mm-472 === Subject: Re: unlimited questionaire> Consider the infinite set S.> 0.3> 0.33> 0.333> 0.3333> 0.33333> ...> S has an infinite amount of rows T/F> F. Sets don't have rows.as its drawn> S has an infinite amount of columns T/F> F. Sets don't have columns either.> S contains only a finite number of successive 3s T/F> F. A set can have at most one 3, but this set, even if I understood> its definition, doesn't have any. Actually, I assume that S = {3/10,> 33/100, 333/1000, 3333/10000, 33333/100000,...} and can be readily> seen to have no 3 in it.as digits in the elements fool> S contains an unlimited number of successive 3s T/F> F. See above.> S contains an infinite number of successive 3s T/F> F.> S contains an infinite number of successive 3s but only a finite number> can be indexed T/F> Meaningless, neither true nor false.that would be F for fool then wouldn't it> Define EQUALS as :> A EQUALS B <=> A draws B on the number line.> What does draw mean?> e.g. N EQUALS 2 since Naturals covers the point 2.> You mean 2 is an element of N. Why not say so?because its just an instance of draw.> Conclusion 1> 0.3 !EQUALS 1/3> 0.33 !EQUALS 1/3> 0.33.. EQUALS 1/3> False since 1/3 is not an element of any of those.0.33.. does not draw the point 1/3 on the number line? you fool> Conclusion 2> S = {0.3}, S !EQUALS 1/3> S = {0.3, 0.33} S !EQUALS 1/3> S = {0.3, 0.33, ...} S EQUALS 1/3> S EQUALS 1/3 T/F> F, I think (probably meaningless too).it not difficult.if you draw the point made from 1/2 + 1/4 + 1/8 +... on the number line,does it mark the point 1 ?> Exam Time : 20 minutes> Closed Book> Marker : Herc> Not on your life. You think I would let such a nincompoop mark me?That's because FOOLS will FAILHerc === Subject: Re: pi, 2 pi and frivolitythe diameter is more fundamental than the radius. consider that one point and the radius fix a circlein the plane; but two points fix its diameter (ortaht of a sphere; if the sphre is spinning,than one diameter is special). again the circumference is piD, andthe area os the sphere is piDD. (actually,I prefer to define the circumference as unit, andthe diameter as pith, and likewisefor the sphere .-) > As I see it pi is good as is. No matter what we defined it as to begin --Give Earth a Trickier Dick Cheeny -- out of office, after GIGA years.http://www.benfranklinbooks.com/http://www.rand.org/ publications/randreview/issues/rr.12.00/http:// members.tripod.com/~american_almanac === Subject: * V Interesting geometry problem *I would'nt insult your intelligence by asking you the area of intersectionof two circles radius r and center distance d. I already know that.Area = r^2 ( theta - sin (theta) )where sin (theta) = (d sqrt (4r^2 - d^2) ) over (2r)What I want you to find is:Area of intersection of two identical squares of unit size. First square center is at (0,0) and is not tilted. Second square center is at (h,k) and is tilted theta CCW.I need a formula or an algorithm in terms of three numbers:h,k,theta, call theta as t for simplicity.Dan Mocharnouk === Subject: Re: * V Interesting geometry problem *Comments: This message probably did not originate from the above address. It was automatically remailed by one or more anonymous mail services. You should NEVER trust ANY address on Usenet ANYWAYS: use PGP !!! Get information about complaints from the URL belowX-Remailer-Contact: http://80.65.224.85/POL/ In case my abuse address is unreachable: It is because it has been flooded by , please contact What I want you to find is:> Area of intersection of two identical squares of unit size. > First square center is at (0,0) and is not tilted. > Second square center is at (h,k) and is tilted theta CCW.> I need a formula or an algorithm in terms of three numbers:> h,k,theta, call theta as t for simplicity.Very easy. It is (1-h)(1-k) if both the squares are one unit as you say.David Baltimore === Subject: Product of these fractions never a power of 2 ?Hi - just came across a funny formula, and can't proceed. can it be shown, that none of the following products can be an integral power of 2 with integer a,b,c >1:(1): 1 (3 + ---) =/= 2^m a(2): 1 1 (3 + ---)( 3 + ---) =/= 2^m a b(3): 1 1 1 (3 + ---)( 3 + ---)( 3 + ---) =/= 2^m a b cand so on ...Note, that using 5 instead of 3 I was given a solution, a=33, b=83, c=13, i=7If that helps, then there are some more restrictions on a,b,c:4) a,b,c are odd integers5) a,b,c are all different6) the smallest can be at least 57) none can be divided by 3I tried a lot of rearranging already, (1 and 2 are accessible to me)but with 3 and more factors I only could manage it by countingthe possible solutions; analytically I'm turning in circles... === Subject: Re: Product of these fractions never a power of 2 ?> (3):> 1 1 1> (3 + ---)( 3 + ---)( 3 + ---) =/= 2^m> a b cWhat about(3+1/8)(3+1/5)(3+1/5) = 32 = 2^5(3 + 1/4)(3+1/5)(3+1/13) = 32 = 2^5(3+ 1/133)(3+ 1/25)(3 + 1/2) = 32 = 2^5(3+ 1/2)(3+ 1/22)(3 + 1/469) = 32 = 2^5plus many others. === Subject: Re: Product of these fractions never a power of 2 ?Am 10.04.04 16:52 schrieb Jason Rodgers:>(3):> 1 1 1> (3 + ---)( 3 + ---)( 3 + ---) =/= 2^m> a b c> What about> (3+1/8)(3+1/5)(3+1/5) = 32 = 2^5> (3 + 1/4)(3+1/5)(3+1/13) = 32 = 2^5> (3+ 1/133)(3+ 1/25)(3 + 1/2) = 32 = 2^5> (3+ 1/2)(3+ 1/22)(3 + 1/469) = 32 = 2^5> plus many others.some lengthy derivations. Good to see, there are some solutionsprincipally. Now, as Rainer mentioned, the question stems ofthe Collatz-problem, so there are more restrictions tothe parameters, for instance, of interest are only solutions,where all numbers a,b,c are odd and are chained with thecollatz-iteration.My original question is/was, to find some analytical toolto access this type of question after I already turnedin circles... But anyway, it is good to *see* some solutions. === Subject: Re: Product of these fractions never a power of 2 ?> can it be shown, that none of the following products can be> an integral power of 2 with integer a,b,c >1:> ...> 1 1 1> (3 + ---)( 3 + ---)( 3 + ---) =/= 2^m> a b c> Note, that using 5 instead of 3 I was given a solution,> a=33, b=83, c=13, i=7This should read m=7. And indeed we have(5 + 1/13) * (5 + 1/83) * (5 + 1/33) = 128 = 2^7. (1)Once upon a time, this formula ended my Collatz-nighmare,to solve the 3x+1 riddle I learned that it was necessaryto find 3^k very near to 2^m for some k and m.Considering the similiar problem 5x+1 instead, we havesuch nearness because of 125 = 5^3 ~ 2^7 = 128.And indeed there is a Collatz-loop:13->66->33->166->83->416->208->104->52->26->13The interesting stepping stones are 13, 33 and 83.I think there is a OEIS sequence lurking behind formula (1).Remember: http://www.research.att.com/~njas/sequences/ === Subject: How to solve an exponential ciphers?How to solve an exponential ciphers?p=2633 , C=423 , e=29plain text: EX (0423)how to solve 423^29 mod 2633 without using calculator? === Subject: Need for with partial differential equationMucho appreciated for any help you might care to provide for solving thisPDE:dc/dt = D*(d2c/dx2)+s*(dc/dx)where c(x,t) and D, s are constants. As you may have guessed, this is aform of the Nernst-Planck equation, but rewritten to express the second(electric) term directly in terms of the drift speed s. D is of course thediffusion coefficient.Gregory === Subject: Re: Need for with partial differential equation> Mucho appreciated for any help you might care to provide for solving this> PDE:> dc/dt = D*(d2c/dx2)+s*(dc/dx)> where c(x,t) and D, s are constants. As you may have guessed, this is a> form of the Nernst-Planck equation, but rewritten to express the second> (electric) term directly in terms of the drift speed s. D is of course the> diffusion coefficient.This may sound dumb, but wouldn't Nernst and/or Planck have solved it?If not, and no one else has either, then it must be a toughie andunlikely to be cracked in all generality on sci.math. But there isone obvious special case: If c = u(x).v(t) where @u/@t = @v/@x = 0then the equation implies the following where C is constant, andu(1) := du/dx and v(1) = dv/dt etc: D.u(2) + s.u(1) + C.u = v(1) + C.v = 0and both these can be solved trivially.---------------------------------------------------- -----------------------John R Ramsden (jr@adslate.com)---------------------------------------------- -----------------------------Eternity is a long time, especially towards the end. Woody Allen === Subject: Re: Need help with partial differential equationIt just occurs to me that I can't write proper titles and I haven't suppliedthe boundary conditions:c(0,t) = C (constant) for all tc(x,0) = 0 for all t > 0c(x,t) = 0 for x, t > 0 (x is semi-infinite)Specifically, I'm looking for an explicit solution, which I've found for theequation written in its original form:dc/dt = D(d2c/dx2 + a*(dc/dx))where a is a constant.Gregory> Mucho appreciated for any help you might care to provide for solving this> PDE:> dc/dt = D*(d2c/dx2)+s*(dc/dx)> where c(x,t) and D, s are constants. As you may have guessed, this is a> form of the Nernst-Planck equation, but rewritten to express the second> (electric) term directly in terms of the drift speed s. D is of course the> diffusion coefficient. === Subject: Re: Is Goodstein sequence is not provable ? charset=Windows-1252> Goodstein sequence is not provable in PA .Is that right?> is there a literature on this> ( well,every Goodstein sequence eventually terminates at 0.> not all undecidable theorems are peculiar or contrived, as> those constructed by Godel's incompleteness theorem> are sometimes considered.)Goodstein's theorem (that 0 is an element of everyGoodstein sequence) is not provable in PA.Seehttp://mathworld.wolfram.com/ GoodsteinsTheorem.htmlBurbanks: Fast-Growing Functions and Unprovable Theoremshttp://www2.maths.bris.ac.uk/~maadb/research/seminars/ online/fgfut/index.htmlSimpson: Unprovable theorems and fast-growing functions--r.e.s. === Subject: Re: How well do the first n terms of Maclaurin's expansion approximate a function?||>assuming that i'm not too badly confused here, then i think that|>it's actually neater to note that at the same rate is a red|>herring here; doesn't it work just as well if each point|>approaches zero at its own idiosyncratically varying pace?|> |> That seems likely to me. (I was about to say no, but the no applies|> to the version I gave, where we assume just that f^(n)(0) exists,|> instead of assuming that f is C^n.)||Right, you assume less but your points tending to 0 are t*{a,b,c,...}|as t -> 0, which is about as same rate as you can get. Just|assuming f^(n)(0) exists is not enough to allow each point|approaches zero at its own idiosyncratically varying pace, as you|indicate.||Here's another proof, assuming f is in C^n, but allowing each point|to -> 0 however it wants to. We have f in C^n on some interval|I. Given distinct x0, x1, ..., xn in I, let L = L_{x0,x1,...,xn} be|the nth-degree Lagrange polynomial that interpolates f at the|xj's. We want to show L_{x0,x1,...,xn} -> P as the xj's -> 0, where P|is the nth degree Taylor polynomial of f at 0. When n = 0 this is|clear; the continuity of f is used here. So assume true for n-1 and|consider the set-up above. By the mean value theorem there are|distinct y1, ..., yn, between the xj's, such that L' = f'. Because P'|is the (n-1)st degree Taylor polynomial of f' at 0, the induction|hypothesis says L' -> P' as the x's go to 0 (becaue the y's are|forced to 0). So now integrate to get L - L(x0) -> P - P(x0), ie, L|-> P - P(x0) + L(x0) as the x's go to 0. But L(x0) = f(x0), and both|f(x0) and P(x0) -> f(0) as the x's go to 0. So L -> P as desired.i guess you're only doing the case where x0reals with u an open subset ofreals^n, consider the partially defined functionlagrange_f:u^[n+1]->{degree n polynomials} assigning to anon-degenerate [n+1]-tuple x=(x0,x1,...xn) the lagrange extrapolant off with respect to x. then f is c^n iff lagrange_f has a totallydefined continuous extension lagrange-taylor_f, which is of courseunique if it exists, and which gives the degree n taylor polynomial off when evaluated at a constant tuple.if this works, then what's the right generalization to the case wheref is a multi-variable function? === Subject: Re: How well do the first n terms of Maclaurin's expansion approximate a function?> |> |>assuming that i'm not too badly confused here, then i think that> |>it's actually neater to note that at the same rate is a red> |>herring here; doesn't it work just as well if each point> |>approaches zero at its own idiosyncratically varying pace?> |> |> That seems likely to me. (I was about to say no, but the no applies> |> to the version I gave, where we assume just that f^(n)(0) exists,> |> instead of assuming that f is C^n.)> |> |Right, you assume less but your points tending to 0 are t*{a,b,c,...}> |as t -> 0, which is about as same rate as you can get. Just> |assuming f^(n)(0) exists is not enough to allow each point> |approaches zero at its own idiosyncratically varying pace, as you> |indicate.> |> |Here's another proof, assuming f is in C^n, but allowing each point> |to -> 0 however it wants to. We have f in C^n on some interval> |I. Given distinct x0, x1, ..., xn in I, let L = L_{x0,x1,...,xn} be> |the nth-degree Lagrange polynomial that interpolates f at the> |xj's. We want to show L_{x0,x1,...,xn} -> P as the xj's -> 0, where P> |is the nth degree Taylor polynomial of f at 0. When n = 0 this is> |clear; the continuity of f is used here. So assume true for n-1 and> |consider the set-up above. By the mean value theorem there are> |distinct y1, ..., yn, between the xj's, such that L' = f'. Because P'> |is the (n-1)st degree Taylor polynomial of f' at 0, the induction> |hypothesis says L' -> P' as the x's go to 0 (becaue the y's are> |forced to 0). So now integrate to get L - L(x0) -> P - P(x0), ie, L> |-> P - P(x0) + L(x0) as the x's go to 0. But L(x0) = f(x0), and both> |f(x0) and P(x0) -> f(0) as the x's go to 0. So L -> P as desired.> i guess you're only doing the case where x0 times?Not sure how you got that idea. I only assumed x0, x1, ..., xn are distinct; I see no problem allowing one of xj's to be 0. I haven't thought about degenerate cases. === Subject: Re: PROOF that Integers are not countable!> You haven't shown how, given any list, you can construct a number not> on the list. There is no such natural number as oo, or oo-1.Assume I have a list of natural numbers., N(i).Let s be a natural number not on the list.s is calculated recursively.s_1 = N(1)+1s_i = N( s_(i-1) ) + s_(i-1) (for i>1)(for finite lists, if N(s_i) doesn't exist then add the last N(i)).s_i is not in the set Union( N( 0 ) - N( s_(i-1) ) ) for all i.Example:N_i = 1, 2, 3, 4, ...s_i = 2, 4, 8, 20, ...Russell- The universe is one dimensional === Subject: Re: PROOF that Integers are not countable!> You haven't shown how, given any list, you can construct a number not> on the list. There is no such natural number as oo, or oo-1.> Assume I have a list of natural numbers., N(i).> Let s be a natural number not on the list.> s is calculated recursively.> s_1 = N(1)+1> s_i = N( s_(i-1) ) + s_(i-1) (for i>1)> (for finite lists, if N(s_i) doesn't exist then add the last N(i)).> s_i is not in the set Union( N( 0 ) - N( s_(i-1) ) ) for all i.> Example:> N_i = 1, 2, 3, 4, ...> s_i = 2, 4, 8, 20, ...> Russell> - The universe is one dimensionalYou have defined, not a natural number, but a sequence of naturalnumbers. And while the n-th member of this sequence is never the n-thnumber on the list, elements of the sequence can still occur in thelist. You haven't given a natural number not in the list. Sometimesthe list will include all natural numbers, and then this won't bepossible. === Subject: Re: Software to visualize Complex Functions>What software can I use to visualize complex functions?>I'd like to see the transformation of say the curve g(t) = 4i t under >the function f(z) = z^2. Is there software available that can easily >help me do that.There was a software packaged called f(z) that was pretty well thoughtof. I'm not sure what happened but the company and software seemed todrop out of sight a year or two ago. I traded email with the authorof Visual Complex Analysis, thinking he might have had some recentcontact with them, since he used that for portions of his fine book,but he was unable to provide any leads.There is a discount software dealer on Ebay who sells very very oldversions of odd and strange software. They usually have one or twocopies of an ancient version of f(z) for sale at any given time. Butthe graphics available in that are extremely poor by current standards. If anyone finds a stash of a recent version f(z) please let me know.For the longer term, if f(z) is really gone but a group of people gottogether and made a persuasive argument for putting a coherent set ofthe features that it had into a future version of Derive there mightbe a reasonable chance of success in getting that done. Derive can dosome things to plot complex functions but it isn't strong in that area.Making a pitch to get the authors to consider incorporating feasibleand consistent features to enhance this might be accepted.