case > where the scalar field of V is the real or complexes. What about the > general case? dan I have never seen a definition of inner product except in the real and complex cases. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== >> 2 >> product with the property = ||u|| always be defined on a normed >> space V which satisfies the parallelogram law? The answer for the case >> where the scalar field of V is the real or complexes. What about the >> general case? >> dan I have never seen a definition of inner product except in the real >and complex cases. The paper with Math. Reviews number 2000i:46076, to wit, Ochsenius, H.(RCH-UCCM); Schikhof, W. H.(NL-NIJM), Banach spaces over fields with an infinite rank valuation, appears to give such a definition for certain p-adic cases. Lee Rudolph ==== There is something about a more general definition of a normed space with a scalar field aother than the real/complex numbers, in Serge Lang, Algebra, Springer 2002 on p.469ff ==== topological space (rather than perhaps the more common defn given in the context of a metric space). My problem is this: Kolmogorov and Fomin (in Introductory Real Analysis) say the following: Given a topological space T = (X,tau), a point x in tau is called a limit point of a set M a subset of T if every nbd of x contains infinitely many points of M My question is whether the use of the phrase infinitely many is necessary.I ask this because when I took basic real analysis (in metric spaces), there were two equivalent definitions of a limit point, one of which was the following: Given a metric space (X,r) where M is a subset of X, then x in X is a limit point of M iff every deleted nbd of x in X contains a point of M. Is it possible to define a limit point in a topological space in the same way and say, Given a topological space T = (X,tau), a point x in tau is called a limit point of a set M a subset of T if every deleted nbd of x contains a point of M. The reason why I ask is that it seems this distinction might arise for finite sets. Let X = {1,2,3} and tau = Power set of X union nullset and M = {nullset, {1},{2},{3}} Using this alternate defn of limit point, 1 is a limit point of M since there are 3 deleted nbds containing 1 (in tau) : nullset, {2}, and {3}. All of these sets are elements of M. However using Kolmogorov and Fomin's defn, 1 is not a limit point of M Matthew Brenneman ==== What is meant by a dense matrix in spectral numerical methods? Marco -- ==== > What is meant by a dense matrix in spectral numerical methods? Marco Most of the entries are non-zero ==== > What is meant by a dense matrix in spectral numerical methods? Marco Most of the entries are non-zero So you could say that the dense matrices are dense in the matrices :-) ==== > What is meant by a dense matrix in spectral numerical methods? Marco (Not strictly a mathematical term, and not only in spectral context:) The opposite of sparse matrix, or in detail, a matrix lacking any useful (for programming purposes) pattern of predictable zero entries. ==== Can anyone point me to a book or website which contains some information on the density of Sophie-Germain primes? GREG ==== While its true that all derivations are technically proofs, most people seem to distinguish proofs that can be accomplished by applying a series of mechanical operations in succession, from something that requires, say, explanation in English. And yes I realize this is all very vague. At what line does a proof go from being a derivation to being a 'proof'-proof? It doesn't seem like there is one (that can defined formally at least). I hope someone knows what I'm talking about. l8r, Mike N. Christoff ==== One of the current threads mentions in passing the relationship, or >>lack thereof, between difference equations and differential equations. >>It points out that a difference equation and a differential equation >>which seems analogous may have quite different solutions. >>I have noticed this many times before and not thought a lot about it. >>But this time it made me wonder whether numerical solutions to >>differential equations have any validity. A naive computer program to >>solve a differential equation will really be solving an analogous >>difference equation. >> I suspect that _most_ computer programs solving de's numerically >> do so by solving difference equations... >>So it may generate completely the wrong answer. >>Any there more sophisticated algorithms with which we can be sure that >>the generated solution is close to a solution of the differential >>equation? Are there ways to verify a particular calculated solution? >>Or do we just have to hope? >> In the example you noted where things went totally bad the step >> size was 1; when one is actually solving de's numerically one >> takes a much smaller step size. There are theorems saying that >> if the coefficients in the de have such and such smoothness, >> we use this method, and the step size is so small, then the >> error in the solution will be less than something. keep the thread to a reasonable size. >Obviously the smaller step size helps but I was wondering how you >would know how small a step was required. That's a long story (and I'm not familiar with the details). But as I said, there _are_ actual _theorems_ that _say_ how small a step size is required for a certain accuracy. I gave a reference, where the name of the book may or may not have spelled correctly. >I guess for some equations >you could calculate likely and maximum errors but for others it may be >difficult or impossible to do so. I will need to look at a few >examples and see how well I can calculate the error as a function of >step size. Will's following post may be the answer that I am looking >for. But at least in the simple example, the difference and differential >equations had solutions with similar behaviour. Is this always the >case of may the solutions sometimes be totally different (e.g. one is >not of the same order as the other). There was a confusing discussion >about factorials and the gamma function which may be what is >concerning me. that came out to exp(x^2/2) would approach gamma(x) instead. >> ************************ >> David C. Ullrich J ************************ David C. Ullrich ==== >> >>One of the current threads mentions in passing the relationship, or >>lack thereof, between difference equations and differential equations. >>It points out that a difference equation and a differential equation >>which seems analogous may have quite different solutions. >>I have noticed this many times before and not thought a lot about it. >>But this time it made me wonder whether numerical solutions to >>differential equations have any validity. A naive computer program to >>solve a differential equation will really be solving an analogous >>difference equation. >> I suspect that _most_ computer programs solving de's numerically >> do so by solving difference equations... Your supposition is not relevant. Right. What's an _example_ of a standard technique for solving de's numerically that does _not_ essentially reduce the thing to some sort of difference equation? >>[...] There's a clear difference between discrete and continuous >mathematics--and discrete mathematics is in general harder. Usually there isn't any reason to suppose that a difference equation >will match up with it's analogous differential equation in terms of >the result of integration. Well that's progress. >And remember that Gauss was intrigued by >the closeness between the prime counting distribution and continuous >functions like li(x) possibly *because* there's not in general a close >relationship between the discrete world and the continuous. Now Chebyshev did a lot of good work that lead to the limits that are >usually posted, and Riemann ultimately came up with a hypothesis that >would give a smaller limit, but the prime distribution and continuous >functions are still different things. However my work gives a prime suspect which may not only explain why >they're so close, but it may lead to a means to check Riemann's >Hypothesis, as well as check certain supposedly proven things like >that the prime count crosses li(x). But here we are back to the same old crap. You've given _no_ indication _why_ anyone should believe that the solution to that pde has anything to do with counting primes. Just saying it's a prime suspect over and over is not going to convince anyone. >Now I'm a discoverer. I get in, get the discovery and expect proper >recognition, so I'm not excited about working through all the >ramifications of my own discoveries. What is clear is that if >mathematicians get away with ignoring my work then you won't know how >much that discovery might reveal, now do you? That's why mathematicians aren't supposed to just sit on things. >They're supposed to put discoveries in journals and textbooks, just in >case. Because the mathematicians alive today do NOT know everything, >and who knows? There could be some kid who doesn't know it, but if she gets the >chance, and comes across my work, it might spark something. She might >have insights that escape everyone else. And then she might shock the entire world with what she discovers. If >mathematicians don't continue to betray the trust of the world, and >give her a chance. and it's not excusable. Many of you may think it ok to let them get >away with it, just because you can't imagine doing anything else. But >you see, someone will pay the price, I guarantee it. >James Harris ************************ David C. Ullrich ==== >One of the current threads mentions in passing the relationship, or >lack thereof, between difference equations and differential equations. >It points out that a difference equation and a differential equation >which seems analogous may have quite different solutions. >I have noticed this many times before and not thought a lot about it. >But this time it made me wonder whether numerical solutions to >differential equations have any validity. A naive computer program to >solve a differential equation will really be solving an analogous >difference equation. > I suspect that _most_ computer programs solving de's numerically > do so by solving difference equations... Partly true. >So it may generate completely the wrong answer. >Any there more sophisticated algorithms with which we can be sure that >the generated solution is close to a solution of the differential >equation? Are there ways to verify a particular calculated solution? >Or do we just have to hope? > In the example you noted where things went totally bad the step > size was 1; when one is actually solving de's numerically one > takes a much smaller step size. But that is not the way it is really done. Given a particular differential >equation, for each stepsize it will result in a difference equation, and >they are all different with (in most cases) different solutions. And DE >solvers use this. What computer programs *do* is use different stepsizes *at the same time* >to aproximate the actual solution. So you may step from f(x) to f(x + 2h) >(where h is the basic stepsize) using for instance two ways, one >directly through the difference equation obtained with 2h as the >difference and once twice through the diffence equation obtained with >h as the difference. This gives two (in general different) estimates >for f(x + 2h). There are two ways programs proceed. The first is to >refine the method when the estimates are not close enough to each other >(e.g. halve the stepsize) the other is to use a weighed value based on >the two values obtained. The latter gives speedy results on some >functions, and if you know your function is in that class, that is the >way to go. The first is a bit more complicated but will ultimately >give good results, except on some pathologically bad functions. Well right. I didn't mean to give the idea that the only thing one ever did was to use Euler's method with a smaller step size, in fact I did point out that that was a very bad method. The methods you allude to above _do_ all involve difference equations, right? >Well, this is the basis. Much refinement and improvement has been going >on in this field. The newsgroup sci.math.num-analysis can give more >information. ************************ David C. Ullrich ==== > >One of the current threads mentions in passing the relationship, or >lack thereof, between difference equations and differential equations. >It points out that a difference equation and a differential equation >which seems analogous may have quite different solutions. >I have noticed this many times before and not thought a lot about it. >But this time it made me wonder whether numerical solutions to >differential equations have any validity. A naive computer program to >solve a differential equation will really be solving an analogous >difference equation. > I suspect that _most_ computer programs solving de's numerically > do so by solving difference equations... > Your supposition is not relevant. This is my thread and the title is Re: Difference equations, > differential equations, computer programs and real life so that last > quoted sentence of mine does seem relevant. The topic may have been > inspired by comments in one of your threads but the intention was to > discuss this particular point by itself. On the other hand any > mention of primes in this thread would not be relevant. That was David Ullrich's comment, and I guess I should have directed it more clearly to him. David Ullrich has a history of talking on subjects where he doesn't have a clue, and I find that annoying. Basically you can toss everything he said as just guesses. Though he's an actual math professor at Oklahoma State University, I've yet to see any evidence that he hangs out in the computer science department of that school, or that he hangs out with the mathematicians who use advanced numerical techniques. Unfortunately though, he *talks* on lots of subjects as if he actually has relevant information, which is a common problem on Usenet: know-it-alls. Here he mentioned his suspicion about how real experts in the area do something, and I note his suspicion is not relevant. > I am aware of the considerable differences between continuous and > discrete mathematics. The purpose of this thread was to try to > explore them and their ramifications. As you say, discrete is usually > harder (analytically) but when it comes to a computer, discrete is > usually the only one possible. Well that's not necessarily true as it's better to think of computers as logic machines, so whatever a human being can do in the area, like integrate without using difference equations or discrete methods, a computer can conceivably be programmed to do as well. That's why it's also important that you don't get waylaid or distracted by posters like David Ullrich, as you will wade through a lot of useless information, when it's better to find good sources. Just remember that just because someone replies and *sounds* credible, it doesn't mean they actually know of what they speak. James Harris ==== >> >>One of the current threads mentions in passing the relationship, or >>lack thereof, between difference equations and differential equations. >>It points out that a difference equation and a differential equation >>which seems analogous may have quite different solutions. >>I have noticed this many times before and not thought a lot about it. >>But this time it made me wonder whether numerical solutions to >>differential equations have any validity. A naive computer program to >>solve a differential equation will really be solving an analogous >>difference equation. >> I suspect that _most_ computer programs solving de's numerically >> do so by solving difference equations... >> Your supposition is not relevant. >> This is my thread and the title is Re: Difference equations, >> differential equations, computer programs and real life so that last >> quoted sentence of mine does seem relevant. The topic may have been >> inspired by comments in one of your threads but the intention was to >> discuss this particular point by itself. On the other hand any >> mention of primes in this thread would not be relevant. That was David Ullrich's comment, and I guess I should have directed >it more clearly to him. David Ullrich has a history of talking on subjects where he doesn't >have a clue, and I find that annoying. Basically you can toss >everything he said as just guesses. Though he's an actual math >professor at Oklahoma State University, I've yet to see any evidence >that he hangs out in the computer science department of that school, >or that he hangs out with the mathematicians who use advanced >numerical techniques. Advanced numerical techniques? Like posting Prime Counting Functions that give the wrong answer because you have no idea what the precision of your square root function is? Guffaw. (What's especially funny about this comment of yours is the timing; most of the time the things I do have nothing to do with numerical analysis, but as it happens (i) right now I _am_ working on a program that involves what one might call advanced numerical techniques, in particular making fixed-point high-precision reals to use for Fourier transforms, which are then to be used to multiply large numbers (ii) there _is_ evidence of this on sci.math - see the thread Sqrt(2) with no division.) >Unfortunately though, he *talks* on lots of subjects as if he actually >has relevant information, which is a common problem on Usenet: >know-it-alls. Here he mentioned his suspicion about how real experts in the area do >something, and I note his suspicion is not relevant. But you didn't seem to notice my request for a _counterexample_ to what I said. Huh. >> I am aware of the considerable differences between continuous and >> discrete mathematics. The purpose of this thread was to try to >> explore them and their ramifications. As you say, discrete is usually >> harder (analytically) but when it comes to a computer, discrete is >> usually the only one possible. Well that's not necessarily true as it's better to think of computers >as logic machines, so whatever a human being can do in the area, like >integrate without using difference equations or discrete methods, a >computer can conceivably be programmed to do as well. Can conceivably be programmed to do such things? Of course computers can be programmed to do symbolic integration and also to do symbolic solutions of differential equations! For a professional programmer you seem quite out of touch with what well-known programs do. You seemed to miss the word numerically in the statement I suspect that _most_ computer programs solving de's numerically do so by solving difference equations. >That's why it's also important that you don't get waylaid or >distracted by posters like David Ullrich, as you will wade through a >lot of useless information, when it's better to find good sources. Just remember that just because someone replies and *sounds* credible, >it doesn't mean they actually know of what they speak. Indeed. So why don't you answer that question I've been asking: If r_1 = 1+2i and r_2 = 1-2i then which of r_1 and r_2 does the number sqrt(5) go with? >James Harris ************************ David C. Ullrich ==== > >One of the current threads mentions in passing the relationship, or >lack thereof, between difference equations and differential equations. >It points out that a difference equation and a differential equation >which seems analogous may have quite different solutions. >I have noticed this many times before and not thought a lot about it. >But this time it made me wonder whether numerical solutions to >differential equations have any validity. A naive computer program to >solve a differential equation will really be solving an analogous >difference equation. > I suspect that _most_ computer programs solving de's numerically > do so by solving difference equations... > Your supposition is not relevant. > This is my thread and the title is Re: Difference equations, > differential equations, computer programs and real life so that last > quoted sentence of mine does seem relevant. The topic may have been > inspired by comments in one of your threads but the intention was to > discuss this particular point by itself. On the other hand any > mention of primes in this thread would not be relevant. That was David Ullrich's comment, and I guess I should have directed > it more clearly to him. Yes, looking again I see it was David's comment that you were referring to. Nonetheless, it does still seem relevant to the thread. My apologies for the mistake. > David Ullrich has a history of talking on subjects where he doesn't > have a clue, and I find that annoying. Basically you can toss > everything he said as just guesses. Though he's an actual math > professor at Oklahoma State University, I've yet to see any evidence > that he hangs out in the computer science department of that school, > or that he hangs out with the mathematicians who use advanced > numerical techniques. That is not my impression. I have exchanged notes with David in a couple of previous threads and read many of his other posts. I have mostly found them interesting and helpful. The few that I did not enjoy were only because they were on subjects that I knew nothing of. I am a computer programmer myself. I do not come to this group for computing advice and tips. Either I come to enjoy maths that is nothing to do with computers or if it does concern computers then it the maths element and not the computing element which brings me here. > Unfortunately though, he *talks* on lots of subjects as if he actually > has relevant information, which is a common problem on Usenet: > know-it-alls. Here he mentioned his suspicion about how real experts in the area do > something, and I note his suspicion is not relevant. > > I am aware of the considerable differences between continuous and > discrete mathematics. The purpose of this thread was to try to > explore them and their ramifications. As you say, discrete is usually > harder (analytically) but when it comes to a computer, discrete is > usually the only one possible. Well that's not necessarily true as it's better to think of computers > as logic machines, so whatever a human being can do in the area, like > integrate without using difference equations or discrete methods, a > computer can conceivably be programmed to do as well. You have a point there. I guess that there is a chance that a computer may be able to analytically solve a differential equation and hence come up with an exact answer. It is quite likely that a good program may solve an equation that I could not (because it was programmed by someone better than me). There is also a chance that a computer may solve an equation that has defeated all humans but only because it could more tirelessely explore numerous possible solutions. However, I doubt that a program would solve an equation that was beyond any human provided the human was given enough time. For example a computer was famously used in solving the four colour problem but the work it did could in principle have been done by a human. The computer just did it a whole lot faster. 