mm-109 Im currently designing a set of interface objects for mac OS X, andone of the possible application would be an interactive spherical trigcourse.The idea is to allow people to mess around with sliders to change thevalues of sides and angles and get immediate feedback, to illustratethat abC always has a solution but abc not, and aAB can have multiplesolutions.Input, especially from those with an interest in teaching sphericaltrig, is appreciated.The OS X InterfaceBuilder palette is available from http://www.afront.be/bin/solid.tar.Zfeedback topaul afront bePaul =I am using hyperlatex (and latex2html in addition) to create web pages. But hyperlatex seems to age without evolving. Do you know a software which could replace I am using hyperlatex (and latex2html in addition) to create web pages. >But hyperlatex seems to age without evolving. Do you know a software >which could replace hyperlatex ?Not that thig ng is really *wrong*, but my general answer to postslike this one is to check comp.text.tex, especially since yourparticular question is not math-related, unlike the one of the posterwho asked, a few days ago, for a particular symbol...However Ive heard *very good* cmts of TeX4ht.Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc =Can someone help me to solve this problem?x+ax=ayy+by+2ay=axI would appreciate some link or directions how to solve it. Can someone help me to solve this problem?>x+ax=ay>y+by+2ay=axI would appreciate some link or directions how to solve it.Its a linear homogenous system so there exists a closed-form solutionfor an initial value problem. Reduce it to a rst-order system offour equations:u = Auwhere u = [x, x, y, y]^T and u = [x, x, y, y]. Then thesolution is:u(t) = e^(tA) * u0where u0 = [x0, x0, y0, y0]^T are the initial values and e^(tA) isthe matrix exponent of the coefcient matrix. =This was a problem given in a againabi == I have a series of Laplace transformed items I need tofind the inverted from of.a. s/(s+1)^2I think this must reduced to something in the form of coskt (k=1),with something added to that, to get rid of the 2s in the denominator,but thats where Im stuck.b. 2/(s^2+1)Same goes for this one, except Im thinking it must be reduced tosint, but here im stuck with the 2...c. 1/s(s+1)^2Dont even know where to start here.. Im having great troublefinding the partial fractions in mentionedexamples. If someone could explain it according to these examples, Iwould greatly appreciate it.. I have a series of Laplace transformed items I need tofind the>inverted from of.a. s/(s+1)^2I think this must reduced to something in the form of coskt (k=1),>with something added to that, to get rid of the 2s in the denominator,>but thats where Im stuck.>It wont give you a cosine.Write it as (s+1-1)/(s+1)^2 = (s+1)/(s+1)^2 - 1/(s+1)^2= 1/(s+1) - 1/(s+1)^2, call this F(s+1) and use the fact that the(s+1) is caused by a factor of e^(-t) in the inverse so thatLinverse(F(s+1)) = e^(-t) Linverse(F(s))= e^(-t) {Linverse(1/s) - Linverse(1/s^2)} etc.>b. 2/(s^2+1)Same goes for this one, except Im thinking it must be reduced to>sint, but here im stuck with the 2...>Remember the transform and its inverse are linear so:Linverse 2*F(s) = 2 * Linverse F(s)>c. 1/s(s+1)^2Dont even know where to start here.. Im having great troublefinding the partial fractions in mentioned>examples. If someone could explain it according to these examples, I>would greatly appreciate it..You should dig out your calculus book and review how to do partialfractions. You should be able to break this into something like:A/s + (Bs + C)/(s+1)^2and use the idea in (a) above.--Lynn Harakiri>>Is there a simple formula for the next Farey fraction?>>Let a/b be any irreducible rational number. Then, next Farey fraction is>>dened as the next element of the Farey sequence F_b. For example, nextI dont understand your comment. Was I sloppy with the denition? Or the>fact that the problem formulation is not general enough upset you?[previous cmt snipped - retrying!]I was not upset by anything. Ifind your request perfectly reasonable.Only I thought that the problem as stated was not completely clearbecause of that use of the word dened.Please do not be concerned by my comment!However quite about the only thing I know about Farey sequences isthat if a/b is the next Farey fraction, then ba-ab=1, so (1) b|ab+1and then a=(ab+1)/b. Now, if b satises (1) in place of b anda is dened accordingly, then(2) a/b < a/b < b(ab+1) < b(ab+1) < b > b.Therefore the smallest a/b is that with the largest b. So,starting with b=b-1 and *decreasing* its value, the rst bsatisfying (1) is the one you need.Dont know how efcient this naive algorithm can be, though (justthink of (b-1)/b).HTH,Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc = Then there is the idea that orthodox quantum theory with signal locality> is only a limited approximation like global Special Relativity and the> more general theory of the quantum information principle, including> quantum gravity, has signal nonlocality in a very essential way and> that all inner consciousness requires signal nonlocality in the sense> dened by Antony Valentini. Whoa! Did I just hear you suggest that orthodox quantum mechanics> just might be bunk?There are some new ideas out there concerning consciousness andthe interaction of quantum and classical motion. Chaos theory andcomplexity science is an entirely different approach, and one thathas far more potential.STUART A. KAUFFMANA TENTATIVE PHYSICAL HYPOTHESIS CONCERNING CONSCIOUSNESShttp://www.santafe.edu/s/People/kauffman/ Epilogue.htmlThus, molecular autonomous Agents are precisely the kind of organization that hasclosed classical causal loops in rich webs. Therefore, if any of the events along suchcausal loops are quantum events - photons hitting rhodopsin or chlorophyll, molecularconformational rearrangements in membranes and membrane buried proteins, etc., thenmolecular autonomous agents hitting against the Heisenberg uncertainty limit in theirdecision subvolumes are precisely the kinds of physical systems which shouldtypically be ne candidates to have closed linked web loops able to propagate quantumcoherent active information :-) Bjacoby =Good morningI was studying Complex Analysis in Ahlfors book and would like helpwith a question. Suppose a function f is dened on an open domain Dof the complex plane C and has values in C. If z= x+ i*y, then f canbe given by f(z)= v(x,y) + i w(x,y), where v and w are functions fromR^2 to R. If f is analytic, then its easy to show that v and wsatisfy the Cauchy-Riemann differential equations. Then Ahlfors saysthey also satisfy Laplace equation and are harmonic conjugate, becausethey have continuous partial derivatives with respect to x and y. Buthe doesnt prove it.Next, Ahlfors shows that, if v and w have continuous partialderivatives, then f exists in the same domain of f. This is not hardto understand, the proof takes into account that, for havingcontinuous partial derivatives, v and w are differentiable in D. Butthen Ahlfors just claims the converse is true. Im confused here. If s analytic in D, then we se all directional derivatives of v and wand so are v and w. But are these conditions enough strong to ensurecontinuity of the rst partial derivatives, or at leastdifferentiability? Its possible that all directional derivatives of afunction exist without the function being differentiable.Im confused here.Artur Good morning>I was studying Complex Analysis in Ahlfors book and would like help>with a question. Suppose a function f is dened on an open domain D>of the complex plane C and has values in C. If z= x+ i*y, then f can>be given by f(z)= v(x,y) + i w(x,y), where v and w are functions from>R^2 to R. If f is analytic, then its easy to show that v and w>satisfy the Cauchy-Riemann differential equations. Then Ahlfors says>they also satisfy Laplace equation and are harmonic conjugate, because>they have continuous partial derivatives with respect to x and y. But>he doesnt prove it.>Next, Ahlfors shows that, if v and w have continuous partial>derivatives, then f exists in the same domain of f. ??? Surely you mean if v and w have continuous partialderivatives and satisfy the Cauchy-Riemann equations.>This is not hard>to understand, the proof takes into account that, for having>continuous partial derivatives, v and w are differentiable in D. But>then Ahlfors just claims the converse is true. Im confused here. If f>is analytic in D, then we se all directional derivatives of v and w>and so are v and w. But are these conditions enough strong to ensure>continuity of the rst partial derivatives, or at least>differentiability? Its possible that all directional derivatives of a>function exist without the function being differentiable.>Im confused here.Im not sure exactly what the question is, but I suspect thatthis may be an answer: It is not obvious, but its true, thatif f has a derivative at every point of D then f (and hencev and w) are innitely differentiable. This follows from theCauchy-Goursat theorem. Im actually not familiar withAhlfors, but Ifind it hard to believe that theres no proofof this fact in there. (Hmm, actually although Im notactually familiar with the book Im almost certain theresa proof of the Cauchy-Goursat theorem, because if Irecall correctly I was talking to someone the other monthabout the fact that the proof in Ahlfors uses subdivisionof a rectangle into smaller rectangles, while the OneTrue Proof uses subdivision of a triangle into smallertriangles.)Are you certain that theres no proof of these things inthe book, or could it be that you just havent got thereyet? Look for Cauchy-Goursat - seems likely to methat there will be a proof of the fact that one derivativeimplies innite differentiability a little later...Sketch:1. C-G: If f exists at every point of D then int_T f = 0for every triangle T (such that T and the interior of T arecontained in D). Clever proof by contradicition: If |int_T f| = d > 0 then there exists T, one of thequarters of T, such that |int_T f| >= d/4. Repeat;now if z is the intersection of the sequence of trianglesit follows that f is not differentiable at T.2. If D is convex and f exists at every point of D thenthere exists F with f = F in D. (Fix p in D, deneF(z) = int_[p,z] f, where [p,z] is the line segment fromp to z, and use the fact that int_T f = 0 for trianglesto show that F = f.)3. If D is convex and f exists at every point of Dthen int_C f = 0 for every closed curve C in D(follows from f = F.)4. If f exists at every point of D then f satisesthe Cauchy Integral Formula for _disks_ containedin D (follows from 3).5. If f exists at every point of D then f is locallygiven by a power series (follows from 4.)(hats the True Proof; using triangles in 1makes the proof of 2 extremely simple. SinceAhlfors uses rectangles in 1 he must do somethingdifferent to nish the proof.)>Artur =Could I suggest - as a courtesy to the group - you use the subject eld(say group theory question) to save us others from reading it if we haveno knowledge. You might also get more group theoriests to respond that way;)> Suppose we have a group G and a homomorphism h exists taking> elements of G into a ring (R, *, +) such that h(x)*h(y) = h(x*y). Next, suppose that for a given ring, the number of such> non-isomorphic Gs could be vast or innite and that, because> of the complexity of the ring, we are initially only aware> of a few. Given one such G and a homomorphism, surely one way to> try ones luck atfinding at least some of the other Gs is to> systematically look for elements x, y of R such that there> exist m, n in naturals: x^m, y^n in h(G). Then we automatically> know the multiplicative inverses of x and y exist, and we> also know that the set> G(x,y) = {z | exists m, n >= 0: z = (x^m)*(y^n) }> -with x^0 dened as h(1)- is a group. In particular G(x) = {z | exists m >= 0: z = x^m } is a subgroup of G(x,y). Does this procedure have a name?> p.s. Im *not* asking this question as a hypothetical (that> would be really quite boring!), since I have used it> a few times to generate a string of new groups branching> out all over the place on a ring and, at the moment,> I dont particularly see and end to these groups in sight.> At rst glance, the method would appear inefcient:> how the hell does one keepfinding new x and ys (and zs> etc. !)? It seems that at least for the ring at hand, there> is a generic method for guessing new ones based on looking> for especially symmetric elements of R. Please remember that I am unfortunately still a novice in your> answers. C. Dement Suppose we have a group G and a homomorphism h exists taking > elements of G into a ring (R, *, +) such that h(x)*h(y) = h(x*y). > Next, suppose that for a given ring, the number of such > non-isomorphic Gs could be vast or innite and that, because> of the complexity of the ring, we are initially only aware > of a few. You ought to require that the homomorphism is injective. Though it is oddto call it a homomorphism really as you arent mapping between two objectsof the same type. Perhaps is an injective homomorphism to the group ofunits of R? Someone must have a better name than that. Anyway, if not injective, then there are innitely many Gs: just take ahomomorphism onto one G from another group (GxG etc)> Given one such G and a homomorphism, surely one way to > try ones luck atfinding at least some of the other Gs is to> systematically look for elements x, y of R such that there> exist m, n in naturals: x^m, y^n in h(G). Then we automatically> know the multiplicative inverses of x and y exist, and we > also know that the set> G(x,y) = {z | exists m, n >= 0: z = (x^m)*(y^n) }> -with x^0 dened as h(1)- is a group. In particular > G(x) = {z | exists m >= 0: z = x^m } is a subgroup of G(x,y).> Does this procedure have a name? > p.s. Im *not* asking this question as a hypothetical (that> would be really quite boring!), since I have used it> a few times to generate a string of new groups branching> out all over the place on a ring and, at the moment,> I dont particularly see and end to these groups in sight.> At rst glance, the method would appear inefcient: > how the hell does one keepfinding new x and ys (and zs > etc. !)? It seems that at least for the ring at hand, there> is a generic method for guessing new ones based on looking > for especially symmetric elements of R.> Please remember that I am unfortunately still a novice in your> answers.> C. Dement If one looks at group algebras rather than rings, then there are thegrouplike elements, but that would require you to look up Hopf algebras(or at least co-algebras) which might be more effort than its worth.Ionly mention it because its a symmetry thing too, it is (I think) the setof elements x in the algebra that get sent to x(tensor)x undercomultiplication. Problem 1: Let G be a simple graph with vertex set V. Let u,v in V. Find a cycle> C in G such that u,v in C and length(C) is minimised.>What if G provides no cycle containing u,v?G = u--v, for example.> My rst thought is tofind the shortest path from u to v, then remove> this andfind the shortest path back. However, this wont always work:> v---------o> /| |> / | |> / | |> 4 5---6 |> | long> | |> | |> 1---2 3 |> | / |> | / |> |/ |> u---------oMinimal cycles containing u,v are u,2,4,v,5,3,uu,long,v,5,2,u; u,2,4,v,long,u; etclong is just a short as u,2,4,v; u,3,5,v; and u,2,5,v> (the path marked long is long enough, or has enough vertices, to> ensure that any cycle containing it is suboptimal). Here the shortest cycle is (u,1,2,4,v,6,5,3,u). However, the shortest> path is (u,2,5,v) which leads to the cycle (u,2,5,v,u).> It is relatively straight forward to show that if the shortest path SLikely this needs be if a minimal path S> from u to v isnt wholly contained by the shortest cycle C (u,P1,v,P2,u), then S must have a common vertex with P1, and a common> vertex with P2. (I used proof by contradiction.) Further, if S = (u,u1,...,v1,v) then C = (u,u1,...,v,v1,...,u). That> is, the second vertex in S must be the second vertex in C (starting from> u), and the penultimate vertex in S must be the vertex following v in C.> (Again I use proof by contradiction.) Now the problem is just tofind paths P1,P2 in V such that P1 and P2> dont intersect, P1 = (u1,...,v), P2 = (v1,...,u) and that minimise> length(P1) + length(P2). Hence... Problem 2:> Let G be a simple graph with vertex set V and let {a,b,c,d} subset V.> Find paths P1 from a to b, P2 from c to d such that P1 and P2 have no> common vertex and length(P1) + length(P2) is minimised. Again I started thinking about shortest paths, but in some cases the> shortest path from a to b (c to d) will not be included...>This graph isnt simple, its weighted.> 2 2> a-------f-------b> / /> 1 1/ 1 /> / /> e * <------ no intersection> / / > 1/ 1 1/ > / / > c-------g-------d> 2 2 Here the optimal (and only) choices for P1 and P2 are (a,f,b) and> (c,g,d), but the shortest paths are (a,e,g,b) and (c,e,f,d).>Huh? Oh, you mean (a,f,b) and c,g,d are the shortestwhile a,e,g,b and c,e,f,d are the lightest or paths with minimal weight.> I really dont know how to proceed with this, any help would be most> appreciated.>Is the problem for simple graphs or weighted graphs?Would you distinguish between shortest, minimal length, lightest, minimalweight?> Problem 3:> Let G be a simple graph with vertex set V and let {a,b,c} be a subset of> V. Find paths P1 from a to b and P2 from b to c such that P1 and P2> have no common vertex and length(P1) + length(P2) is minimised. I think this is just a special case of problem 2, but Im not sure...>Likely same problem with making up your mind. Are you thinkingsimple or weighted graphs? =This is a little novelty isomorphism, one which is probably well known.But here goes anyway.I was wondering about the limit version of the product function, being theequivalent of the integral function for sums. Lets call it Infprod insteadof integral.Integral (f(x)) = lim n=1 to 00 sigma (over k) f(k/n)/nInfprod (f(x)) = lim n=1 to 00 prod (over k) f(k/n)^(1/n)What wonderful new properties would such a thing have?Well none, actually, becauseLog (Inf(f(x)) = Log prod f(k/n)^(1/n) = sum log((k/n) ^ (1/n) = (1/n) sum (log(k/n)) = Integral log(x)ThereforeInf (f(x)) = e ^ (Integral log f(x))How depressing! The whole theory of innite products reduced to the branchof integral calculus that deals with functions of the form log(f(x)).But of course the reverse is equally true. We could have developed ourproduct calculus (Inf) from the analogous denition of the limit in normalcalculus. Then some dumbass posts a message in a newsgroup which inventsthis new form of calculus based upon sums, and develops the new Integraloperator asIntegral (f(x)) = Log (Inf(e^x))So all of this new sum based integral is reduced to a single branch of Infcalculus that deals with functions of the form e^f(x).So normal calculus is isomorphic to Inf calculus, as each is isomorphic (?)to a subset of the other.Though it is interesting that although these two are effectively duals, Ihave never seen an example where innite product integrals appear. I cansee one reason in analysis of physical situations, as innite products donot preserve dimensions (metre, sec, kg) where sums (integrals) do. But ofcourse, calculus isnt just used to measure beam strength.Does anybody know of any applications for Inf calculus?Peter Webb This is a little novelty isomorphism, one which is probably well known.> But here goes anyway.> I was wondering about the limit version of the product function, being the> equivalent of the integral function for sums. Lets call it Infprod instead> of integral.> Integral (f(x)) = lim n=1 to 00 sigma (over k) f(k/n)/n> Infprod (f(x)) = lim n=1 to 00 prod (over k) f(k/n)^(1/n)> What wonderful new properties would such a thing have?> Well none, actually, because> Log (Inf(f(x)) = Log prod f(k/n)^(1/n)> = sum log((k/n) ^ (1/n)> = (1/n) sum (log(k/n))> = Integral log(x)> Therefore> Inf (f(x)) = e ^ (Integral log f(x))> How depressing! The whole theory of innite products reduced to the branch> of integral calculus that deals with functions of the form log(f(x)).> But of course the reverse is equally true. We could have developed our> product calculus (Inf) from the analogous denition of the limit in normal> calculus. Then some dumbass posts a message in a newsgroup which invents> this new form of calculus based upon sums, and develops the new Integral> operator as> Integral (f(x)) = Log (Inf(e^x))> So all of this new sum based integral is reduced to a single branch of Inf> calculus that deals with functions of the form e^f(x). So normal calculus is isomorphic to Inf calculus, as each is isomorphic (?)> to a subset of the other.> Though it is interesting that although these two are effectively duals, I> have never seen an example where innite product integrals appear. I can> see one reason in analysis of physical situations, as innite products do> not preserve dimensions (metre, sec, kg) where sums (integrals) do. But of> course, calculus isnt just used to measure beam strength.> Does anybody know of any applications for Inf calculus?> Peter WebbI dont know of any new applications from this, but I sure used toknow where tofind out. During the sixties I would see ads for FirstNew Calculus in 300 years! For something like 3 dollars you couldsend for enlightment on this great new advance. Around 1972 or 1973 anumber of copies of 2 little booklets showed up at the Berkeley MathDepartment, and they were on the same subject, the work of the GaussFoundation or some such. A read them. Perfectly valid mathematics,about the same idea as yours. They continued to hope for great thingsfrom this. If there have been any, they would know. I dont have aclue how to track them done, but I thought you mightfind thisamusing.Achava This is a little novelty isomorphism, one which is probably well known.> But here goes anyway.> I was wondering about the limit version of the product function, being the> equivalent of the integral function for sums. Lets call it Infprod instead> of integral.> Integral (f(x)) = lim n=1 to 00 sigma (over k) f(k/n)/n> Infprod (f(x)) = lim n=1 to 00 prod (over k) f(k/n)^(1/n)> What wonderful new properties would such a thing have?> Well none, actually, because> For commutative multiplication, as you found, this is not sointeresting. But it IS in the literature for non-commutativemultiplication (so that it does not reduce to the additive case bytaking logarithms). For example matrices. It is known as the productintegral.for example:Product integration with applications to differential equations by JohnDay Dollard and Charles N. FriedmanReading, Mass. : Addison-Wesley Pub. Co., 1979-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ Could a proof of the Goldbach Conjecture be worth of a Fields prize?> Dont you know the regulations for your own prize?> Fields Medals are restricted to those under age 40.