mm-379 === Subject: : Re: Maximal determinant of matrix with elements 1..n^2>In the On-Line Encyclopedia of Integer Sequences we have A085000:>http://www.research.att.com/projects/OEIS?Anum= A085000>Maximal determinant of an n X n matrix using the integers 1 to n^2.>1,10,412,40800>The value 40800 has been found just a few days ago.>Can we find a continuation to n=5 (and further) by better methods>than brute force (or random) search? The best result I've found>after trying more than 10^10 random matrices (don't tell me stories>about insufficient period of RNGs, I used the Mersenne twister ;-)>is det A_5 = 6553521 for the matrix>25 15 10 8 7> 5 24 3 17 16> 6 13 23 18 2>14 1 9 22 19>12 11 20 4 21>Brute force is hopeless, since 25!~=1.55*10^25, even taking into>account the possible reduction by a factor of (5!)^2=14400Tabu search, simulated annealing and genetic algorithms are some possible approaches. The best I've found so far is det A_5 = 6831765 for [25 16 7 10 6] [ ] [ 4 24 5 15 17] [ ] [ 8 13 23 18 3] [ ] [14 2 9 22 19] [ ] [12 11 20 1 21]using a short tabu search.Of course such methods are rather unlikely to find the exact optimalsolution, but they might get pretty close. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: Maximal determinant of matrix with elements 1..n^2>In the On-Line Encyclopedia of Integer Sequences we have A085000:>http://www.research.att.com/projects/OEIS?Anum= A085000>Maximal determinant of an n X n matrix using the integers 1 to n^2.>1,10,412,40800>The value 40800 has been found just a few days ago.>Can we find a continuation to n=5 (and further) by better methods>than brute force (or random) search? The best result I've found>after trying more than 10^10 random matrices (don't tell me stories>about insufficient period of RNGs, I used the Mersenne twister ;-)>is det A_5 = 6553521 for the matrix>25 15 10 8 7> 5 24 3 17 16> 6 13 23 18 2>14 1 9 22 19>12 11 20 4 21>Brute force is hopeless, since 25!~=1.55*10^25, even taking into>account the possible reduction by a factor of (5!)^2=14400Tabu search, simulated annealing and genetic algorithms are some possible > approaches. The best I've found so far is det A_5 = 6831765 for > [25 16 7 10 6]> [ ]> [ 4 24 5 15 17]> [ ]> [ 8 13 23 18 3]> [ ]> [14 2 9 22 19]> [ ]> [12 11 20 1 21]using a short tabu search. > Of course such methods are rather unlikely to find the exact optimal> solution, but they might get pretty close. A more extensive search resulted in det A_5 = 6839492 for [ 4 15 9 25 11] [ ] [19 13 2 10 22] [ ] [17 24 14 7 3] [ ] [ 5 12 23 6 20] [ ] [21 1 18 16 8] Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: Maximal determinant of matrix with elements 1..n^2>In the On-Line Encyclopedia of Integer Sequences we have A085000:>http://www.research.att.com/projects/OEIS?Anum= A085000>Maximal determinant of an n X n matrix using the integers 1 to n^2.>1,10,412,40800>The value 40800 has been found just a few days ago. >Can we find a continuation to n=5 (and further) by better methods>than brute force (or random) search? The best result I've found>after trying more than 10^10 random matrices (don't tell me stories>about insufficient period of RNGs, I used the Mersenne twister ;-)>is det A_5 = 6553521 for the matrix >25 15 10 8 7> 5 24 3 17 16> 6 13 23 18 2>14 1 9 22 19>12 11 20 4 21 >Brute force is hopeless, since 25!~=1.55*10^25, even taking into>account the possible reduction by a factor of (5!)^2=14400 Tabu search, simulated annealing and genetic algorithms are some possible> approaches. The best I've found so far is det A_5 = 6831765 for> [25 16 7 10 6]> [ ]> [ 4 24 5 15 17]> [ ]> [ 8 13 23 18 3]> [ ]> [14 2 9 22 19]> [ ]> [12 11 20 1 21] using a short tabu search.Of course such methods are rather unlikely to find the exact optimal> solution, but they might get pretty close.A more extensive search resulted in det A_5 = 6839492 for [ 4 15 9 25 11]> [ ]> [19 13 2 10 22]> [ ]> [17 24 14 7 3]> [ ]> [ 5 12 23 6 20]> [ ]> [21 1 18 16 8]which can be row/column permuted into25 15 9 11 4 7 24 14 3 17 6 12 23 20 510 13 2 22 1916 1 18 8 21Did you start with the previous solution or from random?One observation: The sum of the matrix elements in the strictly upperright corner (excluding the diagonal) of this matrix is 117. In myfirst matrix it was 115 and in your first improvement it was 116.Is this just coincidence or can one find a relation with the determinantvalues?Could you give some information on the tabu search method you use?Type of exchanges (how many elements), type of memory (short/long) etc.I know the work of Battiti (The Reactive Search (RS) Home Page)at http://rtm.science.unitn.it/~battiti/reactive.html > Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2Hugo=== === Subject: : Re: Maximal determinant of matrix with elements 1..n^2 >> A more extensive search resulted in det A_5 = 6839492 for>>> [ 4 15 9 25 11]>> [ ]>> [19 13 2 10 22]>> [ ]>> [17 24 14 7 3]>> [ ]>> [ 5 12 23 6 20]>> [ ]>> [21 1 18 16 8]>which can be row/column permuted into>25 15 9 11 4> 7 24 14 3 17> 6 12 23 20 5>10 13 2 22 19>16 1 18 8 21>Did you start with the previous solution or from random?Random. >One observation: The sum of the matrix elements in the strictly upper>right corner (excluding the diagonal) of this matrix is 117. In my>first matrix it was 115 and in your first improvement it was 116.>Is this just coincidence or can one find a relation with the determinant>values?Coincidence, I guess.>Could you give some information on the tabu search method you use?>Type of exchanges (how many elements), type of memory (short/long) etc.I interchanged pairs. Here's my Maple program.with(LinearAlgebra): tabulist:= [1,2,3,4,5,6]; bestscore:= 0; currX:=combinat[randperm](25); for iter from 1 do if (iter mod 200) = 0 then currX:= combinat[randperm](25) fi; bestnewscore:= -1; for i from 1 to 24 do if member(i,tabulist) then next fi; for j from i+1 to 25 do if member(j,tabulist) then next fi; newX:= subsop(i=currX[j],j=currX[i],currX); newscore:= abs(Determinant(Matrix([ newX[1..5],newX[6..10],newX[11..15], newX[16..20],newX[21..25]],datatype=integer[4]))); if newscore > bestnewscore then bestnewscore:= newscore; bestnewX:= newX; besti:= i; bestj:= j; fi od od: currX:= bestnewX; currscore:= bestnewscore; tabulist:= [tabulist[3],tabulist[4],tabulist[5],tabulist[6],besti,bestj]; if currscore > bestscore then bestscore:= currscore; bestX:= currX; printf( Iteration %d, New best %d for %an, iter, currscore, bestX); fi od:Interestingly, two separate sessions on different machines, randomized using the system clock, both came to row-and-column permutations of thesame solution.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: Floor-Sumlet c,d positive integers. i try to simplify the following equation ( maple notation ):sum(floor((m/d),m=1..2*c); could anyone help me?> christian ruetherhow changes the result of this simplification, if there isanother positive integer k and the equation is:sum(floor((k*m/d),m=1..n);many thanks!christian ruether=== === Subject: : Re: Floor-Sumsum(floor((k*m/d),m=1..n);> For the case n <= d, the following simplification is very pointlessand actually less efficient. But for the case n > d, the followingsimplification is good.This is more obfuscated than the trivial k=1 version and I haven'tbeen able to completely reduce it to elementary operations like we didfor k=1. But if your objective is to streamline the calculation on acomputer, you should find this to be much more efficient (at leastwhen n is much larger than d):sum(floor(km/d),m=1..n) = (k/d)(n+1)(n)/2 - j*z - sum((mk modd)/d,m=zd+1..n)where:z = floor((n+1)/d), j = sum((mk mod d)/d, m=1,...,d-1)[In case you don't remember, mk mod d is the remainder when youdivide mk by d. you can use the formula mk mod d = d*(mk/d -floor(mk/d)). But most programming languages allow direct modulararithmetic which is much faster than that formula.]This hardly looks like a simplification at all! But if you lookclosely, it reduces one big finite sum to two small finite sums, oneof which has d terms and the other of which has no more than d-1terms. Note also that these two smaller sums involve modulararithmetic, which I believe is much faster than the floor function onmost computers (I may be mistaken).Hope this letter finds you in good health and good fortunes.=== === Subject: : Gaussian limitHow do you computelim (t --> 0 from above) x e^(-x^2/(4t)) / t^(3/2) ? I've been tryingto bound the quotient and simplify things (perhaps use L'Hopital) butto no avail.=== === Subject: : Re: Gaussian limit> How do you compute> lim (t --> 0 from above) x e^(-x^2/(4t)) / t^(3/2) ? I've been trying> to bound the quotient and simplify things (perhaps use L'Hopital) but> to no avail.Let's simplify this to solvinglim(t->0+) [e^(-a/t)]/t^b, a > 0, b > 0take logs -a/t - b.log tNote: lim(t->0) (b.log t)/(a/t) = 0In other words log t goes to -oo slower than 1/t goes to ooThus the -oo + oo form of -a/t - b.log t is biased in favor of -1/t.Take limitslim(t->0) -a/t - b.log t = lim(t->0) -(a/t)(1 + (b.log t)/(a/t)) = lim(t->0) -a/t = -ooThus orginal limit = 0.=== === Subject: : Re: Gaussian limit> How do you computelim (t --> 0 from above) x e^(-x^2/(4t)) / t^(3/2) ? I've been trying> to bound the quotient and simplify things (perhaps use L'Hopital) but> to no avail.as t -> 0+, -x^2/(4t) [as long as x is a nonzero real] tends to +infinityand so exp(-x^2/(4t)) -> 0 rapidly, more rapidly that any polynomial.If this doesn't convince you, then recall that for y > 0exp(y) = 1 + y + y^2/2 + ... so exp(y) > y andthus 0 < exp(-y) < 2/y^2.Henceexp(-x^2/(4t)) < 8t^2/x^4and so0< x exp(-x^2/(4t))/t^(3/2) < 8t^(1/2)/x^3for x > 0.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlHis mind has been corrupted by colours, sounds and shapes. The League of Gentlemen=== === Subject: : Re: Minimize the distance from the elevators> to minimize (total) distance between them (passing through optimizing point).The line between elevators should subtend 120 degrees at the minimizing point. This is seen by differentiation and also followsfrom minimal area physics analogy of soap films, where each path is afilm in equilibrium. See book Science of Soap films by CyrilIsenberg.=== === Subject: : Re: ellipse construction> is there any *practical* solution to construction an ellipse?> something which gives you the compass needle precision you get> when construction a circle. i mean, ok, using a pen and a piece> of rope fixed in the foci works, but if i want to draw an ellipse> in a 3 by 4 inch rectangle, this is far from accurate. any tips?Here's a related puzzle (which I haven't even begun to solve BTW).Do there exist a couple of plane shapes, with cogged edges, suchthat by fixing the position of one shape, with the compass needlethrough a hole in it, and moving the other compass end, with itspencil through a hole in the other shape, and with the latterconstrained by the cogs to rotate about the first shape, thepencil traces an ellipse?Part 2 (for extra marks ;) Can any such shapes be 'universalellipse generators', such that by choosing the initial cogalignment and the compass positions within them any ellipse(within limits of size and eccentricity) can be drawn?I'd guess the keyword 'evolute' is relevant in this problem.Cheers------------------------------------------------ ---------------------------John R Ramsden (jr@aNdOsSlPaAtMe.com)---------------------------------------- -----------------------------------A stockbroker is someone who invests your money until it's all gone. Woody Allen=== === Subject: : Re: mathHandbook.com> SymbMath.com (an abbreviation for Symbolic Mathematics) (former DrMath,> VisualMath) is an online math handbook and computer algebra system.> www.mathHandbook.com=== === Subject: : Re: David Ullrich on Identity> David Ullrich says:> And yes, identity is in _fact_ reflexive. To> refute that statement you need to give an> example of something which is not identical> to itself. The idea that there is something> which is _not_ identical to itself is simply> ludicrous: That's what identity _means_: A> thing is identical to itself and to nothing> For a contrasting standpoint, see> **************> David Ullrich asks:> What's an example of something that's not identical> **************> David Ullrich dares:> Exhibit of proof of Ex~(x=x) from> C1-C4 and someone will point out the error.>>C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1>>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2>>C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)>>Classification>>C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}]> Weak> Would someone be kind enough help David out with a proof?> *************> David Ullrich remonstrates:> I 'blunder' by saying that equality is reflexive by definition?> Huh. Do you have any idea what the word definition means?> Homework for David Ullrich:> 1) What osopher said: > ...definitions are available only for transforming> truths, not for founding them.> 2) In your own words, explain why (or why not) you think> this is true.> --John> True, the provability of Ex~(x=x) will vary from one system to> the next. Nevertheless, there is more to truth than> 'truth-in-a-model';> and just as the truth of Ex~(x is red) depends on how things stand> with individuals and redness so the truth of Ex~(x=x) depends on> how things stand with individuals and identity.> . . . can you offer an example where Ex~(x=x) is true?I'll provide not one but THREE examples. The first> depends on a (very simple) proof that DAVID ULLRICH COULDN'T HACK!A) Take any 'pure' first order logic, that is, any FOL with no> singular terms other than variables (just to keep things simple).B) To your FOL add this formation rule;> If a and b are variables, a=b is an atomic formula.and this axiom scheme: Let be C and D be wff's which differ only in that a occurs free> in C where b occurs free in D. If a=b, then C <-> D.C) Call the resulting logic FOL+.In FOL+, identity is symmetric and transitive and identicals are> indiscernible, but neither Ax(x=x) nor ~Ax(x=x) is a thesis. So,> FOL+ is a subtheory of FOL= (because every thesis of FOL+ is> also a thesis of FOL=), but not conversely (because Ax(x=x)> is a thesis of FOL= but not of FOL+).Example 1: Set TheoryAdd to FOL+ the NBG axiom scheme, C3, and Correy's principle of> extensionality, C4:C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) Classification> C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}]> (Equi-membered classes are identical iff these are sets.)> 1) EyAx(x in y <-> Et(x in t) & ~(x in x)) C3Hence2) Ax(x in r <-> Et(x in t) & ~(x in x)) 1,EIand3) r in r <-> (Et(r in t) & ~(r in r)) 2,UIso that4) ~Et(r in t) 3and 5) ~(set r) 4and 6) ~(r=r) 5,C4so that7) Ex~(x=x) 6,EGTo be continued.--JohnHi John,> Your very interesting proof leads to some questions.1. What is the domain of your variables ? I assume you admit> sets and non-sets. Are physical objects and proper classes and sets> values of the variable x ?2. Do you claim: Et(x in t & y in t) <-> (set y & set x), or,> Et(x in t)& Et(y in t) .<->. set x & set y. ?C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}]> (Equi-membered classes are identical iff these are sets.)C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] Weak3. Is C3 <-> EyAx[x in y <->. Et(x in t) & set y & A] (with y not free> in A) ?4. It seems your C4 entails x=x <-> set x. That is, x=x fails for> empirical values (non sets). Where does this leave us with respect to> 's descriptions?5. How do you distinguish: E!x, x=x, set x, Ey(x in y), Ez(z in x) ?> WittVariables range over classes. Classes that are elements are sets,and classes that are not are proper classes. I'm uncertain whetherphysical objects have any role to play in set theory. But I doagree with G. Frege that proper classes can be taken as *models*of quanta--not proper classes as they are treated by Ullrich andChapman, but proper classes .88 la (C3,C4). (I'm sending you Teller'sespecially his footnote 6.)2. Do you claim: Et(x in t & y in t) <-> (set y & set x), or,> Et(x in t)& Et(y in t) .<->. set x & set y. ?(i) is an instance of C3: (i) EyAx(x in y <-> (Set x & x=w v x x=z)).But (ii) follows from (i,C4) (proof supplied upon request): (ii) EyAx(x in y <-> (x=w v z=z))On the other hand, from (ii,C4) and Ax(Set x <-> x=x) itboth follows that Et(x in t & y in t) <-> (set y & set x),and that Et(x in t)& Et(y in t) .<->. set x & set y. C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}]> (Equi-membered classes are identical iff these are sets.)C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] Weak3. Is C3 <-> EyAx[x in y <->. Et(x in t) & set y & A] (with y not free> in A) ?No. It does not follow that y is a set. To prove that y is a set,it must be established that every class with the same members as yis identical to y. In other words, it must be established thatLeibniz's principle of the identity of indiscernibles holds forevery such class and y. (To establish this, it suffices toestablish that this principle holds for y itself.) If Leibniz'sprinciple does not hold, then y (along with every class equi-memberedwith y) is a proper class. 4. It seems your C4 entails x=x <-> set x. That is, x=x fails for> empirical values (non sets). Where does this leave us with respect to> 's descriptions?The answer to this question is (I think) provided by the followingcritique of Quine by Alberto Cortes:My final comment about variables and names deals with a commentof Quine's: he has claimed that in a logically austere language,names are quite superfluous: Chief among the omitted frills is the *name*. This . . . is a mere convenience and strictly redundant, for the following reason. Think of 'a' as a name, and think of 'Fa' as equivalent to '(Ex(a = x & Fx)'. We see from this consideration that 'a' need never occur except in the context 'a='. But we can as well render 'a=' always as a simple predicate 'A'. thus abandoning the name 'a'. 'Fa' gives way thus to '(Ex)(Ax & Fx)', where the predicate 'A' is *true solely of the object a*. [Cortes's emphasis]. It may be objected that this paraphrase deprives us of an assurance of uniqueness that the name has afforded. It is understood that the name applies to only one object, whereas the preciate 'A' supposes no such condition. However, we lose nothing by this, since we can always stipulate by further sentences, when we wish, that 'A' is true of one and only one thing. (_osophy of Logic_ [Englewood Cliffs, New Jersey: Prentice Hall, 1970]: p. 25)According to Quine, then, one can always eliminate a name byconcocting an appropriate predicate which is true only of theobject designated by that name.But how is one to be assured that the predicate one has concocteddoes not apply simultaneously to more than one object? Onecannot merely stipulate that it be so. One has to presupposethat if two objects are numerically distinct they must notpossess all properties in common. And this is nothing lessthan PII(L) [PII(L): No two substances are completely sim-ilar, or differ *solo numero. -JC] Therefore, it is evidentthat in order to eliminate proper names, *.88 la Quine*, one hasto presuppose PII(L). Needless to say, we shall not admitsuch a presupposition in a critical analysis of PII(L).Therefore, from the point of view of this paper, proper names,or their logical equivalents (e.g., prounouns), are logicallyfundamental ingredients of both ordinary and formal languages.(Leibniz's Principle of the Identity of Indiscernibles: AFalse Principle, _osophy of Science_, Vol.43 Dec., 1976,p. 496)The eliminability of proper names .88 la Quine--which Cortesopposes on the grounds that PII(L) fails to hold for quanta--marches in lockstep with 's Theory of Descriptions.For Cortes's critique of Quine applies to 'sTheory of Descriptions as well.5. How do you distinguish: E!x, x=x, set x, Ey(x in y), Ez(z in x) ?The first three are equivalent. Ez(z in x) and ~Ez(z in x) areboth compatible with the first three.> Witt--John CorreyPS I am pleased to have been dismissed as a crass formalistby the likes of Robin Chapman and David Ullrich. Notuncoincidentally, these are PROFESSORS OF MATHEMATICS!=== === Subject: : Process identification (Control Theory)Hi everybody, I am novice in control theory and I have the followingproblem:Let consider process when step/impulse response involves a longresponse, e.g. 1 hr and maybe more. If I am trying to identify thisresponse, I use for it least square method and I estimate FIRparameters (when I have a sampling time 30 seconds, for 1 hr longprocess I have 120 parameters). But I am sure that instead of 120 FIRparams I can use 1st order curve (3 or 4 parameters) for describingthis process. My problem is that I don't know how to do it.Could anyone of you advice me? Or give me some tips to the book/urlpage in which I can find the requested description.Josef=== === Subject: : Re: Need help with solution: SL Dixon's TextAnother question about the equation referred to in the sol. manual forprob 7.4: Is it a typo where they have on the rhs M.r1^2? Shouldn'tthat be M.r1^3?David> Thank You! Yes I saw the reference to Eqn 7.11, but tried using 7.11a> as 1/[pi*Po1*ao1^3] where ao1 = sqrt(gamma*R*To1).DavidDid you miss the reference to Eqn 7.11? The LHS of the eqn is> mdot*Omega^2/[pi*k*gamma*Po1*sqrt(gamma*R*To1)]. If you take out> mdot*omega^2/k, what's left,> 1/[pi*gamma*Po1*sqrt(gamma*R*To1)], is equal to 0.6598e-8. Did you actually> try to plug in numbers?Pete K.> A solution to problem 7.4 in SL Dixon's Fluid Mechanics and> Thermodynamics of Turbomachinery 4th Ed. has me stumped and I would> appreciate any help understanding the steps that were left out.> The compressible flow relation between mass flow rate, speed of> rotation and the flow parameters at the eye tip is given by eqn.> (7.11),> 0.6598 x 10^-8 x (mdot x rads ^2)/k = ...> I've not figured out yet how we're to get 0.6598 x 10^-8> David=== === Subject: : Re: Need help with solution: SL Dixon's TextYes, it appears to be a typo.Pete K.> Another question about the equation referred to in the sol. manual for> prob 7.4: Is it a typo where they have on the rhs M.r1^2? Shouldn't> that be M.r1^3?> David> Thank You! Yes I saw the reference to Eqn 7.11, but tried using 7.11a> as 1/[pi*Po1*ao1^3] where ao1 = sqrt(gamma*R*To1). David > Did you miss the reference to Eqn 7.11? The LHS of the eqn is> mdot*Omega^2/[pi*k*gamma*Po1*sqrt(gamma*R*To1)]. If you take out> mdot*omega^2/k, what's left,> 1/[pi*gamma*Po1*sqrt(gamma*R*To1)], is equal to 0.6598e-8. Did youactually> try to plug in numbers?> Pete K.> A solution to problem 7.4 in SL Dixon's Fluid Mechanics and> Thermodynamics of Turbomachinery 4th Ed. has me stumped and I would> appreciate any help understanding the steps that were left out.> The compressible flow relation between mass flow rate, speed of> rotation and the flow parameters at the eye tip is given by eqn.> (7.11),> 0.6598 x 10^-8 x (mdot x rads ^2)/k = ...> I've not figured out yet how we're to get 0.6598 x 10^-8> David=== === Subject: : Isoperimetric inequality, some problemRecently I study on Isoperimetric inequality. I try a bit to extendthe inequality to higher dimension, but i am not good at manipulatingmultivariable Fourier series to get across it. The classical way tothe ineauality is consider the envelope of the straight line familyx cos a+y sin b-p(a)=0which represent a closed curve. I dont know how to extend this to 4-Dor more dimension case. Another question: how to confine the type ofclosed surfaces in the proof of this inequality?Here I state the extended inequality:L^n>=n^n*(pi)^(n/2)/Gamma(n/2+1)* A^(n-1), n>=2.where L represents the area of the (n-1)-dimensional sphere, Arepresents the volume of the corresponding n-D ball. When n=2, theinequality reduces to the wellknown isoperimetric inequality:L^2>=4*pi*A.=== === Subject: : Re: Mega Foundation, change in plans> pushed briefly to the>to the shed door:>^ [...]>It has been quite a long time since I last got tempted to poke fun at>you in public, though, hasn't it? It is very tiresome when two people>persist in a public slanging match that should be carried out via>email. So it was unfortunate that you never replied to the email I>sent you at the time. Why in the world would I reply to your email?(You're referring to the email where you suggested that I shouldthink about whether my behavior on sci.math had something todo with my feeling unloved by my father years ago, right?Do people you don't even know _usually_ reply when you sendthem email like that? Just curious. Regardless, the idea thatyou thought I'd be interested in discussing _your_ speculationson a topic like that is simply mind-boggling.I bet you're a lot of fun at parties, eh?)>Had you done so, I doubt we would be persisting>in public slanging now.What a ... never mind. I sure hope I learned my lesson:reply to email you send or you might poke fun at me inpublic. Yes, sir, Mr Spragg, sir. I will reply to your emails.giggle.=== === Subject: : Re: Mega Foundation, change in plansgiggle.Nothing makes a pig happier than a roll in the wallow, as this giggling porker knows! http://www.shop4egifts.com/target.asp?item=/jpg/25112.jpg& width=314&height=400#=== === Subject: : Re: Mega Foundation, change in plans pushed briefly to theto the shed door: ^ > pushed briefly to the^ >to the shed door:^ >^ [...]un-[...]>Just in case you're under any illusion about Dave The Rave being my>creation, this seems like the right place to point out that actually I>just took amused note when someone else used it quite a while ago.> Taylor, I think. I don't know if he coined it or borrowed it in>turn from someone else.Ah, here we are. I was starting to worry for a while when Mr Googlecould find no trace. Boxing Day 2001 - good grief! must have beenhaving a very tedious Yuletide:Hell! - don't worry about old raving Dave Ullrich. He's hadrun-ins with god only knows how many other posters. Basically he's asociopath who can't see a red rag without regarding it as a personalinsult.So apologies all round. Looks like I have to take personalresponsibility for calling you Dave the Rave after all. But 'sversion is better. Good enough to convert into a sig, in fact.Hell! - don't worry about old raving Dave Ullrich ...Basically he's a sociopath who can't see a red ragwithout regarding it as a personal insult. Taylor, sci.math=== === Subject: : Re: Mega Foundation, change in plansDave The Rave Ullrich pushed briefly to theto the shed door:^ ^ > pushed briefly to the^ >to the shed door:^ >^ >^ [...]^ >^ >It has been quite a long time since I last got tempted to poke fun at^ >you in public, though, hasn't it? It is very tiresome when two people^ >persist in a public slanging match that should be carried out via^ >email. So it was unfortunate that you never replied to the email I^ >sent you at the time. ^ ^ Why in the world would I reply to your email?Because you don't have a life and you don't get very many? ^ (You're referring to the email where you suggested that I should^ think about whether my behavior on sci.math had something to^ do with my feeling unloved by my father years ago, right?^ Do people you don't even know _usually_ reply when you send^ them email like that?Don't know, so far you're the only person I've encountered on Usenetwhose perpetual apparent public craving for approval from others hasimpelled me to send such an email.^ Just curious. Regardless, the idea that^ you thought I'd be interested in discussing _your_ speculations^ on a topic like that is simply mind-boggling.I wasn't interested in discussing my speculations either. I wasinterested, genuinely interested, in why you persist so doggedly inflogging the dead horse that is JSH. My speculation was only a way oftrying to get you to examine your own motives. But hey, why bother toexamine your own motives for once when there's endless time to bewasted recycling the same old speculations of yours about those ofJSH? ^ I bet you're a lot of fun at parties, eh?)Yup. Parties are such great occasions for unsolicited psychologicalanalysis. Hint: this is sarcasm. I know you Americans have troublewith the concepts of sarcasm and irony. ^ >Had you done so, I doubt we would be persisting^ >in public slanging now.^ ^ What a ... never mind. I sure hope I learned my lesson:^ reply to email you send or you might poke fun at me in^ public. Yes, sir, Mr Spragg, sir. I will reply to your emails.^ ^ giggle.You are a /professor/, right? Jeesus.Andy-- sparge at globalnet point co point ukThere's no better way to get to know a man's character than to get him tight and see how he talks then Nevil Shute, So disdained=== === Subject: : Re: convolution>Hi. I'm working on a program wherein I need to compute the product of>two multi-precision integers. Numerical Recipes in C suggests using>the Fast Fourier Transform. Then, they go on to state that the problem>can be worked using integer arithmetic (rather than doubles) via>modular arithmetic, but doesn't supply a reference in the same>paragraph. At the end of the section, among other references, I findFast Fourier Transform and Convolution Algorithms by Nussbaumer,>1982. Apparently, this book is OOP and difficult to find, so I'll get>it via Interlibrary Loan. My question is: Does anyone know of anyYou can find a _lot_ about this in Knuth.=== === Subject: : high school mathsI am years past my high school and having a bit of difficulty understandingbasic integration. please help.If we integrate a function 1/(a-x)^2 where a is a constant. Then i shouldget the answer1) 1/(a-x)However i just saw a solution where they say2) x/a(a-x)which seems to be also true(differentiating this does gives the originalfunction). But how on earth do we get the second solution. Please help.=== === Subject: : Re: high school maths complete solution and explanation sent to the e-mail address you hadthe integrity to provide>I am years past my high school and having a bit of difficulty understanding>basic integration. please help.>If we integrate a function 1/(a-x)^2 where a is a constant. Then i should>get the answer>1) 1/(a-x)>However i just saw a solution where they say>2) x/a(a-x)>which seems to be also true(differentiating this does gives the original>function). But how on earth do we get the second solution. Please help.=== === Subject: : Re: high school mathsI am years past my high school and having a bit of difficulty understanding> basic integration. please help.If we integrate a function 1/(a-x)^2 where a is a constant. Then i should> get the answer> 1) 1/(a-x)However i just saw a solution where they say> 2) x/a(a-x)which seems to be also true(differentiating this does gives the original> function). But how on earth do we get the second solution. Please help.Differentiating f(x) + k where k is a constant is the same asdifferentiating f(x). So, do your two solutions differ by a constantonly?-- G.C.=== === Subject: : Re: high school maths> I am years past my high school and having a bit of difficulty> understanding basic integration. please help.If we integrate a function 1/(a-x)^2 where a is a constant. Then i should> get the answer> 1) 1/(a-x)However i just saw a solution where they say> 2) x/a(a-x)which seems to be also true(differentiating this does gives the original> function). But how on earth do we get the second solution. Please help.> === === Subject: : Re: high school maths>I am years past my high school and having a bit of difficulty understanding>basic integration. please help.>If we integrate a function 1/(a-x)^2 where a is a constant. Then i should>get the answer>1) 1/(a-x)>However i just saw a solution where they say>2) x/a(a-x)>which seems to be also true(differentiating this does gives the original>function). But how on earth do we get the second solution. Please help.Subtract the two expressions and simplify. Behold!-- Rouben Rostamian ===>I have begun to teach myself LaTeX. To help myself, I would like to>dealing with number theory, abstract algebra or discrete> mathematics).