(email address is valid, if that matters) === Subject: Re: Software to visualize Complex Functions>There was a software packaged called f(z) that was pretty well thought>of. I'm not sure what happened but the company and software seemed to>drop out of sight a year or two ago. I traded email with the author>of Visual Complex Analysis, thinking he might have had some recent>contact with them, since he used that for portions of his fine book,>but he was unable to provide any leads.By googling f(z) program I got www.lascauxsoftware.com/ so try thatPatrick === Subject: Re: Software to visualize Complex FunctionsAm 10.04.04 01:03 schrieb mr0x:> What software can I use to visualize complex functions?> I'd like to see the transformation of say the curve g(t) = 4i t under > the function f(z) = z^2. Is there software available that can easily > help me do that.You can try WiScy, if it is still available. Extremely simple for a firststart; it's made like a simple calculator. === Subject: Re: Antidiagonal, Infinity> In any integer base you select, for any antidiagonal you select, there> is another integer base where that specific antidiagonal number has> dual representation,That is a huge claim, and would require proof, but even if it were true, it is irrelevant. One only needs an anti-diagonal algorithm able to construct from any list an unlisted number with a unique representation in one base. What happens in other bases is then irrelevant. > Thus EVERY function from N to R, or to any interval in R of positive> length, must have range a _proper_ subset of its codomain.> That, I am trying to understand.> Please present a function between a domain of (each element of) a> closed interval and each element of the set of reals.This has been presented before, and is really qute trivial:(1) Between any two closed and bounded intervals (of more than 1 point) the order preserving linear function carrying endpoints to endpoints is a bijection, so all such intervals are effectively order isomprphic:(2) Similarly for bounded open intervals:(3) The mapping x -> x/sqrt(1+x^2) is an order isomorphism between R and (-1,1)(4) define f:[0,1] -> (0,1) by f(0) = 1/2, f(1/n) = 1/(n+2), for n a positive integer, f(x) = x otherwise.Then this f is a bijection, though not order preserving, map from a closed interval to an open one.Thus there are, by appropriate compostion, bijections, many of then order preserving, between any two real intervals. === Subject: Re: Antidiagonal, Infinity> Thus there are, _by appropriate compostion_, bijections, many of then > order preserving, between any two real intervals. > [sic, emphasis added]I asked about mapping between the closed interval and the set of allreals because I could not think of one that did not use thecomposition of functions to map to a finite open interval andhenceforth to the interval of all reals.The most simple mapping to me between a finite open interval and theset of all reals is (co)tangent, a dimensionless ratio.That is to say, I can think of mappings between closed finiteintervals and open or half open finite intervals, and between open orhalf open finite intervals and open or half open infinite intervals,but not between closed intervals and open or half open infiniteintervals without the (implicit) composition of those functions,although I think they exist. I'm reminded of the herringdomain-constrained inverse.Is it not possible to map a closed finite interval to an open infiniteinterval without compositionally mapping to an open finite interval? I've overlooked obvious counterexamples before.I asked that question to lead to the thought that the notion of anecessarily implicit composition of functions is not so far-fetched indescribing a function between the closed interval and the infiniteinterval (with the point at infinity being an open endpoint). Wherethat is so, I hope to draw a parallel for you that the notion ofmapping between the naturals and a finite interval and compositionallyto an infinite interval is not ridiculous or legerdemain, orunprecedented.I hope to so prove it myself.Please do say if you came to that concept, I think you did.One concept that has come to fore in this and surrounding discussionsis that of the infinitesimal. It is by no means a new concept, fornearly as long as there was the concept of the infinite there was thatof the infinitesimal, they are flip sides of a coin. Where theinfinite is treated institutionally these days as transfinite ordisconnected ordinal, I hope to help reestablish it as a scalar. Thenon-standard there was once quite standard. It may so be again, inrigorous new ways.What do you think about the leading zero(e)s in the matrix of listelement expansions? Why are not there obvious mappings with simpletransformations between a closed interval and the set of all reals? === Subject: Re: Differential equations of periodic functionsMathematica. The solution I always get and the implicit from Fullarton site is symmetrical when all the constants equal one around the line x=y.I added a small Verhulst mortality fraction (e and f small) :x'=a*x+b*x*y -e*x*xy'=-c*y+d*x*y -f*y*yand I get a spirally inward as a result.I posted the picture of the notebook at chaos theory.I also did a decoupled male female 4d version and the3d projections in a notebook are in the file section at chaos theory , too.> In spite of some effort to decouple x and y (to write y'' in terms of> y',y) but I did not succeed. Is this seemingly simple thing> impossible?..I posted this in haste. :)> Let u=ln(x) ; v= ln(y) ; the decoupled equations for VolterraLotka> PredPray are:> u''=(u'-a)(d exp(u)-c) and v''=(v'+c)(a-b exp(v)).> On inspection, they do not appear to be periodic -- but only on> plotting the graph it is evident that 6.6 units time period exists. === Subject: Re: another boring critisism of Cantor's Theorem>But ZFC also has models of arbitrarily large>cardinality in the real world.> You've complained about me saying things that really> make no sense. I think you've just topped anything> I've done.> It makes perfect sense. You yourself said that the models of ZFC in> the countable model of ZFC+Con(ZFC) were of arbitrarily large> cardinality in the countable model, but countable in the real world.> You mean, there are bijections from them to N, not in the countable> model, but in the real world, in the universe, V. I'm simply pointing> out that in V there will be models that really are of arbitrarily> large cardinality.> But that 'real world' could be countable, when looking at it> from an even larger meta-world. And that would mean that> these large cardinals would be 'really-really' countable.> Right? Or do i misunderstand you?> Herman JurjusNo. If ZFC+Con(ZFC) is consistent, and we are uttering theorems ofZFC+Con(ZFC), then we can suppose, without making any of our theoremsfalse, that the semantics of our discourse is such that theinterpretation of it is a countable model of ZFC+Con(ZFC). But that'snot the same as saying V can be countable. V can't be countable. === Subject: Re: space elevator> This sounds like an idea of Physicists, wait till SETI finds something> first.> you might check out the FAQ here http://www.liftport.com/ they have> answered> a lot of questions here that I would have never thought of.> Most especially avoiding debris :) the base of the cable will be mobile.> They also figure the cable will have a natural 7 hour resonance the can> be actively damped and LOTs of other issues that are answered.> Charles> -If Bill Gates had a nickel for every time Windows crashed... oh wait,> he does.->I don't think it is restricted to one orbit, it can come from all>directions. Also even if it came from certain directions more>often then if shaped to handle that an unlucky strike could>severely damage it.>also consider what sort of debris your cable is up against. In>LEO it will be mostly man made and in an equatorial orbit, thus> shaping>your cable would seem to make sense.>How about good ole earth cable? lowest overall cross section to>tension. Ahh I get it, least probability of catastrophic failure>VS least probability of damage.>Surely a simulation would arrive at the answer, 1/4 of an arc is>vulnerable to critical angle of attack, semicircle is vulnerable to> 2>holes, so around 1/3 of an arc.>3 cables would be the best design with spacers, that way at most 2>are knocked>out by a small projectile.>O> o> o> o> O o> o> o o O>composite design>Herc> So you think may be 1/3 of an arc would be optimial? The way I> see> it>the answer is fairly critical. Assuming when they build it for weight>reasons it will be as light as possible relative to the debris so the> risk>of catastrophic damage will be high. Then they would build in safety> margins>but these would be as low as possible so as not to add on too much> weight.> After that point that may decide to make it thicker to carry more> weight>in which case they may be free to consider other shapes. For example> an> arc>is easier for a cable car to grip on than a hollow cylinder.> Separate cables with spacers becomes similar to a mesh where the> fibres>are arranged vertically and horizontally with large holes in between> to>localise the damage from debris.> I'll have to get back to you on the 1/3 figure, my simulator's a bit> slow.> http://www.a1sites.com/spacecable.html> Seems a semicircle gets taken out first, and the 1/4 arc is almost> invisible to> most angles of attack, would require a lot of simulations to detect a> flat> cable is the most catastrophic as on a typical run it seems to performs> the best> i.e. the lower the fraction of the arc the less visible to attack, but> also> the greatest risk of a 'wipeout' in 1 hit.> So my new design is 2 flat cables at say right angles, each with a> small> risk> of total failure during which the single cable left can suffice until> repair.> Herc>vapor ware. Sounds like a gimic to me to get investors money.Eric === Subject: derivation step helpI'm reading a derivation of the 2D wave equation in my engineering mathbook, and there is a step I do not understand.The forces parallel to the xu plane (u is up) areT*dy*sin(B) and -T*dy*sin(a)Summing,T*dy*(sin(B) - sin(a))Since the angles are small they approximate with tangents:T*dy*(tan(B) - tan(a))T*dy*(tan(B) - tan(a)) = T*dy*( u_x(x+dx, y1) - u_x(x, y2))where the subscrip _x denotes partial derivatives, and y1 and y2 are valuesbetween y+dy (an edge of the small element on the membrane we areconsidering)I do not know how they went from tan functions to partial derivatives. Cananyone explain and or tell what section in my calculus book might coverthis. === Subject: Re: derivation step help> T*dy*(tan(B) - tan(a)) = T*dy*( u_x(x+dx, y1) - u_x(x, y2))> where the subscrip _x denotes partial derivatives, and y1 and y2 arevalues> between y+dy (an edge of the small element on the membrane we are> considering)Heh, I think this was actually really easy but I didn't see it right away.u_x(x+dx, y1) gives the slope at the point (x+dx, y1), but tan(B) also givesthe slope at that point, so they are equal. Likewise, u_x(x, y2) gives theslope at (x, y2), but so does tan(a), so they are equal. === Subject: Circular Permutationhello......i know thatThe number of ways to arrange n distinct objects along a fixed circleis (n-1)!and i know the reason (for rotation a round table).but it looks like to view from a table or n-objects.if it's from observer of view, i think that n! is right.suppose that observer sit on a chair in theater,then, n! is right.why is only (n-1)! ?how do you think about it?let me advice, please...thank you very much. === Subject: Re: Circular Permutation> i know that> The number of ways to arrange n distinct objects along a fixed circle> is (n-1)! and i know the reason (for rotation a round table).> but it looks like to view from a table or n-objects.> if it's from observer of view, i think that n! is right.> suppose that observer sit on a chair in theater,> then, n! is right.> why is only (n-1)! ?> how do you think about it?I think that this a matter of language. To be more precise, thisdepends upon what you mean by distinct objects. The answer willbe (n - 1)! if what matters is the relative positions of theobjects, not their absolute position (then the answer would be n!).And if what mattered were the neighbours of each object, but notwhether they were to its left or its right, then the answer wouldbe (n - 1)!/2 (for n > 2, of course).Jose Carlos Santos === Subject: Re: Circular Permutation> I know the number of ways to arrange> n distinct objects along a fixed circle is (n-1)!> and i know the reason (for rotation a round table).> but it looks like to view from a table or n-objects.> if it's from observer of view, i think that n! is right.> suppose that observer sit on a chair in theater,> then, n! is right.> why is only (n-1)! ?> how do you think about it? fix one object anywhere / x 1 n-1 2 ...Seat (or fix) one person or object anywhere at the table.Now there are (n-1) objects left to put around the table.But doing that is isomorphic to lining up (queuing up)(n-1) objects in a line (queue), which can be done in(n-1)! ways. === Subject: Re: Is |Q an open set?>If |Q is open then wouldn't the irrational then also be dense in |R.> Open in what? Q is open as a subspace of itself (in any topology). It> is open in the discrete topology. But it is NOT open in the usual> topology of R.Q open in R with the topology generated including Q as open, viz. { U, U/Q | U open in R }Then RQ isn't dense because Q is now open and Q / RQ = nulset.This space is example of an irregular Hausdorff space.Hausdorff for being finer that usual R.Irregular since 0 not in closed RQ and 0 and RQ can't be separated.If (a,b)/Q open nhood 0 and r in (a,b)Q, then any open cover of r,which has to include base set of the form (s,t) nhood r, will intersect(a,b)/Q> Next: the irrationals ARE dense in R, but it is not because Q is open> or not: it is because Q has empty interior. === Subject: Re: Is |Q an open set?>If |Q is open then wouldn't the irrational then also be dense in |R.> Open in what? Q is open as a subspace of itself (in any topology). It> is open in the discrete topology. But it is NOT open in the usual> topology of R.> Next: the irrationals ARE dense in R, but it is not because Q is open> or not: it is because Q has empty interior. The is not the reason Q is empty. It is because R|Universe=empty. and RvUniverse=1. and NUniverse=1/2, except at the Bing Bang, where it is 2.32%.> -- === ===============> Arturo Magidin> magidin@math.berkeley.edu === Subject: Conversion of Distribution from Plane to Sphereconsidering the case I have a distribution that is uniform on a planein x and y.Mapping this uniform distribution to the surface of a sphere seems tobe simpel - I just have to perform a sinus weighting, lets say alongy, and the data are distributed isotropical over the sphere. Themarginal pdf is thenP=1/2sintheta. (see mathworld.wolfram.com/SpherePointPicking.html)In a more 'advanced' case I have a distribution which is again uniformin x but has a Gaussian distribution in y. Has anyone an idea how tomap such a distribution onto the sphere's surface ?I know that the Fisher distribution is the analogue to the Gaussion onthe sphere, but I have no idea if it is possibel to perform such atransformation for the Gaussian from the plane to the sphere.Dirk === Subject: Re: functions that haltIn sci.logic, ZZBunker In sci.logic, ZZBunker> In sci.logic, ZZBunker> The Ghost In The Machine There was a post in sci.math last year about the class of functions>that are constructed upwards and are guaranteed to halt.>How powerful would they be?>There was no formal answer but the consensus was they would be trivial>functions.>Is this necessarily true?>The halting proof started out as a system to check for errors in programs.>We assume that 'programs' either halt or they don't (good assumption given>our practices).>general function halts or not.>But what if we assume all programs halt? This doesn't>contradict the halting proof, merely denies its construction>since it depends on an infinite loop being programmed.>It is trivial to construct a program that does not halt. There are>plenty of times when a loop is constructed for which it is far from>obvious if it will halt. The loop may even be infinite for some input>and finite for other input.>What if we assume all programs halt? We can then prove anything,>since it is not true.