>>Is this in the rules or is it merely a habit that the prize givers have>>got into?Rules. Heres what it states athttp://www.emis.math.ca/EMIS/mirror/IMU/medals/ At the 1924 International Congress of Mathematicians in Toronto, a> resolution was adopted that at each ICM, two gold medals should be> awarded to recognize outstanding mathematical achievement. Professor> J. C. Fields, a Canadian mathematician who was Secretary of the 1924> Congress, later donated funds establishing the medals which were> named in his honor. Consistent with Fieldss wish that the awards> recognize both existing work and the promise of future achievement,> it was agreed to restrict the medals to mathematicians not over> forty at the year of the Congress. In 1966 it was agreed that, in> light of the great expansion of mathematical research, up to four> medals could be awarded at each Congress.This is all very interesting - but does it answer the original question?I would say that a proof of Goldbachs conjecture by somebody lessthan 40 years old might or might not result in a Fields Medal beingawarded. It would be more likely to do so if the proof involved somefundamental new ideas which opened up major new directionsfor future research.Derek Holt. Paul B. Andersen >The only parameters used in the calculation is the gravitational>>eld of the Earth, and the speed and altitude of the clocks.>> So what is the actual connection between a few maths equations and the physical>> change in clock rates that is observed when they are in free fall?The connection is that the cause of the proper times of the clocks>is to be found in the gravitaional eld of the Earth and the speed>and altitude of the clocks, and that GR correctly accounts for these. Does that mean that the same clocks would not work if they were not in the> Earths eld?When cornered, talk nonsense, eh?> Or does it mean that one free fall is as good as any other, as far as clocks> are concerned?Gravity probe A?HahA!So there!>> Why dont you ever mention clock rates in other orbits? Dont they obey>> Einstein?You know the answer.> GR never got it wrong. HahA!So there!>>But:> It is a correction to compensate for the> fact that clocks circling the Earth in free fall experience mechanical changes> as well as having their characteristics altered due to cutting the magnetic> eld.>>So relativity has OBVIOUSLY nothing to do with the correction.>>Keep it up, Henry. Its fun!>>Paul, amused>> Until you can actually specify how the physical change occurs in these orbiting>> clocks, you are just making a complete fool of yourself.>> It is quite obvious that the clock rates have physically altered, as seen in>> the original Earth frame. Why? An equation? Fairies?Kick and scream all you want, Henry.Claiming that a correction calculated by GR is> NOT a relativistic correction and that the cause>of how the clocks behave is something not accounted>for in the equations that predict the behaviour correctly>to a precision better than 10^-12 is a stupidity beyond belief. If that is true, which I very much doubt because it was never tested properly,> it is purely coincidental. The prediction is wrong for other orbits anyway. For instance, it MUST be wrong for an orbit 1 centimetre above the Earths> surface since the gravitational potential is the same there as o the ground and> since we know that clock rates are not affected by movement. (to discourage> smartarse replies from the idiots here - you know who I mean - assume the earth> is perfectly round and has no atmosphere) So a clock circling the Earth at that height would not run at a different rate> from one at rest and so it woud remain always in synch with the GC.Beautiful, Henry. :-)Henry Wilson knows that the outcome of an unfeasible experimentnever done MUST support Henry Wilsons idea of reality,and thus will falsify GR.What about all the experiments that ARE done which falsiesHenry Wilsons idea of how reality should have been?HahA!So there!>But keep believing in fairies, if you whish.>I will stick to GR until the day it is clear that the GPS>never did work, but is a hoax made by the Great Conspiracy>of GPS Users. Paul, I am aware of you argument that the clock still runs at the same rate> but the length of a second varies with gravity and speed. Do you appreciate how pathetically weak that statement sounds?After all your vivid demonstrations I am getting to knowhow your mind works quite well now.So I can imagine.Paul =Henry, here is the statement of mine you responded to:| The P-code (military) pseudo random pattern is transmitted| with 10.23Mbps, but this bit rate isnt normally measured| from the ground.| The important point is that the 10.23MHz signal is| the frequency standard of the satellite clocks.| That is, the satellite clock counts 10 230 000 cycles for| each second it advances. The clock time is transmitted.| (Coded as a bit pattern modulated on two carriers.)| We know the frequency must be right since the satellite| clocks stay in synch with the GPS time, not because| the frequency is measured from the ground.To this you have responded:| It is all explained by the fact that lioght speed is source dependent.| Unless a GPS clock is directly overhead, it will have a velocity component wrt| the receiver.| To cut a long story short, the doppler method used by the system confuses this| effect with another that it tries to explain using relativity.| The error due to c+v may be rather small when the clocks are near vertical but| generally they are not. The transverse doppler correction assumes light speed| is c and not c+(3770.cos theta)Of course anybody understands that IF you from the groundmeasure the frequency of the carrier or anything modulated onto it,it will be Doppler shifted, and the Doppler shift will change all the timeas the satellite moves relative to the receiver. Of course anybodywill understand that it therefore blazingly obvious is practically impossibleto precisely measure the frequency emitted by the satellite with a receiveron the ground.THATS WHY IT ISNT DONE.So what the heck are you babbling about, and what is the relevance tothe posting of mine you responded to?You are not seriously claiming that source dependency can explainwhy an uncorrected clock in GPS orbit gains 38 us a day every daycompared to the GPS coordinated time, or that this have anythingwhatsoever with Doppler shift to do, are you?If you are, you must have lost your mind completely.Which, BTW, is a possibility I will not exclude.Its up to you.Paul +0100, Paul B. Andersen It refutes relativity.>>Sorry, SR predicts the result precisely.>> Wishful thinking!>> You can twist anything if you have enough faith... which you so vividly demonstrates. :-)You have indeed to twist facts into ction to make laser gyros>support your blind faith in source dependent speed of light.It is actually quite simple.>There are many types of what loosely can be called>laser gyros or ring gyros, the two main types>are the Sagnac ring gyro and the ring laser.>Common to them all is that the light source>is rotating with the ring (in ring lasers the sources>are the atoms in the lasing gas), and that the speed>of light is isotropic (c or c/n) in the non rotating (inertial) frame.>This is as predicted by SR and ether theories, but>blatantly obvious as opposed to the source dependent>light theory, where the speed of light should be isotropic>in the rotating frame (the ring frame) where the source>is stationary. Ah! Paul, rotating frames are very dangerous things. They can lead one up a> blind alley.> Ring gyros are like 4-mirror sagnacs but with an innite number of mirrors.Did you have a point?Thousands of ring lasers are in operation at any time, you willnd several of them in every commercial and military aircraft.If the speed of light were dependent on the velocity of the source,they wouldnt work. They do.>Why do you think Sagnac thought his experiment>supported the ether theory?The Sagnac experiment conrms SR. That isnt the general view.Of course it is.Claiming otherwise is plain stupidity.Or utter ignorance.The Sagnac experiment conrms Michelsons ether theory>The Sagnac experiment falsies source dependent light theory. Every experiment MUST support reality, ie, source dependency.Now, THATS a beauty. :-)And very illustrative of Henry Wilsons way of reasoning: I, Henry Wilson, denes reality. My fantasy world IS reality. All experiments MUST support my idea of reality. If they dont, they are faked. Or misinterpreted. Or awed. Every experiment ever perfomed conrming SR falls in this category.>The MMX conrms SR. Well, indirectly, yes. SR relies on source dependency. It clearly requires that> light travels at c from any source. Any observer at rest with that source will> receive the same light at c.> If the light is returned to the source by a mirror at rest wrt the source that> the return travel time of that light will be constant, irespective of the speed> of the system.It is an indisputable fact that the MMX conrms SR.Thats why it isnt disputed.>The MMX conrms source dependent light theory Absolutely correct. No doubt about that.Thats why it isnt disputed.>The MMX falsies Michelsons ether theory. Not exactly.> As I have previously explained, it is impossible to prove the non-existence of> anything.> Similarly, if something doesnt exist, it is impossible to assign to it> properties that can be tested experimentally.> However, my alternative explanation refutes the conventional argument for the> null result. Mine takes into account the angular departure of the splitting> mirror from 45 degrees due to aetherian length contraction and shows how the> cross beam would actually be deected backwards even though the return travel> time of the beam elements remains constant.Mindless babble.It is an indisputable fact that the MMX falsies Michelsons ether theory.Thats why it isnt disputed.>So which of the three mentioned theories is not>falsied by one of these two experiments? Source dependency of course.Every experiment MUST support [my idea of] reality, ie, source dependency.Thus even the experiments falsifying source dependency MUST support it!So there!You are twisting great, Henry! :-)>But you can twist anything if you have enough faith.>So you will keep believing in your fantasy world, wont you? Well I certainly havent twisted any of the above.The answer to that is so obvious that I wont have to state it. :-)This was possibly the best of them all! Every experiment MUST support reality, ie, source dependency.I think even you will have a problem with topping that one!Go for it!Paul Dishman In the case of ring gyros, the internal reection causes continuous> innitesimal speed change.>>The light is launched at c relative to the material of the>>ring (bearing in mind the effect of refractive index) and>>source-dependent (Ritzian) theory says it will continue to>>do so. There are no innitesimal speed changes relative>>to the material.>> why does the light move around the ring and not straight out the sides?>...>> The ring is just like a four mirror system with an innite number of>mirrors.Correct, the only major difference is that in the gyro the>speed would be c/n immediately so your comments on extinction>distance for binary star systems would not apply. They were not my comments.Sorry, someone talked of a source-dependent model and said thereason it didnt show up in binary star tests was because thespeed adjusted to c over ~1 light second distance. I thought itwas you, sorry if it wasnt.>>Again if the light is launched at c relative to the perimeter,>>it approaches the rst mirror at c+v and the mirror is moving>>at v so thats c relative to the mirror in Ritzian theory.>> The mirror isnt moving at v.Of course it is, its on the rotating table.> It has rotated slightly during the light travel>> time.Yes, thats true but the target will have moved too.> Also the light gets a velocity kick in the direction of mirror>> movement.No, the light approaches at c relative to the mirror so will>either be reected at c or reset to c however you want to>formulate the source dependency part. There is no speed>change relative to the mirrors or bre (always c) or the>lab (always c+v). Nah. ^ very small upward velocity> A> /------------ |> | <---c+v-dv> |> -------------c+v-2dv-> C> BThe velocity of the mirrors is parallel to their surfaces(I cant think of a way to add arrowheads but suppose thetable is rotating anticlockwise in your diagram) and alsoapplies to the source S and detector C. If the horizontalcomponent of v at S is v then the light is moving at c+vto the left. The horizontal component at A is also v soit approaches at c relative. The vertical component at Ais also v so the light moves down at c+v and so on: v / A c+v v / /----------C / v B c+v / vThe other way round it is c-v all the way.> The rays SA and BC are not quite parallel and this causes the fringemovement.. v / A S v / /<----------S | ^ | | | | | | v | ---------->/ / v B C / vIf the light is going the same way as the table, each anglehas to be slightly less than 90 degrees so that the lightgets to S instead of S because S will have moved.Now consider the light going the other way round the table: v / A S v / /---------->S ^ | | | | | | | | v ---------->/ / v B C / vThis time the angles each have to be slightly more than90 degrees for the light to reach S.Since one angle increases and the other decreases, put thetwo together to create fringes and the two beams meet atexactly 90 degrees regardless of the motion of the table.The only thing that affects the fringes is the time delayand in Ritzian theory, there is none since the change ofspeed in the lab frame exactly matches the change of pathlength.This is obvious in the rotating frame where the speed isalways c and the path length is always the circumferenceso the times taken are always equal.>Consider the replacing the usual sine wave of the light>with a narrow pulse waveform, the effect of the phase>change is then obvious. In the c+v model, the two pulses would arrive at different times and be> slightly displaced sideways. Thats what happens.No, the pulses would arrive at the same time. Sidewaysdisplacement is parallel to the wavefront so doesnt movethe fringes.George Let s(G) be the number of conjugacy classes of a group G.1. Prove that if G is nite and non-abelian, then s(G)>|Z(G)|+1, where Z(G) - the center of G. x in Z(G) iff {x} is a conjugacy class of G.>2. Prove that if G is nite and s(G) is even, then |G| is also even.If |G| is odd, then so are the sizes of all conjugacy classes of G.Derek Holt.> Let s(G) be the number of conjugacy classes of a group G.> 1. Prove that if G is nite and non-abelian, then s(G)>|Z(G)|+1, where Z(G) - the center of G.> 2. Prove that if G is nite and s(G) is even, then |G| is also even.> Try http://www.do-your-homework-yourself.comJoachim Let s(G) be the number of conjugacy classes of a group G.> 1. Prove that if G is nite and non-abelian, then s(G)>|Z(G)|+1, where Z(G) - the center of G.> 2. Prove that if G is nite and s(G) is even, then |G| is also even.> Try http://www.do-your-homework-yourself.com> JoachimCuriously, I follwed the link and there was nothing there. I am reallyamazed that the domain name is yet unoccupied. =http://www.geocities.com/actiondevicehttp:// www.geocities.com/v_deviceSo ultimate war between Mind Vs Matter is over.Today I woke up in evening at 05.00 PM. At about 07.00 PM, I went tobicycle shop and purchased two springs and two spokes for Rs.9/-I decided to use only one spring rst because in V-shaped twostretched springs, magnitude of force at vertex point B is given by c= ab*sin(theta). So if theta = 180 degree, c = ab*0 = 0. But if thesetwo springs make 180-degree angle to each other, it resembles onestretched spring. So I decided to use only one spring.But this spring is very stiff. I just cant pull it apart. I will haveto purchase spring of less stiffness.But, Hey, WAIT.....spoke and placed a stone at the center of bended spoke. Mass of stoneis far greater than mass of spoke. And I released both ends of spoke.I amfinding that spoke and stone are propelled in only one direction.No reaction mass is expelled or propellant is used. We have inertialpropulsion. Will some body out there repeat this small niceexperiment?and Arrows.And Welcome To The Universe!-Abhi. http://www.geocities.com/actiondevice> http://www.geocities.com/v_device> So ultimate war between Mind Vs Matter is over.> Today I woke up in evening at 05.00 PM. At about 07.00 PM, I went to> bicycle shop and won. Trolling wog idiot. Serves you right for coming to agunght armed with a putty knife.--Uncle Alhttp://www.mazepath.com/uncleal/qz.pdfhttp:// www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) http://www.geocities.com/actiondevice> http://www.geocities.com/v_device> So ultimate war between Mind Vs Matter is over.> Today I woke up in evening at 05.00 PM. At about 07.00 PM, I went to> bicycle shop and Matter won. Trolling wog idiot. Serves you right for coming to a> gunght armed with a putty knife.I am not using knife. I am using bows and arrows. Tomorrow, I will usefew rubber bands, may be scrap rubber piece of bicycle tube. Damn! Idont have any rubber bands in my room. But tomorrow, let us see,whether your guns win or my bow-arrow win.-Abhi. please click on this link to see this problem.> www.bol.ucla.edu/~khale/spring.gifIn your thread Integral, just one day before you started this new thread,you asked the same question. I dont understand why you thought itnecessary to start a new thread.Anyway, Id love to see a reasonably simple answer to your question. Butthere may not be one. Could you possibly give specic information aboutthe parameters p and q? That might help.BTW:To give you some idea of the problem of getting an antiderivative which isvalid for, say, all real values of the parameters, lets rst think aboutthe trivial case when p = q = 0. So youd be asking for an antiderivativeof Sqrt( cos(t)^2 ) = |cos(t)|. Since that function is continuous for allreal t, youd probably want an antiderivative which is valid for all realt also. Although I could give you such an antiderivative, most computeralgebra systems will provide antiderivatives which are not valid for allreal t, but which rather have spurious jump discontinuities.David Cantrell I am wondering if, parallel to the FAQ, we should not run an ofcial crank> list for newbies. Or at least an ofcious one. Or at least a word of> warning, if only to prevent people to believe this amount of rubbish is> typical of sci.math...Surely, this amount of rubbish *is* typical of sci.math.I doubt I would read it otherwise.-- Jesse F. HughesI thought it relevant to inform that I notied the FBI a couple ofmonths ago about some of the math issues Ive brought up here. -- James S. Harris gives Special Agent Fox a new assignment. =Hallo,Lets assume, Im in vector space |R^4 andhave given the following set of vectors:L( (1,2,2,-1); (-2,-3,-3,3); (2,-1,0,-1); (3,-7,-4,2); (8,1,4,-5) )If I want tofind a basis for |R^4, I use the following method:(1) I reduce a homogeneous equation of the form av_1+bv_2+cv_2+dv_3 = 0 using the Gauss-Jordan-algorithm.(2) By interpreting the result I can say that L(v_1,v_2,v_3) is a basis for |R^4 in my case.Now I read the following denition for the dimension of a vector space:A vector space has the dimension n, if and only if it has n linearlyindependent vectors and (n+1) linearly dependent vectors.Now I wonder, whether my result is correct, because I have 3 vectors inmy basis and not 4. For instance the canonical basis of |R^4 consists of4 vectors.Can anybody explain me my mistake?Karl[P.S. Sorry for my possibly bad English!] Now I read the following denition for the dimension of a vector space:> A vector space has the dimension n, if and only if it has n linearly> independent vectors and (n+1) linearly dependent vectors.I hope you didnt actually read that! It aint right :-(What is the case is that a vector space V has dimension n iff(i) it has a linearly independent subset of size n, and(ii) *all* subsets of size n+1 are linearly dependent.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) Hallo,> Lets assume, Im in vector space |R^4 and> have given the following set of vectors:> L( (1,2,2,-1); (-2,-3,-3,3); (2,-1,0,-1); (3,-7,-4,2);> (8,1,4,-5) )All of which are in R^4.> If I want tofind a basis for |R^4, I use the following method:No.> (1) I reduce a homogeneous equation of the form> av_1+bv_2+cv_2+dv_3 = 0 using the Gauss-Jordan-algorithm.> (2) By interpreting the result I can say that L(v_1,v_2,v_3) is> a basis for |R^4 in my case.No.You are in the spaceL( (1,2,2,-1); (-2,-3,-3,3); (2,-1,0,-1); (3,-7,-4,2); (8,1,4,-5) )all of whose vectors are in R^4.You live in three dimensional space. Take a plane. Take thex and y unit vectors in the plane. You will not get a basisfor 3-space using them (only a basis for the subspace, theplane, that they span).That the vectors are all in R^4 does not mean that you can geta basis for R^4 from them.Consider:L((1,0,0,0);(1,0,0,0);(1,0,0,0);(1,0,0,0))That is a one dimensional subspace of R^4 with basis (1,0,0,0).What is the dimension of L( (1,2,2,-1); (-2,-3,-3,3); (2,-1,0,-1); (3,-7,-4,2); (8,1,4,-5) )?> Now I read the following denition for the dimension of a vector space:> A vector space has the dimension n, if and only if it has n linearly> independent vectors and (n+1) linearly dependent vectors.No.It has n linearly independent vectors and ANY SET of n+1 vectorsis linearly independent.For example, in the plane, we have one independent vector,(1,0). Here is a set of two linearly dependent vectors,(1,0),(2,0). The plane has two dimensions (not EVERY SETof two vectors is dependent, for example, (1,0),(0,1)).The fact that I canfind two linearly dependent vectorsdoes not matter, the question is can Ifind two linearlyINDEPENDENT vectors.> Can anybody explain me my mistake?What is the dimension of: L( (1,2,2,-1); (-2,-3,-3,3); (2,-1,0,-1); (3,-7,-4,2); (8,1,4,-5) )? You live in three dimensional space. Take a plane. Take the> x and y unit vectors in the plane. You will not get a basis> for 3-space using them (only a basis for the subspace, the> plane, that they span).> That the vectors are all in R^4 does not mean that you can get> a basis for R^4 from them.> Consider:> L((1,0,0,0);(1,0,0,0);(1,0,0,0);(1,0,0,0))> That is a one dimensional subspace of R^4 with basis (1,0,0,0).> What is the dimension of > L( (1,2,2,-1); (-2,-3,-3,3); (2,-1,0,-1); (3,-7,-4,2);> (8,1,4,-5) )?> If I understand you right, I have found a basis for a 3D-subspacewithin |R^4. Is that correct?So the dimension of the linear wrapper L, Im searching a basis for,is 4. But if Gauss-Jordan leads to 3 4D-vectors, when where is nobasis for |R^4 from the given linear wrapper. Is that correct?If its correct, how can I build a 3D-space from 3 vectors with 4dimensions? Or did I actually found a basis for a 4D-plane and nota space?Karl[P.S.: Sorry for my possibly bad English!] You live in three dimensional space. Take a plane. Take the> x and y unit vectors in the plane. You will not get a basis> for 3-space using them (only a basis for the subspace, the> plane, that they span).> That the vectors are all in R^4 does not mean that you can get> a basis for R^4 from them. Consider:> L((1,0,0,0);(1,0,0,0);(1,0,0,0);(1,0,0,0))> That is a one dimensional subspace of R^4 with basis (1,0,0,0).> What is the dimension of > L( (1,2,2,-1); (-2,-3,-3,3); (2,-1,0,-1); (3,-7,-4,2);> (8,1,4,-5) )?> If I understand you right, I have found a basis for a 3D-subspace> within |R^4. Is that correct?> So the dimension of the linear wrapper L, Im searching a basis for,> is 4. But if Gauss-Jordan leads to 3 4D-vectors, when where is no> basis for |R^4 from the given linear wrapper. Is that correct?