>Could anyone point me to a website that makes this kind of LaTeX>source code available for download?Some publishers have samples available, e.g. this 80K specimen from> Springer:> ftp://ftp.springer.de/pub/tex/latex/svjour/jems/example/> Here are quite a few other Springer journals:> http://www.springer.de/author/tex/help-journals.html> LHThere's also a link or two under the heading 'Miscellaneous' on thepage: http://www.math.wisc.edu/~propp/math.html(I've just spent a couple of hours reading some facsinating materialat the Grothendieck biography page, which is another link includedon that page, namely: http://www.fermentmagazine.org/home5.html.In due course someone should definitely point that child prodigy,mentioned on sci.math a while back, in the direction of motiviccohomology)Cheers-------------------------------------- -------------------------------------John R Ramsden (jr@adslate`rm -rf *`.com) (remove spambot trap before use)---------------------------------------------------------- -----------------A stockbroker is someone who invests your money until it's all gone. Woody Allen=== === Subject: : Transfinite InductionCan someone please explain in general terms what this means. Also,how is it applicable to the process of obtaining the extension of ameasure from a given field to another larger one. Many thanks.=== === Subject: : Re: Transfinite Inductiontransfinite induction is like ordinary induction, except that it can be appliedto wellordered sets that have limit elements: a limit element in a wellorderedset is any element except the smallest one that doesn't have an (immediate)predecessor| Can someone please explain in general terms what this means. Also,| how is it applicable to the process of obtaining the extension of a| measure from a given field to another larger one. Many thanks.=== === Subject: : Re: Transfinite Induction> transfinite induction is like ordinary induction, except that it can be applied> to wellordered sets that have limit elements: a limit element in a wellordered> set is any element except the smallest one that doesn't have an (immediate)> predecessor> Many thanks to everyone who responded. Mention was made ofapplicability to ordinals and naturals. Can it also apply (or beuseful) to proving statements about sets with other types of binarythe help.=== === Subject: : Re: Transfinite InductionX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS>Can someone please explain in general terms what this means.I'll try the first part, assuming that you understand the concept ofOrdinal numbers.. Let P(i) be a proposition, for each ordinal i. Transfinite Inductionis a rule that allows us to prove (For all i)P(i) if we can prove 1. P(0) 2. (For all j)((For all iP(j))Sometimes it is stated with (For all i)(P(i)=>P(i+1), but that is notactually necessary. Note that for finite ordinals it reduces toordinary induction.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org=== === Subject: : Re: Transfinite Induction Visiting Assistant Professor at the University of Montana.>>Can someone please explain in general terms what this means.>I'll try the first part, assuming that you understand the concept of>Ordinal numbers.. >Let P(i) be a proposition, for each ordinal i. Transfinite Induction>is a rule that allows us to prove (For all i)P(i) if we can prove> 1. P(0)> 2. (For all j)((For all iP(j))>Sometimes it is stated with (For all i)(P(i)=>P(i+1), but that is not>actually necessary. Worse, it is wrong. If you only prove (For all i)(P(i)=>P(i+1)) youhave not necessarily proven that it holds for ANY number. Even if youprove it for some element and you prove this, you will never get pastlimit ordinals, which is the entire point of transfinite induction.Also, your (1) is unnecessary; it is implied by 2 by setting j=0,since For all iCan someone please explain in general terms what this means. Also,>how is it applicable to the process of obtaining the extension of a>measure from a given field to another larger one. Many thanks.Transfinite induction is induction through a general well-ordered set, rather than specifically through the natural numbers. The general theorem is, given a a well-ordered set X and a proposition p(a) with a free,{for all a in X, [p(b) for all b Can someone please explain in general terms what this means. Also,> how is it applicable to the process of obtaining the extension of a> measure from a given field to another larger one. Many thanks.It's like induction, but the index set is an uncountable well-orderedset.=== === Subject: : Re: Transfinite Induction Visiting Assistant Professor at the University of Montana.>> Can someone please explain in general terms what this means. Also,>> how is it applicable to the process of obtaining the extension of a>> measure from a given field to another larger one. Many thanks.>It's like induction, but the index set is an uncountable well-ordered>set.It does not have to be uncountable. You can use transfinite inductionto prove stuff about a set that consists of two copies of thenaturals, a blue and a green copy, with all blue numbers less thanall green numbers, and numbers of the same color ordered in the usualway. And regular induction (proving it for the blue zero, and thenproving that if it holds for k then it holds for k+1) would not proveit for all elements of the set.========================================================== ============It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ================== === Subject: : Algebraic topology questions2 basic questions:1- If K is an n-skeleton of an (n+1)-simplex, prove that |K| is homeomorphicto S^n. Where | | is the standard functor from simplicial complexes toTopological spaces.The previous question confuses me. Does the diameter of a n-simplex haveanything to do with n?? Anyways, a proof of this question will clear upconfusion.2- If g is a function from S^n ---> S^n. How can x and f(x) be tangent??Aren't they just points? If not (obviouisly) how do regrard them asvectors??=== === Subject: : Re: Algebraic topology questions>2 basic questions:>1- If K is an n-skeleton of an (n+1)-simplex, prove that |K| is homeomorphic>to S^n. Where | | is the standard functor from simplicial complexes to>Topological spaces.What do you mean by an in an n-skeleton? Any given(n+1)-simplex has, as I use the language, one and only one n-skeleton, so the n-skeleton of an (n+1)-simplexwould be appropriate.Supposing that the answer to my question is simply thatyou are having trouble with definite and indefinite you will get into mathematical trouble eventuallyif you keep it up), and supposing that what you call the standard functor from simplicial complexesto topological spaces is what I'd call the geometric-realization functor, then I suggest thatyou begin by proving that if L is an (n+1)-simplex,then |L| is homeomorphic to D^{n+1}. After you've finishedthat proof, inspect the homeomorphism you have on hand,and think about the definitions of n-skeleton and S^n.>The previous question confuses me. Does the diameter of a n-simplex have>anything to do with n?? The previous question confuses me. Does it have anything to do withthe question before it? (Where did diameter come from?)>Anyways, a proof of this question will clear up>confusion.I doubt it. More likely, in the presence of confusion, a proof of this question will cause more confusion.>2- If g is a function from S^n ---> S^n. How can x and f(x) be tangent??>Aren't they just points? If not (obviouisly) how do regrard them as>vectors??This is incoherent as stated (and not just because it switchesfrom g to f). What do you really want to know?Lee Rudolph=== === Subject: : Re: Algebraic topology questions>1- If K is an n-skeleton of an (n+1)-simplex, prove that |K| is homeomorphic>to S^n. Where | | is the standard functor from simplicial complexes to>Topological spaces.snip>The previous question confuses me. Does the diameter of a n-simplex have>anything to do with n?? The previous question confuses me. Does it have anything to do with> the question before it? (Where did diameter come from?)confusion isnt the word - im not sure id place much faith in the op if he toldme what day of the week it is. my guess is that hes confusing with the veryunhelpful notation one finds in certain modern books where |S| denotes thecardinality of the set S. to clarify: K is an abstract set with an element {v}for each vertex etc. |K| is a topological space, a subspace of some euclideanspace. it is definitely not a number.>2- If g is a function from S^n ---> S^n. How can x and f(x) be tangent??>Aren't they just points? If not (obviouisly) how do regrard them as>vectors??This is incoherent as stated (and not just because it switches> from g to f). What do you really want to know?my guess is that the op's book identifies the tangent space to S^n at x withthe orthogonal complement of x in R^{n+1}, and by the sound of it the factthat R^n is a vector space is lost on him/her....tom=== === Subject: : Counting symbols in a listHow to count quickly the symbols in a list with Mathematica?For example in {1,1,0,1,0,1} the symbol 1 occurs 4 times.=== === Subject: : Re: Counting symbols in a list> How to count quickly the symbols in a list with Mathematica?> For example in {1,1,0,1,0,1} the symbol 1 occurs 4 times.In[1]:= Count[{1,1,0,1,0,1},1]Out[1]= 4-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.=== === Subject: : polynomial being positive in a closed intervalI'm stuck with having to show that a specific polynomial (with realcoefficients, degree>5) is positive in the interval [0.0, 0.5]. Couldanyone suggest me some method on how to go about it? I wouldappreciate any type of feedback.Finally, I apologize if you found this posting to be inappropriatewith respect to this news group.Georgios=== === Subject: : Re: polynomial being positive in a closed interval>I'm stuck with having to show that a specific polynomial (with real>coefficients, degree>5) is positive in the interval [0.0, 0.5]. Could>anyone suggest me some method on how to go about it? I would>appreciate any type of feedback.Find the zeroes of the polynomial; by a continuity argument, a polynomialmust be always positive (or always negative) between consecutive zeroes.Doug=== === Subject: : Re: polynomial being positive in a closed interval>>I'm stuck with having to show that a specific polynomial (with real>>coefficients, degree>5) is positive in the interval [0.0, 0.5]. Could>>anyone suggest me some method on how to go about it? I would>>appreciate any type of feedback.> Find the zeroes of the polynomial; by a continuity argument, a polynomial> must be always positive (or always negative) between consecutive zeroes.Doug> Look up Sturm's theorem.=== === Subject: : Re: polynomial being positive in a closed interval> I'm stuck with having to show that a specific polynomial (with real> coefficients, degree>5) is positive in the interval [0.0, 0.5]. Could> anyone suggest me some method on how to go about it? I would> appreciate any type of feedback.Do you know the standard algorithm, due to Sturm,for computing the number of real roots of a polynomial in an interval?It is rather complicated to describe,but I'm sure you would find it if you googled for Sturm real roots.-- Timothy Murphy e-mail: tim@birdsnest.maths.tcd.ietel: +353-86-233 6090s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland=== === Subject: : Re: polynomial being positive in a closed interval> I'm stuck with having to show that a specific polynomial (with real> coefficients, degree>5) is positive in the interval [0.0, 0.5]. Could> anyone suggest me some method on how to go about it? I would> appreciate any type of feedback.> Finally, I apologize if you found this posting to be inappropriate> with respect to this news group.The posting is perfectly appropriate.What I would do is to plot of the graph of the polynomial in the range[-0.1, 0.6] and then home in on any areas where it looked as though it_might_ go negative.If you post the coefficients there are probably people here who can give youan answer.-- Clive Toothhttp://www.clivetooth.dk=== === Subject: : Re: polynomial being positive in a closed interval> I'm stuck with having to show that a specific polynomial (with real> coefficients, degree>5) is positive in the interval [0.0, 0.5]. Could> anyone suggest me some method on how to go about it? I would> appreciate any type of feedback.> Finally, I apologize if you found this posting to be inappropriate> with respect to this news group.> GeorgiosIf f(0) and f(0.5) are both positive, show that f'(x) has no zeros in thatinterval.It might be that the zeros of f' are easy to calculate.Note: the existence of zeros does not disprove it..Bob Pease=== === Subject: : Re: Kepler Packing on concrete mixes=== === Subject: : Re: Kepler Packing on concrete mixes Re: theoreticallythe strongest concrete Archimedes Plutonium NOdtgEMAIL whole entire Universe is just one big atom where dots ofthe electron-dot-cloud are galaxies sci.materials, sci.engr, sci.math 1 , 2 , 3 , 4 , 5 , 6 , 7> compressing concrete is quite common, and done with a vibrator, before it> sets.> also, there are additives that help reduce viscosity or increase fluidity,> while keeping initial water content low (but still enough water to react),> similar to what is used in potter / ceramics.> E.VI wonder if sand grains are mostly flat faced and not roundish? TheKeplerPacking deals with spheres and it is the densest packing with 26% voidgaps.Does anyone know what the usual or average sand roundness or flatnessis?I suppose one can get gravel that is mostly roundish. I know thegravel ofquartzite to make a concrete mix is mostly angular with alot offlatlikesurfaces.Perhaps sand is much like this quartzite in having many flat surfaces.So thatwhen one mixes a concrete mix of sand and aggregate gravel that manyof theflat surfaces touch and leaving little if any room for portland cementto gluetogether those flat surfaces. And that the Kepler Packing of sphereswith itsdensity of 74% and 26% gaps for a glue. That in a sand and or gravelmix ofconcrete or asphalt that the 26% gaps need not have a glue becausethey aremostly flat surfaces touching one another. So I suppose the 26% can bedrastically reduced in these mixes because of flatnessvirus toprevent these posts of mine from appearing on all outside ISP, exceptfor myown ISP of dtgnet. So am double posting.Archimedes Plutonium, a_plutonium@hotmail.comwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies=== === Subject: : Re: Kepler Packing on concrete mixes===> === Subject: : > Re: Kepler Packing on concrete mixes Re: theoretically> the strongest concrete> SnipI wonder if sand grains are mostly flat faced and not roundish? The> Kepler> Packing deals with spheres and it is the densest packing with 26% void> gaps.snip> Archimedes Plutonium, a_plutonium@hotmail.com> whole entire Universe is just one big atom where dots> of the electron-dot-cloud are galaxiesArch, The 'Kepler Packing` model is jejune as you are beginning to realize.The nature of concrete would have to be modeled at the microscopiclevel where the needle like hydrate crystals grow and you wouldprobably need fractals to model the surfaces. Too much like work when you have a mature technology with literallyhundreds of years of accumulated data to draw on. The formulas currently in use are the product of experience. They areused because they work best. That is the nature of mature technologiesbuilt on empirical data. The technology teaches you when you try tomodel it. You may learn something, but you are not going to improve onall that experience. Pretending that you are is either arrogance or mere mentalmasturbation.Pragmatist - Use a bigger hammer=== === Subject: : Re: Kepler Packing on concrete mixesreally. I hadn't realized taht AP used the same presentationas JSH, continually restarting the item as a new group. it's a good question, though,just like the realtion to Avagadro's #. > The 'Kepler Packing` model is jejune as you are beginning to realize.> The nature of concrete would have to be modeled at the microscopic> level where the needle like hydrate crystals grow and you would> probably need fractals to model the surfaces.--les ducs d'Enron!=== === Subject: : Re: Kepler Packing on concrete mixes Re: theoretically the strongest concretecompressing concrete is quite common, and done with a vibrator, before itsets.also, there are additives that help reduce viscosity or increase fluidity,while keeping initial water content low (but still enough water to react),similar to what is used in potter / ceramics.E.V> marbles and BBs. Where the marbles are gravel and the BBs are sand. Now the question is, what mix ratio of Portland cement to make a cubicfoot of concrete that has marbles> and BBs and is the strongest cubic foot. I still suspect the final answer to be a cubic foot wherein only BBs areused and the ratio is close to> a 1 :: 1 in terms of volume. Say perhaps 10 liters of portland cementand 10 liters of BBs. Trouble with various sizes of aggregate is that it allows air pocketswhich diminish strength and cause> cracking.> Actually, I'm not sure that this is the case. I can see that the sand> would fill in the gaps between the larger aggregate, and the cement in> turn would fill in the gaps between the sand. OTOH, you probably would> get more air than with sand-sized aggregate alone.> Also, would a marble sized glob of BB's cemented together be stronger> than a steel marble of the same size? I think not. It indicates to me> that the large aggregate is probably responsible for most of the> strength (perhaps in some random almost kissing structure) with the> spaces filled by cement and sand.> So I am guessing that the strongest mix ratio is a 1 to 1 ratio usingonly BBs and portland cement. And I am guessing that the perfect piece of concrete is one in whichthere is a Kepler Packing and where> the portland cement is in between the voids of the Kepler Packing. I wonder if someone can arrange for a Kepler Packing of steel BBs andinfuse the crevasses between the> BBs or take marble size steel balls and Kepler Pack them and then infusethe interstatials with> portlandcement. I have the hunch that such a block of concrete would bethe strongest such block over> any other such block having a different arrangement of those BBs and> marbles. And obviously the ratio is no longer a 1 :: 1 because if memoryserves me (mathematician would> know) that in a Kepler Packing that the gaps> are something like 23% of the volume and so the ration of cement toballs> would be closer to that of 1 :: 4> This seems more reasonable to me. I notice that there appears to be a> sparseness of the larger stronger stuff (by volume). Ie, the large> gravel should be 4 over the smaller stuff and cement combined. Instead> it's slightly better than 1. The sand to cement ratio is a little over> 2-1 with pure sand in cement around 3 to 2.> I see a couple of reasons this theoretical limit isn't reached. First,> the concrete when poured needs to be sufficiently fluid to be usable> (and other people have mentioned the various other attributes that> usable concrete needs depending on the application). That appears to> be a significant constraint and may be the sole reason for the low> ratios of large aggregate to the space filling material. Second, the> material as you note is usually randomly packed. I don't know how> inefficient that is. One way to improving the packing would be to> compress the concrete while it's still wet. That would force the> aggregate into a more optimal packing, and squeeze out air and perhaps> some of the water. I seem to recall that this is done for some> applications (and the weight of a large amount of concrete tends to> compress the buried sections).> Karl Hallowell> khallow@hotmail.com electron-dot-cloud are galaxies=== === Subject: : Re: Kepler Packing on concrete mixes Re: theoretically the strongest concrete> compressing concrete is quite common, and done with a vibrator, before it> sets.> also, there are additives that help reduce viscosity or increase fluidity,> while keeping initial water content low (but still enough water to react),> similar to what is used in potter / ceramics.> E.VI wonder if sand grains are mostly flat faced and not roundish? The KeplerPacking deals with spheres and it is the densest packing with 26% void gaps.Does anyone know what the usual or average sand roundness or flatness is?I suppose one can get gravel that is mostly roundish. I know the gravel ofquartzite to make a concrete mix is mostly angular with alot of flatlikesurfaces.Perhaps sand is much like this quartzite in having many flat surfaces. So thatwhen one mixes a concrete mix of sand and aggregate gravel that many of theflat surfaces touch and leaving little if any room for portland cement to gluetogether those flat surfaces. And that the Kepler Packing of spheres with itsdensity of 74% and 26% gaps for a glue. That in a sand and or gravel mix ofconcrete or asphalt that the 26% gaps need not have a glue because they aremostly flat surfaces touching one another. So I suppose the 26% can bedrastically reduced in these mixes because of flatnessprevent these posts of mine from appearing on all outside ISP, except for myown ISP of dtgnet. So am double posting.Archimedes Plutonium, a_plutonium@hotmail.comwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies=== === Subject: : Re: Kepler Packing on concrete mixes Re: theoretically the strongest concreteIf you use round sand the concrete will be weak. Sand likehttp://www.couger.com/microscope/carl/sand.jpg makes lousy concrete.Gordon> compressing concrete is quite common, and done with a vibrator, beforeit> sets. also, there are additives that help reduce viscosity or increasefluidity,> while keeping initial water content low (but still enough water toreact),> similar to what is used in potter / ceramics. E.V I wonder if sand grains are mostly flat faced and not roundish? The Kepler> Packing deals with spheres and it is the densest packing with 26% voidgaps.> Does anyone know what the usual or average sand roundness or flatness is?> I suppose one can get gravel that is mostly roundish. I know the gravel of> quartzite to make a concrete mix is mostly angular with alot of flatlike> surfaces.> Perhaps sand is much like this quartzite in having many flat surfaces. Sothat> when one mixes a concrete mix of sand and aggregate gravel that many ofthe> flat surfaces touch and leaving little if any room for portland cement toglue> together those flat surfaces. And that the Kepler Packing of spheres withits> density of 74% and 26% gaps for a glue. That in a sand and or gravel mixof> concrete or asphalt that the 26% gaps need not have a glue because theyare> mostly flat surfaces touching one another. So I suppose the 26% can be> drastically reduced in these mixes because of flatnessto> prevent these posts of mine from appearing on all outside ISP, except formy> own ISP of dtgnet. So am double posting.> Archimedes Plutonium, a_plutonium@hotmail.com> whole entire Universe is just one big atom where dots> of the electron-dot-cloud are galaxies=== === Subject: : Re: S^n is not contractible> Proposition: If the orientable manifold X is contractible,> then any map f: X --> X has a fixed point.boy does this statement need a compactness hypothesis....> tom Yeah ... sure does ... Sorry 'bout that -- I was a bit careless. Thinking about this in the shower this morning, I *believe* that I came up with a proof that _does_ only need the Brouwer Fixed Point Theorem: Suppose (for the purpose of reaching a contradiction) that S^n were contractible. Let F : S^n x I --> S^n be a homotopy connecting the identity map on S^n and the map sending everything to the north pole. Collapsing S^n x {1} to a point, we get an induced map F^* : D^{n+1} --> S^n. Composing this map with the inclusion of S^n as the boundary of D^{n+1} and folowing _that_ by a rotation through (say) a right angle, we end up with a map D^{n+1} --> D^{n+1} that has _no_ fixed points, contradicting Brouwer's theorem. []=== === Subject: : Re: S^n is not contractible> Proposition: If the orientable manifold X is contractible,> then any map f: X --> X has a fixed point.boy does this statement need a compactness hypothesis....> tom Yeah ... sure does ... Sorry 'bout that -- I was a bit> careless. Thinking about this in the shower this morning, I *believe*> that I came up with a proof that _does_ only need the Brouwer> Fixed Point Theorem: Suppose (for the purpose of reaching a contradiction) that> S^n were contractible. Let F : S^n x I --> S^n be a homotopy> connecting the identity map on S^n and the map sending> everything to the north pole. Collapsing S^n x {1} to a> point, we get an induced map F^* : D^{n+1} --> S^n.> Composing this map with the inclusion of S^n as the boundary> of D^{n+1} and folowing _that_ by a rotation through (say)> a right angle, we end up with a map D^{n+1} --> D^{n+1} that> has _no_ fixed points, contradicting Brouwer's theorem. [] Oops ... replace that rotation by the antipodal map :-( Then you do end up with a fixed-point-free map D^{n+1} --> D^{n+1}.=== === Subject: : Re: S^n is not contractible> Proposition: If the orientable manifold X is contractible,> then any map f: X --> X has a fixed point.boy does this statement need a compactness hypothesis....> tomNot to mention a continuity hypothesis. Stern=== === Subject: : Re: S^n is not contractible> Proposition: If the orientable manifold X is contractible,> then any map f: X --> X has a fixed point.boy does this statement need a compactness hypothesis....> tomNot to mention a continuity hypothesis.> Nope -- I've already admitted to the compactness blunder, but ... in topologist-speak, maps _are_ continuous ...> Stern=== === Subject: : Find a Group of Unitary MatrixThe question is:I need to find M unitary matrix, U_1 to U_M. The definition of unitarymatrix is:U_i * U^H_i = I, where H is the complex transpose.The objective is to maximum the minimum determinant of U_i-U_j, wherei,j is in[1,M], and i neq j.=== === Subject: : Idempotent matrices over GF(2)Is there a formula for the number of n x n idempotent matrices over GF(2) ?(matrices A such that A^2 = A ) .=== === Subject: : Re: Idempotent matrices over GF(2)> Is there a formula for the number of n x n idempotent matrices> over GF(2) ?> (matrices A such that A^2 = A ) .Such a matrix is similar to a diagonal matrix I_rwith r 1's and (n-r) 0's down the diagonal.Work out the number M of matrices commuting with I_r,and the number N of non-singular matrices.Then the number similar to I_r is N/M .-- Timothy Murphy e-mail: tim@birdsnest.maths.tcd.ietel: +353-86-233 6090s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland=== === Subject: : Sum of integer powersI know that the sum of the integers, call it S1, is given byS1=n*(n+1)/2.Similarly, S2, the sum of the squares of the integers is given byn*(n+1)*(2n+1)/6. And S3, the sum of the cubes of the integers isgiven by n*n*(n+1)*(n+1)/4.My question is: Is there a general expression for SN, the sum of theintegers raised to the power N ?Ian Taylor=== === Subject: : Re: Sum of integer powersIan Taylor escribi.97 en elmensaje|n2eefbf19.0309211106.5c66a88f@posting.google.com:> I know that the sum of the integers, call it S1, is given by> S1=n*(n+1)/2.> Similarly, S2, the sum of the squares of the integers is given by> n*(n+1)*(2n+1)/6. And S3, the sum of the cubes of the integers is> given by n*n*(n+1)*(n+1)/4.> My question is: Is there a general expression for SN, the sum of the> integers raised to the power N ?> Ian TaylorLetS(n, k) = 1^k + 2^k + ... + n^kThen(1+1)^(k+1) = 1^(k+1)+(k+1)*1^k + Comb(k+1, 2)*1^(k-1) + ... + 1(2+1)^(k+1) = 2^(k+1)+(k+1)*2^k + Comb(k+1, 2)*2^(k-1) + ... + 1.....(n+1)^(k+1) = n^(k+1)+(k+1)*n^k + Comb(k+1, 2)*n^(k-2) + ... + 1Adding all the equations and canceling terms, you get(n+1)^(k+1) = 1 + (k+1)Sum(n, k) + Comb(k+1, 2)Sum(n, k-1) + ... + nandS(n, k)=((n+1)^(k+1) - (n+1) - Sum(Comb(k+1, k+1 - i)S(n, i), i, 1,k-1))/(k+1)Sou, you can determine S(n, k) from S(n, 1), S(n, 2), ..., S(n, k-1).--Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com=== === Subject: : Re: Sum of integer powers> I know that the sum of the integers, call it S1, is given by> S1=n*(n+1)/2.> Similarly, S2, the sum of the squares of the integers is given by> n*(n+1)*(2n+1)/6. And S3, the sum of the cubes of the integers is> given by n*n*(n+1)*(n+1)/4.> My question is: Is there a general expression for SN, the sum of the> integers raised to the power N ?> Ian TaylorConduct a search about Bernoulli Polynomials ; they providethe answer.Cheers, ZVK(Slavek).=== === Subject: : Fun with physics on 1000 pages - for free downloadThe free physics text available on http://www.motionmountain.nethas been reworked. It now contains 550 problems, 200 drawings, 70 tables,and a new section on nuclear physics with the story of radioactivity, a MRIscan of two humans making love and the dream of grand unification. It alsoincludes new figures on special relativity, a short explanation ofk-calculus, the paradox of the relativistic submarine, a photograph anddescription of how some caterpillars shoot away their faecal matter, aphotograph of a basilisk running over water, and much more.For all fields of physics the newest research results and the mainunanswered questions are presented. The text is a structured walk throughclassical physics, relativity, quantum theory and unification. The readergets an overview of what motion is and what it can effect, keeping theaccent on surprises and thought-provoking puzzles.Enjoy. Christoph Schiller P.P.S. Any help on (preferably colour postscript) images tobe added to the text (with permission of course) is much appreciated.=== === Subject: : Re: Fun with physics on 1000 pages - for free downloadX-No-Archive: yes>The free physics text available on> http://www.motionmountain.net>has been reworked. It now contains 550 problems, 200 drawings, 70 tables,>and a new section on nuclear physics with the story of radioactivity, a MRI>scan of two humans making love and the dream of grand unification. (snip)Roger=== === Subject: : Re: Counterexample to FLT by modular arithmetic>I found a way to calculate a^n mod c and b^n mod c quickly without computing>the>same Mathcad worksheet to verify that the cycle lengths are always a factor>of>c-1, at least with a, b, and c distinct primes.That is,i = 1 to cal.i = a^n mod c is bad because it can cause an overflow.Butal.0 = 1 andal.i+1 = (al.i * a) mod cHow strange. Yet it works when a and c are distinct primes, and may work when aand c are coprime; neither, one, or both composite but not sharing any factorother than one, which is a legitimate value for a or both a and b. 1^1 + 1^1 =2^1 is the smallest non trivial solution I see, and my argument in FLT bymodular arithmetic was that n < = c, so n < = 2. But this has the status of acoincidence at the moment.Now, it is possible to reduce any Pythagorean triangle and eliminate a commonfactor, but I know 1,1,2 is not a Pythagorean triangle. The smallest is 3,4,5which has a period of four in a^n mod c and b^n mod c. Starting at zero, thistakes us to n =3, so there is no help there.It's interesting that 3,4, and 5 are pairwise coprime, but 4 is not a prime.Also, since (a^n + b^n) ^ (1/n) > (a^(n+1) + b^(n+1)) ^ (1/(n+1)), then c neednot ever be greater than a + b, corresponding to n = 1, since n = 0 has onlyone candidate, 1 + 1 = 1, and this is not a solution to FLT.For example, 5^1 + 7^1 = 12^1, a valid solution in my book, while 5^1 + 7^1 #11^1, a candidate for higher n. But 5^2 + 7^2 < 11 ^ 2, so at n =2, thiscandidate fails irretreivably, not needing any further testing.The value of n for a, b, c, = a, b, b is infinity, so c > b. It is interestingto determine for which a,b,c = a,b,b we find that n < c, since only those casesneed be tested. I haven't done this yet.Yours,Doug Goncz (at aol dot com)Replikon Research Read the RIAA Clean Slate Program Affidavit and Description at http://www.riaa.org/I will be signing an amended Affidavit soon.=== === === Subject: : Re: approximate infinityX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS>anyone know how I can approximate where this function is effectively>zero?What do you mean by effectively zero? And is this a homeworkassignement?>I know at some point its zero though.No, it's never zero.