>What function can your program with an infinite loop calculate that a>special class of programs without infinite loops not calculate?>One example is to return the number of iterations it takes>for the following program to halt. The following is implemented>from a conjecture by L. Collatz, first proposed in 1937.>The notation is Pascal-like and should be fairly obvious.>function CollatzIterations(inp: positive_integer) : positive_integer;>var> numit : positive_integer;> temp: positive_integer;>begin> numit := 0;> temp := inp;> while temp <> 1 do> begin> if(temp mod 2 = 0) then temp := temp div 2;> else temp := 3 * temp + 1;> numit = numit + 1> end> return numit;>end;>The conjecture is that this program will always halt and>give a valid result, for positive integers -- but AFAIK>it has yet to be proven, and may never be provable in>light of a result by Conway in 1972 -- an interesting if>irritating result; darn those Goedel-like formalisms! :-)>http://mathworld.wolfram.com/CollatzProblem.html>(The assumption is that 'positive_integer' can be arbitrarily large, BTW.)>[.sigsnip]>Sci.math added for hopefully obvious reasons; followups reset.> Don't know why you culled sci.logic. Well that's not really a function,> its a simulation. Any simulation could go indefinately. Would an engineer> have any practical objections to use this :> Why do you say it's not a function? It's a function that may or may not > always halt, but it is a function. Put another way, it's a function > that may or may not have domain = N. Also, why bring engineers into a > pure math problem?function CollatzIterations(inp, t: positive_integer) : positive_integer;>var> numit : positive_integer;> temp: positive_integer;>begin> numit := 0;> temp := inp;> while (temp <> 1) & (numit < t) do> begin> if(temp mod 2 = 0) then temp := temp div 2;> else temp := 3 * temp + 1;> numit = numit + 1> end> return temp;>end;> it looks like it will halt anyway, on average 1 in 4 times it will divide by >4,> 1 in 4 times it will divide by 2, 1 in 2 times it will *3.> n/4/2*3*3 = 9/8. OK thats increasing.> n/8/4/2/2*3*3*3*3 = 81/128> A few iterations deep and it should drop like a weight.> Appearances can be deceiving, and it fails to answer the question. Does > CollatzIterations(50912309841029384893201) return a value? If so, what? > Limiting it to a few million iterations does nothing towards answering > the question of interest.> It is impossible for CollatzIterations to return > a value, since *return* is a system call, > not a math call.> A system call to pull a value off the stack and stuffing it> into the PC sounds to me rather pointless. In any event,> 'return' is translated into the single instruction 'RET'> on x86 processors -- and a lot of other computer systems> have similar instructions. (I do know of several exceptions,> however -- none of them modern.)> For those of us fortunate enough to have gcc, one can compile> using the '-S' option to see what the routine translates into> (in symbolic form, anyway).> You are correct in that it is not a math call, though -- but> a function has to get its return value to the requester *somehow*.> Intel has four flavors of return:> RET NEAR [0xC3] -- pops IP from the stack.> RET FAR [0xCB] -- pops CS:IP from the stack.> RET NEAR n [0xC2 nn] -- pops IP and n bytes from the stack (presumably the> n bytes are from a pushed procedure call)> RET FAR n [0xCA nn] -- pops CS:IP and n bytes from the stack (presumably> the n bytes are from a pushed procedure call)> Well, since your example was Intel, they're > simple to figure out, since they're > really just MicroSoft lawyers with a suntan.> Everything that does not come on the chip, > you get from a dynamic libary, > usually a hacked-up Sun Product.> If you wish I can illustrate the point using Sparc or Motorola products.> Well you can, but as far languages are concerned, > it doesn't matter what Intel, Sparcs, or Motorolas> Assemblers do. Since in larger systems, > the assemblers themselves are sometimes both throw away, > and rewritten in something like Perl or Matlab.Microcode is interesting but peripheral to this discussion. :-PAs it is, I'd like for someone to recode CollatzIterations() usingnothing but fixed-count for loops. (The count can be anyinteger computable from the input parameter, of course.)#191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: functions that halt===>Subject: Re: functions that halt === >Subject: Re: functions that halt>The Ghost In The Machine functions>that are constructed upwards and are guaranteed to halt.>How powerful would they be?>There was no formal answer but the consensus was they would be> trivial>functions.>Is this necessarily true?>The halting proof started out as a system to check for errors>in> programs.>We assume that 'programs' either halt or they don't (good> assumption given>our practices).>general function halts or not.>But what if we assume all programs halt? This doesn't>contradict the halting proof, merely denies its construction>since it depends on an infinite loop being programmed.>It is trivial to construct a program that does not halt. There>are>plenty of times when a loop is constructed for which it is far>from>obvious if it will halt. The loop may even be infinite for some> input>and finite for other input.> What if we assume all programs halt? We can then prove>anything,>since it is not true.>What function can your program with an infinite loop calculate>that> a>special class of programs without infinite loops not calculate?>One example is to return the number of iterations it takes>for the following program to halt. The following is implemented>from a conjecture by L. Collatz, first proposed in 1937.>The notation is Pascal-like and should be fairly obvious.>function CollatzIterations(inp: positive_integer) :>positive_integer;>var> numit : positive_integer;> temp: positive_integer;>begin> numit := 0;> temp := inp;> while temp <> 1 do> begin> if(temp mod 2 = 0) then temp := temp div 2;> else temp := 3 * temp + 1;> numit = numit + 1> end> return numit;>end;>The conjecture is that this program will always halt and>give a valid result, for positive integers -- but AFAIK>it has yet to be proven, and may never be provable in>light of a result by Conway in 1972 -- an interesting if>irritating result; darn those Goedel-like formalisms! :-)http://mathworld.wolfram.com/CollatzProblem.html>(The assumption is that 'positive_integer' can be arbitrarily>large,> BTW.)>[.sigsnip]>Sci.math added for hopefully obvious reasons; followups reset.>Don't know why you culled sci.logic. Well that's not really a> function,>its a simulation. Any simulation could go indefinately. Would an> engineer>have any practical objections to use this :>Why do you say it's not a function? It's a function that may or may> not >always halt, but it is a function. Put another way, it's a function>that may or may not have domain = N. Also, why bring engineers into>a >pure math problem? >function CollatzIterations(inp, t: positive_integer) :> positive_integer;>var> numit : positive_integer;> temp: positive_integer;>begin> numit := 0;> temp := inp;> while (temp <> 1) & (numit < t) do> begin> if(temp mod 2 = 0) then temp := temp div 2;> else temp := 3 * temp + 1;> numit = numit + 1> end> return temp;>end;>it looks like it will halt anyway, on average 1 in 4 times it will> divide by >4,>1 in 4 times it will divide by 2, 1 in 2 times it will *3.>n/4/2*3*3 = 9/8. OK thats increasing.>n/8/4/2/2*3*3*3*3 = 81/128>A few iterations deep and it should drop like a weight.>Appearances can be deceiving, and it fails to answer the question.> Does >CollatzIterations(50912309841029384893201) return a value? If so,> what? > Limiting it to a few million iterations does nothing towards> answering >the question of interest.> It is impossible for CollatzIterations to return > a value, since *return* is a system call, > not a math call. *No program* will return a value> for *CollatzIterations(50912309841029384893201)*> Mine does.> But your program is not a program.> It is obviously the 80GByte boot hard disk> on your virus-infected system, o' bright one.> Sounds like someone's jealous of the fact that I've got bigger integers.> Read 'em and weep, short-integer boy:> 2^177149 - 1> 1,531,812 iterations of (n/2) and 854,697 iterations of (3*n+1).> The reason we invented 16 bits processors is to > separate the engineers from the wankers.> If you can't do everything that can be done> in 16 bits, you are quite obviously a > mathematician rather than engineer.How do you manage to get everything ass-backwards?Big Arithmetic is for engineers, not mathematicians.Hell, a real mathematician doesn't even use numbers at all.Mathematicians have given up on the Collatz Conjecture.If it gets solved, it will be by an engineer.Using Big Arithmetic.MensanatorAce of Clubs === Subject: Re: functions that halt===>Subject: Re: functions that halt === >Subject: Re: functions that halt>The Ghost In The Machine functions>that are constructed upwards and are guaranteed to halt.>How powerful would they be?>There was no formal answer but the consensus was they would be> trivial>functions.>Is this necessarily true?>The halting proof started out as a system to check for errors> in> programs.>We assume that 'programs' either halt or they don't (good> assumption given>our practices).>general function halts or not.>But what if we assume all programs halt? This doesn't>contradict the halting proof, merely denies its construction>since it depends on an infinite loop being programmed.>It is trivial to construct a program that does not halt. There> are>plenty of times when a loop is constructed for which it is far> from>obvious if it will halt. The loop may even be infinite for some> input>and finite for other input.>What if we assume all programs halt? We can then prove> anything,>since it is not true.What function can your program with an infinite loop calculate> that> a>special class of programs without infinite loops not calculate?>One example is to return the number of iterations it takes>for the following program to halt. The following is implemented>from a conjecture by L. Collatz, first proposed in 1937.>The notation is Pascal-like and should be fairly obvious.>function CollatzIterations(inp: positive_integer) :> positive_integer;var> numit : positive_integer;> temp: positive_integer;begin> numit := 0;> temp := inp;> while temp <> 1 do> begin> if(temp mod 2 = 0) then temp := temp div 2;> else temp := 3 * temp + 1;> numit = numit + 1> end> return numit;>end;The conjecture is that this program will always halt andgive a valid result, for positive integers -- but AFAIK>it has yet to be proven, and may never be provable in>light of a result by Conway in 1972 -- an interesting if>irritating result; darn those Goedel-like formalisms! :-)http://mathworld.wolfram.com/CollatzProblem.html>(The assumption is that 'positive_integer' can be arbitrarily> large,> BTW.)>[.sigsnip]>Sci.math added for hopefully obvious reasons; followups reset.>Don't know why you culled sci.logic. Well that's not really a> function,>its a simulation. Any simulation could go indefinately. Would an> engineer>have any practical objections to use this :>Why do you say it's not a function? It's a function that may or may> not >always halt, but it is a function. Put another way, it's a function>that may or may not have domain = N. Also, why bring engineers into> a >pure math problem?>function CollatzIterations(inp, t: positive_integer) :> positive_integer;>var> numit : positive_integer;> temp: positive_integer;>begin> numit := 0;> temp := inp;> while (temp <> 1) & (numit < t) do> begin> if(temp mod 2 = 0) then temp := temp div 2;> else temp := 3 * temp + 1;> numit = numit + 1> end> return temp;>end;>it looks like it will halt anyway, on average 1 in 4 times it will> divide by >4,>1 in 4 times it will divide by 2, 1 in 2 times it will *3.>n/4/2*3*3 = 9/8. OK thats increasing.>n/8/4/2/2*3*3*3*3 = 81/128>A few iterations deep and it should drop like a weight.>Appearances can be deceiving, and it fails to answer the question.> Does >CollatzIterations(50912309841029384893201) return a value? If so,> what? > Limiting it to a few million iterations does nothing towards> answering >the question of interest.> It is impossible for CollatzIterations to return > a value, since *return* is a system call, > not a math call. *No program* will return a value> for *CollatzIterations(50912309841029384893201)*> Mine does.> But your program is not a program.> It is obviously the 80GByte boot hard disk> on your virus-infected system, o' bright one.> Sounds like someone's jealous of the fact that I've got bigger integers.> Read 'em and weep, short-integer boy:> 2^177149 - 1> 1,531,812 iterations of (n/2) and 854,697 iterations of (3*n+1).> The reason we invented 16 bits processors is to > separate the engineers from the wankers.> If you can't do everything that can be done> in 16 bits, you are quite obviously a > mathematician rather than engineer.> How do you manage to get everything ass-backwards?> Big Arithmetic is for engineers, not mathematicians.> Hell, a real mathematician doesn't even use numbers at all.> Mathematicians have given up on the Collatz Conjecture.> If it gets solved, it will be by an engineer. We know mathematicans wouldn't use numbers. They would fabricate something like stupid infinity^2 to solve the problem. Physicists would obviously simply claim their usual nonsense that pi is the unique random number. Psychologists would claim their usual drivel that computers are UNCONSCIOUS.> Using Big Arithmetic. Big arithemtic or little arithmetic, engineers don't care. Since as we remind moron scientoids daily, we are obviously NOT in the arithmetic buisness. === Subject: Re: PROOF that numbers are countableIn sci.logic, Will Twentyman:> In sci.logic, Will Twentyman>I answered your questions here. Did you have additional questions?>Just this 1.>Do you understand the difference between:>for any i, there exists a j such that F(i,j) = G(j)>and>there exists a j such that for all i, F(i,j) = G(j)>?>They're quite different.>For that matter, do you understand the difference between:>Ai [F(i,i) /= G(i)]>and>Aj Ei [F(i,j) = G(j)]>The first one is Cantor's argument, the second is what you been trying>to use to disprove it.>Actually I'm using this :>Aj Ei [F(i,j)=G(j) & F(i,j-1)=G(j-1) & F(i,j-2)=G(j-2) ... F(i,1)=G(1)]>Which means whatever number you actually specify computables will match it.>What is your natural language meaning of it? If I did make an infinite list can>you objectively construct a new number? All combinations are present on the>list, your technique is flawed to allow an infinite list and then cite elements at>finite positions in the list. Infinite list by definition means incomplete, not that>you must construct something bigger.>What happens if F is defined as Ai Aj [(i F(i,j)=4) & (i>=j ->F(i,j)=3)], which means for Aj G(j) = 4?>In this case, Ai F_i has only finitely many 4's and infinitely many 3's>in its sequence, but G has infinitely many 4's and no 3's.>F itself contains an infinite number of 4s. Remove F(1) and look at>the diagonal, its 0.4..>F is a function. It does not contain 4's.> F can be modeled as a semiinfinite 2-dimensional matrix,> since its domain is N x N. In fact, there's a rather> amusing ambiguity in the FORTRAN-like notation used here;> F could be an array or a function. (Other languages use> [] for matrices and () for function calls.)> True, but for the moment we're talking about numbers being countable, > and F is pretty clearly a map from N to RAnd F is provably not bijective.>This means when F is mapped to the number line 0.4.. is covered.>Therefore the diagonal failed to demonstrate that F does not contain>that sequence, and I'm using contain in a certain way here. Atleast>with 0 substituted for the 3.>F is not a set, it is a function. .444444444444444... is not in the >image of F.>Perhaps we are not discussing the same thing?>In any case, you haven't shown that your statement has anything to do>with the diagonalization.>All unlimited length finite strings are a subset of all (finite & infinite) strings. The>computables contains the former, diagonalisition indicates the latter. The difference between>the 2 sets is where unspecifiable irrationals exist, its no mans land, a phantom.>It simply DOES NOT PROVE that combinations do not exist on countable lists.>The diagonal was supposed to prove a new sequence, it didn't! It proved a self>referential paradox on handling infinite sequences.>No, the diagonal was supposed to prove the sequence does not list every >number (computable or otherwise) by providing one NOT in the sequence. >No one cares if the sequence covers the number, as that is not the >issue. Is the number, after being computed to the jth digit, the jth >item on the sequence? The answer is always no.