2001 has not arrived yet and we do not seem to have genuinely intelligent computers referring to HAL in the film 2001.) > That's why it's also important that you don't get waylaid or > distracted by posters like David Ullrich, as you will wade through a > lot of useless information, when it's better to find good sources. I will make my own judgements on who I will let distract me. So far I have been quite happy to be distracted by David's posts. > Just remember that just because someone replies and *sounds* credible, > it doesn't mean they actually know of what they speak. I may have been out of university quite a while but I did spend quite a while studying maths. I do not think that I would be too easily duped. > James Harris J ==== >>One of the current threads mentions in passing the relationship, or >>lack thereof, between difference equations and differential equations. >>It points out that a difference equation and a differential equation >>which seems analogous may have quite different solutions. >>I have noticed this many times before and not thought a lot about it. >>But this time it made me wonder whether numerical solutions to >>differential equations have any validity. A naive computer program to >>solve a differential equation will really be solving an analogous >>difference equation. >> I suspect that _most_ computer programs solving de's numerically >> do so by solving difference equations... >>So it may generate completely the wrong answer. >>Any there more sophisticated algorithms with which we can be sure that >>the generated solution is close to a solution of the differential >>equation? Are there ways to verify a particular calculated solution? >>Or do we just have to hope? >> In the example you noted where things went totally bad the step >> size was 1; when one is actually solving de's numerically one >> takes a much smaller step size. There are theorems saying that >> if the coefficients in the de have such and such smoothness, >> we use this method, and the step size is so small, then the >> error in the solution will be less than something. keep the thread to a reasonable size. >Obviously the smaller step size helps but I was wondering how you >would know how small a step was required. That's a long story (and I'm not familiar with the details). But > as I said, there _are_ actual _theorems_ that _say_ how small > a step size is required for a certain accuracy. I gave a reference, > where the name of the book may or may not have spelled > correctly. moment I do not need all the detail to satisfy my curiosity. Are these theorems quite generally applicable or are they restricted to certain classes of equation? If the theorems are quite generally applicable then I can stop worrying about this one. I was worried that a spaceship may go astray due to missing the correct solution to the multiple body problem (*). So far it seems more likely that the spaceship will go astray due to a programming bug. So my profession seems more likely to take the blame than yours. (*) Half joking here. >I guess for some equations >you could calculate likely and maximum errors but for others it may be >difficult or impossible to do so. I will need to look at a few >examples and see how well I can calculate the error as a function of >step size. Will's following post may be the answer that I am looking >for. But at least in the simple example, the difference and differential >equations had solutions with similar behaviour. Is this always the >case of may the solutions sometimes be totally different (e.g. one is >not of the same order as the other). There was a confusing discussion >about factorials and the gamma function which may be what is >concerning me. that came out to exp(x^2/2) would approach gamma(x) instead. Yes, I am still digesting your original post. You can type faster than I can read and understand. I managed to analyse the simple f'(x) = f(x), f(0) = 1 case and as the step size decreases, it clearly converges to the correct solution. I have not worked through the gamma function case yet. >> ************************ >> David C. Ullrich J ************************ David C. Ullrich J ==== > One of the current threads mentions in passing the relationship, or > lack thereof, between difference equations and differential equations. It points out that a difference equation and a differential equation > which seems analogous may have quite different solutions. I have noticed this many times before and not thought a lot about it. > But this time it made me wonder whether numerical solutions to > differential equations have any validity. A naive computer program to > solve a differential equation will really be solving an analogous > difference equation. So it may generate completely the wrong answer. > Any there more sophisticated algorithms with which we can be sure that > the generated solution is close to a solution of the differential > equation? Are there ways to verify a particular calculated solution? > Or do we just have to hope? A search on Runge-Kutta will get you started. I thought that I had already posted a note of thanks but it seems to have gone astray. So thanks now. That name sounds familiar. There may be more knowledge lurking in my brain than I realised but I have lost the index entries to much of it. J ==== Most of the error analysis that I've seen deals only with local error. Global error is a function of how the local errors add up. A good example is a Hamiltonian system (or any chaotic system). Local errors can be very small. But the local errors don't cancel each other out, thus the global error can get to be very large. As far as I can tell, there does not appear to be much in the way of analysis on how the local errors add up to produce a global error. Recently, something called calculus on time scales has been developed where discrete and continuum calculus are unified. I have not had a chance to look into this, but it sounds like it may provide some useful answers. ==== >Obviously the smaller step size helps but I was wondering how you >would know how small a step was required. > That's a long story (and I'm not familiar with the details). But > as I said, there _are_ actual _theorems_ that _say_ how small > a step size is required for a certain accuracy. I gave a reference, > where the name of the book may or may not have spelled > correctly. moment I do not need all the detail to satisfy my curiosity. Are > these theorems quite generally applicable or are they restricted to > certain classes of equation? Classes of equation and classes of methods. The consistency proofs for ODEs are different than for parabolic PDEs and different than for elliptic PDEs. In the elliptic case the proof for finite difference methods are different than for finite element methods. And so on. Still, the theorems are quite useful. They'll either tell you that certain step sizes need to be small enough (search for Courant-Friedrichs-Levy (CFL) condition; it says that given a space step of this much, you can not step more than that much in time) or it tells you how close you get to the solution in terms of order of magnitude of the discretisation step. V. V. -- ==== ... >What computer programs *do* is use different stepsizes *at the same time* >to aproximate the actual solution. So you may step from f(x) to f(x + 2h) >(where h is the basic stepsize) using for instance two ways, one >directly through the difference equation obtained with 2h as the >difference and once twice through the diffence equation obtained with >h as the difference. > Well right. I didn't mean to give the idea that the only thing one > ever did was to use Euler's method with a smaller step size, in > fact I did point out that that was a very bad method. The > methods you allude to above _do_ all involve difference > equations, right? Indeed. But they do *not* solve a single difference equation. There are more equations involved. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > >One of the current threads mentions in passing the relationship, or >lack thereof, between difference equations and differential equations. >It points out that a difference equation and a differential equation >which seems analogous may have quite different solutions. >I have noticed this many times before and not thought a lot about it. >But this time it made me wonder whether numerical solutions to >differential equations have any validity. A naive computer program to >solve a differential equation will really be solving an analogous >difference equation. > I suspect that _most_ computer programs solving de's numerically > do so by solving difference equations... > Your supposition is not relevant. > This is my thread and the title is Re: Difference equations, > differential equations, computer programs and real life so that last > quoted sentence of mine does seem relevant. The topic may have been > inspired by comments in one of your threads but the intention was to > discuss this particular point by itself. On the other hand any > mention of primes in this thread would not be relevant. > That was David Ullrich's comment, and I guess I should have directed > it more clearly to him. Yes, looking again I see it was David's comment that you were > referring to. Nonetheless, it does still seem relevant to the thread. > My apologies for the mistake. No apology necessary as I should have made it clear. It was your post that I was replying to after all. I do admit that I find David Ullrich to be annoyingly toxic. > David Ullrich has a history of talking on subjects where he doesn't > have a clue, and I find that annoying. Basically you can toss > everything he said as just guesses. Though he's an actual math > professor at Oklahoma State University, I've yet to see any evidence > that he hangs out in the computer science department of that school, > or that he hangs out with the mathematicians who use advanced > numerical techniques. That is not my impression. I have exchanged notes with David in a > couple of previous threads and read many of his other posts. I have > mostly found them interesting and helpful. The few that I did not > enjoy were only because they were on subjects that I knew nothing of. Well David Ullrich is a math professor, and apparently has a big ego. Now if you're patiently listening to him and showing what he sees as respect, then why should he attack you? But if you challenge him, he's likely to get vicious. And if you argue with him, he'll reply, reply, reply until you tire of the exchange. search at groups.google.com where David Ullrich is the author, where he uses the word idiot. For another aspect of David Ullrich, check where he's the author and he uses the word rape. > I am a computer programmer myself. I do not come to this group for > computing advice and tips. Either I come to enjoy maths that is > nothing to do with computers or if it does concern computers then it > the maths element and not the computing element which brings me here. The newsgroup isn't bad for getting information. But it shouldn't surprise you that there are certain posters who have a rep for talking to just about everyone as if they actually know the subjects well, when they have a hidden agenda. Don't take my word for it, do the search I recommended on posts authored by David Ullrich where he uses the word idiot and see his true colors. Just remember there's always something in it for the person replying to you. Here I clearly have an axe to grind about Ullrich, so I'm warning you about him in a way that sends a message to a lot of others. > Unfortunately though, he *talks* on lots of subjects as if he actually > has relevant information, which is a common problem on Usenet: > know-it-alls. > Here he mentioned his suspicion about how real experts in the area do > something, and I note his suspicion is not relevant. > > I am aware of the considerable differences between continuous and > discrete mathematics. The purpose of this thread was to try to > explore them and their ramifications. As you say, discrete is usually > harder (analytically) but when it comes to a computer, discrete is > usually the only one possible. > Well that's not necessarily true as it's better to think of computers > as logic machines, so whatever a human being can do in the area, like > integrate without using difference equations or discrete methods, a > computer can conceivably be programmed to do as well. You have a point there. I guess that there is a chance that a > computer may be able to analytically solve a differential equation and > hence come up with an exact answer. It is quite likely that a good > program may solve an equation that I could not (because it was > programmed by someone better than me). There is also a chance that a > computer may solve an equation that has defeated all humans but only > because it could more tirelessely explore numerous possible solutions. > However, I doubt that a program would solve an equation that was > beyond any human provided the human was given enough time. For > example a computer was famously used in solving the four colour > problem but the work it did could in principle have been done by a > human. The computer just did it a whole lot faster. 2001 has not > arrived yet and we do not seem to have genuinely intelligent computers > referring to HAL in the film 2001.) Yet there are people who daily use computers to handle such problems, and I'm not talking about the four color problem. You may *assume* certain things about the state of the art, but how do you know? Now consider how much you know now versus what you might know if you dug in and checked with credible and truly knowledgeable sources. > That's why it's also important that you don't get waylaid or > distracted by posters like David Ullrich, as you will wade through a > lot of useless information, when it's better to find good sources. I will make my own judgements on who I will let distract me. So far I > have been quite happy to be distracted by David's posts. Of coure you will. And besides, I've revealed my agenda which is not about just talking to you. However, the fact is that you've already displayed that you don't seem to have learned much on the subject of this thread, but look at how many times Ullrich has already posted. > Just remember that just because someone replies and *sounds* credible, > it doesn't mean they actually know of what they speak. I may have been out of university quite a while but I did spend quite > a while studying maths. I do not think that I would be too easily > duped. Spoken like you're new to Usenet and the Internet. Welcome to the new Wild Wild West. James Harris ==== >[...] Here I clearly have an axe to grind about Ullrich, so I'm warning you >about him in a way that sends a message to a lot of others. You're sending a message all right, and very clearly. I suspect that it's not exactly the message you mean to be sending, but why should your personal comments be any different from your mathematical statements in that regard? ************************ David C. Ullrich ==== > >One of the current threads mentions in passing the relationship, or >lack thereof, between difference equations and differential equations. >It points out that a difference equation and a differential equation >which seems analogous may have quite different solutions. >I have noticed this many times before and not thought a lot about it. >But this time it made me wonder whether numerical solutions to >differential equations have any validity. A naive computer program to >solve a differential equation will really be solving an analogous >difference equation. > I suspect that _most_ computer programs solving de's numerically > do so by solving difference equations... > Your supposition is not relevant. > This is my thread and the title is Re: Difference equations, > differential equations, computer programs and real life so that last > quoted sentence of mine does seem relevant. The topic may have been > inspired by comments in one of your threads but the intention was to > discuss this particular point by itself. On the other hand any > mention of primes in this thread would not be relevant. > That was David Ullrich's comment, and I guess I should have directed > it more clearly to him. > Yes, looking again I see it was David's comment that you were > referring to. Nonetheless, it does still seem relevant to the thread. > My apologies for the mistake. No apology necessary as I should have made it clear. It was your post > that I was replying to after all. I do admit that I find David > Ullrich to be annoyingly toxic. I don't find him so. > David Ullrich has a history of talking on subjects where he doesn't > have a clue, and I find that annoying. Basically you can toss > everything he said as just guesses. Though he's an actual math > professor at Oklahoma State University, I've yet to see any evidence > that he hangs out in the computer science department of that school, > or that he hangs out with the mathematicians who use advanced > numerical techniques. > That is not my impression. I have exchanged notes with David in a > couple of previous threads and read many of his other posts. I have > mostly found them interesting and helpful. The few that I did not > enjoy were only because they were on subjects that I knew nothing of. Well David Ullrich is a math professor, and apparently has a big ego. > Now if you're patiently listening to him and showing what he sees as > respect, then why should he attack you? I have disagreed with him in previous posts. He defended his view but I did not see this as an attack. I do not expect everyone to think exactly the same as I do. > But if you challenge him, he's likely to get vicious. > And if you argue with him, he'll reply, reply, reply until you tire of > the exchange. If he tired me, I would stop - no problem. > search at groups.google.com where David Ullrich is the author, where > he uses the word idiot. For another aspect of David Ullrich, check where he's the author and > he uses the word rape. > I am a computer programmer myself. I do not come to this group for > computing advice and tips. Either I come to enjoy maths that is > nothing to do with computers or if it does concern computers then it > the maths element and not the computing element which brings me here. The newsgroup isn't bad for getting information. But it shouldn't > surprise you that there are certain posters who have a rep for talking > to just about everyone as if they actually know the subjects well, > when they have a hidden agenda. Don't take my word for it, do the search I recommended on posts > authored by David Ullrich where he uses the word idiot and see his > true colors. Just remember there's always something in it for the person replying > to you. Here I clearly have an axe to grind about Ullrich, so I'm warning you > about him in a way that sends a message to a lot of others. > Unfortunately though, he *talks* on lots of subjects as if he actually > has relevant information, which is a common problem on Usenet: > know-it-alls. > Here he mentioned his suspicion about how real experts in the area do > something, and I note his suspicion is not relevant. > > I am aware of the considerable differences between continuous and > discrete mathematics. The purpose of this thread was to try to > explore them and their ramifications. As you say, discrete is usually > harder (analytically) but when it comes to a computer, discrete is > usually the only one possible. > Well that's not necessarily true as it's better to think of computers > as logic machines, so whatever a human being can do in the area, like > integrate without using difference equations or discrete methods, a > computer can conceivably be programmed to do as well. > You have a point there. I guess that there is a chance that a > computer may be able to analytically solve a differential equation and > hence come up with an exact answer. It is quite likely that a good > program may solve an equation that I could not (because it was > programmed by someone better than me). There is also a chance that a > computer may solve an equation that has defeated all humans but only > because it could more tirelessely explore numerous possible solutions. > However, I doubt that a program would solve an equation that was > beyond any human provided the human was given enough time. For > example a computer was famously used in solving the four colour > problem but the work it did could in principle have been done by a > human. The computer just did it a whole lot faster. 2001 has not > arrived yet and we do not seem to have genuinely intelligent computers > referring to HAL in the film 2001.) Yet there are people who daily use computers to handle such problems, > and I'm not talking about the four color problem. You may *assume* > certain things about the state of the art, but how do you know? I am quite familiar with the state of the art of computers. I visit the labs of a major manufacturer quite frequently and attend numerous conferences and exhibitions. It is part of my job to keep track of the subject. > Now consider how much you know now versus what you might know if you > dug in and checked with credible and truly knowledgeable sources. As I just said, checking knowledgeable sources is part of my job. > That's why it's also important that you don't get waylaid or > distracted by posters like David Ullrich, as you will wade through a > lot of useless information, when it's better to find good sources. > I will make my own judgements on who I will let distract me. So far I > have been quite happy to be distracted by David's posts. Of coure you will. And besides, I've revealed my agenda which is not > about just talking to you. However, the fact is that you've already displayed that you don't seem > to have learned much on the subject of this thread, but look at how > many times Ullrich has already posted. Have I? I thought that I had learned something. > Just remember that just because someone replies and *sounds* credible, > it doesn't mean they actually know of what they speak. > I may have been out of university quite a while but I did spend quite > a while studying maths. I do not think that I would be too easily > duped. Spoken like you're new to Usenet and the Internet. Why do you think that? I am quite a long term user of Usenet and the Internet. > Welcome to the new Wild Wild West. I am a long term resident. > James Harris This will be my last post on the subject of David Ullrich. This is a newsgroup for discussing maths and not David. J ==== prime example of a global error, is the floating-point spec for hardware and software (IEEE-755, -855, I think; is variously implimented on Earth .