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org=== === Subject: : Re: order of integrationtook me a while but finally understood how this works.thanks for the help.why haven't i drawn that picture the first second.. ;-)Zolee> hello again,could please someone explain the steps of swapping the integration order> so that> 2 2x> S S f dy dx> 1 xThe steps are: (1) draw a picture and look at the shape of the integration> area.In this case, the first integration over y, from> x to 2x, means that you are integrating vertical slices> from the line y = x (starts at the origin, goes up and> to the right at 45 degrees) to the line y = 2x (goes> through the origin, rises at a steeper angle).The second integration says that you integrate those> vertical lines from x=1 to x=2. Draw those two lines.So the area of integration is the quadrilateral bounded> by those four lines. It is a trapezoid with parallel> vertical edges on left and right and sloping edges> on top and bottom.(2) See how you would describe the same area first by> the boundaries of a horizontal line (your c and d> below) and then by what the range of y values of those> horizontal lines is (a and b).> becomes b d> S S f dx dy> a c? how do i get a,b,c,d?The second part is easy. Imagine sweeping a horizontal> ruler from bottom to top. What y value is it at the> bottom? What y value is it at the top?The bottom corner is where x=1 crosses y=x, that is> at (1,1). The upper corner is where x=2 crosses y=2x,> that is at (2,4). (There is no way to see this unless> you draw the picture). So that tells you the lowest> and highest values of y,What are the limits on x? Well,if I'm near the bottom,> my horizontal line goes from the edge x=1 to y=x. That> is, for any y, it starts at 1 and continues to x=y.> So the limits are 1 to y.But not at the top. There, the limits are from y/2 to 2> (do you know why?). Since no one description works for> the whole range of y, you have to break this integral> up into chunks of different description. How many chunks?> Draw the picture and write down the coordinates of all> 4 corners. Then we can proceed. - Randy=== === Subject: : Re: Decidability of diophantine equations> I'm trying to get a better understanding of the meaning of> Matijasevic's> negative solution for Hilbert's 10th. Specifically, I'm trying to> understand> the relation between this result and various axiomatizations of> arithmetic.Page 6+ of http://uk.arxiv.org/PS_cache/math/pdf/0210/0210327.pdfcontains what seems like quite an accessible summary of Matijasevic'stheorem and related matters.Cheers------------------------------------------------ ---------------------------John R Ramsden (jr@adslate`rm -rf *`.com) (remove spambot trap before use)---------------------------------------------------------- -----------------A stockbroker is someone who invests your money until it's all gone. Woody Allen=== === Subject: : Re: Decidability of diophantine equations1. Since 2nd-order PA is categorical, if a statement cannot be proved> within it then it does *not* follow that there are two models of 2PA,> one in which it's true and one in which it's false - right? Or is> there a subtlety here that I'm missing?I think the subtlety that your missing is the fact that> there is no analogue of Godel's completeness theorem> for second-order logic.models, but that I shouldn't expect there to be becuase we don't havecompleteness as in first-order logic.So if I work in first-order PA, I get a lot of non-standard models forthe natural numbers and number-theoretic sentences may turn out to beindependent simply because they are true in some models and false inothers. OTOH, if I work in second-order PA there is only one model,but regrettably not everything that is semantically true can besyntactically proven.Is this a fair summary of the situation? In my naive view, it meansthat there are really two slightly different kinds of independence,one which entails multiple incompatible models, and one which doesnot. But perhaps, again, I'm muddling things - I'll be happy to becorrected.=== === Subject: : Re: Decidability of diophantine equations|So if I work in first-order PA, I get a lot of non-standard models for|the natural numbers and number-theoretic sentences may turn out to be|independent simply because they are true in some models and false in|others. OTOH, if I work in second-order PA there is only one model,|but regrettably not everything that is semantically true can be|syntactically proven.It's better to say that there isn't a notion of syntactically provablefor second-order PA. For both first order logic and second order logic,one has a concept of a sentence being satisfied by a model (or moregenerally, a formula being satisfied by some assignment to its freevariables). Being a model of a set of axioms just means that it satisfiesall of them.For first-order logic, it's possible to produce a complete set ofdeductive rules. So given a set of first-order axioms like the axioms ofPA, it provides a notion of formal proof from those axioms, having theproperty that every sentence which is true in all models of the axiomsis provable using those rules from the axioms (and conversely).Second-order logic is a very different animal. One doesn't have such athing as the deductive rules for second-order logic.Keith Ramsay=== === Subject: : Re: Decidability of diophantine equations> It's better to say that there isn't a notion of syntactically provable> for second-order PA. Just to confuse matters further, second order PA may also refer toa theory that does have an associated notion of syntacticallyprovable: the theory obtained from PA by adding variables for setsand a comprehension schema asserting the existence of sets ofnatural numbers definable in the language of the theory, and alsoinduction axioms for all properties of sets and numbers expressiblein the theory. This second order PA is equivalent to a first ordertheory with predicates is a set, is a number, and is a verystrong theory, in which large parts of mathematics can be formalized.Simpson's book Subsystems of Second Order Arithmetic is aboutsubsystems of this theory.=== === Subject: : Re: Decidability of diophantine equationsIt's better to say that there isn't a notion of syntactically provable> for second-order PA. Just to confuse matters further, second order PA may also refer to> a theory that does have an associated notion of syntactically> provable: the theory obtained from PA by adding variables for sets> and a comprehension schema asserting the existence of sets of> natural numbers definable in the language of the theory, and also> induction axioms for all properties of sets and numbers expressible> in the theory. This second order PA is equivalent to a first order> theory with predicates is a set, is a number, and is a very> strong theory, in which large parts of mathematics can be formalized.> Simpson's book Subsystems of Second Order Arithmetic is about> subsystems of this theory.the 2nd order PA I had in mind. This first-orderish version of 2ndorder PA you describe sounds very interesting, and I'll look it up inSimpson's book.=== === Subject: : Euclid's construction of the dodecahedron ...... is one of the most beautiful and accessible achievements of the human mind I can think of, and should be presented in many math (and not only)classes instead of being banished to sourcses such as Heath's Elements(1956 edition, vol. 3, pp. 493-503): while the mathematical details may be intriguing to most students, a hands-on approach might thrill many of them:Start with a cube and built a 'house' on top of each side consisting of *five equal segments*, one 'roof' parallel to two of its edges and four 'beams' attached to its four vertices; just make sure that the 'roofs' of every two adjacent sides are *perpendicular* to each other, and there youget your dodecahedron!The process is illustrated in http://cage.rug.ac.be/~hs/polyhedra/dodeca.htmland also discussed (with icosahedral version) in Peter Cromwell's Polyhedraa p4g tiling recorded in http://mathforum.org/dynamic/one-corona/4.html as3.3.4.3.4(1) and also in http://www.oswego.edu/~baloglou/103/dodeca-2.html &http://www.oswego.edu/~baloglou/103/dodeca-4.html (in color) ... and askwhether or not the Arabs knew this tiling, be it through their Euclid studiesor symmetric/pentagonal tiling explorations. baloglouAToswego.edu=== === Subject: : Re: A function continuous at only one point?X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 10:06 PM, Topy said:>What is an example of a function f : R-->R that is continuous at>precisely one point?F(n) = 0 if n is rationalF(n) = 1/n if n is irrational-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org=== === Subject: : Re: A function continuous at only one point?> at 10:06 PM, Topy said:>What is an example of a function f : R-->R that is continuous at>precisely one point?> F(n) = 0 if n is rational> F(n) = 1/n if n is irrationalAt which point is this function continuous?-- Clive Toothhttp://www.clivetooth.dk=== === Subject: : Re: A function continuous at only one point? at 10:06 PM, Topy said: >What is an example of a function f : R-->R that is continuous at>precisely one point? F(n) = 0 if n is rational F(n) = 1/n if n is irrationalAt which point is this function continuous?My suggestion when we were talking about this before was:f(x) = { x, x E Q } { 1/x, x E P }Although I can see howg(x) = { x, x E Q } { -x, x E P }and as wellf(x) = { x, x E P } { 1/x, x E Q }andg(x) = { x, x E P } { -x, x E Q }would all appear to satisfy a point of continuity at f(x)=1 andg(x)=0.This agrees with that the rationals and irrationals are eacheverywhere discontinuous. Otherwise there would be some point orpoints for a function f(x) = x, x E P, presumably, except for that theirrationals are everywhere discontinuous.One thing that that also agrees with is that the rationals andirrationals alternate in the reals. I'm not saying necessarily thatthe fact that the complement of the rationals in the reals is the (setof) irrationals, and that the union of the rationals and theirrationals is the reals, and that the reals are everywhere continuousand the rationals and irrationals are each nowhere continuous provesthat the rationals and irrationals alternate in the reals, but itmight actually be sufficient logic to imply that conclusion in avariety of models that are useful for other generally acceptedstatements about the reals and continuity.I'm trying to think of a function that is almost continuous. Forexample floor(x) + x for rationals and 1/x for irrationals.Then, the limit from above at x=1, lim_x->1^+ of f(x) = 2 isdifferent than the limit from below and f(x) where it is defined,lim_x->1^- of f(x) = 1. That would have x=1 being an endpoint of f(x)of a segment of f(x), which is otherwise nowhere continuous and has nosegments, as there is not defined another endpoint.What that seems to imply is that for a function to be continuous isthat it requires three points in row, bear with me here, but theabove is a description of two points in a row.Then is there is a single point of continuity at f(1)=1 and g(0)=0,what are the endpoints of the continuous line segment? Are theysimply 1^+ and 1^-, that is infinitely approaching but not reachingthe limit from above and below, (1+- dx, 1/(1+- dx) )?Another thing we've talked about here is mapping [0, 1] to (0, 1).Here's another piecewise functionf(x) = { sin(pi x), x E P } { -sin(pi x), x E Q }It's continuous at each integer, f(x)=0, x E Z.If the rationals and irrationals alternate in the reals, what doesthat imply about their propensity in a uniform random dstribution overthe reals?Ross=== === Subject: : Re: A function continuous at only one point?> Topy said: >What is an example of a function f : R-->R that is continuous at>precisely one point? F(n) = 0 if n is rational> F(n) = 1/n if n is irrationalAt which point is this function continuous?Apparently at plus and minus infinity; which in thex-y plane may be treated as one point, but on the number line looks to me like two points.Show that the following is nowhere continuous:f(x) = x for irrational x f(p/q) = 1/q for integers p,q (q>0)-jiw=== === Subject: : Re: A function continuous at only one point?> Topy said:>What is an example of a function f : R-->R that is continuous at>precisely one point?> F(n) = 0 if n is rational> F(n) = 1/n if n is irrationalAt which point is this function continuous?Apparently at plus and minus infinity; which in the> x-y plane may be treated as one point, but on the > number line looks to me like two points.Show that the following is nowhere continuous:f(x) = x for irrational x > f(p/q) = 1/q for integers p,q (q>0)-jiwIt is not even defined at rationals, other than at zero, without the added condition that p and q be coprime.=== === Subject: : Re: A function continuous at only one point?...> Show that the following is nowhere continuous:f(x) = x for irrational x> f(p/q) = 1/q for integers p,q (q>0)For nowhere continuous instead read continuous at only 1 point :)-jiw=== === Subject: : The malady lingers on...Content-transfer-encoding: 8bitAt certain times, certain news groups have semi-official Cranks inResidence.sci.math recently lost a good one. So it seems, and so we may hope. The group has become a lot quieter. I like that. Don't you?We can make it seem as if He's Still Here, by bickering among ourselvesabout who had a more or less effective strategy of dealing with him. Isuggest it would be more fun, and more useful, to revive and continuethe torrent of comment that accompanied the long-heralded demise of theSeinfeld TV show.-- Chris HenrichThose readers who are unacquainted with the mathematical technicalities willfind that they can manage quite well by ignoring them. -- John Nash=== === Subject: : Re: The malady lingers on...In sci.math, Christopher J. Henrich Residence.sci.math recently lost a good one. So it seems, and so we may hope. > The group has become a lot quieter. I like that. Don't you?We can make it seem as if He's Still Here, by bickering among ourselves> about who had a more or less effective strategy of dealing with him. I> suggest it would be more fun, and more useful, to revive and continue> the torrent of comment that accompanied the long-heralded demise of the> Seinfeld TV show.> Feh. (Or, to keep in character, yada, yada, yada.)Personally, I don't mind a little give-and-take, especially ifone gets into interesting math territory. Unfortunately, forthat Certain Special Someone, it wasn't all that interesting. :-)I'm hoping his Mathematical Prime-Counting Fiasco embarrassedhim enough to at least rethink *not* doing *that* again.(Or at least not to do it half-cocked.)We'll see.As for Seinfeld: never watched it. :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: The malady lingers on...Christopher J. Henrich> At certain times, certain news groups have semi-official Cranks in> Residence.> sci.math recently lost a good one. So it seems, and so we may hope.> The group has become a lot quieter. I like that. Don't you?You miss him. Admit it :) The trouble is, he'll be back. What other forumcan he find? Megalomaniacs.com? That possibility is ruled out by Hammick'sExclusion Principle, which asserts that no group can accomodate more thanone prima donna at the same time.LH=== === Subject: : Re: The malady lingers on...> At certain times, certain news groups have semi-official Cranks in> Residence.> sci.math recently lost a good one. So it seems, and so we may hope.> The group has become a lot quieter. I like that. Don't you?> We can make it seem as if He's Still Here, by bickering among ourselves> about who had a more or less effective strategy of dealing with him. I> suggest it would be more fun, and more useful, to revive and continue> the torrent of comment that accompanied the long-heralded demise of the> Seinfeld TV show.> --> Chris Henrich> Those readers who are unacquainted with the mathematical technicalitieswill> find that they can manage quite well by ignoring them. -- JohnNashDon't worry, that crank has been on and off for over seven years.He will be back with a vengence.He is probably solving the millenium problems:| Birch and Swinnerton-Dyer Conjecture | Hodge Conjecture | Navier-StokesEquations | P vs NP | Poincare Conjecture | Riemann Hypothesis | Yang-MillsTheory |And he will tell us that we are a bunch of cheating, lying, stealing, andkeeping him from his much deserved fame and fortune!Mark my words!=== === Subject: : Re: The malady lingers on...Injector-Info: news.mailgate.org; posting-host=adsl-67-119-172-150.dsl.frsn01.pacbell.net; posting-account=48257; posting-date=1064187623X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ a670d15e1ed60a7191731bbb11a7f4d2.48257%40mygate.mailgate.org>> At certain times, certain news groups have semi-official>> Cranks in Residence.Shouldn't there be a FAQ listing these innews.announce.newusers?I claim talk.bizarre, with 18 year's tenure!>> sci.math recently lost a good one. So it seems, and so>> we may hope. The group has become a lot quieter. I like>> that. Don't you?It would stay that way if when he comes back, no one answershim. He's only here for exactly what you dislike about himbeing here: the long, pointless battle of the forces ofknowledge against the avatar of invincible ignorance.The only difference between you-plural is the opinion on whofills which roles.>> We can make it seem as if He's Still Here, by bickering>> among ourselves about who had a more or less effective>> strategy of dealing with him.But of course. The above is my contribution.>> I suggest it would be more fun, and more useful, to>> revive and continue the torrent of comment that>> accompanied the long-heralded demise of the Seinfeld TV>> show.Who watches television?> Don't worry, that crank has been on and off for over seven> years.Piker.> He will be back with a vengence.> He is probably solving the millenium problems:> | Birch and Swinnerton-Dyer Conjecture> | Hodge Conjecture> | Navier-Stokes Equations> | P vs NP> | Poincare Conjecture> | Riemann Hypothesis> | Yang-Mills TheoryAnd won't it be nice to have all of those taken care of forus to simplify our lives?> And he will tell us that we are a bunch of cheating,> lying, stealingYou dropped this: -+*mathematicians*+-> , and keeping him from his much deserved fame and fortune!> Mark my words!Oh, your I told you so is already credited to you inadvance in Valhalla; of course, you have to die in battle toclaim it.xanthian.-- Posted via Mailgate.ORG Server - http://www.Mailgate.ORG=== === Subject: : Re: Fraud in Computer Science Publishing 3. The fixed point theorem. (for any recursive f there is a value> such that this value and the result of applying f to this value> produce functionally equivalent programs)> (Af e R)(Ex e w)(Ay e w) (Ez e w) [{f(x)}(y) = z & {x}(y) = z] v [(Az> e w) {f(x)}(y) != z / (Az e w) {x}(y) != z]> R is the set of all recursive one-place functions; {x} is the machine> with code x.Yes. (See below.)> And what is the formal representation of a proof of each of these> theorems?> This is left as an exercise to the reader.That is the whole problem. You can give names to concepts like the> recursive one-place functions and Turing Machines, but can you use> that syntax to formally manipulate these concepts? Can you generate> theorems? Can you (partially) decide if a given wff is a theorem? Not really; the Predicate Calculus is too weak to be useful in> practive.> Of course, I could construct a proof of these theorems in the> calculus, but it would too much hassle.That is what is missing from what has been published on the problems> of theoretical computer science. There is no formal derivation of> theorems such as these.I would call that a problem of mathematics, not computer science. If> you need a computer to formalize proofs, you aren't doing it right.You need a computer to verify that your theory isn't BS.The problem is that researchers try to represent metamathematicalassertions in mathematics. The predicate calculus (PC) can be used toexpress mathematical assertions, but a function of PC is needed toexpress metamathematical assertions.Examples: ZMC is really about metamathematics, but PC is used toexpress the axioms. This is a mistake. The result is unnecessarilydifficult to manipulate. Likewise, assertions like the unsolvabilityof the halting problem are sometimes expressed as PC wffs (as you doabove), where a constant is defined to be the halting problem. Butnumbers and problems are at different levels of abstraction(cardinality.) In both cases, this means that a new constant must bedefined for each problem, and we are adding axioms for each newproblem. We cannot axiomatize the theory (computability) at all. Wehave not captured the definition of the concept (the halting problemor various sets referred to in ZMC). We have only stated isolatedaspects of it. We have to add new axioms to develop additionaltheorems.What I did in my papers is to extend predicate calculus to representmetamathematical assertions. (It's very simple - only the semanticswere augmented - the syntax is the same.) ~TRUE(s) is the empty set. To say (in ZMC) there exists an empty set, researchers use theexpression for there is a set such that for all sets the second setis not an element of the first set. But how would we make additionalassertions about the empty set? We would need to declare a Skolemconstant and thus have the problem of defining a constant for eachproblem or set considered (or not being able to link together themultiple references to it except by combining all of the new axiomswith the existing ones in a giant there exists . . . such that . . .and . . . and . . .)What we need to do is to start with ~TRUE(a) and then express the factthat there is a set for this wff. That is what I have done in mysystem (described in my 2 papers.) Then the theorems are easilyderived, without any additional axioms being needed, by referring tothe wff that actually defines the set and the assertion concerning it,rather than a single property of it (which may or may not uniquelydefine it.)BTW: Is there more than one empty set? There is certainly more thanone universal set. How about a penny - is it the empty set? It hasno elements. How do you define set (including the empty set) andexclude a penny?Charlie VolkstorfCambridge, MA=== === Subject: : complex serieshelloif i have a series in the formsum(z^n)where z is a complex expression,can it be that the limit of the sum is a certain complex number as napproaches infinity?z^n should approach zero as n approaches inf. so the series would beconvergent.z is r(cos phi+isin phi)z^n is r^n(cos n*phi+isin n*phi)in case r is 1 and phi is pi/4if abs(r)<1, r^n -> 0 as n -> inf. that's ok.but for every n, phi changes. so there may or may not be a real orimaginary part for a z^n.is it that no matter what real and imaginary parts a complex numberhas, if they're small enough, it is just zero?tiazolee=== === Subject: : Re: complex series> if i have a series in the form> sum(z^n)> where z is a complex expression,> z is r(cos phi+isin phi)> z^n is r^n(cos n*phi+isin n*phi)z^n -> 0 if r < 1 -> oo if r > 1when r = 1, limit doesn't exsist except when phi = 0=== === Subject: : Re: complex seriesi would have one further question.given the seriessum( (p1*i)/(z-p2*i) )where p1, p2 are integershow do i find a complex z so that the series converges?in other words how do i find a z so that(p1*i)/(z-p2*i) is in the unit circle?tried to set up equations, but i just get all messed up..tiazolee=== === Subject: : Re: complex series>given the series>sum( (p1*i)/(z-p2*i) )>where p1, p2 are integers>how do i find a complex z so that the series converges?limit of the terms as i->infinity is -p1/p2, the series divergesfor all z except in the trivial case p1=0.>in other words how do i find a z so that>(p1*i)/(z-p2*i) is in the unit circle?Not the same question at all. And since the limit as i->infinityis -p1/p2, this will be inside the unit circle for all but finitelymany i if |p1|<|p2| and outside the unit circle for all but finitelymany i if |p1|>|p2|. If |p1|=|p2|, |p1*i/(z-p2*i)| < 1 for all i if |arg(z/p2)| >= pi/2. But somehow I doubt that this isthe question you wanted to ask either. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: complex series> if i have a series in the form sum(z^n) where z is a complex expression,> can it be that the limit of the sum is a certain complex number as n> approaches infinity?Yes, that is what it means to be convergent.> z is r(cos phi+isin phi)> if abs(r)<1, r^n -> 0 as n -> inf. that's ok.> but for every n, phi changes. so there may or may not be a real or> imaginary part for a z^n.By the ratio test, your series converges, and the absolute value ofthe sum is less than 1/(1-|r|). > is it that no matter what real and imaginary parts a complex number> has, if they're small enough, it is just zero?That doesn't make any sense. Only 0 is 0.The theory of complex series is hardly any different than the theoryof real series. If you have any intuition about real series, it is agood bet that the same principle will apply to complex series.=== === Subject: : Re: complex series> helloif i have a series in the formsum(z^n)where z is a complex expression,> can it be that the limit of the sum is a certain complex number as n> approaches infinity?z^n should approach zero as n approaches inf. so the series would be> convergent.z is r(cos phi+isin phi)> z^n is r^n(cos n*phi+isin n*phi)in case r is 1 and phi is pi/4if abs(r)<1, r^n -> 0 as n -> inf. that's ok.> That's right, if z is inside the unit circle, r^n-> 0 and so does z^n.But if z is outside the unit circle, r^n goes to infinity and so does z^n. For example z = 1 + i has magnitude sqrt(2), and (1+i)^n goes to infinity. > but for every n, phi changes. so there may or may not be a real or> imaginary part for a z^n.> You're right that each time you multiply by 1 + i, the result moves around and changes the angle. But the magnitude increases each time. The values of z^n spiral outward if |z| > 1. The angle keeps cycling within [0, 2pi) but the radius increases.> is it that no matter what real and imaginary parts a complex number> has, if they're small enough, it is just zero=== === Subject: : Re: complex seriesthanks a lot!!!helloif i have a series in the formsum(z^n)where z is a complex expression,> can it be that the limit of the sum is a certain complex number as n> approaches infinity?z^n should approach zero as n approaches inf. so the series would be> convergent.z is r(cos phi+isin phi)> z^n is r^n(cos n*phi+isin n*phi)in case r is 1 and phi is pi/4if abs(r)<1, r^n -> 0 as n -> inf. that's ok.> That's right, if z is inside the unit circle, r^n-> 0 and so does z^n.But if z is outside the unit circle, r^n goes to infinity and so does > z^n. For example z = 1 + i has magnitude sqrt(2), and (1+i)^n goes to > infinity. but for every n, phi changes. so there may or may not be a real or> imaginary part for a z^n.> You're right that each time you multiply by 1 + i, the result moves > around and changes the angle. But the magnitude increases each time. > The values of z^n spiral outward if |z| > 1. The angle keeps cycling > within [0, 2pi) but the radius increases.is it that no matter what real and imaginary parts a complex number> has, if they're small enough, it is just zero=== === Subject: : Re: Sets.> However, and to paraphrase the title of a movie, Rebellion must be with> a cause - a good cause -> But in this paradigm, the flexibility has a *limit*Cause enough.:-)=== === Subject: : Re: Sets.>>However, and to paraphrase the title of a movie, Rebellion must be with>>a cause - a good cause ->> >>But in this paradigm, the flexibility has a *limit*>>Cause enough.Since the limit is intrinsic to human (in)ability to formalize infinity in its entirety, in the current paradigm, I take it thatyou're proposing to make a revolutionary change to the paradigm. If so, what would be your suggestion as an alternative to ourcurrent mathematics/logic paradigm?---Nam>:-)>> === === Subject: : Re: Sets.>However, and to paraphrase the title of a movie, Rebellion must be with>>a cause - a good cause ->But in this paradigm, the flexibility has a *limit*Cause enough. Since the limit is intrinsic to human (in)ability to formalize infinity> in its entirety, in the current paradigm, I take it that> you're proposing to make a revolutionary change to the paradigm. If so,> what would be your suggestion as an alternative to our> current mathematics/logic paradigm?I know you know what mereology is.Look up Pervin quasi-uniformities. You will find that they relate the(part-whole) and (part-excluded part) relationships according to acoordinatization (Look up the relevance of that term in Birkhoff's LatticeTheory.). Moreover, the specific form of the entourages suggests whywell-founded sets can be specified without reference to a whole.The reason logic has compactness theorems is because language istopological. What makes anyone think that any lower logical order could befoundational?:-)=== === Subject: : Sets -the alternativeThe set of:'Two vegeburgers with medium fries please.And some of those sauce sauchets, the nice ones.No. I gave you a tenner.' 's.JJ=== === Subject: : Re: Sets.> So the issue of expressing infinity, imho, is not exactly> the reason why we should make a radical change to> the notion of mathematics, and logic, whether or not> we're happy with the current way of doing mathematics.Namduc======Yes - a good point. Revolutions should, indeed, be made of sternerstuff!So long as we treat the concept of infinity as essentiallyindicating the presence of a well-defined non-terminating process, andprovide effective, and unambiguous, methods for determining which ofthese processes we define as mathematical objects (e.g., Cauchysequences), the fact that the outpput of the process itself may becompletely representable individually, but not as a completedtotality, within a mathematical language should not, by itself, beconsidered sufficient grounds for treating the mathematical object, orits introduction, as intuitionistically unacceptable within thelanguage.Bhup=== === Subject: : Re: Sets.> indicating the presence of a well-defined non-terminating process,Its not that 'defined non-terminating' diviner again? Into the ring with it.JJ> So the issue of expressing infinity, imho, is not exactly> the reason why we should make a radical change to> the notion of mathematics, and logic, whether or not> we're happy with the current way of doing mathematics.> Namduc> ======> Yes - a good point. Revolutions should, indeed, be made of sterner> stuff!> So long as we treat the concept of infinity as essentially> indicating the presence of a well-defined non-terminating process, and> provide effective, and unambiguous, methods for determining which of> these processes we define as mathematical objects (e.g., Cauchy> sequences), the fact that the outpput of the process itself may be> completely representable individually, but not as a completed> totality, within a mathematical language should not, by itself, be> considered sufficient grounds for treating the mathematical object, or> its introduction, as intuitionistically unacceptable within the> language.> Bhup=== === Subject: : Re: MS in Math or Comp Sci??...