>Note: if you do not accept Cantor's argument, do you accept Godel's >incompleteness theorem? Please understand that they are closely related >concepts.> Not to mention proved in much the same fashion, although the diagonal> in the Godel theorem is a little harder to see. :-)> My point exactly.>No new sequence = Herc is not convinced!>You apparently *still* do not understand how diagonalization works or >what it is attempting to prove.> Or, for that matter, the difference between> [Ai][Ej][F(i,j) != G(j)]> [Ej][Ai][F(i,j) != G(j)]> (The first is guaranteed true, because of the way G()> is defined, based on F()'s definition. The second,> depending on how F() is defined, is most likely false.)>Remember what sequences look like?>0.12345>Thats computable!>0.12345..>.. can be ANYTHING! That's computable!>If it can be anything, it's not a number, and therefor not computable.> One has to be a little careful here. What does 0.12345 mean? Under> standard definitions it refers to the value 12345/100000, a> rational number. 0.12345... may not be well enough specified> to be sure, but most would agree that it's somewhere in the> interval [12345/100000, 12346/100000). One gets into some interesting> issues of computability here, for the same argument applies to pi> (which is a semi-computable number, since one can approximate it to> any desired precision even though one cannot compute it exactly):> 3.14159265...> Any finite digit sequence is by definition computable, but the trick> is to find a series of digit sequences that converge to a number. :-)> I would agree that 0.12345... is in [12345/100000, 12346/100000] (note > change to closed since it could be trailing 9's). The question is, > Which of those is it?A good point, although it can lead to minor problems. :-)> 3.14159265... is normally understood to represent pi, so it's a little > clearer. I could argue that 0.12345... is> 0.1234567890123456789012345... or 0.123456789101112131415161718192021...Without context it could be any of those. There might be a conventionthat what I repeat 3 times is endless -- e.g., 1/3 = 0.333...,1/7 = 0.142857142857142857... or 1/11 = 0.090909... . However, that'sonly a convention, and one can be led astray: e = 2.718281828... onlylooks like a rational number (the next 4 digits are 4590).It is possible if slightly pointless to define a new notation (onemight call it the ASCII-representable overbar) such as1/3 = 0.3[3], and then define formal rules of addition, subtraction,multiplication, and division thereon. (One might be better off simplyusing fractions, of course -- 1/7 = 142857/999999, for example. Givenany integer coprime to 2 and 5 there exists an integer consisting solelyof 9's which it divides.)Numbers lead an interesting life... :-)#191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: PROOF that numbers are countableIn sci.logic, |-|erc:> oo> ____|mn> / /_/ / _> / K-9/ /_/ - www.YeOldeCoffeeShoppe.com -> /____/_____> --------------> Am I correct in assuming that you are stating that the set> {0,> 0.1, 0.2, 0.3, 0.4, ..., 0.9,> 0.01, 0.02, ... 0.09, 0.11, 0.12, ..., 0.19, ... 0.91, 0.92, ... 0.99,> 0.001, 0.002, ..., 0.009, 0.011, 0.012, ... 0.019, ... 0.999,> 0.0001, 0.0002, ..., 0.0009, 0.0011, 0.0012, ... 0.0019, ... 0.9999,> ...}> contains every real number?> (The set is basically an enumeration of all finite fractional digit> sequences, with no duplicates -- e.g., I don't include 0.10 since> 0.1 is already in the set. This set is provably dense.)> Yes, but not as elements of that set.Correct.> This set will score the entire> real number line but not be able to index it.Also correct if I understand your usage of index correctly. (I mightsubstitute the term define, approach, or lim sup here -- the sup ofa set is the smallest real which equals or exceeds all elements of that set.For example, 1/3 is the lim sup of the infinite set {0.3, 0.33, 0.333, ...}.)> To index 1/3, we would take an element of the power set containing> a member from each row (sorry!).I'm assuming you're referring to the element {0.3, 0.33, 0.333, ... }.> pi/10 = = {0.3, 0.31, 0.314,...}> As long as the power set remains 'uncountable' you should have> no real objection to this.I have no objection at present, but your direction is clear as mud.> Herc#191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: PROOF that numbers are countable> RIGHT!> So you can see why I say rationals COVERS the number line. I don't need> to define Cover look up a dictionary if all the terms are deliberately miss mashed.Yes, you mean as I suspected that the rationals are DENSE in thereals, not that they cover the reals.> By map I mean draw each point/element/number from the set onto the number line.> Number line is obviously the real number line.> To you I need the *set* itself to mark1/3, to me I just have to map the *infinite* number of> points and I'll get there. Much like 0.33.. recurring infinitely reaches 1/3.Ah not entirely true, I said you either need a set, or a collection ofpointers, my collection of pointers is your map. I still don't seewhat this has to do with the constructable number, and essentially Ifail to see the point of this line of argumentation. What exactly areyou trying to prove here? > There is certainly no point on the number line between the infinite set of marks made> by S and 1/3!! As #marks ->oo, Smarks -> 1/3. And here we allow the> theoretical construct of infinite marks.You've now introduced two terms you haven't used before, #marks ->ooans Smarks -> 1/3 as well as the idea of infinite marks. I don't seehow you're going from the power set to these objects. You must moreclearly define what you mean here. So far all it looks to me thatyou've proven nothing, except that that subset of the rational numbersis countable. === Subject: Re: PROOF that numbers are countable> RIGHT!> So you can see why I say rationals COVERS the number line. I don't need> to define Cover look up a dictionary if all the terms are deliberately miss mashed.> Yes, you mean as I suspected that the rationals are DENSE in the> reals, not that they cover the reals.> By map I mean draw each point/element/number from the set onto the number line.> Number line is obviously the real number line.> To you I need the *set* itself to mark1/3, to me I just have to map the *infinite* number of> points and I'll get there. Much like 0.33.. recurring infinitely reaches 1/3.> Ah not entirely true, I said you either need a set, or a collection of> pointers, my collection of pointers is your map. I still don't see> what this has to do with the constructable number, and essentially I> fail to see the point of this line of argumentation. What exactly are> you trying to prove here?> There is certainly no point on the number line between the infinite set of marks made> by S and 1/3!! As #marks ->oo, Smarks -> 1/3. And here we allow the> theoretical construct of infinite marks.> You've now introduced two terms you haven't used before, #marks ->oo> ans Smarks -> 1/3 as well as the idea of infinite marks. I don't see> how you're going from the power set to these objects. You must more> clearly define what you mean here. So far all it looks to me that> you've proven nothing, except that that subset of the rational numbers> is countable.I'm going from these infinite objects, to the powerset, to the real.The number made from 1/2 + 1/4 + 1/8 +... = 1.Can you describe a process using this sequence that marks the position 1 on the number line?Herc === Subject: Re: PROOF that numbers are countable> I'm going from these infinite objects, to the powerset, to the real.> The number made from 1/2 + 1/4 + 1/8 +... = 1.> Can you describe a process using this sequence that marks the position 1 on the number line?> HercWhat does this have to do with the computable numbers then? What areyou proving here? All I see is that you're defining elements of thepowerset of your herc set as real numbers. What are you attemptingto prove now? Are you trying to say that this powerset is computable?What's the point Herc? Are you suggesting that there exists an elementof this power set that can represent each number on the real line? Ifso you've just proven Cantor's argument in a longwinded way. I seeyour definitions, I just don't see what you're concluding. === Subject: Re: PROOF that numbers are countableIn sci.logic, |-|erc:> RIGHT!> So you can see why I say rationals COVERS the number line. I don't need> to define Cover look up a dictionary if all the terms are deliberately miss mashed.> Yes, you mean as I suspected that the rationals are DENSE in the> reals, not that they cover the reals.> By map I mean draw each point/element/number from the set onto the number line.> Number line is obviously the real number line.> To you I need the *set* itself to mark1/3, to me I just have to map the *infinite* number of> points and I'll get there. Much like 0.33.. recurring infinitely reaches 1/3.> Ah not entirely true, I said you either need a set, or a collection of> pointers, my collection of pointers is your map. I still don't see> what this has to do with the constructable number, and essentially I> fail to see the point of this line of argumentation. What exactly are> you trying to prove here?> There is certainly no point on the number line between the infinite set of marks made> by S and 1/3!! As #marks ->oo, Smarks -> 1/3. And here we allow the> theoretical construct of infinite marks.> You've now introduced two terms you haven't used before, #marks ->oo> ans Smarks -> 1/3 as well as the idea of infinite marks. I don't see> how you're going from the power set to these objects. You must more> clearly define what you mean here. So far all it looks to me that> you've proven nothing, except that that subset of the rational numbers> is countable.> I'm going from these infinite objects, to the powerset, to the real.> The number made from 1/2 + 1/4 + 1/8 +... = 1.> Can you describe a process using this sequence that marks the> position 1 on the number line?Yes:S = {1/2, 3/4, 7/8, 15/16, ... }will have the lim sup of 1. If one prefers, one can write S as:S = {1/2, 1/2 + 1/4, 1/2 + 1/4 + 1/8, 1/2 + 1/4 + 1/8 + 1/16, ...}.> Herc#191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: PROOF that numbers are countable oo ____|mn / /_/ / _ / K-9/ /_/ - www.YeOldeCoffeeShoppe.com -/____/_____--------------> RIGHT!> So you can see why I say rationals COVERS the number line. I don't need> to define Cover look up a dictionary if all the terms are deliberately miss mashed.> Yes, you mean as I suspected that the rationals are DENSE in the> reals, not that they cover the reals.> By map I mean draw each point/element/number from the set onto the number line.> Number line is obviously the real number line.> To you I need the *set* itself to mark1/3, to me I just have to map the *infinite* number of> points and I'll get there. Much like 0.33.. recurring infinitely reaches 1/3.> Ah not entirely true, I said you either need a set, or a collection of> pointers, my collection of pointers is your map. I still don't see> what this has to do with the constructable number, and essentially I> fail to see the point of this line of argumentation. What exactly are> you trying to prove here?> There is certainly no point on the number line between the infinite set of marks made> by S and 1/3!! As #marks ->oo, Smarks -> 1/3. And here we allow the> theoretical construct of infinite marks.> You've now introduced two terms you haven't used before, #marks ->oo> ans Smarks -> 1/3 as well as the idea of infinite marks. I don't see> how you're going from the power set to these objects. You must more> clearly define what you mean here. So far all it looks to me that> you've proven nothing, except that that subset of the rational numbers> is countable.> I'm going from these infinite objects, to the powerset, to the real.> The number made from 1/2 + 1/4 + 1/8 +... = 1.> Can you describe a process using this sequence that marks the> position 1 on the number line?> Yes:> S = {1/2, 3/4, 7/8, 15/16, ... }> will have the lim sup of 1. If one prefers, one can write S as:> S = {1/2,> 1/2 + 1/4,> 1/2 + 1/4 + 1/8,> 1/2 + 1/4 + 1/8 + 1/16,> ...}.So if you draw a point at each of these numbers.1/23/47/8..you would draw the number 1?number 1 has been referenced via the infinite set of points!So this set{0, 0.1, 0.2, 0.3, 0.4, ..., 0.9, 0.01, 0.02, ... 0.09, 0.11, 0.12, ..., 0.19, ... 0.91, 0.92, ... 0.99, 0.001, 0.002, ..., 0.009, 0.011, 0.012, ... 0.019, ... 0.999, 0.0001, 0.0002, ..., 0.0009, 0.0011, 0.0012, ... 0.0019, ... 0.9999, ...}would completely cover the number line, color it in, no gaps!Herc === Subject: Re: PROOF that numbers are countable> So this set> {0,> 0.1, 0.2, 0.3, 0.4, ..., 0.9,> 0.01, 0.02, ... 0.09, 0.11, 0.12, ..., 0.19, ... 0.91, 0.92, ... 0.99,> 0.001, 0.002, ..., 0.009, 0.011, 0.012, ... 0.019, ... 0.999,> 0.0001, 0.0002, ..., 0.0009, 0.0011, 0.0012, ... 0.0019, ... 0.9999,> ...}> would completely cover the number line, color it in, no gaps!> HercYou mean color in the points, or color in the INTERVALS betweenpoints? Once again you're stating the property of DENSITY of therationals as a subset of the reals, not that these numbers are coloredin. Take a square who's sides are length one, and construct it on yournumber line, then take it's diagonal, obviously this line has alength, and should be on your number line, But the square root of twois NOT covered by the set you've defined. No matter how close you get,you never reach the square root of two. Are you trying to argue thatthe real number line doesn't contain any irrational numbers? Unlessyou had some special meaning for the word color it in then yourargument is wrong. === ===Subject: New book announcementNew book announcementZubov I.V. Methods of control systems dynamics analysis. Moscow,with the support of the Russian fund of fundamental research,Project # 03-01-14025The book is published in Russian.The author is from St. Petersburg State Univ. Russia, faculty ofApplied Math and Control Processes http://www.apmath.spbu.ruPublisher's website http://www.fml.ruThe book is devoted to the development of mathematical instrumentof control systems dynamics research. The methods based onanalysis of sufficiently nonlinear models with the aim to discoverexistence and features of limit sets. The strict mathematicalproof of the existing of global attracting set in practicallysignificant models with several unstable fixed points is given.Thus Smale's 14th problem is solved without any computer modeling.Numerical algorithms for the most exact system dynamics predictingare offered. The book will be useful for control systems designspecialists as far as for students and postgraduate students inthe field of applied math. and computer sciences.The ContentsIntroductionChap.1 The structural features of control systems1. Stability and optimality of control systems2. The state of chaos3. Topological methods4. Dissipativity and auto-oscillationsChap. 2 Analysis of control systems features5. The first integral and Jacobi's last multiplier6. Jacobi's last multiplier and integral invariant7. Equations for the last multiplier8. Finding integral. Exceptional integral.Chap.3 Simple structure systems9. 3-d quadratic systems10. Equation for a regular integral11. Sequential localization of a invariant set12. Building a system with the strange attractorChap.4 Systems with the invariant manifolds13. Systems dynamics predicting14. Numerical integration15. Stability of integral manifolds16, Stability of the conservative systems with cyclic coordinatesChap. 5 Distributed parameter systems research17. Compatible partial differential systems18. Finding coefficients in Poison's brackets19. Automodelling solutions20. Stabilization of the distributed parameter systemsChap, 6 Analysis of the dynamics of the systems with unboundedsolutions21. Unbounded solutions research. Stability of unbounded solutions.22. Stability of the invariant sets of dynamical periodic systems.Chap.7 Lyapunov's functions in computational practice23. The uniform system of numerical algorithms24. Fast matrix inversion and linear systems solution method25. Solution of nonlinear equation and optimization problems26. Polar matrix decomposition27. Large system numerical analysisThe bibliography(160 items) === Subject: Map distribution from plane to sphereHi all,I have a question regarding the tranformation of a probability densitydistribution (pdf) taken at a plane to the surface of a sphere.An exampel : A distribution is uniform (isotropically) distributed on a x,y-planewith marginal pdf'sp(x) = 1; p(y) = 1;On a sphere, the marginal pdf for the isotropic distribiution isp(theta) = 1/2 sin(theta) according tomathworld.wolfram.com/SpherePointPicking.htmlMy question: For a pdf on the plane that has a Gaussian as marginal pdf in xdirection and a uniform distribution in y direction, how do I map sucha distributon onto the sphere ?I know that the Fisher distribution on the sphere is the analogue tothe Gaussian in the plane, but I have no idea if I can perfoirm atransformation and how.All hints are very welcome,Dirk === Subject: Re: Map distribution from plane to sphereThats why I repeated it.Dirk> Hi all,> I have a question regarding the tranformation of a probability density> distribution (pdf) taken at a plane to the surface of a sphere.