mm-65 === 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 overa set of stationary atoms! Making a computer from theset.And so the person is left with the proper means ofprogramming the little atom computer. So a good projectis to relate the set state to the computer state. Meaningthe abstract computer as Turings Theory appears the correctrelation in this example.And the idea of Turings innite state requirment is then onlya sets abstract necessity and not really necessary in a usefulworking computer.And the simple relation of the set to its cause is denedby the little atom computers design.So the shifted atom computes!Meaning a project is to copy this new atom computer ina software and learn to write the dened shift causedset, without relation to the number of places theatom jumped.That is all the darn thing is! And your special sequencemay then be programmed correctly without mathematicaltheory. Meaning, how is your special sequence related toTurings machine? Related independent of its mathematicaldenition.And so cryptography appears another subject.How that?Douglas EaglesonGaithersburg,MD USA === Here is description of a Turing Machine which computes Fibonacci numbers. %%= %% === Turing Machine === %%== Computing a Fibonacci number == %%== Machine Denition : 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 1Input : number n Sample (n = 7) : 1 1 1 1 1 1 1Output : Fibonacci#n Sample (Fibonacci#7) : 1 1 1 1 1 1 1 1 1 1 1 1 1 = States =Initial state : q0Halting state : qfInternal states : q101 q102 q103 q104 q105 q106 q107 q108 q109 q201 q202 q203 q204 q301 q302 q303 q304 q305 q306 q307 q308 q309 q310q311 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 : bInput alphabet : 1Internal 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 Denition : 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 nd out half rangefourier series, we will carry out in Maple like thisa0 := 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 twohalf range series like thisplot((S1+S2), x = -2..2);or it is just meaningless. === > Suppose function f(x)=x^2 is given and we have to nd 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 off(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 xed positive real numbers)(3) p(f) is continuous on [0,B]LetQ: S-> R such thatQ(p(f)) = int_0^B log( p(f) + u(f) ) dfwhere u(f) is a xed continuous function which is positive for all f in [0,B]It is easy to show that Q is bounded on S, so theres 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 thatQ(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 xed positive real numbers)(3) p(f) is continuous on [0,B]LetQ: S-> R such thatQ(p(f)) = int_0^B log( p(f) + u(f) ) dfwhere u(f) is a xed continuous function which is positive for all f in >[0,B]It is easy to show that Q is bounded on S, so theres 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 thatQ(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 makesu + p constant, and this is the u that gives theinequality shows that for any other u we haveint 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 whichmakes u + p as close as possible to constant, in somesense (and then depending on what sense that is itshould 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 xed 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 xed continuous function which is positive for all f in >[0,B]>It is easy to show that Q is bounded on S, so theres 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 thatQ(p(f)) = sup_S Q.Note: just an existence proof is needed!It seems to me fairly clear that the maximum is obtained withp(f) = { c - u(f) if u(f) < c { 0 otherwisewhere the constant c is chosen so that int_0^B p(f) df = P.The simplest existence proof may be just to show that thisis the maximum. A main ingredient is the concavity of ln.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israelUniversity of British ColumbiaVancouver, 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 xed 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 xed continuous function which is positive for all f in >[0,B]It is easy to show that Q is bounded on S, so theres 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 thatQ(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 1Z2Call your function p_0(x) (I cant bring myself to use f for theindependent variable!). For any other xed admissible function p(x),denep_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)) dxNote that p_0(x) + u(x) >= c for all x in [0, B]. This implies thatG(0) <= int_0^B dp(x) / c = 0Furthermore,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_0is 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 wont 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 cant bring myself to use f for the> independent variable!). For any other xed admissible function p(x),> dene 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 dened 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 lastquestion 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 ingeneral is no, but I cant nd a counterexample. Clearly, (since itis understood that R is commutative), if one of I or J is a principalideal, the set A is an ideal, so a counterexample has to consist of anon-PID and to ideals generated by at least two elements each.My strategy was to construct I and J so that I could nd ab and abin A with ab + ab = 1, and then argue that 1 could not be written asa product of elements from I and J (because then each would contain aunit, so I = J = R), but so far my search has failed. Does anybodyhave a hint (maybe Im looking the wrong places, or even after thewrong thing; I am not 100% certain that the answer _is_ no, but thenagain, I see no reason why it should be yes).Rasmus-- === > In my algebra textbook, the product of two ideals I,J is dened 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 cant nd 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 nd ab and ab> in A with ab + ab = 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.Its bound to. If a, a in I and ab + a b = 1, then 1 in Ias I is an ideal. Hence I = R is principal (and also J = R).0You need to look at nonprincipal ideals. Two rings whereits easy to nd nonprincipal ideals are the polynomialrings 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 dened 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 cant nd 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 whetherthe proof is valid. Dont try to guess whether theanswer is yes or no just by the pattern ofyes-es and no-s that precedes it.(BTW, though Im no expert on this stuff, I think theanswer is yes by an elementary argument. Thats whyI suggest going through your more complex proof for IJ.) - Randy === >In my algebra textbook, the product of two ideals I,J is dened 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 cant nd 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 nd ab and ab>in A with ab + ab = 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.Wouldnt you have a rather hard time doing that? Note that the set Ais a subset of IJ, and also note that IJ is contained in theintersection of I and J; if you could nd I and J with a,a in I,b,b in J, and with ab + ab = 1, then 1 would be an element of BOTH Iand J, and hence you would not be able to have a contradiction.> Does anybody>have a hint (maybe Im 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 nonprincipalideals; the ideals I=(x,2) and J=(x,3), for example, are easy provenon-principal.I consists of the polynomials with integer coefcients whose constantterm is a multiple of 2; J consists of the polynomials whose constantterm 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 rst 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), ifA were an ideal then x=3x-2x would also have to be in A.Is that possible? If so, we would havex = (x*p(x) + 2a)(x*q(x)+3b) = p(x)q(x)*x^2 + (2aq(x)+3bp(x))*x + 6abfor 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. Ifa=0, then we have x = p(x)q(x)*x^2 + 3bp(x)*x; since the linear termof the right hand side is equal to 3bp(0)*x, we see that this cannever 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 thisis also never equal to x. So we cannot have b=0.In conclusion, x cannot be written as a product of an element ofI=(2,x) and an element of J=(3,x), even though it lies in the smallestideal which contains A. Therefore, A cannot be an ideal in this case.The answer to the question you have, then, is not necessarily (sincesometimes A is an ideal, and sometimes it is not).== === Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)= Arturo Magidinmagidin@math.berkeley.edu === In my algebra textbook, the product of two ideals I,J is dened 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 cant nd 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 dened 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 cant nd 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 nd ab and ab> in A with ab + ab = 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 Im 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 letI = <2,x> the ideal generated by 2 and x,J = <3,x> the ideal generated by 3 and x,A as what youve dened earlier.Then 6 = 2*3 is in A, and x^2 is also in A. But x^2 + 6 is irreducible inZ[x] and hence cannot be written as a product of elements in I and J. Hencex^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, withan invertible derivative.Is there necessarily an open neighbourhood, U, of x, such that f isinjective on U, and such that the inverse of f is differentiable onf(U)?The answer is probably no, but I think counterexamples are hard tocome 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 Epsteinf(x) = x + x^2*sin(1/x) on R^1You can check that f(0) = 1, but f(1/2PI*n) = 0 for all positiveintegers n, so the inverse function is not differentiable on anyneighborhood of 0.John Mitchell <9v03gvo0r862lt018ko75vh2m1q9r6f1hr@4ax.com> === > Im just being too humble again.) Im going to be famous. And rich:> theyll probably give me a big fat prize, maybe a Clay prize: this> is like P=NP ... yeah! Thats it! Without the need to divide, prime> factorization becomes simple, so P=NP, but the Riemann Hypothesis too!> Its about primes! So thats 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 wasnt 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 ashlight> and compass, yet never get more than 100 feet in. You must stride boldly with> your eyes closed. Use the force. There. Its 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 its 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 dened?> AH HA!!! The object IS the object, so all the zeroes have real part equal to the> spin of the fundamental object! And you nd 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 youve 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 dont evenknow where my ISP is. (Im in San Antonio using a different ISPtemporarily, in case you decide to trace my headers). Its quite commonwith broadband that your DSL service provider is nowhere close to you.Anyway, Im not giving credit to the post, just pointing out that where theISP 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 dont even> know where my ISP is. (Im in San Antonio using a different ISP> temporarily, in case you decide to trace my headers). Its quite common> with broadband that your DSL service provider is nowhere close to you.> Anyway, Im not giving credit to the post, just pointing out that where the> ISP is has nothing to do with where the person is. Wheres Trinity University?This years Crypto conference is in Santa BarbaraAugust 17-21. The early registration deadline isJuly 14th. Full program information is available at:Itll be great, both technically and socially!Greg.(General Chair)-- Greg Rose INTERNET: ggr@qualcomm.comQualcomm Australia VOICE: +61-2-9817 4188 FAX: +61-2-9817 5199Level 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.tori3A1=(- 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 im 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 im 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 cyclicallypermutes the coefcients of a polynomial. Then, S(P*Q) = S(P)*S(Q). Actually I know this is generally false, because I tried thefollowing:P = x+1Q = x+2P*Q = x^2 + 3x + 2S(P) = x+1S(Q) = 2x+1S(P*Q) = 2x^2 + x + 3S(P)*S(Q) = 2x^2 + 3x + 1So, my question is, what conditions do the polynomials need to satisfyin order for this to be true? It seems to be true if either of thepolynomials contains only one non-zero coefcient (in other words,its true if, for P a polynomial of degree n with coefcients 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 its not true that would mean my proos incorrect. ;-)Anyway, if someone can help me Id 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 coefcients 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 coefcient (in other words,>its true if, for P a polynomial of degree n with coefcients 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 mustbe 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 coefcients 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 areall mth roots of unity), so there is no Q of degree >= 1 for whichthis 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>=1so 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 ofP and of Q cancel.Robert Israel israel@math.ubc.caDepartment 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 Ihad the chanceto study it properly and feel so ignorant about it. I am currently messingwith 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 nished part a), but am struggling with part b). This is what I havetried: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. (Inthe following, I will useZ rather than the typographically more complicated Z(G).x, y elms of G => xZ, yZ elms of G/Z.Since G/Z is cyclic, xZ = (aZ)^j = (a^j)Z and yZ = (aZ)^k = (a^k)Z for someints 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)ZThis seems to be making some progress toward G being Abelian, but can Iconclude here thatxy = yx giving commutativity. I realize that (xy)Z and (yx)Z are cosetsthat have the sameelements, but I dont think this gives me what I need. Do I have to bringin some other fact atthis 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 chanceto 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 nished 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 elms of G => xZ, yZ elms of G/Z.Since G/Z is cyclic, xZ = (aZ)^j = (a^j)Z and yZ = (aZ)^k = (a^k)Z for some>ints 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)ZThis 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 dont 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 isan 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?Heres a more general result:Let G be a group, let Z(G) be the center of G. Let H be a subgroup ofZ(G).