-) > Most of the error analysis that I've seen deals only with local error. > Global error is a function of how the local errors add up. A good > example is a Hamiltonian system (or any chaotic system). Local errors > can be very small. But the local errors don't cancel each other out, > thus the global error can get to be very large. As far as I can tell, > there does not appear to be much in the way of analysis on how the > local errors add up to produce a global error. Recently, something > called calculus on time scales has been developed where discrete and > continuum calculus are unified. I have not had a chance to look into > this, but it sounds like it may provide some useful answers. --UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?... La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto: (FOSSILISATION [McCainanites?] (TM/sic))/ BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm. Http://www.tarpley.net/bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81 23 -- Le FIN d'HISTOIRE 24 -- L'ORDEUR du MONDE NOUVEAU 25 -- THYROID STORK !?! ==== Well David Ullrich is a math professor, and apparently has a big ego. > Now if you're patiently listening to him and showing what he sees as > respect, then why should he attack you? I have disagreed with him in previous posts. He defended his view but > I did not see this as an attack. I do not expect everyone to think > exactly the same as I do. Nor do I. That's not my point. > But if you challenge him, he's likely to get vicious. > > And if you argue with him, he'll reply, reply, reply until you tire of > the exchange. If he tired me, I would stop - no problem. Well consider my experience. When I stopped replying to him, he kept replying to me, stalking me in threads, and even went so far as to go to newsgroups like alt.writing and others to do the stalking. Let's imagine that I started posting in reply to you anywhere you went on Usenet, don't you think that'd be a bit odd? And remember I'd *stop* replying to him, but he has kept up his stalking to this day. > search at groups.google.com where David Ullrich is the author, where > he uses the word idiot. > For another aspect of David Ullrich, check where he's the author and > he uses the word rape. > I am a computer programmer myself. I do not come to this group for > computing advice and tips. Either I come to enjoy maths that is > nothing to do with computers or if it does concern computers then it > the maths element and not the computing element which brings me here. > The newsgroup isn't bad for getting information. But it shouldn't > surprise you that there are certain posters who have a rep for talking > to just about everyone as if they actually know the subjects well, > when they have a hidden agenda. > Don't take my word for it, do the search I recommended on posts > authored by David Ullrich where he uses the word idiot and see his > true colors. > Just remember there's always something in it for the person replying > to you. > Here I clearly have an axe to grind about Ullrich, so I'm warning you > about him in a way that sends a message to a lot of others. Yet there are people who daily use computers to handle such problems, > and I'm not talking about the four color problem. You may *assume* > certain things about the state of the art, but how do you know? I am quite familiar with the state of the art of computers. I visit > the labs of a major manufacturer quite frequently and attend numerous > conferences and exhibitions. It is part of my job to keep track of > the subject. That's not the subject. My guess is that you're annoyed at my implication that you've been taken in by David Ullrich. Rather than consider the evidence, you're reacting as if you personally feel attacked. However, doesn't it even bother you slightly that you just fibbed about your own subject line for a thread *you* created? > Now consider how much you know now versus what you might know if you > dug in and checked with credible and truly knowledgeable sources. As I just said, checking knowledgeable sources is part of my job. Then why did you create the thread asking questions? Was it as a ploy so that you can start sharing your knowledge with the group after first behaving as if you were ignorant? I think you've become defensive. That point is key: Now consider how much you now know versus what you might know if you dug in and checked with credible and truly knowledgeable sources. You answer is basically that you already have such sources, which makes me wonder why you created the thread. > That's why it's also important that you don't get waylaid or > distracted by posters like David Ullrich, as you will wade through a > lot of useless information, when it's better to find good sources. > I will make my own judgements on who I will let distract me. So far I > have been quite happy to be distracted by David's posts. > Of coure you will. And besides, I've revealed my agenda which is not > about just talking to you. > However, the fact is that you've already displayed that you don't seem > to have learned much on the subject of this thread, but look at how > many times Ullrich has already posted. Have I? I thought that I had learned something. Well it didn't sound like it to me, which is subjective, but your reply sounded like you still didn't know of routine programs for doing integrations, when from what I've picked up in posts, much of the math world uses them routinely, and they don't necessarily rely on difference techniques. You see that's part of the problem, the newsgroup readers don't necessarily feel a responsibility to save you from people like David Ullrich, a math professor, who has repeatedly demonstrated a lack of knowledge of even basic computer tools for math. > Just remember that just because someone replies and *sounds* credible, > it doesn't mean they actually know of what they speak. > I may have been out of university quite a while but I did spend quite > a while studying maths. I do not think that I would be too easily > duped. > Spoken like you're new to Usenet and the Internet. Why do you think that? I am quite a long term user of Usenet and the > Internet. You don't sound like it to me. > Welcome to the new Wild Wild West. I am a long term resident. Good for you. > James Harris This will be my last post on the subject of David Ullrich. This is a > newsgroup for discussing maths and not David. Yet somehow he does manage to come up as the subject quite often. I'd be interested in someone with basic knowledge on the subject giving an overview of the actual state of the art in the area of computer analysis with regard to differential equation, and difference equations, including some of the known math experts in the field. James Harris ==== > Well David Ullrich is a math professor, and apparently has a big ego. > Now if you're patiently listening to him and showing what he sees as > respect, then why should he attack you? > I have disagreed with him in previous posts. He defended his view but > I did not see this as an attack. I do not expect everyone to think > exactly the same as I do. Nor do I. That's not my point. > > But if you challenge him, he's likely to get vicious. > > And if you argue with him, he'll reply, reply, reply until you tire of > the exchange. > If he tired me, I would stop - no problem. Well consider my experience. When I stopped replying to him, he kept > replying to me, stalking me in threads, and even went so far as to go > to newsgroups like alt.writing and others to do the stalking. Let's imagine that I started posting in reply to you anywhere you went > on Usenet, don't you think that'd be a bit odd? And remember I'd *stop* replying to him, but he has kept up his > stalking to this day. As I said already, I do not intend to discuss David anymore. > search at groups.google.com where David Ullrich is the author, where > he uses the word idiot. > For another aspect of David Ullrich, check where he's the author and > he uses the word rape. > I am a computer programmer myself. I do not come to this group for > computing advice and tips. Either I come to enjoy maths that is > nothing to do with computers or if it does concern computers then it > the maths element and not the computing element which brings me here. > The newsgroup isn't bad for getting information. But it shouldn't > surprise you that there are certain posters who have a rep for talking > to just about everyone as if they actually know the subjects well, > when they have a hidden agenda. > Don't take my word for it, do the search I recommended on posts > authored by David Ullrich where he uses the word idiot and see his > true colors. > Just remember there's always something in it for the person replying > to you. > Here I clearly have an axe to grind about Ullrich, so I'm warning you > about him in a way that sends a message to a lot of others. > Yet there are people who daily use computers to handle such problems, > and I'm not talking about the four color problem. You may *assume* > certain things about the state of the art, but how do you know? > I am quite familiar with the state of the art of computers. I visit > the labs of a major manufacturer quite frequently and attend numerous > conferences and exhibitions. It is part of my job to keep track of > the subject. That's not the subject. No it is not the subject of the post but nor is anything you write. You appeared to be questioning my knowledge of computers. That is why I made those comments. > My guess is that you're annoyed at my implication that you've been > taken in by David Ullrich. Rather than consider the evidence, you're > reacting as if you personally feel attacked. No, I do not feel that I have been taken in by him. If I am annoyed at anyone, it is you for all this off topic stuff you are posting to my thread. > However, doesn't it even bother you slightly that you just fibbed > about your own subject line for a thread *you* created? > Now consider how much you know now versus what you might know if you > dug in and checked with credible and truly knowledgeable sources. > As I just said, checking knowledgeable sources is part of my job. Then why did you create the thread asking questions? The knowledge that I am claiming is of computers and not mathematics. It is the mathematics that I was asking about. > Was it as a ploy so that you can start sharing your knowledge with the > group after first behaving as if you were ignorant? No. I have some knowledge of computers and a rather rusty knowledge of mathematics. I come here for fun and to try and clean some of the rust of my maths knowledge. > I think you've become defensive. This comment of yours above You may *assume* certain things about the state of the art, but how do you know? seemed rather offensive. Especially with the asterisks. > That point is key: Now consider how much you now know versus what you > might know if you dug in and checked with credible and truly > knowledgeable sources. You answer is basically that you already have such sources, which > makes me wonder why you created the thread. I have easy access to reliable information on computers but not for mathematics. If it became important, I could drive 30 or so miles to my old university and look in the library. > That's why it's also important that you don't get waylaid or > distracted by posters like David Ullrich, as you will wade through a > lot of useless information, when it's better to find good sources. > I will make my own judgements on who I will let distract me. So far I > have been quite happy to be distracted by David's posts. > Of coure you will. And besides, I've revealed my agenda which is not > about just talking to you. > However, the fact is that you've already displayed that you don't seem > to have learned much on the subject of this thread, but look at how > many times Ullrich has already posted. > Have I? I thought that I had learned something. Well it didn't sound like it to me, which is subjective, but your > reply sounded like you still didn't know of routine programs for doing > integrations, when from what I've picked up in posts, much of the math > world uses them routinely, and they don't necessarily rely on > difference techniques. See below. > You see that's part of the problem, the newsgroup readers don't > necessarily feel a responsibility to save you from people like David > Ullrich, a math professor, who has repeatedly demonstrated a lack of > knowledge of even basic computer tools for math. > Just remember that just because someone replies and *sounds* credible, > it doesn't mean they actually know of what they speak. > I may have been out of university quite a while but I did spend quite > a while studying maths. I do not think that I would be too easily > duped. > Spoken like you're new to Usenet and the Internet. > Why do you think that? I am quite a long term user of Usenet and the > Internet. You don't sound like it to me. I may be more modest and polite than the average Usenet user. I am also British rather than American. Maybe you confuse these things with naivety. > Welcome to the new Wild Wild West. > I am a long term resident. Good for you. > > James Harris > This will be my last post on the subject of David Ullrich. This is a > newsgroup for discussing maths and not David. Yet somehow he does manage to come up as the subject quite often. > I'd be interested in someone with basic knowledge on the subject > giving an overview of the actual state of the art in the area of > computer analysis with regard to differential equation, and difference > equations, including some of the known math experts in the field. Others have also posted to this thread. > James Harris This is now my last post on the subject of me. I will no longer reply to you unless your post is relevant to the thread and interesting. I do not wish to discuss me, David or anyone else any further. This is a maths newsgroup. J ==== I thought that a good pseudo random number generator could generate two >uncorrelated sequences of random numbers with two different seeds. Let >the random number generator be rand() and it takes a seed to initialize >it. So I thought rand(-1) and rand(-2) should produce two sequences of >uncorrelated random numbers. But my professor told me that it was not >guaranteed and the two sequences were possibly correlated or even highly >correlated. He said only two sequences of random number generated with >the same seed could be uncorrelated. What do you think about this? I >doubt what he said. If you doubt what he said why didn't you ask him to explain? Anyway, almost every random number generator in the universe generates a sequence of numbers by saying x[0] = the seed and x[n+1] = f(x[n]) for some function f. Now think about what happens if you use two seeds, and the second seed happens to be f(the first seed)... (are the two sequences 1, 5, 8, 5, 6, 0, 6, 8, 0, ... 5, 8, 5, 6, 0, 6, 8, 0, 3, ... uncorrelated?) > If a random number is good enough, such as ran1 and >ran2 in Numerical Recipes, it should produce two uncorrelated sequences of >random numbers with two different seeds. David ************************ David C. Ullrich ==== Hauke Reddmann escribi.97 en el mensaje|nbe6im3$ks5$2@rzsun03.rrz.uni-hamburg.de: > Using the a/g mean ineqality, it should be provable > that x=y=z=9 is the largest solution possible, and > x<=y<=z<=9 isn't that big an area to be searched > even by hand :-) Using the a/g inequality was my first attempt, but I only get n >= 9/a and n >= 9/g. totally unusefull to upper bound n... And x = y = z = 9 isn't the largest solution in any sense. By example, these are the solutions with 1000 <= x + y + z <= 2000, x <= y <= z: x y z n = = = = 25 45 980 1 16 72 968 1 1 324 845 5 8 27 1225 6 9 75 1176 2 1 169 1156 9 4 200 1156 2 3 294 1089 2 4 81 1445 5 12 150 1458 1 1 363 1352 6 9 225 1521 1 2 121 1681 8 5 81 1849 5 1 289 1682 8 Althought there are solutions with x, y z >= 9, with smaller values. By example 9 9 36 1 related with 1 1 4 9 3 3 12 3 As Will said, there are multiple solutions with n = 1, 2, 3, 4, 5, 6, 8 or 9, but as far as x + y + z = 4000 there aren't solutions for n = 7 or n > 9. But I still didn't found any theorethicall reasons for the lack n = 7 and the limit n = 9 ... -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com For n = 7, any variable and the sum of the others two must be multiiple of 7. But it is also true for n = 2, 3 and 5 ... ==== Context: (x+y+z)^2 = nxyz. (x,y,z must be positive integers) >Hauke Reddmann escribi.97 en el >mensaje|nbe6im3$ks5$2@rzsun03.rrz.uni-hamburg.de: >> Using the a/g mean ineqality, it should be provable >> that x=y=z=9 is the largest solution possible, and >> x<=y<=z<=9 isn't that big an area to be searched >> even by hand :-) Using the a/g inequality was my first attempt, but I only get n >= 9/a and >n >= 9/g. totally unusefull to upper bound n... As Will said, there are multiple solutions with n = 1, 2, 3, 4, 5, 6, 8 or >9, but as far as x + y + z = 4000 there aren't solutions for n = 7 or n > 9. >But I still didn't found any theorethicall reasons for the lack n = 7 and >the limit n = 9 ... I was able to show that n <= 12, and furthermore that the problem reduces to a finite number of two-variable quadratic equations (and is therefore decidable). Maybe someone can refine the argument: Fix n > 0, and assume WLOG that 0 < x <= y <= z and x+y+z is minimum. Now, rewrite (x+y+z)^2 = nxyz as: z^2 - (nxy-2x-2y)z + (x+y)^2 = 0. This is a quadratic in z with integer coefficients, and thus if we substitute z <- (x+y)^2/z we get another solution. Since x+y+z is chosen to be minimum, we must have z^2 <= (x+y)^2, i.e. z <= x+y. Now, nxyz = (x+y+z)(x+y+z) <= (z+z+z)(x+y+x+y) = 6z(x+y), so that nxy <= 6(x+y), which we can rewrite as (nx-6)(ny-6) <= 36. Clearly we cannot have nx-6 > 6, so n <= 12/x. Since x >= 1, we have the desired upper bound on n. Furthermore, for any given n, we must have 1 <= x <= 12/n, which means there are only finitely many values of x to check. Each pair of (n,x) yields a quadratic Diophantine in y and z, which is algorithmically decidable. Maybe someone can see a simple way to characterise all solutions with x = 1? Then for any other solution we'd have n <= 6, which nicely rules out n = 7. -- Erick ==== >Context: (x+y+z)^2 = nxyz. (x,y,z must be positive integers) Maybe someone can see a simple way to characterise all solutions >with x = 1? Then for any other solution we'd have n <= 6, which >nicely rules out n = 7. When in doubt, try another descent. If (y+z+1)^2 = nyz, then by a very similar argument we have WLOG either z = y or z = y+1, which leaves only the solutions (y,z,n) = (1,1,9) or (1,2,8). So together with the known solutions for n <= 6 the problem is solved. -- Erick ==== Erick Bryce Wong escribi.97 en el mensaje >> Context: (x+y+z)^2 = nxyz. (x,y,z must be positive integers) >> Maybe someone can see a simple way to characterise all solutions >> with x = 1? Then for any other solution we'd have n <= 6, which >> nicely rules out n = 7. When in doubt, try another descent. If (y+z+1)^2 = nyz, then by a > very similar argument we have WLOG either z = y or z = y+1, which > leaves only the solutions (y,z,n) = (1,1,9) or (1,2,8). Also (y, z, n) = (4, 5, 5) and (2, 3, 6) although tis change nothing .. > So together > with the known solutions for n <= 6 the problem is solved. Good! -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== What follows is elaborated from the solution of Eryck Bryce Wong Find all positive integers n such that there exists at least one solution to the diophantine equation: (x+y+z)^2 = nxyz (x,y,z must be positive integers) Let us consider, without loss of generallity, 1 < = x < = y < = z, and, fixed n, look for the solution with x + y + z minimum. Writing the equation like one of second degree in z, it is z^2 - (nxy - 2x - 2y) + (x + y)^2 = 0 If z and z', z < = z', are the solutions of this equation, we have zz' = (x + y)^2 ==> z < = x + y. Then we have nxyz = (x + y + z)(x + y + z) <= (z + z + z)(x + y + x + y) = 6z(x + y) nxy < = 6(x + y) ==> n^2xy - n(x + y) + 36 < = 36 ===> (nx - 6)(ny - 6) < = 36 ===> nx - 6 < = 6 ===> n < = 12/x (since we are supposing y > = x). Therefore, if x > 1, has to be 1 < = n < = 6. If x = 1, we have (1 + and + z)^2 = nyz. Let do like before, z^2 - (ny - 2 - 2y) + (1 + y)^2 = 0 It must be y < = z < = y + 1, which only leaves two possibilities: i) z = y (2y + 1)^2 = ny^2 ==> (n - 4)y^2 - 4y - 1 = 0 y = (4 +/- rq(16 + 4(n-4)))/(2(n - 4)) = (2 +/- rq(n))/(n - 4) what only it leaves the possibility n = 9, y = z = 1 ==> (1 + 1 + 1)^2 = 9*1*1*1 ii) z = y + 1 (2y + 2)^2 = ny(y + 1) ==> (n - 4) y^2 + (n - 8)y - 4 = 0 y = (- (n - 8) +/- rq((n - 8)^2 + 16(n - 4)))/(2(n - 4) = (- n + 8 +/- n)/(2(n - 4)) ==> y = - 1 or y = 4/(n - 4) that it provides the solutions n = 5, y = 4, z = 5 ==> (1 + 4 + 5)^2 = 5*1*4*5 n = 6, and = 2, 3 z = ==> (1 + 2 + 3)^2 = 6*1*2*3 n = 8, and = 1, 2 z = ==> (1 + 1 + 2)^2 = 8*1*1*2 Therefore, the only possible values of n are 1, 2, 3, 4, 5, 6, 8 and 9. Only leaves to show solutions for n = 1, 2, 3 and 4. Greetings, Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== I have difficulty with your approach. This is how I look at the problem: First allow negative integers as well and make y = -x, z = nx^2. (Or find other solutions allowing negative integers.) Now ask, given one solution, when are there other solutions such that x'+y'+z' = x+y+z and n stays the same? Why, starting from here, would making all positive integers be impossible for n > 9? This is a very big jump for me, and I can't follow your logic very easily. Ignacio Larrosa Ca.96estro Find all positive integers n such that there exists at least one solution >to >the diophantine equation: >(x+y+z)^2 = nxyz (x,y,z must be positive integers) Let us consider, without loss of generallity, 1 < = x < = y < = z, and, >fixed n, look for the solution with x + y + z minimum. Writing the equation >like one of second degree in z, it is z^2 - (nxy - 2x - 2y) + (x + y)^2 = 0 If z and z', z < = z', are the solutions of this equation, we have zz' = >(x + y)^2 ==> z < = x + y. Then we have nxyz = (x + y + z)(x + y + z) <= (z + z + z)(x + y + x + y) = 6z(x + y) nxy < = 6(x + y) ==> n^2xy - n(x + y) + 36 < = 36 === >(nx - 6)(ny - 6) < = 36 ===> nx - 6 < = 6 ===> n < = 12/x (since we are supposing y > = x). Therefore, if x > 1, has to be 1 < = n < >= 6. If x = 1, we have (1 + and + z)^2 = nyz. Let do like before, z^2 - (ny - 2 - 2y) + (1 + y)^2 = 0 It must be y < = z < = y + 1, which only leaves two possibilities: i) z = y (2y + 1)^2 = ny^2 ==> (n - 4)y^2 - 4y - 1 = 0 y = (4 +/- rq(16 + 4(n-4)))/(2(n - 4)) = (2 +/- rq(n))/(n - 4) what only it leaves the possibility n = 9, y = z = 1 ==> (1 + 1 + 1)^2 = >9*1*1*1 ii) z = y + 1 (2y + 2)^2 = ny(y + 1) ==> (n - 4) y^2 + (n - 8)y - 4 = 0 y = (- (n - 8) +/- rq((n - 8)^2 + 16(n - 4)))/(2(n - 4) = (- n + 8 +/- n)/(2(n - 4)) ==> y = - 1 or y = 4/(n - 4) that it provides the solutions n = 5, y = 4, z = 5 ==> (1 + 4 + 5)^2 = 5*1*4*5 n = 6, and = 2, 3 z = ==> (1 + 2 + 3)^2 = 6*1*2*3 n = 8, and = 1, 2 z = ==> (1 + 1 + 2)^2 = 8*1*1*2 Therefore, the only possible values of n are 1, 2, 3, 4, 5, 6, 8 and 9. Only leaves to show solutions for n = 1, 2, 3 and 4. >(9 + 9 + 9)^2 = 1*9*9*9, solution for n = 1. Multiplying by 3^2, (3 + 3 + 3)^2 = 3*3*3*3, solution for n = 3. >multiplying by 4^2, (4 + 4 + 8)^2 = 2*4*4*8, solution for n = 2. multiplying by 2^2, (2 + 2 + 4)^2 = 4*2*2*4, solution for n = 4. So, the equation has solution if and only if n = 1, 2, 3, 4, 5, 6, 8 or 9. >Greetings, Ignacio Larrosa Ca.96estro >A Coru.96a (Espa.96a) >ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== > Using the a/g inequality was my first attempt, but I only get n >= 9/a and > n >= 9/g. totally unusefull to upper bound n... But if you first eliminate n by the trick of my followupee (to get (x'+y'+z')^2=x'y'z') then it could work?! -- Hauke Reddmann <:-EX8 For our chemistry workgroup,remove math from the address For spamming, remove anything else ==== Jeffrey H escribi.97 en el mensaje|n3f098b32.1350813@news.mindspring.com: > I have difficulty with your approach. This is how I look at the > problem: First allow negative integers as well and make y = -x, z = nx^2. > (Or find other solutions allowing negative integers.) But the OP enunciate prescribe positive integers. And observe that the LHS is always positive; it must be a even number of negative variables. > Now ask, given > one solution, when are there other solutions such that > x'+y'+z' = x+y+z and n stays the same? But the enunciate ask for the values of n that give at least one solution. > Why, starting from here, would making all positive integers be > impossible for n > 9? This is a very big jump for me, and I can't > follow your logic very easily. It there are solutions for a fixed n, it must be a solution (or more, don't care) with minimum sum x + y + z. Eryck Bryce proved that only happen for n = 1, 2, 3, 4, 5, 6, 8 and 9. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com Ignacio Larrosa Ca.96estro > What follows is elaborated from the solution of Eryck Bryce Wong >> Find all positive integers n such that there exists at least one >> solution to >> the diophantine equation: >> (x+y+z)^2 = nxyz (x,y,z must be positive integers) >> Let us consider, without loss of generallity, 1 < = x < = y < = z, >> and, fixed n, look for the solution with x + y + z minimum. Writing >> the equation like one of second degree in z, it is >> z^2 - (nxy - 2x - 2y) + (x + y)^2 = 0 >> If z and z', z < = z', are the solutions of this equation, we have >> zz' = (x + y)^2 ==> z < = x + y. Then we have >> nxyz = (x + y + z)(x + y + z) <= (z + z + z)(x + y + x + y) = 6z(x + >> y) >> nxy < = 6(x + y) ==> n^2xy - n(x + y) + 36 < = 36 ===> (nx - 6)(ny - 6) < = 36 ===> nx - 6 < = 6 ===> n < = 12/x >> (since we are supposing y > = x). Therefore, if x > 1, has to be 1 >> < = n < = 6. >> If x = 1, we have (1 + and + z)^2 = nyz. Let do like before, >> z^2 - (ny - 2 - 2y) + (1 + y)^2 = 0 >> It must be y < = z < = y + 1, which only leaves two possibilities: >> i) z = y >> (2y + 1)^2 = ny^2 ==> (n - 4)y^2 - 4y - 1 = 0 >> y = (4 +/- rq(16 + 4(n-4)))/(2(n - 4)) = (2 +/- rq(n))/(n - 4) >> what only it leaves the possibility n = 9, y = z = 1 ==> (1 + 1 + >> 1)^2 = 9*1*1*1 >> ii) z = y + 1 >> (2y + 2)^2 = ny(y + 1) ==> (n - 4) y^2 + (n - 8)y - 4 = 0 >> y = (- (n - 8) +/- rq((n - 8)^2 + 16(n - 4)))/(2(n - 4) >> = (- n + 8 +/- n)/(2(n - 4)) ==> y = - 1 or y = 4/(n - 4) >> that it provides the solutions >> n = 5, y = 4, z = 5 ==> (1 + 4 + 5)^2 = 5*1*4*5 >> n = 6, and = 2, 3 z = ==> (1 + 2 + 3)^2 = 6*1*2*3 >> n = 8, and = 1, 2 z = ==> (1 + 1 + 2)^2 = 8*1*1*2 >> Therefore, the only possible values of n are 1, 2, 3, 4, 5, 6, 8 and >> 9. >> Only leaves to show solutions for n = 1, 2, 3 and 4. >> (9 + 9 + 9)^2 = 1*9*9*9, solution for n = 1. >> Multiplying by 3^2, >> (3 + 3 + 3)^2 = 3*3*3*3, solution for n = 3. >> multiplying by 4^2, >> (4 + 4 + 8)^2 = 2*4*4*8, solution for n = 2. >> multiplying by 2^2, >> (2 + 2 + 4)^2 = 4*2*2*4, solution for n = 4. >> So, the equation has solution if and only if n = 1, 2, 3, 4, 5, 6, 8 >> or 9. >> Greetings, >> Ignacio Larrosa Ca.96estro >> A Coru.96a (Espa.96a) >> ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== > What follows is elaborated from the solution of Eryck Bryce Wong > [... Details of Erick's solution ...] Good job, Erick! Ignacio, thanks for the expanded version with details about the case x=1. ==== >Erick Bryce Wong escribi.97 en el mensaje >> When in doubt, try another descent. If (y+z+1)^2 = nyz, then by a >> very similar argument we have WLOG either z = y or z = y+1, which >> leaves only the solutions (y,z,n) = (1,1,9) or (1,2,8). Also (y, z, n) = (4, 5, 5) and (2, 3, 6) and for the very nice elaboration! -- Erick ==== >Erick Bryce Wong escribi.97 en el mensaje >> When in doubt, try another descent. If (y+z+1)^2 = nyz, then by a >> very similar argument we have WLOG either z = y or z = y+1, which >> leaves only the solutions (y,z,n) = (1,1,9) or (1,2,8). Also (y, z, n) = (4, 5, 5) and (2, 3, 6) and for the very nice elaboration! -- Erick I know the problem is over, but I'm always the last one to leave the movie theater... :-) I was just interested in finding all the solutions in some special case. I chose n=6, x=1, so I wanted all the solutions to (1+y+z)^2 = 6yz with y < z I found these: y z 2 3 3 8 8 27 27 98 And that made me look closer. As a quadratic in y, the equation has two solutions. Let y continue to represent the smaller one, and w the larger, so also (1+z+w)^2 = 6zw The sum of the roots is the negative of the coefficient of y, so y + w = -(2-4z), or w = 4z - y - 2. This gives us a recurrence relation w_n = 4w_(n-1) - w_(n-2) - 2 whose solution, starting with w_1 = 2 and w_2 = 3, is w_n = 1 + [(2+s)^(n-1) + (2-s)^(n-1)]/2 where s stands for sqrt(3) So any two successive terms of the sequence 2 3 8 27 98 363 1352 5043 18818 ... give a solution to (1+y+z)^2 = 6yz I'm pretty sure this work for all the cases, since all it needs is to have a symmetric quadratic polynomial in y and z. ==== Given integer n>=2 , what is the maximal number of pairs of points with Euclidean distance 1 in a set of n points in the plane ? ==== > Given integer n>=2 , what is the maximal number of pairs of points with > Euclidean distance 1 in a set of n points in the plane ? > I think the best you can do is pack the points in as vertices of a tesselation of the plane by equilateral triangles. If that is the case, then the maximum number of pairs would be floor(3n - sqrt(12n-3)). For n of the form 3k^2 + 3k + 1, the points can be arranged in a hexagonal shape, and in this case 3n - sqrt(12n-3) is an integer. ==== > Given integer n>=2 , what is the maximal number of pairs of points with > Euclidean distance 1 in a set of n points in the plane ? > I think the best you can do is pack the points in as vertices of a > tesselation of the plane by equilateral triangles. If that is > the case, then the maximum number of pairs would be > floor(3n - sqrt(12n-3)). For n of the form 3k^2 + 3k + 1, the points can be arranged in a > hexagonal shape, and in this case 3n - sqrt(12n-3) is an integer. This is a famous problem of Erd.9as. The best known lower bound is Erd.9as', n^{1+c/loglog n} for some c, and comes from packing points in a square lattice with spacing 1/q where the factorization of q involves many primes congruent to 1 mod 4. The best upper bound is quite far from this, O(n^{4/3}). See http://cs.smith.edu/~orourke/TOPP/P39.html for a more detailed discussion. If you required distance 1 to be the smallest distance in the point set you would get answers more similar to the one you posted. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science ==== > Given integer n>=2 , what is the maximal number of pairs of points with > Euclidean distance 1 in a set of n points in the plane ? > I think the best you can do is pack the points in as vertices of a > tesselation of the plane by equilateral triangles. If that is > the case, then the maximum number of pairs would be > floor(3n - sqrt(12n-3)). For n of the form 3k^2 + 3k + 1, the points can be arranged in a > hexagonal shape, and in this case 3n - sqrt(12n-3) is an integer. When you said packing, does this discriminate between two different types of equivilent packing, which involve shifting the planes upon which the vertices were placed differently with respect to eachother? Or is it the same result every time? Also, what if this is only an asymptotic value and not a real value? (...Starblade Riven Darksquall...) ==== > Given integer n>=2 , what is the maximal number of pairs of points with > Euclidean distance 1 in a set of n points in the plane ? > I think the best you can do is pack the points in as vertices of a > tesselation of the plane by equilateral triangles. If that is > the case, then the maximum number of pairs would be > floor(3n - sqrt(12n-3)). For n of the form 3k^2 + 3k + 1, the points can be arranged in a > hexagonal shape, and in this case 3n - sqrt(12n-3) is an integer. Do you assume that the points are distinct ? Tim ==== See David Eppstein's reply in this thread. It appears that I am somewhere out in left field, if I'm in the ball park at all. > Given integer n>=2 , what is the maximal number of pairs of points with > Euclidean distance 1 in a set of n points in the plane ? I think the best you can do is pack the points in as vertices of a > tesselation of the plane by equilateral triangles. If that is > the case, then the maximum number of pairs would be > floor(3n - sqrt(12n-3)). For n of the form 3k^2 + 3k + 1, the points can be arranged in a > hexagonal shape, and in this case 3n - sqrt(12n-3) is an integer. Do you assume that the points are distinct ? Tim ==== >> Given integer n>=2 , what is the maximal number of pairs of points with >> Euclidean distance 1 in a set of n points in the plane ? >> I think the best you can do is pack the points in as vertices of a >> tesselation of the plane by equilateral triangles. If that is >> the case, then the maximum number of pairs would be >> floor(3n - sqrt(12n-3)). Do you assume that the points are distinct ? Yes, if you don't assume the points are distinct then the answer is very well-known. Turan's theorem tells us that the number of pairs is at most the number of edges of a complete tripartite graph on n vertices which is about n^2/3. If the points are not required to be distinct, this bound is trivially achievable: split the n points equally at the corners of an equilateral triangle, and we get a complete tripartite graph. -- Erick ==== I have the following setup. I have a particular test function f in D(R^n) and a distribution g (I don't know that g is L^1_loc). I'll denote the action of the distribtion on a test function phi as [g,phi]. I know that the integral int gf exists. Is this therefore the action of the distribution g on f? ie does int gf = [g,f]? Keep in mind that I don't know that g is in L^1_loc. All I know is that the integral exists and is finite. I hope this isn't a silly question. T ==== > I have the following setup. I have a particular test function f in > D(R^n) and a distribution g (I don't know that g is L^1_loc). I'll > denote the action of the distribtion on a test function phi as > [g,phi]. I know that the integral int gf exists. Is this therefore the action of the distribution g on f? ie does int gf = [g,f]? Keep in mind that I don't know that g is in > L^1_loc. All I know is that the integral exists and is finite. I hope > this isn't a silly question. The only way I know that functions become distributions is through integration, so I can't make any sense of this. Let's get precise here: What is g - a continuous linear functional on D(R^n) or a function on R^n? ==== I have the following setup. I have a particular test function f in >D(R^n) and a distribution g (I don't know that g is L^1_loc). I'll >denote the action of the distribtion on a test function phi as >[g,phi]. I know that the integral int gf exists. If g is not a locally integrable function then what sort of thing _is_ it, in order that there should even _be_ a function gf which has an integral? >Is this therefore the action of the distribution g on f? ie does int gf = [g,f]? Keep in mind that I don't know that g is in >L^1_loc. All I know is that the integral exists and is finite. I hope >this isn't a silly question. It's not a silly question. But it is a question about the technical details, so it has to be stated much more precisely - neither WWW nor I have any idea what you mean when you say the integral of gf exists although g is not locally integrable. T ************************ David C. Ullrich ==== > neither > WWW nor I have any idea what you mean when you say > the integral of gf exists although g is not locally integrable. Well actually we can make perfectly good sense out of that. We are told f is one particular test function; suppose it has compact support away from 0. Now set g(x) = 1/x on R {0}. Then fg is Lebesgue integrable over R although g is not locally integrable over R. My gripe is the OP hasn't specifed how g is on one hand a function and on the other a distribution. We now enter into the realm of: Guess what the OP is asking. (Why do we play this game? Isn't it enough that we answer the damned questions? LOL.) transform); in that case there is a function g lurking about. There's also a limiting process involved. And in that case the answer to the OP's question is obviously yes. But until we have the question posed more clearly we can't know for sure. ==== neither > WWW nor I have any idea what you mean when you say > the integral of gf exists although g is not locally integrable. Well actually we can make perfectly good sense out of that. We are told f > is one particular test function; suppose it has compact support away from > 0. Now set g(x) = 1/x on R {0}. Then fg is Lebesgue integrable over R > although g is not locally integrable over R. My gripe is the OP hasn't specifed how g is on one hand a function and on > the other a distribution. We now enter into the realm of: Guess what the OP > is asking. (Why do we play this game? Isn't it enough that we answer the > damned questions? LOL.) transform); in that case there is a function g lurking about. There's also > a limiting process involved. And in that case the answer to the OP's > question is obviously yes. But until we have the question posed more > clearly we can't know for sure. Hey Wade, good job of second-guessing the OP. I wonder if we'll hear again. ==== > neither >> WWW nor I have any idea what you mean when you say >> the integral of gf exists although g is not locally integrable. Well actually we can make perfectly good sense out of that. We are told f >is one particular test function; suppose it has compact support away from >0. Now set g(x) = 1/x on R {0}. Then fg is Lebesgue integrable over R >although g is not locally integrable over R. Yes, I missed the fact that we were talking about only one particular f. >My gripe is the OP hasn't specifed how g is on one hand a function and on >the other a distribution. We now enter into the realm of: Guess what the OP >is asking. (Why do we play this game? Isn't it enough that we answer the >damned questions? LOL.) transform); in that case there is a function g lurking about. There's also >a limiting process involved. And in that case the answer to the OP's >question is obviously yes. But until we have the question posed more >clearly we can't know for sure. Yup. ************************ David C. Ullrich ==== > Whether the first equality is legal is something we can't say > without more information, it seems to me. (I mean, at the > top you say that ^ is the _distribution_ FT. If that's what the > ^ in g^ means then int phi_j g^ does not _officially_ make > any sense, just as statements like h**1/2 g^ in L^2(R^n) > official make no sense... > I'm confused. Why doesn't this make sense? All I'm saying is that I have a function g that can also be viewed as a distribution. When I take the distributional FT of g then granted i get a distribution back. However, in this work, I know that h**1/2 g^ coincides with an L^2 function. > say something about how _if_ this and that distribution > is actually given by integration against this or that function... > now it turns out that you're not sure whether it is or not. > We can't say whether it is or not - whether or not g is > given by integration against some function is part of > the _definition_ of g.) >The >second equality here is legal by definition of the FT. The 3rd >equality is legal here as in this work I also know that g in L^1_loc. Aargh. Here's a quote from your first post: Keep in mind that I > don't know that g is in L^1_loc... > Sorry for this confusion but you should notice that the g in my first post is not the g of the second post. In fact in my first post you'll remember I didn't want to put too many details in and so steered clear of the FT. So, the g of the first post is actually the g^ of my second post. So there is no difficulty in the last sentence. The seemingly contradictory statement should read I do not know that g^ is L^1_loc. Hope that's cleared that up. Ta, T ==== >> Whether the first equality is legal is something we can't say >> without more information, it seems to me. (I mean, at the >> top you say that ^ is the _distribution_ FT. If that's what the >> ^ in g^ means then int phi_j g^ does not _officially_ make >> any sense, just as statements like h**1/2 g^ in L^2(R^n) >> official make no sense... I'm confused. Why doesn't this make sense? >All I'm saying is that I have a function g that can also be viewed as >a distribution. >When I take the distributional FT of g then granted i get a >distribution back. However, >in this work, I know that h**1/2 g^ coincides with an L^2 function. Exactly what does that last statement mean? One guess is that the distribution g^ is in fact given by a function, also denoted g^, such that h**1/2 g^ is in L^2. But if that's what you mean then the question that you snipped for our inconvenience, whether or not int phi_j g^ = [g^,phi_j] holds, doesn't make much sense - if g^ is given by integration against some function then int phi_j g^ = [g^,phi_j] _does_ hold. So that can't be what you mean. Otoh if you don't know that g^ is given by integration against some function then I don't know what function g^ you're referring to when you say that h**1/2 g^ coincides with an L^2 function. >> say something about how _if_ this and that distribution >> is actually given by integration against this or that function... >> now it turns out that you're not sure whether it is or not. >> We can't say whether it is or not - whether or not g is >> given by integration against some function is part of >> the _definition_ of g.) >The >>second equality here is legal by definition of the FT. The 3rd >>equality is legal here as in this work I also know that g in L^1_loc. >> Aargh. Here's a quote from your first post: Keep in mind that I >> don't know that g is in L^1_loc... Sorry for this confusion but you should notice that the g in my first >post is not >the g of the second post. For heaven's sake, how was one supposed to notice this? I give up. >In fact in my first post you'll remember I >didn't want to put too >many details in and so steered clear of the FT. So, the g of the first >post is actually the g^ of my second post. So there is no difficulty >in the last sentence. The seemingly contradictory statement should >read I do not know that g^ is L^1_loc. Hope that's cleared that up. Ta, T ************************ David C. Ullrich ==== >Sorry for this confusion but you should notice that the g in my first >post is not >the g of the second post. For heaven's sake, how was one supposed to notice this? I give up. > I concur, I've made quite a mess of things in my explanation. So, I think I'll back out of this thread gracefully now. I assure you that my intention wasn't to irritate you David. It's back to Rudin's Functional Analysis for me to get to grips with the subtleties of distributions which I clearly lack. Ta, T ==== >> My gripe is the OP hasn't specifed how g is on one hand a function and on >> the other a distribution. We now enter into the realm of: Guess what the OP >> is asking. (Why do we play this game? Isn't it enough that we answer the >> damned questions? LOL.) Apologies. I didn't want to give too many unnecessary details about my >problem, but looks like I gave away too little! Here is my problem. The actual integral is int_{R^n} h f^ g^ Here, ^ means (distributional) Fourier transform. I know that both f >and g are continuous functions that can be viewed as distributions. >They satisfy h**1/2 f^ in L^2(R^n) and h**1/2 g^ in L^2(R^n) so the integral certainly exists. What I'd really like to do is 'unhat >the g' so that I can rewrite the integral as int_{R^n} (h f^)check g. Here, check denoted the inverse Fourier transform (IFT). I am >assuming in this that (h f^) is L^2 so that I can take it's IFT in >fact in this work (h f^)check is assumed to have compact support. >However, the unhating is not legal (at first site at least) because g >is not L^2! I'm confident there is some distributional argument that >will make this move legal. My first thoughts were to approximate (h f^) by a sequence of test >functions phi_j then hope to write: int phi_j g^ = [g^,phi_j] = [g,phi_j^] = int phi_j^ g and then 'go to the limit'. Here, [ , ] is the action of a >distribution on a test function. I don't know if the first equality is >legal - this was intended to be the substance of my first post. Whether the first equality is legal is something we can't say without more information, it seems to me. (I mean, at the top you say that ^ is the _distribution_ FT. If that's what the ^ in g^ means then int phi_j g^ does not _officially_ make any sense, just as statements like h**1/2 g^ in L^2(R^n) official make no sense... say something about how _if_ this and that distribution is actually given by integration against this or that function... now it turns out that you're not sure whether it is or not. We can't say whether it is or not - whether or not g is given by integration against some function is part of the _definition_ of g.) >The >second equality here is legal by definition of the FT. The 3rd >equality is legal here as in this work I also know that g in L^1_loc. Aargh. Here's a quote from your first post: Keep in mind that I don't know that g is in L^1_loc... >I hope things have been made clearer. Am I on the right tracks? Ta, T ************************ David C. Ullrich ==== >>Sorry for this confusion but you should notice that the g in my first >>post is not >>the g of the second post. >> For heaven's sake, how was one supposed to notice this? >> I give up. I concur, I've made quite a mess of things in my explanation. So, I >think I'll >back out of this thread gracefully now. I assure you that my intention >wasn't to irritate you David. Don't worry about the irrittation - I guess I was in fact irritated, but just a little. But do note the part you snipped about why I don't see how various things you're saying make any sense: If you're assuming one thing then the question answers itself while if you're not assuming that then I honestly don't see what something else _means_... it's very possible I'm just being stupid there, but. >It's back to Rudin's Functional Analysis >for me to get to grips with the subtleties of distributions which I >clearly lack. Ta, T ************************ David C. Ullrich ==== This is a torus with 8 triangles at each vertex, 2 holes and 12 vertices. Are there coordinates for it in 3-space? (1 5 6) (1 6 7) (1 7 8) (1 8 9) (1 9 10) (1 10 11) (1 11 12) (1 12 5) (2 6 5) (2 7 6) (2 8 7) (2 9 8) (2 10 9) (2 11 10) (2 12 11) (2 5 12) (3 10 9) (3 9 12) (3 12 11) (3 11 6) (3 6 5) (3 5 8) (3 8 7) (3 7 10) (4 9 10) (4 12 9) (4 11 12) (4 6 11) (4 5 6) (4 8 5) (4 7 8) (4 10 7) ==== Oops, messed up. ==== This i think works. This is a torus with 8 triangles at each vertex, 3 holes and 12 vertices. Can this one sit in 3-space? > (1 5 6) > (1 6 7) > (1 7 8) > (1 8 9) > (1 9 10) > (1 10 11) > (1 11 12) > (1 12 5) (2 11 10) (2 10 5) (2 5 12) (2 12 7) (2 7 6) (2 6 9) (2 9 8) (2 8 11) > (3 10 9) > (3 9 12) > (3 12 11) > (3 11 6) > (3 6 5) > (3 5 8) > (3 8 7) > (3 7 10) (4 9 6) (4 6 11) (4 11 8) (4 8 5) (4 5 10) (4 10 7) (4 7 12) (4 12 9) ==== I have been going through some old statistics books, and trying to find the standard error of the estimate. I believe that I have Pinpointed it. I believe although that the equations that I came up with equal zero. Could someone tell me if they do equal zero? These equations would be to hairy to write out in text form, so here is a link to a pdf I created. http://mywebpages.comcast.net/spmoya/SE.pdf ==== > Suppose we denote by Omega(n,q) the set of all matrices of order nxn > over the field of cardinality q that have all their row and column > sums equal to 1. A natural name for Omega(n,q) will be, of course, > the doubly stochastic group over GF(q). (It's easy to check that > it's indeed a group). [...] > After deep and heavy thinking I realized that for Omega(n,q) you probably > meant only the invertible doubly stochastic matrices. With this understanding, and with the help of Mathematica and a > remarkable ability to do anything except what I am supposed to be doing, > I took a fairly close look at G = Omega(3,5), the 3 by 3 matrices over > Z_5. It has 480 elements, broken down as follows: [...] would appear that you have a sizeable job to figure out general things about > these Omega groups. The group Omega(n,q) obviously stabilizes the vector v=(1,...,1), and the hyperplane V={x | x_1+...