> An exampel : > A distribution is uniform (isotropically) distributed on a x,y-plane> with marginal pdf's> p(x) = 1; p(y) = 1;> On a sphere, the marginal pdf for the isotropic distribiution is> p(theta) = 1/2 sin(theta) according to> mathworld.wolfram.com/SpherePointPicking.html> My question: > For a pdf on the plane that has a Gaussian as marginal pdf in x> direction and a uniform distribution in y direction, how do I map such> a distributon onto the sphere ?> I know that the Fisher distribution on the sphere is the analogue to> the Gaussian in the plane, but I have no idea if I can perfoirm a> transformation and how.> All hints are very welcome,> Dirk === Subject: Re: More about binary trees> I am reading book on Intro to algorithms by CLR, 2nd edition.> Chapter 6 describes heap sort and heap structure used in this sort which> is binary tree constructed in such a way that parent is always greater> than its children and relation between children does not matter as longas> they are smaller than their parent.> The authors also say that each child's subtree has size of at most 2n/3> elements , I can't see how it could have this size. Maximum size when the> tree is full and each child subtree would have (n-1)/2 elements.Perhaps,> I can't> see something here. It's important to understand this because this> information is used to set up recurrence relation which I use analyse> the algorithm.Remember, the tree that you are considering has the property that the pathlength from the root to some leaf differs by at most one from the pathlength from the root to any other leaf.For example, the 11 numbers 10 20 29 30 35 45 50 60 70 72 80 might bearranged in such a tree as follows (read this with a fixed-width font): 80 / / / / / / / 60 72 / / / / / / 45 50 30 70 / / 29 10 35 20Note that the left subtree for 80 has 7 elements while the right subtree hasonly 3 elements. This most unbalanced behavior happens when one subtreehas 2^(k+1)-1 elements and the other has 2^k-1 elements, for somenon-negative integer k. Let m=2^k.If n is the total number of elements in such a tree, thenn=2^(k+1)-1+2^k-1+1=2m-1+m-1+1=3m-1.So m=(n+1)/3 and the size of the largest subtree is2^(k+1)-1=2m-1=2((n+1)/3)-1=(2n-1)/3. So the largest subtree will alwayshave less than 2n/3 elements.http://www.clivetooth.dk === Subject: Re: Lebesgue Integration curious question> so this definition would say that the integral> of f should be the inf of the empty set, which is +infinity (note the> empty set has the curious property that its inf is larger than its> sup).I'll believe you, but I just can't see it. Is this because everystatement of the form Ax(Fx -> Ey(y so this definition would say that the integral> of f should be the inf of the empty set, which is +infinity (note the> empty set has the curious property that its inf is larger than its> sup).>I'll believe you, but I just can't see it. Is this because every>statement of the form Ax(Fx -> Ey(yinterpreted to mean x is an element of the empty set?Well yes, it's because any number is a lower bound for theempty set, so the greatest lower bound is +infinity.You got the formalism wrong, though: If Fx means thatx is in the empty set then y is a lower bound for theempty set is (Fx -> y <=x).Another way to say why the inf of the empty set _should_be +infinity, without anything about vacuous truths:If A and B are nonempty sets and A is a subset of Bthen inf(A) >= inf(B); if you want that fact to be true whenA is empty you need to define inf(A) = infinity.>'cid 'ooh === Subject: Re: Lebesgue Integration curious question> so this definition would say that the integral> of f should be the inf of the empty set, which is +infinity (note the> empty set has the curious property that its inf is larger than its> sup).>I'll believe you, but I just can't see it. Is this because every>statement of the form Ax(Fx -> Ey(yinterpreted to mean x is an element of the empty set?>Well yes, it's because any number is a lower bound for the>empty set, so the greatest lower bound is +infinity.>You got the formalism wrong, though: If Fx means that>x is in the empty set then y is a lower bound for the>empty set is (Fx -> y <=x).Typo - I meant Ax(Fx -> y <=x). (What I was trying tosay was that the Ey you included defintely doesn'tbelong there.)>Another way to say why the inf of the empty set _should_>be +infinity, without anything about vacuous truths:>If A and B are nonempty sets and A is a subset of B>then inf(A) >= inf(B); if you want that fact to be true when>A is empty you need to define inf(A) = infinity.>'cid 'ooh>************************> === Subject: Question on Functional EquationIt is known that the function f(x) satisfying the functional equation,f(x+y) = F(f(x), f(y))follows certain ordinary differential equation when the functional relation F satisfies some conditions (E. Picard.) Does anyoneknow the recent progress on this matter? I tried to deal with specific F in the following site,http://139.134.5.123/tiddler2/cauchy/ cauchyequation.htmUsing that specific functional relation, we can obtain the functional relation explicitly on nonlinear ordinary differentialequation such as that for elliptic function(!?) === Subject: Does differentiability imply continuity of partial derivatives?Hi thereGiven some function z=f(x,y). Does the fact that f is differentiableat (a,b) imply that fx and fy (partial derivatives) are continuous at(a,b)?Seems plausible since given a linearization (plane) of f at (a,b) -moving in fromthe x direction (y constant = b) to a, we can't have a break in f, sofx must be continuous. Likewise for fy. Am I on the right track here?I'm asking because that will give me what I'm really looking for: thatgiven a function where fx and fy are NOT continuous at (a,b) YET aredefined at (a,b) [piece-wise defined function e.g.] implies that f isnot differentiable at (a,b).Or is there a nice counter-example to this?RC === Subject: Re: Does differentiability imply continuity of partial derivatives?> Hi there> Given some function z=f(x,y). Does the fact that f is differentiable> at (a,b) imply that fx and fy (partial derivatives) are continuous at> (a,b)?Not even for functions of one variable does this hold. One counterexample isf(x) = x^2 * sin(1/x). This is differentiable everywhere, even at x=0, butthe derivative does not have a limit as x->0 and hence is not continuous atx=0.-Michael. === Subject: Re: Does differentiability imply continuity of partial derivatives?> Given some function z=f(x,y). Does the fact that f is differentiable> at (a,b) imply that fx and fy (partial derivatives) are continuous at> (a,b)?No.> Seems plausible since given a linearization (plane) of f at (a,b) -> moving in from> the x direction (y constant = b) to a, we can't have a break in f, so> fx must be continuous. Likewise for fy. Am I on the right track here?> I'm asking because that will give me what I'm really looking for: that> given a function where fx and fy are NOT continuous at (a,b) YET are> defined at (a,b) [piece-wise defined function e.g.] implies that f is> not differentiable at (a,b).> Or is there a nice counter-example to this?Take, for instance, f(x,y) = x^2 sin(1/x) if x is different from 0and f(0,y) = 0. The function f is differentiable at (0,0), butthe partial derivatives aren't continuous at (0,0).Jose Carlos Santos === Subject: Re: Does differentiability imply continuity of partial derivatives?> Given some function z=f(x,y). Does the fact that f is differentiable> at (a,b) imply that fx and fy (partial derivatives) are continuous at> (a,b)?no === Subject: Re: Very simple question.> 60 is 40% of what number? And how do I go about answering it.60 40-- = --- x 10040 * x = 100 * 6040x = 6000--- ---- 40 40x = 15060 is 40 percent of 150. === Subject: Basic question please explain wont take longHi I am 17 years of age and a first year student of higher maths pastrequired national level this question is rattling my brains any helpgiven, will be greatly appreciated thank you.The sum of Sn of the first n terms of an arithmetic progression isgiven by Sn=3n+2n^2.i)Write down the first and second terms of the progression and find aformula for the Kth term === Subject: Re: Basic question please explain wont take long> Hi I am 17 years of age and a first year student of higher maths past> required national level this question is rattling my brains any help> given, will be greatly appreciated thank you.> The sum of Sn of the first n terms of an arithmetic progression is> given by Sn=3n+2n^2.> i)Write down the first and second terms of the progression and find a> formula for the Kth termYou Have S_nso substitute n=1 === 3.(1) + 2.(1^2) = 5 =S_1 == p_1so substitute n=2 === 3.(2) + 2.(2^2) = 14=S_2so p_1 = 5p_2 = S_2-S_1 = 14 - 5 = 9also p_k = S_k -S_(k-1)so substituteWhy?S_(n+1)= S_n + p_(n+1) CarlThe world is flat it's pi that's round!There is only one number.about me htttp://cparkes.actewagl.net.au === Subject: Re: Basic question please explain wont take long>The sum of Sn of the first n terms of an arithmetic progression is>given by Sn=3n+2n^2.>i)Write down the first and second terms of the progression and find a>formula for the Kth termsum of one term: 3*1 + 2*1^2 = 5sum of two terms: 3*2 + 2*2^2 = 14What is the first term? What is the second term? How is one termrelated to the next in an arithmetic progression? === Subject: Re: Complex Residues>I need to calculate the residue of f(z)=e^(e^(1/z)) at 0.>When I try solving this on paper I expand e^(1/z) to ** = ( 1 + 1/z +>1/(2*z^2) + 1/(6*z^3)...>And then I expand e^(e^(1/z)) to 1 + ** + (**)^2/2 + (**)^3/3! ....>And it's easy to see that c_-1 = 1>But on the other hand when I put the question to Mathematica (the>computer program) it says that the residue is e.>What am I doing wrong?I believe you are on the right track, but you have forgotten to take outthe constant before exponentiating the second time: 1/z 1 1 1 e = 1 + ( - + ----- + ----- + ... ) z 2 z^2 6 z^3 = 1 + f(z) 1+f(z) f(z) e = e e 1 2 = e ( 1 + f(z) + - f(z) + ... ) 2Note that f(z)^2 consists only of terms 1/z^2 and higher, so the only1/z contribution comes from f(z). Thus, the coefficient of 1/z is e. === Subject: Re: Infinitesimal generators of the 6x6 symmetric matrices Lie algebra> Hi to everybody,> I'm a physicist and I need all the 21 6x6 matrices which form a basis> for the Lie Algebra of the 6x6 real symmetric matrices,> or, equivalently, the 21 infinitesimal generators of the Sp(6,R)> symplectic group in the representation of 6x6 matrices (they are> related to the symmetric matrices through matrix multiplication by the> symplectic form).> Please can you tell me a web link (or a book) where I can find them?> Or kindly post me them directly?It's fairly easy to construct explicitly a basis for the real Liealgebra sp(6,R) of the the Lie group Sp(6,R).Let Y be an element of Sp(6,R) and J be the symplectic form, soJ = Y^t J Y.Take a 1-parameter subgroup {Y(t)} of Sp(6,R) that has Y(0) = 1,differentiate the above equation, and evaluate at t = 0 to find acorresponding equation for the elements of the Lie algebra sp(6,R):0 = X^t J + J X, (1)where X = dY/dt evaluated at t = 0.Write everything in terms of 3x3 blocks: _ _ _ _ | A B | | 0 1 | X = | |, J = | |. |_C D_| |_-1 0_|Substituting these into (1) gives _ _ _ _ | -C^t A^t | | C D | 0 = | | + | |, |_-D^t B^t_| |_ -A -B_|so B^t = B, C^t = C, and D^t = -A.B a 3x3 symmetric matrix gives 6 basis elements; C a 3x3 symmetricmatrix gives 6 basis elements; A an arbitrary 3x3 gives nine basiselements. This accounts for the 21 element basis of the Lie algebrasp(6,R). Just put ones and zeros in the appropriate places.George === Subject: Re: Infinitesimal generators of the 6x6 symmetric matrices Lie algebraI would like to see this work of yours as commented on in a cross section ofthe photon.. and in particular, development of the theory for a Woodgertesseract study using the new Optics.orgphotonic fiber.Doug GillmanAAAS, MADD, ACS, IEEE, IEE, TPAS, FSEEEUnion of Concerned ScientistsDream on nacelles' wind, Caught in tepid Tome, an urge, Gin in from the cold. === Subject: Re: continuous function on a bounded set === Subject: Re: continuous function on a bounded set > Every compact, connected, locally connected metric space X > has a continuous, onto function g:[0,1]->X. For [0,1]^2, there > are well-known examples (Peano curves). Mazurkiewicz and Hahn > have credit for the general result. >Hm. If compact metric, then totally bounded. >Does connected, locally connected imply path connected? >Thus locally path connected. >Need compact metric also.Indeed. Also implies locally path connected.Assume S connected, locally connected. Pick on some a,bCover S with open connected base sets contained in balls of radius 1.There's an over lapping chain V0,V1,..V_(k-1) of those sets from a to bAssign 0 to a, j/k to a point in V_(j-1) / Vj and 1 to bNow as each Vj is open, its locally connected. Also it's connected.So do the same stunt from the point assign with j/k to the point assignwith (j+1)/k with a finer scale and so forth on ad infinitum nasium.Where have I used compact? It's needed for completeness.Next show the assignment preserves Cauchy sequences and assign to alimit of a Cauchy sequence, the limit of the assigned Cauchy sequence.Whence voile, have I constructed continuous map from a grainy map? >Then make series of paths p_n thru each of the finite number of > balls of radius 1/n. >Show p_n uniformily converges to a super path p that's dense >in the space. >As the path p is compact, hence closed and also dense, >it can meet everybody in the compact district. >That's close to the idea of the proof, yes.Hurrah. Any other details? >That's quite a trip, how flats and how many times did I run out of > gas? Better I assume compact metric path-connected. >Will I get to the finish line then? >You don't have to assume path connected-ness, you prove it!Now to establish convergence. >Amazing isn't it?Indeed, and it's amazing this time I've even a notion how it's done.Let S be connected, locally path connected, compact metric space.S path connected since S connected, locally path connected.Some B(x1,1),.. B(x_n1,1) cover S.Let p1 be path x1,.. x_n1Each cl B(xj,1) is covered by B(xj1,1/2),...B(xj_n2j,1/2).Let p2 be path x1, x11,.. x1_n21, x1, x2, x21,.. x2_n22, x2, ... x_n1, x_n11,.. x_n1_n2.n1Continuing so indefinitely unto utmost notational nightmare.Now pointwise p_n(x) is Cauchy sequence, hence converges.Now to get p_j's to uniformly converge.Make p1 and p2 have same arguement when the get to x1,x2,.. x_n1Make adjustment to the chosen balls to be local path connected balls.Require the path from xj, xj1, xj2,.. xj_n2j, xj to be within the path connected cl B(xj,1)Oh oh. Does path connected A ==> path connected cl A ?No. However in locally path connected space it does.Ok, stage has been set for a continuous grand path p = lim(n->oo) p_n.For a dense set D, choose all the centers of the chosen balls.Now let B(x,r) be any ball. Out of the previous selection of balls,find a set with radius < r/2. x will be in one of those balls and thatball will be contained within B(x,r), thus providing a point in D alsoin B(x,r). Ok, denseness established.Thus as before, continuous p:[0,1] -> S, with dense p([0,1]) which is alsocompact, hence closed to conclude p([0,1]) = cl p([0,1]) = S, surjection.Yes, I'm amazed Hahn & Mazurkiewicz could get it to work.I'm still beset with the details, wherein doesth not the devil reside?Is it a corollary that path-connected Hausdorff S ==> S arc-connectedor is metric or compact metric require?---- === Subject: Re: continuous function on a bounded set> Is it true that if f:[a,b]-->R is continuous on [a,b] and that the> preimage f-1(y) of every y in the range of f consists of infinitely> many elements, then f must be constant on [a,b]?> Clearly not. Consider the arctan function on (-pi, pi).> Your function is an injective function defined on an open interval.> What does it have to do with the question?