(1) Prove that H is normal in G.(2) Prove that if G/H is cyclic, then G is abelian. =Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)= Arturo Magidinmagidin@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 whenI> 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 nished 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 elms of G => xZ, yZ elms of G/Z. Since G/Z is cyclic, xZ = (aZ)^j = (a^j)Z and yZ = (aZ)^k = (a^k)Z forsome> ints 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 dont 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 satises the parallelogram law,> an inner product can be dened;> 2 2> = 1/4 ||u+v|| - 1/4 ||u-v|| that satises:> 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 denition of normedspace, or of linear space), then you have grounds for complaint.For example, the inner product in a COMPLEX linear space has somenon-real values, but the one dened above has only real values.There is a corresponding formula for complex vector spaces, similar tothis one, but with 4 terms instead of 2. After you prove the realversion, 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 satises the parallelogram law,> an inner product can be dened;> 2 2> = 1/4 ||u+v|| - 1/4 ||u-v|| > that satises:> 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 denition 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 dened 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 acomplex vector space into an inner product space. My question is now:can an inner 2 product with the property = ||u|| always be dened on a normedspace V which satises the parallelogram law? The answer for the casewhere the scalar eld of V is the real or complexes. What about thegeneral case?dan === > 2 > product with the property = ||u|| always be dened on a normed> space V which satises the parallelogram law? The answer for the case> where the scalar eld of V is the real or complexes. What about the> general case? danI have never seen a denition of inner product except in the realand complex cases.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === > 2 > product with the property = ||u|| always be dened on a normed> space V which satises the parallelogram law? The answer for the case> where the scalar eld of V is the real or complexes. What about the> general case?> danI have never seen a denition 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 elds with an innite rank valuation,appears to give such a denition for certain p-adic cases.Lee Rudolph === There is something about a more general denition of a normed spacewith a scalar eld aother than the real/complex numbers, inSerge Lang, Algebra, Springer 2002 on p.469ff === topological space (rather than perhaps the more common defn given in thecontext 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 limitpoint of a set M a subset of T if every nbd of x contains innitely manypoints of MMy question is whether the use of the phrase innitely many isnecessary.I ask this because when I took basic real analysis (in metricspaces), there were two equivalent denitions of a limit point, one ofwhich was the following:Given a metric space (X,r) where M is a subset of X, then x in X is a limitpoint of M iff every deleted nbd of x in X contains a point of M.Is it possible to dene a limit point in a topological space in the sameway and say,Given a topological space T = (X,tau), a point x in tau is called a limitpoint of a set M a subset of T if every deleted nbd of x contains a point ofM.The reason why I ask is that it seems this distinction might arise fornite 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 sincethere are 3 deleted nbds containing 1 (in tau) : nullset, {2}, and {3}.All of these sets are elements of M.However using Kolmogorov and Fomins defn, 1 is not a limit point of MMatthew Brenneman === What is meant by a dense matrix in spectral numerical methods?Marco-- === > What is meant by a dense matrix in spectral numerical methods? MarcoMost of the entries are non-zero === > What is meant by a dense matrix in spectral numerical methods? Marco Most of the entries are non-zeroSo 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 spectralcontext:)The opposite of sparse matrix, or in detail, a matrix lackingany useful (for programming purposes) pattern of predictablezero entries. === Can anyone point me to a book or website which contains some information onthe density of Sophie-Germain primes?GREG === While its true that all derivations are technically proofs, most people seemto distinguish proofs that can be accomplished by applying a series ofmechanical operations in succession, from something that requires, say,explanation in English. And yes I realize this is all very vague. At whatline does a proof go from being a derivation to being a proof-proof? Itdoesnt seem like there is one (that can dened formally at least). I hopesomeone knows what Im 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 des 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 des numerically one> takes a much smaller step size. There are theorems saying that> if the coefcients 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. Thats a long story (and Im not familiar with the details). Butas I said, there _are_ actual _theorems_ that _say_ how smalla step size is required for a certain accuracy. I gave a reference,where the name of the book may or may not have spelledcorrectly.>I guess for some equations>you could calculate likely and maximum errors but for others it may be>difcult 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. Wills 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. UllrichJDavid 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 des numerically> do so by solving difference equations...Your supposition is not relevant.Right. Whats an _example_ of a standard technique for solving desnumerically that does _not_ essentially reduce the thing to some sortof difference equation?>[...]Theres a clear difference between discrete and continuous>mathematics--and discrete mathematics is in general harder.Usually there isnt any reason to suppose that a difference equation>will match up with its analogous differential equation in terms of>the result of integration. Well thats progress.>And remember that Gauss was intrigued by>the closeness between the prime counting distribution and continuous>functions like li(x) possibly *because* theres 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>theyre so close, but it may lead to a means to check Riemanns>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. Youve given _no_indication _why_ anyone should believe that the solution to thatpde has anything to do with counting primes. Just sayingits a prime suspect over and over is not going to convinceanyone.>Now Im a discoverer. I get in, get the discovery and expect proper>recognition, so Im not excited about working through all the>ramications of my own discoveries. What is clear is that if>mathematicians get away with ignoring my work then you wont know how>much that discovery might reveal, now do you?Thats why mathematicians arent supposed to just sit on things. >Theyre 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 doesnt 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 dont continue to betray the trust of the world, and>give her a chance.and its not excusable. Many of you may think it ok to let them get>away with it, just because you cant imagine doing anything else. But>you see, someone will pay the price, I guarantee it.>James HarrisDavid 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 des 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 des 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 rst is to>rene 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 rst is a bit more complicated but will ultimately>give good results, except on some pathologically bad functions.Well right. I didnt mean to give the idea that the only thing oneever did was to use Eulers method with a smaller step size, infact I did point out that that was a very bad method. Themethods you allude to above _do_ all involve differenceequations, right?>Well, this is the basis. Much renement and improvement has been going>on in this eld. 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 des 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 Ullrichs comment, and I guess I should have directedit more clearly to him.David Ullrich has a history of talking on subjects where he doesnthave a clue, and I nd that annoying. Basically you can tosseverything he said as just guesses. Though hes an actual mathprofessor at Oklahoma State University, Ive yet to see any evidencethat he hangs out in the computer science department of that school,or that he hangs out with the mathematicians who use advancednumerical techniques.Unfortunately though, he *talks* on lots of subjects as if he actuallyhas relevant information, which is a common problem on Usenet:know-it-alls.Here he mentioned his suspicion about how real experts in the area dosomething, 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 ramications. As you say, discrete is usually> harder (analytically) but when it comes to a computer, discrete is> usually the only one possible.Well thats not necessarily true as its better to think of computersas logic machines, so whatever a human being can do in the area, likeintegrate without using difference equations or discrete methods, acomputer can conceivably be programmed to do as well.Thats why its also important that you dont get waylaid ordistracted by posters like David Ullrich, as you will wade through alot of useless information, when its better to nd good sources.Just remember that just because someone replies and *sounds* credible,it doesnt 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 des 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 Ullrichs comment, and I guess I should have directed>it more clearly to him.David Ullrich has a history of talking on subjects where he doesnt>have a clue, and I nd that annoying. Basically you can toss>everything he said as just guesses. Though hes an actual math>professor at Oklahoma State University, Ive 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 CountingFunctions that give the wrong answer because you have no ideawhat the precision of your square root function is? Guffaw.(Whats especially funny about this comment of yours is the timing;most of the time the things I do have nothing to do with numericalanalysis, but as it happens (i) right now I _am_ working on a programthat involves what one might call advanced numerical techniques,in particular making xed-point high-precision reals to use forFourier transforms, which are then to be used to multiply largenumbers (ii) there _is_ evidence of this on sci.math - see thethread 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 didnt seem to notice my request for a _counterexample_ towhat 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 ramications. As you say, discrete is usually> harder (analytically) but when it comes to a computer, discrete is> usually the only one possible.Well thats not necessarily true as its 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 coursecomputers can be programmed to do symbolic integration andalso to do symbolic solutions of differential equations! For aprofessional programmer you seem quite out of touch withwhat well-known programs do.You seemed to miss the word numerically in the statementI suspect that _most_ computer programs solving des numericallydo so by solving difference equations.>Thats why its also important that you dont get waylaid or>distracted by posters like David Ullrich, as you will wade through a>lot of useless information, when its better to nd good sources.Just remember that just because someone replies and *sounds* credible,>it doesnt mean they actually know of what they speak.Indeed. So why dont you answer that question Ive been asking:If r_1 = 1+2i and r_2 = 1-2i then which of r_1 and r_2 does thenumber sqrt(5) go with?>James HarrisDavid 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 des 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 Ullrichs comment, and I guess I should have directed> it more clearly to him.Yes, looking again I see it was Davids comment that you werereferring 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 doesnt> have a clue, and I nd that annoying. Basically you can toss> everything he said as just guesses. Though hes an actual math> professor at Oklahoma State University, Ive 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 acouple of previous threads and read many of his other posts. I havemostly found them interesting and helpful. The few that I did notenjoy 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 forcomputing advice and tips. Either I come to enjoy maths that isnothing to do with computers or if it does concern computers then itthe 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 ramications. As you say, discrete is usually> harder (analytically) but when it comes to a computer, discrete is> usually the only one possible. Well thats not necessarily true as its 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 acomputer may be able to analytically solve a differential equation andhence come up with an exact answer. It is quite likely that a goodprogram may solve an equation that I could not (because it wasprogrammed by someone better than me). There is also a chance that acomputer may solve an equation that has defeated all humans but onlybecause it could more tirelessely explore numerous possible solutions. However, I doubt that a program would solve an equation that wasbeyond any human provided the human was given enough time. Forexample a computer was famously used in solving the four colourproblem but the work it did could in principle have been done by ahuman. The computer just did it a whole lot faster. 