> piggybacki have some bad news for you.unless you really love math and or comp.sci for their own sake, i> advice you keep away from them. advanced degrees in either of those> subjects, especially mathematics, are bound to lead you nowhere> careerwise.I find this quite harsh... My long term goal is to either join a> reputable company in their research department or join academia as a> professor. I disagree that a graduate degree in Comp Sci or> Mathematics will lead me nowhere.i hope you understand that i did not imply that 100 percent ofthose who get master degrees in math and/or comp.sci go nowhere.of course there are notable exceptions.there have been reports that jobs in comp.sci are beingout-sourced overseas. if these reports are accurate, you can restassured that current and future comp.sci degrees are heading tounemployment or new careers. i have personally witnessed asignificant number of unemployed comp.sci people with post bacdegrees going for new careers.now, as far as math goes, it's worse. master degrees in math areessentially worthless - even in academia. those few who make itinto academic positions do so in mediocre positions formediocre salaries under mediocre conditions. do you know whatan adjunct instructor is? if you do not, find out. a lot, i meanA LOT, of math masters end up as such...make of this info what you want. bottom line is that the immediateoutlook for higher degrees in comp.sci and math isn't bright. electron-dot-cloud are galaxies=== === Subject: : censorship of Internet posts Re: blacktop ratio tied to concrete ratio?? Re: Kepler Packing on concrete mixes Re: theoretically the strongest concrete>Luckly some roadwork nearby of blacktop is going on and was able to observe that the underlayer is a pure tar and>then the upper layers seems to be a mix.Jay, I am doubtful of this 4-6% figure as binder. Because looking at asphalt in that it is black in color indicates>that there had to be much more tar constituents than a mere 4-6%. Maybe the binder is a small fraction of the>tar-constituents.So I am wondering if in a cubic centimeter of blacktop that the tar elements comprise much more than 4-6%. Perhaps>close to 30% with the 70% of sand>and gravel.Just the appearance of fresh blacktop suggests more than a mere 4-6%.Perhaps the 4-6% is a tiny fraction of the tar elements.> The 4-6% value I gave you was based on actual data from an asphalt> extraction of asphaltic concrete. I just had the data on 1 mix,> however. You might want to go to www.asphaltinstitute.org for> information straight from the horse's mouth.>I suppose if one heats the tar hot enough that the diffusion of the tar throughout the sand and gravel is much>easier than the diffusion of portlandcement in the concrete matrix. So I am guessing that it is far easier to>insure uniformity of mix with tar than with portland cement.> Personally I would think it would be the other way around, since the> asphaltic emulsion has a higher viscosity than portland cement paste.>Jay, has anyone worked out the cost difference because the cost in energy of oil to heat up limestone in order>>to produce portland cement compared to the cost of getting tar of 4-6%. It seems that the binder of blacktop is>>so much cheaper because of the quantity needed.One of the biggest cost differences between asphalt and concrete is>> that you can put traffic on the asphalt almost immediately, but the>> concrete needs to cure and harden, sometimes for as much as a month.>> I'm not certain, but I think that on a per ton or per cubic yard>> basis, concrete is usually cheaper than asphalt.Well I was asking more on the lines of theory than on the lines of present day commerce. Something of theory that a>cubic meter of portlandcement requires>the heat from petroleum of say (I am guessing) 4 barrels of oil. Whereas the oil required to make a cubic meter of>tar requires 3 barrels of oil and where the tar>of a cubic meter can surface one kilometer of highway yet a cubic meter of portland cement could only pave 1/100 of>a kilometer of highway. I think people have done that, but I don't know who.>>However, there is a constant and that constant is the fact of packing aggregate as per Kepler Packing and as>>per a 3-d chessboard. And after packing them in a Kepler Packing is to fill the voids with either>>portlandcement or with asphalt-tar and then compare the strength of each.Remember that in asphalt not all the voids are filled. This is part of>> the asphalt design.I can accept that but I cannot seem to accept the idea that 4-6% of volume to>hold together all that gravel and sand.>Do you mind telling me what your particular interest in this is? Is it>> an academic question or are you trying to address a specific>> application? If you are responding to an academic problem, you can>> probably assume away many of the variables. If you are responding to a>> real-life problem, I would be happy to refer you to sources that could>> provide you with a lot better education in asphalt than I can.I find much satisfaction and enjoyment in tackling theoretical issues. I go from>one problem to another and often come back to a problem. Some how I landed>on concrete, blacktop and Kepler Packing in the last few days. I am building a>concrete block garage and that has focused my attention.>Jay ShilstonePerhaps the Kepler Packing Problem is entwined in this concrete and blacktop in the idea that KPP is the minimum>amount of binder to make a solid mass. If one can imagine a KPP with its 74% touching balls and its 26% holes or>gaps in between, and if those 26% holes were filled with tar or portland cement then the entire material is one>piece solid and very strong.But ordinary concrete and ordinary blacktop are never uniform balls that touch at kissing points but at all sorts>of angles and flat pieces with air pockets and no binder.If one could take BBs and pack them Keplerian with a density of 74% and fill the remaining 26% with tar or with>portlandcement then they would achieve a solid that is very strong. Such Keplerian Packed concrete or Keplerian>Packed>blacktop would be a minimum solid because the Kepler Packing is a maximum>density.Experiment: to make a Kepler Packing of BBs and fill the gaps with tar and do the same with a concrete of Kepler>Packing. I wonder what the characteristics of the two would be. And of course neither one of these materials have>ever been created. And it should be easier to create the KPP of blacktop for I do not see how to infuse a KPP with>portland cement. It wouldn't do to infuse the Keplerian Packing with just portland> cement. You need to infuse it with cement and water paste. That should> be very easy to do, depending on the ratio of water to cement. More> water means lower viscosities, but also lower strength of the paste.> Jay ShilstoneHi Jay, I am having a hard time of believing this 4-6% tar binding in typical road asphalt. Hard believing because itis so black that it must have more than 6% tar. And due to the theoretical notion that the densest of packing whetherthey be oranges in a grocery store or baseballs or sand grains that are somewhat roundthat the densest packing is a Kepler Packing and so the percentage of volume ofgaps to glue together those sand grains would be 26%. Now I realize that a concrete mix of portlandcement and sand thatmany of those gaps in the sand will have no portlandcement.What I am saying is the the Minimum Strongest Perfect concrete of sand that is round would be of a 74% sand and 26%portland cement filling each of those gaps of the Kepler Packing.So I am having a difficult time of envisioning highway asphalt as being only a mere 6% by volume of a tar binder.Granted that the gravel used in asphalt and concrete is not round shaped but having many flat surfaces where one flatsurface rests upon another relatively flat face surface and so you would need less tar than the Kepler Packing amountof 26%. But I cannot envision that the flat surfaces would reduce the 26% down to a mere 6%.Perhaps there is something about the term binder of a 6% binder that I am missing in that the tar consists of abinder plus some other tar ingredients.P.S. I am having trouble in getting posts to the Internet in that they appear on my ISP but not on any other ISP.Archimedes Plutonium, a_plutonium@hotmail.comwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies=== === Subject: : Re: blacktop ratio tied to concrete ratio?? Re: Kepler Packing on concrete mixes Re: theoretically the strongest concrete=== === Subject: : censorship of Internet posts Re: blacktop ratio tied toconcrete ratio?? Re: Kepler Packing on concrete mixes Re: theoretically the strongest concrete Archimedes Plutonium NOdtgEMAIL whole entire Universe is just one big atom where dots ofthe electron-dot-cloud are galaxies sci.materials, sci.engr, sci.math 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13>Luckly some roadwork nearby of blacktop is going on and was able to observe that the underlayer is a pure tar and>then the upper layers seems to be a mix.Jay, I am doubtful of this 4-6% figure as binder. Because looking at asphalt in that it is black in color indicates>that there had to be much more tar constituents than a mere 4-6%. Maybe the binder is a small fraction of the>tar-constituents.So I am wondering if in a cubic centimeter of blacktop that the tar elements comprise much more than 4-6%. Perhaps>close to 30% with the 70% of sand>and gravel.Just the appearance of fresh blacktop suggests more than a mere 4-6%.Perhaps the 4-6% is a tiny fraction of the tar elements.> The 4-6% value I gave you was based on actual data from an asphalt> extraction of asphaltic concrete. I just had the data on 1 mix,> however. You might want to go to www.asphaltinstitute.org for> information straight from the horse's mouth.>I suppose if one heats the tar hot enough that the diffusion of the tar throughout the sand and gravel is much>easier than the diffusion of portlandcement in the concrete matrix. So I am guessing that it is far easier to>insure uniformity of mix with tar than with portland cement.> Personally I would think it would be the other way around, since the> asphaltic emulsion has a higher viscosity than portland cement paste.>Jay, has anyone worked out the cost difference because the cost in energy of oil to heat up limestone in order>>to produce portland cement compared to the cost of getting tar of 4-6%. It seems that the binder of blacktop is>>so much cheaper because of the quantity needed.One of the biggest cost differences between asphalt and concrete is>> that you can put traffic on the asphalt almost immediately, but the>> concrete needs to cure and harden, sometimes for as much as a month.>> I'm not certain, but I think that on a per ton or per cubic yard>> basis, concrete is usually cheaper than asphalt.Well I was asking more on the lines of theory than on the lines of present day commerce. Something of theory that a>cubic meter of portlandcement requires>the heat from petroleum of say (I am guessing) 4 barrels of oil. Whereas the oil required to make a cubic meter of>tar requires 3 barrels of oil and where the tar>of a cubic meter can surface one kilometer of highway yet a cubic meter of portland cement could only pave 1/100 of>a kilometer of highway. I think people have done that, but I don't know who.>>However, there is a constant and that constant is the fact of packing aggregate as per Kepler Packing and as>>per a 3-d chessboard. And after packing them in a Kepler Packing is to fill the voids with either>>portlandcement or with asphalt-tar and then compare the strength of each.Remember that in asphalt not all the voids are filled. This is part of>> the asphalt design.I can accept that but I cannot seem to accept the idea that 4-6% of volume to>hold together all that gravel and sand.>Do you mind telling me what your particular interest in this is? Is it>> an academic question or are you trying to address a specific>> application? If you are responding to an academic problem, you can>> probably assume away many of the variables. If you are responding to a>> real-life problem, I would be happy to refer you to sources that could>> provide you with a lot better education in asphalt than I can.I find much satisfaction and enjoyment in tackling theoretical issues. I go from>one problem to another and often come back to a problem. Some how I landed>on concrete, blacktop and Kepler Packing in the last few days. I am building a>concrete block garage and that has focused my attention.>Jay ShilstonePerhaps the Kepler Packing Problem is entwined in this concrete and blacktop in the idea that KPP is the minimum>amount of binder to make a solid mass. If one can imagine a KPP with its 74% touching balls and its 26% holes or>gaps in between, and if those 26% holes were filled with tar or portland cement then the entire material is one>piece solid and very strong.But ordinary concrete and ordinary blacktop are never uniform balls that touch at kissing points but at all sorts>of angles and flat pieces with air pockets and no binder.If one could take BBs and pack them Keplerian with a density of 74% and fill the remaining 26% with tar or with>portlandcement then they would achieve a solid that is very strong. Such Keplerian Packed concrete or Keplerian>Packed>blacktop would be a minimum solid because the Kepler Packing is a maximum>density.Experiment: to make a Kepler Packing of BBs and fill the gaps with tar and do the same with a concrete of Kepler>Packing. I wonder what the characteristics of the two would be. And of course neither one of these materials have>ever been created. And it should be easier to create the KPP of blacktop for I do not see how to infuse a KPP with>portland cement. It wouldn't do to infuse the Keplerian Packing with just portland> cement. You need to infuse it with cement and water paste. That should> be very easy to do, depending on the ratio of water to cement. More> water means lower viscosities, but also lower strength of the paste.> Jay ShilstoneHi Jay, I am having a hard time of believing this 4-6% tar binding intypical road asphalt. Hard believing because itis so black that it must have more than 6% tar. And due to thetheoretical notion that the densest of packing whetherthey be oranges in a grocery store or baseballs or sand grains thatare somewhat roundthat the densest packing is a Kepler Packing and so the percentage ofvolume ofgaps to glue together those sand grains would be 26%. Now I realizethat a concrete mix of portlandcement and sand thatmany of those gaps in the sand will have no portlandcement.What I am saying is the the Minimum Strongest Perfect concrete of sandthat is round would be of a 74% sand and 26%portland cement filling each of those gaps of the Kepler Packing.So I am having a difficult time of envisioning highway asphalt asbeing only a mere 6% by volume of a tar binder.Granted that the gravel used in asphalt and concrete is not roundshaped but having many flat surfaces where one flatsurface rests upon another relatively flat face surface and so youwould need less tar than the Kepler Packing amountof 26%. But I cannot envision that the flat surfaces would reduce the26% down to a mere 6%.Perhaps there is something about the term binder of a 6% binder thatI am missing in that the tar consists of abinder plus some other tar ingredients.P.S. I am having trouble in getting posts to the Internet in that theyappear on my ISP but not on any other ISP.and perhaps other posts.Archimedes Plutonium, a_plutonium@hotmail.comwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies=== === Subject: : Recursive DefinitionMime-version: 1.0Content-transfer-encoding: 7bitCould some kind soul please help me with the following?Give a recursive definition of the set of odd positive integers.Give a recursive definition of the set of positive integer powers of 3.How do I go about solving these? The text book is little to no help. Its thelast question form the section so hopefully there are no more in my future.=== === Subject: : Re: Recursive DefinitionIn sci.math, Jelly:> Could some kind soul please help me with the following?Give a recursive definition of the set of odd positive integers.> An integer is odd and positive if it is either '1' or if itis two more than another odd positive integer.Or, one can define it:An integer is odd and positive if it is either '1' or '3',or if it is four more than another odd positive integer.(This one is not as elegant. :-) However, both are correct.)Give a recursive definition of the set of positive integer powers of 3.An integer is a positive integer power of 3 if it's either '3' orif it is three times more than another positive integer power of 3.As you can see all answers call themselves -- it's asomewhat common method in computer software engineering.For example, one can define a tree as a single node, or anode with N children (N an integer > 0), each of which aretrees. (For trees, one could go with a simpler approach,even: a node with 0 children is automatically a leaf. Orone can specialize on binary trees, which can only have leftand right children.)A classic recursively-defined series is the Fibonaccinumbers. F(0) = 0, F(1) = 1, F(2) = 1, F(n+1) = F(n) + F(n-1).The series starts out 1,1,2,3,5,8,13,21, ...and can be iteratively defined, although it's notquite as pretty as the recursive form. (It turns outto depend on the Golden Mean, if I remember correctly.Of course now I can't reproduce the precise formula. Grr.Fortunately, at least one webpage does:F(n) = ( ((sqrt(5) + 1) / 2) ^ n - ((1 - sqrt(5)) / 2) ^ n) / sqrt(5))I would hope your textbook at least mentioned Fibonacci(to be precise, Leonardo Pisano Fibonacci, sun of Guiliemo,born 1170 A.D. into the Bonacci family, died 1250.)Some of the better ones should mention Lucas, as well:2,1,3,4,7,11,18,...How do I go about solving these? The text book is little> to no help. Its the last question form the section so> I'm not sure one can solve these as such; it appears to bemore of an issue of going back and forth between the symbolicform of a problem, and the English form.The English form can be a tad tricky. :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: Recursive Definition Visiting Assistant Professor at the University of Montana.>Could some kind soul please help me with the following?>Give a recursive definition of the set of odd positive integers.>Give a recursive definition of the set of positive integer powers of 3.>How do I go about solving these? The text book is little to no help.Recursive means that you say how to start, and then you say how to getthe next number from the one you have.For the odd positive integers, OP, then we have OP(1)=1, the first oddpositive integer; then, if you have an odd positive integer, you getthe next odd positive integer by adding 2 to the one you have, soOP(n+1) = OP(n) +2.So a recursive definition of the first set is:OP(1)=1.OP(n+1) = OP(n)+1.For the positive powers of 3, PT, what is the first positive power of3? If you have one, how do you get the next one?========================================================== ============It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ================== === Subject: : Prime number sequence, randomnessx_{n+1} = a x_n + d_n, d_in in {-2, +2}Is it possible to find a sequence (d_n) and two numbers a and x_1 suchthat (x_n) contains only prime numbers? Clearly, this is not possiblewith a=3 and x_1=1:x_1=1x_2=5x_3=13x_4=37x_5=113x_6=337x_7=1013x_8=3037x_9= 9109x_10=27329x_11 is NOT a primeany other sequence with a=3 and x_1=1 would end even before x_11.If we find a sequence (d_n) and two integers a and x_1 that work, thenwould the sequence (d_n) would necessraly look like pseudo-randomnumbers?--Vincent Granville, Ph.D.www.datashaping.com=== === Subject: : Re: Prime number sequence, randomness>x_{n+1} = a x_n + d_n, d_in in {-2, +2}>Is it possible to find a sequence (d_n) and two numbers a and x_1 such>that (x_n) contains only prime numbers? Clearly, this is not possible>with a=3 and x_1=1:Try a = 1. I think it's unlikely that there is a solution with a > 1.>If we find a sequence (d_n) and two integers a and x_1 that work, then>would the sequence (d_n) would necessraly look like pseudo-random>numbers?Not if a = 1...Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Problem with a vector functionMime-version: 1.0Content-type: text/plain; charset=ISO-8859-1Content-transfer-encoding: 8bitFirst, I dont know how to spell the word for the sign one uses for partialderivatives, so I just call it dye.Let theta(r,t) + t^n*e^(-r^2/(4t)). Im trying to find for which n is ittrue that(dye theta)/(dye t) = (1/r^2) (dye[r^2 * (dye theta)/(dye r)]) / (dye r).My approach involved quite a bit of algebraic work (and hence prone tomistakes): First I differentiated theta with respect to t to put on the LHSof the equation. Then I differentiated theta with respect to r (the innerdifferentiation), and then I differentiated r^2*(result of differentiationof theta with respect to r). I then multiplied this result with 1/r^2. Ikinda worked myself from inside and outwards.With further manipulations, I ended up with(n+(3/2))*t^(n-1) + ((1/4)r^2 - (r/2))*t^(n-2) = 0, which, by dividing byt^(n-1), becomes(n+(3/2)) + ((1/4)r^2 - (r/2))*t^(-1) = 0Now Im stuck... maybe my method didnt work, or maybe I made some mistakesin previous calculations, but I have no idea of how to go further from here.I think there must be an easier way to solve this problem, and Idapprechiate any help possible...Tomas=== === Subject: : Re: Problem with a vector function> First, I dont know how to spell the word for the sign one uses for partial> derivatives, so I just call it dye.> Let theta(r,t) + t^n*e^(-r^2/(4t)).Do you mean = or +? Such sloppy work habits can cause problem.>Im trying to find for which n is it> true that> (dye theta)/(dye t) = (1/r^2) (dye[r^2 * (dye theta)/(dye r)]) / (dye r).f(r,t) = t^n * e^((-r^2)/4t)f_t(r,t) = (1/r^2) g_r(r,t) where g(r,t) = r^2 * f_r(r,t)f_t(r,t) = nt^(n-1) * e^((-r^2)/4t) + t^n * e^((-r^2)/4t) * (r^2 /(4t^2)f_r(r,t) = t^n * (-2r/4t) e^((-r^2)/4t) = t^n * (-r/2t) e^((-r^2)/4t)g(r,t) = t^n * (-r^3 /2t) e^((-r^2)/4t)g_r(r,t) = t^n * (-3r^2 /2t) e^((-r^2)/4t) + t^n * (-r^3 /2t) (-2r/4t) e^((-r^2)/4t) = t^n * (-3r^2 /2t) e^((-r^2)/4t) + t^n * (r^4 /4t^2) e^((-r^2)/4t)g_r(r,t)/r^2 = t^n * (-3 /2t) e^((-r^2)/4t) + t^n * (r^2 /4t^2) e^((-r^2)/4t)> (n+(3/2)) + ((1/4)r^2 - (r/2))*t^(-1) = 0No, n + r^2 /4t = constant + r^2 /4t> Now Im stuck... maybe my method didnt work, or maybe I made some mistakes> in previous calculations, but I have no idea of how to go further from here.You made error. Be more methodical and take tiny steps at a time.> I think there must be an easier way to solve this problem, and Id> apprechiate any help possible...There isn't. Just be more methodical and don't try to take steps that domore than one thing to a term at a time. Take tiny steps as big onesassure slip ups and always check and recheck what you've done.=== === Subject: : Math Question: path of ballThis is a problem from my precalc book that I don't know how to solve:The path of a ball is given by:y = ((-1/20)(x^2)) + 3x + 5where y is the height of the ball (in feet) and x is the horizontal distance (in feet) from where the ball was thrown.a) find the maximum height of the ballb) find the distance the ball travelsI don't know where/how to begin with this (without just plugging in the numbersfor trial and error).Math Learner.=== === Subject: : Re: Math Question: path of ball> This is a problem from my precalc book that I don't know how to solve:The path of a ball is given by:> y = ((-1/20)(x^2)) + 3x + 5where y is the height of the ball (in feet) and x is the > horizontal distance (in feet) from where the ball was thrown.a) find the maximum height of the ball> b) find the distance the ball travels> I don't know where/how to begin with this (without just plugging in the numbers> for trial and error).Math Learner.Here's my two cents worth...Re-arrange your equation with a LHS of zero, and a quadratic on theRHS with the constant term = (5-y)Apply the quadratic formula.Ask yourself: what is x if y = 0? One answer is behind the thrower,the other is the range.Ask yourself: How large can y be in the quadratic formula and stillhave a non-negative discriminant. That's your max height...=== === Subject: : Re: Math Question: path of ball> This is a problem from my precalc book that I don't know how to solve:> The path of a ball is given by:> y = ((-1/20)(x^2)) + 3x + 5> where y is the height of the ball (in feet) and x is the> horizontal distance (in feet) from where the ball was thrown.> a) find the maximum height of the ball> b) find the distance the ball travels> I don't know where/how to begin with this (without just plugging in thenumbers> for trial and error).> Math Learner.I assume they mean the Horizontal distance the ball travels before hittingthe ground.Without using Calculus, I would suggest finding both roots of the equation .The positive root gives the horizontal difference you need.and then their average is the axis of the parabola. .Plug this in to get y max.RJ Pease=== === Subject: : Re: Math Question: path of ball> This is a problem from my precalc book that I don't know how to solve:The path of a ball is given by:> y = ((-1/20)(x^2)) + 3x + 5where y is the height of the ball (in feet) and x is the > horizontal distance (in feet) from where the ball was thrown.a) find the maximum height of the ball> b) find the distance the ball travels> I don't know where/how to begin with this (without just plugging in the > numbers> for trial and error).Math Learner.Do you know how to graph this function?=== === Subject: : Re: Math Question: path of ball> This is a problem from my precalc book that I don't know how to solve:The path of a ball is given by:> y = ((-1/20)(x^2)) + 3x + 5where y is the height of the ball (in feet) and x is the > horizontal distance (in feet) from where the ball was thrown.a) find the maximum height of the ball> b) find the distance the ball travels> I don't know where/how to begin with this (without just plugging in the numbers> for trial and error).You should know how to factorise the quadratic in x. This will give you something like y = c(x-a)(x-b). You'll observe then that y = 0 when x = a and when x = b. These are the x values of the points where the path intersects the ground, i.e. one is the point it was thrown from (I think you have to assume it was thrown from the ground) the other is where it lands.Because the trajectory is a parabola, it is symmetric about the line x = (a+b)/2. By symmetry, the highest point lies on this line.Gib=== === Subject: : Re: Math Question: path of ballInjector-Info: news.mailgate.org; posting-host=adsl-67-119-172-150.dsl.frsn01.pacbell.net; posting-account=48257; posting-date=1064187623X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 11d35514472f8102216c27e95ab69369.48257%40mygate.mailgate.org= This is a problem from my precalc book that I don't know=>> how to solve:=>> The path of a ball is given by:=>> y = ((-1/20)(x^2)) + 3x + 5=>> where y is the height of the ball (in feet) and x is the=>> horizontal distance (in feet) from where the ball was=>> thrown.=>> a) find the maximum height of the ball=>> b) find the distance the ball travels=>> I don't know where/how to begin with this (without just=>> plugging in the numbers for trial and error).=> You should know how to factorise the quadratic in x. This=> will give you something like y = c(x-a)(x-b). You'll=> observe then that y = 0 when x = a and when x = b. These=> are the x values of the points where the path intersects=> the ground, i.e. one is the point it was thrown from (I=> think you have to assume it was thrown from the ground)=> the other is where it lands.=> Because the trajectory is a parabola, it is symmetric=> about the line x = (a+b)/2. By symmetry, the highest=> point lies on this line.Well, almost. The throwing point is at x=0, y=5, so theother x axis intercept than the point of impact is _behind_the thrower, not part of the actual path the ball takes, buta projection of that path backwards from the point oflaunch, and will have some negative value for x.Otherwise your advice is correct.xanthian.-- Posted via Mailgate.ORG Server - http://www.Mailgate.ORG=== === Subject: : Re: Math Question: path of ball> =>> This is a problem from my precalc book that I don't know> =>> how to solve:=>> The path of a ball is given by:> =>> y = ((-1/20)(x^2)) + 3x + 5=>> where y is the height of the ball (in feet) and x is the> =>> horizontal distance (in feet) from where the ball was> =>> thrown.=>> a) find the maximum height of the ball> =>> b) find the distance the ball travels=>> I don't know where/how to begin with this (without just> =>> plugging in the numbers for trial and error).=> You should know how to factorise the quadratic in x. This> => will give you something like y = c(x-a)(x-b). You'll> => observe then that y = 0 when x = a and when x = b. These> => are the x values of the points where the path intersects> => the ground, i.e. one is the point it was thrown from (I> => think you have to assume it was thrown from the ground)> => the other is where it lands.=> Because the trajectory is a parabola, it is symmetric> => about the line x = (a+b)/2. By symmetry, the highest> => point lies on this line.Well, almost. The throwing point is at x=0, y=5, so the> other x axis intercept than the point of impact is _behind_> the thrower, not part of the actual path the ball takes, but> a projection of that path backwards from the point of> launch, and will have some negative value for x.I read too fast and missed the info that x is the distance the ball travels from where it is thrown, which as you point out tells us that it was thrown from a height of 5 feet.Gib=== === Subject: : What keywords that I should search for references for this kind of problems - vector comparison? Set partitions?I have two vectors of 1 by 3 dimension and each cell can only take value of either 0 or 1. For example, vector A can be (0,1,1), (1,0,1) or (1,1,0) and vector B can be (0,1,1), (1,0,1) or (1,1,0).My question is that how many ways (permutations?) that vector B is greater than vector A?I need the definition of comparing vectors. In my example above, (1,1,0) > (0,1,1) and (1,0,1) > (0,1,1). Therefore, there are two situations that vector B > vector A.Couple you please tell me what keywords (vector comparison, set partition, combinatorial, etc) that I should use to search for the references of this kind of problem?Thank you for your help.John=== === Subject: : Re: What keywords that I should search for ...John> I have two vectors of 1 by 3 dimension and each cell can only take value> of either 0 or 1. For example, vector A can be (0,1,1), (1,0,1) or> (1,1,0) and vector B can be (0,1,1), (1,0,1) or (1,1,0).> My question is that how many ways (permutations?) that vector B is> greater than vector A?Write down how many A's have n 1's in them, then how many B's have thesame 1's plus at least one more, and take the sum....> Couple you please tell me what keywords (vector comparison, set> partition, combinatorial, etc) that I should use to search for the> references of this kind of problem?I think you're looking for notions and terminology in combinatorics. Onegood trick is to search for a suitable MSC code (mathematics subjecthttp://www.ams.org/msc/LH=== === Subject: : GCD of two reals in irrational-ratioThis is just a light musin'...No need to interpret this (necessarily)as trolling...here, perhaps.)If we have two positive reals, x and y, such thatx/y is irrational,then maybe we can justifiably defineGCD(x,y) = 0.Why?n|m can be defined as:m/n = integer.So, let us say that x and y are such thatx/y is a positive RATIONAL.Then: x/y = k/j, where j and k are positive integers andGCD(k,j) =1.So, if the GCD(x,y) = z,then: x/y = (z*k)/(z*j),where x = z*k and y = z*j.So, z = x/k = y/j.Now, if we have x/y = positive rational, then x/y can be approximatedby, say, the m_th continued-fraction convergent:k(m)/j(n), k(m) and j(m) = positive coprime integers.So, we can have, because the numerators and denominators of eachconvergent (to an irrational) increase without bound:GCD(x,y) = limit{m->oo} x/k(m) = limit{m->oo} y/j(m) = ....0.(yay!)And, as a bonus, zero is an integer!Leroy Quet=== === Subject: : (typo fixed)GCD of two reals in irrational-ratioA small fix below to my reposted message.----This is just a light musin'...No need to interpret this (necessarily)as trolling...here, perhaps.)If we have two positive reals, x and y, such thatx/y is irrational,then maybe we can justifiably defineGCD(x,y) = 0.Why?n|m can be defined as:m/n = integer.So, let us say that x and y are such thatx/y is a positive RATIONAL.Then: x/y = k/j, where j and k are positive integers andGCD(k,j) =1.So, if the GCD(x,y) = z,then: x/y = (z*k)/(z*j),where x = z*k and y = z*j.So, z = x/k = y/j. [edited below]Now, if we have x/y = positive IRRATIONAL, then x/y can be approximatedby, say, the m_th continued-fraction convergent:k(m)/j(n), k(m) and j(m) = positive coprime integers.So, we can have, because the numerators and denominators of eachconvergent (to an irrational) increase without bound:GCD(x,y) = limit{m->oo} x/k(m) = limit{m->oo} y/j(m) = ....0.