> Jose Carlos SantosI think this comes down to differing conventions. (And mine may infact be wrong).The function is clearly continuous on the interval, and it's pre-imageis periodic (so arctan(x) = arctan(x + 2pi). So, as I read thequestion, the function satisfies the hypothesis.Perhaps I'm confusing the issue by thinking about branch cuts,however.'cid 'ooh === Subject: Re: continuous function on a bounded set>Is it true that if f:[a,b]-->R is continuous on [a,b] and that the>preimage f-1(y) of every y in the range of f consists of infinitely>many elements, then f must be constant on [a,b]?>Clearly not. Consider the arctan function on (-pi, pi).>Your function is an injective function defined on an open interval.>What does it have to do with the question?> The function is clearly continuous on the interval, and it's pre-image> is periodic (so arctan(x) = arctan(x + 2pi). So, as I read the> question, the function satisfies the hypothesis.Jose Carlos Santos === Subject: Re: continuous function on a bounded set> Is it true that if f:[a,b]-->R is continuous on [a,b] and that the> preimage f-1(y) of every y in the range of f consists of infinitely> many elements, then f must be constant on [a,b]?> Clearly not. Consider the arctan function on (-pi, pi).> Your function is an injective function defined on an open interval.> What does it have to do with the question?> Jose Carlos Santos>I think this comes down to differing conventions. (And mine may in>fact be wrong).>The function is clearly continuous on the interval, But (-pi,pi) is not [a,b].>and it's pre-image>is periodic (so arctan(x) = arctan(x + 2pi). Huh? Maybe you're confusing arctan and tan.If f(x) = arctan(x) _or_ tan(x) on (-pi,pi) then f^(-1}(y)is _finite_ for every y. Hence people's confusion onwhat this has to do with the question.Now, if you let f(x) = tan(x), defined on the set of allpoints where tan is finite, then f^{-1}(y) is infinite forall y. But that has nothing at all to do with the question,because the domain of the function is not [a,b]; infact the domain is the union of infinitely many disjointopen intervals, making the existence of such a functionobvious (and totally irrelevant to the question.)> So, as I read the>question, the function satisfies the hypothesis.>Perhaps I'm confusing the issue by thinking about branch cuts,>however.>'cid 'ooh === Subject: Re: continuous function on a bounded set> Is it true that if f:[a,b]-->R is continuous on [a,b] and that the> preimage f-1(y) of every y in the range of f consists of infinitely> many elements, then f must be constant on [a,b]?> Clearly not. Consider the arctan function on (-pi, pi).> Your function is an injective function defined on an open interval.> What does it have to do with the question?>(...)>The function is clearly continuous on the interval, and it's pre-image>is periodic (so arctan(x) = arctan(x + 2pi).Confusion ! Correct is: tan(x)=tan(x+pi) [no need for the factor 2,needed for sin and cos]. And you must be thinking about tan itselfas f, which is defined on all |R minus the m*pi/2 (m odd integer)where the tangent becomes infinite. But the problem is that thisset (where tan is defined) is not even an interval; and if you addthe poles x=m*pi/2 (m odd) to the set (with f(x)=oo) then the intervalwhere f is defined is all |R therefore not compact (as is [a,b]) andf has a value oo not in |R !>(...)>'cid 'ooh === Subject: Re: continuous function on a bounded set> Is it true that if f:[a,b]-->R is continuous on [a,b] and that the> preimage f-1(y) of every y in the range of f consists of infinitely> many elements, then f must be constant on [a,b]?> Clearly not. Consider the arctan function on (-pi, pi).> Your function is an injective function defined on an open interval.> What does it have to do with the question?> Jose Carlos Santos>I think this comes down to differing conventions. (And mine may in>fact be wrong).Yes it's wrong.>The function is clearly continuous on the interval, and it's pre-image>is periodic (so arctan(x) = arctan(x + 2pi).Huh??? Perhaps you meant to write tan, otherwise the equality doesnot hold at all.I might be making too much of a typo, but then again, it's hard to seewhy in your previous post you took the trouble to specify pi in the*domain* of arctan -- a rather pointless thing to do unless you wereconfused then in a similar way. Am I right, or what? ;-) So, as I read the>question, the function satisfies the hypothesis.But I suggest that you rethink whether you really wanted to proposetan or arctan as your candidate for f; and in either case, really lookcarefully at whether your f satisfies *both* parts of the hypothesis.(It won't.) In particular, note that the hypothesis says nothingabout a periodic preimage; it says that the preimage is *multivalued*,indeed infinitely so, everywhere. That's not the same thing as aperiodic function. === Subject: Re: Hille's lower bound for all primes of ~ 1.72864 (ordered factorisation)Further information added to post below.Can anyone assist?Ta.> http://mathworld.wolfram.com/OrderedFactorization.html> Hille provided a lower bound for all primes ~ 1.72864 > I'm confused here. That web page does not contain that sentence. In> fact, it contains neither the word lower nor the word bound.> G'Day Jose, sorry about the confusion. I provided the above reference> in order to uniquely specify the type of factorisation involved. I have no> idea whether the following item of information is well-known or whether> it is considered trivial, as I have not delved into combinatorial maths> theory for more than 35 years.> Essentially, I obtained the reference for Hille from a number of papers> including this one.....> Discrete Mathematics 214 (2000) p 123 - 133.> On the number of ordered factorizations of natural numbers> Benny Chor, Paul Lemke, Ziv Mador> I have now tracked the Hille reference to its source in:> Hille (Acta Arith. 2 (1)(1936) p134-144> from it.... where H(n) is the ordered factorisation of n:> Hille established a relation between H(n) and the Reimann> zeta function (Z). This relation was used by Hille to provide> tight asymptotic upper and lower bounds on H(n). In particular,> Hille showed an existential lower bound on H(n):> For any t < p = (Z)^-1 (2) ~ 1.73> there are infinitely many n which satisfy H(n)< n^p> === =====[end quote] === =================> My problem is understanding what the above means.> Can this be expressed in words?> Unfortunately I am not familiar with the zeta function.> Could easily imply that I must be, but was wondering> whether someone who could understand the nature of> these bounds could explain them simply to me without> reference to the zeta function.> Pete Brown> Falls Creek> Oz> www.mountainman.com.au === Subject: Work around of using 1D-LUT instead of 2D-LUTI originally have a look up table of size UInt8[256][256].However, there is a restriction that just 1D look up table of sizeUInt8[256] can be use. The no. of UInt8[256] tables to be used is notlimited.Are there any method for me to constructing many 1D look up tables ofUInt8[256] so that the same result can be obtained as if the 2D-LUTUInt8[256][256] is used?Note that the 1D-LUT can just have max. no. of element of 256.Jason === Subject: Re: Non-Noether Conserved Quantity?> Are x and y perhaps typos/shorthand for v_x, v_y?> right Hamiltonian function to get the answers I've quoted...!That was mainly a sanity check for myself. I-) === Subject: Re: CardinalityCorrection and new questions: 1.- I just want to know if the class of all sets having fixcardinality is a set...or if there exist a set A sucht that for each cardinality x thereexist a set a in A such that |a| = x . ... Ok.. What is a Proper class??2.- Talkin about TOTAL order sets, There existe a TOTAL order set A withan injection F:R--->A of the reals to A preserving orders sucht thatF(R) is dense in A (with the topology induced by de order) but suchthat |R| < |A| but not =.3. If the set in 2. existe... can we construct a non finit secuencesof sets s.t. |R| < |A1| < ... and always F(A1) dense in Ai+1 ? dose ithave a maximum?and what about if |R| < |A1| < ... and F(R) dense in Ai for all i?sorry about my englis> 1.- I just want to know if the class of all sets having fix> cardinality is a set...> or if there exist a set A sucht that for each cardinality x there> exist a set a in A such that |a| = x .> 2.- Talkin about well order sets, There existe a well order set A with> an injection F:R--->A of the reals to A preserving orders sucht that> F(R) is dense in A (with the topology induced by de order) but such> that |R| < |A| but not =.> 3. If the set in 2. existe... can we construct a non finit secuences> of sets s.t. |R| < |A1| < ... and always F(A1) dense in Ai+1 ? dos it> have a maximum?> sorry about my englis. === Subject: Re: Cardinality Adjunct Assistant Professor at the University of Montana.>1.- I just want to know if the class of all sets having fix>cardinality is a set...No. It is a proper class.>or if there exist a set A sucht that for each cardinality x there>exist a set a in A such that |a| = x .No. This would make the collection of all cardinals a set byreplacement, and it is known that the collection of all cardinals is aproper class.>2.- Talkin about well order sets, There existe a well order set A with>an injection F:R--->A of the reals to A preserving ordersDoes R have the usual ordering? If so, no.> sucht that>F(R) is dense in A (with the topology induced by de order) but such>that |R| < |A| but not =.If f: R->A respects ordering, then the order in A would necessarilysatisfy that { f(r): r in R, r<0} is a nonempty set; by well-orderingof A, this set would have a first element, which would give you aleast negative real number. No such thing exists. === ============Subject: Re: CardinalitySorry, In my last mesage I didn't want to say well order if nottotal order ... do you have any answer for thise...>1.- I just want to know if the class of all sets having fix>cardinality is a set...> No. It is a proper class.>or if there exist a set A sucht that for each cardinality x there>exist a set a in A such that |a| = x .> No. This would make the collection of all cardinals a set by> replacement, and it is known that the collection of all cardinals is a> proper class.>2.- Talkin about well order sets, There existe a well order set A with>an injection F:R--->A of the reals to A preserving orders> Does R have the usual ordering? If so, no.> sucht that>F(R) is dense in A (with the topology induced by de order) but such>that |R| < |A| but not =.> If f: R->A respects ordering, then the order in A would necessarily> satisfy that { f(r): r in R, r<0} is a nonempty set; by well-ordering> of A, this set would have a first element, which would give you a> least negative real number. No such thing exists.> -- === ===============> Arturo Magidin> magidin@math.berkeley.edu === Subject: Re: Cardinality>1.- I just want to know if the class of all sets having fix>cardinality is a set...> No. It is a proper class.Except when the given fixed cardinality is 0. === Subject: Re: Why do math folks tend to end their letters with cheers?> In my experience with email correpsondence with mathematicians, a> great number end their letters with the salutation cheers. In my> experience with email correspondence with non-mathematicians, I don't> think I have ever gotten a letter with a salutations cheers.Because, in base-36: million+cheers = mudbath+pommyIf you have Mathematica you can check this by entering: 36^^million + 36^^cheers == 36^^mudbath + 36^^pommyhttp://www.clivetooth.dk === Subject: Re: Why do math folks tend to end their letters with cheers?> cheers is an extremely common salutation in Britain and Australia. Maybe> you correspond with commonwealth mathematicians, or Britophile ones? I've> never noticed cheers to be more used by mathematicians, but is more used> by academic types in general, who are often Britophile or at least not> Brit-phobic. definitely a British term. As with all things, the shorter form isoften used.http://dictionary.reference.com/search?q=cheerioSo on that note...cheers === Subject: Re: Why do math folks tend to end their letters with cheers?> In my experience with email correpsondence with mathematicians, a> great number end their letters with the salutation cheers. In my> experience with email correspondence with non-mathematicians, I don't> think I have ever gotten a letter with a salutations cheers.> Why is this?This is a good example of small sample size leading to faulty conclusions.Doug === Subject: Re: Why do math folks tend to end their letters with cheers?> In my experience with email correpsondence with mathematicians, a> great number end their letters with the salutation cheers. In my> experience with email correspondence with non-mathematicians, I don't> think I have ever gotten a letter with a salutations cheers.> Why is this?>This is a good example of small sample size leading to faulty conclusions.There is even a name for that. The law of small numbers.>Doug === Subject: Re: Why do math folks tend to end their letters with cheers?> Nothing to do with mathematics: cheers is very commonly used in the> British Isles (and possible some of the countries that used to be part> of the extinct British empire) in the way you describe.> Is that right? I've always used it (well, for a suitable definition of> always), but I simply picked it up from the net... I always thought it> Oh well, I hope people haven't thought that I was being pedantic by using> an expression that doesn't fit naturally with a non-native English speaker> like me :-)> Never mind what people think; use it if you like. That's what I do>myself, and I am also not a native English speaker.But you don't have a .sig at all!BELANGERBe seeing you. === Subject: Re: Why do math folks tend to end their letters with cheers?> In my capacity as network administrator, I was helping a guy with an e-mail> problem and noticed he used the salutation chow.> Sometimes there just aren't any explanations.That sounds to me like the Italian (but internationalized)word ciao -- a mix of pronunciation rules from Englishand Spanish makes chow sound like ciao (well, veryclose, at least) :-)Carlos === Subject: Re: Why do math folks tend to end their letters with cheers?:> In my experience with email correpsondence with mathematicians, a:> great number end their letters with the salutation cheers. In my:> experience with email correspondence with non-mathematicians, I don't:> think I have ever gotten a letter with a salutations cheers.:> :> Why is this?:> :> Craig: In my capacity as network administrator, I was helping a guy with an e-mail: problem and noticed he used the salutation chow.: Sometimes there just aren't any explanations.I think the explanation for that one is that he meant 'ciao'.Stephen === Subject: Re: Why do math folks tend to end their letters with cheers? === >Subject: Re: Why do math folks tend to end their letters with cheers?>Message-id: :> In my experience with email correpsondence with mathematicians, a>:> great number end their letters with the salutation cheers. In my>:> experience with email correspondence with non-mathematicians, I don't>:> think I have ever gotten a letter with a salutations cheers.>:>:> Why is this?>:>:> Craig>: In my capacity as network administrator, I was helping a guy with an e-mail>: problem and noticed he used the salutation chow.>: Sometimes there just aren't any explanations.>I think the explanation for that one is that he meant 'ciao'.Yeah, I meant an explanation for how someone can be that dumb.>StephenMensanatorAce of Clubs === Subject: Re: Why do math folks tend to end their letters with cheers? === :>Subject: Re: Why do math folks tend to end their letters with cheers?:>I think the explanation for that one is that he meant 'ciao'.: Yeah, I meant an explanation for how someone can be that dumb.I am not sure I would call it dumb, but rather ignorant, and unawarethat all words are not spelled phonetically. My favorite sucherror is could of for the contraction could've. Someday Ihope to see somebody write I dove, for the contraction I'd've(I would have). :)Of course, maybe that Austin Powers explanation has some merit too.Stephen === Subject: Re: Why do math folks tend to end their letters with cheers?>I think the explanation for that one is that he meant 'ciao'.