2001 has notarrived yet and we do not seem to have genuinely intelligent computersreferring to HAL in the lm 2001.) > Thats why its also important that you dont get waylaid or> distracted by posters like David Ullrich, as you will wade through a> lot of useless information, when its better to nd good sources.I will make my own judgements on who I will let distract me. So far Ihave been quite happy to be distracted by Davids posts. > Just remember that just because someone replies and *sounds* credible,> it doesnt mean they actually know of what they speak.I may have been out of university quite a while but I did spend quitea while studying maths. I do not think that I would be too easilyduped. > James HarrisJ === >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 des 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 des numerically one> takes a much smaller step size. There are theorems saying that> if the coefcients 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. Thats a long story (and Im 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. Arethese theorems quite generally applicable or are they restricted tocertain classes of equation? If the theorems are quite generallyapplicable then I can stop worrying about this one. I was worriedthat a spaceship may go astray due to missing the correct solution tothe multiple body problem (*). So far it seems more likely that thespaceship will go astray due to a programming bug. So my professionseems 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>difcult 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. Wills 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 fasterthan I can read and understand.I managed to analyse the simple f(x) = f(x), f(0) = 1 case and as thestep size decreases, it clearly converges to the correct solution. Ihave not worked through the gamma function case yet. > David C. UllrichJ David C. UllrichJ === > 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 tohave gone astray. So thanks now.That name sounds familiar. There may be more knowledge lurking in mybrain than I realised but I have lost the index entries to much of it.J === Most of the error analysis that Ive seen deals only with local error. Global error is a function of how the local errors add up. A goodexample is a Hamiltonian system (or any chaotic system). Local errorscan be very small. But the local errors dont 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 thelocal errors add up to produce a global error. Recently, somethingcalled calculus on time scales has been developed where discrete andcontinuum calculus are unied. I have not had a chance to look intothis, 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. Thats a long story (and Im 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 forODEs are different than for parabolic PDEs and different than forelliptic PDEs. In the elliptic case the proof for nite differencemethods are different than for nite element methods. And so on.Still, the theorems are quite useful. Theyll either tell you thatcertain step sizes need to be small enough (search forCourant-Friedrichs-Levy (CFL) condition; it says that given a space stepof this much, you can not step more than that much in time) or it tellsyou how close you get to the solution in terms of order of magnitude ofthe 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 didnt mean to give the idea that the only thing one > ever did was to use Eulers 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 aremore equations involved.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: 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 des 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 Ullrichs comment, and I guess I should have directed> it more clearly to him. Yes, looking again I see it was Davids 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 postthat I was replying to after all. I do admit that I nd DavidUllrich to be annoyingly toxic. > David Ullrich has a history of talking on subjects where he doesnt> have a clue, and I nd that annoying. Basically you can toss> everything he said as just guesses. Though hes an actual math> professor at Oklahoma State University, Ive 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 youre patiently listening to him and showing what he sees asrespect, then why should he attack you?But if you challenge him, hes likely to get vicious.And if you argue with him, hell reply, reply, reply until you tire ofthe exchange.search at groups.google.com where David Ullrich is the author, wherehe uses the word idiot.For another aspect of David Ullrich, check where hes the author andhe 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 isnt bad for getting information. But it shouldntsurprise you that there are certain posters who have a rep for talkingto just about everyone as if they actually know the subjects well,when they have a hidden agenda.Dont take my word for it, do the search I recommended on postsauthored by David Ullrich where he uses the word idiot and see histrue colors.Just remember theres always something in it for the person replyingto you.Here I clearly have an axe to grind about Ullrich, so Im warning youabout 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 ramications. As you say, discrete is usually> harder (analytically) but when it comes to a computer, discrete is> usually the only one possible.> Well thats not necessarily true as its 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 lm 2001.)Yet there are people who daily use computers to handle such problems,and Im 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 youdug in and checked with credible and truly knowledgeable sources.> Thats why its also important that you dont get waylaid or> distracted by posters like David Ullrich, as you will wade through a> lot of useless information, when its better to nd 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 Davids posts.Of coure you will. And besides, Ive revealed my agenda which is notabout just talking to you.However, the fact is that youve already displayed that you dont seemto have learned much on the subject of this thread, but look at howmany times Ullrich has already posted.> Just remember that just because someone replies and *sounds* credible,> it doesnt 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 youre 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 Im warning you>about him in a way that sends a message to a lot of others.Youre sending a message all right, and very clearly. I suspectthat its not exactly the message you mean to be sending, butwhy should your personal comments be any different from yourmathematical 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 des 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 Ullrichs comment, and I guess I should have directed> it more clearly to him.> Yes, looking again I see it was Davids 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 nd David> Ullrich to be annoyingly toxic.I dont nd him so. > David Ullrich has a history of talking on subjects where he doesnt> have a clue, and I nd that annoying. Basically you can toss> everything he said as just guesses. Though hes an actual math> professor at Oklahoma State University, Ive 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 youre 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 butI did not see this as an attack. I do not expect everyone to thinkexactly the same as I do. > But if you challenge him, hes likely to get vicious. > And if you argue with him, hell 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 hes 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 isnt bad for getting information. But it shouldnt> 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. Dont 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 theres always something in it for the person replying> to you. Here I clearly have an axe to grind about Ullrich, so Im 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 ramications. As you say, discrete is usually> harder (analytically) but when it comes to a computer, discrete is> usually the only one possible.> Well thats not necessarily true as its 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 lm 2001.) Yet there are people who daily use computers to handle such problems,> and Im 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 visitthe labs of a major manufacturer quite frequently and attend numerousconferences and exhibitions. It is part of my job to keep track ofthe 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. > Thats why its also important that you dont get waylaid or> distracted by posters like David Ullrich, as you will wade through a> lot of useless information, when its better to nd 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 Davids posts. Of coure you will. And besides, Ive revealed my agenda which is not> about just talking to you. However, the fact is that youve already displayed that you dont 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 doesnt 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 youre new to Usenet and the Internet.Why do you think that? I am quite a long term user of Usenet and theInternet. > Welcome to the new Wild Wild West.I am a long term resident. > James HarrisThis will be my last post on the subject of David Ullrich. This is anewsgroup for discussing maths and not David.J === prime example of a global error,is the oating-point spec for hardware and software(IEEE-755, -855, I think;is variously implimented on Earth .-)> Most of the error analysis that Ive 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 dont 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 unied. 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 dExploitation de la Protocolsdes 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 -- LATTEMPTER de COUP DETAT, 3/30/81 23 -- Le FIN dHISTOIRE 24 -- LORDEUR du MONDE NOUVEAU 25 -- THYROID STORK !?! === Well David Ullrich is a math professor, and apparently has a big ego. > Now if youre 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. Thats not my point. > But if you challenge him, hes likely to get vicious.> And if you argue with him, hell 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 keptreplying to me, stalking me in threads, and even went so far as to goto newsgroups like alt.writing and others to do the stalking.Lets imagine that I started posting in reply to you anywhere you wenton Usenet, dont you think thatd be a bit odd?And remember Id *stop* replying to him, but he has kept up hisstalking 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 hes 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 isnt bad for getting information. But it shouldnt> 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.> Dont 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 theres always something in it for the person replying> to you.> Here I clearly have an axe to grind about Ullrich, so Im 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 Im 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.Thats not the subject.My guess is that youre annoyed at my implication that youve beentaken in by David Ullrich. Rather than consider the evidence, yourereacting as if you personally feel attacked.However, doesnt it even bother you slightly that you just bbedabout 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 thegroup after rst behaving as if you were ignorant?I think youve become defensive.That point is key: Now consider how much you now know versus what youmight know if you dug in and checked with credible and trulyknowledgeable sources.You answer is basically that you already have such sources, whichmakes me wonder why you created the thread.> Thats why its also important that you dont get waylaid or> distracted by posters like David Ullrich, as you will wade through a> lot of useless information, when its better to nd 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 Davids posts. Of coure you will. And besides, Ive revealed my agenda which is not> about just talking to you.> However, the fact is that youve already displayed that you dont 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 didnt sound like it to me, which is subjective, but yourreply sounded like you still didnt know of routine programs for doingintegrations, when from what Ive picked up in posts, much of the mathworld uses them routinely, and they dont necessarily rely ondifference techniques.You see thats part of the problem, the newsgroup readers dontnecessarily feel a responsibility to save you from people like DavidUllrich, a math professor, who has repeatedly demonstrated a lack ofknowledge of even basic computer tools for math.> Just remember that just because someone replies and *sounds* credible,> it doesnt 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 youre new to Usenet and the Internet. Why do you think that? I am quite a long term user of Usenet and the> Internet.You dont 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.Id be interested in someone with basic knowledge on the subjectgiving an overview of the actual state of the art in the area ofcomputer analysis with regard to differential equation, and differenceequations, including some of the known math experts in the eld.James Harris === Well David Ullrich is a math professor, and apparently has a big ego. > Now if youre 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. Thats not my point.> But if you challenge him, hes likely to get vicious.> And if you argue with him, hell 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. Lets imagine that I started posting in reply to you anywhere you went> on Usenet, dont you think thatd be a bit odd? And remember Id *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 hes 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 isnt bad for getting information. But it shouldnt> 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.> Dont 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 theres always something in it for the person replying> to you.> Here I clearly have an axe to grind about Ullrich, so Im 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 Im 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. Thats 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 whyI made those comments. > My guess is that youre annoyed at my implication that youve been> taken in by David Ullrich. Rather than consider the evidence, youre> reacting as if you personally feel attacked.No, I do not feel that I have been taken in by him. If I am annoyedat anyone, it is you for all this off topic stuff you are posting tomy thread. > However, doesnt it even bother you slightly that you just bbed> 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 rst behaving as if you were ignorant?No. I have some knowledge of computers and a rather rusty knowledgeof mathematics. I come here for fun and to try and clean some of therust of my maths knowledge. > I think youve become defensive.This comment of yours above You may *assume* certain things about thestate 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 formathematics. If it became important, I could drive 30 or so miles tomy old university and look in the library. > Thats why its also important that you dont get waylaid or> distracted by posters like David Ullrich, as you will wade through a> lot of useless information, when its better to nd 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 Davids posts.