+x_n=0}. And every matrix that stabilizes v and V is in Omega. As the only scalar matrix in Omega is the identity, it acts faithfully on the projective space PG(n-1,q). If p (here p is a prime such that p^m=q) does not divide n (i.e. v is not in V) then Omega is isomorphic to GL(n-1,q). (quick check: GL(2,5)=4S_5 is indeed of order 480) Otherwise it is q^(n-1):H, with H the stabilizer of a point in the natural action of PGL(n-1,q) on PG(n-2,q). Writing down generators, as asked here, is trivial in another basis, where v=(1,0,...,0) and V={x | =0} with u=v when p|n and u=(0,0,...0,1) otherwise. (and then conjugating by appropriate matrix) HTH, Dmitrii http://www.thi.informatik.uni-frankfurt.de/~dima/ PS. Just as another poster in the thread, quite proud of my ability to do things I'm not paid for, either :) ==== [...] > Suppose we denote by Omega(n,q) the set of all matrices of order nxn > over the field of cardinality q that have all their row and column > sums equal to 1. A natural name for Omega(n,q) will be, of course, > the doubly stochastic group over GF(q). (It's easy to check that > it's indeed a group). [...] > The group Omega(n,q) obviously stabilizes the vector v=(1,...,1), > and the hyperplane V={x | x_1+...+x_n=0}. And every matrix > that stabilizes v and V is in Omega. I assume you mean the hyperplane V={x | x_1+...+x_n=1}. > As the only scalar matrix in Omega is the identity, it > acts faithfully on the projective space PG(n-1,q). If p (here p is a prime such that p^m=q) > does not divide n (i.e. v is not in V) then Omega is isomorphic to > GL(n-1,q). (quick check: GL(2,5)=4S_5 is indeed of order 480) > Otherwise it is q^(n-1):H, with H the stabilizer of > a point in the natural action of PGL(n-1,q) on PG(n-2,q). Writing down generators, as asked here, is trivial in another > basis, where v=(1,0,...,0) and V={x | =0} with > u=v when p|n and u=(0,0,...0,1) otherwise. > (and then conjugating by appropriate matrix) HTH, > Dmitrii > http://www.thi.informatik.uni-frankfurt.de/~dima/ PS. Just as another poster in the thread, quite > proud of my ability to do things I'm not paid for, either :) Maybe you could post again and verify my correction. Also more details would be appreciated. ==== Is a continued fraction that extends in both directions a Mathematical object? -6 -5 -4 -3 -2 -1 0 1 2 3 4 <- --- --- --- --- --- --- --- --- --- --- --- -> -4 -3 -2 -1 0 1 1 2 3 5 8 The right side is the irrational of the relationship of Fibonacci count to the discrete time of the dynamical systems of [A(x)+y,x/2] ( sorry if the notation is wrong; I mean the family the Collatz problem belongs to ) Discrete time over Fibonacci count T/F is what I am trying to say. I am guessing that expresses the valid relationship. So is it a real object? any references to such works? Way back in October 2001 I was fooling around with T/F By skipping the three {-2/0,-1/1,0/1} and adding the opposite systems together a nifty result is the infinite series of 3/2, 4/3, 5/5, 6/8, 7/13 and so on... It's an interesting idea to me that it might be a ternary system structure using null infinity, negative infinity and positive infinity; however, I know I ought to get the opinion of those more skilled than I on this. So is it a possible that two irrational fractions are somehow a part of ternary system? Yes I did assume the irrational fraction on the left. That was my guess of how it might be. You might try a different fraction. It's just that I like that one. They all form systems of their own as well.. So Am I fooling myself with this? Ernst ==== I'm not following your lingo very well, but i tried some thing like that, some time ago: just reverse the operation of finding F-numbers, by subtraction. on the other side, the ration (as i recall) tends to 0.618... with an alternating sign. > Is a continued fraction that extends in both directions a > Mathematical object? -6 -5 -4 -3 -2 -1 0 1 2 3 4 > <- --- --- --- --- --- --- --- --- --- --- --- - -4 -3 -2 -1 0 1 1 2 3 5 8 > The right side is the irrational of the relationship of Fibonacci > count to the discrete time of the dynamical systems of [A(x)+y,x/2] ( > sorry if the notation is wrong; I mean the family the Collatz problem > belongs to ) --Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac ==== Is a continued fraction that extends in both directions a > Mathematical object? .....[sniped] I do not know what you want, exactly, since all else in your post is outside my understanding, perhaps. But as for continued fractions going in both directions: Yes, these can indeed exist, I would guess. If we have such a CF which converges in both directions, [...a(-3),a(-2),a(-1),a(0),a(1),a(2),a(3),...], We can let x = [a(0),a(1),a(2),a(3),...]. Then we can let c(m) = [a(-m),a(m-1),...,a(-1),x], where, for convergence, I would guess that the a's with negative index could not always all be positive integers. (all a's = 1 works, however) Then, if c(m) approaches a limit, it is the value of the two-way CF.... Maybe. I have not seen this kind of CF before, so my assumptions are completely unrigorous. If we arbirarily let a(0) be any term within the CF, then will the same final evaluation of the CF occur? And under which restrictions on the sequence of a's does this kind of CF converge, and converge to the same value whatever term we arbitrarily decide will be a(0)?? Leroy Quet ==== > [...a(-3),a(-2),a(-1),a(0),a(1),a(2),a(3),...], > We can let x = [a(0),a(1),a(2),a(3),...]. > Then we can let c(m) = [a(-m),a(m-1),...,a(-1),x], They look fun but they get you nothing new. Just recursive sequences. > And under which restrictions on the sequence of a's does this kind of > CF converge, Simple enough. The CF converges iff the RH part converges (i.e. x exists) and also a(-n) --> some finite limit. ---------------------------------------------------------------------------- -- Bill Taylor W.Taylor@math.canterbury.ac.nz ---------------------------------------------------------------------------- -- The three stages of sexual activity: (1) Tri-weekly; (2) try weekly; (3) try weakly. ---------------------------------------------------------------------------- -- ==== for the solution of ODE y''(x) - 7y'(x) + 12y(x) = 0, after the substitution z=dy/dx, we have z(dz/dy)-7z+12y = 0. For z=/=0 we have (dz/dy) - 7 + 12/(z/y) = 0 and after the substitution w=z/y we have y(dw/dy) = 7- 12/w -w which by separation of variables gives (w/(7w-12-w^2))(dw/dy) = 1/y and by intergration leads to the expression (w-3)^3 = cy((w-4)^4), where c=/=0 is constant. So the problem is that for the solution I have to solve a non-easy quadratic equation... The question: is there any easier way to solve this ODE? Grigorios Kostakos ==== > for the solution of ODE > y''(x) - 7y'(x) + 12y(x) = 0, > after the substitution z=dy/dx, we have > z(dz/dy)-7z+12y = 0. > For z=/=0 we have > (dz/dy) - 7 + 12/(z/y) = 0 > and after the substitution w=z/y we have > y(dw/dy) = 7- 12/w -w > which by separation of variables gives > (w/(7w-12-w^2))(dw/dy) = 1/y > and by intergration leads to the expression > (w-3)^3 = cy((w-4)^4), where c=/=0 is constant. > So the problem is that for the solution I have to solve a non-easy > quadratic equation... > The question: is there any easier way to solve this ODE? > assume that y = e^(rx), substitute this into your original equation, and see what conditions r must satisfy... --bill ==== > for the solution of ODE > y''(x) - 7y'(x) + 12y(x) = 0, This is a linear DE with constant coefficients. The solutions should be of the form y = exp(ax). Substitute this into the equation and solve for a. That will give you two roots a1 and a2. So the general solution will be y = C1 exp(a1 * x) + C2 exp (a2 * x) where C1 and C2 are constants usually determined by the initial conditions. > after the substitution z=dy/dx, we have > z(dz/dy)-7z+12y = 0. > For z=/=0 we have > (dz/dy) - 7 + 12/(z/y) = 0 > and after the substitution w=z/y we have > y(dw/dy) = 7- 12/w -w > which by separation of variables gives > (w/(7w-12-w^2))(dw/dy) = 1/y > and by intergration leads to the expression > (w-3)^3 = cy((w-4)^4), where c=/=0 is constant. > So the problem is that for the solution I have to solve a non-easy > quadratic equation... > The question: is there any easier way to solve this ODE? Grigorios Kostakos ==== >Message-id: <3c65f87.0307071855.427a0d67@posting.google.com How many of you gave a damn about the bombs that dropped on a wing of >a hospital in Iraq for pregnant women? Two birds with one stone. No, you'd just care if it happened in your town, You're not from New York, are you? But you still pray, now don't you? WHEN I WAS A CHILD, I spake as a child, I understood as a child, I thought as a child: but when I became a man, I put away childish things. James Harris -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm ==== >> Theorem. Let >> f(x) = a_n x^n + a_n-1 x^(n-1) + ... + a_0 >> be a polynomial with integer coefficients. If there exists a prime >> number p such that >> a_n-1 = a_n-2 = ... = a_0 = 0 (mod p) >> but a_n not congruent 0 (mod p) and a_0 not congruent 0 (mod p^2), >> then f(x) is irreducible over the field of rational numbers. Ok, not so hard to prove. > Corollary. If p is prime, then the polynomial >> phi(x) = x^(p-1) + ... + x + 1 >> is irreducible over the field of rational numbers. Egads, how does the corollary follow from the theorem? Hard to say :-) -- Fundamental Theorem of Algebra >> Any polynomial of positive degree with complex coefficients >> has a complex root. What method is used to show the fundamental theorem? Lots are possible (Gauss found 4, essentially distinct) Simplest use Liouville's theorem (any bounded analytical function is constant) > The related material below seems inadequate. DeMoivre's Observation. >> For any positive integer n, the equation z^n = 1 >> has n distinct roots in the set of complex numbers. > cos 2k.pi/n + i sin 2k.pi/n, for 1 <= k <= n. Kronecker's Theorem. >> Let F be a field and f(x) a nonconstant polynomial in F[x]. >> Then there exists an extension field E of F >> and an element u in E such that f(u) = 0. Show any such extension field of the complex field C is C? > Ouch, doesn't seem the way to go. Well, it is possible (if you add the idea that any odd degree polynomial of R[X] has a real root : Euler-Lagrange, with a twist) but certainly not obvious (as the fist was well-known since deMoivre :-), and the second was admitted as obvious in Euler and Gauss times) ---- ==== I am planning to teach Algebra for my daughter and am planning to buy the following book from an online used book shop (B&N, Abe etc.) Elementary Algebra for Schools By Hall & Knight. --- It was recommended to me by a friend as one of the best books on the subject. After doing a little bit of search, I found that there is a 3 volume set by the same author as given below: Elementary Algebra for Schools Vol I, II, III by Hall & H.S. Wood. I have no access to the table of contents of either of the above two books and I would like to know if they are essentially the same in content. If not, which one is better for starting on Algebra? TIA, rajanish... ==== Check out amazon.com, half.com, and bn.com cause they normally have good reviews, prices, and sometimes a preview of the book. Good luck. -drew I am planning to teach Algebra for my daughter and am > planning to buy the following book from an online > used book shop (B&N, Abe etc.) Elementary Algebra for Schools By Hall & Knight. > --- It was recommended to me by a friend as one of the > best books on the subject. After doing a little > bit of search, I found that there is a 3 volume > set by the same author as given below: Elementary Algebra for Schools Vol I, II, III > by Hall & H.S. Wood. I have no access to the table of contents of either > of the above two books and I would like to know > if they are essentially the same in content. If not, > which one is better for starting on Algebra? > TIA, rajanish... > ==== Has anyone translated or attempted to translate English sentences into math equations? For example, sky is blue translated into sky = blue. Yes, I know that equation isn't accurate. It's just to give an idea of what I mean. If someone has done so, I'd appreciate knowing the URLs on this topic -- Like a cure for A.I.D.S., Alzheimer, Parkinson, & Mad Cow Disease? Volunteer your computer for folding-protein research for when it's idle. Go to http://www.distributedfolding.org/ to sign up your computer. ==== > Has anyone translated or attempted to translate English sentences into math > equations? For example, sky is blue translated into sky = blue. Yes, I > know that equation isn't accurate. It's just to give an idea of what I > mean. If someone has done so, I'd appreciate knowing the URLs on this topic > This happens all the time, but not as numerical algebra but as predicates in formal languages or set theory. Do note I've crossed post this thread into sci.logic where you may get some chatter about mathematical logic. Thus we may write Blue(sky), sky has the property Blue or sky in Blue, sky is member of the set of blue things. ==== >Has anyone translated or attempted to translate English sentences into math >>equations? For example, sky is blue translated into sky = blue. Yes, I >>know that equation isn't accurate. It's just to give an idea of what I >>mean. If someone has done so, I'd appreciate knowing the URLs on this topic >> >This happens all the time, but not as numerical algebra but as predicates >in formal languages or set theory. Do note I've crossed post this thread >into sci.logic where you may get some chatter about mathematical logic. Thus we may write > Blue(sky), sky has the property Blue >or > sky in Blue, sky is member of the set of blue things. > > Kind of related, but I wonder if anybody has ever attempted to *model* the universe through time by: 1) let U = {(x,y,z,t,p)}, where x,y,z,t,p are reals, and (x,y,z) are points in 3-dimension Euclidean space for a given t, and p is just the property of the corresponding (x,y,z,t) 2) axiomatize x,y,z,t,p in a meaningful way, if not close to a famliliar physics model, say Newton physics. ? ==== This is a special case of a general theorem, an easy consequence but noteworthy. Let A = {a(k)} and B ={b(k)} be two infinite sequences of DISTINCT POSITIVE integers. A and B and {c(k)} are also such that the limit(s) below converge (and the infinite sum {*} below converges absolutely). Then: m --- --- limit m -> oo / / c(a(j)) *j --- --- k=1 j|k a(j) is element of B = m --- --- limit m -> oo / / c(b(j)) *j --- --- k=1 j|k b(j) is element of A (a(j) is already understood as being part of A, and b(j) of B, by definition.) In linear-mode: limit{m->oo} sum{k=1 to m} sum{j|k, a(j) is element of B} c(a(j)) *j = limit{m->oo} sum{k=1 to m} sum{j|k, b(j) is element of A} c(b(j)) *j This is because both = {*} sum{k= element of {A and B}} c(k), if this sum converges absolutely. (This seems more trivial than it is, perhaps, until you notice those 'j's multiplied by the c's in the averages' general terms.) A specific example: If a(j) = j^2, b(j) = j, c(j) = 1/j, then: we get that: limit{m->oo} (1/m) sum{k=1 to m} sum{j|k} 1/j = limit{m->oo} (1/m) sum{k=1 to m} d'(k) = pi^2/6, where d'(k) is the number of divisors of k which are perfect-squares. Leroy Quet ==== Equivalent random variables implies that P[X neq Y] = 0 However I don't understant completely this definition. Do you have a different one? Does it implies that... P[x = y] = 0 for some distributions? Diego Andr.8es ==== Equivalent random variables implies that P[X neq Y] = 0 However I don't understant completely this definition. Do you have a > different one? Does it implies that... P[x = y] = 0 for some distributions? > Suppose X is uniformly distributed in [1,2] and Y is uniformly distributed in [3,4]. Then of course P[X=Y] = 0 because X=Y is never true. Or let X be uniformly distributed in [0,1] and let Y=-X. Then the event [X=Y] is the event [X=0] which has probability zero, and in this case P[X=Y]=0 again. Even though that event is not empty, it still has probability zero. ==== Greetings, Could anyone give me an example of a stochastic process (or random process) that is ergodic but not stationary? Sincerely, rjc. -- Ryan Cassidy Electrical Engineering Graduate Student Stanford University ==== the ellipsoid-method was the first polynomial algorithm to solve linaear programmming problems. I have some problems estimating the worst-case running time of the algorithm: The initialization (a) k:=1; (b) N:= 2n( (2n+1) C + n b -n^3); (c) A0:= R^2I mit R:= n^(1/2) 2^(C,b -n^2); (d) a0:= 0; (k index, N maximumm number of iterations, n number of variables, C mxn-matrix, encoding length, A0 matrix defining the shape of the ellipsoid, a0 center of the ellipsoid, R radius, I Einheitsmatrix) needs according to one book O(n^2), according to another one O(m*n) elementary steps. Which one is right and why?. I have also problem with C,b, the encoding length of C and b. Is that the number of cells needed to binary represent C and b (in the sense of a sum)? (A small example would be great.) I think it's O(m*n) because for every entry of the mxn-matrix C the encoding length has to be calculated, right?. This means O(n^2) would be wrong because m>=n. Eva ==== [snip] > It seems to me that clearing you out of the discipline is necessary. Harrigance - the guy on the balcony strikes again: http://users.pandora.be/vdmoortel/dirk/Stuff/Arrogance1.jpg http://users.pandora.be/vdmoortel/dirk/Stuff/Arrogance2.jpg Dirk Vdm ==== If we choose randomly and uniformly two points p,q in the n-dimensional unit cube in R^n what is the expected Euclidean distance d(p,q) ? ==== >If we choose randomly and uniformly two points p,q in the n-dimensional >unit cube in R^n what is the expected Euclidean distance d(p,q) ? I can give you a method of computing it numerically with one integration and a reasonable amount of computing. The moment generating function of the square of the distance is m(t) = (6(exp(t) - 1 - t - t^2/2)/t^3)^n, and the Laplace transform is just m(-t). Now compute int -m'(-t)/sqrt(t) dt/sqrt(pi), the integral going from 0 to infinity, and this is the answer. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Deptartment of Statistics, Purdue University ==== If we choose randomly and uniformly two points p,q in the n-dimensional > unit cube in R^n what is the expected Euclidean distance d(p,q) ? If you only need an approximate value, note that a Taylor expansion around the mean gives: E sqrt(Z) =~ sqrt(mu) - 1/(8*mu^(3/2)) * sigma^2 for a random variable Z with mean mu and variance sigma^2. The sum Z of squared distances in each component has mean n/6 and variance 7n/180 giving an approximate expected distance of: sqrt(n/6) - (6/n)^(3/2) * 7*n/1440 Simulation shows this approximation appears to be good to within about 1% for all n. -- Kevin ==== I'm looking for a function g such that for all positive integer a, b, c we have that Integral(0 -> 1) { g(ax)*g(bx)*g(cx) } dx = 1 if a = b = c = 0 otherwise To clarify, the integrand is the product of the three quantities g(ax), g(bx) and g(cx) (where, as usual, ax denotes the product a*x) and the limits of the integral are 0 and 1. ==== > Integral(0 -> 1) { g(ax)*g(bx)*g(cx) } dx = 1 if a = b = c > = 0 otherwise Are you familiar with the Dirac delta function? This is an improper function (strictly speaking a funcional) that satisfies: Integral(-infinity -> infinity) { f(x) delta(x) } dx = f(0) Informally, delta(x) is zero everywhere except for an infinite spike at zero. Let w be an irrational number greater than zero and less than one. w = 1 / sqrt(2) will do as an example. Let h(x) = Sum(k = 1 -> infinity) { delta(x-k) } Informally, h has a spike at every positive integer, and is zero elsewhere. Now g(x) = cuberoot( h(x - w) ) For any positive integers a and b, g(ax) g(bx) = 0. For any positive integer k, g(kx)^3 has k spikes between zero and one, but each is only 1/k times as wide as the original delta. Now, if you don't like delta functions, you can do the same thing with very narrow rectangular spikes, provided a, b and c are not too large. ==== After completing a degree in Computer Science, I realised I absolutely sucked at Math. I know I have the ability, so I want to learn. Can anyone point me in the direction of any online material before I go a-hunting at the library? Any idea's where the FAQ for this group went? ==== > After completing a degree in Computer Science, I realised I absolutely > sucked at Math. I know I have the ability, so I want to learn. Can anyone point me in > the direction of any online material before I go a-hunting at the > library? Any idea's where the FAQ for this group went? > What math do you already know? The FAQ is quack and won't be back. ==== > After completing a degree in Computer Science, I realised I absolutely > sucked at Math. I know I have the ability, so I want to learn. Can anyone point me in > the direction of any online material before I go a-hunting at the > library? Any idea's where the FAQ for this group went? I think Wikipedia, an on-line encyclopedia of sorts, is pretty good. It is searcheable, and also many pages have links to definitions, related areas, etc. The main page of Wikipedia is at: http://www.wikipedia.org/wiki/Main_Page David Bernier ==== [...] >> I know I have the ability, so I want to learn. Can anyone point me in >> the direction of any online material before I go a-hunting at the >> library? Any idea's where the FAQ for this group went? > I think Wikipedia, an on-line encyclopedia of sorts, is pretty > good. It is searcheable, and also many pages have links to > definitions, related areas, etc. The main page of Wikipedia is at: http://www.wikipedia.org/wiki/Main_Page For example, you might want to start here: http://www.wikipedia.org/wiki/Mathematics David Bernier ==== > After completing a degree in Computer Science, I realised I absolutely > sucked at Math. I know I have the ability, so I want to learn. Can anyone point me in > the direction of any online material before I go a-hunting at the > library? Any idea's where the FAQ for this group went? Online stuff is so untrustworthy. For example, as much as I like Eric Weisstein's MathWorld, there are lots of things in it that are marginal, and lots of things that are just plain wrong. Same with Wikipedia, I have found Have you got an idea of what kind of math you would like to start with? If you do, I would recommend finding a book in your chosen area and concentrating on it. If you're not sure where you ought to start, let me recommend linear algebra. ==== > I hope you don't mind my thinking out loud, it helps me keep this > stuff straight. Not at all. > So I think we basically are doing the same thing (assuming I haven't made > some mistake in the above), just from different directions. Yes. Another way of looking at it is to choose all possible binary numbers of length 11 containing exactly 7 1's (thus also 4 0's). There are C(11,7) of them (i.e. I am allowing the high-order bit to be 0). Suppose that we number the positions from 0 to 10 in the natural way, i.e. the rightmost bit corresponding to position 0 and the leftmost to position 10. The vector of length 7 which lists in ascending order the positions of the 7 1's in the original vector will be a reduced vector in the sense in which I am using it. Another way of viewing the succ() function is that with val(f) = succ(val(e),b), if we expand e and f into binary numbers of length 11 containing 1's at the positions designated by the components of e and f, then the operation of succ() is to rotate the binary number one position to the right, with the rightmost bit being shifted into the leftmost position. This accounts for the periodicity of the succ() function for val(e). The binary number here is equivalent to the parity vector obtained by reducing the elements of the sequence generated by succ(val(e),b) mod 2. I came across this characterization of the loops, or attractors, more or less the same way you did. When I searched for loops for b = -139, I found 30 of them, which was far more than I had found for any smaller value of b. Also, the numbers in the 30 loops were all in the range 2059 (= 3^7 - 2^7) to 16*2059 = 32944. Analyzing these numbers, I came up with the observations in my original message. Now, consider any y and c such that the sequence generated starting with succ(y,c) is periodic with length s. This means that if we define y[0] = y, y[i+1] = succ(y[i],c) for i=0..s-1, then y[s] = y[0] = y. Consider the numbers z[i] = y[i]*2^i. These satisfy a recursion related to succ(y,c), except that at even steps, i.e. those with y[i+1] = y[i]/2, we will have z[i+1] = z[i], while at odd steps, i.e. those with y[i+1] = (3*y[i]+c)/2, we will have z[i+1] = 3*z[i] + c*2^i. Suppose next that there are t odd steps and (s-t) even steps, and suppose that the odd steps occur at i = e[1], i = e[2], ..., i = e[t] Then we will have z[s] = 3^t*z[0] + c*(2^e[1]*3^(t-1) + ... + 2^e[t]) But we also have z[s] = y[s]*2^s = y[0]*2^s = z[0]*2^s, thus y[0]*(2^s - 3^t) = c*(2^e[1]*3^(t-1) + ... + 2^e[t]) We can then substitute y = y[0], and we recognize the second multiplicand on the right as val(e), which gives y = c*val(e)/(2^s - 3^t) This is the most general solution to the question, for which values y is succ(y,c) periodic? The particular solution that I gave is for c = 2^s - 3^t, y = val(e). ==== > I hope you don't mind my thinking out loud, it helps me keep this > stuff straight. Not at all. > This is the most general solution to the question, for which values y > is succ(y,c) periodic? The particular solution that I gave is for c = > 2^s - 3^t, y = val(e). > Hey what's up? Are you guys looking for an equation for all numbers go to 1 ? If so how do you guys frame such a question? I'm still planning my next Hailstone generator. So I cound add things to it and follow what you guys are doing if yall will share. Here is a quip :) Dynamical Blind Men exploring infinities of infinite Elephants may not be sure how many of the same Elephants they have touched in the past week if they were alert. The ever learning Ernst :) ==== >Original Format >> I hope you don't mind my thinking out loud, it helps me keep this >> stuff straight. >> Not at all. >> This is the most general solution to the question, for which values y >> is succ(y,c) periodic? The particular solution that I gave is for c = >> 2^s - 3^t, y = val(e). Hey what's up? We're discussing how to find attractors. I assumed you were following along, after all, you asked the original question. If you need to, ask questions. Are you guys looking for an equation for all numbers go to 1 ? > I can't speak for Bill Daly, but I will say: no. There is no single equation. What we're talking about are specific equations that result in loops. If you can find an equation of a loop for some positive number other than b in a 3a+b system, then the Collatz Conjecture is false for that system. For example, 3a+1 has only one known loop in the posiitve numbers: 4 2 1 4 2 1 4 2 1 ... so the conjecture is still open for 3a+1. Remember that not proven false does not mean something is true. But 3a+5, in addition to it's b loop (which every system has) 5 20 10 5 ... also has positive loops 1 8 4 2 1 ... 