(yay!)And, as a bonus, zero is an integer!Leroy Quet=== === Subject: : Re: (typo fixed)GCD of two reals in irrational-ratio> A small fix below to my reposted message.----This is just a light musin'...No need to interpret this (necessarily)> as trolling...here, perhaps.)> If we have two positive reals, x and y, such thatx/y is irrational,then maybe we can justifiably defineGCD(x,y) = 0.Why?> I would define the GCD(x,y) to be the supremum of all non-negative numbers d such that x/d and y/d are both integers. With this definition (and maybe defining infinity to be an integer), your definition would fit nicely.-- Stephen Montgomery-Smithstephen@math.missouri.eduhttp:// www.math.missouri.edu/~stephen=== === Subject: : Re: (typo fixed)GCD of two reals in irrational-ratio> ...> If we have two positive reals, x and y, such that> x/y is irrational,> then maybe we can justifiably define> GCD(x,y) = 0....> I would define the GCD(x,y) to be the supremum of all non-negative numbers d> such that x/d and y/d are both integers. With this definition (and maybe> defining infinity to be an integer), your definition would fit nicely.That is not well-defined since the supremum of an empty setis undefined. To see that the set of d as above is emptymore often than not, consider y=x*z where x is any real number and z is any irrational number. If x/d and y/d are integers iand j, then z = y/x = j/i, a contradiction.-jiw=== === Subject: : Re: (typo fixed)GCD of two reals in irrational-ratio> ...>If we have two positive reals, x and y, such that>x/y is irrational,>then maybe we can justifiably define>GCD(x,y) = 0....>>I would define the GCD(x,y) to be the supremum of all non-negative numbers d>>such that x/d and y/d are both integers. With this definition (and maybe>>defining infinity to be an integer), your definition would fit nicely.> That is not well-defined since the supremum of an empty set> is undefined. To see that the set of d as above is empty> more often than not, consider y=x*z where x is any real number > and z is any irrational number. If x/d and y/d are integers i> and j, then z = y/x = j/i, a contradiction.> -jiw1. I think that the supremum of the empty set can make sense if one understands the context - so if the empty set is thought of as a subset of the non-negative numbers, then I would normally define the supremum of the empty set as 0.Think of it this way - the supremum satisfies the relationsup(A union B) = max{sup(A),sup(B)}.If all the sets are subsets of the non-negative reals, then sup(emptyset)=0 is the only one that works.2. I added the proviso that we define infinity to be an integer, so in the case that x/y is irrational means that the set under consideration is {0}.I admit that it is not exactly super rigorous, just feely type math that makes it reasonable that GCD(x,y)=0 when x/y is irrational.-- Stephen Montgomery-Smithstephen@math.missouri.eduhttp:// www.math.missouri.edu/~stephen=== === Subject: : Re: (typo fixed)GCD of two reals in irrational-ratio> ...>If we have two positive reals, x and y, such that>x/y is irrational,>then maybe we can justifiably define>GCD(x,y) = 0.> ...>>I would define the GCD(x,y) to be the supremum of all non-negative numbers d>>such that x/d and y/d are both integers. With this definition (and maybe>>defining infinity to be an integer), your definition would fit nicely. That is not well-defined since the supremum of an empty set> is undefined. To see that the set of d as above is empty> more often than not, consider y=x*z where x is any real number> and z is any irrational number. If x/d and y/d are integers i> and j, then z = y/x = j/i, a contradiction.> -jiw1. I think that the supremum of the empty set can make sense if one understands> the context - so if the empty set is thought of as a subset of the non-negative> numbers, then I would normally define the supremum of the empty set as 0.Think of it this way - the supremum satisfies the relation> sup(A union B) = max{sup(A),sup(B)}.> If all the sets are subsets of the non-negative reals, then sup(emptyset)=0 is> the only one that works....Let A = (-2,-1), ie, the open interval between -2 and -1.Obviously sup(A) = -1. But if you define sup(phi) as you suggest, you would have -1 = sup(A) = sup(A U phi) = 0, a contradiction because -1 is not 0.-jiw=== === Subject: : Re: (typo fixed)GCD of two reals in irrational-ratio>...>If we have two positive reals, x and y, such that>x/y is irrational,>then maybe we can justifiably define>GCD(x,y) = 0.>...>>I would define the GCD(x,y) to be the supremum of all non-negative numbers d>>such that x/d and y/d are both integers. With this definition (and maybe>>defining infinity to be an integer), your definition would fit nicely.>That is not well-defined since the supremum of an empty set>is undefined. To see that the set of d as above is empty>more often than not, consider y=x*z where x is any real number>and z is any irrational number. If x/d and y/d are integers i>and j, then z = y/x = j/i, a contradiction.>-jiw>>1. I think that the supremum of the empty set can make sense if one understands>>the context - so if the empty set is thought of as a subset of the non-negative>>numbers, then I would normally define the supremum of the empty set as 0.>>Think of it this way - the supremum satisfies the relation>>sup(A union B) = max{sup(A),sup(B)}.>>If all the sets are subsets of the non-negative reals, then sup(emptyset)=0 is>>the only one that works....Let A = (-2,-1), ie, the open interval between -2 and -1.> Obviously sup(A) = -1. But if you define sup(phi) as you > suggest, you would have -1 = sup(A) = sup(A U phi) = 0, > a contradiction because -1 is not 0.> -jiwHere's what I think the definition of supremum is: sup(A) = min{x in [universal set] | forall a in A, x >= a}where [universal set] means the original set containing allcandidate sets A as subsets.For subsets of the set No of non-negative integers, sup( emptyset ) = min{ x in No | forall a in emptyset, x>=a } = min{ x in No }i.e., since there is no a in emptyset, the condition holds for allx in No. Thus, one obtains sup( emptyset ) = min( x in No ), or 0.In your case, you've got numbers outside No, so that's not anappropriate version of sup( emptyset ). If you mean to allow allintegers, or all reals for that matter, sup( emptyset ) shouldbe min( x in Z ), (or min( x in R ), respectively). That wouldbe -infinity.EXERCISE: find a suitable definition for inf(...) to determineinf( emptyset ) in Z, N, R, N- (negative integers).Dale=== === Subject: : Re: (typo fixed)GCD of two reals in irrational-ratio> Think of it this way - the supremum satisfies the relation> sup(A union B) = max{sup(A),sup(B)}.> If all the sets are subsets of the non-negative reals, then sup(emptyset)=0> is> the only one that works.> ...Let A = (-2,-1), ie, the open interval between -2 and -1.> Obviously sup(A) = -1. But if you define sup(phi) as you > suggest, you would have -1 = sup(A) = sup(A U phi) = 0, > a contradiction because -1 is not 0.> -jiwThat's why he put in that bit about subsets of the non-negative reals.=== === Subject: : Re: (typo fixed)GCD of two reals in irrational-ratio>Think of it this way - the supremum satisfies the relation>sup(A union B) = max{sup(A),sup(B)}.>If all the sets are subsets of the non-negative reals, then sup(emptyset)=0>is>the only one that works.>>...>>Let A = (-2,-1), ie, the open interval between -2 and -1.>>Obviously sup(A) = -1. But if you define sup(phi) as you >>suggest, you would have -1 = sup(A) = sup(A U phi) = 0, >>a contradiction because -1 is not 0.>>-jiw> That's why he put in that bit about subsets of the non-negative reals.Yes, if you are considering all subsets of the reals, the sensible definition of sup(emptyset) is -infinity.-- Stephen Montgomery-Smithstephen@math.missouri.eduhttp:// www.math.missouri.edu/~stephen=== === Subject: : Re: question about a really easy problem...> I've been out of school for a while and have only started to go back> to nursing school, this is probably going to sound like a really> stupid question, but there was a problem in one of my classes and the> math dosage portion of the question was:> 16-4+3 = ?> A common problem is explaining that 2 + 3 x 4 is not 20. Many will> prove that it is by using their calculators. How could the> calculator be wrong?. I sometimes respond by showing mine which does> give 14 and their are surprised that two calculators are giving> different answers.> Heck, the two skins for the one Windoze calculator, common andscientific, give two different answers to 2 + 3 * 4=== === Subject: : Re: question about a really easy problem...> go back to nursing school, this is probably going to sound like a> really stupid question, but there was a problem in one of my> classes and the math dosage portion of the question was: 16-4+3 => ?> I'd answered 9 cause I used PEDMAS, addition before subtraction,> but everyone else answered 15 which was apparently the right> anwer. Do people no longer use the order or do I have a wrong> conception of it???> Jeeves> >> Most people would probably answer 15, but don't feel too bad about>> answering 9. There are all kinds of different ways to parse>> mathematical formulas, resulting in different orders of evaluation.> > > >> As you point out, 9 is the correct answer if the + operator has> precedence over -. It is also the correct answer if formulas are>> evaluated from right to left.> >> N> But reasons there are good for using conventions standard regarding> ordering.> John ellYes, there are good reasons for using the standard ordering (or one of> the standard orderings, since there is more than one standard). And> there are also good reasons for NOT using the standard ordering (or> using a standard other than the usual one). It all depends on the> context and purpose.For example, the J language evaluates from right to left, and the J> designers had good reasons for defining it that way.http://www.jsoftware.com/books/help/primer/precedence.htm> http://www.jsoftware.com/books/help/primer/order_of_ eval.htmNemoThe subject of discussion was arithmetic, not the J-language, whateverthat is. Do you know any good reasons for abandoning the standard,well-known conventions of arithmetic?John=== === Subject: : Re: question about a really easy problem... > > to go back to nursing school, this is probably going to sound > like a really stupid question, but there was a problem in one > of my classes and the math dosage portion of the question > was: 16-4+3 = ? > I'd answered 9 cause I used PEDMAS, addition before > subtraction, but everyone else answered 15 which was > apparently the right anwer. Do people no longer use the > order or do I have a wrong conception of it??? > Jeeves >>> Most people would probably answer 15, but don't feel too bad >> about answering 9. There are all kinds of different ways to >> parse mathematical formulas, resulting in different orders of >> evaluation. >>> As you point out, 9 is the correct answer if the + operator has >> precedence over -. It is also the correct answer if formulas >> are evaluated from right to left. >>> N > But reasons there are good for using conventions standard > regarding ordering. > John ell >>> Yes, there are good reasons for using the standard ordering (or >> one of the standard orderings, since there is more than one >> standard). And there are also good reasons for NOT using the >> standard ordering (or using a standard other than the usual one). >> It all depends on the context and purpose. >>> For example, the J language evaluates from right to left, and the J >> designers had good reasons for defining it that way. >>> http://www.jsoftware.com/books/help/primer/precedence.htm >> http://www.jsoftware.com/books/help/primer/order_of_eval.htm >>> Nemo > > The subject of discussion was arithmetic, not the J-language, > whatever that is. Do you know any good reasons for abandoning the > standard, well-known conventions of arithmetic? > JohnMy point is: There is more than one convention for parsing andevaluating arithmetic expressions.The J language deals with arithmetic, among other things. The designersof J adopted an evaluation order different from the multiplicationbefore addition order. The URL above explains their reasons.I don't advocate abandoning anything. My point is that other conventionsalso make sense, depending on the context and purpose.Nemo=== === Subject: : Re: question about a really easy problem...>> For example, the J language evaluates from right to left, and the J>> designers had good reasons for defining it that way.>> http://www.jsoftware.com/books/help/primer/precedence.htm>> http://www.jsoftware.com/books/help/primer/order_of_eval.htm>> Nemo> The subject of discussion was arithmetic, not the J-language, whatever> that is. Do you know any good reasons for abandoning the standard,> well-known conventions of arithmetic?I think the reason the J designers wanted their language to behave thatway was to make it look as much as possible like APL, but without thefunny character set.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.=== === Subject: : Re: question about a really easy problem...> I've been out of school for a while and have only started to go back> to nursing school, this is probably going to sound like a really> stupid question, but there was a problem in one of my classes and the> math dosage portion of the question was:> 16-4+3 = ? I'd answered 9 cause I used PEDMAS, addition before subtraction, but> everyone else answered 15 which was apparently the right anwer. Do> people no longer use the order or do I have a wrong conception of> it??? Jeeves> Over here in the UK, I have never heard of PEDMAS but I have heard of> BODMAS. Brackets, Of, Division, Multiplication, Addition,> Subtraction. There seems to be a UK / US language difference here.> I guess that the P of PEDMAS is parentheses. We rarely use that> word here and tend to call all of ( ) [ ] { } brackets. If it is> necessary to distinguish the types we usually qualify them as round,> square and curly. If brackets is used unqualified (as in> BODMAS) then the round ones are usually meant.> Anyway, I have never liked BODMAS and feel it hinders as much as it> helps. As you found the suggestion that division has priority over> multiplication and addition over subtraction is not standard. When> teaching maths, I have rarely found trouble with the idea that the> contents of brackets (in UK sense) should be evaluated first (*).> Then the only non-obvious rule in simple arithmetic is:> multiplication and division before addition and subtraction. It is> probably no harder to remember that than BODMAS.> (*) Except in a few cases but then the individual operations were also> a challenge and things were not much worse because of a failure to> understand brackets.> A common problem is explaining that 2 + 3 x 4 is not 20. Many will> prove that it is by using their calculators. How could the> calculator be wrong?. I sometimes respond by showing mine which does> give 14 and they are surprised that two calculators are giving> different answers.PEDMAS should be written PE(DM)(AS)RJ Pease=== === Subject: : Re: Simple question about non-standard analysis <3F69BCB8.4090806@no.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWSsaid:>There was a little (well, very little) friction because well, it>always gives the right answer, so what's wrong with it? Okay,>there's nothing wrong with using infinitesimals.That's true Post-Robinson. Naive use of Infinitesimals back inNewton's day didn't always give the right answer.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org=== === Subject: : Re: Help with product and box topologiesX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 12:16 PM, Topy said:>But why won't this reasoning work for the product topology? I don't understand the question.If B_Alpha = X_Alpha for all but finitely many Alpha, and C_Alpha =X_Alpha for all but finitely many Alpha, then B_Alpha intersectC_Alpha = X_Alpha for all but finitely many Alpha; hence, theintersection of two basis elements is a basis element.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org=== === Subject: : Possible Errors in Diehard Test ResultsI have been running several RN's through Marsaglia's Diehard package. I findabsolutely no consistency in the KS test p values to the D+, D- and Dn teststatistics. On a graph, just a scatter of points, with some generalcorrelation of p values to Dn values.There is a precise mathematical relationship between Dn values and p valuesfor a uniform distribution.David Heiser=== === Subject: : Re: Possible Errors in Diehard Test Results> I have been running several RN's through Marsaglia's Diehard package. Ifind> absolutely no consistency in the KS test p values to the D+, D- and Dntest> statistics. On a graph, just a scatter of points, with some general> correlation of p values to Dn values.> There is a precise mathematical relationship between Dn values and pvalues> for a uniform distribution.> David HeiserThe term KS test is generic for the wide class of tests usedto measure one's confidence in whether a set n real numbersin [0,1) can be considered the outcome of n independentuniform [0,1) random variables.I am so accustomed to viewing KS as a class of testsfrom which one of several can be selected, that I didnot amplify on the particular choice for Diehard;perhaps I should have.The K,K+,K- ( in some circles, D,D+,D-) are the oldest,and still widely used, but other methods, particularly thosebased on int(w(x)*(F(x)-x)^2,x=0..1), where F(x) is the sampleCDF (staircase function), are considered better, particularly ifthe weight function w(x) is chosen so as to equalize variances.This was proposed by Savage and is incorporated in theAnderson-Darling version of the KS test (more properly,the Anderson-Darling version of the class of Cramer-von Misesapproaches to the KS test).The A-D version of the C-vM class of K-S tests is the oneI chose for application in the Diehard tests.For large n, there seems no strong reason for choosing aparticular version of the K-S tests, but for smaller n, then+1 spacings induced by the ordered set of uniform variatesshould form a point s= (s_1,...,s_{n+1}) uniformly distributed over the simplex S={(x_1,...,x_{n+1}), x_1+...+x_{n+1}=1, x's>=0}.If you look at the projections of sets of points for which K The K,K+,K- ( in some circles, D,D+,D-) are the oldest,> and still widely used, but other methods, particularly those> based on int(w(x)*(F(x)-x)^2,x=0..1), where F(x) is the sample> CDF (staircase function), are considered better, particularly if> the weight function w(x) is chosen so as to equalize variances.> This was proposed by Savage and is incorporated in the> Anderson-Darling version of the KS test (more properly,> the Anderson-Darling version of the class of Cramer-von Mises> approaches to the KS test).> The A-D version of the C-vM class of K-S tests is the one> I chose for application in the Diehard tests.> For large n, there seems no strong reason for choosing a> particular version of the K-S tests, but for smaller n, the> n+1 spacings induced by the ordered set of uniform variates> should form a point s= (s_1,...,s_{n+1}) uniformly distributed> over the simplex> S={(x_1,...,x_{n+1}), x_1+...+x_{n+1}=1, x's>=0}.> If you look at the projections of sets of points for which K using Kolmogorov's K (or Smirnov's K+), where the projections> are from, say, 10-space onto various 3-dimensional faces of the> 9-dimensional simplex, you will see that certain regions seem to> be favored---others neglected--- in the original K,K+,K- versions,> contrary to the presumed test of uniformity of s over the simplex S.I would try to show why I think your implementation is not good for sometest (for example the overlapping sum) and I really hope you don't takeoffence.In the overlapping sum test you get 10 p-values (no_obs = 10) and then youcalculate the overall p-value with AD test.Suppose to get a perfect distribution of p-values. With 10 p-values thismeans:1 02 0.1111111111111113 0.2222222222222224 0.3333333333333335 0.4444444444444446 0.5555555555555567 0.6666666666666678 0.7777777777777789 0.88888888888888910 1With the plain KS test I get the overall p-value = 0.999897950683125, butwith your AD test I get 0.00946405272470502.The distribution is perfect, but with AD test it seems very bad.My thought is that you should use the plain KS for all the tests and AD*only* if you want to calculate the overall p-value over more than 200p-values (although I think this is useless). In this case (with 200p-values), AD with a perfect distribution gives 0.995636593372975, there isonly a little error (the exact value should be 1; I get 0.99999999752982with KS).I don't know if I'm wrong somewhere (I'm not a methematician), but this iswhat I get in my experiments.> That is one of the reasons that I chose the Savage-Anderson-Darling> version rather than the original Kolmogorov K, or Smirnov K+,K-.> Another is that Savage-Anderson-Darling gives equal weight to> variations around 0, around 1 and around the center, while the> original K,K+,K- give greater weight to variations around the center.In my experiments I don't see that. I see, instead, that AD gives a very bigweight to the tails while around the center it gives the same weight of KS.What you said about AD I seen for KS: KS gives equal weight to variationsaround 0, around 1 and around the center.> And it is often bad fit in the tails that suggest a poor RNG.You have, beyond any doubt, much more experience in testing RNG, but I thinkalso a bad fit in the center suggests a bad RNG. The difference is that badtails can be seen also with our eyes, while our eyes don't see a bad center.Best regardsCristiano=== === Subject: : Re: first quartile> Why do you think that the algorithms in these computer packages differ?> Why aren't they universal?> Any thoughts or comments?> Mark Any thoughts?> Mark> Nat Silver If I have the following Example Data Set:> 14, 29, 32, 38, 41, 45, 68, 71, 74, 83, 86, 87, 94, 99> How would I find the first quartile of that data set?> What is the procedure?> Find the median = 69.5> Then find the median of the lower half,> which is 38, the first quartile.I agree with Nat that 38, the median of the lower half, is the first> quartile. I have heard about computer packages giving different answers> before. For example, Excel (the least trustworthy for statistics) says> 38.75! Got me how the packages come up with these.This is a standard value (computed by Splus and R for e.g.).It results from taking the order statistics to correspond toproportions equally spaced from 0 to 100%. Other quantilesare obtained by linear interpolation between these.=== === Subject: : Re: first quartile> Why do you think that the algorithms in these computer packages differ?> Why aren't they universal?> Any thoughts or comments?When it comes to distributions that are discrete,there is not one single definition of 'percentile' that meetseveryone's needs. ('Percentiles' include quartile). There is not any textbook or organization that sets a standard that folks feel a need to follow, for a great many statistical terms, so no single definition has won out.I think you can find some computer-program definitions collected in the last three years, by searching the Usenet groups,sci.stat.* , for centiles or percentiles.Here is an excerpt from what I posted in August, 2002 -======== [ in a post from 18 months previous ] I don't know the answer to your question, . But while digging around, I found this interesting collection of comments on quartiles: http://exploringdata.cqu.edu.au/ticktack.htm ======= posted by Warren Sarle of SAS, same day. SAS provides at least six ways to estimate quantiles (none of which is what Excel does). Five of them are described at http://sasdocs.ats.ucla.edu/proc/zormulas.htm . A sixth one is provided in PROC MEANS for very large data sets and is considerably more complicated than the first five.========= end of excerpt .-- Rich Ulrich, wpilib@pitt.eduhttp://www.pitt.edu/~wpilib/index.htmlTaxes are the price we pay for civilization. === === Subject: : Re: first quartile> Why do you think that the algorithms in these computer packages differ?> Why aren't they universal?> Any thoughts or comments?Well, it is easy to define a quantile for a continuous variable,say x, for which there exists a smooth positive density function, say f(x): a quantile x_q is the unique value such that P(x <= x_q) = q,i.e., q = integral(-infinity,x_q) f(x) dx, and x_q exists for all q in [0,1].However, problems arise as soon as some part of the above assumptionis broken. In particular, if the density has point masses (i.e.,integral(x,x) f(x) dx > 0 for some x) then there may be many x_q suchthat P(x <= x_q) = q, or there may be no such x_q. A simple example of a density with point masses is the empirical density on a sample, i.e., the density that assigns mass 1/n to each datum.Consider a sample with two elements, x0 and x1. If the densityis assumed to put mass 1/2 on x0 and 1/2 on x1, there is no x_{0.25}such that P(x <= x_{0.25}) = 0.25, and any value x0 <= x_{0.5} < x1has the property that P(x <= x_{0.5}) = 0.5. In order to get a unique x_q for all q in [0,1], it is essentiallynecessary to work with some density other than the empirical density.Which one you work with is pretty arbitrary; there aren't any fundamental constraints, so some combination of convenience and justification governs. That's why different software packages willgiven different results.You could deduce the density assumed in the quantile formulaused by some package: tabulate the pairs (x_q, q) for a large numberof q, make divided differences (q(i+1) - q(i))/(x_q(i+1) - x_q(i)),and plot the result; that's an approximation to the assumed density.For what it's worth,Robert Dodier--A lofty boast, a shattered statue; Ozy, the desert's laughing at you. -- Ozymandias (abridged version)=== === Subject: : Re: first quartile> Why do you think that the algorithms in these computer> packages differ? Why aren't they universal?> Any thoughts or comments? Sample Quantiles in Statistical Packages Rob J. Hyndman; Yanan Fan The American Statistician, Vol. 50, No. 4. (Nov., 1996), pp. 361-365.Your university *may* provide online access to it athttp://links.jstor.org/sici?sici=0003-1305%28199611%2950% 3A4%3C361%3ASQISP%3E2.0.CO%3B2-G-- Karl Ove Hufthammerhttp://blogg.huftis.org/=== === Subject: : Re: first quartile boundary=----=_NextPart_000_000F_01C37FB4.71BF2920------------ --------------------------------------------------------- charset=iso-8859-1 Why do you think that the algorithms in these computer packages differ? Why aren't they universal? Any thoughts or comments? Mark > Any thoughts? > Mark > Nat Silver If I have the following Example Data Set: > 14, 29, 32, 38, 41, 45, 68, 71, 74, 83, 86, 87, 94, 99 > How would I find the first quartile of that data set? > What is the procedure? > Find the median = 69.5 > Then find the median of the lower half, > which is 38, the first quartile. I agree with Nat that 38, the median of the lower half, is the first quartile. I have heard about computer packages giving different answers before. For example, Excel (the least trustworthy for statistics) says 38.75! Got me how the packages come up with these. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu -------------------------------------------------------------- -------------------------------- Actually you are wrong. Excel uses Gumbel's method. This is one of many accepted methods. Are you accusing Gumbel that he is wrong? DAHeiser=== === Subject: : Error in notes on physically based modelling?HiI'm sticking my neck out a bit but I think there is an error in the lecture notes on constrained dynamics here http://www-2.cs.cmu.edu/afs/cs.cmu.edu/user/baraff/www/ sigcourse/.On page 4: T' = m*x''.x' = m*f.x' + m*f_hat.x'but on the previous page x'' = (f + f_hat)/m.I think that the equation on pg 4 should read: T' = f.x' + f_hat.x'Is that right?-- sashanhttp://www.cs.auckland.ac.nz/~sgov008/=== === Subject: : Re: Error in notes on physically based modelling?> HiI'm sticking my neck out a bit but I think there is an error in the > lecture notes on constrained dynamics here > http://www-2.cs.cmu.edu/afs/cs.cmu.edu/user/baraff/www/ sigcourse/.On page 4:> T' = m*x''.x' = m*f.x' + m*f_hat.x'but on the previous page x'' = (f + f_hat)/m.I think that the equation on pg 4 should read:> T' = f.x' + f_hat.x'You're right. Work is force times distance, and thatis what is being calculated here. f*x has the unitsof energy. - Randy=== === Subject: : Differential Equation HelpHi there,I've been trying to solve this different via numerical methods, and it'sdriving me crazy....Anyone know how to pop this into matlab preferably and solve it.......orsome other means, via C, language.d^2y/dx^2 - 1/(L^2)*y .... where L is just some arbitrary number......L =0.2045....I cannot figure this out, I was trying ODE45 function in matlab to nosuccess. Maybe theres an easier way in C.Any help is appreciated,gs=== === Subject: : Re: Differential Equation Help> Hi there,> I've been trying to solve this different via numerical methods, and it's> driving me crazy....> Anyone know how to pop this into matlab preferably and solve it.......or> some other means, via C, language.> d^2y/dx^2 - 1/(L^2)*y .... where L is just some arbitrary number......L => 0.2045....The solution is in terms of simple sines and cosines. The general solutionisy = A cos (x/L) + B sin (x/L)You can verify this quite easily by plugging this solution into theequation. A and B are constants determined by initial conditions which youdon't seem to list.> I cannot figure this out, I was trying ODE45 function in matlab to no> success. Maybe theres an easier way in C.> Any help is appreciated,> gs=== === Subject: : Re: Differential Equation Help> Hi there, I've been trying to solve this different via numerical methods, and it's> driving me crazy.... Anyone know how to pop this into matlab preferably and solve it.......or> some other means, via C, language. d^2y/dx^2 - 1/(L^2)*y .... where L is just some arbitrary number......L => 0.2045....> The solution is in terms of simple sines and cosines. The general solution> is> y = A cos (x/L) + B sin (x/L)> You can verify this quite easily by plugging this solution into the> equation. A and B are constants determined by initial conditions which you> don't seem to list. I cannot figure this out, I was trying ODE45 function in matlab to no> success. Maybe theres an easier way in C. Any help is appreciated, gs(1) It was not a differential equation but a differentialexpression. Assuming that it is meant to equal to 0, here iswhat MATLAB can do (cut and pasted from command screen):>> dsolve('D2y-(1/L^2)*y=0')ans =C1*sinh(1/L*t)+C2*cosh(1/L*t)(2) Watch out for signs: the trigonometric solution wouldcorrespond tod^2y/dx^2 - 1/(L^2)*y = 0(3) To get any answer from ode45 or similar, you have to (a) re-write the second-order ODE into a first-order system, as described in standard textbooks, (b) supply initial conditions and required domain, as specified in the help part, displayed by>> help ode45(4) numerical software, in contrast with symbolic software, will not provide the structure of the solution, only a bunch of numbers.The purpose of computation is insight, not numbers.(R.W. Hamming, 1973 or maybe 1962)The purpose of computing is not yet in sight(R.W. Hamming (?), later)Cheers, ZVK(Slavek).