> Of coure you will. And besides, Ive revealed my agenda which is not> about just talking to you. However, the fact is that youve already displayed that you dont 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 didnt sound like it to me, which is subjective, but your> reply sounded like you still didnt know of routine programs for doing> integrations, when from what Ive picked up in posts, much of the math> world uses them routinely, and they dont necessarily rely on> difference techniques.See below. > You see thats part of the problem, the newsgroup readers dont> 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 doesnt 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 youre new to Usenet and the Internet. Why do you think that? I am quite a long term user of Usenet and the> Internet. You dont sound like it to me.I may be more modest and polite than the average Usenet user. I amalso British rather than American. Maybe you confuse these thingswith 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. > Id 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 eld.Others have also posted to this thread. > James HarrisThis is now my last post on the subject of me. I will no longer replyto you unless your post is relevant to the thread and interesting. Ido not wish to discuss me, David or anyone else any further. This isa 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 didnt you ask him to explain?Anyway, almost every random number generator in the universegenerates a sequence of numbers by saying x[0] = the seedand x[n+1] = f(x[n]) for some function f. Now think aboutwhat happens if you use two seeds, and the second seedhappens to be f(the rst seed)... (are the two sequences1, 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.DavidDavid C. Ullrich === Hauke Reddmann escribi.97 en elmensaje|nbe6im3$ks5$2@rzsun03.rrz.uni-hamburg.de:> Using the a/g mean ineqality, it should be provable> that x=y=z=9 is the largest solution possible, and> x<=y<=z<=9 isnt that big an area to be searched> even by hand :-)Using the a/g inequality was my rst attempt, but I only get n >= 9/a andn >= 9/g. totally unusefull to upper bound n...And x = y = z = 9 isnt the largest solution in any sense. By example, theseare the solutions with 1000 <= x + y + z <= 2000, x <= y <= z:x y z n= = = = 25 45 980 1 16 72 968 1 1 324 845 5 8 27 1225 6 9 75 1176 2 1 169 1156 9 4 200 1156 2 3 294 1089 2 4 81 1445 5 12 150 1458 1 1 363 1352 6 9 225 1521 1 2 121 1681 8 5 81 1849 5 1 289 1682 8Althought there are solutions with x, y z >= 9, with smaller values. Byexample9 9 36 1related with1 1 4 93 3 12 3As Will said, there are multiple solutions with n = 1, 2, 3, 4, 5, 6, 8 or9, but as far as x + y + z = 4000 there arent solutions for n = 7 or n > 9.But I still didnt found any theorethicall reasons for the lack n = 7 andthe limit n = 9 ...-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.comFor n = 7, any variable and the sum of the others two must be multiiple of7. But it is also true for n = 2, 3 and 5 ... === Context: (x+y+z)^2 = nxyz. (x,y,z must be positive integers)>Hauke Reddmann escribi.97 en el>mensaje|nbe6im3$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 isnt that big an area to be searched> even by hand :-)Using the a/g inequality was my rst 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 arent solutions for n = 7 or n > 9.>But I still didnt 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 reducesto a nite number of two-variable quadratic equations (and is thereforedecidable). Maybe someone can rene 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 coefcients, and thus if wesubstitute z <- (x+y)^2/z we get another solution. Since x+y+z ischosen 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 thatnxy <= 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, wehave the desired upper bound on n. Furthermore, for any given n,we must have 1 <= x <= 12/n, which means there are only nitelymany values of x to check. Each pair of (n,x) yields a quadraticDiophantine in y and z, which is algorithmically decidable.Maybe someone can see a simple way to characterise all solutionswith x = 1? Then for any other solution wed have n <= 6, whichnicely 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 wed have n <= 6, which>nicely rules out n = 7.When in doubt, try another descent. If (y+z+1)^2 = nyz, then by avery similar argument we have WLOG either z = y or z = y+1, whichleaves only the solutions (y,z,n) = (1,1,9) or (1,2,8). So togetherwith 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 wed 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.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === What follows is elaborated from the solution of Eryck Bryce WongFind all positive integers n such that there exists at least one solutiontothe 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,xed n, look for the solution with x + y + z minimum. Writing the equationlike one of second degree in z, it isz^2 - (nxy - 2x - 2y) + (x + y)^2 = 0If z and z, z < = z, are the solutions of this equation, we have zz =(x + y)^2 ==> z < = x + y. Then we havenxyz = (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 = 0It 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 = 0y = (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*1ii) z = y + 1(2y + 2)^2 = ny(y + 1) ==> (n - 4) y^2 + (n - 8)y - 4 = 0y = (- (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 solutionsn = 5, y = 4, z = 5 ==> (1 + 4 + 5)^2 = 5*1*4*5n = 6, and = 2, 3 z = ==> (1 + 2 + 3)^2 = 6*1*2*3n = 8, and = 1, 2 z = ==> (1 + 1 + 2)^2 = 8*1*1*2Therefore, 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.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === I have difculty with your approach. This is how I look at theproblem: First allow negative integers as well and make y = -x, z = nx^2. (Or nd other solutions allowing negative integers.) Now ask, givenone 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 beimpossible for n > 9? This is a very big jump for me, and I cantfollow 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,>xed n, look for the solution with x + y + z minimum. Writing the equation>like one of second degree in z, it isz^2 - (nxy - 2x - 2y) + (x + y)^2 = 0If z and z, z < = z, are the solutions of this equation, we have zz =>(x + y)^2 ==> z < = x + y. Then we havenxyz = (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 = 0It 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 = 0y = (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*1ii) z = y + 1(2y + 2)^2 = ny(y + 1) ==> (n - 4) y^2 + (n - 8)y - 4 = 0y = (- (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 solutionsn = 5, y = 4, z = 5 ==> (1 + 4 + 5)^2 = 5*1*4*5n = 6, and = 2, 3 z = ==> (1 + 2 + 3)^2 = 6*1*2*3n = 8, and = 1, 2 z = ==> (1 + 1 + 2)^2 = 8*1*1*2Therefore, 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 rst attempt, but I only get n >= 9/a and> n >= 9/g. totally unusefull to upper bound n...But if you rst eliminate n by the trick of my followupee(to get (x+y+z)^2=xyz) then it could work?!-- Hauke Reddmann <:-EX8 For our chemistry workgroup,remove math from the addressFor spamming, remove anything else === Jeffrey H escribi.97 en elmensaje|n3f098b32.1350813@news.mindspring.com:> I have difculty 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 nd other solutions allowing negative integers.)But the OP enunciate prescribe positive integers. And observe that the LHSis 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 cant> follow your logic very easily.It there are solutions for a xed n, it must be a solution (or more, dontcare) 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.96estroA 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, xed 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 Ericks 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! -- ErickI know the problem is over, but Im always the last one to leave the movietheater... :-)I was just interested in nding 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 < zI found these:y z2 33 88 2727 98And that made me look closer.As a quadratic in y, the equation has two solutions. Let y continue torepresent the smaller one, and w the larger, so also (1+z+w)^2 = 6zwThe sum of the roots is the negative of the coefcient of y, soy + w = -(2-4z), or w = 4z - y - 2. This gives us a recurrencerelationw_n = 4w_(n-1) - w_(n-2) - 2whose solution, starting with w_1 = 2 and w_2 = 3, isw_n = 1 + [(2+s)^(n-1) + (2-s)^(n-1)]/2 where s stands for sqrt(3)So any two successive terms of the sequence2 3 8 27 98 363 1352 5043 18818 ...give a solution to(1+y+z)^2 = 6yzIm pretty sure this work for all the cases, since all it needs is tohave a symmetric quadratic polynomial in y and z. === Given integer n>=2 , what is the maximal number of pairs of points withEuclidean 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 atesselation of the plane by equilateral triangles. If that isthe case, then the maximum number of pairs would beoor(3n - sqrt(12n-3)).For n of the form 3k^2 + 3k + 1, the points can be arranged in ahexagonal 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> oor(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> oor(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 differenttypes of equivilent packing, which involve shifting the planes uponwhich 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> oor(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 Eppsteins reply in this thread. It appears that I amsomewhere out in left eld, if Im in the ball park at all.> Given integer n>=2 , what is the maximal number of pairs of pointswith> 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> oor(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> oor(3n - sqrt(12n-3)).Do you assume that the points are distinct ?Yes, if you dont assume the points are distinct then the answer is verywell-known. Turans theorem tells us that the number of pairs is at mostthe number of edges of a complete tripartite graph on n vertices which isabout n^2/3. If the points are not required to be distinct, this boundis trivially achievable: split the n points equally at the corners of anequilateral triangle, and we get a complete tripartite graph. -- Erick === I have the following setup. I have a particular test function f inD(R^n) and a distribution g (I dont know that g is L^1_loc). Illdenote 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 dont know that g is inL^1_loc. All I know is that the integral exists and is nite. I hopethis isnt 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 dont know that g is L^1_loc). Ill> 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 dont know that g is in> L^1_loc. All I know is that the integral exists and is nite. I hope> this isnt a silly question.The only way I know that functions become distributions is through integration, so I cant make any sense of this. Lets 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 dont know that g is L^1_loc). Ill>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 ofthing _is_ it, in order that there should even _be_ afunction 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 dont know that g is in>L^1_loc. All I know is that the integral exists and is nite. I hope>this isnt a silly question.Its not a silly question. But it is a question about the technicaldetails, so it has to be stated much more precisely - neitherWWW nor I have any idea what you mean when you saythe integral of gf exists although g is not locally integrable.TDavid 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 hasnt 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? Isnt it enough that we answer the damned questions? LOL.)transform); in that case there is a function g lurking about. Theres also a limiting process involved. And in that case the answer to the OPs question is obviously yes. But until we have the question posed more clearly we cant 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 hasnt specifed how g is on one hand a function and on> the other a distribution. We now enter into the realm of: Guess what theOP> is asking. (Why do we play this game? Isnt it enough that we answer the> damned questions? LOL.) transform); in that case there is a function g lurking about. Theres also> a limiting process involved. And in that case the answer to the OPs> question is obviously yes. But until we have the question posed more> clearly we cant know for sure.Hey Wade, good job of second-guessing the OP. I wonder if well hearagain. === > 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 particularf.>My gripe is the OP hasnt 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? Isnt it enough that we answer the >damned questions? LOL.)transform); in that case there is a function g lurking about. Theres also >a limiting process involved. And in that case the answer to the OPs >question is obviously yes. But until we have the question posed more >clearly we cant know for sure.Yup.David C. Ullrich === > Whether the rst equality is legal is something we cant say> without more information, it seems to me. (I mean, at the> top you say that ^ is the _distribution_ FT. If thats what the> ^ in g^ means then int phi_j g^ does not _ofcially_ make> any sense, just as statements like h**1/2 g^ in L^2(R^n)> ofcial make no sense...> Im confused. Why doesnt this make sense? All Im saying is that I have a function g that can also be viewed asa distribution.When I take the distributional FT of g then granted i get adistribution 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 youre not sure whether it is or not.> We cant say whether it is or not - whether or not g is> given by integration against some function is part of> the _denition_ of g.) >The>second equality here is legal by denition of the FT. The 3rd>equality is legal here as in this work I also know that g in L^1_loc. Aargh. Heres a quote from your rst post: Keep in mind that I > dont know that g is in L^1_loc...