> Yeah, I meant an explanation for how someone can be that dumb.That was a joke that was going around for a while. Lots of folk were signing chow. I think it camefrom one of the Austin Powers movies, where one ofthe many bad puns is when Austin says Chow, babyvery pointedly to the character named Fat Bastardwho is constantly pigging out.Not that I would watch such a stupid movie.As to cheers, it's funny how a group of people willadopt an idosyncracy. The clots of of Trekkies and D&D players when I was an undergrad had all adoptedwhat they've infected each other with by now. Felicitation?I have a friend who fell in with a group of pentacostalChristians for a while who had Bible study nearly everynight. This group prided themselves on not having anyrituals, as opposes to us nasty Catholics, Lutherans, etc.who merely parrot meaningless prayers. Oh no, these guyswere always sincere and ex corde and would never do anythingout of habit or repetition.So my friend noticed that everyone, himself included, whilepraying during their Bible studies, made this clacking sound I just want to praise Your great left the system yet.Nice post.Reminds me of the effect, noted by Tom Wolfe in _The_Right_Stuff_,whereby airline pilots all over the world make their announcements inthe folksy and understated west-Texas style of Chuck Yeager. Even ifit's a wing falling off or something. Now *there's* something for thelaw of small numbers to chew on.Or should I saychaw. === Subject: Re: Why do math folks tend to end their letters with cheers?> ...>left the system yet.> Nice post.> Reminds me of the effect, noted by Tom Wolfe in _The_Right_Stuff_,> whereby airline pilots all over the world make their announcements in> the folksy and understated west-Texas style of Chuck Yeager. Even if> it's a wing falling off or something. Now *there's* something for the> law of small numbers to chew on.> Or should I say> chaw.Or chow. === Subject: we are the world (worthless writing)we are the world.we are the children.here is peace. but iraq is at war.i don't like war.i wish that iraq problem settled well.because we are the world. === Subject: Re: Partial sums of probabilities of multinomial variables> Your help in this problem is highly appreciated.> Suppose we are given a multinomial probability distribution > N!snipped out an expression mucked up by word wrap> How to compute an approximation of the following> probability, where {X_i}s are distributed as above:> P( X_1 = max{X_1, X_2,..., X_K} )?> if p>q or if p > q*(K-1)> For example, when K=2 we simply have (assume N is even):> N N!> P( X_1 = max{X_1, X_2} ) = P( X_1 >= X_2 ) = Sum -------- p^i*(1-p)^(N-i) > i=N/2 i!(N-i)!> Are any approximations known for such sums?> What about the case of general K? > In my applications N is on order of several hundreds,> so exact computation is not feasible.I haven't checked your logic, but that summation looks like a right tail probability for the negative binomial. There are good approximations for N! that should work if you want to roll your own function, but there is a multitude of canned functions out on the Web.David WinsemiusIf the statistics are boring, then you've got the wrong numbers. -Edward Tufte === Subject: textbooks - limits & continuityThe sections on limits and continuity using epsilon-delta proofs in theundergraduate Calculus textbooks are very short. Does some book exist whichgoes into more deeper detail that someone trying to study at the lower divisionundergraduate level can understand? Alternatively, are these epsilon-deltaproofs really not too important at this lower division undergraduate level ofstudy? Examing such proofs more thoroughly should maybe wait until moreadvanced coursework? (As if don't study from too advanced a book because onecould hurt himself...)G C === Subject: Re: textbooks - limits & continuity> The sections on limits and continuity using epsilon-delta proofs in the> undergraduate Calculus textbooks are very short. Does some book exist which> goes into more deeper detail that someone trying to study at the lower division> undergraduate level can understand?Spivak, Calculus> Alternatively, are these epsilon-delta> proofs really not too important at this lower division undergraduate level of> study?Proofs are considered not too important for engineering and science students.> Examing such proofs more thoroughly should maybe wait until more> advanced coursework?There is probably a course for juniors or seniors that does this. === Subject: Milman-Pettis theoremcould someone please give me a short outline of the proof of Milman-Pettis' theorem? (every uniformly convex banach space is reflexive)Cannot find it in any book on functional analysis at hand.Mark === Subject: what is the inverse of a monomialwhat kind of object is 1/(a+bx) can it be reduced to a simpler form kiran chandrauniversity of kerala === Subject: Re: what is the inverse of a monomial> what kind of object is 1/(a+bx) > can it be reduced to a simpler form A minor point: a + bx is a binomial, not a monomial.As far as your question is concerned, what is it thatyou're hoping for? For instance, one could rewritethe expression 1/(a + bx)as (1/a) 1/(1 + bx/a),and express the second factor as 1/(1 + bx/a) = 1 - bx/a + (bx/a)^2 - (bx/a)^3 + ... = sum((-bx/a)^j, j=0, ..., infinity).leading to this: 1/(a + bx) = sum( (-bx)^j / a^(j+1) , j=0, ..., infinity).The series will converge as long as |bx/a| < 1. It willdiverge whenever |bx/a| > 1. In fact, it also divergeswhen |bx/a| = 1, by inspection of the series obtained.A previous poster, G1, noted that your expression isn't reallyreducible. I assume he means that it isn't a polynomial, or anyexpression that's simpler than the one you already have.To see that it can't be a polynomial, note that if x = -a/b,then your expression is undefined (that is, it reduces to 1/0,the value of which is not defined).To coin a horribly convoluted phrase, a polynomial is neverundefined, that is, if you are given a polynomial P(x) inthe variable x, and a value x = A, there is always a definedvalue P(A) obtained by setting x = A and doing the arithmeticthat the polynomial suggests.Dale. === Subject: Re: what is the inverse of a monomialIt can't be reduced> what kind of object is 1/(a+bx)> can it be reduced to a simpler form === Subject: Re: finite index?do you mind to point out where the thread of leo's is?> How about something like this?> Let > L = the set of left cosets of H intersect K> L_1 = the set of left cosets of H> L_2 = the set of left cosets of K> Define a function> f : L --> L_1 x L_2> by> f( x (H intersect K)) = ( x H, x K)> You'd have to show that f is well-defined and 1 -1 for the theorem to> follow.> Another person, Leo, posted the same problem here, and his proof looks> to be just like this.> Brian>How to go about this proof?>Suppose that G is a (possibly infinite) group, and that G has two>subgroups H and K of finite index. Prove that H intersect K is a>group of finite index in G, and that>[G:H intersect K] <= [G:H] [G:K] === Subject: Re: finite index?|How to go about this proof?||Suppose that G is a (possibly infinite) group, and that G has two|subgroups H and K of finite index. Prove that H intersect K is a|group of finite index in G, and that|[G:H intersect K] <= [G:H] [G:K]this is one in an endless series of silly problems that are for nogood reason written so as to appear as trick questions, but whichbecome transparently trivial when re-written in more sensible form asquestions about finite g-sets instead of about subgroups of finiteindex. it's all part of the evil of teaching people about abstractgroups without teaching them about the concrete groups that motivatethe entire subject of group theory. === Subject: Re: Resistance to Change > Very true, if the proposed great improvement is just silly.> No, quite the opposite. If a proposal is patently without merit, then> it is very easy to point out the flaws and avoid the impression that> the above gives: that the person making this response has no valid> objection and is just avoiding the question.Well if it was that easy, why didn't the author do it themselves? Orat least ask one other person before putting the result in public. IfI walked into a room full of vegitarians eating a steak and sayinglook I'm a vegitarian! he'd likely recieve a rude response toobecause he's mocking the audience, or if nothing else wasting theretime. Analogously if someone comes in with a weak argument here, witheasy to spot errors claiming they already have a sound argument, thenthey're likely to be mocked for the same reason, this isn't a place tomake claims contrary to the norm.Now if you *suggested* a meathod would improve on the status quo, ANDyou were able to back it up, having done a bit of research and thencame here using our language to describe a theroem you want this bodyto accept, then you're going to get a better response. I think thisdoesn't have anything to do with resistance to change as much as ithas to do with the thought rude people are treated rudely. === Subject: Re: Resistance to Change>[...]>. . ., then you are apt to get a much different response, one that:>(1) lacks mathematical content, (2) is emotional, e.g. expresses>anger or sarcasm, and (3) is resistant to making any meaningful>analysis or comments.> Very true, if the proposed great improvement is just silly.You seem to be suggesting that such a response is justified if onefeels that the proposal is silly (devoid of merit.) But doesn'tsuch a standard run the risk that:1. The person who deems the proposal as being without merit is simplywrong, for any number of possible reasons. (1) They misunderstood theproposal. (2) Simple human fallibility. By giving a technicaljustification of their opinion, the cause of the misevaluation can beuncovered.2. It opens the door to unscrupulous people who might claim that theiropinion need not be justified because of this policy, even though theyknow that the proposal is in fact worthwhile. In other words, itallows them to avoid the truth and make unsubstantiated assertions inresponse to an earnest proposal.I also wonder if the principle of feeling that one is of such a highstature that they need not justify their assertions might also play arole in advocating such a policy.And, overall, might all of these justifications merely mask the factthat the person responding simply doesn't want to acknowledge themerit of the proposal? In that light, the question becomes: Why theresistance?As always, your thoughts on these points would be appreciated.Charlie Volkstorf> ************************> === Subject: Re: Resistance to Change> Of course, Bozo the Clown never axiomatized the Theory of Computation>Who knows?! :-)> Bozo the Clown was no mathematician, but he did work with Einstein> in the early days of quantum mechanics. He helped develop Bozo-Einstein> his honor.> 8^)I would hope that we can all see the differences between aMathematician being laughed at (e.g. Cantor) and a comedian. To quoteonly the fact that laughter is involved is to miss the point.In fact, recall the delightful song by the 3 Stooges:B-A Bay. B-E Be. B-I-Biddy By B-O Bow. Biddy Bye Bow Be You BooBiddy By Bow Boo.C-A Say. C-E See. C-I-Siddy Sigh C-O-So Siddy Sigh So See You SueSiddy Sigh So Sue.D-A Day . . .How many of those who only laughed at Moe, Larry and Curly Joerealized that they were the first to publish an enumeration of thestrings over the English alphabet? (Seehttp://www.personal.psu.edu/faculty/t/2/t2l/warmup.htm for details ofthe algorithm.) (This is a generalization of Peano's Axioms, which,as I discovered a few years ago, are reducible to the equivalentproposition that the universal set is recursively enumerable, one ofthe three axioms in my axiomitization of the Theory of Computation.http://www.arxiv.org/html/cs.lo/0003071 Sections VI and VII. )Charlie VolkstorfCambridge, MA === Subject: Re: Resistance to Change> Of course, Bozo the Clown never axiomatized the Theory of Computation> Who knows?! :-)> F.You're right. I just made the mistake of those who judge people basedon superficialities! See http://www.imdb.com/name/nm0512357/bio:Although Lindsey has pretty much made a career out of playingvariations of his most famous character, Goober from the AndyGriffith Show--a genial, good-hearted but somewhat slow-wittedhick--Lindsey has a Bachelors Degree in BioScience from the Universityof Alabama and was a science teacher before deciding to become anactor.Perhaps I should have said, Bozo the Clown wasn't laughed at foraxiomatizing the Theory of Computation.Charlie VolkstorfCambridge, MA === Subject: Re: Resistance to Change> They laughed at Einstein. They laughed at the Wright Brothers. But they> also laughed at Bozo the Clown. -- Carl Sagan> Carl Sagan never made the idiotic assertion They laughed at> Einstein. Since Sagan's dead, any old idiocy can be attributed to> him.I'm at a loss as to why you refer to the very idea that Einstein waslaughed at as being idiotic. Are you thinking about the validity orimportance of his work, or perhaps his great popularity and admirationby the public? (One joke: When he was a child his parents said,Well, aren't you the little Einstein!)I would point out that the degree to which great thinkers areinitially minimalized by their peers has no bound. There are alsopeople who laugh at anything (and have yet to be fitted with theirstraightjacket.)Ironically, it was Einstein himself who once said, Great spirits havealways encountered violent opposition from mediocre minds.Charlie Volkstorf === Subject: Re: Resistance to Change> You always did believe everything you read. [ Written by one of those> eminent guys... ]> Indeed! :-) [ Well, not *everything*, but some/many things. ]Are you serious? Do you really? This is quite interesting.Charlie Volkstorf> F. === Subject: Re: Resistance to Change> . . ., then you are apt to get a much different response, one that:> (1) lacks mathematical content, (2) is emotional, e.g. expresses> anger or sarcasm, and (3) is resistant to making any meaningful> analysis or comments.> In other words, the response is reminiscent of the group of> politicians wearing suits smashing a little electronic device to bits> with sledgehammers.> I wonder who has noticed this and if anyone might have an explanation> as to why many people behave this way (or perhaps has beliefs contrary> to the above.) I will withhold my own conjectures for the time being.> Charlie Volkstorf> Cambridge, MAThere is a strong prior belief that it is very unlikely that thesolution to a famous problem such as P=NP will be solved by apreviously unknown internet poster. Evenmoreso, there is an almost asstrong presumption that such an attempt will contain quite basicerrors. The response is an expression of irritation with someone whoattempts to tackle an elephant with a pea shooter. The thought is,Do you really think somebody who knows essentially nothing is goingto stumble over a solution to a problem that someone much betterprepared might spend their life on? They find the chutzpah excessiveto the point of being insulting.On the other hand, right now there is a thread going about this veryquestion of P=NP. Someone has proposed a solution which doesn't lookterribly promising but does show a basic competence in the matter. This man is being treated with a certain amount of respect and hiserrors pointed out gently, the idea is that he is a student and canlearn something. The student shows ability to learn and the solutionimproves. So such things can be proposed without problems. === Subject: Re: Resistance to Change> as to why many people behave this way (or perhaps has beliefs contrary> to the above.) I will withhold my own conjectures for the time being.That's elementary, Mr. Charlie-Boo. They are all plotting against you.By 'they' I mean the same 'they' who are plotting against J. Harris.You two should join together, to fight against that plot. === Subject: Re: Resistance to Change> I wonder who has noticed this and if anyone might have an explanation> as to why many people behave this way (or perhaps has beliefs contrary> to the above.) I will withhold my own conjectures for the time being.> That's elementary, Mr. Charlie-Boo. They are all plotting against you.> By 'they' I mean the same 'they' who are plotting against J. Harris.> You two should join together, to fight against that plot.Don't be silly. History is replete with great thinkers who wereridiculed during their times (as discussed throughout this thread.)You seem to be saying that it is a manifestation in the minds of theauthors. I am afraid that is more of the same (a response of sarcasmwith no mathematical content) although I am glad that you expressedyour opinion. (Perhaps it will serve as another example andcontribute to understanding this phenomenon.)Charlie Volkstorf === Subject: Re: Factoring idea, off M> Now I rather arbitrarily set a conditional statement, which is> x^2 + ax + by + y^2 = z^2,> which is basically a way for me to pick any x, y and z that I want,> since there will always exists integers 'a' and 'b' that will allow> that to work.As is your usual practice, you flippantly post an assertion for which youprovide neither proof nor supporting evidence. You haven't learnedanything!