=== === Subject: : Volume of RevolutionA storage shed is the shape of a hemisphere. The base is circular withradius r =1. Cross sections of the shed which are perpendicular to adiameter of the base are square. Find the volume of the shed. This isnot difficult for a hemisphere but I'm having a problem with ahemisphere with a square cross section. Anyone? Holman=== === Subject: : Re: Volume of RevolutionIn sci.math, Holman:> A storage shed is the shape of a hemisphere. The base is circular with> radius r =1. Cross sections of the shed which are perpendicular to a> diameter of the base are square. Find the volume of the shed. This is> not difficult for a hemisphere but I'm having a problem with a> hemisphere with a square cross section. Anyone?You've specified the (top of the?) shed as a hemisphere.The cross section must include a semicircle on top,regardless of cutting-plane orientation; the plane needn'tcut the shed straight up and down.If one assumes a plane going straight up and down, onegets a hemisphere sitting on a cylinder, if the problemis suitably corrected. However, this may not be the onlysolution, although I'm not entirely certain what the bottomwill look like if the plane cuts at a specified angle --probably a modified frustrum of some sort.If one assumes the former, the problem is completely specified,although I'm not sure how correctly. Holman> -- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: Volume of Revolution> In sci.math, Holman> :> A storage shed is the shape of a hemisphere. The base is circularwith> radius r =1. Cross sections of the shed which are perpendicular to a> diameter of the base are square. Find the volume of the shed. Thisis> not difficult for a hemisphere but I'm having a problem with a> hemisphere with a square cross section. Anyone?> You've specified the (top of the?) shed as a hemisphere.> The cross section must include a semicircle on top,> regardless of cutting-plane orientation; the plane needn't> cut the shed straight up and down.> If one assumes a plane going straight up and down, one> gets a hemisphere sitting on a cylinder, if the problem> is suitably corrected. However, this may not be the only> solution, although I'm not entirely certain what the bottom> will look like if the plane cuts at a specified angle --> probably a modified frustrum of some sort.> If one assumes the former, the problem is completely specified,> although I'm not sure how correctly.Lets see...Volume = Hemisphere + Cylinder = Sum Pi*y^2 dx + Pi*r^2*h = Sum Pi*Sqrt(1-x^2)^2 dx + Pi*r^2*h (for asquare x section r=h=1) = Pi [x-1/3x^3] + Pi ...........x=1 = 2/3Pi + Pi = 5/3 PiThe answer give at the bottom of the page where I read this from is16/3. Other problem solutions are given in terms of Pi and even though5/3 Pi is not too far from 16/3, it looks like there is an error in theway the question is stated. Holman=== === Subject: : Re: Volume of RevolutionInjector-Info: news.mailgate.org; posting-host=adsl-67-119-172-150.dsl.frsn01.pacbell.net; posting-account=48257; posting-date=1064187623X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 5ac6df3507e9b9b93665d9bdc48be13e.48257%40mygate.mailgate.org> Volume of RevolutionOh, mostly, they're loud.> A storage shed is the shape of a hemisphere. The base is circular with> radius r =1. Cross sections of the shed which are perpendicular to a> diameter of the base are square. Find the volume of the shed. This is> not difficult for a hemisphere but I'm having a problem with a> hemisphere with a square cross section. Anyone?Anyone would, since the cross section of a hemisphere/hemiballperpendicular to a diameter is a semi-circle/semi-disk, depending onyour viewpoint.Perhaps you had some other shaped shed in mind?xanthian.-- Posted via Mailgate.ORG Server - http://www.Mailgate.ORG=== === Subject: : Re: Volume of RevolutionOriginator: gheston@hiwaay.net (Gary Heston)>> Volume of Revolution>Oh, mostly, they're loud.>> A storage shed is the shape of a hemisphere. The base is circular with>> radius r =1. Cross sections of the shed which are perpendicular to a>> diameter of the base are square. Find the volume of the shed. This is>> not difficult for a hemisphere but I'm having a problem with a>> hemisphere with a square cross section. Anyone?>Anyone would, since the cross section of a hemisphere/hemiball>perpendicular to a diameter is a semi-circle/semi-disk, depending on>your viewpoint.>Perhaps you had some other shaped shed in mind?The cross-section stuff is BS to distract you; a storage shed in the shape of a hemisphere has a volume of 1/2 the sphere with a radius of 1.The formula for the volume of a sphere is easily located; the answer ismore than two, but less than three.Gary-- Gary Heston gheston@hiwaay.net VeriSign resolves http://www.unscrupulousbastards.com correctly,click and see...=== === Subject: : [meta]-museInjector-Info: news.mailgate.org; posting-host=adsl-67-119-172-150.dsl.frsn01.pacbell.net; posting-account=48257; posting-date=1064187623X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ b8d06d611fcb0ef2bca904b4bcf01c5a.48257% 40mygate.mailgate.orgWhen does X theory is fundamentally flawed differ fundamentally from: I obdurately and fundamentally refuse to understand X theoryfor small whole values of X?xanthian.-- Posted via Mailgate.ORG Server - http://www.Mailgate.ORG=== === Subject: : Re: [meta]-museIn sci.math, Kent Paul Dolan:> When does X theory is fundamentally flawed differ fundamentally from: I obdurately and fundamentally refuse to understand X theoryfor small whole values of X?xanthian.> This might be proven by induction.[1] Assume X = 1. The two are quite different, although I'm not sure what 1 theory is.[2] Assume X = N, and look at the following: (a) N+1 theory is fundamentally flawed. (b) I obdurately and fundamentally refuse to understand N+1 theory.Of course, I'm not entirely sure what N theory is, so I'mnot sure I can go much farther here.... :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : A Hyperpower/Hyperfactorial family of seriesI stumbled upon these while working on the hyperpower function.The hyperfactorial sequence grows much faster than the correspondinghyperpower function and a couple of lower bounds (which are not optimal,Using the hyperfactorial sequence on the denominator of a series canbring down even the hyperpower function on the numerator and force theentire series to converge.Unfortunately, Maple code for all this is out of the question, as thenumbers involved are really huge, so even Maple cannot handle them.in my Math section:Enjoy the read.-- Ioannishttp://users.forthnet.gr/ath/jgal/_____________________ ______________________Eventually, _everything_ is understandable.=== === Subject: : Re: Polya & Alkanes> I am looking into the ways of working out the number of isomers of> alkanes. I understand there is a way of doing this using Polya's> Counting Theory - can anyone enlighten me as to how this is applied to> this situation??To recast the problem into mathematical language: Given a positive> integer C, you want to count the isomorphism classes of trees that> have C vertices of degree 4, 2C+2 leaves, and no other vertices.This will not distinguish stereoisomers. That seems to be a harder> problem. Here's an idea for the stereoisomers. Someone who knows> more chemistry might be able to tell me if I am overcounting (or maybe> even undercounting): For each isomorphism class, count the number of> vetices of degree four that have the property that when they are> removed the resulting four trees are pairwise non-isomorphic. Call> this n. Then there are 2^n stereoisomers of that isomer. For> example,> CC C C> CCCCCC CCCCCC CCCCCC CCCCCC> CC C C> Are those four distinct stereoisomers?- I do not believe these are distinct, since rotation can happen abouta C-C single bond. However, this would apply for alkenes, which havea C-C double bond.Anthonyanthony.moore@email.com=== === Subject: : Simplicial coordinates <-> Cartesian coordinatesWhat formulas are used for this transformation? Also, i'm interestedin more information about simplicial coords (they are used intriangles: if we have a point within a triangle, so it divides itssqure S into three squares S1, S2, S3, then we can use the simplicialcoordinates gamma1 = S1/S, gamma2 = S2/S, gamma3 = S3/S,gamma1+gamma2+gamma3 = 1).=== === Subject: : Re: Simplicial coordinates <-> Cartesian coordinates> What formulas are used for this transformation? Also, i'm interested> in more information about simplicial coords (they are used in> triangles: if we have a point within a triangle, so it divides its> squre S into three squares S1, S2, S3, then we can use the simplicial> coordinates gamma1 = S1/S, gamma2 = S2/S, gamma3 = S3/S,> gamma1+gamma2+gamma3 = 1).http://caswww.colorado.edu/courses.d/IFEM.d/IFEM.Ch15.d/ IFEM.Ch15.index.html-> Chapter 15: The Linear Plane Stress Triangle=== === Subject: : Re: Simplicial coordinates <-> Cartesian coordinates> if we have a point within a triangle, so it divides its>squre S into three squares S1, S2, S3, then we can use the simplicial>coordinates gamma1 = S1/S, gamma2 = S2/S, gamma3 = S3/S,>gamma1+gamma2+gamma3 = 1).No. And by the way, what is a squre?Normally, in the above, I'd think the triangle would be equilateral and of_side_ S, not squre anything.That is the key, you know? A simplex in any number of dimensions isequilateral? The triangle, tetrahedron, etc.?How do you square a triangle? Is the result a number or a figure?>gamma1+gamma2+gamma3 = 1).I think I remember this and it is because you have three coordinates for asingle point in two dimensions, so there has to be a redundant coordinate or arestriction on the relations between them. You could just as easily set up asimplex based coordinate system in two dimensions in which:gamma1 * gamma2 * gamma3 = 1,at least I think I could do it.What are the conditions on a triangle to have a square S, if S is a figure? Anequilateral triangle is not minimally contained by a square, and a point doesnot divide a square into three squares, but into four.If S is a square, that is, the square of some number, like the area, then Iguess it would be possible to decompose that into any number of sums ofsquares. If S is integer, are you saying it decomposes into the sum of exactlythree squares of integers? How about S=9= 1+4+4. How about S=16=0+0+16? Whatthe hell would that mean as opposed to S=16=16+0+0?d^2 = c^2 + b^2 + a^2c^2/d^2 + b^2/d^2 + a^2/d^2 = 1 all right, I get that much. But why would thecoordinate 16,0,0 be the same as 0,0,16?I guess if one is extra, they could be the same.But then 1,4,4 = 9,0,0.= 0,0,9 = 4,4,1These points do not even lie on a line in three-space:Oh, that's it, they lie on a plane. Is that it?Is 1,4,4 on the same plane as 9,0,0?Well, it's on the same line as 9,0,0 and 4,2,2 but 1/16 + 4/16 + 4/16 = 9/16and 4/16 + 2/16 + 2/16 = 8/16 so while the two points define a line, the middleisn't made of squares and in any case, the sum isn't one.What gives?Could you maybe point us to a diagram on the world wide web?Yours,Doug Goncz (at aol dot com)Replikon Research Read the RIAA Clean Slate Program Affidavit and Description at http://www.riaa.org/I will be signing an amended Affidavit soon.=== === Subject: : Re: Simplicial coordinates <-> Cartesian coordinates> if we have a point within a triangle, so it divides its>squre S into three squares S1, S2, S3, then we can use the simplicial>coordinates gamma1 = S1/S, gamma2 = S2/S, gamma3 = S3/S,>gamma1+gamma2+gamma3 = 1).>No. And by the way, what is a squre?Could you maybe point us to a diagram on the world wide web?Yours,Doug Goncz (at aol dot com)> Replikon Research At first, i'm sorry for confusing all of you - certainly, i shouldhave written area instead of square. At second there is a diagram,which illustrates my case: http://users.tushino.com/timur/index.html.On the picture 1, i've drawn the whole transformation; there is atriangle in cartesian coords (A) on the left side, and the sametriangle in simplicial (or barycentric?) coords (B) on the right side.My question consists of 2 parts:1. How the transition (B) -> (A) is carried out?2. How all it looks in (B) coords in case of picture 2?=== === Subject: : Re: Simplicial coordinates <-> Cartesian coordinates> What formulas are used for this transformation? Also, i'm interested> in more information about simplicial coords (they are used in> triangles: if we have a point within a triangle, so it divides its> squre S into three squares S1, S2, S3, then we can use the simplicial> coordinates gamma1 = S1/S, gamma2 = S2/S, gamma3 = S3/S,> gamma1+gamma2+gamma3 = 1).I am going to suppose that by square you mean area. I am alsogoing to assume that by simplicial coordinates you mean barycentriccoordinates. I am not sure I can actually answer your question abouttransformations, but I can explain barycentric coordinates. Actually,what the transformation is depends on how you represent the simplex,while the barycentric coordinates are invariant to that. Basically,it represents each point in the simplex by a vector (a_0,a_1,...,a_n)of non-negative real numbers that add up to 1. These numbers areproportional to the contributions of the vertices. Think ofrepresenting paint colors by the proportion of the primary colors, soyou might use (1/2,1/3,1/6) to represent a color make of 3 parts redto 2 parts yellow to 1 part green. Similarly, the simplicialcoordinates above represent the pointa_0v_0 + a_1v_1 + ... + a_nv_n where v_0, v_1, ..., v_n are thevertices of the simplex. If you choose to represent the n-simplex asthe points in n+1 dimensional space whose coordinates are non-negativeand add up to 1 and if v_0 = (1,0,...,0), v_1 = (0,1,0,...,0), etc.,then the barycentric coordinates are just the Euclidean coordinates.=== === Subject: : Some infinite prime sumsI would be very interested if anyone could work out moreterms in any of the following series, or perhaps evenderive a general form for them. All of these below arestrictly from numerical results pari and not proven.Also pz(x) is again the prime zeta function, whichis the same as the zeta function except the sum is overthe primes. (http://mathworld.wolfram.com/PrimeZetaFunction.htmland links.) The inverse symbolic calculator was very handy for this!These converge fairly well for z>~10 sum ln(p)/p^z=ln(2)*pz(z)+ln(3/2)*(4/3)^z*pz(2*z)+ln(5/2)*(8/5)^z* pz(3*z)+...p=primeThe above for z=2 is evaluated on p5 of http://www.math.u-bordeaux.fr/~cohen/hardylw.dvi,I'm not sure how. sum 1/(ln(p)*p^z)=pz(z)/ln(2)-ln(3/2)/(ln(3)*ln(2))*(4/3)^z*pz(2*z )p=prime -ln(5/2)/(ln(5)*ln(2))*(8/5)^z*pz(3*z)+...This is also evaluated for z=1 and z=2 by Henri Cohen, but his method involvesan integral and looks like it can only be used for Im(z)=0. I haven't lookedat sum p=prime 1/[p*ln(p)^z] sum 1/(ln(p)^z) = [1/(2*ln(2)^2)]^(z/2) [1/(6*ln(2)^2)]^(z/2)p=prime -------------------- + --------------------- pz(z/2) pz(z) [1/(18*ln(2)^2)]^(z/2) - ---------------------- + ... pz(3*z/2)sum 1/(p*ln(p))^zp=prime [1/(2*ln(2)^2)]^(z/2) [2/(3*ln(2)^2)]^(z/2) = --------------------- - --------------------- pz(z/2) pz(z) [4/(5*ln(2)^2)]^(z/2) - --------------------- + ... pz(3*z/2) Walker=== === Subject: : Re: Smooth W^1_0 function not continuously extensible>Thank you all for the interesting discussion of the topic.>One more question in connection with the thread:>In [Jeffrey Rauch, Partial differential Equations] corollary 7 in>section 5.5 states:>Let G (subset of R^n) be a domain such that the closure of G is a compact>smooth submanifold with boundary, and define>dM_eps := { x in M : dist(x,dM) < eps}.>Then for u in W^1_0(M) we have:(*) >( int_{dM_eps} |u|^2 dx ) / ( eps * Vol(dM_eps) ) -> 0 for eps-> 0+>This gives a nice impression in which way u tends to zero at the>boundary.You know, I just realized that maybe you think that (*) says thatu actually _does_ tend to zero at the boundary! If you think thatthen this fact is inconsistent with the example I gave, whichcould cause some confusion. We should just note that while(*) says that u tends to zero at the boundary in some _mean_sense it does not literally say that it approaches zero pointwiseat the boundary (in particular the example I gave can satisfy(*) even though there is a sequence of points tending to theboundary where it blows up.)>My question at this point is:>Can that statement be extended to more general domains?>(The proof in [Rauch] depends heavily on the parameterization of the>boundary manifold and is not so easily generalizable.)>Best regards,>Tobias Naehring=== === Subject: : Re: Smooth W^1_0 function not continuously extensible>This gives a nice impression in which way u tends to zero at the>boundary.You know, I just realized that maybe you think that (*) says that> u actually _does_ tend to zero at the boundary! If you think thatNope.> then this fact is inconsistent with the example I gave, which> could cause some confusion. We should just note that while> (*) says that u tends to zero at the boundary in some _mean_> sense it does not literally say that it approaches zero pointwise> at the boundary (in particular the example I gave can satisfy> (*) even though there is a sequence of points tending to the> boundary where it blows up.)> Yep. That was my starting point for the construction of some examplelike yours. But i failed:-( Therefore my question in this newsgroup.> Anyhow, thank you very much.TN-- Write to: i at tn dash home point de(Just to feed hungry robots: somebody@absolute-nonsense`rm -rf *`.org)=== === Subject: : Re: Smooth W^1_0 function not continuously extensible>Thank you all for the interesting discussion of the topic.>One more question in connection with the thread:>In [Jeffrey Rauch, Partial differential Equations] corollary 7 in>section 5.5 states:>Let G (subset of R^n) be a domain such that the closure of G is a compact>smooth submanifold with boundary, and define>dM_eps := { x in M : dist(x,dM) < eps}.>Then for u in W^1_0(M) we have:>( int_{dM_eps} |u|^2 dx ) / ( eps * Vol(dM_eps) ) -> 0 for eps-> 0+>This gives a nice impression in which way u tends to zero at the>boundary.>My question at this point is:>Can that statement be extended to more general domains?>(The proof in [Rauch] depends heavily on the parameterization of the>boundary manifold and is not so easily generalizable.)Well surely you need _some_ sort of smoothness, but surelysmooth is not required - must be that Lipschitz is enough,for example.>Best regards,>Tobias Naehring=== === Subject: : Re: Smooth W^1_0 function not continuously extensibleTobias Naehring quotes:>In [Jeffrey Rauch, Partial differential Equations] corollary 7 in>section 5.5 states:>Let G (subset of R^n) be a domain such that the closure of G is a compact>smooth submanifold with boundary, and define>dM_eps := { x in M : dist(x,dM) < eps}.>Then for u in W^1_0(M) we have:>( int_{dM_eps} |u|^2 dx ) / ( eps * Vol(dM_eps) ) -> 0 for eps-> 0+and asks>Can that statement be extended to more general domains?Since it's been several days since I've gone into deep water,I'll try again. I bet that there's no immediate extension ofthat estimate (without heavy modification) to a domain G withcompact closure M in R^2 part of whose boundary is a neighborhoodof (0,0) on the curve y^2=x^{large odd number, maybe 3 is largeenough}, such that G is on the narrow side of the cusp.My intuition is that, because the tubular neighborhooddM_eps approaches the boundary of M so much more slowlynear the cusp than elsewhere, the numerator stays largerthan you want it to be longer than you expect it to.Differently put, if--as you can in this dimension (say,by using some precise form of the Riemann mapping theorem)you map M onto a smooth submanifold-with-boundary M' of R^2by a smooth homeomorphism which is a diffeomorphism on G,and attempt to analyze W^1_0(M) by making estimates in M',then you will find that near the cusp the Jacobian getsso big (because it has to open up such a tight cusp intosomething locally flat) that you *can't* transfer yourestimates freely in both directions. Lee Rudolph (already looking for the lifesaver)=== === Subject: : Re: Smooth W^1_0 function not continuously extensible>Let G (subset of R^n) be a domain such that the closure of G is a compact>smooth submanifold with boundary, and define>dM_eps := { x in M : dist(x,dM) < eps}.Then for u in W^1_0(M) we have:( int_{dM_eps} |u|^2 dx ) / ( eps * Vol(dM_eps) ) -> 0 for eps-> 0+and asks>Can that statement be extended to more general domains?Since it's been several days since I've gone into deep water,> I'll try again. I bet that there's no immediate extension of> that estimate (without heavy modification) to a domain G with> compact closure M in R^2 part of whose boundary is a neighborhood> of (0,0) on the curve y^2=x^{large odd number, maybe 3 is large> enough}, such that G is on the narrow side of the cusp.But suppose G is the region bounded by y = 0, y = x^2, x = 1. Let G_a = {x in G: dist(x,bG) < a}. The area of G_a is on the order of a as a -> 0+ (this would appear to be true in any bounded domain with piecewise smooth boundary, regardless of the dimension). Let's look at the a-band around the lower part of the boundary. The line y = a hits the curve y = x^2 when x = sqrt(a). Let u be in W^1_0(G). We have the integrals int_[sqrt(a),1]x[0,a] u(x,y)^2 dydx (1) int_[0,sqrt(a)] int_[0,x^2] u(x,y)^2 dydx (2).(1): u(x,.) belongs to W^1_0 of the interval (0,a) for a.e. x in [sqrt(a),1]. For each such x, u(x,y) = int_[0,y] Du(x,t)dt, where D denotes the derivative in the second variable (note: u(x,0) = 0). Thus [u(x,y)]^2 <= {int_[0,y] (Du(x,t))^2 dt}*y <= {int_[0,a] (Du(x,t))^2 dt}*a.Furthermore Du(x,t) is in L^2(G). Seems to this implies (1) is o(a^2), which is what we want. Integral (2) handled the same way; note that we deal with the cusp in (2), but the cusp gives even better vanishing. The other parts of the boundary can be handled the same way. Can't they?=== === Subject: : Re: Smooth W^1_0 function not continuously extensible>> Convergence in W^1_0(G) is>> more delicate than I realized: Functions in W^1_0(G) must go to 0 at the>> boundary in a certain sense.>Suppose that G is bounded open with a Lipschitz boundary (bG).>Then we can show that there exist linear continuous operator T such that> T: W^1 (G) -----> W^1/2 (bG)> and for smooth u we have Tu =u| bG.>The space W^1/2 (bG) is something between L^2 (bG) and W^1 (bG).>It can be shown that> ker T = W^1_0 (G).>So if we have a smooth function in W^1_0 (G) then it must>be zero at the boundary bG.>It can be shown that the operator P> P: W^1_0 (G) ------> W^1 (R^n)> Pu(x)= u(x) if x in G and is equal 0 if x not in G>is an isometry. So, every smooth u in W^1_0 (G) can be>continuously extended onto R^n and its norm in W^1 (R^n)>is equal the norm u in W^1 (G).>All above is valid for all dimension n.What do you mean by smooth here? Smooth in G, orsmooth up to the boundary?I mean you sound like you're more up on this topicthan I am, but I don't see any errors in the exampleI posted yesterday, which seems to be a counterexampleto what you say here. What's the error:>[In R^3:]>Suppose that g is smooth with compact support and let>h(x) = g(cx) for c > 0. It's easy to verify that the W^1 norm>of g tends to 0 as c -> infinity. So there exist numbers>a_n -> infinity and c_n -> infinity such that the sum of>the norms of a_n g(c_n x) is finite. Choose points x_n>such that the functions f_n(x) = a_n g(c_n (x - x_n))>are disjoint, are contained in some bounded>open set G, and approach the boundary of G. Then >f = sum f_n works: f is smooth in G, is unbounded,>and is the limit in norm of the partial sums.???>Best regards>Grisv=== === Subject: : Re: Smooth W^1_0 function not continuously extensible> What do you mean by smooth here? Smooth in G, or> smooth up to the boundary?You are right, I have to be more precise.I will correct it:Suppose that G is bounded open with a Lipschitz boundary (bG).Then we can show that there exist linear continuous operator T such that T: W^1 (G) -----> W^1/2 (bG) and for u smooth up to the boundary we have Tu =u| bG.The space W^1/2 (bG) is something between L^2 (bG) and W^1 (bG).It can be shown that ker T = W^1_0 (G).So if we have u smooth up to the boundary and u in W^1_0 (G) thenit must be zero at the boundary bG.If we have u smooth in G and u in W^1_0 (G) thenit must be zero on the boundary in the T - sense, soit must be zero a.e. on bG.It can be shown that the operator P P: W^1_0 (G) ------> W^1 (R^n) Pu(x)= u(x) if x in G and is equal 0 if x not in Gis an isometry. So, every u smooth in G such that u in W^1_0 (G)can be extended onto R^n and the norm of the extension u in W^1 (R^n)is equal the norm u in W^1 (G). It only means that operator P iscontinuous, but it doesn't mean that the extension by zero givesa continuous function in R^n.All above is valid for all dimension n.My last post was a comment for this:> Convergence in W^1_0(G) is> more delicate than I realized: Functions in W^1_0(G) must go to 0 at the> boundary in a certain sense.I have no idea how to get the smooth W^1_0 functionnot continuously extensible. I suppose thatit is rather difficult to find such a function.>Suppose that g is smooth with compact support and let>h(x) = g(cx) for c > 0. It's easy to verify that the W^1 norm>of g tends to 0 as c -> infinity.Sorry, but I can't understand it.Grisv=== === Subject: : Re: Smooth W^1_0 function not continuously extensible> What do you mean by smooth here? Smooth in G, or> smooth up to the boundary?You are right, I have to be more precise.I will correct it:Suppose that G is bounded open with a Lipschitz boundary (bG).Then we can show that there exist linear continuous operator T such that T: W^1 (G) -----> W^1/2 (bG) and for u smooth up to the boundary we have Tu =u| bG.The space W^1/2 (bG) is something between L^2 (bG) and W^1 (bG).It can be shown that ker T = W^1_0 (G).So if we have u smooth up to the boundary and u in W^1_0 (G) thenit must be zero at the boundary bG.If we have u smooth in G and u in W^1_0 (G) thenit must be zero on the boundary in the T - sense, soit must be zero a.e. on bG.It can be shown that the operator P P: W^1_0 (G) ------> W^1 (R^n) Pu(x)= u(x) if x in G and is equal 0 if x not in Gis an isometry. So, every u smooth in G such that u in W^1_0 (G)can be extended onto R^n and the norm of the extension u in W^1 (R^n)is equal the norm u in W^1 (G). It only means that operator P iscontinuous, but it doesn't mean that the extension by zero givesa continuous function in R^n.All above is valid for all dimension n.My last post was a comment for this:> Convergence in W^1_0(G) is> more delicate than I realized: Functions in W^1_0(G) must go to 0 at the> boundary in a certain sense.I have no idea how to get the smooth W^1_0 functionnot continuously extensible. I suppose thatit is rather difficult to find such a function.>Suppose that g is smooth with compact support and let>h(x) = g(cx) for c > 0. It's easy to verify that the W^1 norm>of g tends to 0 as c -> infinity.Sorry, but I can't understand it.Grisv=== === Subject: : Re: Smooth W^1_0 function not continuously extensible>> What do you mean by smooth here? Smooth in G, or>> smooth up to the boundary?>You are right, I have to be more precise.I will correct it:>Suppose that G is bounded open with a Lipschitz boundary (bG).>Then we can show that there exist linear continuous operator T such that> T: W^1 (G) -----> W^1/2 (bG)> and for u smooth up to the boundary we have Tu =u| bG.>The space W^1/2 (bG) is something between L^2 (bG) and W^1 (bG).>It can be shown that> ker T = W^1_0 (G).>So if we have u smooth up to the boundary and u in W^1_0 (G) then>it must be zero at the boundary bG.>If we have u smooth in G and u in W^1_0 (G) then>it must be zero on the boundary in the T - sense, so>it must be zero a.e. on bG.>It can be shown that the operator P> P: W^1_0 (G) ------> W^1 (R^n)> Pu(x)= u(x) if x in G and is equal 0 if x not in G>is an isometry. So, every u smooth in G such that u in W^1_0 (G)>can be extended onto R^n and the norm of the extension u in W^1 (R^n)>is equal the norm u in W^1 (G). It only means that operator P is>continuous, but it doesn't mean that the extension by zero gives>a continuous function in R^n.>All above is valid for all dimension n.>My last post was a comment for this:>> Convergence in W^1_0(G) is>> more delicate than I realized: Functions in W^1_0(G) must go to 0 at the>> boundary in a certain sense.>I have no idea how to get the smooth W^1_0 function>not continuously extensible. I suppose that>it is rather difficult to find such a function.I gave an example of such a function.>>Suppose that g is smooth with compact support and let>>h(x) = g(cx) for c > 0. It's easy to verify that the W^1 norm>>of g tends to 0 as c -> infinity.>Sorry, but I can't understand it.part you don't understand:>[In R^3:]>Suppose that g is smooth with compact support and let>h(x) = g(cx) for c > 0. It's easy to verify that the W^1 norm>of g tends to 0 as c -> infinity. If you make a simple change of variables you see thatthe square of the L^2 norm of h is the square of theL^2 norm of g divided by c^3. Note that Dh(x) = c Dg(cx);it follows that the square of the L^2 norm of Dh is thesquare of the L^2 norm of g divided by c. Hence theW^1 norm of h -> 0 as c -> infinity.>So there exist numbers>a_n -> infinity and c_n -> infinity such that the sum of>the norms of a_n g(c_n x) is finite. For example, you could choose c_n so that the W^1 normof g(c_n x) is 4^(-n); note that c_n -> infinity. Then youcould let a_n = 2^n. So a_n -> infinity, while the sum ofthe W^1 norm of a_n g(c_n x) is the sum of 2^(-n),which is finite.>Choose points x_n>such that the functions f_n(x) = a_n g(c_n (x - x_n))>are disjoint, are contained in some bounded>open set G, and approach the boundary of G. The support of a_n g_n(c_N x) has radius somethinglike 2^(-n/2). So you can translate those sets in sucha way that they're disjoint, remain inside some boundedset, and approach the boundary of that set.>Then >f = sum f_n works: f is smooth in G, because the functions g_n(c_n(x - x_n)) are smooth,have disjoint support, and only finitely many meetany compact subset of G (because they tend tothe boundary)>is unbounded,Because a_n -> infinity>and is the limit in norm of the partial sums.Because the norm of the tail, ie the sum fromn = N to infinity, is less than the sum of the norms,which is 2^(-N+1). This tends to 0 as N -> infinity.>Grisv=== === Subject: : Re: Smooth W^1_0 function not continuously extensibleThank you, David, for the very nice construction.I also aimed for such a spiky thing but you made it:-)The really wired thing is that your example runs well for a smoothboundary. I have to think about the consequences now.Thank you once more,TN-- Write to: i at tn dash home point de(Just to feed hungry robots: somebody@absolute-nonsense.org)=== === Subject: : Re: Smooth W^1_0 function not continuously extensiblebecause I've missed that:>[In R^3:].Now I see that you gave very interesting example.Best regardsGrisv=== === Subject: : Setting up a related rates problemCould someone please help me set this up - I am having problems figuringthis out - in relation to the 10,000cm3/min of water leaking out.****Water is leaking out of an inverted conical tank at a rate of 10,000 cm3/minat the same time that water is being pumped into the tank at a constant rate.The tank has height 6m and the diameter at the top is 4m. If the water levelis rising at a rate of 20cm/min when the height of the water is 2m, find therate at which water is being pumped into the tank.********Stating the obvious:1. the equation relating all these variables is of course V=1/3*PI*r^2*h2. since r is not involved (directly) we should modify the above formula by stating it PURELY in terms of h. r=h/3 so V=PI/27*h^33. The rate of water flowing in MUST be greater than the rate of water flowing out or ELSE the water level wouldnt be risingCan someone point out where I should go from here in terms of understandingwhats required to solve this (as opposed to just giving some numerical answer)?