19 62 31 98 49 152 76 38 19 ... 23 74 37 116 58 29 92 46 23 ... so the conjecture is false for 3a+5. At this point, you may well wonder what makes 3a+5 different from 3a+1? So we study the equations of known loops to see if they provide some insight that we can apply towards 3a+1. Maybe we haven't found another loop for 3a+1 because all we've checked are small loops. Someone says that the smallest loop would contain a quarter million numbers. So we're dealing with numbers that can't be reached by brute force. That's where induction helps. As an example, I asked myself can two Mersenne Numbers appear in a 3a+1 sequence? (Mersenne Numbers are numbers of the form 2^n - 1. In binary, these are numbers that are all 1s). By induction, I can show that - all Mersenne numbers with an even number of bits == 0 (mod 3) - all Mersenne numbers with an odd number of bits == 1 (mod 3) Knowing that a 3a+1 sequence can only have a single odd number that is == 0 (mod 3), the above proves that you can't have a sequence with two even-bit Mersenne Numbers. There is no need to do a brute force test of all the numbers. Note that that is only half a proof, it doesn't say anything about having two odd-bit Mersenne Numbers in a sequence. Or an odd-bit following an even-bit. But you also can't have an even-bit follow an odd-bit, because every predecessor of every number == 0 (mod 3) is an even number and all Mersenne Numbers are odd. If we can't find the counter-example by brute force, maybe we can construct it. But to do that, we need to understand how loops are formed, hence, the interest in Hailstone Function Generators, parity vectors, etc. But is it likely that a huge loop exists? I don't know, but some very simple questions can lead to very large numbers. For instance, take the 3a+1 sequence 27 82 41 124 62 31. Note that this sequence - ends in a Mesenne Number (2^5 - 1 or 11111 in binary) - has a parity vector 10100 Let's call 31 the Hailstone (H) and 27 the Seed (S). We say that 27 is a solution to the function given by S10100M. There are an infinite number of such solutions, 31 just happens to be the first. Now if we treat the parity vector as a rep-unit, then S10100M is a First Generation Type 10100 Mersenne Hailstone Function. A second generation function has two rep-units: S1010010100M. Now here is the simple question What is the first 6th Generation Type 10100 Mersenne Hailstone? In other words, find M for S101001010010100101001010010100M. Simple question, very difficult answer. I put together my Hailstone Function Generator to answer that very question. M is gazillions of times larger than any number that can be reached by brute force, but can easily be computed from the Hailstone Function Generator: M = 2^177149 - 1 This number has over 53,000 decimal digits, and yes, I did verify it by running it through the Collatz process (if I remember correctly, over 2 million iterations). So just because every number has been tested up to 10^15 and just because the smallest possible loop would have a quarter million iterations, the brute force approach has not proved that a gigantic loop is impossible. Another interesting critter I found is what I call a 6th Generation Quadracycle Pulsar with 4-bit Rep-unit and 2-bit Resonator. This requires 4098 bits to construct and is also beyond the range of brute force attacks. What makes a Pulsar interesting is that the binary pattern of the rep-units is regenerated during the Collatz run. Alas, the rep-units are consumed faster than they are regenerated, otherwise I would have found the other way to falsify the conjecture: a number that goes to infinity. By studying the Hailstone Functions, crossover points, attractors, etc., there may be a simple formula with REALLY large constants that form a loop in 3a+1. Or maybe there's a REALLY large Pulsar that diverges. And no, I don't have any delusion that I'll ever find any, but I'm enjoying the journey. If so how do you guys frame such a question? I'm still planning my >next Hailstone generator. So I cound add things to it and follow what >you guys are doing if yall will share. It's not a question of being unwilling to share. I can't tell from your replies whether you 'get' what I write. The above is a summary. If you want more details, you only have to ask. I'm not going to waste time writing out detailed explanations if you're not interested. Here is a quip :) Dynamical Blind Men exploring infinities of infinite Elephants may >not be sure how many of the same Elephants they have touched in the >past week if they were alert. The ever learning Ernst :) ==== >Original Format > I hope you don't mind my thinking out loud, it helps me keep > this >> stuff straight. >> Not at all. >> It's not a question of being unwilling to share. I can't tell from > your replies whether you 'get' what I write. The above is a summary. > If you want more details, you only have to ask. I'm not going to waste > time writing out detailed explanations if you're not interested. > Oh.. Cool... . Oh I'm interested in learning. I can say I don't understand your goal but, I don't have the training you do. Note : Trigger words and phrases.. Like 'get' suck... Do you know about trigger words and phrases? I will press on with the refit. I have a guess. Ernst ==== >Original Format > >> I hope you don't mind my thinking out loud, it helps me keep > this >> stuff straight. >> Not at all. >> It's not a question of being unwilling to share. I can't tell from > your replies whether you 'get' what I write. The above is a summary. > If you want more details, you only have to ask. I'm not going to waste > time writing out detailed explanations if you're not interested. > Oh.. Cool... . Oh I'm interested in learning. Good. I am not claiming that I know anything profound, but as you can see, that doesn't stop me from going on endlessly about it. > I can say I don't understand your goal Well, my goals probably wouldn't interest you, and they are not always obvious. For example, you might think that this web page has something to do with the Collatz Conjecture: http://members.aol.com/mensanator666/collatz/bang0010.htm But it does not. The goal here was an excercise in writing a perl program to generate slides used to create .gif animations. It just so happened that I was learning about binary patterns in the Collatz Conjecture at the time, so it became the subject of my animation study. Note that there is nothing on that page that gives you any clue as to what my real motivation was. > but, I don't have the training you do. I have no training. I am entirely self-taught. As I have said, I am just a hobbyist. A well read hobbyist. Note : Trigger words and phrases.. Like 'get' suck... Do you know > about trigger words and phrases? Do you mean - a word that calls up a repressed experience? - or something else? I will press on with the refit. I have a guess. > Ernst ==== > You know I am with you. I see more questions than answers but oh what > questions they are. :) > I'm now at over load point. I'll need time to recharge. > others. > I missed some of your points and made errors due to my hurry. > Yet, since I have risked talking to others I have gone from thinking > 3x+1 was the only system to x+y and x-y, GCD's, rings, dynamical > systems and more. Not bad even if made me look foolish. The fools are those who don't learn from their mistakes. > Ernst >>Hey would you check [5x+7,x/2] That is what is called Divergent right? >Ok, but it will be tomorrow night before I can get the web page updated. > > Forget everything I've previously said. > I need to take a closer look at my program, it may not be doing what I think > it's doing. > I didn't check _every_ one of the 2765 attractors I found in 3a+b, but every > one I checked seemed legit. But then I noticed some funny things. > For 3a+5, three of the attaractors found were, with parity vector: > 19 10101000 > 31 10100010 > 49 10001010 > but these three are actually part of the same loop > 19 62 31 98 49 152 76 38 19 > If we actually extend this loop to more than one cycle > 19 62 31 98 49 152 76 38 19 62 31 98 49 152 76 38 19 62 31 98 49 152 76 38 > 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 > 1 0 1 0 0 > we get a different parity vector depending on which odd number we start on. And > in the course of testing every permution of parity vector, the program > encountered this loop from each of the three possible starting numbers. And > since it really is part of a loop, that must be why I got an integer crossover. I've changed my program. After finding an integer crossover point, I calculate the rest of the odd numbers in the sequence and record the minimum one. I get a record that looks like: system: [3x+5] crossover: 31 attractor: 19 parity vector: [1, 3, 1] (The parity vector list just shows the count of 0s in the vector. It is implied that there is a single 1 to the left of each block of 0s.) The 19 62 31 98 49 152 76 38 19 sequence will also produce two other integer crossover points: system: [3x+5] crossover: 19 attractor: 19 parity vector: [1, 1, 3] system: [3x+5] crossover: 49 attractor: 19 parity vector: [3, 1, 1] This gives me the best of both worlds. I've three different crossovers/parity vectors if I'm looking for patterns there, but a single attractor if that's what I'm interested in. > Even stranger things are happening in 5a+b. I'm finding attractors that lead > into a loop without actually being part of it > In 5a+5 I have 5 and 31 as attractors, but the sequence is > 31 160 80 40 20 10 5 30 15 80 40 20 10 5 30 15 80 40 20 10 5 > I'm not sure what to make of that. What happened here was that I inadvertently recorded 3a+b data as 5a+b. Naturally, when I put the 3a+b attractors into the 5a+b Collatz process, strange things resulted. Strange because 5 just happens to be an attractor in BOTH 3a+5 AND 5a+5, so the above sequence almost works. Good lesson here on the importance of verifying your computer programs. It was an operator error, not the program's. I told it to compute 3a+b and that's what it did. It wasn't the program's fault that I directed the output to the 5a+b file. > I won't be putting up anymore charts and will remove the ones that are there > now. ==== I have found a structure to the attractors. There is more to do but the basic structure I see. Ernst ==== I want to find a 3d point (x, y, z) in the 3d Cartesian space using the following data that I have: - Cylinder parameters: two 3d points that describes cylinder axis, let call these p0 and p1; - Radius of the cylinder, let call R; Here is some sort of visualization - the view of the cylinder from the side and the 3 points and radius: p2------------ | | R R | | p0-----------p1 | | R R | | -------------- I hope that this characters makes the picture more clear ;-))) I have p0, p1 and R, and I want to find coordinates of p2 - this point should be on the surface of the cylinder and on the edge. I know I have to supply the angle of revolution of the vector from p0 to p2. How can I find this vector - it should be tangent to vector( p0, p1 ), so the dot product should be equal to 0. Here is some of my equations: (1) (x2 - x0)*(x1 - x0) + (y2 - y0)*(y1 - y0) + (z2 - z0)*(z1 - z0) = 0 This is dot product of the tangent vectors. (2) (x2 - x0)^2 + (y2 - y0)^2 + (z2 - z0)^2 = R This is distance in 3d from p2(the unknown quantity) I have two equations with 3 unknown variables - coordinates of p2: x2, y2, z2. ==== > I have a list of 8 points (not necessarily integers) and the value of a >> function at those 8 points, taken modulo 17. I'm trying to recreate the >> function (which has degree 8). Is this possible or do i have to do the >> evaluations without the mod? >> What ring are you in? >> Modulo 17 and not necessarily integers clash somewhat. >> You can perform division modulo 17, so if your values are rationals, then >> they're actually just integers (or undefined). Is that necessarily true? Things aren't true out of necessity. Which bit do you disagree with? The terminology's outragiously slack, but if you're prepared to identify 0 with 0 (mod 17) it should be obvious what's meant. > Can't you work in, say, R/17Z and end up with > values in [0, 17)? Take care with that structure - What's half of 7? Is it 3.5 or is it 12, as 2*12=7? OP didn't say it was R, only that it was not Z. I offered some insight if it were in fact Q, which does not have the problem that the structure you mention above has. Phil ==== William, Be careful, 2^9 /= 2^10 = 1024 number by asking yes/no questions of Bob. Mary knows that Bob always > tells the truth. If Mary uses an optimal strategy then she will > determine that answer at the end of exactly how many questions on the > worst case? This is same as searching an alphabetized table of size n, namely > ceiling log_2 n. In your case 2^9 = 1024, ceiling log_2 1000 = 9. ==== > number by asking yes/no questions of Bob. Mary knows that Bob always > tells the truth. If Mary uses an optimal strategy then she will > determine that answer at the end of exactly how many questions on the > worst case? > This is same as searching an alphabetized table of size n, namely > ceiling log_2 n. In your case 2^9 = 1024, ceiling log_2 1000 = 9. 2^9 = 1024? Hmm, got to fix my calculator! -- For suitable values of 9. ==== > William, > Be careful, 2^9 /= 2^10 = 1024 Whoops lost count, 2*2*2*..*2 = 1024. Recalling log 2 log_2 1000 = log 1000 / log 2 = 3/.301030 = 9.966 > number by asking yes/no questions of Bob. Mary knows that Bob always > tells the truth. If Mary uses an optimal strategy then she will > determine that answer at the end of exactly how many questions on the > worst case? > This is same as searching an alphabetized table of size n, namely > ceiling log_2 n. In your case 2^9 = 1024, ceiling log_2 1000 = 9. > ==== > Yes, and an r-generator finite abelian group is a product of > r cyclics. This drops out immediately from the theory of the Smith > normal form. Which Smith? -- Timothy Murphy tel: +353-86-233 6090 ==== > Yes, and an r-generator finite abelian group is a product of >> r cyclics. This drops out immediately from the theory of the Smith >> normal form. Which Smith? Is there more than one Smith? It's Henry John Stephen Smith of Balliol College. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== >> Yes, and an r-generator finite abelian group is a product of > r cyclics. This drops out immediately from the theory of the Smith > normal form. >> Which Smith? Is there more than one Smith? Apparently yes. D.A. Smith, `A basis algorithm for finitely generated abelian groups', Math. Algorithms, 1 (1966), 13-26. But that's the wrong Smith! H.J.S. Smith, `On systems of linear indeterminate equations and congruences' Philos. Trans. Royal Soc. London, CLI (1861), 293-326. is the correct Smith. (I didn't know he was of Balliol though!) Derek Holt. >It's Henry John Stephen Smith of Balliol College. -- >Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > The League of Gentlemen ==== >>> Yes, and an r-generator finite abelian group is a product of >> r cyclics. This drops out immediately from the theory of the Smith >> normal form. >> Which Smith? >>It's Henry John Stephen Smith of Balliol College. It's a rather disgusting way of expressing the Structure Theorem for Finite Abelian Groups, isn't it? Who showed one could express such a group as the product of cyclic groups of prime-power order? Maybe it was Jones ... -- Timothy Murphy tel: +353-86-233 6090 ==== >This is a rather unusual function for which I'd like to get a >closed-form integral, if one exists: {exp(a x)}{exp(b exp(c x))} or, rewritten, exp (a x + b exp (c x)) I suspect that somehow this might include the exponential integral but >so far I've been unsuccessful in finding a closed form solution. > > I did it quickly so there is possibility of mistakes. Introduce new variable y = exp(c x) => x = 1/c ln(y). You get I = int( exp(b*y)*y^(a/c-1)/c, dy) The resulting integral can be solved in terms of incomplete gamma function I = -(-b)^(-a/c)/c*gamma(a/c, -b*y) Finally I = -(-b)^(-a/c)/c*gamma[a/c,(-b)*exp(cx)) Good luck, Leszek Sczaniecki ==== understand the number zero today. I'm remodelling a bathroom, and to make the old, sagging, cieling appear flat, instead of unsagging it with heavy machinery, I used the common, more pragmatic approach called furring: I needed to make six wedges of lumber, all the same corresponding dimensions, in inches: the length of the wedge, the thicknesses of the wedge on the heavy side, and the thickness of the wedge on the light side: three numbers: respectively, 56, 1, and 0. I thought, 0?? am I doing this right?? How can there be zero inches of wood on this side??? I was momentarily confused, checked my measurements, and then I thought. Oh, ZERO wood! And I realized that there was NO (that is, ZERO) wood there. And that it, there, as well, existed: ZERO stone, ZERO cars, ZERO elephants. And what that little point on the end of the wedge had in common with so many other, including unimaginable things, was this: it didn't exist there. So, technically, in the diagram, there was as much wood there, as there was iron, refrigerators, angels, and dinosours. And that it was upon this mute point to be contemplating singularities, ie. that they can't exist, and the distinguishing characterization of the inverse of zero, is that it means EVERYTHING, including refrigerators, which brought, to me, a new meaning to the term undefined that screams for a boundary condition on the distance force is exerted by gravitation. ==== > >> Can anyone please point me to a good basic textbok or reference on >> computer algebra algorithms. I am mainly interested in writing C code >> for simple algebraic simplication. ... >> I am really very interested in understanding algorithms to simplify >> and solve algebraic equations. I want to to enter a string like >> (x^2+x*(x+1))/(2^x+x) and output the simplified string 1 ... >> I am not interested in advanced >> concepts, like solving integral equatons, just simple algebra for a >> start. > You do not specify what you mean by simple algebra and > algebraic equations and solve. Translation: the field is very complex, some things are easy, some things are hard. Some algorithms are obvious (you learned them in grade school: multiplying polynomials), some are not obvious (those are the ones you saw in Knuth, and you probably learned in grade school but didn't realize it: long division and gcd of polynomials), some are hard/impossible/undecidable/unknown (undecidable means we know we can't do it always) and hundreds of years of mathematical research has been spent on them: factoring/solving polys/solving PDEs/..can't think of an unknown one right now). > If you mean: > constants: integers > variables: just one, x. > operations +, *, /, integer power, This selection of components (the alphabet over which you will construct the possible algebraic statements that you will simplify/solve) is the one that Richard believes is manageable given your request. I agree very much with his specific choices. Any additions/extensions will pose ... uh... difficulties. OK, add - and maybe rationals. and we'd only be doing simplification, -not- solving (notice = is not in that set). > then things are simple, at least with the right basic > language support. (Not C, which cannot natively represent > long numbers, necessary to do this: > 1234567812345678123456781234567812345678+1. > ) There are libraries to do unlimited precision integer and rational arithmetic in C. google for those keywords. But manipulation of mathematical statements is (comparatively) a pain in C (doable with trees). Therefore, the suggestion that... > But it would be simple in Lisp, ML, ... . ...which have very easy to use constructs for dealing with symbolic constructs (trees). If you expect to do solve 2^x=y for x, then you > have logarithms. And algebraic functions for solve x^5=y^2+1 for x. these are the extensions that will make things not so simple anymore. For your first attempt, don't try to solve for vars yet. or use more than one var, or allow vars in the exponent. Mitch ==== I've never posted before, but as it is, it's 4:48am and for some stupid reason, I've been trying to solve a double integral for almost 3 hours now. It's just personal