> If its correct, how can I build a 3D-space from 3 vectors with 4> dimensions? Or did I actually found a basis for a 4D-plane and not> a space?> Karl> [P.S.: Sorry for my possibly bad English!]>The set of all linear combinations of a set of vectors forms a vector space called the _span_ of the set. The dimension of this space is the largest number of linearly independent vectors in the original set, which may be less than the total number of vectors in the set. In your example, the 4 vectors only span a 3 dimensional space. And any 3 indiependent vecors in that 3-space form a _basis_ for that space. The set of all linear combinations of a set of vectors forms a vector> space called the _span_ of the set. The dimension of this space is the largest number of linearly> independent vectors in the original set, which may be less than the> total number of vectors in the set. In your example, the 4(?) vectors only span a 3 dimensional space.> And any 3 independent vectors in that 3-space form a _basis_ for that> space.So in my case this could be L((1,0,0),(0,1,0),(0,0,1)) (without the fourthcoordinates, because they are useless in |R^3.) Right?Karl[P.S. (?) Shouldnt it be 5, or did I misunderstand something? By the way, I know this is another theme, but I didnt want to start anextrathread for this: Do you, or does anybody know a good internet-site aboutthe basics of descriptive matrices and basis-transformation matrices? Imlookingfor a site with lots of examples but didntfind anything so far.] = 1. For every nite set A1, A2, ......An of axioms of ZFC, it is> provable in ZFC that these axioms are consistent.> The statement is provable by very elementary means: we> show how to construct a proof in ZF of A -> there is a model of A> for every sentence A in the language of set theory. This can be done> by nitistic reasoning, and thus certainly in ZF. In particular, by> choosing A to be a conjunction of axioms of ZF, we conclude that> for every nite subtheory A1&...&An of ZF, it is provable in ZF> that the theory is consistent. What we cannot do in ZF is to conclude> from this that every nite subtheory of ZF *is* consistent.Id love to see a couple of specic examples of formal proofs ofconsistency of subsets of ZFC in ZFC. (Actually, I just want to testout my simplication of ZFC and formal proofs within ZFC are so hardtofind.)BTW: How could a set of axioms of ZFC not be nite (since there areonly nite many ZFC axioms in the rst place)? And what if we useall of the ZFC axioms for the subset? (unconfuse me)Charlie Volkstorf = > 1. For every nite set A1, A2, ......An of axioms of ZFC, it is> provable in ZFC that these axioms are consistent.> The statement is provable by very elementary means: we> show how to construct a proof in ZF of A -> there is a model of A> for every sentence A in the language of set theory. This can be done> by nitistic reasoning, and thus certainly in ZF. In particular, by> choosing A to be a conjunction of axioms of ZF, we conclude that> for every nite subtheory A1&...&An of ZF, it is provable in ZF> that the theory is consistent. What we cannot do in ZF is to conclude> from this that every nite subtheory of ZF *is* consistent.> Id love to see a couple of specic examples of formal proofs of> consistency of subsets of ZFC in ZFC. (Actually, I just want to test> out my simplication of ZFC and formal proofs within ZFC are so hard> tofind.)> BTW: How could a set of axioms of ZFC not be nite (since there are> only nite many ZFC axioms in the rst place)? And what if we use> all of the ZFC axioms for the subset? (unconfuse me)> Charlie VolkstorfEach axiom is a rst-order sentence. ZFC contains innitely many of them(some are schemas). BTW: How could a set of axioms of ZFC not be nite (since there are>only nite many ZFC axioms in the rst place)? And what if we use>all of the ZFC axioms for the subset? (unconfuse me)In spite of Torkels response, the correct answer is that there are innitely many axioms (the axioms of separation and collections are both actually axiom schemes: there are innitely many of each, as each is dependent on a set theoretic formula, and there are innitely many such formulae).David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. > 1. For every nite set A1, A2, ......An of axioms of ZFC, it is>> provable in ZFC that these axioms are consistent.> The statement is provable by very elementary means: we>> show how to construct a proof in ZF of A -> there is a model of A>> for every sentence A in the language of set theory. This can be done>> by nitistic reasoning, and thus certainly in ZF. In particular, by>> choosing A to be a conjunction of axioms of ZF, we conclude that>> for every nite subtheory A1&...&An of ZF, it is provable in ZF>> that the theory is consistent. What we cannot do in ZF is to conclude>> from this that every nite subtheory of ZF *is* consistent.Id love to see a couple of specic examples of formal proofs of>consistency of subsets of ZFC in ZFC. (Actually, I just want to test>out my simplication of ZFC and formal proofs within ZFC are so hard>tofind.)BTW: How could a set of axioms of ZFC not be nite (since there are>only nite many ZFC axioms in the rst place)? And what if we use>all of the ZFC axioms for the subset? (unconfuse me)Actually ZFC has innitely many axioms. Those innitely manyaxioms can be described by nitely many axiom _schemata_.(Schema? Oh hell, schemes?)>Charlie Volkstorf************************David C. Ullrich BTW: How could a set of axioms of ZFC not be nite (since there are> only nite many ZFC axioms in the rst place)? A mystery! The conclusion is inescapable: ZFC is inconsistent. =Exercise 13, chapter 6, Real and Complex Analysis (Rudin):Prove that L^inf([0,1],m) is not isometric to L^1([0,1],m) by showingthat there exists a bounded lineal nonzero functional T on the rstspace such that Tf=0 for all f continuous on [0,1].Any help appreciated!nojb. =On 17 Jan OjedaExercise 13, chapter 6, Real and Complex Analysis (Rudin):Prove that L^inf([0,1],m) is not isometric to L^1([0,1],m) by showing>that there exists a bounded lineal nonzero functional T on the rst>space such that Tf=0 for all f continuous on [0,1].No, thats not what the exercise says. I cant tell you what it_does_ say, because the book is at the ofce and Im at home(and its raining), but the exercise doesnt say that, I guaranteeyou.>Any help appreciated!What does the exercise actually ask?>nojb.************************David C. Ullrich Exercise 13, chapter 6, Real and Complex Analysis (Rudin):> Prove that L^inf([0,1],m) is not isometric to L^1([0,1],m) by showing> that there exists a bounded lineal nonzero functional T on the rst> space such that Tf=0 for all f continuous on [0,1].> Any help appreciated!> you can construct such a linear funcional. Then extend it toall of L^inf by the Hahn-Banach theorem. =may you give me a sequence (obviously in a non-Hausdorff topological space)with two distinct limits?Tern may you give me a sequence (obviously in a non-Hausdorff topologicalspace)> with two distinct limits?>Take X any non-empty topological space and (Y,{{},Y}) then any f: X->Y willadmit any point y in Y as a limit at any point x in X. (I took thedenition of limit where x must be in the closure of X; the almostidentical denition where x is supposed to be an accumulation point of Xcan be treated the same way (take X = R with the usual metric).--J.S may you give me a sequence (obviously in a non-Hausdorff topological> space) with two distinct limits? Take X any non-empty topological space and (Y,{{},Y}) then any f: X->Y will> admit any point y in Y as a limit at any point x in X. (I took the> denition of limit where x must be in the closure of X; the almost> identical denition where x is supposed to be an accumulation point of X> can be treated the same way (take X = R with the usual metric).>Lets answer the question instead of beating about the bush.Let x1,x2,.. be a sequence in X with indiscrete topology.Let X have more that one point.Show for all x in X, x1, x2, x3, ... has limit of x.That is easy, let U be any open set containing x.As the topology of X is the indiscrete topology { nulset, X }necessarily U = X and thus x1, x2, .... is eventually in U.Hence for any open U nhood of x, x1,x2,... is eventually in U;showing x limit of x1,x2,... and this was show for every x in Xthat x is a limit of x1,x2,...Now apply requirement X has more than one point.--Exercises. Let X = {a,b} with topology { nulset, {a}, {a,b} }Show both a and b are limit to sequence a,a,a....and onlly b is limit to seauence b,b,b...Let X be innite set with conite topology, U open in X iff complement of U in X is niteDiscuss when a sequence x1,x2,... has a limit x. and when S is nite. =I am reading an example in my diff. geom. book, and I dont reallyunderstand why the one-sheeted cone isnt a regular surface.Let C be the one sheeted cone given by z = sqrt(x^2 + y^2). Use theproposition that if S in R^3 is a regular surface and p is in S, then thereis a neighborhood V of p in S such that V is the graph of a differentiablefunction which has one of the following forms: z = f(x,y), y = g(x,z), x =h(y,z).A graph of a function f : U --> R where U is a subset of R^2 is the subsetof R^3 given by (x,y,f(x,y)), (x,f(x,z),z), or (f(y,z),y,z).Now, if C (the onesheeted cone) were a regular surface, it would be, in aneighborhood of (0,0,0) the graph of a differentiable function having one ofthree forms: y = h(x,z), x = g(y,z), z = f(x,y). This means that theneighborhood could be written as (x,y,f(x,y)), (x,h(x,z),z), or(g(y,z),y,z), for some f,g,h : U --> R for all (x,y), (x,z), or (y,z) in Uwhich is a subset of R^2, right? The example states that the rst twoforms can be discarded by the fact that the projections of C over the xz andyz planes are not one to one. (I understand why the projections arent oneto one, but why does this imply that the rst two forms can be discarded?)Moshe I am reading an example in my diff. geom. book, and I dont really> understand why the one-sheeted cone isnt a regular surface.It goes funny at (0,0,0).I like to see it like this. Each point on a regular surfacehas a neighbourhood in the surface homeomorphic to an open disc.But the point (0,0,0) doesnt. Each connected neighbourhood of (0,0,0)in the cone, can be disconnected by removing (0,0,0). The open dischasnt the property of being disconnectable by removing one point.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) Chapman> I am reading an example in my diff. geom. book, and I dont really>> understand why the one-sheeted cone isnt a regular surface.It goes funny at (0,0,0).I like to see it like this. Each point on a regular surface>has a neighbourhood in the surface homeomorphic to an open disc.>But the point (0,0,0) doesnt. Each connected neighbourhood of (0,0,0)>in the cone, can be disconnected by removing (0,0,0). The open disc>hasnt the property of being disconnectable by removing one point.I suspect youre thinking of the surface dened by z^2 = x^2 + y^2;the problem asked about the surface z = sqrt(x^2 + y^2), which isvery different. (In particular it _is_ a continuous manifold, just nota differentiable one.) degrees of freedom in n-dimensional space (with n> 1,> and neglecting time as a dimension included in n).> v x a = 0 *is* too draconian, requiring only colinear motion/action.> But> I cannot gure out how to express deceleration *exactly* as anything> other> than d|v|/dt < 0. Perhaps you can put your massive intellect to the> task. David A. Smith Here is the proof that v.a < 0 = deceleration. let v = velocity vector let a = acceleration vector We know that decleration occurs when d|v|/dt is a negative vector.Not true. d|v|/dt is a *scalar*. So unless you have qualms, I will assumethe above sentence says is negative.> If vector a0 is the projection of a onto v, then a0 = d|v|/dt.Not true. *All* the change in v is not a *limited* change in v. You haveconstrained a to be colinear with v, for all further consideration. Itamounts to v x a = 0, as I had done. It is no more general.Your proof is DOA.d|v|/dt < 0 is still deceleration, is elegant, and sufcient. And youcant patent it.David A. Smith = John Schoenfeld: >> There are too many degrees of freedom in n-dimensional space (with n > 1, >> and neglecting time as a dimension included in n). > v x a = 0 *is* too draconian, requiring only colinear motion/action. >> But >> I cannot gure out how to express deceleration *exactly* as anything >> other >> than d|v|/dt < 0. Perhaps you can put your massive intellect to the >> task. > David A. Smith >Here is the proof that v.a < 0 = deceleration. An obvious counter-example is v(t) = v_0 sin(wt). In thefourth quadrant, the velocity is obviously increasing, sinceits going from more negative to less negative values becausethe acceleration is positive. Since v is negative and a ispositive, v.a is obviously negative. let v = velocity vector >let a = acceleration vector >We know that decleration occurs when d|v|/dt is a negative vector. >If vector a0 is the projection of a onto v, then a0 = d|v|/dt. >a0 = [(a.v) / |v|] Youve attempted to obtain a simple result by starting with alot of nonsense about vector-valued functions and obtained onewhich is incorrect (or at best, stated awkwardly and inconict with your rst assertion about v.a). Since v^2 = v.v, (1/2) dv^2/dt = v . dv/dt = v.a Since the kinetic energy is proportional to v^2, if v^2 isbecoming smaller, the kinetic energy is getting smaller.The rst derivative of v^2 is 2 v.a. 2v.a is the slope ofv^2 and if the slope is negative, v^2 is decreasing. Thatsays _nothing_ about v, because v is a vector and v.v ispositive regardless of whether v is positive or negative. You cant discard the sign if you are talk about vector-valuedfunctions. Obviously its possible to have a large negativevelocity and apply an acceleration which increases its velocitytoward more positive values, making the vector-valued functionan increasing function. Your absolute value signs are an attempt to use v^2 withoutacknowledging that you are really talking about v.a representingthe slope of the scalar, v^2. +0200, Alexander Pambook>Really ! Who? Who is inventor of multiplication table?Almost certainly a Mr, Mrs or Miss Table, just as the inventor of thegeiger counter was a Mr Counter.>What exactly is the multiplication table?-->glenn>-- G.C. For the real proof of the non-existence of Odd Perfect Numbers see:> http://www.bearnol.pwp.blueyonder.co.uk/Math/ perfect.htmlwhich says: There exist no odd perfect numbers Proof: Suppose sigma(n)=2n [n perfect] sigma(pn)=2n(p+1) [if hcf(n,p)=1] p+1==0 [mod 2] [if p odd] sigma(pn)==0 [mod 4] Let p->99999... [Euclid] sigma(pn)/pn->2 pn==0 [mod 2] n==0 [mod 2] [since p odd] Copyright 1997 James Wanlessoops! I copied it :-)So what can we tell from this?That we can conclude that as p -> 99999... [Euclid](dunno what 99999... [Euclid] is but never mind) andsigma(pn)/pn -> 2 then pn is even. Hmmm. I ask Mr Wanless this?Does (6p+1)/3p -> 2 as p -> 99999... [Euclid], and doesit follow that 3p is even (even if its odd?)?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) TeXnicCenters Online Help. (TeXnicCenter being a free TeX for MS>Windows.)> tried it but heard very good cmts about it...> True, my apologies. If you are ever forced to use MS Windows (yuk) andTeX (also yuk) Id recommend it. While there are alternatives to MSWindows, there dont seem to be any serious ones to TeX--or are there?-- G.C. =Trivial question:I am reading a paper, and I see the following formula:exp(-at^c) =summation (from j = 0 to innity) of(-1)^j (at^c)^j-----------------Gamma(j+1)This is clearly incorrect (right?) The bottom should read j! How couldGamma even be introduced into the taylor series of e^y for some function y?Moshe Trivial question: I am reading a paper, and I see the following formula: exp(-at^c) = summation (from j = 0 to innity) of (-1)^j (at^c)^j> -----------------> Gamma(j+1) This is clearly incorrect (right?) The bottom should read j! How could> Gamma even be introduced into the taylor series of e^y for some functiony?> Trivial question:> I am reading a paper, and I see the following formula:> exp(-at^c) summation (from j = 0 to innity) of> (-1)^j (at^c)^j> -----------------> Gamma(j+1)> This is clearly incorrect (right?) The bottom should read j!No. It is correct. For integer arguments, Gamma( j + 1 ) = j!For some info on the Gamma function, see: -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question t perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau Trivial question:> I am reading a paper, and I see the following formula:> exp(-at^c) summation (from j = 0 to innity) of> (-1)^j (at^c)^j> -----------------> Gamma(j+1)> This is clearly incorrect (right?) The bottom should read j! How could> Gamma even be introduced into the taylor series of e^y for some function y?> MosheIts the gamma function not distribution A SUMMARY OF CONCLUSIONS OF THE THREAD>1st CONCLUSION: > * The transnite construction N <-> R is not possible (See PROOF >1 below)I note that your PROOF 1 below is identical to the awed proof thatyou initially offered in order to demonstrate the existence of a trivialproof that there is no bijection between N and R. You have repeated theproof in spite of the fact that I have already debunked it bydemonstrating that your so-called proof relies on an unstated assumptionwhich is KNOWN TO BE FALSE. This means that your PROOF 1 below isactually worthless. I also note that there was no attempt by you to showanything wrong with my debunking of your proof.>That means that the naturals fail in counting the reals because they >cannot undertake the innity of R. Noting that your PROOF 1 is worthless, and that something moresophisticated will have to be adopted.>Then, we cannot assume N <-> R as >possible. If you want to prove it by reductio ad absurdum, then it is legitimate to assume that there is such a bijection, and then to show that theassumption leads to a contradiction.Note that you have not yet proven that no such bijection exists, so youwill have to look around for a legitimate proof.>On the other hand, senders have sent at the moment three kinds of >proofs, trying to refute my arguments>a) Proofs by contradiction o reductio ad absurdum.Which is a legitimate method of proof. If the assumption ~P leads toa contradiction, then P must be true.>b) Direct proof (supposedly the Cantors rst proof).>c) Others (supposedly the Cantors rst proof too).>a) As far as Ive seen, all the proofs by contradictions need the >premise *lets assume that f: N <-> R is true*. And if the assumption that there exists a bijection between N andR leads to a contradiction, the only logical conclusion is that thereis no such bijection. No problem so far.>If we carry on with a >proof by contradiction in spite of the 1st conclusion, Since your PROOF 1 is worthless, this statement holds no water.>then the >proofs outcome must be proved that is true, since the premise (raw >material in the proof) is false. In fact it is a silly thing to go >with a false premise to a proof, It is not a silly thing to assume a false premise and to demonstrate that it leads to a contradiction. This technique allows one to prove that the premise must be false because it cannot possibly be true.That is the essence of reductio ad absurdum.>but if you insists, then you must >prove that your result is reliable.Since you are the only person who knows what you mean by reliable,might as well be hand waving.>2nd CONCLUSION: >* The results obtained by contradiction are not reliable, Yes, they are. If assuming P leads to a contradiction, then that can only happen if P is false, and so it can only happen if ~P istrue. So demonstrating that the assumption of P leads to a contradiction IS a proof of ~P.>if the >conclusion has not been obtained by the method of the diagonal >argument.What is so special about the method of the diagonal argument?>b) Direct proof>This proof does not assume anything in advance, but the mapping f: N ->> R, trying to prove that it is not a surjection. So, there is not >false premise, and then it goes directly to its goal applying the >diagonal argument, getting a wrong conclusion. False.>Why? Have a look to my >own impartial version of this argument, Your own impartial version being the thoroughly discredited PROOF 1,and therefore being absolutely worthless.>and compare it with the >partial version given by Cantor. (See PROOF 2 below)This PROOF 2 relying on the furphy that time must come into any innite process in mathematics (or any process at all), so thatthe generation of Cantors diagonal number supposedly takes time.Cantors argument has NO dependence on time.>3rd CONCLUSION:>As this proof uses the diagonal argument, the result is false.So why does making use of the diagonal argument make the proof false?>c) Others>The only proof sent to the thread that ts in this section states >that R is uncountable because N <-> P(N) it is not possible, By which you mean that there is no bijection between N and P(N). Please use the proper terminology, rather than not bothering.>since >the cardinality of P(N) is bigger than the cardinality of N (Cantors >theorem). On the other hand, it can be rigorously proved that it is >possible the bijection P(N) <-> R. By which you mean that there is a bijection between P(N) and R.>The logical conclusion is that R >is uncountable, and that there are more reals than naturals. However, >Proof 3 shows that N <-> P(N) is possible. (See PROOF 3 below)Your PROOF 3 is worthless for the reasons detailed below.>4th CONCLUSION:>The reals are countable.Only because you want them to be. None of your proofs are valid. Allof your proofs are awed. It is easy to prove that there is no surjection from N to P(N), by explicitly constructing, for any function g : N -> P(N), a subset of N which is not in the range of g.>5th FINAL CONCLUSIONSince PROOF 1 is discredited, all conclusions drawn from it are worthless.>it is obvious that the naturals cannot undertake the >counting of the reals using the decimal expansion. Why? Because in >the list the reals clearly show a two-dimensional innity, one of >them horizontal with innitely many digits, and the other vertical >with innitely many reals. On the other side, naturals just show a >one-dimensional innity (the vertical) and therefore they cannot >deal with the horizontal innity of the reals.The above is just a lot of meaningless jargon, designed to befuddle, rather than provide any clear thought.>As a consequence of this fact, all the decimal-expansion based proofs >do not work properly, as we have already seen.Maybe, if you spent some time learning logic, you might actually learnwhat is a legitimate proof and what is not. That would be better foryou than proclaiming many legitimate proofs are nonlegitimate, whiletrying to force all your nonlegitimate proofs on us.> Finally, although the naturals fail counting the reals by means of >the decimal expansion, it does not imply that R may not be counted by >other mathematical means. Make up your mind. You claim to have a proof that there is no bijectionbetween N and R, and also a proof that there is a bijection between N and R. You cant have both.>Effectively, the elements of the power set >of N *P(N) * can do it, No, it cant. It is very easy to prove that there is no bijection betweenN and P(N).