> Sorry for this confusion but you should notice that the g in my rstpost is notthe g of the second post. In fact in my rst post youll remember Ididnt want to put toomany details in and so steered clear of the FT. So, the g of the rstpost is actually the g^ of my second post. So there is no difcultyin the last sentence. The seemingly contradictory statement shouldread I do not know that g^ is L^1_loc.Hope thats cleared that up.Ta,T === > Whether the rst equality is legal is something we cant say> without more information, it seems to me. (I mean, at the> top you say that ^ is the _distribution_ FT. If thats what the> ^ in g^ means then int phi_j g^ does not _ofcially_ make> any sense, just as statements like h**1/2 g^ in L^2(R^n)> ofcial make no sense...Im confused. Why doesnt this make sense? >All Im 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 thats whatyou mean then the question that you snipped for our inconvenience,whether or not int phi_j g^ = [g^,phi_j] holds, doesnt make muchsense - if g^ is given by integration against some function thenint phi_j g^ = [g^,phi_j] _does_ hold. So that cant be what youmean.Otoh if you dont know that g^ is given by integration againstsome function then I dont know what function g^ yourereferring to when you say that h**1/2 g^ coincides with an L^2function.> say something about how _if_ this and that distribution> is actually given by integration against this or that function...> now it turns out that youre not sure whether it is or not.> We cant say whether it is or not - whether or not g is> given by integration against some function is part of> the _denition_ of g.)>The>second equality here is legal by denition of the FT. The 3rd>equality is legal here as in this work I also know that g in L^1_loc.> Aargh. Heres a quote from your rst post: Keep in mind that I > dont know that g is in L^1_loc...Sorry for this confusion but you should notice that the g in my rst>post is not>the g of the second post. For heavens sake, how was one supposed to notice this?I give up.>In fact in my rst post youll remember I>didnt want to put too>many details in and so steered clear of the FT. So, the g of the rst>post is actually the g^ of my second post. So there is no difculty>in the last sentence. The seemingly contradictory statement should>read I do not know that g^ is L^1_loc.Hope thats cleared that up.Ta,TDavid C. Ullrich === >Sorry for this confusion but you should notice that the g in my rst>post is not>the g of the second post. For heavens sake, how was one supposed to notice this? I give up.> I concur, Ive made quite a mess of things in my explanation. So, Ithink Illback out of this thread gracefully now. I assure you that my intentionwasnt to irritate you David. Its back to Rudins Functional Analysisfor me to get to grips with the subtleties of distributions which Iclearly lack.Ta,T === > My gripe is the OP hasnt 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? Isnt it enough that we answer the > damned questions? LOL.)Apologies. I didnt 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 satisfyh**1/2 f^ in L^2(R^n)and h**1/2 g^ in L^2(R^n)so the integral certainly exists. What Id really like to do is unhat>the g so that I can rewrite the integral asint_{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 its IFT in>fact in this work (h f^)check is assumed to have compact support.>However, the unhating is not legal (at rst site at least) because g>is not L^2! Im condent there is some distributional argument that>will make this move legal.My rst 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 dont know if the rst equality is>legal - this was intended to be the substance of my rst post. Whether the rst equality is legal is something we cant saywithout more information, it seems to me. (I mean, at thetop you say that ^ is the _distribution_ FT. If thats what the^ in g^ means then int phi_j g^ does not _ofcially_ makeany sense, just as statements like h**1/2 g^ in L^2(R^n)ofcial make no sense...say something about how _if_ this and that distributionis actually given by integration against this or that function...now it turns out that youre not sure whether it is or not.We cant say whether it is or not - whether or not g isgiven by integration against some function is part ofthe _denition_ of g.)>The>second equality here is legal by denition of the FT. The 3rd>equality is legal here as in this work I also know that g in L^1_loc.Aargh. Heres a quote from your rst post: Keep in mind that I dont know that g is in L^1_loc...>I hope things have been made clearer. Am I on the right tracks? Ta,TDavid C. Ullrich === >Sorry for this confusion but you should notice that the g in my rst>post is not>the g of the second post. > For heavens sake, how was one supposed to notice this?> I give up.I concur, Ive made quite a mess of things in my explanation. So, I>think Ill>back out of this thread gracefully now. I assure you that my intention>wasnt to irritate you David. Dont worry about the irrittation - I guess I was in fact irritated,but just a little. But do note the part you snipped about why Idont see how various things youre saying make any sense:If youre assuming one thing then the question answersitself while if youre not assuming that then I honestly dontsee what something else _means_... its very possible Imjust being stupid there, but.>Its back to Rudins Functional Analysis>for me to get to grips with the subtleties of distributions which I>clearly lack.Ta,TDavid 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 tryingto nd the standard error of the estimate. I believe that I have Pinpointed it. I believe although that the equations that I came upwithequal 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 eld 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). (Its easy to check that> its 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 gure 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 matrixthat stabilizes v and V is in Omega.As the only scalar matrix in Omega is the identity, itacts 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 toGL(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 ofa point in the natural action of PGL(n-1,q) on PG(n-2,q).Writing down generators, as asked here, is trivial in anotherbasis, where v=(1,0,...,0) and V={x | =0} withu=v when p|n and u=(0,0,...0,1) otherwise.(and then conjugating by appropriate matrix) HTH,Dmitriihttp://www.thi.informatik.uni-frankfurt.de/~dima/ PS. Just as another poster in the thread, quiteproud of my ability to do things Im not paid for, either :) === [...]> Suppose we denote by Omega(n,q) the set of all matrices of ordernxn> over the eld 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). (Its easy to check that> its 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 Im not paid for, either :)Maybe you could post again and verify my correction. Also more detailswould be appreciated. === Is a continued fraction that extends in both directions aMathematical 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 Fibonaccicount 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 problembelongs 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 systemstogether a nifty result is the innite series of 3/2, 4/3, 5/5, 6/8,7/13 and so on... Its an interesting idea to me that it might be a ternary systemstructure using null innity, negative innity and positiveinnity; however, I know I ought to get the opinion of those moreskilled than I on this. So is it a possible that two irrational fractions are somehow a partof ternary system? Yes I did assume the irrational fraction on the left. That was myguess of how it might be. You might try a different fraction. Itsjust that I like that one. They all form systems of their own as well.. So Am I fooling myselfwith this?Ernst === Im not following your lingo very well, buti tried some thing like that, some time ago:just reverse the operation of nding 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 ONeill, reelectedto 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 isoutside my understanding, perhaps.But as for continued fractions going in both directions: Yes, thesecan 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 as with negative indexcould not always all be positive integers. (all as = 1 works,however)Then, if c(m) approaches a limit, it is the value of the two-wayCF....Maybe.I have not seen this kind of CF before, so my assumptions arecompletely unrigorous.If we arbirarily let a(0) be any term within the CF, then will thesame nal evaluation of the CF occur?And under which restrictions on the sequence of as does this kind ofCF converge, and converge to the same value whatever term wearbitrarily 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 as 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 nite 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 ODEy(x) - 7y(x) + 12y(x) = 0,after the substitution z=dy/dx, we havez(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 havey(dw/dy) = 7- 12/w -wwhich 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-easyquadratic 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 originalequation, 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 coefcients. The solutions should be ofthe 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 bey = 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.comHow 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, youd just care if it happened in your town, Youre not from New York, are you?But you still pray, now dont 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--Mensanator2 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 coefcients. 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 eld 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 eld 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 coefcients> has a complex root. What method is used to show the fundamental theorem?Lots are possible (Gauss found 4, essentially distinct) Simplest useLiouvilles theorem (any bounded analytical function is constant)> The related material below seems inadequate. DeMoivres 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. Kroneckers Theorem.> Let F be a eld and f(x) a nonconstant polynomial in F[x].> Then there exists an extension eld E of F> and an element u in E such that f(u) = 0. Show any such extension eld of the complex eld C is C?> Ouch, doesnt seem the way to go.Well, it is possible (if you add the idea that any odd degree polynomial ofR[X] has a real root : Euler-Lagrange, with a twist) but certainly notobvious (as the st was well-known since deMoivre :-), and the second wasadmitted as obvious in Euler and Gauss times) ---- === I am planning to teach Algebra for my daughter and amplanning to buy the following book from an onlineused book shop (B&N, Abe etc.)Elementary Algebra for Schools By Hall & Knight.---It was recommended to me by a friend as one of thebest books on the subject. After doing a littlebit of search, I found that there is a 3 volumeset by the same author as given below:Elementary Algebra for Schools Vol I, II, IIIby Hall & H.S. Wood.I have no access to the table of contents of eitherof the above two books and I would like to knowif 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 goodreviews, 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 mathequations? For example, sky is blue translated into sky = blue. Yes, Iknow that equation isnt accurate. Its just to give an idea of what Imean. If someone has done so, Id 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 its 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 isnt accurate. Its just to give an idea of what I> mean. If someone has done so, Id appreciate knowing the URLs on this topic>This happens all the time, but not as numerical algebra but as predicatesin formal languages or set theory. Do note Ive crossed post this threadinto sci.logic where you may get some chatter about mathematical logic.Thus we may write Blue(sky), sky has the property Blueor 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 isnt accurate. Its just to give an idea of what I>mean. If someone has done so, Id 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 Ive 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 butnoteworthy.Let A = {a(k)} and B ={b(k)} be two innite sequences of DISTINCTPOSITIVE integers.A and B and {c(k)} are also such that the limit(s) below converge (andthe innite 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, bydenition.)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)) *jThis 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 thosejs multiplied by the cs in the averages general terms.)A specic 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 thatP[X neq Y] = 0However I dont understant completely this denition. 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 dont understant completely this denition. 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 uniformlydistributed in [3,4]. Then of course P[X=Y] = 0 because X=Yis 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, andin this case P[X=Y]=0 again. Even though that event is not empty, itstill 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 CassidyElectrical Engineering Graduate StudentStanford University === the ellipsoid-method was the rst polynomial algorithm to solvelinaear programmming problems. I have some problems estimating theworst-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, Cmxn-matrix, encoding length, A0 matrix dening the shape of theellipsoid, 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 problemwith C,b, the encoding length of C and b. Is that the number ofcells needed to binary represent C and b (in the sense of a sum)? (Asmall example would be great.)I think its O(m*n) because for every entry of the mxn-matrix C theencoding length has to be calculated, right?. This means O(n^2) wouldbe 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. jpgDirk Vdm === If we choose randomly and uniformly two points p,q in the n-dimensionalunit 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 withone integration and a reasonable amount of computing.The moment generating function of the square of the distanceis m(t) = (6(exp(t) - 1 - t - t^2/2)/t^3)^n, and the Laplacetransform is just m(-t). Now compute int -m(-t)/sqrt(t) dt/sqrt(pi),the integral going from 0 to innity, and this is the answer.-- This address is for information only. I do not claim that these viewsare 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 expansionaround the mean gives: E sqrt(Z) =~ sqrt(mu) - 1/(8*mu^(3/2)) * sigma^2for a random variable Z with mean mu and variance sigma^2.The sum Z of squared distances in each component has mean n/6 andvariance 7n/180 giving an approximate expected distance of: sqrt(n/6) - (6/n)^(3/2) * 7*n/1440Simulation shows this approximation appears to be good to within about1% for all n.