now (not sure of notation here so i'm going to display it twice): / 1 / arccos y | | exp (sin(x)) dx dy / 0 / 0 or integrate relative to y {0..1} integrate relative to x {0..arccos(y)} the equation: exp(sin(x)) dx dy My trusty calculator can give me an answer, but for some reason (fatigue for example) I can't figure out how to come up with the answer. any help? ==== > [...] / 1 / arccos y > | | exp (sin(x)) dx dy > / 0 / 0 My trusty calculator can give me an answer, but for some reason > (fatigue for example) I can't figure out how to come up with the > answer. any help? Look up Fubini's theorem. In Maple: int(int(exp(sin(x)),x=0..arccos(y)),y=0..1); # returns unevaluated plot(arccos(y),y=0..1); # strictly monotonic int(int(exp(sin(x)),y=0..cos(x)),x=0..Pi/2); # returns exp(1)-1 -- Thomas Richard Maple Support Scientific Computers GmbH http://www.scientific.de ==== >I've never posted before, but as it is, it's 4:48am and for some >stupid reason, I've been trying to solve a double integral for almost >3 hours now. It's just personal now (not sure of notation here so i'm going to >display it twice): >/ 1 / arccos y >| | exp (sin(x)) dx dy >/ 0 / 0 or integrate relative to y {0..1} >integrate relative to x {0..arccos(y)} >the equation: exp(sin(x)) dx dy >My trusty calculator can give me an answer, but for some reason >(fatigue for example) I can't figure out how to come up with the >answer. any help? Interchange the order of integration. -- http://www.math.fsu.edu/~bellenot bellenot math.fsu.edu +1.850.644.7189 (4053fax) ==== Perhaps surprising, this is likely to be false. CAS internal > programs tend not to evaluate numerical objects, but manipulate > data. Most of the arithmetic is likely to be in iterating over > the index of an array. In the whole program to multiply polynomials, > you probably see one occurrence of multiplication of the coefficients, > and one occurrence of addition of exponents. You see much > more data allocation and checking. > Depend on what you call often, I would not say that polynomial multiplication is representative, integration might be more. You will not have arithmetic on each line of code, but from time to time you have to translate a non-trivial arithmetic expression, like for reducing the degrees in the square-free factorization of the denominator by integration by part. > Maxima, along with most CAS, has a user command language intended to > mimic mathematical notation. It is loosely based on Algol 60. For > Maxima, some internal programs are written in this language > (translated into Lisp automatically). It doesn't give convenient > access to all aspects of the data, but sometimes that doesn't matter. > It makes interfacing with Maxima evaluation and simplification > more automatic, and (especially for the novice) it indeed provides very > generic operations of +, *, / ... in an infix notation. I think a better example may be Maple or Mathematica each of which > has large portions of its system routines written in the user > language. Does that mean those languages are best for implementing > a CAS? Probably not, because if that were the case, the Mathematica > people would stop writing any more in C. Not the case. I believe that the reason is efficiency since (unlike maxima which can translate to Lisp) the user langage is interpreted. Unfortunately, when mathematica/maple started, C++ was in its infancy or not born at all, hence they had to choose C, and as I said earlier I find C++ more adequate than C for a CAS. Now, if people could write mathematica or maple code with the same efficiency than kernel code and with a similar syntax they would probably do it. That's why I find C++ a good language for CAS, because I believe it will be possible to write CAS-user language code directly in C++ with only minimal adaptation. ==== No, mathematicians (tend to) use 1 *glyph* symbols, drawn from an > arbitrary large alphabet of symbols that borrows from latin, greek, > hebrew and adding customizations as necessary to make them sufficiently > distinctive. Further, they combine them with a graphical syntax that is > nearly 2-dimensional. Using only letters and linear strings, correct equivalents are > many-letters identifiers like zeta, hbar or aleph. This goes as well for > single-stroke calculator symbols like sin, cos or exp. > Of course sometimes you will need to use 2 or more letters identifiers, because you don't have alpha or beta on your keyboard and you want to use standard notation (like eps or epsilon for a small number). But you would use the 1 letter symbols if you had them on the keyboard. Now why did mathematicians decide to use the greek or other alphabets? Because they do not want to use 2 or more letters identifiers, to have a more readable notation. So your argumentation does not contradict my point that using 1 letter identifiers for local variables is a good idea. sin, cos, exp are functions, not variable identifiers, they have standard notation, I don't see why you spoke of them in this context. ==== Of course sometimes you will need to use 2 or more letters identifiers, >because you don't have alpha or beta on your keyboard and >you want to use standard notation (like eps or epsilon for >a small number). But you would use the 1 letter symbols if you had them >on the keyboard. Now why did mathematicians decide to use the >greek or other alphabets? >Because they do not want to use 2 or more letters identifiers, >to have a more readable notation. So your argumentation does >not contradict my point that using 1 letter identifiers for >local variables is a good idea. >sin, cos, exp are functions, not variable identifiers, they have >standard notation, I don't see why you spoke of them in this context. > Didn't Mathematicians decide to use greek, because that was the language they spoke? Sorry I couldn't resist. Modern mathematicians use lots of alphabets, because they learn with lots of alphabets. But why do they continue to use them, even as textbooks try to remove them from the low level math courses? I think Mathematicians use greek letters and many alphabets because it gives them clues as to the type of object they are thinking about. In trig, alpha, beta, gamma are the angles, a, b, c are the lengths, A, B, C are the vertices. This `type' is different from programming types, alpha and a are both real numbers and hence the same programming type, but are mathematically different types. Some C programmers, mostly m$, had an ugly and IMHO awful system called Hungarian notation. Each variable name had an ugly wart, a prefix which indicated its type. Even local variables had to use this system, so loops were done using iI or i_i instead of i. While for many modest subroutines one can get away with using only one character names, it is IMHO ridiculous to require all local variables to have only one character. Is t time or temp? Is a area or just the first variable? Programming has made it clear that even limiting the number of characters to a dozen or so is too limiting. It is easy to juggle a few one letter variables, but silly to do so when there are a dozen. As a mathematician, I often use the identifier `mess' for a common subexpression. It always gets a laugh the first time I use in a class. Other short words are also effective identifiers, `area', `len', `dist', for example. Some identifiers are 2 letters by default. In calculus dx, Delta x, have not been stamped out by using the letter h. As a programmer, I use one letter local variables more than I should, but I don't do it when I know it would be noticably better with longer identifiers. -- http://www.math.fsu.edu/~bellenot bellenot math.fsu.edu +1.850.644.7189 (4053fax) ==== I think Mathematicians use greek letters and many alphabets because > it gives them clues as to the type of object they are thinking about. > In trig, alpha, beta, gamma are the angles, a, b, c are the lengths, > A, B, C are the vertices. This `type' is different from programming > types, alpha and a are both real numbers and hence the same programming > type, but are mathematically different types. > I agree for the choice of alphabet, or even inside the alphabet. n, m is commonly used for integers, v, w for vectors, a,b etc for parameters, f,g,h for functions, t for time, x,y,z for unknowns, A,B,C,M for matrices, etc. While for many modest subroutines one can get away with using only one > character names, it is IMHO ridiculous to require all local variables > to have only one character. I would never require that of course. I use myself a lot identifiers like eps, tmp or args for example. But in my experience, for many variables inside a math subprogram, there is a local 1-letter identifier whose name is appropriate (see list above, sometimes a 2-letter identifier like df for the derivative of a function). ==== Your presumptions are both totally wrong. Mais comme on dit, il n'y a > pire sourd que celui qui ne veut pas entendre, et qui veut tuer son > chien l'accuse de la rage. Je vous laisse .88 vos petits jeux d'autiste. > Je me contenterai de vous mettre sous le nez que la syntaxe d'expression > C est en fait, elle-m.90me, fort loin de la notation mathematique standard. > (sorry for english reader). Je ne vois pas pourquoi vous prenez ce ton meprisant. Je programme actuellement un logiciel de calcul formel en C++ et j'en ai programm.8e un en RPL (en gros du lisp a l'envers) qui est int.8egr.8e sur les calculatrices HP4xG, donc je pense avoir un minimum d'experience sur la programmation d'un logiciel de calcul formel. J'enseigne les maths, j'ai donc aussi quelque exp.8erience dans le domaine. Si vous trouvez que la syntaxe C a*b*c*(d+e+f) est fort .8eloign.8ee de la notation math.8ematique standard, alors que dire de la notation lisp! De plus c'est la notation standard utilis.8ee dans toutes les calculatrices et logiciels de calcul formel, et c'est la plus proche des notations 1-d qu'on puisse trouver. ==== Having written a program does not make you an expert on all programs or all programming languages. This is a mistake many people make, but you could still learn. Having taught math does not make you an expert on math (or math notation). Writing in RPL for a calculator certainly does not make you conversant with Scheme or ANSI Common Lisp. >> Your presumptions are both totally wrong. Mais comme on dit, il n'y a >> pire sourd que celui qui ne veut pas entendre, et qui veut tuer son >> chien l'accuse de la rage. Je vous laisse .88 vos petits jeux d'autiste. >> Je me contenterai de vous mettre sous le nez que la syntaxe >> d'expression C est en fait, elle-m.90me, fort loin de la notation >> mathematique standard. >> (sorry for english reader). > Je ne vois pas pourquoi vous prenez ce ton meprisant. Je programme > actuellement un logiciel de calcul formel en C++ et j'en ai > programm.8e un en RPL (en gros du lisp a l'envers) qui est int.8egr.8e > sur les calculatrices HP4xG, donc je pense avoir un minimum > d'experience sur la programmation d'un logiciel de calcul formel. > J'enseigne les maths, j'ai donc aussi quelque exp.8erience dans > le domaine. Si vous trouvez que la syntaxe C a*b*c*(d+e+f) est > fort .8eloign.8ee de la notation math.8ematique standard, alors que > dire de la notation lisp! De plus c'est la > notation standard utilis.8ee dans toutes les calculatrices et > logiciels de calcul formel, et c'est la plus proche des notations > 1-d qu'on puisse trouver. > ==== > Having written a program does not make you an expert on all > programs or all programming languages. I never said that. I said I had some experience in writing CAS application in two different languages and I have clearly a preference, I explained why and I said that each programmer could have a different choice based on different priorities. It seems some people in this newsgroup are absolutely convinced than lisp is the only language that should be used to write a CAS, they used arguments that I don't find convincing and I explained why. > This is a mistake > many people make, but you could still learn. > It's not with this kind of sentence that you will convince other people that you are right. > Having taught math does not make you an expert > on math (or math notation). > I would probably say I am an expert in my research field and I have certainly a larger experience of math notation than people who do not do math professionnally. > Writing in RPL for a calculator > certainly does not make you conversant with Scheme or > ANSI Common Lisp. Something I never said. There are however some common points especially regarding notation, something I consider important. Well, I think I have lost enough time discussing these topics, I just hope people using non-Lisp languages were convinced by (some of) my arguments that there is a life for CAS outside of Lisp, which was in fact my main objective entering this thread. ==== > Well, I think I have lost enough time discussing these topics, > I just hope people using non-Lisp languages were convinced by > (some of) my arguments that there is a life for CAS outside of Lisp, > which was in fact my main objective entering this thread. Unfortunately you were arguing quite something else. You were saying, based on your knowledge of C++ and based on your lack of knowledge of other languages, that C++ is better than (Lisp, or any other language?) for writing a CAS. Most people using C++ hardly need your encouragement to write in C++. Most people using Lisp don't need my encouragement to write in Lisp (or other similar languages). What is missing is a set of tests to compare languages, which is what I asked for initially. RJF > ==== >> Well, I think I have lost enough time discussing these topics, >> I just hope people using non-Lisp languages were convinced by >> (some of) my arguments that there is a life for CAS outside of Lisp, >> which was in fact my main objective entering this thread. > Unfortunately you were arguing quite something else. You were > saying, based on your knowledge of C++ and based on your lack of > knowledge of other languages, that C++ is better than > (Lisp, or any other language?) for writing a CAS. > No, I didn't say that. I said that I liked the possibility to overload arithmetic operators in C++ because I think it is very important for code readability inside a CAS. I wanted to give another bell sound because this thread started with someone asking to write a program in C to do some CAS tasks, and you answered promptly trying to convince him to learn Lisp, then I said C++ was a better choice for him in my opinion, I explained why and the thread began to derive as it seems is the rule for any thread when someone recommend another language than lisp to program CAS (it's a little bit like some mathematicians with algebraic geometry:-)) > Most people using C++ hardly need your encouragement to write in C++. > Most people using Lisp don't need my encouragement to write in Lisp > (or other similar languages). What is missing is a set of tests to compare languages, which is > what I asked for initially. > I'm not particularly interested in tests to compare languages, tests give a biaised view on what is the best language for the tests. I'm interested in real implementations of CAS. ==== (* a b c (+ d e f)) ==== >(* a b c (+ d e f)) makes me want to puke. Addition and Multiplication >are *binary* operators. Writing * a b c for >a*b*c flaunts ALL standard mathematical notation. > I would not agree, mathematicians use sum_{i=1}^n a_i and sum A, where A is a set. So sum {a, b, c} seems to be outside the puke class. Admittly 3 elements is small to switch to sum notation, but I would have no trouble parsing prod {a,b,c,sum{d,e,f}} once I was warned that the sum over a set was being used. This also handles your binary objection, the sum or prod is either defined inductively or alternately order doesn't matter because of commutativity. -- http://www.math.fsu.edu/~bellenot bellenot math.fsu.edu +1.850.644.7189 (4053fax) ==== Actually, Bob's complaint holds just as much for the notation a*b*c*(d+e+f). Which do you do first, compute s :=e+f, then t:= d+s then .... or do you add d+e first? If you don't think it matters, note that (1.0e-20 +1.0)+ (-1.0) = 0 in double precision. 1.0e-20 + (1.0 - 1.0) = 1.0e-20. if a*b*c = 1.0e20, then one ordering gives you an answer of 1.0, and the other gives you 0.0. Now you may assume that only one of these is correct, and will always be computed according to the definition of the programming language. Last I looked, C optimizing compilers could do it either way, and in fact could do it different ways depending on the setting of optimization flags. Now Bob says these operators + and * are BINARY. So he would require that you write a*(b*(c*(a+(b+c)))) which is not very pretty. And worse, some language processors will disregard the () and rearrange things in the name of optimization. Lisp's standard defining (+ a b c) explicitly does not tell you the ordering; if it matters, do (+ (+ a b) c) or the other way... I don't know if this cures Bob's nausea. :) > (* a b c (+ d e f)) makes me want to puke. Addition and Multiplication > are *binary* operators. Writing * a b c for > a*b*c flaunts ALL standard mathematical notation. One can not expect that a computer language > is totally conformant to mathematical notation > (i.e. I don't like = for assignment and == for > equality either) but * a b c d e etc. is too far > out in left field. (IMO) > You can lead a horse's ass to knowledge, but you can't make him think. ==== I have published papers on both Fermat'a Last Theorem and Goldbach's Conjecture. They are listed in drectories throughout the Internet and specifically at: web.math.fsu.edu/Science/Specialized, under In Defense of Mr. Fermat, and Goldbach Proof. I will try to address X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCEOIP12898; ==== i am verymuch interested i prime numbers and divisibility tests . but i dont know much of it. ok, my question is how can you say that a.b-(a+b) is either a prime of twin primes or multiple of 3 , can u prove it? please do reply me help me with the doubt i got. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBD7ZlG26151; ==== I've been using CAS for sometime (my dissertation involved the use of CAS with high school students). Although I realize Maple is more popular, I prefer MuPAD. I have both apps (MuPAD 2.5.2 and Maple 8 on my mac). 1. I find it easier to write short programs in MuPAD. 2. I appreciate the fact that MuPAD is freely downloadable as a demo. Try before you buy. 3. I like the fact that MuPAD started as an open-source project. It MuPAD has never failed to perform for me . . . I've been using it lately to investigate partitions of integers. The various partitioning and permutation functions are actually more extensive in MuPAD than in either Maple or Mathematica (at least as far as I know). I find the language and syntax more intuitive, too - easier for young kids to understand. -MTE ==== > Can you solve this equation for x? x log(kx) = (x - 1)log(1 - x) the answer might involve Lambert's W function, but I'm not sure It's not solved for x, but it looks nice this way... > log(k)=Int(1+log(s),s=1/x-1 .. 1/x); 1/x > / > | > ln(k) = | ln(s) + 1 ds > | > / > 1/x - 1 I was playing with the Q, stumbled into probability integrals and now give up. It seems the Q is equivalent to ask for the inverse of the fct (x+1)^(x+1)/(x^x) [ = (1+1/x)^(x+1)*x and so is seen to exist by looking at the derivative], 0 <= x and fctValues = [1,infinity[ . --- remove the no ==== I need to run a C code on an Alpha station here and I need to link it against clapack. My sys admin seems not to be of much help and so I'm looking for any suggestion. Here it is the configuration I have: % cc -V DEC C V5.9-005 on Digital UNIX V4.0 (Rev. 1229) % uname -a -p OSF1 ....... V4.0 1229 alpha % man lapack LAPACK(3DXML) LAPACK( 3DXML) Digital Digital Name lapack - A library of linear algebra routines Description LAPACK (Linear Algebra Package) is a new library of dense linear and eigen- problem solvers that supercedes LINPACK and EISPACK, offering better per- formance and accuracy. DXML includes a compiled and optimized version of LAPACK 2.0, which was released in September 1994. As far as I can understand I have a C compiler and the original lapack library already installed, now how can I get clapack working on this machine? May I use original (Fortran) lapack from within my C code? If so, how? Andrea. ==== a big piece of my Ph.D. thesis will concern writing numerical C code to solve some numerical problems. Provided that I know there are very good numerical libraries outside there and that Press & Teukolsky et al ( Numerical Recipes) is a good start point to improve my knowledge, I feel the needing of something more about do & dont's of numerical programming. What I really need to learn are the tricks of the trade such as optimization, solution approach, data structures, etc... and I don't want to spend most of the time re-disovering what skilled people already knows, so I hope someone here can point me in the right direction. What books can I read about this topic? Could you suggest to me any kind of references? Andrea. ==== > I feel the needing of something more about do & dont's of > numerical programming. What I really need to learn are the tricks of > the trade such as optimization, solution approach, data structures, > etc... What books can I read about this topic? Could you suggest to me any kind > of references? author = David Goldberg, title = What Every Computer Scientist Should Know About Floating-Point Arithmetic, journal = ACM Computing Surveys, volume = 23, number = 1, pages = 5--48, year = 1991 } is a worthwhile read, IMHO, for some general background knowledge about floating point and IEEE arithmetic. The paper also available in different formats various places on the web. ==== > I feel the needing of something more about do & dont's of > numerical programming. What I really need to learn are the tricks of > the trade such as optimization, solution approach, data structures, > etc... What books can I read about this topic? Could you suggest to me any kind > of references? author = David Goldberg, > title = What Every Computer Scientist Should Know About Floating-Point Arithmetic, > journal = ACM Computing Surveys, > volume = 23, > number = 1, > pages = 5--48, > year = 1991 } is a worthwhile read, IMHO, for some general background knowledge > about floating point and IEEE arithmetic. The paper also available in different formats various places on the > web. For example, there is a link to a *.pdf version on my numerical methods page: http://www.phys.virginia.edu/classes/551.jvn.fall01/ -- Julian V. Noble Professor Emeritus of Physics jvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ God is not willing to do everything, and thereby take away our free wiil and that share of glory that rightfully belongs to ourselves. -- N. Machiavelli, The Prince. ==== You may find worthwile to use the GNU GMP and GSL librairies. J.