>because it is possible the one-to-one >correspondence between P(N) <-> R, as it has been rigorously proved >by several mathematicians and in different ways. Perhaps the simpler >is the one that considers the binary expansion of the reals.>On the other side, the Cantors theorem proof fails because it >analyzes a counting process (the bijection N <-> P(N)) taking a >timeless approach. To count is a process, the one-to-one >correspondence is a kind of counting process, and time is inherent to >any kind of process. No. To count is to establish a bijection with a subset of N. Time doesnot enter into it.>So, it is very dangerous, if possible, to carry >out any process into a timeless context. Justify this claim rigourously. All that I have seen from you is hand waving.>As we have seen in PROOF 3, PROOF 3 is wrong. I show this below.>P(N) is countable with naturals because after doing P(N) <-> R we get >rid of the horizontal innity of the reals, i. e., P(N) has just a >one-dimensional innity, as N, so that it works the bijection N <->P(N).Since PROOF 3 is wrong, the above statements are unsupported.>I think it is a conclusive summary for everybody, Including the fact that you are still trying to use an argument in PROOF 1 which I have discredited, and you did not address my discrediting.>except for some >mathematicians that, I am completely sure, they will try to discredit >my arguments, but not to refute them, because simply they cannot.This is the arrogance of the crank who is so thoroughly convinced that heis right that he refuses to even consider the possibility that he may bewrong, except to dismiss it immediately. The question is: Is it possibleto get such a crank to see the error of their ways, or are they so xedon how right they are that even the most rigourous rebuttals will not swaythem?>****************>PROOF 1> A METHOD TO PROVE THAT NATURALS CANNOT UNDERTAKE THE COUNTING OF THE >REALS DIRECTLY FROM ITS DECIMAL EXPANSION>It is possible to represent any real number using a given positional >system of numeration. For ease we will use the decimal notation.>Proposition: >The transnite construction N <-> R is not possible>Proof: >Every real number within the interval (0, 1) has as rst decimal As can be seen from below, 0 should be an element of the interval since you are listing it below. The correct interval is [0,1).>digit one of the ten digits of the decimal system of numeration, i. >e. it must be of the form 0.0, 0.1, 0.2, ..., 0.9. As you can see for >this rst digit there are 10 ^ 1 possibilities. For the second digit >we have 0.00, 0.01, 0.02,..., 0.99. Consequently, there are 10 ^2 >possible combinations for the two rst digits, 10 ^3 for the rst >three digits, and so on. When the number of digits is innite we get >R.When the number of digits is innite, we get the elements of R with a nonterminating decimal expansion.>Now, when we have only a digit of the decimal expansion of the reals, >we make the one-to-one correspondence 0 <-> 0.0, 1 <-> 0.1, 2 <->0.2, ..., 9 <-> 0.9. Next, with two digits we begin the 1-1 >correspondence 0 <-> 0.00, 1 <-> 0.01,..., 10 <-> 0.10, 11 <-> 0.11, ...,>99 <-> 0.99. With three digits, we continue doing the same, and so on.>And now is when the impossible transnite construction appears. >While we keep doing the 1-1 correspondence within the nite decimal >expansion, everything will go ne. As I pointed out previously, what you have written above will NOT go ne.You will not end up with a bijection between N and the terminatingdecimals in the interval [0,1). Think carefully about it, and askyourself what natural number corresponds to 0.1 in your intended limitingbijection between N and the terminating decimals in the interval [0,1), orwhat terminating decimal corresponds to 1 in your intended limitingbijection. The answer is that neither of these questions has a legitimateanswer since there is no limiting bijection. Did you even bother reading<-> 0, 1 <-> 0.1, ..., 9 <-> 0.9, 10 <-> 0.01, 11 <-> 0.11, 12 <-> 0.21,...., 254987631 <-> 0.136789452, .... Note that I am REVERSING the orderof the digits, and placing them after the decimal place. THIS willdetermine a bijection such as you require. Your way certainly wont.And note that I can immediately tell you what natural number corresponds to a given terminating decimal, and vice versa.>But, what natural number would >correspond to the decimal expansion of the number e? NO natural number would correspond to e. e is between 2 and 3, and so eis not an element of [0,1). It follows that e is not in the range. Abetter question would have been: What natural number would correspond tothe decimal expansion of the number e-2? The reason why this would have been a better question is that e-2 DOES fall in the interval [0,1).>By induction it >is obvious that it should be 71828182..., With your ill-dened, and nonexistent, bijection. With my corrected bijection, the answer would be ...828182817, i.e. not a natural number.>i. e., a number with innite >nonzero digits, which doesnt belong to N (Point 1). This is true, but does not lead where you think it does.>Conclusion: >The transnite construction N <->R is not possible, i. e. the naturals>cannot undertake the counting of the reals directly from its decimal>expansions. Here, you would use the fact that there exists a bijection between N andthe terminating decimals (specically, one specic bijection supplied byME, by the way, since you supplied no such bijection at all), to assertthat there is no bijection between N and R. Your argument is based on thestatement that there cannot be a bijection between N and a subset of [0,1)which properly contains as a subset the set of all terminating decimals in[0,1). In other words, if A is a subset of [0,1), and A contains all theterminating decimals in [0,1) as elements, and A contains at least onenonterminating decimal as an element, then you would have us believe thatthe existence of my bijection above forbids the existence of a bijectionbetween N and A. So let us take a look at a set B whose elements are theterminating decimals in [0,1), and also e-2. Can Ifind a bijectionbetween N and B? Since you expect us to buy your argument, then you wouldpresumably answer No. But the correct answer is Yes. Take thecorrespondence 0 <-> e-2, 1 <-> 0, 2 <-> 0.1, ..., 9 <-> 0.8, 10 <-> 0.9,11 <-> 0.01, 12 <-> 0.11, 13 <-> 0.21, ...., 12356987 <-> 0.68965321, .... Note the rule: 0 corresponds to e-2, and for natural numbers greater thanzero, subtract 1, reverse the order of the digits, and place after thedecimal point.The point that I am making is that just because I can come up with abijection between N and the terminating decimals in [0,1), that does notof itself forbid any bijections between N and [0,1), since we have noteliminated the possibility that I can manipulate the bijection so that itis a bijection between N and [0,1). Of course, I cannot so manipulate thebijection in this manner, but more sophisticated tools are required toprove it than you have used here.In short, you relied on the unstated assumption that N is not bijective with a proper subset of itself, an assumption which is known to be false.Your PROOF 1 is now worthless.>PROOF 2>AN IMPARTIAL VERSION OF THE DIAGONAL ARGUMENT>uses the one that it is useful for his proof. Here we will take all >the information the diagonal argument supplies.>Proposition: >is not possible to conclude whether the synthesised number is in the >list or not.>Proof:>Lets suppose we have the mapping f: N -> R, being then an arbitrary >list of reals in the form>f(1) =0.a1 a2 a3 a4 ...>f(2) = 0.b1 b2 b3 b4 ...>f(3) = 0.c1 c2 c3 c4 ...>f(4) = 0.d1 d2 d3 d4 ...>.>.>.>Moreover, let S be the synthesized number, being S = 0.s1 s2 s3 ... In >addiction, Thats in addition, not in addiction, which has a completely different meaning.>*t1* will be the moment when s1 is generated, *t2*, when >s2 is generated and so forth.Time again. What is this obsession about time that you have?>Working in decimal notation, a simple combinatory tell us that in t1 >there can be in the list 10^1 sets of reals beginning with a >different decimal digit. Obviously f(1) necessarily belongs to one of >these sets, and S to another. So, in t1 there is one set that no >contains S and (10^1 - 1) sets to which S could belong. In t2 we can >be sure that S cant belong to 2 sets of reals out of the 10^2 >possible sets that we can form with the elements beginning with two >specic digits; so, there are (10^2 - 2) sets left where S could be >an element. Finally, in tn it will be n sets where S cannot be a >member, and (10^n - n) sets where S could be. Therefore, if we >suppose that in every moment tk the sets are more or less of the same >size (cardinality), how we can be so sure that S is not a member in >one of the (10^n - n) sets?That is not how the argument works. Take f(n). After the n-th decimal place is established, then S and f(n) disagree in the n-th place, and are therefore not equal.>In other terms, in t1 we can be sure that S != f(1), but we cannot >guess whether S is in the list or not, because there are innitely >many reals that begin with s1.>In t2 we can be sure that S != f(1) and S != f(2) but we cannotfind >out whether S is in the list, because there are innitely many reals >that begin with s1 s2, and so on.>Therefore, may someone tell us in what precise moment (tn) and why we >begin to be sure that S is not in the list?The argument has nothing to do with your articial impositions of time.The argument is simple enough for anybody except the most wilfully obstinate to comprehend.>Conclusion:>It is not possible to conclude from the diagonal argument whether the >synthesised diagonal number is in the list or not.This is because you have introduced the illogical idea that time must play a part in the proof. If you had ANY understanding, you wouldunderstand what the proof is REALLY saying.>PROFF 3>A TRANSFINITE METHOD TO COUNT THE POWER SET OF N>Proposition: >The power set of N is countable.>Proof:>Being P(N) the set of the subsets of N (power set of N), we will >count its elements in an orderly way. First we will count the subset >with 0 elements, and then the subsets with one element (singletons), >then the subsets with two elements, and so on.>1) 1 <-> {}>2) In order to count the subsets with 1 element {n}, we will use the >naturals that are multiple of 2. We have 2 <-> {1}, 4 <-> {2}, 6 <->{3} ... 2k <-> {k}>3) In order to count the subsets with 2 elements {n1, n2} we will use >the naturals multiple of 3 that they are not divisible by 2. We have >3 <-> {1, 2}, 9 <-> {1, 3}, 15 <-> {1, 4}, ...What numbers correspond to {2,3}, {4,11}, {5,10}, etc.? At present,you have not accounted for ANY of these subsets of N to appear in your list.>4) In order to count the subsets with 3 elements {n1, n2, n3} we will >use the naturals multiple of 5 that they are not divisible by 2 and >3. We have 5 <-> {1, 2, 3}, 25 <-> {1, 2, 4}, 35 <-> {1, 2, 5}, ...What numbers correspond to {1,4,10}, {1,8,9}, {65,478,10000}, etc.?At present, you have not accounted for ANY of these subsets of N toappear in your list.>4) In order to count the subsets with 4 elements {n1, n2, n3, n4} we >will use the naturals multiple of 7, that they are not divisible by >2, 3 and 5, and so on.And similar nite subsets of N of all nite cardinalities greater than 3will be left out of your list.>Conclusion:>As the naturals are innite and the prime numbers too, the bijection >N <-> P(N) is possible, so P(N) is countable.You have not proven this since you have left many subsets of N out of your list. Specically, for any nite cardinality greater than 1, you have left a large number of subsets of N of that cardinality off your list.Further, you have left ALL innite subsets of N off your list. It follows that you have NOT proven a bijection between N and P(N).The standard proof that there is no bijection between N and P(N) still holds, and you proof has not challenged the statement that there is no such bijection.David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. A TRANSFINITE METHOD TO COUNT THE POWER SET OF N> Proposition: > The power set of N is countable.> Proof:> Being P(N) the set of the subsets of N (power set of N), we will > count its elements in an orderly way. First we will count the subset > with 0 elements, and then the subsets with one element (singletons), > then the subsets with two elements, and so on.> 1) 1 <-> {}> 2) In order to count the subsets with 1 element {n}, we will use the > naturals that are multiple of 2. We have 2 <-> {1}, 4 <-> {2}, 6 <-> {3} 2k <-> {k}> 3) In order to count the subsets with 2 elements {n1, n2} we will use > the naturals multiple of 3 that they are not divisible by 2. We have > 3 <-> {1, 2}, 9 <-> {1, 3}, 15 <-> {1, 4}, > 4) In order to count the subsets with 3 elements {n1, n2, n3} we will > use the naturals multiple of 5 that they are not divisible by 2 and > 3. We have 5 <-> {1, 2, 3}, 25 <-> {1, 2, 4}, 35 <-> {1, 2, 5}, > 4) In order to count the subsets with 4 elements {n1, n2, n3, n4} we > will use the naturals multiple of 7, that they are not divisible by > 2, 3 and 5, and so on.> Conclusion:> As the naturals are innite and the prime numbers too, the bijection > N <-> P(N) is possible, so P(N) is countable.> Nicolas de la Foz What about the sets with innitely many elements in them? When do theyget counted? > A TRANSFINITE METHOD TO COUNT THE POWER SET OF N>> Proposition:>> The power set of N is countable.> Proof:> Being P(N) the set of the subsets of N (power set of N), we will>> count its elements in an orderly way. First we will count the subset>> with 0 elements, and then the subsets with one element (singletons),>> then the subsets with two elements, and so on.>> 1) 1 <-> {}>> 2) In order to count the subsets with 1 element {n}, we will use the>> naturals that are multiple of 2. We have 2 <-> {1}, 4 <-> {2}, 6 <-> {3} 2k <-> {k}>> 3) In order to count the subsets with 2 elements {n1, n2} we will use>> the naturals multiple of 3 that they are not divisible by 2. We have>> 3 <-> {1, 2}, 9 <-> {1, 3}, 15 <-> {1, 4}, >> 4) In order to count the subsets with 3 elements {n1, n2, n3} we will>> use the naturals multiple of 5 that they are not divisible by 2 and>> 3. We have 5 <-> {1, 2, 3}, 25 <-> {1, 2, 4}, 35 <-> {1, 2, 5}, >> 4) In order to count the subsets with 4 elements {n1, n2, n3, n4} we>> will use the naturals multiple of 7, that they are not divisible by>> 2, 3 and 5, and so on.> Conclusion:>> As the naturals are innite and the prime numbers too, the bijection>> N <-> P(N) is possible, so P(N) is countable.>> Nicolas de la Foz> What about the sets with innitely many elements in them? When do they> get counted?They never do. This is the standard crank bijection between N and P(N)neglecting, of course, the innite subsets of P(N).I should point out that the superior cranks often count the niteand conite subsets of P(N) :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) > A TRANSFINITE METHOD TO COUNT THE POWER SET OF N> Proposition:> The power set of N is countable.> Proof:> Being P(N) the set of the subsets of N (power set of N), we will> count its elements in an orderly way. First we will count the subset> with 0 elements, and then the subsets with one element (singletons),> then the subsets with two elements, and so on.> 1) 1 <-> {}> 2) In order to count the subsets with 1 element {n}, we will use the> naturals that are multiple of 2. We have 2 <-> {1}, 4 <-> {2}, 6 <->> {3} 2k <-> {k}> 3) In order to count the subsets with 2 elements {n1, n2} we will use> the naturals multiple of 3 that they are not divisible by 2. We have> 3 <-> {1, 2}, 9 <-> {1, 3}, 15 <-> {1, 4}, > 4) In order to count the subsets with 3 elements {n1, n2, n3} we will> use the naturals multiple of 5 that they are not divisible by 2 and> 3. We have 5 <-> {1, 2, 3}, 25 <-> {1, 2, 4}, 35 <-> {1, 2, 5}, > 4) In order to count the subsets with 4 elements {n1, n2, n3, n4} we> will use the naturals multiple of 7, that they are not divisible by> 2, 3 and 5, and so on.> Conclusion:> As the naturals are innite and the prime numbers too, the bijection> N <-> P(N) is possible, so P(N) is countable.> Nicolas de la Foz> What about the sets with innitely many elements in them? When do they>> get counted?> They never do. This is the standard crank bijection between N and P(N)> neglecting, of course, the innite subsets of P(N).> I should point out that the superior cranks often count the nite> and conite subsets of P(N) :-)I was expecting Mr le Foz to say after all the nite ones.My favourite crank explanation is that the power set is in bijection withthe p-adic integers and therefore countable: that shows a lot ofcommitment to misunderstanding. mathematician: the one that prefers to analyse all the >mathematical concepts by himself. The ratio of odd mathematicians is >more or less of one in ten thousand, what justies this adjective. >Cantor was an outstanding odd mathematician.Normal mathematician: the one that accept the settled concepts >because they have been there for long time, and because there are >many outstanding mathematicians that think that these concepts are >right. The ratio of normal mathematicians is of twenty in the dozen. >For your writing I guess that you are a normal mathematician.Crank: the one that doesnt understand mathematical concepts or knowhow to analyze them so deduces they must be wrong. Strives to justifyhis faulty reasoning using silly thought experiments. Uses ridiculeand accusations of mathematical orthodoxy to portray himself as anoriginal thinker and victimized genius. The ratio of cranks to saneposters in sci.math is worryingly high. nico80@jazzfree.com (Nicolas> Thats a lie. Its clear from your behavior in other threads that you have> no interest in having your mathematical errors corrected.>>Cantor seems to attract a lot of odd people. Denitions: >Odd mathematician: the one that prefers to analyse all the >mathematical concepts by himself. The ratio of odd mathematicians is >more or less of one in ten thousand, what justies this adjective. >Cantor was an outstanding odd mathematician.Normal mathematician: the one that accept the settled concepts >because they have been there for long time, and because there are >many outstanding mathematicians that think that these concepts are >right. Uh, no. A normal mathematician accepts standard facts aboutsettled concepts because he knows how to _prove_ them. help mefind the length of L1 and L2 in the attached photo and>explain how it was done! thank you very much!The picture isnt completely clear. Do the 75s represent the lengthsfrom the ends of L1 and L2 to the end of the 100 segment? If so, youcan just use the law of cosines:L1 = sqrt(75^2 + 100^2 - 2*75*100*cos(96)) = 131.1218020and similarly for L2.--Lynn While playing with the Fibonacci sequence in a spreadsheet, the> following interesting pattern appeared.> When the Fibonacci number is divided by its position in the list, the> result is a whole number when the position corresponds to a left> truncatable prime*. I ran out of accuracy to test this past the 67th> number and I know little about left truncatable primes. Does anyone> know if this bonacci and primes numbers returns a few things but nothing of> great depth.> If anyone has more info about Fibonacci and primes I would appreciate> anything you care to share.> The numbers from the spreadsheet are below. Primes(*) Column> A=Fibonacci Column B=Position Column C=A/B Also notice the primes> appear the intervals of 6,4,6,14,6,4,6,14... wonder if that would> continue? Do I have too much time on my Position > A B A/B> 1 0 > 1 1 1.000000> 2 2 1.000000> 3 3* 1.000000 -----> 5 4 1.250000> 8 5 1.600000> 13 6 2.166667> 21 7* 3.000000-------> 34 8 4.250000> 55 9 6.111111> 89 10 8.900000> 144 11 13.090909> 233 12 19.416667> 377 13* 29.000000------> 610 14 43.571429> 987 15 65.800000> 1597 16 99.812500> 2584 17* 152.000000-----> 4181 18 232.277778> 6765 19 356.052632> 10946 20 547.300000> 17711 21 843.380952> 28657 22 1302.590909> 46368 23* 2016.000000-----> 75025 24 3126.041667> 121393 25 4855.720000> 196418 26 7554.538462> 317811 27 11770.777778> 514229 28 18365.321429> 832040 29 28691.034483> 1346269 30 44875.633333> 2178309 31 70268.032258> 3524578 32 110143.062500> 5702887 33 172814.757576> 9227465 34 271396.029412> 14930352 35 426581.485714> 24157817 36 671050.472222> 39088169 37* 1056437.000000-----> 63245986 38 1664368.052632> 102334155 39 2623952.692308> 165580141 40 4139503.525000> 267914296 41 6534495.024390> 433494437 42 10321296.119048> 701408733 43* 16311831.000000-----> 1134903170 44 25793253.863636> 1836311903 45 40806931.177778> 2971215073 46 64591632.021739> 4807526976 47* 102287808.000000----> 7778742049 48 162057126.020833> 12586269025 49 256862633.163265> 20365011074 50 407300221.480000> 32951280099 51 646103531.352941> 53316291173 52 1025313291.788460> 86267571272 53* 1627690024.000000----> 139583862445 54 2584886341.574070> 225851433717 55 4106389703.945450> 365435296162 56 6525630288.607140> 591286729879 57 10373451401.386000> 956722026041 58 16495207345.534500> 1548008755920 59 26237436541.016900> 2504730781961 60 41745513032.683300> 4052739537881 61 66438353080.016400> 6557470319842 62 105765650320.032000> 10610209857723 63 168416029487.667000> 17167680177565 64 268245002774.453000> 27777890035288 65 427352154389.046000> 44945570212853 66 680993488073.530000> 72723460248141 67* 1085424779823.000000 ----> 117669030460994 68 1730426918544.030000Bob,Using your same list above, where A is the Fibonacci sequence in the rst column and B is the second column of positions thenwhere if any B is (ODD) then B = prime if A == 0 or 1 (mod B).At least up to B = 67. ;-)I dont know if this fails for any higher (ODD) values of B.Dan =Are you trading the markets? 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Click on http://www.traderiskmanagement.comor http://www.forextrm.comand reduce your stress in trading down to your sleeping level.Remember: There is no other site on or off the web that has charts andanalysis like Trade Risk Management! I have one number, zero, that I like better than others.> 0s the solution to every algebraic equation. > Thats why 0s my favorite number.> Whats your favorite?> And why?> Garry Denke, Geologist> Denoco Inc. of Texas> The reason garry Dense loves zero so much is that it tells you > everything there is to know about him.A bit off-topic, but at zero density of the closed universe1/c^4 = 123456790.1234567... x 10^-42 time^4/length^4the expanding universe will reach a maximum expansion and beginto collapse. On the other hand, at zero density of the open universe1/c^4 = 123456789........... x 10^-42 time^4/length^4the universe will continue to expand forever. Of course c wouldhave to be dened at 3 x 10^8 length/time to demonstrate it, andthat will never mathematics, is 0/0 = 0, and0/0 = 1, with 0 divided by any number as 0, and any number dividedby itself as 1. It is as logical as your posts.Nothings the solution to every algebraic equation. Thats why nothings my favorite number.Whats your favorite?And why?Garry Denke, GeologistDenoco Inc. of Texas =Nowadays,some are seeking and seeking for the highest possible prime. So Iguess that this means lots of work to know how to discard some of the bignumbers... Those problems are quiet a thrill and they even could have some niceapplications in cryptography. So, I just wonder, is there any well knowntheorem about the density of easily decomposable numbers. By this I thinkto numbers that we could easly discard, in some prime search... A connected problem, is the following is there a chance to know for sure,that arround some big prime there are some easily decomposable numbers(like powers of little primes for instance...) ?Well, just another little question about the big world of primes.nico =Nowadays,some are seeking and seeking for the highest possible prime.They are?Funny, since Euclid proved over 2000 years ago that there is no suchthing.They ->are<- searching for ->big<- primes, though, for any number ofreasons. Perhaps thats what you mean?