-- Kevin === Im looking for a function g such that for all positive integer a, b,c we have thatIntegral(0 -> 1) { g(ax)*g(bx)*g(cx) } dx = 1 if a = b = c = 0 otherwiseTo clarify, the integrand is the product of the three quantitiesg(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 otherwiseAre you familiar with the Dirac delta function?This is an improper function (strictly speaking a funcional) that satises:Integral(-innity -> innity) { f(x) delta(x) } dx = f(0)Informally, delta(x) is zero everywhere except for an innite 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 -> innity) { 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/ktimes as wide as the original delta.Now, if you dont like delta functions, you can do the same thingwith very narrow rectangular spikes, provided a, b and c are not too large. === After completing a degree in Computer Science, I realised I absolutelysucked at Math.I know I have the ability, so I want to learn. Can anyone point me inthe direction of any online material before I go a-hunting at thelibrary? Any ideas 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 ideas where the FAQ for this group went?>What math do you already know?The FAQ is quack and wont 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 ideas where the FAQ for this group went?I think Wikipedia, an on-line encyclopedia of sorts, is prettygood. It is searcheable, and also many pages have links todenitions, related areas, etc.The main page of Wikipedia is at:http://www.wikipedia.org/wiki/Main_PageDavid 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 ideas 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> denitions, related areas, etc. The main page of Wikipedia is at: http://www.wikipedia.org/wiki/Main_PageFor example, you might want to start here:http://www.wikipedia.org/wiki/MathematicsDavid 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 ideas where the FAQ for this group went?Online stuff is so untrustworthy. For example, as much as I like EricWeissteins MathWorld, there are lots of things in it that are marginal, andlots of things that are just plain wrong. Same with Wikipedia, I have foundHave you got an idea of what kind of math you would like to start with? Ifyou do, I would recommend nding a book in your chosen area andconcentrating on it. If youre not sure where you ought to start, let merecommend linear algebra. === > I hope you dont 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 havent made > some mistake in the above), just from different directions.Yes. Another way of looking at it is to choose all possible binarynumbers of length 11 containing exactly 7 1s (thus also 4 0s). Thereare 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 toposition 10. The vector of length 7 which lists in ascending order thepositions of the 7 1s in the original vector will be a reduced vectorin 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 nto binary numbers of length 11 containing 1s at the positionsdesignated by the components of e and f, then the operation of succ()is to rotate the binary number one position to the right, with therightmost bit being shifted into the leftmost position. This accountsfor the periodicity of the succ() function for val(e). The binarynumber here is equivalent to the parity vector obtained by reducingthe elements of the sequence generated by succ(val(e),b) mod 2.I came across this characterization of the loops, or attractors, moreor 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 anysmaller value of b. Also, the numbers in the 30 loops were all in therange 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 startingwith succ(y,c) is periodic with length s. This means that if we deney[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 recursionrelated to succ(y,c), except that at even steps, i.e. those withy[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 secondmultiplicand 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 yis succ(y,c) periodic? The particular solution that I gave is for c =2^s - 3^t, y = val(e). === > I hope you dont 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 whats up? Are you guys looking for an equation for all numbersgo to 1 ? If so how do you guys frame such a question? Im still planning mynext Hailstone generator. So I cound add things to it and follow whatyou guys are doing if yall will share. Here is a quip :) Dynamical Blind Men exploring innities of innite Elephants maynot be sure how many of the same Elephants they have touched in thepast week if they were alert.The ever learning Ernst :) === >Original Format > I hope you dont mind my thinking out loud, it helps me keepthis> stuff straight.> Not at all.> This is the most general solution to the question, for which valuesy> is succ(y,c) periodic? The particular solution that I gave is for c=> 2^s - 3^t, y = val(e).Hey whats up? Were discussing how to nd attractors. I assumed you were followingalong, after all, you asked the original question. If you need to, askquestions.Are you guys looking for an equation for all numbers go to 1 ?>I cant speak for Bill Daly, but I will say: no. There is no singleequation. What were talking about are specic equations that resultin loops. If you can nd an equation of a loop for some positivenumber other than b in a 3a+b system, then the Collatz Conjectureis 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 provenfalse does not mean something is true.But 3a+5, in addition to its b loop (which every system has)5 20 10 5 ...also has positive loops1 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 wellwonderwhat makes 3a+5 different from 3a+1?So we study the equations of known loops to see if they provide someinsight that we can apply towards 3a+1. Maybe we havent foundanother loop for 3a+1 because all weve checked are small loops.Someone says that the smallest loop would contain a quarter millionnumbers. So were dealing with numbers that cant be reached by bruteforce. Thats where induction helps.As an example, I asked myself can two Mersenne Numbers appear in a3a+1 sequence? (Mersenne Numbers are numbers of the form 2^n - 1. Inbinary, these are numbers that are all 1s). By induction, I can showthat- 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 cant have a sequence with twoeven-bit Mersenne Numbers. There is no need to do a brute force testof all the numbers. Note that that is only half a proof, it doesntsay anything about having two odd-bit Mersenne Numbers in a sequence.Or an odd-bit following an even-bit. But you also cant have aneven-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 cant nd the counter-example by brute force, maybe we canconstruct it. But to do that, we need to understand how loops areformed, hence, the interest in Hailstone Function Generators, parityvectors, etc. But is it likely that a huge loop exists? I dont 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 thatthis sequence- ends in a Mesenne Number (2^5 - 1 or 11111 in binary)- has a parity vector 10100Lets call 31 the Hailstone (H) and 27 the Seed (S). We say that 27 isa solution to the function given by S10100M. There are an innitenumber of such solutions, 31 just happens to be the rst. Now if wetreat the parity vector as a rep-unit, then S10100M is a FirstGeneration Type 10100 Mersenne Hailstone Function. A second generationfunction has two rep-units: S1010010100M. Now here is the simplequestionWhat is the rst 6th Generation Type 10100 Mersenne Hailstone?In other words, nd M for S101001010010100101001010010100M.Simple question, very difcult answer. I put together my HailstoneFunction Generator to answer that very question. M is gazillions oftimes larger than any number that can be reached by brute force, butcan easily be computed from the Hailstone Function Generator:M = 2^177149 - 1This number has over 53,000 decimal digits, and yes, I did verify itby running it through the Collatz process (if I remember correctly,over 2 million iterations). So just because every number has beentested up to 10^15 and just because the smallest possible loop wouldhave a quarter million iterations, the brute force approach has notproved that a gigantic loop is impossible.Another interesting critter I found is what I call a 6th GenerationQuadracycle Pulsar with 4-bit Rep-unit and 2-bit Resonator. Thisrequires 4098 bits to construct and is also beyond the range of bruteforce attacks. What makes a Pulsar interesting is that the binarypattern of the rep-units is regenerated during the Collatz run. Alas,the rep-units are consumed faster than they are regenerated, otherwiseI would have found the other way to falsify the conjecture: a numberthat goes to innity.By studying the Hailstone Functions, crossover points, attractors,etc., there may be a simple formula with REALLY large constants thatform a loop in 3a+1. Or maybe theres a REALLY large Pulsar thatdiverges. And no, I dont have any delusion that Ill ever nd any,but Im enjoying the journey. If so how do you guys frame such a question? Im still planning my>next Hailstone generator. So I cound add things to it and follow what>you guys are doing if yall will share.Its not a question of being unwilling to share. I cant tell fromyour replies whether you get what I write. The above is a summary.If you want more details, you only have to ask. Im not going to wastetime writing out detailed explanations if youre not interested. Here is a quip :) Dynamical Blind Men exploring innities of innite 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 dont mind my thinking out loud, it helps me keep> this> stuff straight.> Not at all.> Its not a question of being unwilling to share. I cant 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. Im not going to waste> time writing out detailed explanations if youre not interested.> Oh.. Cool... . Oh Im interested in learning. I can say I dont understand yourgoal but, I dont have the training you do. Note : Trigger words and phrases.. Like get suck... Do you knowabout trigger words and phrases? I will press on with the ret. I have a guess. Ernst === >Original Format > I hope you dont mind my thinking out loud, it helps me keep> this> stuff straight.> Not at all.> Its not a question of being unwilling to share. I cant 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. Im not going to waste> time writing out detailed explanations if youre not interested.> Oh.. Cool... . Oh Im interested in learning. Good. I am not claiming that I know anything profound, but as you cansee, that doesnt stop me from going on endlessly about it.> I can say I dont understand your goal Well, my goals probably wouldnt interest you, and they are not alwaysobvious.For example, you might think that this web page has something to dowith the Collatz Conjecture:http://members.aol.com/mensanator666/collatz/ bang0010.htmBut it does not. The goal here was an excercise in writing a perlprogram to generate slides used to create .gif animations. It just sohappened that I was learning about binary patterns in the CollatzConjecture at the time, so it became the subject of my animationstudy. Note that there is nothing on that page that gives you any clueas to what my real motivation was.> but, I dont have the training you do.I have no training. I am entirely self-taught. As I have said, I amjust 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 ret. I have a guess. > Ernst === > You know I am with you. I see more questions than answers but oh what> questions they are. :)> Im now at over load point. Ill 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, GCDs, rings, dynamical> systems and more. Not bad even if made me look foolish.The fools are those who dont 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 Ive previously said.> I need to take a closer look at my program, it may not be doing what I think> its doing.> I didnt 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.Ive changed my program. After nding an integer crossover point, Icalculate the rest of the odd numbers in the sequence and record theminimum one. I get a record that looks like:system: [3x+5]crossover: 31attractor: 19parity vector: [1, 3, 1](The parity vector list just shows the count of 0s in the vector. Itis 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 otherinteger crossover points:system: [3x+5]crossover: 19attractor: 19parity vector: [1, 1, 3]system: [3x+5]crossover: 49attractor: 19parity vector: [3, 1, 1]This gives me the best of both worlds. Ive three differentcrossovers/parity vectors if Im looking for patterns there, but asingle attractor if thats what Im interested in.> Even stranger things are happening in 5a+b. Im nding 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> Im not sure what to make of that.What happened here was that I inadvertently recorded 3a+b data as5a+b. Naturally, when I put the 3a+b attractors into the 5a+b Collatzprocess, strange things resulted. Strange because 5 just happens to bean attractor in BOTH 3a+5 AND 5a+5, so the above sequence almostworks. Good lesson here on the importance of verifying your computerprograms. It was an operator error, not the programs. I told it tocompute 3a+b and thats what it did. It wasnt the programs faultthat I directed the output to the 5a+b le.> I wont 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 butthe basic structure I see.Ernst === I want to nd a 3d point (x, y, z) in the 3d Cartesian space usingthe 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 theside 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 nd coordinates of p2 - this pointshould 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 p0to p2. How can I nd 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) =0This is dot product of the tangent vectors.(2) (x2 - x0)^2 + (y2 - y0)^2 + (z2 - z0)^2 = RThis 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. Im 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> theyre actually just integers (or undened). Is that necessarily true? Things arent true out of necessity.Which bit do you disagree with? The terminologys outragiously slack, butif youre prepared to identify 0 with 0 (mod 17) it should be obviouswhats meant.> Cant you work in, say, R/17Z and end up with> values in [0, 17)?Take care with that structure - Whats half of 7? Is it 3.5 or is it 12, as 2*12=7?OP didnt 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 structureyou 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 x my calculator! --For suitable values of 9. === > William,> Be careful, 2^9 /= 2^10 = 1024Whoops 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 nite 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 nite 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?Its 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 nite 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 nitely generated abelian groups,Math. Algorithms, 1 (1966), 13-26.But thats the wrong Smith!H.J.S. Smith, `On systems of linear indeterminate equations and congruencesPhilos. Trans. Royal Soc. London, CLI (1861), 293-326.is the correct Smith. (I didnt know he was of Balliol though!)Derek Holt.