> James Often in error, but never in doubt! HarrisWacky, isn't it? But hey, it's only math. Yup, yup, yup.There are two things you must never attempt to prove: the unprovable --and the obvious.Democracy: The triumph of popularity over principle.http://www.crbond.com === Subject: Re: Factoring idea, off MJames Harris> ...Check out JSH in action here:http://mathforum.org/discuss/sci.math/a/m/526906/ 527128Excerpt: === ==I will not show mercy going forward. I was trained as a soldierin the United States Army after all.And when the US Army plays a game, we play to win.Yup, you guessed it. If worse comes to worse, I *will*turn to the Army to help me with mathematicians. And thenmathematicians don't think the NSA or CIA can save yourasses, as generals LIKE me.And I think I know the CIA and NSA better than anymathematician.When push comes to shove, they'll throw you out withthe garbage.They'd personally shoot you themselves, if it were necessary.If I have to sic Army generals on you, I will be really pissed. === =====Subject: Re: Factoring idea, off M> Does it work? I don't know. I toss ideas like this out to see if> someone else will find out.OTOH, when the chimp tosses its poo, it always works ...(hint, hint) === Subject: Re: Factoring idea, off Mwer were getting worried about you. and taht's all I have to say, except,What in Hell is a Tautological Space? and, congratulation on your renewed sobriety.> Now then, on to my latest idea, which is an attempt at finding a> simple way to factor a positive integer M.> I use a tautological space. > Does it work? I don't know. I toss ideas like this out to see if> someone else will find out. --Give Earth a Trickier Dick Cheeny -- out of office, after GIGA years.http://www.benfranklinbooks.com/http://www.rand.org/ publications/randreview/issues/rr.12.00/http:// members.tripod.com/~american_almanac === Subject: Re: Factoring idea, off M| Now I rather arbitrarily set a conditional statement, which is| | x^2 + ax + by + y^2 = z^2,| | which is basically a way for me to pick any x, y and z that I want,| since there will always exists integers 'a' and 'b' that will allow| that to work.How about x = 2, y = 2, z = 3? I don't think there there are anyintegers a and b that make x^2 + ax + by + y^2 = z^2 true in thatcase. === Subject: Hey guys, isolated singularities?I've got this question,When are singularities of analytic functions necessarily isolated?I think poles always have to be isolated (since f(z) = g(z)/h(z), so1/f(z) has must have isolated zeroes else it will be all zero -- andby the way, what's the convention for 1/h(z) if h(z)=0 identically ina domain D? is 1/h(z) infinity? hehe)But what about essential singularities? For example,sin (1/z) has a sequence of zeroes with a limit at the originbut sin (1/z) is not analytic at the origin so it has the right tobe not identically zerobut what about 1/(sin (1/z)) ... we get an essential singularity, atthe origin, right?Am I correct in saying that essential singularities are always notisolated? (e.g. because of Picard's theorem) === Subject: what to calculate them?[IMG]http://www.geocities.com/i141802596/Image1.txt[/IMG] [IMG]http://www.geocities.com/i141802596/Image2.txt[/IMG] [IMG]http://www.geocities.com/i141802596/Image3.txt[/IMG] === Subject: Scientific Group Dealing With Theoretical PhysicsOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)This is the best place for discussion about some Quantum theories likeQuantum Electrodynamics(QED), Quantum Chromodynamics(QCD), QuantumField Theories(QFT), Loop Quantum Gravity(LQG) and some other newtheories in theoretical high energy physics(hep) like Superstringtheories, M-Theory (11 dimensional supergravity at low energies),F-Theory (12 spacetime dimensions), Grand Unified Theories (GUTs),....Hope to see you as an active and useful member in this club.K.N------------------------------------------------------ --------------------For more information and joining you can see below URL:---------------------------------------------------------- ----------------http://groups.yahoo.com/group/kiarashniknejad/ -------------------------------------------------------------- ------------ === Subject: problem on matrices from Mat(3,Z_p) = 3x3 matrices over the p-adic whole numbers .Originator: bergv@math.uiuc.edu (Maarten Bergvelt)hello to all,let us call A from Mat(3,Z_p) good, if its reduction mod p has aminimal polynomial (from (Z/p)[x]) of degree 3 . put G := set ofgood matrices in Mat(3,Z_p). my question is: is there a cannocial wayto express matrices that have mod p^k a minimal polynomial of degree< 3 but have mod p^(k+1) a minimalpolynomial of degree 3 in terms ofelements from G? for example the matrices that are mod p^k skalar andhave full degree minimal polynomial mod p^(k+1) are given by theproduct of the sets a*1 + p^k*G where a runs through the scalars of Z_p.but it is much more difficult to discribe the set of matrices, thathave a minimal polynomial of degree 2 mod p^k and have a full defgreeminimal polynomial mod p^(k+1). my question is: is there any theory,that deals with such geometric problems, are there cannonicalmappings or more precisely polynomials, so that we can express thosematrices as f(g) with a g from G? thank you for every answertim === Subject: Re: Question related to factoringX-mailer: epigoneEpigone-thread: therdsmemquauOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)>For a primes p and q,>1) If p=q=3 mod 4, then n=p*q=1 mod 4.>2) If p=q=1 mod 4, then n=p*q=1 mod 4.>If you only know n, and don't know p or q,>then can you distinguish these two cases ? >Is there any algorithm or reference for this problem ? >I don't know if this is a (computationally) good method, but youcould>check to see if -1 is a square modulo n. But, yet another question remains. How can you check whether -1 is a square modulo n without knowing pand q ? It seems not clear to me... === Subject: Re: Question related to factoringOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)>For a primes p and q,>1) If p=q=3 mod 4, then n=p*q=1 mod 4.>2) If p=q=1 mod 4, then n=p*q=1 mod 4.>If you only know n, and don't know p or q,>then can you distinguish these two cases ?>Is there any algorithm or reference for this problem ?> I don't know if this is a (computationally) good method, but you could> check to see if -1 is a square modulo n.How do you do that? === Subject: Re: Question related to factoring Adjunct Assistant Professor at the University of Montana.Originator: bergv@math.uiuc.edu (Maarten Bergvelt)>For a primes p and q,>1) If p=q=3 mod 4, then n=p*q=1 mod 4.>2) If p=q=1 mod 4, then n=p*q=1 mod 4.>If you only know n, and don't know p or q,>then can you distinguish these two cases ?>Is there any algorithm or reference for this problem ?> I don't know if this is a (computationally) good method, but you could> check to see if -1 is a square modulo n.>How do you do that?I ->did<- say I didn't know if it was a computationally good method;of course, you can always just evaluate x^2+1 mod n forx=1,....,floor(n/2)+1 and see. I do realize that something like the Jacobi symbol will do not good,since you will get 1 in both cases. === Subject: Lower bound on probabilityOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)Reposted from sci.stat.math, where it got no replies...Hope somebody will be able to help out with this.For a random variable X and a constant a, I'm looking for a bound of theform P[X > a] > b, where b is some function of the moments of X (but not theLaplace transform of X). So similar to a Markov/Chebyshev type inequalitybut 'the other way around'.Hope this makes sense - and any references would be more than welcome!Tronc === Subject: Re: Lower bound on probabilityOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)>Reposted from sci.stat.math, where it got no replies...>Hope somebody will be able to help out with this.>For a random variable X and a constant a, I'm looking for a bound of the>form P[X > a] > b, where b is some function of the moments of X (but not the>Laplace transform of X). So similar to a Markov/Chebyshev type inequality>but 'the other way around'.>Hope this makes sense - and any references would be more than welcome!The Chebyshev inequalitieS provide the precise bounds onthe cdf as a function of the moments. There is a detailedaccount in Shohat and Tamarkin, and probably in many otherbooks on the problem. These bounds are both upper andlower, and in fact the same computation gets both.There are also bounds for distributions on the half-lineand on finite intervals given there.This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Lower bound on probabilityOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)>For a random variable X and a constant a, I'm looking for a bound of the>form P[X > a] > b, where b is some function of the moments of X (but not the>Laplace transform of X). So similar to a Markov/Chebyshev type inequality>but 'the other way around'.For any even positive integer n and any c > a, consider f(x) = 1 - (x-c)^n/(c-a)^n. Since f(x) <= {0 for x <= a, 1 for x > a}we have P[X > a] >= E[f(X)], which of course can be expressed in terms ofthe first n moments of X. For example, with n = 2,P[X > a] >= (a^2 - 2ac + 2 c E[X] - E[X^2])/(c-a)^2.The bounds are tight in the sense that equality holds for measures supported on {a,c}. === Subject: Re: Lower bound on probabilityOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)>For a random variable X and a constant a, I'm looking for a bound of the>form P[X > a] > b, where b is some function of the moments of X (but not the>Laplace transform of X). So similar to a Markov/Chebyshev type inequality>but 'the other way around'.>For any even positive integer n and any c > a, consider >f(x) = 1 - (x-c)^n/(c-a)^n. Since f(x) <= {0 for x <= a, 1 for x > a}>we have P[X > a] >= E[f(X)], which of course can be expressed in terms of>the first n moments of X. For example, with n = 2,>P[X > a] >= (a^2 - 2ac + 2 c E[X] - E[X^2])/(c-a)^2.>The bounds are tight in the sense that equality holds for measures >supported on {a,c}.I should have optimized with respect to c. Let m_j = E[X^j].In the case n=2, assuming m_1 > a, take c = (m_2 - a m_1)/(m_1 - a)to get the bound P[X > a] >= (m_1 - a)^2/(m_2 - 2 a m_1 + a^2)For n > 2 things seem to get complicated. === Subject: Re: Lower bound on probabilityOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)sort of material, what would it be? Some sort of Introduction to LargeDeviations perhaps?Tronc>For a random variable X and a constant a, I'm looking for a bound of the>form P[X > a] > b, where b is some function of the moments of X (but notthe>Laplace transform of X). So similar to a Markov/Chebyshev typeinequality>but 'the other way around'.>For any even positive integer n and any c > a, consider>f(x) = 1 - (x-c)^n/(c-a)^n. Since f(x) <= {0 for x <= a, 1 for x > a}>we have P[X > a] >= E[f(X)], which of course can be expressed in terms of>the first n moments of X. For example, with n = 2,>P[X > a] >= (a^2 - 2ac + 2 c E[X] - E[X^2])/(c-a)^2.>The bounds are tight in the sense that equality holds for measures>supported on {a,c}.> I should have optimized with respect to c. Let m_j = E[X^j].> In the case n=2, assuming m_1 > a, take c = (m_2 - a m_1)/(m_1 - a)> to get the bound> P[X > a] >= (m_1 - a)^2/(m_2 - 2 a m_1 + a^2)> For n > 2 things seem to get complicated.> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2 === Subject: exponentiating vector spacesOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)given a complex vector space V. is there a natural action of U(n) on the n-fold tensor product of V with itself that extends the action of the symmetric group on n letters (seen as a subgroup of permutation matrices in U(n))?similar question for real vector spaces and O(n).best, Andreas Thom === Subject: Proximic MappingsX-mailer: epigoneEpigone-thread: wimfrahtwingOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)I'm looking for information about proximic mappings. A mapping p is*proximic* iff it preserves nearestness. This means p:U->V formetricspaces U and V in such a way that the following holds:For any points x,y and z in U, distance(x,y)<=distance(x,z)if and only if distance(p(x),p(y))<=distance(p(x),p(z)).For instance, take U and V to be 3-space. Proximic mappings includetranslations, rotations, flipping through 4-space, andmag/minificationtopic?I'm particularly interested in work done in classification of proximicmappings.Kerry Soileau === Subject: Re: Proximic MappingsOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)> I'm looking for information about proximic mappings. A mapping p is> *proximic* iff it preserves nearestness. This means p:U->V for> metric> spaces U and V in such a way that the following holds:> For any points x,y and z in U, > distance(x,y)<=distance(x,z)> if and only if> distance(p(x),p(y))<=distance(p(x),p(z)).> For instance, take U and V to be 3-space. Proximic mappings include> translations, rotations, flipping through 4-space, and> mag/minification> topic?> I'm particularly interested in work done in classification of proximic> mappings.Hmmm, I sent you more email on the matter. But it appears to me thatthese maps are (in the case of Riemann manifolds with dimensiongreater than one) are exactly isometries times some constant scaling.I've been able to show that geodesics are preserved, and it appears tome that the metric is preserved up to scalar constant. If this istrue, then you probably can get all you want about this fairly broadcategory of objects from reading up on the isometries of these maps tothemselves. Homogenous Riemann manifolds (every point can be maps viaan isometry to an arbitrary point) like the n-sphere, flat space (eg,R^n and the n-torus), and hyperbolic spaces (all derived from thehyperbolic plane and have finite volume) are particularly interestingdue to their rich isometry structure.Karl Hallowellkhallow@hotmail.com === Subject: Re: Proximic MappingsX-mailer: epigoneEpigone-thread: wimfrahtwingOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)>I'm looking for information about proximic mappings. A mapping p is>*proximic* iff it preserves nearestness. This means p:U->V for>metric>spaces U and V in such a way that the following holds:>For any points x,y and z in U, >distance(x,y)<=distance(x,z)>if and only ifdistance(p(x),p(y))<=distance(p(x),p(z)).>For instance, take U and V to be 3-space. Proximic mappings include>translations, rotations, flipping through 4-space, and>mag/minification>topic?>I'm particularly interested in work done in classification ofproximic>mappings.What about Lipschitz-continuous mappings with upper and lowerLipschitz constant equal to 1? === Subject: matrix inequalityOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)HiI'm having a little difficulty showing that (if it is indeed true)Tr_A[ Tr_B(PU^*) Tr_B(UP) ] <= Tr_A[ (Tr_B P)^2 ] where P is a projector and U is unitary (U^* is its Hermitian conjugate), and both act on a tensor product space C^(n_A) otimes C^(n_B). Any ideas?Seems as though it shouldn't be too hard with a little help from the singular value decomposition, but I'm having trouble dealing with the tensor product structure.If it helps: Tr_A[ Tr_B(X) Tr_B(Y) ] = Tr[ (X otimes Y) T_A ]where T_A acts on (C^n_A otimes C^n_B)^(otimes 2) by swaping vectors in subsystem A. e.g. T_A u_A otimes u_B otimes v_A otimes v_B = v_A otimes u_B otimes u_A otimes v_BOr if it is well known, a reference would be helpful.andrew. === Subject: mapping-class groupsOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)I have a presentation for the mapping-class (a/k/a homeotopy) group of thesphere with n boundary components. I also have a set of generators, andsome relators (enbough for my purposes) of the mapping-class group of anyclosed orientable surface with zero boundary components. What I need is aset of generators and relators (as many relators as possible, since Idon't know what I need until I see it) for the mapping-class group of agenus-g orientable surface with n boundary components, n>0. Does anyonehave a reference (quick and easy, preferably). (J. Birman's book doesn'thave it, I don't think.)Tia,Michael Hamm NB: Of late, my e-mail address is beingmsh210@math.wustl.edu e-mail that seems to be from me. Pleasehttp://math.wustl.edu/~msh210/ realize that no spam is in fact from me. === Subject: Weyl group of type affine E6Originator: bergv@math.uiuc.edu (Maarten Bergvelt)How much is known about the representations of the affine Weyl group of type E6and its Hecke algebra? (I mean type E6-tilde, not the finite group of type E6.)I'm particularly interested in any direct connection between this infinite groupand the exceptional Jordan algebra, and in anything interesting that is knownabout 36 dimensional representations of the infinite group.Richard Green