-- I accept as reality. --- Calvin (Calvin and Hobbes) yes, or in another word, there are true statement that cannot be> proved, and false statement that cannot be disproved. since if one> believe one statement, we can say it is true for him or her.I agree with your position, however human believe systems are not soarbitrary, if we share close models/axiomatic we can discuss what istrue and what is false (depends on how close axiomatics are). The bestcandidate for common model/axiomatic is a reality perception. I mayassume that you are part of my model of reality (a predicate of it)and I am a predicate of your model/axiomatic. You may see that ouraxiomatics are remarkably close to formulating the similar truestatements. The true since they are observable by both of us. Thelink below is a good formalization the existence as a true predicate.However it is presented as a pure formalism, you may see how generalit is.http://www-formal.stanford.edu/jmc/mccarthy-buvac-98/ context/node12.htmlGeorgeP.S.>1+1=2Seems a predicate for me (a trivial one but predicate) =An apple + Another apple = two apples.Two apples are absolute truth, becauseYou cannot derive it from an other apple.The truth is absolute truth, when You cannotderive it from something. The absolute truthis always the eternal truth. Nothing can treaten it.It is eternal truth that one apple + one apple = two apples. The truth is absolute truth, when You cannot> derive it from something. The absolute truth> is always the eternal truth. Nothing can treaten it.so i have 100 legs must be absolute truth,since you cannot derive it from anything. An apple + Another apple = two apples.> Two apples are absolute truth, because> You cannot derive it from an other apple. The truth is absolute truth, when You cannot> derive it from something. The absolute truth> is always the eternal truth. Nothing can treaten it. It is eternal truth that one apple + one apple = two apples.>Applesauce.*plonk* > 1+1=2 isnt an axiom. It is a denition. 2 is dened as the sucessor of>> 1.> of course this is a axiom. well, 1,2 are just short hand> notations for s(0) and s(s(0)), where s is the successor function.> or f.x.fx and f.x.f(fx) in lambda calculus. so the statement> 1=f.x.fx is what you (we) believe, and cannot be proofed. you> can call them denition if want. i just happen to learn that as an> axiom. and some other people learned that as presumed information.> what ever we call it, the important fact is we cannot proof them,> we can only believe them.> The axioms of the system are :1. If a is an element of S and b is an element of S, the a+b is an elementof S2. If a,b,c are elements of S then a+(b+c) = (a+b)+c3. There is a unique element, e, of S such that for all a in S, a+e = e+a =a.4. Each element a of S has a unique inverse ( !a) such that !a is an elementof S and a + !a = !a + a = e5. For all a,b elements of S, a+b = b+aWe can add more if we wish to do multiplication, but addition seems to bethe topic under discussion. > P(x) is true iff:> * P(x) is axiom in the system.> * P(x) can be derived from axioms in the system.>> Again no. If P(x) is an axiom it *cant* be derived from axioms in the>> system. If P(x) can be derived from axioms in the system it is a theorem.> here im obviously meaning OR. let me rewrite it:> P(x) is true iff:> * P(x) is axiom in the system.> Or,> * P(x) can be derived from axioms in the system.> more clear?> OK. But still wrong because something can be true but neither of the above.That is what Godel said.>> If you dont breathe you die.> you are talking about reality here, not truth. truth is dened in the> realm of logic while reality deal with physical world. 1+2=3 is true.> but it does not mean that some where in the universe there are physical> objects called 1, 2, and 3, nor any phenomenon exist called + that> make them into 3 whenever 1 and 2 met. 1,2,3 and + are all objects> that only exist in our mind, nothing to do with the reality.> Yes, counting and math are human tools. The value of math lies not in itbeing some universal law of nature, but in its utility as a countingmechanism for humans.> now lets see this your statement (i will die if i dont breathe).> before i test it, it is only true if you believe it. isnt it?> think why are you so sure about it. perhaps because you have seen> people die due to lack of oxygen. perhaps you had learned experiment> done in labs. but no matter how many experiment we have done, we still> cannot be sure what result will come out when next we do the same> experiment. if you dont agree, show me how can you proof this statement> is true without believe anything. we believe this universe is consistent,> and this is the fundamental philosophy of science. however, this is what> you believe, you cannot proof it.> It does not matter if I believe or dont believe in the necessity ofbreathing. The statement has a truth value. > so what if i test it? suppose i stop breathe now, most likely after> a few minute, i will die, and you will say: see, i was right.> are you really right? nope. suppose i told you: roll two dices,> you will get 6+6. obviously this statement is not so true, it has> only 1/36 probability to be true, assume the dice has no defect.> you dont agree with me, and you would like to test it. so you roll> the dices, you got both 6. and i said: see, i was right. may be> now you may think i am just lucky. yes i was just luck. but if i am> luck enough, i can roll the dices a billion times, and all got> 6+6. so for this statement it is true if i believe it (or i setup> some axiom system that able to proof it), and for you this statement> will be false since you dont believe it. testing it does not make> any difference, because no matter how many times you have test it,> one can always say its because of luck. for the same reason, no matter> how many people has died because of lack of oxygen, one can still> say it is because you just happen to look at people who cannot survive> without oxygen. well... but in either case, i will never dare to test> it myself.> Godel proved that for any system complex enough to dene multiplication,>> there are undecidable theorems.> yes, or in another word, there are true statement that cannot be> proved, and false statement that cannot be disproved. since if one> believe one statement, we can say it is true for him or her.-- Russ Lyttlelyttlec(@)earthlink.net Both of us avoided using goto. My faith in programmerdomjust>went up a notch. :) (Dijkstra would be proud). I would be more impressed if you *had* used goto in a program written> in Java.Why? goto creates spaghetti code, and is completely unnecessary in anystructured or higher imperative language. The only time goto makes sense isin languages like C where there is no expception handling, ie: goto can beused to escape from deeply nested loops in the case of an error. But otherthan that, its been shunned by professional programmers for decades withgood reason.l8r, Mike N. Christoff Michael N. Christoff >Pretty Cool. Both of us avoided using goto. My faith in programmerdom> just>went up a notch. :) (Dijkstra would be proud). I would be more impressed if you *had* used goto in a program written> in Java.>By the way, if this is your way of starting an anti-Java languageame-war - dont. Language ame-wars are always pointless. Everylanguage has its pluses and minuses. There is no perfect language that is agood t for every application. Java, C++, C, Python, Smalltalk, Fortran,C#, Haskell, Lisp etc... are all great languages in their own way.l8r, Mike N. Christoff > On Cool. Both of us avoided using goto. My faith in programmerdom> just>>went up a notch. :) (Dijkstra would be proud).>> I would be more impressed if you *had* used goto in a program written>> in Java.> Why? goto creates spaghetti code, and is completely unnecessary in any> structured or higher imperative language. The only time goto makes sense is> in languages like C where there is no expception handling, ie: goto can be> used to escape from deeply nested loops in the case of an error. But other> than that, its been shunned by professional programmers for decades with> good reason.I think his point was that the Java language does not have a goto statement.-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling. Michael N. Christoff>>Pretty Cool. Both of us avoided using goto. My faith inprogrammerdom> just>>went up a notch. :) (Dijkstra would be proud).>> I would be more impressed if you *had* used goto in a program written>> in Java. Why? goto creates spaghetti code, and is completely unnecessary in any> structured or higher imperative language. The only time goto makessense is> in languages like C where there is no expception handling, ie: goto canbe> used to escape from deeply nested loops in the case of an error. Butother> than that, its been shunned by professional programmers for decades with> good reason. I think his point was that the Java language does not have a gotostatement.>Youre right. Doh! It is a reserved word, but it is not part of thelanguage. Sorry about that Tony. :pl8r, Mike N. Christoff =A related question:We would expect to throw 1 six-sided die 6(1 + 1/2 + 1/3 + 1/4 + 1/5 +1/6) = 14.7 times before throwing all the numbers 1 to 6 at least once.How many times would we expect to roll 2 six-sided dice before we havethrown all the numbers from 2 to 12 at least once?-- We would expect to throw 1 six-sided die 6(1 + 1/2 + 1/3 + 1/4 + 1/5 +>1/6) = 14.7 times before throwing all the numbers 1 to 6 at least once.How many times would we expect to roll 2 six-sided dice before we have>thrown all the numbers from 2 to 12 at least once?>This is combinatorally a much harder question, since the distribution is no longer uniform. However, the space is not so big here that one cannot program it.Let p_k = min(k-1, 13-k) /36, the probability of rolling k. Let v(S) = the expected remaining number of rolls until you roll numbers, where you have already rolled the numbers in S, a subset of {2, 3, 4,..., 12}.v(S) = 1 + sum(k in S, p_k) v(S) + sum(k not in S, p_k v(S U {k}))= [1 + sum(k not in S, p_k v(S U {k}))] / sum(k not in S, p_k)v({2, 3, 4,..., 12}) = 0We want v({}).Note that there are 2^11 = 2048 states, but it can be done. If I had access to the computer (which is down) with Mathematica right now, I could write up a little program fast.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu A related question:> We would expect to throw 1 six-sided die 6(1 + 1/2 + 1/3 + 1/4 + 1/5 +> 1/6) = 14.7 times before throwing all the numbers 1 to 6 at least once.> How many times would we expect to roll 2 six-sided dice before we have> thrown all the numbers from 2 to 12 at least once? Thats a pig of a problem, since it requires a summation over 11! terms (the number of rolls expected for each of the outcomes 2 to 12 in every possible order, multiplied by the probability of that particular order actually occurring).Let us tackle rst the simpler case of two coins, and the number of tosses required to obtain two heads, two tails, and one head and tail, at least once each, Only three outcomes on each toss so only 3! (= 6) terms to add.Once you have tried that, consider the second case of two loaded coins which have a probability 2/3 of showing heads and 1/3 of showing tails.Answers below under the spoilersSPOILER 5SPOILER 4SPOILER 3SPOILER 2SPOILER 1Fair coins: 6.333... tossesLoaded coins: 9.775 tossesWorkings available on request-- Paul TownsendI put it down there, and when I went back to it, there it was GONE!Interchange the alphabetic elements to reply =You can calculate the probabilities of each roll pretty quickly by using recursion. Call the number of ways to roll s from n dice, nu(s,n). Eg. the number of ways to roll a 7 from 3 dice, nu(7,3) is just the number of ways to roll a 6 with 2 dice, nu(6,2) + nu(5,2) + nu(4,2) + nu(3,2) + nu(1,2)likewise, nu(6,2) = nu(5,1) + nu(4,1) + nu(3,1) + nu(2,1) + nu(1,1) = 5in general, nu(s, n) := Sum[nu(s - j, n - 1) , {j, 1, 6}];and nu(s,1) = 1 for all s<=6and nu(s,n) = 0 if s3.5n So, you can build up an array of all the numbers of ways to get s from n die with s ranging from 1..120 and n from 1..20 pretty quickly.In order to get the probability that player A wins, you can sum over the entire disributions of nu(s,n(A)) and nu(s,n(B)). In your specic case of A with 3 dice and B with 5,Probability A=3 beats B=5: 50987/839808 = 0.0607127Probability A=3 ties B=5: 3143/209952 = 0.0149701> dice - you probably hear this sort of thing a lot, Id imagine. Anyway, it> goes like this: Each player rolls an arbitrary number of six-sided dice and> totals the results; the player with the higher total wins. My problem is> guring out the odds. For example, lets say that Player A rolls three> dice, and Player B rolls ve dice; what is the probability that Player A> wins? That Player B wins? That they tie?> Ive already written a brute-force algorithm, but the running time becomes> prohibitive when the total number of dice gets much higher than twelve;> doing 6d6 versus 8d6, for example, took twenty-four minutes to process on> my home computer, and the running time of the algorithm increases> exponentially with added dice. Ideally I need to go all the way up to 20d6> versus 20d6, and that could take until the heat-death of the universe to> churn through the way Im doing it.> Could anyone point me in the direction of a more efcient way to go about> this?> - Sir Bob.> goes like this: Each player rolls an arbitrary number of six-sided dice and>totals the results; the player with the higher total wins. My problem is>guring out the odds. For example, lets say that Player A rolls three>dice, and Player B rolls ve dice; what is the probability that Player A>wins? That Player B wins? That they tie?The probability that A wins is: 101974/(6^8) (which is about 0.0607).Was that what you came up with also? (That would at least mean yourand my algorithm do the same thing :) )>Ive already written a brute-force algorithm, but the running time becomes>prohibitive when the total number of dice gets much higher than twelve;>doing 6d6 versus 8d6, for example, took twenty-four minutes to process on>my home computer, and the running time of the algorithm increases>exponentially with added dice. Ideally I need to go all the way up to 20d6>versus 20d6, and that could take until the heat-death of the universe to>churn through the way Im doing it.What are you using to calculate this??? a Vic20? My algorithm (on an1Ghz Athlon) takes less time than it takes me to press enter.>Could anyone point me in the direction of a more efcient way to go about>this?Use tables (arrays). Let the row number of one table be the number of the sum of the diceand the colum number be the number of dice. You can use theprobabilities of the colum before the one you are calculating to getthe probabilities of the colum you are calculating.Once you calculated that table, you can also make one that contain theprobability that a player throws less than a certain sum.When you have these two tables, you can easily calculate victoryprobabilities. =In sci.logic, Garry Still made more sense than Garrys. :-)> Your logic is 0/0 = 0, and 0/0 = 1, which makes no sense. lim(x->0) x/x = 0/0 = 1.lim(x->0) 2*x/x = 0/0 = 2.lim(x->0) x/(2*x) = 0/0 = 1/2 (or {1}over{2} for TeX-users, perhaps).lim(x->0) x^2/x = 0/0 = 0.lim(x->0) x/x^2 = 0/0 = oo (or infty for TeX-users).You were saying?>> To be sure, one has to be careful.> lim (x->+oo) x = +oo.>> lim (x->+oo) 1/x = 0.>> lim (x->+oo) x * 1/x = 1 but>> lim (x->+oo) x * lim(x->+oo) 1/x = +oo * 0> which equals 0 if one follows Rudin, and AFAIK is undened>> if one follows others.> Fortunately, if lim(x->+oo) a(x) = A, and lim(x->+oo) b(x) = B,>> where A and B are nite, one can prove lim(x->+oo) a(x)b(x) = AB.> My logic is 0/0 = 0/0, which makes sense.> Go gure?Yeah, go gure. Is 0.000...1 [*] equal to innity? Zero?Is 1.000...1 equal to innity? One? Zero? Do suchquestions even make sense?> Garry Denke, Geologist> Denoco Inc. of Texas[*] the ellipses presumably refer to an innite sequence of digits, in this case. Obviously for a nite sequence the answer is clearly no in all cases.-- #191, ewill3@earthlink.netIts still legal to go .sigless. =In sci.math, Aunty Uncle Al seems to picture himself as *the* omnipotent> self-declared garbage collector on this forum.Omnipotent, denitely not. Omniscient, maybe. Admittedly,I was under the impression his regular haunts were morealong the lines of sci.physics, where he routinelyexcoriates cranks -- but he can be quite pleasant withintelligent questions. Of course they do have to beintelligent questions -- something along the lines ofhow does a rocket work? would not quite qualify as theanswers are readily available on Google for those willingto search.But if hes here, I for one welcome him, even if he isirascible at times. :-) Just dont ask him regardingthe continuance of the Space Shuttle...> Two points:> 1. Glad your not my uncle.> 2. Saying your brain is actually full of might be> overdoing it a little, but the next time you take a big> nasty one and look down and see all that red globby stuff in > there- please scoop it out and send the vile off to the lab to > check for traces of brain tissue.> Thats vial or phial. Vile is what the stuff is. :-)> uncle-al.gc() > Aunty Allyson> Undened variable: uncleSubtraction with void operand not supported.Function call has no effect, as operand isnt garbage.:-)-- #191, ewill3@earthlink.netIts still legal to go .sigless. =In Hey, you forgot C#, Visual Basic, ASP, most versions of assembly,>> Modula-2, Modula-3, Ruby, SQL, WATFOR and WATFIV (though theyre>> really dialects of Fortran), Postscript (yes, its a programming>> language!)> Look, if you are going to include things that werent intended to be> used as a general purpose programming language but can be so used> we have to include most text editors, most word processors, most> spreadsheets, etc. etc. etc. well never nish!. Of course> well never nish even if we stick to general purpose programming> languages but this is even worse. Its sort of like the difference> between the innity of the integers and the innity of the reals> (note the valiant attempt to get back on topic).Theres a topic?! :-)Of course Java can only represent 2^64 of the real numbers anyway,minus Inf, NaN, and the underowed zeroes. Id have to work it out.(Its an IEEE thang; C, C++, etc. etc. have the same problem.COBOL might be able to work around it with appropriate PIC statements;in fact, in my youth one programmer was using COBOL for nance workand made it a point not to use oating-point. She was kinda cutebut too old for me at the time...)> - William Hughes> P.S. Have I explained the self evident error in logic made> by the evil cantorians?Does it involve Java? :-) and FORTH... > :-)> C#: Tomorrows technology today. Maybe. Wait until the bugpatch,>> which will be out sometime, erm, next month. Yeah, next month.>> Oh, and its sort of like Java except not really. Were not>> sure yet.>> Visual Basic: Its pretty basic, and it only can be used visually.>> ASP: It bites. Hard.>> Assembly: Sometimes its required. Most times it should sit there>> in little pieces on the oor, as thats what it looks like.>> Modula-2: Pascal squared.>> Modula-3: Pascal cubed.>> Ruby: Is it a gem of a language? I dont know, really.>> SQL: Please, someone, write one.>> WATFOR: ... and give it what for. At least FORTRAN has an excuse.>> WATFIV: A SQL to WATFOR. And you know what sequels do...>> Postscript/FORTH: Both of these are backwards, too, although Ill admit>> theres a certain elegance to>> 1 5 + .>> as it doesnt need parentheses. Of course LISP took them>> all anyway...>> TeX: An interesting entry, actually, and I think it does qualify as>> a computer language although thats a bit like saying an automobile>> qualies as a guerney. (Or vice versa.)>> JCL: You dont want to know. Fortunately, Unix usurped dd>> and made it a lot clearer -- which for Unix is amazing... :-)>> DCL: VMSs answer to shell scripts, complete with F$PARSE().>> Bourne: Unixs answer to how do I do this as cryptically as possible?>> REXX: IBMs answer to shell scripts, and probably someones dogs name.>> There are times it looks like the end product thereof.>> HTML: LISP with angle brackets.>> XML: LISP with angle brackets and question marks.>> XHTML: The worst of both.>> MathML: Like XHTML only mathematical.>> XSL: Denitely in excess.>> ASN.1: Otherwise known as Abstruse, Strange, and Non-comprehensible.>> Even parsing the MIB can give one a headache. The sad thing>> is: this might very well be the most efcient, absent>> compression.>> ASN.2: I hope not.>> SOAP: Was it ever simple? How is XML an object? How does one>> access an object using XML? This one should be put out of its misery.-- #191, ewill3@earthlink.netIts still legal to go .sigless. 19:29:52 JCL: You dont want to know. Fortunately, Unix usurped dd>> and made it a lot clearer -- which for Unix is amazing... :-) I was exposed to JCL once. My therapist says I am making> good progress.> - William HughesSo was I, way back in my school daze. Perhaps *I* need a therapist...:-)-- #191, ewill3@earthlink.netIts still legal to go .sigless. > On Bushnell, > Note that the following question is a _question_. Im not disputing>> your statement that Python is not good for programming large>> systems On the other hand I will dispute this statement.>Python is good for programming large systems.I certainly wouldnt know, having no experience with largesystems. But Im curious to see _why_ Thomas says itsno good for large systems. (Also why you say it _is_ okfor such things: this is based on your experience or thingsyouve heard or what?)>>Is Python better than C? I have no idea. > The statements Python is better than C and C is better than Python>> both strike me as ridiculous; asking whether Python is better>> than C is like asking whether a hammer is better than a screwdriver.>> For many purposes Python is much better than C. For many other>> purposes trying to use Python instead of C would be utterly stupid.>> Hey, I have an idea. Lets combine Python and C. Use Python>for the complicated high level stuff where speed of development is important>but execution speed isnt, and C for the simpler low level stuff where>execution speed is important. Wow, what a concept, combine an interpreted>language with a compiled language. Im brilliant,>Im going to be famous, Im going to be rich ....