>Its 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 nite abelian group is a product of> r cyclics. This drops out immediately from the theory of the Smith> normal form.> Which Smith?>Its Henry John Stephen Smith of Balliol College.Its a rather disgusting way of expressing the Structure Theoremfor Finite Abelian Groups, isnt it?Who showed one could express such a groupas 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 Id 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 Ive been unsuccessful in nding 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. Im remodelling a bathroom, and tomake the old, sagging, cieling appear at, instead of unsagging itwith heavy machinery, I used the common, more pragmatic approachcalled furring: I needed to make six wedges of lumber, all the samecorresponding dimensions, in inches: the length of the wedge, thethicknesses of the wedge on the heavy side, and the thickness of thewedge on the light side: three numbers: respectively, 56, 1, and 0.I thought, 0?? am I doing this right?? How can there be zero inchesof wood on this side??? I was momentarily confused, checked mymeasurements, and then I thought. Oh, ZERO wood! And I realized thatthere was NO (that is, ZERO) wood there. And that it, there, aswell, existed: ZERO stone, ZERO cars, ZERO elephants. And what thatlittle point on the end of the wedge had in common with so many other,including unimaginable things, was this: it didnt exist there. So,technically, in the diagram, there was as much wood there, as therewas iron, refrigerators, angels, and dinosours.And that it was upon this mute point to be contemplatingsingularities, ie. that they cant exist, and the distinguishingcharacterization of the inverse of zero, is that it means EVERYTHING,including refrigerators, which brought, to me, a new meaning to theterm undened that screams for a boundary condition on the distanceforce 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 simplied 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 eld is very complex, some things are easy, somethings are hard. Some algorithms are obvious (you learned them in gradeschool: multiplying polynomials), some are not obvious (those are theones you saw in Knuth, and you probably learned in grade school but didnt realize it: long division and gcd of polynomials), some arehard/impossible/undecidable/unknown (undecidable means we know we cantdo it always) and hundreds of years of mathematical research has beenspent on them: factoring/solving polys/solving PDEs/..cant think of anunknown 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 specic choices. Any additions/extensions will pose ...uh... difculties. OK, add - and maybe rationals. and wed only be doing simplication, -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 inC (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 rst attempt, dont try to solve for vars yet. or use more than one var, or allow vars in the exponent.Mitch === Ive never posted before, but as it is, its 4:48am and for somestupid reason, Ive been trying to solve a double integral for almost3 hours now.Its just personal now (not sure of notation here so im going todisplay it twice):/ 1 / arccos y| | exp (sin(x)) dx dy/ 0 / 0orintegrate relative to y {0..1}integrate relative to x {0..arccos(y)}the equation: exp(sin(x)) dx dyMy trusty calculator can give me an answer, but for some reason(fatigue for example) I cant gure out how to come up with theanswer.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 cant gure out how to come up with the> answer. any help?Look up Fubinis theorem. In Maple:int(int(exp(sin(x)),x=0..arccos(y)),y=0..1); # returns unevaluatedplot(arccos(y),y=0..1); # strictly monotonicint(int(exp(sin(x)),y=0..cos(x)),x=0..Pi/2); # returns exp(1)-1-- Thomas RichardMaple SupportScientic Computers GmbHhttp://www.scientic.de === >Ive never posted before, but as it is, its 4:48am and for some>stupid reason, Ive been trying to solve a double integral for almost>3 hours now.Its just personal now (not sure of notation here so im going to>display it twice):>/ 1 / arccos y>| | exp (sin(x)) dx dy>/ 0 / 0orintegrate 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 cant gure out how to come up with the>answer.any help?Interchange the order of integration.-- http://www.math.fsu.edu/~bellenotbellenot 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 coefcients,> 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 thatpolynomial multiplication is representative, integrationmight be more. You will not have arithmetic on each line of code,but from time to time you have to translate a non-trivialarithmetic expression, like for reducing the degrees in thesquare-free factorization of the denominator by integrationby 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 doesnt give convenient> access to all aspects of the data, but sometimes that doesnt matter.> It makes interfacing with Maxima evaluation and simplication> more automatic, and (especially for the novice) it indeed provides very > generic operations of +, *, / ... in an inx 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 efciency since (unlike maximawhich can translate to Lisp) the user langage is interpreted.Unfortunately, when mathematica/maple started, C++ was in itsinfancy or not born at all, hence they had to choose C, andas I said earlier I nd C++ more adequate than C for a CAS.Now, if people could write mathematica or maple code with thesame efciency than kernel code and with a similar syntaxthey would probably do it. Thats why I nd C++ a good languagefor CAS, because I believe it will be possible to write CAS-userlanguage 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 sufciently > 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 identiers 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 identiers,because you dont have alpha or beta on your keyboard andyou want to use standard notation (like eps or epsilon fora small number). But you would use the 1 letter symbols if you had themon the keyboard. Now why did mathematicians decide to use thegreek or other alphabets?Because they do not want to use 2 or more letters identiers,to have a more readable notation. So your argumentation doesnot contradict my point that using 1 letter identiers forlocal variables is a good idea.sin, cos, exp are functions, not variable identiers, they havestandard notation, I dont see why you spoke of them in this context. === Of course sometimes you will need to use 2 or more letters identiers,>because you dont 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 identiers,>to have a more readable notation. So your argumentation does>not contradict my point that using 1 letter identiers for>local variables is a good idea.>sin, cos, exp are functions, not variable identiers, they have>standard notation, I dont see why you spoke of them in this context.>Didnt Mathematicians decide to use greek, because that was the languagethey spoke? Sorry I couldnt resist. Modern mathematicians use lotsof alphabets, because they learn with lots of alphabets. But why dothey continue to use them, even as textbooks try to remove them fromthe low level math courses?I think Mathematicians use greek letters and many alphabets becauseit 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 programmingtypes, alpha and a are both real numbers and hence the same programmingtype, but are mathematically different types. Some C programmers, mostly m$, had an ugly and IMHO awful system calledHungarian notation. Each variable name had an ugly wart, a prex whichindicated its type. Even local variables had to use this system, soloops were done using iI or i_i instead of i. While for many modest subroutines one can get away with using only onecharacter names, it is IMHO ridiculous to require all local variablesto have only one character. Is t time or temp? Is a area or just therst variable? Programming has made it clear that even limiting thenumber of characters to a dozen or so is too limiting. It is easy tojuggle a few one letter variables, but silly to do so when there area dozen.As a mathematician, I often use the identier `mess for a commonsubexpression. It always gets a laugh the rst time I use in a class.Other short words are also effective identiers, `area, `len, `dist,for example. Some identiers 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 dont do it when I know it would be noticably better with longeridentiers.-- http://www.math.fsu.edu/~bellenotbellenot 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 forparameters, 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 identierslike eps, tmp or args for example.But in my experience, for many variablesinside a math subprogram, there is a local 1-letter identierwhose name is appropriate (see list above, sometimes a 2-letteridentier like df for the derivative of a function). === Your presumptions are both totally wrong. Mais comme on dit, il ny a > pire sourd que celui qui ne veut pas entendre, et qui veut tuer son > chien laccuse de la rage. Je vous laisse .88 vos petits jeux dautiste. > Je me contenterai de vous mettre sous le nez que la syntaxe dexpression > 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 programmeactuellement un logiciel de calcul formel en C++ et jen aiprogramm.8e un en RPL (en gros du lisp a lenvers) qui est int.8egr.8esur les calculatrices HP4xG, donc je pense avoir un minimumdexperience sur la programmation dun logiciel de calcul formel.Jenseigne les maths, jai donc aussi quelque exp.8erience dansle domaine. Si vous trouvez que la syntaxe C a*b*c*(d+e+f) estfort .8eloign.8ee de la notation math.8ematique standard, alors quedire de la notation lisp! De plus cest lanotation standard utilis.8ee dans toutes les calculatrices etlogiciels de calcul formel, et cest la plus proche des notations1-d quon puisse trouver. === Having written a program does not make you an expert on allprograms or all programming languages. This is a mistakemany people make, but you could still learn.Having taught math does not make you an experton math (or math notation).Writing in RPL for a calculatorcertainly does not make you conversant with Scheme orANSI Common Lisp.> Your presumptions are both totally wrong. Mais comme on dit, il ny a > pire sourd que celui qui ne veut pas entendre, et qui veut tuer son > chien laccuse de la rage. Je vous laisse .88 vos petits jeux dautiste. > Je me contenterai de vous mettre sous le nez que la syntaxe > dexpression 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 jen ai> programm.8e un en RPL (en gros du lisp a lenvers) qui est int.8egr.8e> sur les calculatrices HP4xG, donc je pense avoir un minimum> dexperience sur la programmation dun logiciel de calcul formel.> Jenseigne les maths, jai 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 cest la> notation standard utilis.8ee dans toutes les calculatrices et> logiciels de calcul formel, et cest la plus proche des notations> 1-d quon 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 writingCAS application in two different languages and I have clearlya preference, I explained why and I said that each programmercould have a different choice based on different priorities.It seems some people in this newsgroup are absolutely convincedthan lisp is the only language that should be used to write aCAS, they used arguments that I dont nd convincing and Iexplained why.> This is a mistake> many people make, but you could still learn.> Its not with this kind of sentence that you will convinceother 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 eldand I have certainly a larger experience of math notationthan 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 pointsespecially 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 weresaying, 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 dont need my encouragement to write in Lisp(or other similar languages).What is missing is a set of tests to compare languages, which iswhat 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 didnt say that. I said that I liked the possibility to overloadarithmetic operators in C++ because I think it is very importantfor code readability inside a CAS. I wanted to give another bellsound because this thread started with someone asking to writea program in C to do some CAS tasks, and you answered promptlytrying to convince him to learn Lisp, then I said C++ was a betterchoice for him in my opinion, I explained why and the thread beganto derive as it seems is the rule for any thread when someonerecommend another language than lisp to program CAS (its a little bitlike some mathematicians with algebraic geometry:-))> Most people using C++ hardly need your encouragement to write in C++.> Most people using Lisp dont 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.> Im not particularly interested in tests to compare languages,tests give a biaised view on what is the best language for the tests.Im 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 aunts 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 beoutside the puke class. Admittly 3 elements is small to switch to sum notation, but I would have no trouble parsingprod {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 prodis either dened inductively or alternately order doesntmatter because of commutativity.-- http://www.math.fsu.edu/~bellenotbellenot math.fsu.edu +1.850.644.7189 (4053fax) === Actually, Bobs complaint holds just as much for the notation a*b*c*(d+e+f).Which do you do rst, compute s :=e+f, then t:= d+s then ....or do you add d+e rst?If you dont 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 answerof 1.0, and the other gives you 0.0.Now you may assume that only one of these is correct, andwill always be computed according to the denition ofthe programming language. Last I looked, C optimizingcompilers could do it either way, and in fact could do itdifferent ways depending on the setting of optimizationags.Now Bob says these operators + and * are BINARY. Sohe would require that you writea*(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.Lisps standard dening (+ a b c) explicitly does not tellyou the ordering; if it matters, do (+ (+ a b) c) or theother way... I dont know if this cures Bobs 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 aunts ALL standard mathematical notation. One can not expect that a computer language> is totally conformant to mathematical notation> (i.e. I dont like = for assignment and == for> equality either) but * a b c d e etc. is too far> out in left eld. (IMO) > You can lead a horses ass to knowledge, but you cant make him think. === I have published papers on both Fermata Last Theorem and GoldbachsConjecture. They are listed in drectories throughout the Internet andspecically at: web.math.fsu.edu/Science/Specialized, under InDefense of Mr. Fermat, and Goldbach Proof. I will try to addressX-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 thata.b-(a+b) is either a prime of twin primes or multiple of 3 , can uprove it?please do reply mehelp 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; === Ive been using CAS for sometime (my dissertation involved the use ofCAS with high school students).Although I realize Maple is more popular, I prefer MuPAD. I have bothapps (MuPAD 2.5.2 and Maple 8 on my mac). 