-M. > a big piece of my Ph.D. thesis will concern writing numerical C code to > solve some numerical problems. Provided that I know there are very good > numerical libraries outside there and that Press & Teukolsky et al ( > Numerical Recipes) is a good start point to improve my knowledge, I feel > the needing of something more about do & dont's of numerical > programming. What I really need to learn are the tricks of the trade > such as optimization, solution approach, data structures, etc... and I > don't want to spend most of the time re-disovering what skilled people > already knows, so I hope someone here can point me in the right > direction. What books can I read about this topic? Could you suggest to me any kind > of references? Andrea. ==== a big piece of my Ph.D. thesis will concern writing numerical C code to > solve some numerical problems. Provided that I know there are very good > numerical libraries outside there and that Press & Teukolsky et al ( > Numerical Recipes) is a good start point to improve my knowledge, I feel > the needing of something more about do & dont's of numerical > programming. What I really need to learn are the tricks of the trade > such as optimization, solution approach, data structures, etc... and I > don't want to spend most of the time re-disovering what skilled people > already knows, so I hope someone here can point me in the right > direction. What books can I read about this topic? Could you suggest to me any kind > of references? Andrea. It is not hard to learn enough C to write decent numerical-intensive code. Most C's are pretty well optimized so it should not be necessary to descend into assembler to achieve sufficient speed. Jon Bentley's 2-volume Programming Pearls is a good place to learn some philosophy of writing decent programs. My personal problem with C (I generally develop programs in Forth, and yes, there are some good native-code optimized Forths around these days, so no need to sneer) is that it is hard to write factored code. Factoring is the art of defining small subroutines that do simple things (so they don't have any built-in bugs) and then combining them into larger subroutines. Here is an example in Forth (it is the subroutine that computes an integral adaptively): : )integral ( f: xa xb err -- I[xa,xb]) ( xt --) initialize BEGIN N 0> WHILE subdivide converged? IF interpolate THEN REPEAT f finalI ( f: I[xa,xb] ) Ntimes @ . . function calls optional line ; You can see immediately the structure of the program, since all the lower-case names are appropriately named subroutines that do the individual tasks. Forth is optimized for this fine-grained decomposition, and that is my chief reason for liking it. (Forth purists don't like my Forth style, BTW, since I use a formula translator to embed formulas in subroutines with f ... --they get translated into Forth automatically.) The Forth program, and two C equivalents (unfactored and factored) can be found at http://www.phys.virginia.edu/classes/551.jvn.fall01/CiSE_progs/Cprogs.htm under item 5. The final subroutine of the factored C version looks just like that of the Forth version: double integral( double a, double b, double eps, double (*pn) (double) ) { initialize( &i, a, b, eps, pn); while ( i.n > 0 ) { subdivide( &i); MoveDown( &i); NewPoints( &i, pn); converged( &i); } return i.finI ; } However, in order to achieve this I had to define a struct to hold all sorts of data; the pointers into it make the other subroutines look sort of baroque. I do not consider the result very readable. Finally, I disagree with the people who have bad things to say about Numerical Recipes. The C versions of their routines may not be elegant, but they are _clear_ which is a big plus. They are good enough for most purposes. (Read what Bentley has to say about optimization before worrying about whether they run fast enough.) -- Julian V. Noble Professor Emeritus of Physics jvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ Science knows only one commandment: contribute to science. -- Bertolt Brecht, Galileo. ==== > It is not hard to learn enough C to write decent numerical-intensive > code. Most C's are pretty well optimized so it should not be necessary > to descend into assembler to achieve sufficient speed. Might the increasing prevalence of Single Instruction Multiple Data operations (e.g. 3DNow, SSE*, Altivec) be an exception to this? If you have an algorithm that can (at least in part) use 32 bit floating point numbers in parallel, you might be able to double or quadruple its speed using one of those instruction sets; however your C/C++ compiler may not generate SIMD instructions at all and IIRC even the ones that do aren't very good at it yet. --- Roy Stogner ==== > It is not hard to learn enough C to write decent numerical-intensive > code. Most C's are pretty well optimized so it should not be necessary > to descend into assembler to achieve sufficient speed. Might the increasing prevalence of Single Instruction Multiple Data > operations (e.g. 3DNow, SSE*, Altivec) be an exception to this? If you > have an algorithm that can (at least in part) use 32 bit floating point > numbers in parallel, you might be able to double or quadruple its speed > using one of those instruction sets; however your C/C++ compiler may not > generate SIMD instructions at all and IIRC even the ones that do aren't > very good at it yet. > --- > Roy Stogner Certainly. Also the MMX instructions, or any specialized instructions that generic compilers won't employ (because they aren't on all target machines). And there are certain things that can be done MUCH faster in assembler than in optimized C, Fortran, whatever. Case in point: I am writing a library for quad-precision floating point, wherein the exponent field is the same as in IEEE 64-bit, but the mantissa has an additional 64 bits. There are lots of C or Fortran multiple precision libraries around that give arbitrary precision. But if you want maximum speed, you have to roll your own in assembler. When it's done I'll report on it. -- Julian V. Noble Professor Emeritus of Physics jvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ God is not willing to do everything, and thereby take away our free wiil and that share of glory that rightfully belongs to ourselves. -- N. Machiavelli, The Prince. ==== >a big piece of my Ph.D. thesis will concern writing numerical C code to >solve some numerical problems Use Fortran. A.L. ==== |> a big piece of my Ph.D. thesis will concern writing numerical C code to |> solve some numerical problems. Provided that I know there are very good |> numerical libraries outside there and that Press & Teukolsky et al ( |> Numerical Recipes) is a good start point to improve my knowledge, I feel |> the needing of something more about do & dont's of numerical |> programming. What I really need to learn are the tricks of the trade |> such as optimization, solution approach, data structures, etc... and I |> don't want to spend most of the time re-disovering what skilled people |> already knows, so I hope someone here can point me in the right |> direction. Well, if you think that Numerical Recipes is a good place to start, you should start by unlearning what you know that ain't so. It isn't. Nick Maclaren. ==== Ok, if NR is not a good point to start from, could you show me the right direction? Andrea. > Well, if you think that Numerical Recipes is a good place to start, > you should start by unlearning what you know that ain't so. It > isn't. > Nick Maclaren. > ==== Write some code and find out why it's not. > Do comparison studies. Write a suite that reports differences > that are useful. Ie. do your PhD on it!?! Remember that usenet advice is mostly worth what you pay for it. ==== |> Ok, if NR is not a good point to start from, could you show me the right |> direction? Choose a few representative routines, put data into them that is close to their (numerical) capabilities, and see what they do. Then look at NAG's documentation and compare the number of error returns. To read the text, chase up a good book or two on numerical analysis and compare what it says with what Numerical Recipes says, concentrating on where the latter glosses over potential problems. Nick Maclaren. ==== > a big piece of my Ph.D. thesis will concern writing numerical C code to > solve some numerical problems. Provided that I know there are very good > numerical libraries outside there and that Press & Teukolsky et al ( > Numerical Recipes) is a good start point to improve my knowledge, I feel > the needing of something more about do & dont's of numerical > programming. What I really need to learn are the tricks of the trade > such as optimization, solution approach, data structures, etc... and I > don't want to spend most of the time re-disovering what skilled people > already knows, so I hope someone here can point me in the right > direction. What books can I read about this topic? Could you suggest to me any kind > of references? An excellent start in the philosophy of programming is probably If you take a look at chapter 14 of that book, you'll see that there are numerous reasons *NOT* to chose C as a programming language unless you are doing systems level programming (which isn't numerical stuff at all). In fact, when writing software, your primary focus should be on a proper design. Optimization should come at the end, and only after profiling reveals a bottleneck. Chose good algorithms instead of creating messy code. I'll recommend you take a look at either Python or Java. Python has a good selection of numerical software which you can use, and Java is a good choice if you need to create high-performance applications which are as portable as possible. Also, Java features a very restricted set of pointer operations, which insulates you against most memory corruption problems possible in C. -- Bj¿rn-Ove Heimsund Centre for Integrated Petroleum Research University of Bergen, Norway ==== I know Python and I actually work in a mixed environment with both Python and C. Nevertheless sometime I need C for CPU-intesive tasks. What I'm asking for is to improve my numerical programming skillness to be able to write better (I mean more efficient) code. Any help will be appreciated. Andrea > An excellent start in the philosophy of programming is probably If you take a look at chapter 14 of that book, you'll see that there > are numerous reasons NOT to chose C as a programming language unless > you are doing systems level programming (which isn't numerical stuff > at all). In fact, when writing software, your primary focus should be on a > proper design. Optimization should come at the end, and only after > profiling reveals a bottleneck. Chose good algorithms instead of > creating messy code. I'll recommend you take a look at either Python or Java. Python has a > good selection of numerical software which you can use, and Java is a > good choice if you need to create high-performance applications which > are as portable as possible. Also, Java features a very restricted set > of pointer operations, which insulates you against most memory > corruption problems possible in C. > ==== >Dear All, > >I am trying to solve a system of nonlinear equations which is very >complicated. The resulting function is a stochastic variable, which I >have to approximate using the Monte Carlo methods. Therefore >evaluation of the function is very time consuming (one evaluation is >about one minute). My problem is to find an appropriate algorithm >which solves the nonlinear system of equations without approximation >of the Jacobian/Hessian matrix. Due to the complexity of the >stochastic function it does not make sense to use methods like Newton, >Newton-Raphson, etc. > >I was thinking about something like bisections or regula falsi, >eventually secant method. They work fine in one dimensional space. But >I do not know about higher dimensions. > > >Pshem > methods SIAM J. Numer Anal.14,(1977),1051-1065. it describes some bisection like method in R^n. there are generalizations of the classical sequential secant method (Gragg worked on it in the seventies) and so called Quasi-Newton methods which only use function values (for general nonlinear systems Broydens good method) but these will not work if the evaluation of the function is noisy. but these methods cannot be globalized like Newtons method by damping. so they will work only with truly good initial guesses I assume your zero problem is not noisy in itself, since I have problems in reading something like F(x)=0 with the stochastic F. hth peter ==== About a year ago, somebody posted some information regarding selecting optimal subsets of combinations. I seem to recall a paper on a Web site and some comuter-based method for making the selections, but I've lost/misplaced everything I had about the subject. My apologies for the vagueness of my question, but I can't recall the paper's title or subject. ==== >Is it true that for every n>=3 there is a finite non-abelian group >with exactly n conjugacy classes ? > For every n>=3 there is a dihedral group with n conjugacy classes . But is the same assertion you asked (for n>=2) true also for infinite non-abelian groups ? TIA, Jadek ==== Is there a formula for the number of non-congruent solutions to x^2 + y^2 == z^2 mod n for n>=1 ? TIA Bill ==== > Is there a formula for the number of non-congruent > solutions to x^2 + y^2 == z^2 mod n for n>=1 ? > If there is one it should be added to the OEIS entry: ID Number: A062775 URL: http://www.research.att.com/projects/OEIS?Anum=A062775 Sequence: 1,4,9,24,25,36,49,96,99,100,121,216,169,196,225,448,289,396, 361,600,441,484,529,864,725,676,891,1176,841,900,961,1792, 1089,1156,1225,2376,1369,1444,1521,2400,1681,1764,1849,2904, 2475,2116,2209,4032,2695,2900 Name: Number of Pythagorean triples mod n; i.e. number of non-congruent solutions to x^2 + y^2 = z^2 mod n. Comments: a(n) is multiplicative and for a prime p: a(p) = p^2. ==== A few years ago I think I have found a paper or book with a proof of the > following statement: > A field is a field with product formula if and only if it is one of > the following > 1. a number field > 2. a function field in one variable or a finitely algebraic > extension of such a field > Is this true and - if yes - can anyone post me a reference to that proof Expecting the statement is true I have an additional problem: > In the book Fundamentals of Diophantine Geometry by Serge Lang on page > that there are product formulas arising from more general situations. > This is a little bit astonishing for me - and so I think there is > something wich I haven't understood right! I would be very pleased if you can help me with those two questions. You may find your answers, along with much more of interest, in http://google.com/groups?selm=y8zu1ahl2nl.fsf%40nestle.ai.mit.edu -Bill Dubuque ==== *********************** second announcement ******************** The MAT seminar of Montpellier Arithmetic Quantum Chaos 23-24 January 2004 The Mediterranean Seminar of Algebra and Topology (aka. MAT seminar) of Montpellier 2004 will consist of a series of introductory lectures on Arithmetic Quantum Chaos. The invited speakers are J. P. KEATING and P. SARNAK University of Bristol University of Princeton & Courant Institute The session will take place on the 23-24 January 2004 at the University Montpellier II (France). Organisers: Ph. Elbaz-Vincent (pev@math.univ-montp2.fr) Ph. Michel (michel@math.univ-montp2.fr) R.T. Ramadas (ramadas@math.univ-montp2.fr) More details, and also an abstract, are available at the following address http://www.math.univ-montp2.fr/~pev/MAT2004 and will also be available on the web page of the seminar MAT; http://www.math.univ-montp2.fr/MAT Ms Lacan, lacan@math.univ-montp2.fr or FAX +33 4.67.14.35.58 Registration deadline: 9 January 2004. When you send your registration please give your full name, your dates of arrival and departure, your host institution with address. NB: There are no fees, but registration is necessary in order to provide enough packages for the participants. NB2: prices of the rooms will range from 30 to 45 euro per night. ---------------------------------------------- I3M, UMR CNRS 5149, CC51, Universit,bi(B Montpellier II. ==== Please let me know if this is the wrong group to post this question. I am interested in finding the blocks that correspond to this inclusion relation table I found in a publication at http://www.math.colostate.edu/~betten/pub/99/steiner36.ps Should be 62832 blocks,however my experience in combinatorics is limited to self instruction. If I could only find some insructional websites that will help me to better understand. Any help would be greatly appreciated. Tony Pope ==== (1) Are there available any criteria for a Banach space (over reals, if it matters) to have a predual? Sufficient conditions (of any kind) are also welcome. (2) Let Y be a Banach subspace of the dual of a Banach space X. What properties of X and Y imply that X contains a predual of Y? -- Alexander E. Gutman gutman@math.nsc.ru ==== >> I would like some info on the equation >> y''(t) = w(t) * y(t) (1) >> where w is a smooth C^infty function and t is in R. >> In particular I would like to know under what >> conditions (on w) the solution of (1) posses >> at most one zero. >I assume you mean, every solution that is not identically 0 >has at most one zero. A necessary and sufficient condition is that there is a positive solution. That is, the w which have your property are all y''/y for (strictly) positive C^infinity functions y. Proof: If w has the property, let y(t) = Y(v,t) be the solution of y''=w y with y(0) = 1 and y'(0) = v. Let A = {v in R: Y(v,t) < 0 for some t < 0} B = {v in R: Y(v,t) < 0 for some t > 0}. Clearly A and B are nonempty open sets, and if w has the property their intersection is empty. Since R is connected, their union can't be all of R, so there is some v such that Y(v,t) >= 0 for all t. In fact Y(v,t) > 0 for all t, because if y(t) >= 0 and y(t_0) = 0 we'd have y'(t_0) = 0, and the solution with y(t_0) = 0 and y'(t_0) = 0 is identically 0. Conversely, if y_1 is a positive solution of y'' = w y, the general solution is y(t) = (c_1 + c_2 u(t)) y_1(t) where u'(t) = 1/y_1(t)^2. In order for y(t) = 0 we need c_1 + c_2 u(t) = 0, and since u is increasing there is at most one such t except in the case c_1 = c_2 = 0 (in which case y is identically 0). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== Let e_B(A) be the erosion of A respect to the structuring element B (in a group). It is easy to see that if 0 in B then e_B(A)subseteq A. I have read that also the converse is true and some experiment confirm this but I wasn't able to prove. Could someone help me in proving it? Homer ==== slightly OT > Hey all, > I have the following (apparently common) problem, that is slightly different > than as described under the MS knowledge base. The symptoms are the same > except for the ones marked with (X). None of these solutions worked of > course.. 322097 Slow Browsing to Windows 98 or Windows Me Clients from Windows XP > http://support.microsoft.com/default.aspx?scid=kb;en-us;322097&Product=winxp > SYMPTOMS > If the following conditions exist, you may experience slow browsing for file > shares and printers that are on Microsoft Windows 98-based or Microsoft > Windows Millennium Edition-based computers: > a.. You are browsing from a Windows XP-based client computer. > b.. Both computers are part of a workgroup. > c.. (X: No, passwords are disabled - I even reinstalled File and Printer > Sharing services to make sure that they would be)The Windows 98 or Windows > Millennium Edition (Me) client has a password-protected file share. > d.. (X: I don't think so, at least I have no idea what this is talking > about, but the only time I enter a poassword in XP is when I log into the > administrator account) You are using the Save this password feature on > your Windows XP client computer. > I will also say this...everything worded extra fine and dandy until PC#2's > HD crashed. I bought a new HD and re-installed all and now things are > extremely slow (no matter whether connecting to the PC#2 or my laptop). I > have a 10MB card in my laptop (win98 OSR2) and a 100MB in my (PC#2 98 OSR2). > Or at least, I only noticed this poblem after fixing PC#2. This didn't > happen before this, and as far as I can recall, I didn't make any major > changes to my XP machine (which is PC#1 and from which browsing any other > PC's on the network lags initially). > Also, this problem doesn't exist when browsing any other machine from any of > the win98 machines. The last thing I'd like to mention is that browsing the folders marked as > shared folders on my XP machine is slow via my XP machine, if these are > accessed through the My Network Places. Any and all help is greatly appreciated. I'm really running out of ideas > here Doug ==== > slightly OT LOL -- Even with X scattered throughout? And look at all those numbers! That is a pretty bizarre choice of groups to cross-post to. -- - relic - ==== I sent this yesterday with crossposting but I didn't see it posted. I am sorry if it is repeated. Mersenne prime has been discovered. Today is the first day I heard of this. I haven't seen this announced in UseNet so I am spreading the news. 2^(20,996,011) -1 is the largest known prime. It was discovered by Michael Shafer using the GIMPS program. For those interested in a web URL try http://www.utm.edu/research/primes/mersenne/ Dan in NY -- dKlinkenberg at hvc dot rr dot com ==== Dan in NY schrieb in Nachricht ... I sent this yesterday with crossposting but I didn't see it posted. I am >sorry if it is repeated. Mersenne prime has been discovered. Today is the first day I heard of this. >I haven't seen this announced in UseNet so I am spreading the news. >2^(20,996,011) -1 is the largest known prime. It was discovered by Michael >Shafer using the GIMPS program. >For those interested in a web URL try >http://www.utm.edu/research/primes/mersenne/ Yes, and the announcement had already appeared in -- >Dan in NY >-- >dKlinkenberg at hvc dot rr dot com > ==== Would anyone be able to help with the following question? How do you get from integrate sec 3x tan 3x to 1/3 sec 3x + c? If anyone is able to help, I would be eternally grateful. B. ==== > Would anyone be able to help with the following question? > How do you get from integrate sec 3x tan 3x to 1/3 sec 3x + c? > If anyone is able to help, I would be eternally grateful. B. It helps if you know that the derivative of sec(x) is sec(x)*tan(x). ==== > Would anyone be able to help with the following question? > How do you get from integrate sec 3x tan 3x to 1/3 sec 3x + c? > If anyone is able to help, I would be eternally grateful. B. Let u=3x so du=3dx. This makes the integral sec u tan u du. We note that dx=1/3 du therefore we multiply by 1/3 and the integral of sec u tan u is sec u. Therefore you get 1/3 sec 3x + C -- David Moran Chief Meteorologist Oklahoma Storm Team ==== trial-and-error method? I haven't yet learnt the substitution method yet. B. > Would anyone be able to help with the following question? > How do you get from integrate sec 3x tan 3x to 1/3 sec 3x + c? > If anyone is able to help, I would be eternally grateful. B. Let u=3x so du=3dx. This makes the integral sec u tan u du. We note that > dx=1/3 du therefore we multiply by 1/3 and the integral of sec u tan u is > sec u. Therefore you get 1/3 sec 3x + C -- > David Moran > Chief Meteorologist > Oklahoma Storm Team ==== > trial-and-error method? I haven't yet learnt the substitution method yet. B. > Would anyone be able to help with the following question? > How do you get from integrate sec 3x tan 3x to 1/3 sec 3x + c? > If anyone is able to help, I would be eternally grateful. > B. > > Let u=3x so du=3dx. This makes the integral sec u tan u du. We note that > dx=1/3 du therefore we multiply by 1/3 and the integral of sec u tan u is > sec u. Therefore you get 1/3 sec 3x + C -- > David Moran > Chief Meteorologist > Oklahoma Storm Team Well, think about it like this. What trig function has a derivative of sec x tan x? sec x. Now since there's a 3 in front of the x, you have sec 3x tan 3x. When you take the derivative, you'll have a 3 out front. Since you don't have a 3 out in front of the integrand, you need to multiply by 1/3. -- David Moran Chief Meteorologist Oklahoma Storm Team