>What do you mean someone else thought of this rst?Heh. Yes, for the benet of those who are not in on the joke, oneof the things that people who like Python like about it is that itsspecically designed to make it easy to extend Python withmodules compiled in C (or whatever), and also to embed Pythonin C (or whatever) programs, allowing exactly what you suggest,using Python for the complicated high-level stuff and C (orwhatever) for the inner loop.>Oh well, send back the Rolls.>> (For example, trying to write a record-breaking PrimePi(n) calculator>> in Python would probably be stupid (at least doing it in straight>> Python would be, having no experience with numpy I couldnt>> say how well suited it would be for something like that.Numpy will allow you to do a lot of matrix stuff quickly without>having to go to C (or equivalent). However, if you have something>specialized that needs to be done quickly, Python/numpy is not>the way to go. - William Hughes Only an absolute moron would design a language>>(e.g. Python) where white space is signicant.> Well of course your post was not meant to be taken>> seriously (or rather it seems you may have a serious>> point, but these comments are not meant to be>> taken literally).I didnt have a serious point, I just threw out>a bunch of languages and insulted each one. In>some cases I used the canonical insult for the languague.>For Python the canonical insult is the use of signicant>white space.I am a Python fan. I dont particularly like the signicant>white space, but it doesnt bother me either.It only bothers people who havent tried it.>The biggest problem is that the signicant white space issue>distracts from other much more important issues.Does it? I used to read comp.lang.python, and there wasplenty of discussion of actual issues, with no distractionthat I can recall.>(Its a bit like what would happen if the ANSI committee>decided to enforce the One True Brace Style>in standard C). - William HughesMaybe we should switch this to comp.lang.c People here just>dont get upset enough about off topic posts.************************David C. Ullrich =In 04:40:26 is the Denke head that is at, here!> Here are some more algebraic equations whose solution, in accordance> with the rules of the mathematicians here, is zero (0) the number:> --> quadratic equation;> x^2 + bx + c = 0You *are* familiar with completing the square and/or theQuadratic Formula, arent you?Given the equation A * x^2 + B * x + C = 0, the solutionis not 0, but -B sqrt(B^2 - 4*A*C)x = method of deriving it isntdifcult, either:[1] A * x^2 + B * x + C = 0[2] x^2 + B/A * x + C/A = 0[3] x^2 + B/A * x = - C/A[4] x^2 + B/A * x + B^2/(4*A^2) = -C/A + B^2/(4*A^2) = (B^2 - 4*A*C)/(4*A^2)[5] (x + B/(2*A))^2 = (B^2 - 4*A*C)/(4*A^2)[6] x + B/(2*A) = sqrt((B^2 - 4*A*C)/(4*A^2)) = sqrt(B^2 - 4*A*C) / (2*A)The usual term is completing the square, which is an apt descriptionfor step [4].Of course Mathworld has an entry:http://mathworld.wolfram.com/ QuadraticEquation.htmlwhich basically reprises and expands what Ive said here,then goes into some interesting sidelights; for example,one can also represent the solution as 2Cx = ---------------------- -B sqrt(B^2 - 4*A*C)which would not have occurred to me personally.> cubic equation; x^3 + bx^2 + cx + d = 0http://mathworld.wolfram.com/CubicEquation.html> quartic equation;> x^4 + bx^3 + cx^2 + dx + e = 0http://mathworld.wolfram.com/QuarticEquation.html> quintic equation;> x^5 + bx^4 + cx^3 + dx^2 + ex + f = 0http://mathworld.wolfram.com/QuinticEquation.htmlDare I mention I took some Galois theory in college? :-P(Admittedly, Ive forgotten most of it... :-) )> --> Algebra is easy because 0 is an now a number.0s been a number ever since someone discovered it way back.That someone had to discover it is in itself interesting,however; a fairly large number of ancient cultures simplyhad no representation therefor. The Romans in particularhad no zero; neither did the Greeks -- although the ancientBabylonians, from whence we derive our admittedly ratherweird hour-minute-second time and degree-minute-second circlesubdivisions, had a double-wedge zero symbol apparently.The modern positional notation we got from the Indians.> Garry Denke, Geologist> Denoco Inc. of Texas-- #191, ewill3@earthlink.netIts still legal to go .sigless. In sci.math, Garry Denke> Denke head that is at, here! Here are some more algebraic equations whose solution, in accordance> with the rules of the mathematicians here, is zero (0) the number: -- quadratic equation; x^2 + bx + c = 0 You *are* familiar with completing the square and/or the> Quadratic Formula, arent you? Given the equation A * x^2 + B * x + C = 0, the solution> is not 0, but -B sqrt(B^2 - 4*A*C)> x = ----------------------> 2A Most people know this solution. The method of deriving it isnt> difcult, either: [1] A * x^2 + B * x + C = 0> [2] x^2 + B/A * x + C/A = 0> [3] x^2 + B/A * x = - C/A> [4] x^2 + B/A * x + B^2/(4*A^2) = -C/A + B^2/(4*A^2) = (B^2 -4*A*C)/(4*A^2)> [5] (x + B/(2*A))^2 = (B^2 - 4*A*C)/(4*A^2)> [6] x + B/(2*A) = sqrt((B^2 - 4*A*C)/(4*A^2)) = sqrt(B^2 - 4*A*C) /(2*A) The usual term is completing the square, which is an apt description> for step [4]. Of course Mathworld has an entry: http://mathworld.wolfram.com/QuadraticEquation.html which basically reprises and expands what Ive said here,> then goes into some interesting sidelights; for example,> one can also represent the solution as 2C> x = ----------------------> -B sqrt(B^2 - 4*A*C) which would not have occurred to me personally.> cubic equation; x^3 + bx^2 + cx + d = 0 http://mathworld.wolfram.com/CubicEquation.html> quartic equation; x^4 + bx^3 + cx^2 + dx + e = 0 http://mathworld.wolfram.com/QuarticEquation.html> quintic equation; x^5 + bx^4 + cx^3 + dx^2 + ex + f = 0 http://mathworld.wolfram.com/QuinticEquation.html Dare I mention I took some Galois theory in college? :-P> (Admittedly, Ive forgotten most of it... :-) )> -- Algebra is easy because 0 is an now a number. 0s been a number ever since someone discovered it way back.> That someone had to discover it is in itself interesting,> however; a fairly large number of ancient cultures simply> had no representation therefor. The Romans in particular> had no zero; neither did the Greeks -- although the ancient> Babylonians, from whence we derive our admittedly rather> weird hour-minute-second time and degree-minute-second circle> subdivisions, had a double-wedge zero symbol apparently. The modern positional notation we got from the Indians.>Are you having a problem with the equals sign or 0 the number to the rightof the equals sign?Just kidding. But seriuously Ghost, youre wasting your time. He isobviously a lost cause.l8r, Mike N. Christoff =In sci.math, Garry Consider the equation: x - 3 = 0 >> The solution for x - 3 is the number 0 > No, its x = 3.> Are you having a problem with the equals sign> or 0 the number to the right of the equals sign?The equationx - 3 = 0cannot be solved by declaring that the solution is x - 3 = 0or just 0. Thats just stupid.Of course for such a simple equation its fairly obvious anywaythat the solution is x = 3; that sort of thing is probablylearned in the third grade -- although Im not entirely surewhen simple algebra is rst taught anymore. (*I* learned itin third grade, but thats me...and Im not entirely sure ifthe taught material in the third grade is where I learned it,as I have an old algebra book which now I cant nd.)>> For example:> Consider the equation x^3 - 3x - 4 = 0>> The solution for x^3 - 3x - 4 is the number 0> Id have to work it out, but x = 0 wont work here (0^3 - 3*0 - 4 = -4).> Are you having a problem with the equals sign> or 0 the number to the right of the equals sign?Are *you* having a problem with the notion of solvingan arbitrary equation f(x) = 0, f(x) = K, or f(x) = g(x),for the variable x?>> For example:> Consider the equation x - log x = 0 >> The solution for x - log x is the number 0> That has no solution at all in the reals. Im not sure regarding>> the complex plane minus the origin.> Are you having a problem with the equals sign> or 0 the number to the right of the equals sign?Polly wanna cracker?>> For example:> Consider the equation y - 4 x + 1 = 0>> The solution for y - 4 x + 1 is the number 0> That solution is an innite line (x,y) which passes through>> the points (0, -1), (1/4, 0) and (1, 3), among uncountably>> innite others. Or, if you prefer, the slope is 4 and the>> y-intercept is -1.> Are you having a problem with the equals sign> or 0 the number to the right of the equals sign?Polly wanna cracker?> Consider the equation x^2 + 1 = 0> The solution for x^2 + 1 is the number 0Polly wanna cracker?ObSheesh: Sheesh.> Write again if you need help.I got a degree in this stuff, Rocks-In-Head-Boy. :-PI would hope to have at least half a clue.> Garry Denke, Geologist> Denoco Inc. of Texas-- #191, ewill3@earthlink.netIts still legal to go .sigless. =isnt the formula for planar gona,pi(n-2) ??> what are the sums of the measures of the angles in a convex 29-gon?- I was once part of a conspiracy, but I didnt inhale. There is no dimension without time. --RBF (Synergetics, 527.01)(Brian Hutchings, Living Space Programs, Santa Monica College) =I rest my case; 26/50 of baked is medium-well! > the entire paradox lies in ignoring et sequentia,> that is, time. > Consider your point now .52 baked.-- I was once part of a conspiracy, but I didnt inhale.There is no dimension without time. --RBF (Synergetics, 527.01)(Brian Hutchings, Living Space Programs, Santa Monica College) =I was just referring to Russels bogus paradoxii;I dont follow your thing, there. > You have an excellent (albeit halfbaked) point there. I sometimes> think that the difference between Logic (which is instantaneous) and> Computer Science (which involves processes that take time)* is> woefully underexploited. Paradox may be one casualty of this lack of> attention to detail. > http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/ 20021008.1/1> http://www.arxiv.org/html/cs.lo/0003071--Give the Gift of Dick Cheeny -- out of ofce, at last!http://www.rand.org/publications/randreview/issues/rr .12.00/http://members.tripod.com/~american_almanachttp:// = Meglicki isnt claiming that a Turing Machine can both halt yes and> halt no (an absurd proposition), despite his misleading title. Hes> describing an alternative to the [original] Turing Machine. Right, he is describing a Quantum Turing Machine, not a classical> TM. :-)> And s u c h a TM machine can both halt yes and halt now (at the same> time).> I hope you got it THIS time. :-)> F.There is only one Turing Machine, dened by Turing in 1937. Theseother bases of computing are equivalent but use different rules.Think of a bag full of red marbles and blue marbles. There are redmarbles and there are blue marbles. Now think of a bag full of TuringMachines. It has only one thing in it. (Im talking about the model,not its instantiations dened by sets of 5-tuples.)I dont think Quantum Turing Machine is a good name for a base ofcomputing, but if we ignore the linguistic implication (that it is aspecial case of a Turing Machine), you can use it. Just dont thinkthat it is a special case of a Turing Machine. It isnt. ReadJust because someone calls it a Quantum Turing Machine doesnt meanits a Turing Machine. Youre not allowed to change Turingsdenition. A QTM is another Base of Computing, not another TuringMachine.You (and a few others) seem to have a habit of placing signicance onsyntax that isnt there. An axiom is not a true statement. It isquite arbitrary. An axiom is not a construction, either. (So anaxiom that says that a Quine Atom exists is not a construction of aQuine Atom, no more than an axiom that says that a solution to FLTexists is a construction of a solution to FLT.) That seems to berelated to the idea that if something appears in print then it must betrue. Or if one gives a citation then it must contain what they claim(and they need not show where it actually occurs.)Remember: Easier said than done. - Talk is cheap. (AT & Tadvertisement?)Charlie Volkstorfnuff said =Thats about as fungible as a pi-dollar bill. -- me,circa The Nineties my favorite relation is that the areaof a sphere is commensurable to the areaot its great circle (equitorial e.g.). >> PI can be (has been?) quantied.-- I was once part of a conspiracy, but I didnt inhale.There is no dimension without time. --RBF (Synergetics, 527.01)(Brian Hutchings, Living Space Programs, Santa Monica College) =Why should I care that physicists and other scientists have adopted the metricsystem as their ofcial system of weights and measures? It was invented forthe international trade of commercial goods - so that merchants all over theworld could have a common system of units to buy and sell their products. Itis also the ofcial system of weights and measures for science and physicstoo?Just because the majority of people dont have the wherewithall or commonsense enough to know that the quantity of matter contained in a body ismeasured by the (net) force exerted on, and/or by it, divided by theacceleration (a) that it causes: That this ratio of the quantity of a bodysmass is the same ratio that measures its inertia.If they dont know that force and acceleration are essential to science andphysics let them use mass. Even though they dont know that the algebraicsymbol for mass (m) contains units of force, divided by units ofacceleration: That these units of mass are expressed mathematiclly as [m = f/a = ft/(s/t) = ft^2/s :: which is equal to 2w/g.Newton tryed to teach us that; three hundred years ago!While its alright for mechants and their customers not to know that kilosare units of mass, and pounds and newtons are units of mechanical force it_isnt_ alright for any self respecting scientist, engineer, or physicist.Like I say though, why should I care? Im long retired from engineering, andwhile it was important to me then, why should I care now(;^) http://newsone.net/ -- Free reading and anonymous posting to 60,000+ groups NewsOne.Net prohibits Why should I care that physicists and other scientists have adopted the metric> system as their ofcial system of weights such as yourself should becontent with beer, Survivor, and rapid breeding. You are govenrmentsubsidized, arent you?Dumb Donny Head is in on an important NASA mission. Dumb Donnywill be designing external ornamental wrought ironwork for the nextgeneration NASA space vehicle. It will be powered by studies andcapable of lofting expectations.--Uncle Alhttp://www.mazepath.com/uncleal/qz.pdfhttp:// www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) =On a supercalifragilisticexpialidocious day, after dancing about singing Bibbety bobbety boo!, Saint Isidore of Laytonville ishkabibbled:^Ill call his sectary also and tell her to draw up^a check and send it on to you ASAP.^^Whats your name and address?^ The Psychedelick Pope ^ Saint Isidore of Laytonville^^.85^ Patron Saint of the Internet ^.85^^ ^.85^ ^ http://apple2.org.za/gswv/me^ ^ AOXOMOXOA and ENESSA QUA ONNICA^^^^^In the spirit of the thing, have her draw him a VIRTUAL check, cashable on any planet in the Universe.Might as well make it easy for him!!-- The Queen of DXers, as well asQueen of the Commonwealth of Virginia, as well asThe Ruler of A.D.P., as well asSaint Debbe, as well asOur Lady of the Black Hole Exploratory Input Services as OhFishAlly Appointed by the Psychedelic Pope, a/k/a Saint Isidore of SevilleAn Ointed Minister of the Universal Life ChurchReverant of the Church of the SubGenius, UnOrthodoxSuperior Mutha Superior of the Little Sistahs of the Politically IncorrectWorshipper of Eris, Goddess of DiscordI WONT grow up!! -- Peter Pan =On a supercalifragilisticexpialidocious day, after dancing about singing Bibbety bobbety boo!, reotpreeoj ishkabibbled:^> The Straight Dope and its author, Cecil, totally ROCK! I really dig the ^> kind of obscure info he digs up.^^aye, i enjoy it as well. its a shame they banned me about 8(my favorite^number for now on!) times. silly-skeptics dont realize they are part^of a single entity - the vast quantum machine that includes humanity^and all sentient beings, presumably. of course, now that i have proved^that normal is merely a function of our collective idea of how^the world is perhaps i wont have to break another one of their ridiculous^rules to enjoy their very ne site. ^^the vast majority of humanity believes in one God or another. He/She/It^cannot be seen, touched, tasted, smelled or heard. skeptics assume that^He does not exist and is of the paranormal. since normal is determined by^collective perspective GOD EXISTS. therefore.. THE PARANORMAL EXISTS, at^least as dened by the silly-skeptics. ^I agree. That which we call magick is simply those abilities for which we havent yet gured out the mechanisms. For me to stand with a small silver object in my hand and communicate, was something that, as little as 200 years ago, would have had one burned at the stake. Now its commonplace and easily understood.-- The Queen of DXers, as well asQueen of the Commonwealth of Virginia, as well asThe Ruler of A.D.P., as well asSaint Debbe, as well asOur Lady of the Black Hole Exploratory Input Services as OhFishAlly Appointed by the Psychedelic Pope, a/k/a Saint Isidore of SevilleAn Ointed Minister of the Universal Life ChurchReverant of the Church of the SubGenius, UnOrthodoxSuperior Mutha Superior of the Little Sistahs of the Politically IncorrectWorshipper of Eris, Goddess of DiscordI WONT grow up!! -- Peter Pan =On a supercalifragilisticexpialidocious day, after dancing about singing Bibbety bobbety boo!, ClueStickMan ishkabibbled:^^The aluminum foil deector beanie.^^http://zapatopi.net/afdb.html^^Its your only defense. Get one now, before the eeeevillll Bush/Gates^conspiracy assimilates you.^^^^>On a supercalifragilisticexpialidocious day, after dancing about singing ^>Bibbety bobbety boo!, MiRo ishkabibbled:^>^^>^>Im a cock sucker, I suck so much cock.^>^^>^Im sure you do Ryan, Im sure you do, boy.^>^^>^^>^Ryan - the badtrip dude^>^he sucks cocks^>^his mom is nude^>^shes a whore^>^thats why hes rude^>^^>^and now chorus:^>^^>^And his father sucks cocks in hell,^>^^>^^>^^>^^>^^>Got a melody for that, sos we can sing along???^^Oy, I better get to work! Fortunately I have my ofshal Alien Detector which glows reddish in the presence of AYLEENNNS.-- The Queen of DXers, as well asQueen of the Commonwealth of Virginia, as well asThe Ruler of A.D.P., as well asSaint Debbe, as well asOur Lady of the Black Hole Exploratory Input Services as OhFishAlly Appointed by the Psychedelic Pope, a/k/a Saint Isidore of SevilleAn Ointed Minister of the Universal Life ChurchReverant of the Church of the SubGenius, UnOrthodoxSuperior Mutha Superior of the Little Sistahs of the Politically IncorrectWorshipper of Eris, Goddess of DiscordI WONT grow up!! -- Peter Pan root password.Type: rm -rf /Problem solved :)-- Boris = This is the fth Peano Axiom from Mathworld: If a set S of numbers contains zero and also the successor of everynumber> in S, then every number is in S. We started with ZF because we used the Axiom of Innity to show> that there exists a set, A, that contains zero and is closed under the> successor function. We can assume A contains every natural number and is an inductive set> as dened by fth Peano Axiom. The question at hand is what else> does A contain? It could contain members that are not natural numbers. But consider the intersection of all inductive sets.If every inductive set also includes a member that is not a natural numberthen the intersection could contain a member that is not a natural number.> We can exclude any non-natural number, and all its successors other than> zero and still have the inductive property.But, you would lose the ability to show this set is closed under thesuccessorfunction. Without this you can not prove the set is inductive.> So how that intersection contain anything except natural numbers? THAT> is the set we call N, and its members are the objects we call natural> numbers, and it cant contain anything else.I havent seen a proof of this that doesnt assume what it is trying toprove.Russell- 2 many 2 count = > This is the fth Peano Axiom from Mathworld:> If a set S of numbers contains zero and also the successor of every> number> in S, then every number is in S.> We started with ZF because we used the Axiom of Innity to show that there exists a set, A, that contains zero and is closed under the> successor function.> We can assume A contains every natural number and is an inductive set> as dened by fth Peano Axiom. The question at hand is what else> does A contain? It could contain members that are not natural numbers. But consider the intersection of all inductive sets.> If every inductive set also includes a member that is not a natural number> then the intersection could contain a member that is not a natural number.But then there is a subset of it which excludes that non-natural, and that subset is at least as inductive as its containing set.The intersection of all inductive sets thus excludes all these non-natural members.> We can exclude any non-natural number, and all its successors other than> zero and still have the inductive property.> But, you would lose the ability to show this set is closed under the> successor> function. Without this you can not prove the set is inductive.How do we lose that? Such a constrained set still contains zero, and still contains the successor of each of its members, since no natural has zero as a successor, so it is still inductive.You are bating a dead horse.> So how that intersection contain anything except natural numbers? THAT> is the set we call N, and its members are the objects we call natural> numbers, and it cant contain anything else.> I havent seen a proof of this that doesnt assume what it is trying to> prove.We asssume only:(1) there is an empty set.(2) for each set, there is a set containing it as a member.(3) any union of sets is a set.(4) any intersection of sets is a set.Where is your vaunted circularity here? = > Many people have complained that> my method will never end.> This same complaint can be made> of the diagonal proof.> But for the diagonal proof, I can, as required, described my> algorithm> exactly. That is ALL I need to do to validate the theorem.> I described my algorithm. Which part did you not understand? Why you think it proves anything.> Because you just said that is ALL I need to do to validate the theorem.> By your dention, my theorem must be valid.You misread. I said that all I need do is to show that for any listing of reals there is some real not listed.But you have NOT shown that for EVERY listing of rationals there is an unlisted rational, you have only shown that for SOME listings that is the case.But for every set with more than one member, your situation is true, i.e., for every set with more than one element there are mappings from the set to itself that are not surjective.So if your proof were valid, you would also have proved that only the empty set and one element sets are countable.> Ending, whatever that means, is not a requirement, constructing the> algorithm is.> Bull.> The algorithm in the diagonal proof requires an innite number of> computations. If you continue to insist my method will never end> then the diagonal method never ends, either. For any listing of the reals, the diagonal algorithm desribes a real> number that cannot be in the list. No construction steps are needed> since the number described must already exist as soon as the list exists.