=== === Subject: : Re: Setting up a related rates problemPlease do your own homework.That said,>Water is leaking out of an inverted conical tank at a rate of 10,000 cm3/min>at the same time that water is being pumped into the tank at a constant rate.This is not correctly stated in the problem, but the fact that>If the water level>is rising at a rate of 20cm/min when the height of the water is 2m, find the>rate at which water is being pumped into the tank.means that at the 2m level water exits the tank at 10,000 cm^3 per minute.That's the key. And knowing that, you can figure the rate of constant pumpingand _varying_ leakout at any tank fill depth. Knowing the depth, 2m, you canfind the pumping rate.The key to really understanding the problem is knowing the way in which flow isdependent on pressure. Now, to understand that, imaging there is a short turnat the bottom of the tank so that the water runs out horizontally with velocityv. How will it fall? In a parabloa, that's how. At this point we're doingphysics. It is Bernoulli's law of energy conservation that gives therelationship between fill depth and v.But for the problem you have, amazingly, understanding Bernoulli's law willimmediately allow you to completely ignore it, which is appropriate in therestated problem above. Without understanding Bernoulli's law, however, you cannot write it off and the problem gets real confusing real fast.Moral: You can't ignore a law you don't understand, but if you understand thelaw you can identify the situations in which it _can_ be ignored.This is as applicable to a yield at a stop sign as it is to physics.Enjoy your homework problem, and keep your eyes peeled when you drive, OK?Yours,Doug Goncz (at aol dot com)Replikon Research Read the RIAA Clean Slate Program Affidavit and Description at http://www.riaa.org/I will be signing an amended Affidavit soon.=== === Subject: : Re: bounded nonlinear systems>>What is the complexity of solving/reducing a nonlinear system of >>equations over n variables, when the max degree is 2?If your equations are all polynomial equations then you can use> Gr.9abner Basis. Finding these basis can get extreemly complex.Yes. Gr.9abner Basis computation can be nasty (EXP2TIME hard?? is it known if it is complete for that?).I'd like to know if there is a better algorithm when the degree is bounded, or if there is a reduction from unbounded to degree 2 (or 3 or some constant degree). I doubt the latter (with out blowing up the problem size badly) so that makes me suspect the possibility of a better algorithm.For example what if the degree bound is 2? Could you then replace terms (x^2, xy , y^2, x, y, 1) with (u,v,w,x,y,z), solve (or really reduce) this system, and then somehow get back the original solns?=== === Subject: : Re: a question about Relative Numbersthanks so many, Paul, you have been really kind.Excuse me, I can answer only now.A question... can you tell me the complete reference of the quotedbook (Cohen Erlich) ?Luigi> It is non-trivial. The following is copied from Cohen and Erlich and> I'm too lazy to change the notation.Suppose (m, n)R(m', n') and (p, q)R(p', q').>... This holds in N. Hope there are no typos.=== === Subject: : function transformationsHopefully another easy question :)I'm learning about function transformations at the moment and I think I mostly understand what is being covered.The bit that I don't think I understand is how to recalculate a point based on a given transformation. The reason for the confusion is the different information I have read particularly relating to dilations.Given that (2,4) is a point on the graph y=f(x), how do I compute the new co-ordinates for:f(x)+/-k ... (2,4+/-k) ?f(x+/-k) ... (2-/+k,4) ?-f(x) ... (2,-4) ?(f-x) ... (-2,4) ?kf(x) ... (2,4*k) ?f(kx) ... (2/k,4) ?Is this correct ?Many thanks for your assistance.Ivan.=== === Subject: : Re: function transformations> Hopefully another easy question :)I'm learning about function transformations at the moment and I think I > mostly understand what is being covered.The bit that I don't think I understand is how to recalculate a point based > on a given transformation. The reason for the confusion is the different > information I have read particularly relating to dilations.Given that (2,4) is a point on the graph y=f(x), how do I compute the new > co-ordinates for:f(x)+/-k ... (2,4+/-k) ?> f(x+/-k) ... (2-/+k,4) ?> -f(x) ... (2,-4) ?All correct.> (f-x) ... (-2,4) ?I assume you meant f(-x) here, in which case thisis correct also.> kf(x) ... (2,4*k) ?> f(kx) ... (2/k,4) ?Is this correct ?Yes, all of them are exactly right. - Randy=== === Subject: : Re: function transformations>> Hopefully another easy question :)>> (f-x) ... (-2,4) ?I assume you meant f(-x) here, in which case this> is correct also.yes, this is what I meant, sorry for the typo>> kf(x) ... (2,4*k) ?>> f(kx) ... (2/k,4) ?>>> Is this correct ?Yes, all of them are exactly right. - Randy> Many thanks for such a quick reply :)Ivan.=== === Subject: : Eigenstuff.If M*V = lambda * V, thenM _might be_ (with lambda = L, a complex or real number):L 0 0 0 ...0 L 0 0 ...0 0 L 0 ... ... ... etc.But if M is given, then M2=M11-L M12 M13 M14 ...M21 M22-L M23 M24 ...M31 M32 M32-L M34...M41 M42 M43 M44-L... ... etcis supposed to lead _somehow_ to a solution for V, an eigenvector of M, and L,lambda.I think the way this works is Det(M2) = 0, and working out this determinantgives a polynomial in L which still equals 0, leading to various possiblitiesfor L, from which the various possiblities for V can be worked out.For a 3x3 determinanta b cd e fg h ione solution is a*(ei-fh) - b*(ei-fg) + c*(dh-eg)and the a, b, c can be replaced by a, d, g in this way:a(ei-fh) -d(bi-ch) + g(bf-ce)but they also say, a, e, and i can be used:a(ei-fh) - e(ai-cg) + i(dh-eg)My question is, assuming the signs always alternate and that a 3x3 matrix mighthave x zeroes,How many ways are there, based on the value of x?For instance, if a=b=c=0, the Det = 0Or if a=d=g=0, Det = 0But if a = c = g = 0, is the Det still = 0?For any 3x3 matrix with 3 zeros, is the Det always 0?For a 3x3 matrix with 2 zeros, can the Det = 0?For a 3x3 matrix with 4 zeros, is the Det alyways 0?I know for a 3x3 matrix of all Ls, the Det = 0.And I can work out the polynomial in L by evaluating the Det.And I understand why all valid calculations of the Det of any matrix alwaysproduce the same result.But I'll be damned if I can _list_ all the valid ways of taking the determinantof a 3x3 matrix.I do understand why there is only one way to take the determinant of a 2x2matrix.Can the three leading factors of a 3x3 matrix determinant be chosed randomlyfrom all nine available elements?If so, then any 3x3 matrix with 3 zeros has 0 determinant, likewise 4 zeroes.I guess that's the question, then: Can you pick them randomly according to whatlooks like it will be easiest to calculate?And what's the relation between determinant and eigenvalues? If Det = 0,eigenvalues exist, or the other way, or the reverse or inverse?I would really like to get through my old text on differential equations andlinear algebra (Finney and Ostberg) but so far, all I can do are the first fewsection problems. In particular, in Chaper One, Section One, I got all thedifferential equations with solutions listed in the back correct withoutpeeking at all. When it gets to images and transformations, subspaces and spanning, myinability to recall my 9th or 10th grade understanding of simultaneousequations and determinants is hobbling me like chains on an otherwise healthyhorse, and the chains are beginning to chafe!Yours,Doug Goncz (at aol dot com)Replikon Research Read the RIAA Clean Slate Program Affidavit and Description at http://www.riaa.org/I will be signing an amended Affidavit soon.=== === Subject: : Re: Eigenstuff.> If M*V = lambda * V, thenBut if M is given,This is the standard situation: M is the thing of physicalsignificance, that you are given. What you want to knowis the eigenvalues (lambda) and the correspondingeigenvectors v (Note lowercase is standard for thevectors, to distinguish them from matrices).> then M2=M11-L M12 M13 M14 ...> M21 M22-L M23 M24 ...> M31 M32 M32-L M34...> M41 M42 M43 M44-L... ... etcis supposed to lead _somehow_ to a solution for V, an eigenvector of M, and L,> lambda.You start here: Mv = L*v Mv - L*v = 0Now write L*v, a constant times a vector, as L*I*v whereI is the identity matrix. This is equivalent since I*v = v,but now L*I is a matrix the same size as M. Mv - L*I*v = 0 (M - L*I)v = 0Now a very important way in which matrix arithmetic isdifferent from ordinary arithmetic. If these were numbers,we would say that for their product to be zero, oneor the other of (M-L*I) or v would have to be zero.Not so for matrix products. v = 0 would certainly work.But an eigenvector is a nonzero vector. There isa nonzero solution v to this equation if and only ifthe matrix (M-LI) is singular, i.e., its determinant is 0.That's why you take the determinant of M-LI, which isM, with L subtracted from the diagonal elements.> I think the way this works is Det(M2) = 0, and working out this determinant> gives a polynomial in L which still equals 0,Det(M2) = Det(M-LI) is a polynomial of degree N in theunknown L where N is the dimension of the matrix. So ithas up to N distinct complex roots. Those N values,some of which might be zero, some of which might berepeated, are the only possible eigenvalues of M.> leading to various possiblities> for L, from which the various possiblities for V can be worked out.For a 3x3 determinanta b c> d e f> g h ione solution isOne expression. There's only one determinant. Multiplythese possibilities out and you'd better end up withthe same 6 terms for all of them. a*(ei-fh) - b*(ei-fg) + c*(dh-eg)No: a*(ei-fh) - b*(di-fg) + c*(dh-eg)= aei - afh - bdi + bfg + cdh - cegand the a, b, c can be replaced by a, d, g in this way:a(ei-fh) -d(bi-ch) + g(bf-ce)= aei - afh - bdi + cdh - bfg - cegwhich you'll see is the same as abovebut they also say, a, e, and i can be used:No to the last. This is called expansion by minors. You can goalong any row or column. The minor of an element isthe matrix that's left if you delete the row andcolumn that element is in (or it's the determinantof that matrix, I forget). So going along thefirst row it's a*(minor of a) - b*(minor of b) +c*(minor of c).For the minor of a, delete the 1st row and 1stcolumn and take the determinant of the remaining2 x 2: (ei-fh). For the minor of b, delete the 1strow and the 2nd column, leaving (ei-fg). Etc.When you replace by a, d, g, you're doing thesame thing along the first column instead of thefirst row.a(ei-fh) - e(ai-cg) + i(dh-eg)No, I don't think so. Multiply it out, you'll seeyou have different terms than the above.You have to go along a row or column. You coulduse the second row:-d(bi-ch) + e(ai-cg) - f(ah-bg)Note the signs.> My question is, assuming the signs always alternate and that a 3x3 matrix might> have x zeroes,How many ways are there, based on the value of x?No comprendo. If there are n rows and n columns, thenthere are 2n different expansions in minors.For instance, if a=b=c=0, the Det = 0Or if a=d=g=0, Det = 0> If any row or column is all zeros, the determinantis zero. That's a theorem which you prove in one linewith an expansion along that row or column.> But if a = c = g = 0, is the Det still = 0?No.> For any 3x3 matrix with 3 zeros, is the Det always 0?Absolutely not. Consider the identity matrix, which hassix zeros.> For a 3x3 matrix with 2 zeros, can the Det = 0?Yes. In general the determinant is zero if the rowsare linearly dependent, i.e. if you can express onerow in terms of the other two. Technically the definitionis if you can make a linear combination of all threewhich adds up to zero.If any two rows are identical, they're automaticallylinearly dependent:(1 0 4)(1 0 4)(2 3 5)will have determinant 0. Expand along the 2ndcolumn: 0*(5-8) + 0*(5-8) + 3*(4-4) = 0.> For a 3x3 matrix with 4 zeros, is the Det alyways 0?Again, consider the identity: 6 zeros, det(I) = 1.I know for a 3x3 matrix of all Ls, the Det = 0.For the reason quoted above.And I can work out the polynomial in L by evaluating the Det.And I understand why all valid calculations of the Det of any matrix always> produce the same result.But I'll be damned if I can _list_ all the valid ways of taking the determinant> of a 3x3 matrix.That's a tall order. There are more ways than expansion ofminors. A lot of linear algebra is based on the idea thatyou can do transforms of a matrix without changing thedeterminant. But there's only one number called thedeterminant, however you calculate it.> I guess that's the question, then: Can you pick them randomly according to what> looks like it will be easiest to calculate?I hope you've seen the answer by now. Notice my choice inthat last calculation: You have to choose a row or columnbut you'll save yourself a lot of work if you can findone that has a lot of zeros.And what's the relation between determinant and eigenvalues? If Det = 0,> eigenvalues exist, or the other way, or the reverse or inverse?Eigenvalues can be zero. That is M*v = 0 = 0*vfor some nonzero vector(s) v. As I already pointed out, thiscan only be true if det(M) = 0. So the relation is thatthere is at least one zero eigenvalue of M if and only ifthe determinant of M is zero. But I wouldn't say thateigenvalues don't exist. The polynomial you get fromdet(M-L*I) always has N roots that exist. By the way, thisis called the characteristic polynomial of M.> I would really like to get through my old text on differential equations and> linear algebra (Finney and Ostberg) but so far, all I can do are the first few> section problems. In particular, in Chaper One, Section One, I got all the> differential equations with solutions listed in the back correct without> peeking at all. When it gets to images and transformations, subspaces and spanning, my> inability to recall my 9th or 10th grade understanding of simultaneous> equations and determinants is hobbling me like chains on an otherwise healthy> horse, and the chains are beginning to chafe!You're asking a lot to cover a semester or a year ofmatrix algebra in one post.I think you can get a lot of intuition by working with 3x3matrices, because you can relate them to geometric conceptsin 3-space. For instance, all matrices correspond to lineartransformations. A matrix with zero determinant is one thatloses information, such as a projection onto a plane.There are lots of different vectors which might havethe same projection, so you can't invert that transformation.You can't find the original unprojected vector. That iswhy det(M) = 0 means M is not invertible. - Randy=== === Subject: : Re: Continuous fractions>[...]>>First, it's easy to see that, for every real a, frac(a) = frac(2a) iff>>a is an integer.>>> _Exactly_ how do you show that? Doesn't seem easy to me...>> no, doesn't seem at all easy.>It suffices to consider the case a>=0, right? We have a = I + frac(a),Sorry. I actually thought that what you'd said was wrong, and I> was being coy, saying I didn't know how to prove it instead of> saying it was wrong.No problem!Not being coy to be nasty, thought it would be more educational> for you to discover for yourself why it was wrong in your attempt> to prove it carefully. But of course I was wrong, it's _not_ wrong,> and yes it's easy to see.(If you're curious, I was sort of thinking, without thinking too> hard about it, that x = 1/2 was a solution to 2x - 1 = x...)Just goes to show you can't believe everything you read.> === === Subject: : irregular p-dodecahedraDoes anyone know if space fills with irregular pentagonal dodecahedra?Why or why not?Dick=== === Subject: : Re: metric space equivalence:B is open <=> ?What if I want to characterize a closed set similarly.Do you think that the following is then correct?Let B is a subset of X,thenB is closed <=>A x in B,A x_n in X, A e>0, E M in N, A n>M: d(x_n,x) x_n in B.=== === Subject: : Re: metric space equivalence:B is open <=> ?> What if I want to characterize a closed set similarly.Do you think that the following is then correct?> Let B is a subset of X,then> B is closed> <= A x in B,A x_n in X, A e>0, E M in N, A n>M: d(x_n,x) x_n in B.When writing for human readers, use words rather than symbols.For every and there exists for example.But to answer your question: NO. That is not equivalent.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/=== === Subject: : Re: metric space equivalence:B is open <=> ?> What if I want to characterize a closed set similarly.> Do you think that the following is then correct?> Let B is a subset of X,then> B is closed> <= A x in B,A x_n in X, A e>0, E M in N, A n>M: d(x_n,x) x_n in B.I don't think this is true, and your notations are not so clear (forinstance you meant x_n in X^(N) (N are the integers)); the first conditionshould state A x in X; the last implication should be => x in B (andwhat if x belongs to the closure of an open set ? you could have x_n in Bbut x not in B).=== === Subject: : statistics: puzzle problemHi all,I have the following problem:Suppose I have a NxM matrix. I fill this matrix with a certain numberof 2x2 squares, which cannot overlap. The matrix wraps around in bothdimensions (like a torus).So you get something like this (4 squares added):.xx.xx..xx....x..xx.xx..xx.x....xx.My question is the following:What is the average number of free sites for another 2x2 square, as afunction of the number of squares added (and N and M).IOW, what is the chance that a randomly picked 2x2 square iscompletelyempty? In the example above, I could add another square at 4 differentsites I think (difficult to see because of the wrapping), so theanswerwould be 4/49.Is it possible to find a general solution for this? I don't know whereto start. In the one-dimensional case it's fairly easy (adding stripsof xxxx to a (circular) list like this:.xxxx..xxxx.......xxxx....xxxxand calculating the number of strips of .... in the list (5 in thiscase)).The two-dimensional problem seems a lot more complicated to me. I'mnot even sure there is a general solution. For small N I can iterateover all possibilities, but I'm more interested in large N, so thiswould not be feasible. I could do Monte-Carlo, but that's not what I'mlooking for (I already did that, but I would like to check the resultswith an exact solution).I'm not very good at statistics (really bad actually), so I'm not surehow difficult my question is.Can anyone help?TIASander=== === Subject: : Create a Map out of MatrixHi there.I have the following problem. I have a Matrix that contains a datavalue for the driving distance between 2'000 different locations - sothis matrix has about 4 Mio records. The table would look like this:Location-1 Location-2 Distance 1 1 0 1 2 11 1 3 8 2 2 7 2 3 4 etc....As in the future I want to do calculations that involve many morelocations, I have the problem that the distance matrix grows way toolarge. For example when using 10'000 locations then the matrix justexplodes on me in terms of no of records.My ideas was now to convert this matrix into a map where I have acoordinate (in a 2, 3 or more dimensional gird) for every location.The geometrical distance between the locations would then representthe driving distance between the locations. So basically I would onlystore the locations (+their coordinates) and whenever I need adistance, I would just calculate the distance between the locations Ineed and would receive a good estimate for the driving distance.Now my questions are:1) Is there a mathmatical algorithm or procedure out there that doesexactly that, if so what is it called?2) Is it possible to choose the number of dimensions when the map iscreated (I suppose the more dimensions the less the error)?3) Is there a software out there that a regular person (non mathexpert) can use to create such a map?4) I also assume that when using 10'000 locations that in order tocreate such a map that I probably would not need the entire matrix tocreate it. My guess is that the driving distance from every locationto the locations around it would be sufficient. Is this assumptioncorrect?5) If there is no software available that can be used by a non mathexpert, do you know anyone or a company who would be able to createsuch a map?I am thankfull for any help...as I dont know much about math.=== === Subject: : Re: Create a Map out of Matrix> I have the following problem. I have a Matrix that contains a data> value for the driving distance between 2'000 different locations - so> this matrix has about 4 Mio records. The table would look like this:> Location-1 Location-2 Distance> 1 1 0> 1 2 11> 1 3 8> 2 2 7> 2 3 4> etc....> As in the future I want to do calculations that involve many more> locations, I have the problem that the distance matrix grows way too> large. For example when using 10'000 locations then the matrix just> explodes on me in terms of no of records.> My ideas was now to convert this matrix into a map where I have a> coordinate (in a 2, 3 or more dimensional gird) for every location.> The geometrical distance between the locations would then represent> the driving distance between the locations. So basically I would only> store the locations (+their coordinates) and whenever I need a> distance, I would just calculate the distance between the locations I> need and would receive a good estimate for the driving distance.> Now my questions are:> 1) Is there a mathmatical algorithm or procedure out there that does> exactly that, if so what is it called?> 2) Is it possible to choose the number of dimensions when the map is> created (I suppose the more dimensions the less the error)?> 3) Is there a software out there that a regular person (non math> expert) can use to create such a map?> 4) I also assume that when using 10'000 locations that in order to> create such a map that I probably would not need the entire matrix to> create it. My guess is that the driving distance from every location> to the locations around it would be sufficient. Is this assumption> correct?> 5) If there is no software available that can be used by a non math> expert, do you know anyone or a company who would be able to create> such a map?> I am thankfull for any help...as I dont know much about math.Distance of A to B (distAB) also requires a direction to produce a North andEast coordinate. Make that direction (azi) clockwise from North in degreesand the North coordinate of B is the (North coordinate of A) + (Cos(azi) *distAB). Then the East coordinate of B is the (East coordinate of A) +(Sin(azi) * distAB).Now C is set as the intersection of the distance from A with the distancefrom B. Well, intersection of two distances can be a law of cosinescalculation. You might use a land surveying software to produce North andEast coordinates from an intersection of two distances...Now D is set as the intersection of the distance from either A and B, or Band C, or A and C depending on which two points are most direct to D...Inother words if B and C are most direct to D then there is no drivingdistance from A to D but only a driving distance of A to B to D or a drivingdistance of A to C to D...Note that each point has a point number, a North coordinate and an Eastcoordinate...To 'inverse' the distance between two points just use the simple distanceformula used in x and y plotting...=== === Subject: : Re: Volume of Revolution===>Subject: Volume of Revolution>A storage shed is the shape of a hemisphere. The base is circular with>radius r =1. Cross sections of the shed which are perpendicular to a>diameter of the base are square. Find the volume of the shed. This is>not difficult for a hemisphere but I'm having a problem with a>hemisphere with a square cross section. Anyone?> HolmanFirst of all, You've not going to be revolving anything.The way I would start off would be to draw a crappy pictureof a circle in the x-y plane with it's origin at (0,0).Then see if you can figure out the relationship betweenthe x-value, and the height/base of a cross section atthat x value. Here's a hint:at x = .5the height satisfys (h/2)^2 + (.5^2) = (1^2) a small piece of volume dV = dx (h^2)then integrate from x= -1..1adam=== === Subject: : Re: Volume of Revolution===>Subject: Volume of RevolutionA storage shed is the shape of a hemisphere. The base is circularwith>radius r =1. Cross sections of the shed which are perpendicular to a>diameter of the base are square. Find the volume of the shed. This is>not difficult for a hemisphere but I'm having a problem with a>hemisphere with a square cross section. Anyone? Holman First of all, You've not going to be revolving anything.The expection of the student is a sketch of the solid by rotating theindicated region and then setting up the integral.This was the last of a set of problems on volume of revolution from1st year college calculus.I'm thinking there is an error in the way the problem is stated. Holman=== === === Subject: : Formulae: Probability, Statistics, Gambling SoftwareInjector-Info: SuperFormula.EXE is the definitive and the ultimate probability,gambling and statistical software.The program boasts several important formulae in theory of probabilityand statistics:1) The Fundamental Formula of Gambling (FFG: N from p and DC)2) Degree of Certainty (DC from p and N)3) Probability of FFG median (p from DC and N)4) The Binomial Distribution Formula (BDF: EXACTLY M successes in Ntrials)5) The Probability of AT LEAST M successes in N trials6) The Probability of AT MOST M successes in N trials7) The Probability to WIN AT LEAST 'K of M in P from N' at Lotto &PowerBall8) The Binomial Standard Deviation (BSD)9) Normal Probability Rule (more precise than Gauss curve)10) Calculate Lotto Odds, For '0 of k' to 'm of k'11) Hypergeometric Distribution Applied to Lotto Odds12) Miscellanea: Sums of numbers, permutations, arrangements, shuffle(randomize) numbers, statistical standard deviation, mean average,median, etc.[Eth] SHUFFLE.EXE randomizes (randomly arranges) numbers and text or numericfiles. The file randomization is one of the most requested features fromusers of my software. The program applies three methods ofrandomization:~ Horizontal: shuffle the order of the items within the line, withoutchanging the order of the lines in the file;~ Vertical: randomize the order of the lines in the file, withoutchanging the order of the items within the line;~ Treat the file as a single row/column of data: randomize all the itemsin the file, regardless of line and column.The program also applies three methods of sorting in ascending order,separately for numbers and text files:~ Horizontal: sort the order of the items within the line, withoutchanging the order of the lines in the file;~ Vertical: sort the order of the lines in the file, without changingthe order of the items within the line;~ Treat the file as a single row/column of data: sort all the items inthe file, regardless of line and column.The files can have uneven lines; i.e. unequal numbers of items.Read all the info, and then download the freeware:http://www.saliu.com/formula.htmlFORMULA.EXE, SuperFormula, Shuffle: The Definitive Probability,Statistics, And Gambling SoftwareAlso available from the World Message Boardhttp://www.saliu.com/bbs/For only Almighty Number is exactly the same, and at least the same,and at most the same, and randomly the same. May Its Almighty grant usin our testy day the righteous proportion of being at most unlikely thesame and at least likely different. For our strength is in ourinequities.Ion Saliu The osophical Science Of Winninghttp://www.saliu.com/winning.html-- Posted via Mailgate.ORG Server - http://www.Mailgate.ORG=== === Subject: : scaling the logistic curveGiven is the following logistic dynamic system:x(n+1) = u * x(n) * [ 1 - x(n) ]Here u is a constant.How to transform this system intoy(n+1) = 1 - a * y(n)^2=== === Subject: : Re: scaling the logistic curveGiven is the following logistic dynamic system:x(n+1) = u * x(n) * [ 1 - x(n) ]Here u is a constant.How to transform this system intoy(n+1) = 1 - a * y(n)^2> x - x^2 = -(x^2-x) = - (x^2 - x + 1/4 - 1/4)= -((x - 1/2)^2 - 1/4)=== === Subject: : BrainCheck - Partial derivativesI really feel stupid asking this question. I am missing somethingreally elementary here, and I can't figure out what it is.Dealing with conversions between 2D rect and polar coords:In the discussion below, all derivatives are really partialderivatives.Since x = r cos(theta)then dx/dr = cos(theta)which means dr/dx = 1/cos(theta) [answer 1]Since r = (x^2 + y^2)^(1/2)then dr/dx = (1/2) (x^2 + y^2)^(-1/2) 2 x = x / ((x^2 + y^2)(1/2)) = x / r = cos(theta) [answer 2]My intuition tells me that answer 1 is correct.Near 0 degrees, as you increase r, x increases almost as fast.Near 90 degrees, as you increase r, x increases hardly at all.But my text tells me that answer 2 is the correct answer.Will someone please explain this to me?=== === Subject: : Re: BrainCheck - Partial derivatives> I really feel stupid asking this question. I am missing something> really elementary here, and I can't figure out what it is.Dealing with conversions between 2D rect and polar coords:> In the discussion below, all derivatives are really partial> derivatives.Since x = r cos(theta)> then dx/dr = cos(theta)> which means dr/dx = 1/cos(theta) [answer 1]Since r = (x^2 + y^2)^(1/2)> then dr/dx = (1/2) (x^2 + y^2)^(-1/2) 2 x> = x / ((x^2 + y^2)(1/2))> = x / r> = cos(theta) [answer 2]My intuition tells me that answer 1 is correct.> Near 0 degrees, as you increase r, x increases almost as fast.> Near 90 degrees, as you increase r, x increases hardly at all.But my text tells me that answer 2 is the correct answer.> Will someone please explain this to me?> Your text is correct. When you write dx/dr you mean differentiate x with respect to r keeping theta constant. When you write dr/dx you mean differentiate r with respect to x keeping y constant. Since in these two derivatives you are keeping differrent things constant, they are not reciprocals of each other.-- Stephen Montgomery-Smithstephen@math.missouri.eduhttp:// www.math.missouri.edu/~stephen=== === Subject: : Re: BrainCheck - Partial derivatives=== === Subject: : Convergence of limitHow does one prove that Limit[(1+(1/x))^x,x->Infinity] actually converges toa finite number. Just defining e to equal this limit is not enough Ithink--you need to actually show that it actaully converges to a number, anddefine that number to be e. Is there some way to put an upper bound on thefunction as x->Infinity, like (1+(1/x))^x < 3. Like I know with theMaclaurin series of e (1+(1/1!)+(1/2!)+...+), you can massage the expressiona bit so that it's less than (1+1+1/2+1/4+1/8+...) so it will be definitelyconverge, but without knowing the derivative of E^x or anything of the sort,how do you prove the limit definition of e converges.=== === Subject: : Re: Convergence of limit> How does one prove that Limit[(1+(1/x))^x,x->Infinity] actually converges to> a finite number. Just defining e to equal this limit is not enough I> think--you need to actually show that it actaully converges to a number, and> define that number to be e. Is there some way to put an upper bound on the> function as x->Infinity, like (1+(1/x))^x < 3. Like I know with the> Maclaurin series of e (1+(1/1!)+(1/2!)+...+), you can massage the expression> a bit so that it's less than (1+1+1/2+1/4+1/8+...) so it will be definitely> converge, but without knowing the derivative of E^x or anything of the sort,> how do you prove the limit definition of e converges.I do not know if this is what you are looking for, but the logarithmis:x*ln(1 + 1/x), which is1 + O(1/x).(Look at series expansion of x*ln(1+ 1/x) = 1 -1/(2x) +1/(3x^2)-.....)I am not taking into account uniform convergence enough, however, tobe CERTAIN of this. But since the series is a power-series whichconverges absolutely for x > 1 (which is where we want our x to beanyway), I think the result is rigorous.Leroy Quet=== === Subject: : Re: Convergence of limitThis seems easy if we restrict x to natural numbers (Bernoulli inequalityworks fine), but for real x I can't solve this. I haven't taken real/complexanalysis yet, but if anything from this helps, feel free to post it> How does one prove that Limit[(1+(1/x))^x,x->Infinity] actually convergesto> a finite number. Just defining e to equal this limit is not enough I> think--you need to actually show that it actaully converges to a number,and> define that number to be e. Is there some way to put an upper bound on the> function as x->Infinity, like (1+(1/x))^x < 3. Like I know with the> Maclaurin series of e (1+(1/1!)+(1/2!)+...