1. I nd 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. ItMuPAD has never failed to perform for me . . . Ive been using itlately to investigate partitions of integers. The variouspartitioning and permutation functions are actually more extensive inMuPAD than in either Maple or Mathematica (at least as far as I know).I nd the language and syntax more intuitive, too - easier for youngkids to understand.-MTE === > Can you solve this equation for x? x log(kx) = (x - 1)log(1 - x) the answer might involve Lamberts W function, but Im not sure Its 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 - 1I was playing with the Q, stumbled into probability integrals and nowgive 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 lookingat the derivative], 0 <= x and fctValues = [1,innity[ .---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 Im looking for any suggestion. Here it is the conguration I have:% cc -VDEC C V5.9-005 on Digital UNIX V4.0 (Rev. 1229)% uname -a -pOSF1 ....... V4.0 1229 alpha% man lapackLAPACK(3DXML) LAPACK(3DXML)Digital DigitalName lapack - A library of linear algebra routinesDescription 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 & donts 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 dont 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 & donts 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 knowledgeabout oating point and IEEE arithmetic.The paper also available in different formats various places on theweb. === > I feel the needing of something more about do & donts 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 oating 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. NobleProfessor Emeritus of Physicsjvn@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 nd 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 & donts 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 > dont 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 & donts 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> dont 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-intensivecode. Most Cs are pretty well optimized so it should not be necessaryto descend into assembler to achieve sufcient speed.Jon Bentleys 2-volume Programming Pearls is a good place to learnsome 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 aroundthese days, so no need to sneer) is that it is hard to write factoredcode. Factoring is the art of dening small subroutines that dosimple things (so they dont have any built-in bugs) and thencombining them into larger subroutines.Here is an example in Forth (it is the subroutine that computes anintegral adaptively):: )integral ( f: xa xb err -- I[xa,xb]) ( xt --) initialize BEGIN N 0> WHILE subdivide converged? IF interpolate THEN REPEAT f nalI ( f: I[xa,xb] ) Ntimes @ . . function calls optional line;You can see immediately the structure of the program, since all thelower-case names are appropriately named subroutines that do theindividual tasks. Forth is optimized for this ne-grained decomposition,and that is my chief reason for liking it. (Forth purists dont like myForth style, BTW, since I use a formula translator to embed formulasin subroutines with f ... --they get translated into Forth automatically.)The Forth program, and two C equivalents (unfactored and factored) canbe found at http://www.phys.virginia.edu/classes/551.jvn.fall01/CiSE_ progs/Cprogs.htmunder item 5. The nal subroutine of the factored C version looks just likethat 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.nI ; }However, in order to achieve this I had to dene a struct to hold allsorts of data; the pointers into it make the other subroutineslook sort of baroque. I do not consider the result very readable.Finally, I disagree with the people who have bad things to say aboutNumerical 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 mostpurposes. (Read what Bentley has to say about optimization before worryingabout whether they run fast enough.)-- Julian V. NobleProfessor Emeritus of Physicsjvn@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 Cs are pretty well optimized so it should not be necessary> to descend into assembler to achieve sufcient speed.Might the increasing prevalence of Single Instruction Multiple Dataoperations (e.g. 3DNow, SSE*, Altivec) be an exception to this? If youhave an algorithm that can (at least in part) use 32 bit oating pointnumbers in parallel, you might be able to double or quadruple its speedusing one of those instruction sets; however your C/C++ compiler may notgenerate SIMD instructions at all and IIRC even the ones that do arentvery good at it yet.---Roy Stogner === > It is not hard to learn enough C to write decent numerical-intensive> code. Most Cs are pretty well optimized so it should not be necessary> to descend into assembler to achieve sufcient 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 oating 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 arent> very good at it yet.> ---> Roy StognerCertainly. Also the MMX instructions, or any specialized instructions thatgeneric compilers wont employ (because they arent on all target machines).And there are certain things that can be done MUCH faster in assembler thanin optimized C, Fortran, whatever. Case in point: I am writing a library forquad-precision oating point, wherein the exponent eld is the same asin IEEE 64-bit, but the mantissa has an additional 64 bits. There are lotsof C or Fortran multiple precision libraries around that give arbitraryprecision. But if you want maximum speed, you have to roll your own inassembler. When its done Ill report on it. -- Julian V. NobleProfessor Emeritus of Physicsjvn@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 problemsUse 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 & donts 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 |> dont 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 aint so. Itisnt.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 aint so. It> isnt. > Nick Maclaren.> === Write some code and nd out why its 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 isclose to their (numerical) capabilities, and see what they do.Then look at NAGs documentation and compare the number of errorreturns.To read the text, chase up a good book or two on numerical analysisand 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 & donts 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 > dont 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, youll see that thereare numerous reasons *NOT* to chose C as a programming language unlessyou are doing systems level programming (which isnt numerical stuffat all).In fact, when writing software, your primary focus should be on aproper design. Optimization should come at the end, and only afterproling reveals a bottleneck. Chose good algorithms instead of creatingmessy code.Ill recommend you take a look at either Python or Java. Python has agood selection of numerical software which you can use, and Java is agood choice if you need to create high-performance applications whichare as portable as possible. Also, Java features a very restricted setof pointer operations, which insulates you against most memory corruptionproblems possible in C.-- Bjrn-Ove HeimsundCentre for Integrated Petroleum ResearchUniversity 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 Im asking for is to improve my numerical programming skillness to be able to write better (I mean more efcient) 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, youll see that there> are numerous reasons NOT to chose C as a programming language unless> you are doing systems level programming (which isnt 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> proling reveals a bottleneck. Chose good algorithms instead of > creating messy code. Ill 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 nd 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 ne 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 methodswhich only use function values (for general nonlinear systems Broydensgood 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.hthpeter === About a year ago, somebody posted some information regardingselecting optimal subsets of combinations. I seem to recalla paper on a Web site and some comuter-based method formaking the selections, but Ive lost/misplaced everything Ihad about the subject. My apologies for the vagueness of myquestion, but I cant recall the papers title or subject. === >Is it true that for every n>=3 there is a nite 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 innite non-abelian groups ?TIA,Jadek === Is there a formula for the number of non-congruentsolutions to x^2 + y^2 == z^2 mod n for n>=1 ?TIABill === > 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: A062775URL: http://www.research.att.com/projects/OEIS?Anum= A062775Sequence: 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, 2475,2116,2209,4032,2695,2900Name: 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 eld is a eld with product formula if and only if it is one of> the following> 1. a number eld> 2. a function eld in one variable or a nitely algebraic> extension of such a eld> 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 havent understood right! I would be very pleased if you can help me with those two questions.You may nd 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 2004The Mediterranean Seminar of Algebra and Topology (aka. MAT seminar) ofMontpellier 2004 will consist of a series of introductorylectures on Arithmetic Quantum Chaos. The invited speakers are J. P. KEATING and P. SARNAK University of Bristol University of Princeton & Courant InstituteThe session will take place on the 23-24 January 2004 at the UniversityMontpellier 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 theseminar MAT; http://www.math.univ-montp2.fr/MATMs Lacan, lacan@math.univ-montp2.fror FAX +33 4.67.14.35.58Registration deadline: 9 January 2004.When you send your registration please give your full name,your dates of arrival and departure, your host institutionwith 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 nding the blocks that correspond to thisinclusion relation table I found in a publication athttp://www.math.colostate.edu/~betten/pub/99/steiner36. psShould be 62832 blocks,however my experience in combinatorics islimited to self instruction. If I could only nd some insructionalwebsites that will help me to better understand. Any help would begreatly appreciated.Tony Pope === (1) Are there available any criteria for a Banach space (over reals, if it matters) to have a predual? Sufcient 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. Gutmangutman@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 sufcient condition is that there is a positivesolution. That is, the w which have your property are all y/yfor (strictly) positive C^innity functions y.Proof: If w has the property, let y(t) = Y(v,t) be the solutionof y=w y with y(0) = 1 and y(0) = v. LetA = {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 propertytheir intersection is empty. Since R is connected, their union cantbe 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) = 0wed have y(t_0) = 0, and the solution with y(t_0) = 0 and y(t_0) = 0is identically 0.Conversely, if y_1 is a positive solution of y = w y, the generalsolution 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 isincreasing 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.caDepartment of Mathematics http://www.math.ubc.ca/~israelUniversity of British ColumbiaVancouver, 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 conrmthis but I wasnt able to prove.Could someone help me in proving it?Homer === slightly OT> Hey all,> I have the following (apparently common) problem, that is slightlydifferent> 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 forle> 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 le share.> d.. (X: I dont 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 ne and dandy until PC#2s> 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 98OSR2).> Or at least, I only noticed this poblem after xing PC#2. This didnt> happen before this, and as far as I can recall, I didnt make any major> changes to my XP machine (which is PC#1 and from which browsing any other> PCs on the network lags initially).> Also, this problem doesnt exist when browsing any other machine from anyof> the win98 machines. The last thing Id 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. Im really running out of ideas> here Doug === > slightly OTLOL -- 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 didnt see it posted. I amsorry if it is repeated.Mersenne prime has been discovered. Today is the rst day I heard of this.I havent 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 MichaelShafer using the GIMPS program.For those interested in a web URL tryhttp://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 didnt see it posted. I am>sorry if it is repeated.Mersenne prime has been discovered. Today is the rst day I heard of this.>I havent 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 thatdx=1/3 du therefore we multiply by 1/3 and the integral of sec u tan u issec u. Therefore you get 1/3 sec 3x + C-- David MoranChief MeteorologistOklahoma Storm Team === trial-and-error method? I havent 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 havent 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 uis> sec u. Therefore you get 1/3 sec 3x + C -- > David Moran> Chief Meteorologist> Oklahoma Storm TeamWell, think about it like this. What trig function has a derivative of sec xtan x? sec x. Now since theres a 3 in front of the x, you have sec 3x tan3x. When you take the derivative, youll have a 3 out front. Since you donthave a 3 out in front of the integrand, you need to multiply by 1/3.-- David MoranChief MeteorologistOklahoma Storm Team