> If no construction steps are needed why bother to describe an> algorithm? Why not just say the list is incomplete and be done?Because such claims require proof, at least in mathematics.> Are you saying my number doesnt exist because it has to be> constructed? The diagonal number has to be constructed> and it requires an innite number of computations.The so-called diagonal number already exists because every innite sequence of decimal digits represents an actual real number beteen 0 and 1 in the usual fashion, so it doesnt need to be constructed, it only needs to be identied. Since it is clear that there does exist a number with the diagonal number properties, the proof is complete at that point.I do not see that your unlisted rational exists. = > One can always assume there is a set of all natural numbers.> One can also assume the moon is made of green cheese. Mathematical assumptions and physical ones are of quite different> natures.> They share at least one property.> Both types of assumptions can be false.> Russell> - 2 many 2 count> But the nature of the truth or falshood of the two types of assumption is different. Mathematical assumptions can only be false because they contradict those other assumptions that we take as axioms, whereas physical assumptions can be faslied by failing to conform to physical reality.As it happens, assuming that there is a set of all natural numbers does not contrdict the mathematical or logical axioms of the ovedrwhelming majority of mathematical systems, whereas the assumption that the moon is made of green cheese is contradicted by the evidence of the moon rocks brought back to earth. Either N is a proper subset of A or it is exactly equal to A, it>> doesnt matter which. Moreover, it is easy to see that N> satises>> the usual dening axioms for the natural numbers. In particular,> it>> is not hard to show that N dened as above is inductive.>> N may be inductive, but does it contain members that are not natural> numbers?>> No, it doesnt, by the inductive property of the natural numbers.> Is there an axiom or something that says only natural numbers>> have this inductive property. Sounds like another assumption>> to me.>> The fth Peano axiom.>> Whether its an assumption depends on your starting point. If you view> the>> Peano axioms as your starting point, then its one of your ve> assumptions.>> If you view ZF as your starting point, then the Peano axioms become> theorems>> regarding the natural numbers (= nite ordinals).> This is the fth Peano Axiom from Mathworld:> If a set S of numbers contains zero and also the successor of every number> in S, then every number is in S.This seems to talk about the unknown set S, but it really is telling ussomething important about the set N of natural numbers. It says that ifS is any inductive set, then N subseteq S. In particular, if a certainx is a member of every inductive set S, then x is also a member of N;that is, x must be a natural number.> We started with ZF because we used the Axiom of Innity to show> that there exists a set, A, that contains zero and is closed under the> successor function.> We can assume A contains every natural number and is an inductive set> as dened by fth Peano Axiom. The question at hand is what else> does A contain? It could contain members that are not natural numbers.> You dene a set, I, that contains every inductive subset (not necessarily> proper) of A. You then dene N as the intersection of every member> of I. You claim that N contains only natural numbers and that it> is an inductive set (ie, it contains every natural number).> Lets call any set that contains zero and is closed under the successor> function a closed set. Obviously, A is a closed set.When I say its an inductive set, I am using the second denition at. This denition,according to that other guy named Russell, says that an inductive set isa nonempty partially ordered set in which every element has a successor.That is, the set is closed with respect to the successor operation.There are lots of inductive sets around. For example, every limitordinal is an inductive set, but only the smallest one, w, is a model forthe natural numbers.> Here are a few questions I have. Is the intersection of two closed> set guaranteed to be a closed set?in A and also x is in B, then x is in the intersection.> For the sake of argument, lets say that a closed set must contain some> member> that is not a natural number. Lets say it must contain an Aleph to be> closed.> Let set x contain all of the natural numbers and Aleph_0Thats not an inductive set because it doesnt contain the successor ofaleph_0. We usually call that number w when we are treating it as anordinal, rather than a cardinal. The smallest inductive set containingall the natural numbers and w is w+w.> and set y contain all of the natural numbers and Aleph_1.Same comment applies. The smallest inductive set containing y as asubset is w U w = { 0, 1, 2, ... } U { w_1, w_1+1, w_1+2, ... }. Thisset is not an ordinal, but its a perfectly respectable inductive set.> Let z be the intersection of x and y. z is not a closed set> using our (temporary) denition of a closed set.Why not? The intersection is exactly w.> How would we prove that z contains every natural number> if we can not prove z is closed?Maybe you cant prove it, but I can.> I do not see how your denition of N guarantees that N> is closed under the successor function. If N is not closed> under the successor function, how can you prove that> N contains every natural number?The intersection of any collection of inductive sets is inductive.Proof. Let S = intersection_i A_i, where each A_i is inductive. Then0 in A_i for each i, implying 0 in S. Suppose x in S. We are to showthat s(x) in S. By hypothesis, we have x in A_i for each i, and sinceeach A_i is inductive, it follows that s(x) in A_i for each i, hence s(x)in S.-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling. =Again, for the sake of argument, lets assume that a setclosed under the successor function must contain someelement that is not a natural number.PA and ZF do not rule out this possibility.Let @ and # represent two unnatural numbers.@ and # do not have to have successors becausethey are not natural numbers.Dene two sets:x = (0,1,2,...,@)y = (0,1,2,...,#)Both x and y are closed using my denition.Let z be the intersection of x and y.z = (0,1,2,...)z is not closed under the denition I give above.It is impossible to prove z contains every natural number.Russell- 2 many 2 count Again, for the sake of argument, lets assume that a set> closed under the successor function must contain some> element that is not a natural number.> PA and ZF do not rule out this possibility.> Let @ and # represent two unnatural numbers.> @ and # do not have to have successors because> they are not natural numbers.> Dene two sets:> x = (0,1,2,...,@)> y = (0,1,2,...,#)> Both x and y are closed using my denition.Inductively closed sets must contain the successor of every memnber, and that successor must be different from the member itself.What are the successors of @ and #?So that your denition is awed.> Let z be the intersection of x and y.> z = (0,1,2,...)> z is not closed under the denition I give above.> It is impossible to prove z contains every natural number.As your sets are not inductive, they are also not relevant. Again, for the sake of argument, lets assume that a set> closed under the successor function must contain some> element that is not a natural number.> PA and ZF do not rule out this possibility. Let @ and # represent two unnatural numbers.> @ and # do not have to have successors because> they are not natural numbers. Dene two sets:> x = (0,1,2,...,@)> y = (0,1,2,...,#)> Both x and y are closed using my denition. Inductively closed sets must contain the successor of every memnber, and> that successor must be different from the member itself. What are the successors of @ and #? So that your denition is awed.The induction axiom only says that an inductive set containsevery natural number. It says nothing about membersthat are not natural numbers. @ and # dont have to havesuccessors.> Let z be the intersection of x and y.> z = (0,1,2,...) z is not closed under the denition I give above.> It is impossible to prove z contains every natural number. As your sets are not inductive, they are also not relevant.Yes they are. Sets x and y contain every natural number.(Set z can not be shown to have this property.)I am not trying to prove there are unnatural numbers.I am just trying to show why the proofs given that Nonly contains natural numbers are not sufcient.Russell- 2 many 2 count <97adneZXNdeRbWCi4p2dnA@comcast.com> <40009fc8$16$fuzhry+tra$mr2ice@news.patriot.net> <40044b66$29$fuzhry+tra$mr2ice@news.patriot.net> <4005a65f$29$fuzhry+tra$mr2ice@news.patriot.net> <87k73s8fsk.fsf@phiwumbda.org> <-PKdnRWH4KtqKprdRVn-iQ@comcast.com> Actually, we can take ANY inductive set with any initial element and any > compatible successor operation and DEFINE it as N. That is, if it > satises the Peano postulates, it is a suitable N. If the axiom of innity guarantees us a Peano set, we are home free.There is evidently some confusion in this thread and not all of it isdue to Ross.Most of us would not regard just any ol model of Peano to be the setof natural numbers... especially not with any ol successoroperation. Non-standard models of PA are not the natural numbers, Idguess.Id use the second-order version of induction to characterize thenatural numbers. Better, the natural numbers are (up to isomorphism)the initial <0,s>-structure. That means that, for any other set Stogether with a distinguished element x in S and a map f:S -> S, thereis a unique map h:N -> S such that h(0) = x and h(s(n)) = f(h(n)).-- All intelligent men are cowards. The Chinese are the worlds worstghters because they are an intelligent race[...] An average Chinesechild knows what the European gray-haired statesmen do not know, thatby ghting one gets killed or maimed. -- Lin Yutang =A couple of times, this problem from the rst set of problems in HersteinsTopics in Algebra has been discussed here: In a group G, if for all a, b, (ab)^i = a^i b^i for three consecutive values of i, show that G is abelian.This problem is given a star by Herstein, although most people seem to thinkit is not that hard. Here was Per Hjeltmans solution, posted on2000/02/01: By assumption, it is true that 1a) (ab)^i = a^i*b^i 2a) (ab)^(i+1) = a^(i+1)*b^(i+1) 3a) (ab)^(i+2) = a^(i+2)*b^(i+2) If (ab)^n = a^n*b^n for some integer n, it is true that (ba)^(n-1) = a^(n-1)*b^(n-1). Using this, equations 2a) and 3a) can be written as: 2b) (ba)^i = a^i*b^i 3b) (ba)^(i+1) = a^(i+1)*b^(i+1) Now, equations 1a) and 2b) together imply that (ab)^i = (ba)^i. Similarly, eqns. 2a) and 3b) imply that (ab)^(i+1) = (ba)^(i+1). Thus, (ab)^i*ab = (ba)^i*ba = (ab)^i*ba and so ab = ba as was to be shown.I think thats more elegant than mine: a^(i+1) b^(i+1) = (ab)^(i+1) = ab (ab)^i = ab a^i b^i, which means a^i b = b a^i I.e., any ith power in G commutes with every element of G a^2 b^2 a^i b^i = a^2 a^i b^2 b^i = a^(i+2) b^(i+2) = (ab)^(i+2) = ab ab (ab)^i = ab ab a^i b^i, which means a^2 b^2 = abab, therefore ab = baOK, that is pretty easy.The problem *after* that one, however, has me totally stumped. That is toshow that the conclusion is not true if we only assume (ab)^i = a^i b^iholds for two consecutive integers, rather than three.Its easy to see how both of the above proofs make use of all threeconsecutive i, but invalidating specic proofs isnt enough. The only wayI can think of to show that two consecutive i is not sufcient is toexhibit a non-abelian group where (ab)^i = a^i b^i holds for 2 consecutivei. I cant think of any such group that a student at that point in Hersteincould reasonably come up with. Is there some easy-to-nd group Iveoverlooked? Or some other way to show this? Or did Herstein miss and putthe star on the wrong problem? :-)-- --Tim Smith =Well, trivially any non-abelian group works with i=0. So, are we to assumei not zero?Tyler> A couple of times, this problem from the rst set of problems inHersteins> Topics in Algebra has been discussed here: In a group G, if for all a, b, (ab)^i = a^i b^i for three consecutive> values of i, show that G is abelian. This problem is given a star by Herstein, although most people seem tothink> it is not that hard. Here was Per Hjeltmans solution, posted on> 2000/02/01: By assumption, it is true that> 1a) (ab)^i = a^i*b^i> 2a) (ab)^(i+1) = a^(i+1)*b^(i+1)> 3a) (ab)^(i+2) = a^(i+2)*b^(i+2) If (ab)^n = a^n*b^n for some integer n, it is true that (ba)^(n-1) a^(n-1)*b^(n-1). Using this, equations 2a) and 3a) can be written as: 2b) (ba)^i = a^i*b^i> 3b) (ba)^(i+1) = a^(i+1)*b^(i+1) Now, equations 1a) and 2b) together imply that (ab)^i = (ba)^i.> Similarly, eqns. 2a) and 3b) imply that (ab)^(i+1) = (ba)^(i+1). Thus, (ab)^i*ab = (ba)^i*ba = (ab)^i*ba and so ab = ba as was to be> shown. I think thats more elegant than mine: a^(i+1) b^(i+1) = (ab)^(i+1) = ab (ab)^i = ab a^i b^i, which means> a^i b = b a^i> I.e., any ith power in G commutes with every element of G a^2 b^2 a^i b^i = a^2 a^i b^2 b^i = a^(i+2) b^(i+2) = (ab)^(i+2)> = ab ab (ab)^i = ab ab a^i b^i, which means> a^2 b^2 = abab, therefore ab = ba OK, that is pretty easy. The problem *after* that one, however, has me totally stumped. That is to> show that the conclusion is not true if we only assume (ab)^i = a^i b^i> holds for two consecutive integers, rather than three. Its easy to see how both of the above proofs make use of all three> consecutive i, but invalidating specic proofs isnt enough. The onlyway> I can think of to show that two consecutive i is not sufcient is to> exhibit a non-abelian group where (ab)^i = a^i b^i holds for 2 consecutive> i. I cant think of any such group that a student at that point inHerstein> could reasonably come up with. Is there some easy-to-nd group Ive> overlooked? Or some other way to show this? Or did Herstein miss and put> the star on the wrong problem? :-) -- > --Tim Smith =A second more of thought. Pick any non-abelian group of nite order n.Then (ab)^n = 1 = a^n b^n and (ab)^(n+1) = ab = a^(n+1)b^(n+1).Tyler Smith> Well, trivially any non-abelian group works with i=0. So, are we toassume> i not zero? Tyler A couple of times, this problem from the rst set of problems in> Hersteins> Topics in Algebra has been discussed here: In a group G, if for all a, b, (ab)^i = a^i b^i for threeconsecutive> values of i, show that G is abelian. This problem is given a star by Herstein, although most people seem to> think> it is not that hard. Here was Per Hjeltmans solution, posted on> 2000/02/01: By assumption, it is true that> 1a) (ab)^i = a^i*b^i> 2a) (ab)^(i+1) = a^(i+1)*b^(i+1)> 3a) (ab)^(i+2) = a^(i+2)*b^(i+2) If (ab)^n = a^n*b^n for some integer n, it is true that (ba)^(n-1) a^(n-1)*b^(n-1). Using this, equations 2a) and 3a) can be writtenas: 2b) (ba)^i = a^i*b^i> 3b) (ba)^(i+1) = a^(i+1)*b^(i+1) Now, equations 1a) and 2b) together imply that (ab)^i = (ba)^i.> Similarly, eqns. 2a) and 3b) imply that (ab)^(i+1) = (ba)^(i+1). Thus, (ab)^i*ab = (ba)^i*ba = (ab)^i*ba and so ab = ba as was to be> shown. I think thats more elegant than mine: a^(i+1) b^(i+1) = (ab)^(i+1) = ab (ab)^i = ab a^i b^i, which means> a^i b = b a^i> I.e., any ith power in G commutes with every element of G a^2 b^2 a^i b^i = a^2 a^i b^2 b^i = a^(i+2) b^(i+2) = (ab)^(i+2)> = ab ab (ab)^i = ab ab a^i b^i, which means> a^2 b^2 = abab, therefore ab = ba OK, that is pretty easy. The problem *after* that one, however, has me totally stumped. That isto> show that the conclusion is not true if we only assume (ab)^i = a^i b^i> holds for two consecutive integers, rather than three. Its easy to see how both of the above proofs make use of all three> consecutive i, but invalidating specic proofs isnt enough. The only> way> I can think of to show that two consecutive i is not sufcient is to> exhibit a non-abelian group where (ab)^i = a^i b^i holds for 2consecutive> i. I cant think of any such group that a student at that point in> Herstein> could reasonably come up with. Is there some easy-to-nd group Ive> overlooked? Or some other way to show this? Or did Herstein miss andput> the star on the wrong problem? :-) -- > --Tim Smith <40088251.7040704@rutcor.rutgers.edu> <4008AEF2.5080304@rutcor.rutgers.edu> =gotta be the most helpful newsgroup Ive ever been to.-Ed> Youre right. This is not the pdf per se. This partial pdf (for lack of a better word) describes probability density in the Quadrant IV in a Cartesian system (as evidenced by the fact that X is positive and Y is negative). So if you double-integrate f(x,y) for 0 given the density of (X,Y) restricted to Q. IV.> Maybe this is a good time to ask this question which Ive been >pondering for a while: if f(x,y) is a partial pdf for Quadrant IV only, >can I apply it to derive a partial pdf of Z dened as Z=arctan(Y/X)? >Intuitively, I would think that whatever partial pdf of Z that may be >obtained describes only in the range of 3*pi/2 <= Z <= 2*pi. Am I correct in making the above claim? Your partial pdf quadrupled is the conditional pdf given {(X,Y) in Q. > IV}. You could work from there, I suppose. However, this concept of > partiality is quite nonstanrard in probability theory. I suppose you > could make better sense of it if you open your discussion to more > general measures, but I am unsure that you want to do so.> change of variable approach. Could you also clarify a bit on your rst >suggested method? For instance, what integral region are you referring to here? Given any continuous random variable V, the density g(v) of V is > the derivative of its cdf G(v) = P{V <= v}; also, g(v) = -d/dv P{V > v}.> Given X and Y independent;> Z = arctan(Y/X);> 0 < X < 1;> -1 < Y < 0, and> -pi/2 < Z < 0.> For z < 0,> P{Z > z} = P{Y/X > tan z} = P{Y > (tan z) X}> = int(x= 0..1, y= (tan z) x..0, f(x) g(y)) when -pi/4 < z < 0.> P{Z <= z} = int(y= -1..0, x = 0..(cot z) y, f(x) g(y)) when -pi/2 < > z < -pi/4> Use Leibniz rule, the fundamental theorem of calculus, and the chain > rule to determine the densitiy of Z.> When> f(x) = 2(1-x) for 0 < x < 1 and> g(y) = 2(1+y) for -1 < y < 0> (i.e., X and -Y each have Beta(0,1) distribution), I get (with the > assistance of Mathematica(R)) the following density for Z:> h(z) = (csc^2 z) (2 + cot z) / 3 for -pi/2 < z < -pi/4> h(z) = (sec^2 z) (2 + tan z) / 3 for -pi/4 < z < 0> The pdfs of X and Y are as follows:>> f(x)=-1/(2b)^2*x+1/(2b) 0 f(y)=1/(2b)^2*y+1/(2b), -2b>where b is a positive constant. And the joint pdf of X and Y is given >as:>> f(x,y)=-1/(2b)^4*x*y-1/(2b)^3*x+1/(2b)^3*y+1/(2b)^2,>>for 0>So again, I want tofind the pdf of Z which is dened as Z=arctan(Y/X).>>First, note that since you are just considering the ratio of Y and X, you can assume WLOG that b = 1/2. However, we immediately see that your densities do not integrate to 1. Should your marginals be doubled and the joint density quadrupled?> correctly pointed out that X and Y cannot be identically distributed since >they operate in different ranges. I was dead tired when I typed out my >question last night and so didnt give the complete picture. Sorry about >the confusion. And also to clarify a point Stephen brought up, I use X, Y >to refer to random variables and x and y to refer to their sample values >that X and Y can take respectively. Thats my way of the convention. The pdfs of X and Y are as follows: f(x)=-1/(2b)^2*x+1/(2b) 0 f(y)=1/(2b)^2*y+1/(2b), -2bas: f(x,y)=-1/(2b)^4*x*y-1/(2b)^3*x+1/(2b)^3*y+1/(2b)^2,for 0Your help on this greatly appreciated. Hope to hear from you soon.> I was about to post the answer below with each of the x and y density> functions doubled, but I read in another subthread that this is just> the density for one of four quadrants, so I have removed the doubling.> Since you are dividing Y by X, we can change b without affecting Z.> I will use b = 1/2:> f(x) = 1-x 0 < x < 1> f(y) = 1+y -1 < y < 0> Consider the cumulative probabilities.> When -pi/4 < z < 0:> F(z)> |1 |tan(z)x> = | | (1-x)(1+y) dy dx [1]> | 0 | -1> When -pi/2 < z < -pi/4> F(z)> |0 |cot(z)y> = | | (1-x)(1+y) dx dy [2]> |-1 | 0> The density functions are the derivatives of the cumulatives:> When -pi/4 < z < 0:> f(z)> d |1 |tan(z)x> = -- | | (1-x)(1+y) dy dx> dz | 0 | -1> 2 |1> = sec (z) | x(1-x)(1+tan(z)x) dx> | 0> 2> = sec (z) (2+tan(z))/12 [3]> When -pi/2 < z < -pi/4> f(z)> d |0 |cot(z)y> = -- | | (1-x)(1+y) dx dy> dz |-1 | 0> 2 |0> = -csc (z) | y(1-cot(z)y)(1+y) dy> |-1> 2> = csc (z) (2+cot(z))/12 [4]> Notice that [4] is complementary to [3]. That is, the distribution is> symmetric about z = -pi/4 (as one would expect).> Rob Johnson =How can you show that exp takes B_eps = {X in M_n(C) such that ||X|| < eps}injectively into M_n(C), where eps < log(2)?M_n(C) denotes the n x n matrices with complex entries, and exp denotesexponential, eps denotes epsilon.Is it like this? :Of course we need eps < log(2) because this is where log is analytic. Now,to show injectivity show that if e^A = I, then A = 0.Well, take log of both sides and done.Moshe =I yet again forgot the command to block messages by a certain person, asrarely do people annoy me. This guy obviously understands a few interestingfeatures of quantum mechanics and string theory, as well as concepts of asinularity-based collective conciousness, but this juvenile continuedobsessing on trivial details is closer to psychosis than enlightenment, inmy opinion. did not envision it to be a question, now i now TRUTH. the> answer to this equation is 1=0 . .> - - - -> the limits are this the limits are imposed by the creation by the rules> that exist since conception the things that form perception. it began with> a specic density beyond which nothing could escape besides radiation> besides nothing could escape except what was detected. it was blackness> unknown it was nothing it was not black black was not created black was> not yet a concept. this form composed of a specic formulation from a> certain calculation a simple equation that is equal to you and is equal to> me it is me a represented by that thing that thing that grouping of matter> it can be expressed mathematically. we are the tip the point the beginning> pyramid like it grows larger encompassing more until it is close to> innity but is not because it cannot know what continues to grow. there> are only so many deviations so many complications until everything is> right until it equals me or it equals you it approaches one it becomes> clearer more like us more like this another existence in the distance that> could be us parallel running alongside it is the same but for its> differences it could be similar almost close to approximately you almost> us. > the river time ows it goes forward the photons continue we cannot> understand what has not reached us cannot be perceived as useless as what> cannot be grasped known by the mind. here i am here you are look for> another me another she another he another tree if you can travel faster> than light you could know the future because it is circumstance it can be> known because all that is possible is likely it is innity with no end> the possible becomes runs closer nearer to fact to existence. hit the bong> and sing along come with me to another place just another face live or die> only you should care so only you should prepare since time cannot be> repaired you will die and so will i. we are nothing nothing it is> circumstance possibility that which could happen is required to happen the> bigger the number. prophesy is prediction it can be known by a pattern it> tries to be shown in the mind fact ltered through layers of tissue> distorted by tears death joy anger hate sadness black death despair the> end. it is fr away the twin that which only trivial contrasts that with> only the differences that cannot be seen tasted touched smelled seen no> understanding it becomes alike is a mirror. he types there he types there> mocking in imitation suffers a thousand times repeated mirror mirrors> tricks of the light. GOOGLE will remain #1 . . I ran across this little puzzle that has me hung up. Id> like to gure it out, but a hint would be nice.> hats and three black hats, the teacher places a hat on each> childs head. The third child sees that hats of the rst> two, the second child sees the hat on the rst, and the> rst child sees no hats. The children who reason> carefully, are told to speak out as soon as they can> determine the color of the hat they are wearing. After 30> seconds, the front child correctly names the color of her> hat. Which color is it, and why? > David> As I think Ive worked this out I have included a great big white space so that people dont see the (possibly wrong) answer inadvertantly.It seems to me that if the 3rd child can see 2 red hats then he/she knows that their own hat is black. Therefore, the 3rd child can see either 2 black hats or 1 black and 1 red hat and so cannot deduce his/her own colour.If the second child can see a red hat then he/she knows that their own hat must be black and therefore must be able to see a black hat.This makes sense to me but is it actually correct?Ivan. =In sci.math, David Serias<98dcfc8c7ba95821859a051f40c948fd@ news.teranews.com>:> I ran across this little puzzle that has me hung up. Id like to gure it> out, but a hint would be nice.> black hats, the teacher places a hat on each childs head. The third child> sees that hats of the rst two, the second child sees the hat on the rst,> and the rst child sees no hats. The children who reason carefully, are> told to speak out as soon as they can determine the color of the hat they> are wearing. After 30 seconds, the front child correctly names the color of> her hat. Which color is it, and why?> David> Well, lessee. The following scenarios are possible. Assume allchildren are facing leftward. Presumably the children can see theleftovers still in the hatbox, know there are 3 Bs and 2 Rsin total, and that the teacher didnt do something goofy like usea hidden dunce cap instead. It is also assumed that each childcan tell theres a hat on his or her head, and that each child canhear the others responses and is being honest -- children beingwhat they are they might shout Green for lots of giggles. :-)BBB RR - All three children shout out Black.BBR RB - The third child states Red. The rst and second then state Black.BRB RB - The third child states Black. The second states Red. The rst states Black, by the process of elimination.BRR BB - The third child states Red. The second child states Red. The rst states Black.RBB BR - The third child states Black. The second child states Black. The rst child states Red, by elimination.RBR BB - The third child states Red. The second child states Black. The rst child states Red.RRB BB - The third child states Black. The other two state Red.If the children cant see the hatbox the problem becomes much moredifcult.BBB - #3 is completely ummoxed, and no one says anything.BBR - #3 is completely ummoxed, and no one says anything.BRB - #3 is completely ummoxed, and no one says anything.BRR - #3 is completely ummoxed, and no one says anything.RBB - #3 is completely ummoxed, and no one says anything.RBR - #3 is completely ummoxed, and no one says anything.RRB - #3 states Black. After a bit of thought #1 and #2 state Red.Im not sure the problem is correctly specied. Of courseIm assuming all the children here can reason carefully;that will introduce 6 new variations of the problem,where only one or two of the children can think. (The 7thone isnt horribly interesting; the children basicallyjust giggle a lot. The 8th one Ive already enumerated above,where all three are budding logicians.)The way Ive heard it, all three children can see each other.However, Im not that familiar with the problem to recollectit fully in that form.-- #191, ewill3@earthlink.netIts still legal to go .sigless. = [.snip.]>Lets consider a similar statement to Lemma 1 >in a general ring R.> Lemma A: Let f,a and b be elements of R and let f divide ab in R.> Then there exist s and t in R, such> that s divides a in R, t divides b in R and f=st. [.snip.]>P.S. Is there a known characterization of rings R for which Lemma A holds?They are the pre-Schreier rings. If, in addition, R is integrallyclosed in its eld of fractions, then you have a Schreier ring.There is a fair amount of work on them, as was pointed out to me byBill Dubuque; see his post with fsf%40nestle.ai.mit.eduP.M. Cohns paper is particularly interesting (it is mentioned inBills post, but I thought I would repeat it here): P.M. Cohn: Bezout rings and their subrings. Proc. Cambridge Philos. Soc. 65 (1968), pp. 251-264. MR 36#5117.-- I accept as reality. --- Calvin (Calvin and Hobbes) [.snip.]> Proofs that your denition of coprime and Virgils denition >are equivalent in the algebraic integers have been known since the >time of Dedekind. You have never shown the slightest awareness of the Euclidean>algorithm, one of the absolutely basic tools in algebraic number >theory and the real reason that the denition that Virgil uses >for coprime is the standard accepted denition.No, the Euclidean algorithm is not really at issue here. For theEuclidean algorithm to apply, you must have a division algorithm,which means the domain is an Euclidean domain; but Euclidean impliesPID, so the ring of all algebraic integers is certainly notEuclidean. In fact, few number elds are Euclidean, even fewer thanare PIDs. (In fact, thats one reason whyfinding gcds is hard in the ring ofall algebraic integers: if we had the Euclidean algorithm, it wouldnot be so hard...)-- I accept as reality. --- Calvin (Calvin and Hobbes) > If you think your denition in terms of factors is not > equivalent to Virgils in the algebraic integers, by all means> give an example. That is all it would take to prove Virgil > wrong here. Produce it or produce a reference or (more logically)> concede.> To be absolutely fair, you ought to produce a reference for the fact> you claim: that Virgils denition of coprime *is* equivalent to the> denition JSH uses. You say that its well-known and has been known> for some time, and I believe you. But I wouldnt know where to look> for this fact and neither would James. So, it is only fair that you> give some reference for this fact or that you give the proof itself.> Arturo Magidin provided as good a reference as it is possible toget: Dedekinds book. Here is an outline of a proof: Denition A: a and b are relatively prime if the only common divisors of a and b are units. Denition B: a and b are relatively prime if there exist s and t such that a*s + b*t = 1. 1. Known Fact: Algebraic integers are a Bezout domain. This means that any ideal of the form is principal. This in turn means that there exist algebraic integers s and t such that = . 2. Let d = a*s + b*t. Since = , d must be a divisor of both a and b. 3. Let c be another divisor of both a and b. Then c divides a*s + b*t. Therefore c divides d. 4. Now assume a and b have no common divisors which are not units. This implies that d (as dened in 2. above) is a unit. That is, 1/d is an algebraic integer. This means that (1/d)*(a*s + b*t) = a*(s/d) + b*(t/d) = 1. This proves that Denition A implies Denition B. 5. To prove that Denition B implies Denition A: Assume there exist s and t such that a*s + b*t = 1. Let c be a common divisor of a and b. Then (by Harriss beloved Distributive Property), c is a divisor of 1. Therefore c is a unit. The difcult core of this, where you need a major theorem,is (1.) the Known Fact: that the algebraic integers are aBezout domain. Arturo shows how this follows from the class number theorem, which is the really nontrivial part of this. > Oddly, I think that a published reference impresses James more than a> careful presentation of the proof --- especially if the reference says> explicitly that Dedekind or some other of Jamess heroes (of whom he> knows nothing) rst proved the fact.> Arturo has provided such.> Of course, James will never look up the reference unless its a web> page, but his argument will be that much more blatantly silly once a> clear reference is given.> I dont know that its on a web page. Lots of important thingsare not.> (My apologies if someone has already provided a reference or two and I> missed it.)> No one needs further evidence. This is past mere mathematical> error. Deep inside, I think you know it. This is dishonesty, even> unto yourself. This is denial of reality. This is sickness.> You sound like youve been reading Kierkegaard or something. I havent. You disagree with the statements? Nora B. [.snip.]>The theorem correctly states that, using these standard denitions and >in ring of algebraic integers, or any other ring with identity, if f >divides a*b and is coprime to a then f divides b.> I agree with you, but as Bill Dubuque has pointed out many times,> coprime does have other standard meanings (though the one you use> is, as far as I am aware, the gold standard in algebraic number> theory). We know that James uses coprime to mean any common> divisors are units; in the ring of algebraic integers the two notions> are equivalent, but that is a hard fact to establish. It is, of> course, provable; but it ->is<- hard to do so.> Its funny that so many posted as if something were so obvious only tohave Magidin come in and claim that its not.My guess is that other posters will now back off at least a littlebit, but some, like Rick Decker, who supposedly is something of anexpert, have just lost a lot of their luster.Decker is, as I guessed, just another amateur.Posters are clearly *social* creatures, who act together often as aband, pushing certain truths as obvious, only to obediently backaway if certain people challenge them.Ifind that at least somewhat interesting.James Harris [.snip.]>The theorem correctly states that, using these standard denitions and >in ring of algebraic integers, or any other ring with identity, if f >divides a*b and is coprime to a then f divides b.> I agree with you, but as Bill Dubuque has pointed out many times,> coprime does have other standard meanings (though the one you use> is, as far as I am aware, the gold standard in algebraic number> theory). We know that James uses coprime to mean any common> divisors are units; in the ring of algebraic integers the two notions> are equivalent, but that is a hard fact to establish. It is, of> course, provable; but it ->is<- hard to do so.> Its funny that so many posted as if something were so obvious only to> have Magidin come in and claim that its not.While Magidin says that it is not obvious that the standard denition and your denition of coprime are equivalen, he also says that it is true.Strange how JSH ignores that part of Magidins posting. > My guess is that other posters will now back off at least a little> bit, but some, like Rick Decker, who supposedly is something of an> expert, have just lost a lot of their luster.Whereas JSH, by contrast, has no luster to lose.> Decker is, as I guessed, just another amateur.Whareas JSH, by contrast, is still a rank amateur. In fact, the rankest, in all senses of that word. [.snip.]>The theorem correctly states that, using these standard denitions and >in ring of algebraic integers, or any other ring with identity, if f >divides a*b and is coprime to a then f divides b.> I agree with you, but as Bill Dubuque has pointed out many times,> coprime does have other standard meanings (though the one you use> is, as far as I am aware, the gold standard in algebraic number> theory). We know that James uses coprime to mean any common> divisors are units; in the ring of algebraic integers the two notions> are equivalent, but that is a hard fact to establish. It is, of> course, provable; but it ->is<- hard to do so.> Its funny that so many posted as if something were so obvious only to> have Magidin come in and claim that its not.> It is well-known. No one said it was trivial. The main point here is not that its obvious or difcult.The main point is, its true and it has been known for a longtime. And it implies that your claims are wrong. Are you trying to divert attention from that? > My guess is that other posters will now back off at least a little> bit, but some, like Rick Decker, who supposedly is something of an> expert, have just lost a lot of their luster.> The only posters who should back off are those whose claims have been proven incorrect. What Arturo has done is to backup, not refute, what others have said regarding your error. Yet it seems you would like to imply that it casts doubt.If your own claim had had any luster to start with (it didnt)Arturos post, and his more detailed post in another thread,would have tarnished it.> Decker is, as I guessed, just another amateur.> Just another amateur ? You described yourself very recently asan *admitted amateur* (emphasis *yours*). Are you thus implying that Deckers math is incorrect, like that of some other amateur? Sounds like a classic ad hominem argument: if you cant refute the math, pin a stupid label on the person.> Posters are clearly *social* creatures, Some are, yes - here it is clearly *you* who is trying to use a socialargument [if you cant argue the math, call the person an amateur] to further your cause.> who act together often as a> band, pushing certain truths as obvious, only to obediently back> away if certain people challenge them.> Who has backed away?> Ifind that at least somewhat interesting.> Yes, again, The Uebermensch, the sci- mutant intellect from on high looks down and condescends to note that we poorly-wired mere mortals are squabbling over the details (even though were not). However The Uebermensch in this case is desperately trying to get people to ignore the big fat mistake that Virgil pointed out. Your insipid little habit of saying that something is interestingor fascinating is wearing very thin. Its transparent that it is just a mealy-mouthed way of trying cast aspersions without anycontent. Whatever you are, you have little in common with the Mr. Spock (fascinating, Dr. McCoy) except possibly (1) yourcapabilities are purely ctional, (2) you have pointy ears,and (3) you have green blood. Nora B.> James Harris = [.snip.] >The theorem correctly states that, using these standard denitionsandin ring of algebraic integers, or any other ring with identity, if f>divides a*b and is coprime to a then f divides b. I agree with you, but as Bill Dubuque has pointed out many times,> coprime does have other standard meanings (though the one you use> is, as far as I am aware, the gold standard in algebraic number> theory). We know that James uses coprime to mean any common> divisors are units; in the ring of algebraic integers the two notions> are equivalent, but that is a hard fact to establish. It is, of> course, provable; but it ->is<- hard to do so.> Its funny that so many posted as if something were so obvious only to> have Magidin come in and claim that its not. My guess is that other posters will now back off at least a little> bit, but some, like Rick Decker, who supposedly is something of an> expert, have just lost a lot of their luster. Decker is, as I guessed, just another amateur. Posters are clearly *social* creatures, who act together often as a> band, pushing certain truths as obvious, only to obediently back> away if certain people challenge them.*shakes head* Everyone who posts here is just like you James in that wereall human, though I wonder about your humanity from time to time. I thinkyou have a grossly inaccurate view of mathematics and such. As you seem toclaim, mathematics is not built on dogma. Grow up, James. Sad to see a manact like this.David Moran I nd that at least somewhat interesting.> James Harris > Theorem:>>In any commutative ring with identity, (A,+,*), if>>(1) f divides a*b (i.e., f*g = a*b, for some g in A), and >>(2) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A),>Youre relying on the very thing youre claiming to prove.>The conclusion to Virgils argument is f divides b.> conclusion. > But (2), in the ring of algebraic integers (though not in all rings)> may be taken to be the *denition* of coprime.> Its interesting how often I hear that a *denition* is whatsimportant.If thats how you dene coprime then being coprime is an*additional* condition which is basically producing the desiredoutcome.Take it away, and youre stuck.Now then, what do *you* Rick Decker assume if your denition forcoprime doesnt apply?I need to understand your logic here, what actually do you think ishappening mathematically? > Its like the parallel postulate, not provable for algebraic integers> f, g, a, and b, which isnt really subtle.> That is, given algebraic integers f, g, a, and b, where fg = ab, its> not provable *in the ring of algebraic integers* that if f is not a> factor of b, and doesnt share non-unit factors with b that it must be> a factor of a. > Of course its provable. You have the proof in front of you.> No. You seem to be clinging to trying to use a *denition* claim,and Ive taken that away from you.Now then, with coprime taken away, what do you think now must betrue?James Harris Theorem:>>In any commutative ring with identity, (A,+,*), if>>(1) f divides a*b (i.e., f*g = a*b, for some g in A), and >>(2) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A),>Youre relying on the very thing youre claiming to prove.>The conclusion to Virgils argument is f divides b.> conclusion.> But (2), in the ring of algebraic integers (though not in all rings)> may be taken to be the *denition* of coprime.> Its interesting how often I hear that a *denition* is whats> important.> If thats how you dene coprime then being coprime is an> *additional* condition which is basically producing the desired> outcome.> Take it away, and youre stuck.Arturo Magidin has provided evidence that the JSH denition of coprime and the denition I used of coprime are equivalent , at least in the ring of algrbraic integers.So JSHs argument is no longer with my proof, but with the evidence Magidin provided.> Now then, what do *you* Rick Decker assume if your denition for> coprime doesnt apply?That Magidinhas shown that the denition DOES apply in the ring of algebraic integers.> I need to understand your logic here, what actually do you think is> happening mathematically?IO think that JSH is again between a rock and a hard place, increased again.> Its like the parallel postulate, not provable for algebraic integers> f, g, a, and b, which isnt really subtle.> That is, given algebraic integers f, g, a, and b, where fg = ab, its> not provable *in the ring of algebraic integers* that if f is not a> factor of b, and doesnt share non-unit factors with b that it must be> a factor of a.> Of course its provable. You have the proof in front of you.> No. You seem to be clinging to trying to use a *denition* claim,> and Ive taken that away from you.And Magidin has given it back to us.> Now then, with coprime taken away, what do you think now must be> true?Now with coprime given back, what do you now think must be true?> James Harris =its always been my impression, althoughlaggard in my work on Beilers book, that ifyou know the meaning of primality, thencoprimality is a simple corrolary (even thoughthe two compared numbers may not be prime,themselves with respect to the whole eld of interest). if thats the case, thenHarris constant jackanapes about this simple thing is either a)sophistry, or b)marketing. or, c)BS. or even z)all of the below! > Now then, what do *you* Rick Decker assume if your denition for> coprime doesnt apply?> I need to understand your logic here, what actually do you think is> happening mathematically?--Give the Gift of Dick Cheeny -- out of ofce, at last!http://www.rand.org/publications/randreview/issues/rr .12.00/http://members.tripod.com/~american_almanachttp:// >Theorem:>>In any commutative ring with identity, (A,+,*), if>>(1) f divides a*b (i.e., f*g = a*b, for some g in A), and>>(2) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A),Youre relying on the very thing youre claiming to prove.>The conclusion to Virgils argument is f divides b.> conclusion.> But (2), in the ring of algebraic integers (though not in all rings)> may be taken to be the *denition* of coprime. > Its interesting how often I hear that a *denition* is whats> important.> If thats how you dene coprime then being coprime is an> *additional* condition which is basically producing the desired> outcome. Take it away, and youre stuck.No, take it away, and youre talking about some other statement.See earlier in this thread::> If f divides g_1*g_2 and f is coprime to g_1 then f divides g_2.:> Duh!::Not necessarily. You see it keeps coming back to the same thing,:which is your apparent inability to comprehend a certain possibility.You see, James, it keeps coming back to the same thing, whichis your apparent inability to comprehend a certain possibility:that you are wrong.Jim Burns <40080921.2060303@hamilton.edu> <87smifkkwf.fsf@phiwumbda.org (My apologies if someone has already provided a reference or two and I>> missed it.)> The following was posted by Arturo Magidin indicating that JSHs > denition and the one I used are equivalent in the ring of algebraic > integers, though the proof is not trivial.Perfect! Includes a reference to Dedekind, so it cannot be wrong (inJamess eyes).-- Jesse HughesSuch behaviour is exclusively conned to functions invented bymathematicians for the sake of causing trouble. -Albert Eagles _A Practical Treatise on Fouriers Theorem_ > Just a question: Couldntfinding GCD of pairs of the>> numbers and repeating the process be a better procedure?>> It could be, depending on how you divided the numbers up into pairs. If you>> get it wrong, you take just as long as you would with the naive method,>> plus you use additional storage space for the intermediate GCDs.> Any suggestions for how to divide up the numbers?> I am not yet sure of that. You have much less number of> calls of GCD. As to intermediate results, they could be> stored at where the pairs originally were.> If you have n numbers, then you need exactly n-1 calls of GCD, no matter> how you arrange the grouping. Some of them may be carried out in> parallel, but it doesnt decrease the number of calls that need to be> made.round of GCD computation of the pairs to enter theGCD computation of the next round will become however less and less in magnitude. This could be an advantagein the general case, I would think. M. K. Shen