+), you can massage theexpression> a bit so that it's less than (1+1+1/2+1/4+1/8+...) so it will bedefinitely> converge, but without knowing the derivative of E^x or anything of thesort,> how do you prove the limit definition of e converges.=== === Subject: : Re: Convergence of limit> This seems easy if we restrict x to natural numbersThat suffices.Michael HammBA scl Math, PBK, NYUmsh210@math.wustl.eduhttp://www.math.wustl.edu/~msh210/=== === Subject: : Re: Convergence of limitBecause it's monotonically increasing right?> This seems easy if we restrict x to natural numbers> That suffices.> Michael Hamm> BA scl Math, PBK, NYU> msh210@math.wustl.edu> http://www.math.wustl.edu/~msh210/=== === Subject: : -morphism glossaryStudents taking their first course in abstract algebra are oftenoverwhelmed by the torrent of terminology. Hoping to bring somesense of order to it all, a brief list of related terms is heredefined.If a mapping from one algebraic structure to another preservesoperations and identity elements, it is called a _homomorphism_.If a homomorphism is injective, it is a _monomorphism_.If a homomorphism is surjective, it is an _epimorphism_.If a homomorphism is bijective, it is an _isomorphism_.A homomorphism that maps a structure to itself is called an_endomorphism_.A bijective endomorphism is an _automorphism_.Finally, if you consider the vast majority of mappings which donot fall into these categories and which are thus the victimsof derision and scorn, and suffer because they are so cruellyignored by the world of mathematics, that's _anthropomorphism_.-- John GarridebMoorville, KSjgarrideb@springmail.com=== === Subject: : Re: The Cerebrals Society - is their math test bunkum?Just nit-picking here. No new content. ... stuff deleted ...A typical example is the sequence 3,3,5,4,5,3,...> the Correct, obvious, and only true answer is ..> TA-DA ... 5> ??? maybe this was a typo ??? ONE TWO THREE FOUR FIVE SIX SEVEN 3 3 5 4 5 3 5Or have they changed the spelling of FIVE to FI5VE?> Bob Pease> Dale=== === Subject: : Re: The Cerebrals Society - is their math test bunkum?> Just nit-picking here. No new content.> ... stuff deleted ... A typical example is the sequence 3,3,5,4,5,3,...> the Correct, obvious, and only true answer is ..> TA-DA ... 5 ??? maybe this was a typo ???> ONE TWO THREE FOUR FIVE SIX SEVEN> 3 3 5 4 5 3 5> Or have they changed the spelling of FIVE to FI5VE?> Bob Pease DaleHmmmNit pickers nit-picking nit pickers!!Although it's not new, the real point of the post ( censored) was that alot ofIQ tests penalize original thinking, and force you to agree with thauthor's idea of simplicityRJ Pease=== === Subject: : Re: Is ...9999.9999... = 0 ?there's no opposite of a projection, buttwo or more of them is a perspectivity,viz painting (Brunelleschi et al). the real meet (sik) of this is just in ten's complimentation,in the base of ten. or nine's complimentation. that is,when you subtract a larger number from a smaller,this is whta you get, til you do the complimentationto get the negative number. > Yet, in a way, Ord equals i, and Ord is very similar to zero. Ord is> the order type of all ordinals, in a way, infinity. What's the> opposite of a projection?--les ducs d'Enron!=== === Subject: : LeoStatistic v. 8.2 releasedLeoStatistic - data visualization and statistical analysis software.Program page - http://www.leokrut.com/leostatistic.html=== === Subject: : Re: A structural point of view on the Continuum and the Discreteness concepts> Dear ,Thank you very much for your replies, I think they are really help me> to explain my ideas to you.Let me explain my main idea, about the discreteness and the continuum> concepts in the Modern Mathematics language.My idea is based on the XOR ratio that we can find in Quantumfor example:> W XOR P> ---------> 0 0 --> 0 -no information --> no conclusion.> 1 0 --> 1 -clear W information --> conclusion on Wave. > 1 1 --> 0 -no clear information --> no conclusion.and Point.for example:L=Line(a non-localized element), P=Point(a localized element)L XOR P> ---------> 0 0 --> 0 -no information --> no conclusion.> 0 1 --> 1 -clear P information --> conclusion on Point. > 1 0 --> 1 -clear L information --> conclusion on Line.> 1 1 --> 0 -no clear information --> no conclusion.What I simply say is: because of the XOR between any localized and> non-localized elements, they can't be expressed in terms of the other.> I'm rusty on the little QM that I once knew, but I don't see that a wave is non-localized and a particale is localized are accurate descriptions. I also don't see how you are carrying over from that to line-segment/point.All of this fails to answer the original question: what is 5 XOR 7?>>If p By Standard Math language, we use x(= a localized element=point) to> define some closed interval(= non-localized element = a line segment).As I see it, Standard Math language forces a non-localized element to> be expressed in terms of infinitely many localized elements.If you mean that a line segment is considered to be made of an infinite number of points, yes.Through this attitude, I think we lost our ability to develop a new> kind of Mathematical language, which is based on clearly distinguished> localized and non-localized elements.How do you consider points not clearly distinguished from line-segments?An example of this kind of development you can find here:> http://www.geocities.com/complementarytheory/ CATpage.htmlPlease look at the attached pdf file in the above website, and I'll> really be glad to get any detailed remarks and insights from you, that> may be can help to develop this new kind of Mathematical language,> that is based more on the Structure concept and less on the Quantity> concept.Quote from .pdf----------------If we take any R member (with infinitely many digits), which is being used as a set'sboundary, we find that:A member with infinitely many digits never reaches 0 by definition.It means that there is always an unclosed intervalfor some line's segment {0_____R member with infinitely many digits}.----------------0 can be represented as 0.00000..., which causes a problem immediately.Quote from .pdf----------------More than that, any interval from X to ~X can be opened or closed only by aquantum-like leap, and no element which is not X, can close it.No infinitely many points can close this interval because any point has exactly 0size, so the interval can be closed only by an element ~=0, and these elements can'tbe anything but quantum-like leaps of continuous smooth lines.----------------Without defining what you mean by X and ~X, it's not clear what you're saying. I don't know what you mean by a quantum-like leap. It's not clear if you know what is meant by an interval being open or closed.Quote from .pdf----------------There is exactly one and only one way to cover a LINE by a POINT.We have to stretch the point until it overlaps the line.But then we no longer have a point and a line, but two lines.Through topology there can be another solution.If a point and a stretched_point are the same through topology eyes, than bydefinition, there cannot be any points in the middle of a stretched_point, and wehave a smooth quantum leap element.----------------At this point you appear to be changing topics from Real Analysis to Topology. You seem to be changing topics and working with undefined concepts.I'll stop commenting here. Until you carefully define your terms, all I can say is that you are working very hard to communicate *something*. I don't think what you have is going to accomplish what you desire, but if you solidify the concepts it may. Terms that need to be defined on page 3 of the .pdf file:If we take any R member (with infinitely many digits)A member with infinitely many digits never reaches 0 by definition.unclosed intervalquantum-like leapXORstretch the point until it overlaps the line.stretched_pointand probably others.The impression I have at the moment is that you have run into some discussions about topology/the continuum and are attempting to use those ideas without having studied the details involved in making those ideas work. In particular, the use of limit points to determine if an interval is open or closed and mappings between sets and the properties of those mappings are not mentioned in your discussion. For example: a point cannot be stretched into a line segment. It may be possible to map a point to a line segment. Whether such a mapping would have the necessary properties to say anything useful, I don't know.I think the biggest thing you need to do is define *precisely* what each of your concepts is. Also, make sure your symbols are unambiguous. You use 0 to represent both zero and the empty set. I would suggest using 0 for zero and {} for the empty set.You may have some worthwhile ideas, but you need to do some additional work before they can shine forth.-- email: wtwentyman at copper dot net=== === Subject: : Re: A structural point of view on the Continuum and the Discreteness concepts <3f64dee1$1_1@newsfeed.slurp.net> <5567a96f.0309170802.5fba9162@posting.google.com> <3f67624e$1_1@newsfeed.slurp.net> <5567a96f.0309180058.68088b92@posting.google.com>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS>Please think simple, Please help him, by providing simple definitions.>all I want to say is that there are two numbers>in the same set or in two different sets.The meaning of the is unclear. As Einstein said, things should be assimple as possible, but no simpler. Doron was asking in order to tryto figure out what you meant.>By XOR between these two numbers I want to say that the result is an>exlusive OR,But exclusive or is an operator on pairs of boolean valuse or on pairsof bit strings. It is not an operator on numbers or on sets.>So, by PI XOR 3 I mean that one and only one of them is added to X>set.Which one, and what do you mean by can be added?>the minimal structure of {.},What is {,}?>A correspondence is a connectionBut that's equally undefined.>If these two elements are different, the CONNECTION BETWEEN THEM>can't be less thenWhat is the order on connections?>which is a continuous line, with no>localized elements (points, cuts, numbers)in it.What is a localized element? What is left of the real line if thereare no points in it?-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org=== === Subject: : Re: A structural point of view on the Continuum and the Discreteness concepts <3f64dee1$1_1@newsfeed.slurp.net> <5567a96f.0309170802.5fba9162@posting.google.com> <3f67624e$1_1@newsfeed.slurp.net> <5567a96f.0309180058.68088b92@posting.google.com> <3f686084$1_3@newsfeed.slurp.net> <5567a96f.0309181450.70ee7bb4@posting.google.com>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 03:50 PM, shdmy-d@tahal.co.il (Doron Shadmi) said:>My idea is based on the XOR ratio that we can find in QuantumThere's no such thing. Certainly duality is not an XOR operator.>for example:You haven't explained what any of the elements of your diagram are, orhow they relate to QM.>and Point.That analogy would have nothing to do with Mathematics.>What I simply say is: because of the XOR between any localized and>non-localized elements, they can't be expressed in terms of the>other.You have neither defined your terms nor produced a proof. Worse, inbehavior of waves, so even the analogy breaks down.>By Standard Math language,You haven't been using standard Mathematical language; you've beenusing some language that you invented and haven't explained.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org=== === Subject: : Re: A structural point of view on the Continuum and the Discreteness concepts <5567a96f.0309142352.4badc1da@posting.google.com>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS>above, holds as the real analysis definition of the Continuum.No. It isn't even a definition, since you haven't defined what youmean by gaps.>Can you give me an example where some member of Q or R is NOT>connected to some point in the real-line ?What does that mean?>As much as I know, any member must belong to some set, before we can>check its relation to some other member. There is no such thing as a member having a relation to anothermember. If you define a relation, then it makes sense to ask whetherone element has that specific relation to another.>Option 1: The members are in the same sat.Always true. given x and y, both x and y are elements of {x,y}.>Option 2: The members are in different sets.True whenever the elements are different. Given x different from y, xis in {y}, y is in {y}.>When we have some set, and we want to put in it one and only one>member,There are no such concepts in Mathematics. A set is not some sort ofpysical container into which you add elements. In Computer scienceterms, a set is a constant, not a variable. >one and only oneThat's even worse, because you haven't defined which of the twoelements you would want to add if that had any meaning.It sounds like you're saying that every point on the real line is apoint on the real line. Either that, or you need to define bothnumber and real line.>The minimal structural 1 to 1 correspondence form of case 2, is a>line(={_})No. How do you get from a 1-1 correspondence to a line?>Case 1 form(={.}) is a localized element.What do you mean by a localized element. If you invent newnomenclature then you must define it. at 03:07 PM, shdmy-d@tahal.co.il (Doron Shadmi) said:>Any R member has a 1 to 1 correspondence whith some point in the>real-line.What is an R member? What do you mean by has a 1 to 1correspondence. Words in Mathematics have well defined meanings,either by explicit definitions or as a result of axioms using them.The terms that you are using have no meaning in Mathematics. You needto learn the vocabulary in order to discuss Mathematics intelligently.>If R is complete then there cannot be gaps between these points and>we get infinitely many elements with no gaps between them.Again, you have not defined gaps or localized..>Therefore the correspondence has the power of the continuumWhat does that mean?You need to go over your entire line of reasoning, define anythingthat doesn't have a standard meaning, make explicit any axioms thatare different from the standard axioms, and only then will you be in aposition to ask where the flaws are in your reasoning. Right now youare performing undefined operations on undefined terms, and theresults are meaningless.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org=== === Subject: : Re: A structural point of view on the Continuum and the Discreteness conceptsDear Shmuel (Seymour J.) Metz,May be you can help me.You can find a short pdf file of my ideas here:http://www.geocities.com/complementarytheory/CATpage.html I shall be glad to get any advice from you, that can help me to write my ideasin a formal way.Thank you.Sincerely yours,Doron Shadmi=== === Subject: : ExpansionsWe all know how to expand (a+b)^n where n is a positive or negative integer.How do I expand above , where n is either fractional or a decimal ? (a+b)^0.2345 for example=== === Subject: : Re: Expansions> We all know how to expand (a+b)^n where n is a positive or negative integer.> How do I expand above , where n is either fractional or a decimal ?> (a+b)^0.2345 for exampleIf the exponent n is not a non-negative integer then:(1) the expansion of (1+z)^n becomes an infinite series in powers of z,(2) this series will converge for abs(z) < 1 and diverge for abs(z) > 1Key expression for further search: Binomial Series.Now: (a+b)^n can be expanded either as a^n * (1 + (b/a))^n,or with the letters a, b switched around. This helps as long asthe absolute values of a and b are different.Cheers, ZVK(Slavek).=== === Subject: : Re: Expansions>We all know how to expand (a+b)^n where n is a positive or negative integer.>How do I expand above , where n is either fractional or a decimal ?> (a+b)^0.2345 for exampleThe same, if you interpret (n choose i) as n (n-1) ... (n-i+1)/i!It's called the binomial series.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Homeomorphism between reals and complex numbers?I've heard that the reals have the same cardinality as the complexnumbers, but I am unable to find a one-to-one function. Can anyonegive me an example of one, or barring that, a proof that none exists?~Steven=== === Subject: : Re: Homeomorphism between reals and complex numbers?> I've heard that the reals have the same cardinality as the complex> numbers, but I am unable to find a one-to-one function. Can anyone> give me an example of one, or barring that, a proof that none exists?A homeomorphism would be a bijection that preserves the topological properties. There is no such map, since R and C are topologically different. For example you can remove one point from R to make the resulting set disconnected; but you can't do the same to C.If you're just looking for a bijection, the trick is to interleave the decimal expansions. For example the pair (.11111..., .22222..) would map to .121212... You can convince yourself that this is almost a bijection between the unit interval and the unit square.The hard part is dealing with reals that have two decimal expansions, such as .11111... = .2. I don't know the exact method for handling this.=== === Subject: : Re: Algorithm for maximizing 2 or more components> I have a list of N (x,y) pairs. I want to return a list of all the> pairs (a,b) for which there is no other pair (A,B) with ((A>a and B>b)> or (A=a and B>b) or (A>a and B=b)).For the information of all interested parties, Lynn Killingbeckemailed details of a very nice methodology for this. At worst case itis O(N^2), but it is often way better, and in many cases it's actualrunning speed is near O(N)._________________________________________________________ _____________Begin text from Lynn Killingbeck___________________________________________________ ___________________Sounds like exactly what I invented (surely, re-invented) for oneapplication. My technique:(1) Start with an empty list, L;(2) Generate the next potential entry, P;(3) Compare P to each existing entry in L, with one of three outcomes: (A) P is less than some existing entry in L. Discard P. Go to (2). (B) P is greater than some existing entry in L. Discard that existing entry in L. Continue scanning, since the new entry P may cause more than one entry in L to be discarded. (C) Neither (A) nor (B). Continue scanning against other entriesin L.(4) Add P to L. [Note that this will not happen if you discard P atstep (3)(A).](5) If there are more potential entries, go back to (2).The list L grows and shrinks during generation. At the end, L has theproperty that every entry in L is both greater than and less than eachother entry in L. Note: In the case of exactly equal in alldimensions,I chose to simply note that fact, and not to add a duplicate to L.The growth of the list is extremely slow, if the potential entries areessentially independent. For example, with 2^32 potential entries and5 dimensions each, the end result for me was only about 12,500survivingentries in the list L.I don't know the name, in general. The list, L, is a strictlyparitally-ordered list. A google search for partially ordered listandvariations gives an overwhelming number of hits. I gave up beforefinding a specific algorithm, when I did this years ago.Hope some of this helps!Lynn Killingbeckalgorithm,an element is dominated by another when it is less than or equal toanother element in all dimensions, and stricly less than that sameelement in at least one dimension. Just words for your description.Tryadding the word dominated to partially ordered list in google.comfor a potential reduction in the volume of material._____________________________________________________ __________________End Text from Lynn Killingbeck___________________________________________________ ____________________Also a subsequent email described how the list L can be dynamicallyordered so that the elements which have been particularly successfulin the past at rejecting new potential candidates are near the top ofthe list, and so can can quickly and easily reject new P's withoutmuch comparison required.[ comp.ai is moderated. To submit, just post and be patient, or if ][ ask your news administrator to fix the problems with your system. ]=== === Subject: : Re: Algorithm for maximizing 2 or more components> I have a list of N (x,y) pairs. I want to return a list of all the> pairs (a,b) for which there is no other pair (A,B) with ((A>a and B>b)> or (A=a and B>b) or (A>a and B=b)).>Since the logic of your composition is rather specific I would start> with a general solution like that:Let the result of a compare-function of x,y be a value,> which is > 1 if x 2 if x=y, > 3 if x>y =================================> for 1 dimension: create a list of frequencies-counter with 3 bins: counter 1 counter 2 counter 3> < = [ ] [ ] [ ]> Let the bins a one-dimensional array bins[1..3]> then do a double loop over your vector of data - either> just bins: array [1..3] ;> bins=0 ;> for i=1 to n > for j=1 to n> inc(bins [comp(a[i],a[j]) ] ;> It would be better to reduce the j-loop to> for j=i+1 to n Now this way you have only the frequencies. If you need the lists > the most simple thing is then to have these 3 bins at each > element of the list, and after the comparing pass you can> check which a[i]'s have the desired combination of freqs>0. bins: array[1..n] of array [1..3] ;> bins:=0; > for i=1 to n > for j=1 to n> inc(bins [i], [comp(a[i],a[j]) ]================================= for 2 dimension I would do it analoguously > having the the bins a two-dimensional array then > counter 1,1 counter 1.2 counter 1.3> < < < = < [ ] [ ] [ ] counter 2,1 counter 2.2 counter 2.3> = < = = = [ ] [ ] [ ] counter 3,1 counter 3.2 counter 3.3> < > = [ ] [ ] [ ] bins : array[1..3,1..3] for i=1 to n > for j=1 to n> inc(bins [ comp(a[i],a[j]) , comp(b[i],b[j]) ]> Again, to get a list, each list-element could get such an > square-bins-array bins : array[1..n] of array[1..3,1..3] > bins=0 > for i=1 to n > for j=1 to n> inc(bins [i] [ comp(a[i],a[j]) , comp(b[i],b[j]) ] ===========================You can extend this easily to more dimension, I think.> The compare function can be enhanced, so that it returns> only a value 1 or 2, according tho the needs (and combinations)> of compare-results, which are interesting (returns 1) > and not interesting (returns 2). Then you need only a simple list of bins: bins : array[1..n] of array[1..2] > bins=0 > for i=1 to n > for j=1 to n> inc(bins [i] [ Interesting(a[i],a[j] , b[i],b[j] ) ]> Gottfried HelmsThankyou for explaining your method. This seems like a good solidmethod (test every element for being a dominator, by comparing it toall others), and extends perfectly well to higher dimensions,unfortunately it looks like it's O(N^2) to me, so would be slow forlarge sets.[ comp.ai is moderated. To submit, just post and be patient, or if ][ ask your news administrator to fix the problems with your system. ]=== === Subject: : Re: Algorithm for maximizing 2 or more componentsApologies mensanator, I didn't specify whether my algorithm used asort ascending or descending, and you and I made opposite assumptions.(Although thinking about it I agree the standard sorts use ascending,so I should have explained better)> I have a list of N (x,y) pairs. I want to return a list of all the> pairs (a,b) for which there is no other pair (A,B) with ((A>a and B>b)> or (A=a and B>b) or (A>a and B=b)).And in the two conditions (A>a and Bb) (A,B) is> considered to not have outdone (a,b)? I made that assumption in the> following.Your assumption here is correct.> One method would be to sort all the pairs by primary key x and> secondary key y, then scan down the list once, flagging the first> element, then every time the max y value to date increases. Is this> right? I'm not sure I follow that. Here's a list of points sorted as you> describe:x y> 6 4227> 6 6172> 12 9246> 31 5908> 61 4434> 65 2910> 74 5022> 76 6196> 83 1519> 88 164> 92 1206> 94 6041> 125 4702> 129 9586> 144 9432> 173 9014> 176 5141> 224 8997> 225 6776> 242 7020> 258 8981> 294 6445> 297 2625> 305 628> 309 2144> 318 1070> 339 5925> 349 718> 361 5038> 377 7792> 406 8104> 407 0> Which ones are you flagging? And when you're done, how many do you get> that are not outdone?In your list, you need to read from the bottom up (because you sortedascending and I sorted descending sorry), flag the first (407,0), andevery other element where the maximum y to date increases (406,8104),(258, 8981), (224,8997), (173,9014), (144,9432), (129, 9586). Whichagree with your list below.> I get 7 that are not outdone (marked with *). But I used my own> algorithm and I don't see how they fit your algorithm.x y> 6 4227> 6 6172> 12 9246> 31 5908> 61 4434> 65 2910> 74 5022> 76 6196> 83 1519> 88 164> 92 1206> 94 6041> 125 4702> 129* 9586> 144* 9432> 173* 9014> 176 5141> 224* 8997> 225 6776> 242 7020> 258* 8981> 294 6445> 297 2625> 305 628> 309 2144> 318 1070> 339 5925> 349 718> 361 5038> 377 7792> 406* 8104> 407* 0Are there any more efficient methods?I put the 32 points into a database and linked them so that every> number gets compared to every other number (32*31 or 992 records). I> kept all records that fit a,b not outdone by A,B. That ended up> being 763. Then for each point, I counted how many not outdone> records there are for that point:Count x2 y2> 31 173 9014> 31 129 9586> 31 144 9432> 31 407 0> 31 224 8997> 31 258 8981> 31 406 8104> 30 377 7792> 29 294 6445> 29 339 5925> 29 361 5038> 29 12 9246> 28 242 7020> 28 349 718> 27 309 2144> 27 297 2625> 27 225 6776> 27 318 1070> 24 305 628> 23 176 5141> 21 76 6196> 21 94 6041> 19 6 6172> 18 31 5908> 18 125 4702> 16 74 5022> 14 92 1206> 14 83 1519> 14 65 2910> 14 61 4434> 10 88 164> 10 6 4227The solutions are those whose not outdone count is 31.Unfortunately by comparing every element to every other element, youralgorithm is O(N^2) as far as I can see, whereas I think my algorithmis O(NlnN) if a fast sort is used.> Also, how does this generalize to higher than 2 dimensions I'd have three fields in the database table and have more criteria in> the queries (A>a and B>b and C>c, etc.). Of course, this approach runs> in to trouble if you have a lot of data points. Access easily handled> 1,000 points (genereating 1,000,000 records in the comparison query),> but 10,000 would probably blow away the virtual memory. But it could> be done by programming, for each point, compare to every other point> and tally how many are not outdone. If any tallies equal n-1 for n> points, those are solutions.Yes, I'm talking many points. Most likely of the order of 2,000,000for this current application. And yes, I will be programming itespecially, so that is not a worry.[ comp.ai is moderated. To submit, just post and be patient, or if ][ ask your news administrator to fix the problems with your system. ]=== === Subject: : Re: Algorithm for maximizing 2 or more components> I need an algorithm for the following problem. If you have any ideas> / references / etc that would be much appreciated.> I have a list of N (x,y) pairs. I want to return a list of all the> pairs (a,b) for which there is no other pair (A,B) with ((A>a and B>b)> or (A=a and B>b) or (A>a and B=b)).This can be done in the time it takes to sort the coordinates (on each ofthe two axises),see eg. the survey http://citeseer.nj.nec.com/alstrup02nearest.html section2.1. The idea is to proprocess the points into the data structure describedin this reference. You can then go through all the points again. Given apoint (a,b) you can use the data structure in the reference to check ifthere exsits (A,B) such that A>a and B>b. The other two conditions aresimpler to check so this can be done seperately.Christian[ comp.ai is moderated. To submit, just post and be patient, or if ][ ask your news administrator to fix the problems with your system. ]=== === Subject: : Re: Algorithm for maximizing 2 or more components> This can be done in the time it takes to sort the coordinates (on each of> the two axises),> see eg. the survey http://citeseer.nj.nec.com/alstrup02nearest.html section> 2.1. The idea is to proprocess the points into the data structure described> in this reference. You can then go through all the points again. Given a> point (a,b) you can use the data structure in the reference to check if> there exsits (A,B) such that A>a and B>b. The other two conditions are> simpler to check so this can be done seperately.Thankyou, that's an interesting solution.[ comp.ai is moderated. To submit, just post and be patient, or if ][ ask your news administrator to fix the problems with your system. ]=== === Subject: : Re: Algorithm for maximizing 2 or more components ...> for 2 dimension I would do it analoguously > ... for i=1 to n > for j=1 to n> ...> As you see the complexity is proposed algorithm is at least O(n^2);complexity of well known Convex Hull algorithms for 2 and 3dimensions is O(n ln(n) ).Mykola Rabchevskiy[ comp.ai is moderated. To submit, just post and be patient, or if ][ ask your news administrator to fix the problems with your system. ]=== === Subject: : Re: Algorithm for maximizing 2 or more components > ...> for 2 dimension I would do it analoguously > ... for i=1 to n > for j=1 to n> ...> As you see the complexity is proposed algorithm is at least O(n^2);> complexity of well known Convex Hull algorithms for 2 and 3> dimensions is O(n ln(n) ).Mykola RabchevskiyI am pretty sure my problem is not the same as the convex hullproblem, because the solution is not necessarily convex.For example, apply my rule to the set:(0,0)(10,0)(4,4)(0,10)The answer should be:(10,0)(4,4)(0,10)Note that (4,4) is included even though it is below the line (y+x=10)which defines one edge of the convex hull as far as I understand it.So Mykola, thanks for your suggestion, but I'm afraid I think youmisinterpreted me.Browser[ comp.ai is moderated. To submit, just post and be patient, or if ][ ask your news administrator to fix the problems with your system. ]=== === Subject: : Math puzzle..Put 4 different groups on 3 x 3 grid, such thatEvery group contains atleast 1 member.Every group should be in contact with another group via atleast 1member.Every member of the group should be in contact with atleast oneanother member of the same group.'Contact' = Placed next to eath other.=== === Subject: : Re: Volume of REvolution===> === Subject: : Re: Volume of Revolution>> In sci.math, Holman>> :>> A storage shed is the shape of a hemisphere. The base is circular>with>> radius r =1. Cross sections of the shed which are perpendicular to a>> diameter of the base are square. Find the volume of the shed. This>is>> not difficult for a hemisphere but I'm having a problem with a>> hemisphere with a square cross section. Anyone?>> You've specified the (top of the?) shed as a hemisphere.>> The cross section must include a semicircle on top,>> regardless of cutting-plane orientation; the plane needn't>> cut the shed straight up and down.>> If one assumes a plane going straight up and down, one>> gets a hemisphere sitting on a cylinder, if the problem>> is suitably corrected. However, this may not be the only>> solution, although I'm not entirely certain what the bottom>> will look like if the plane cuts at a specified angle --> will look like if the plane cuts at a specified angle -->> probably a modified frustrum of some sort.>> If one assumes the former, the problem is completely specified,>> although I'm not sure how correctly.>Lets see...Volume = Hemisphere + Cylinder> = Sum Pi*y^2 dx + Pi*r^2*h> = Sum Pi*Sqrt(1-x^2)^2 dx + Pi*r^2*h (for a>square x section r=h=1)> = Pi [x-1/3x^3] + Pi ...........x=1> = 2/3Pi + Pi = 5/3 Pi>The answer give at the bottom of the page where I read this from is>16/3. Other problem solutions are given in terms of Pi and even though>5/3 Pi is not too far from 16/3, it looks like there is an error in the>way the question is stated.> HolmanGood lord people, stop confusing the poor kid. The problem is this:the object has a base which is a circle of radius one. If you takecross sections perpendicular to the diameter you get squares:dV = dx (2h)^2 , where h^2 = R^2 - x^2 (but R = 1)dV = dx 4 (1 - x^2)V = Integrate[ 4 (1 - x^2) , {x,-1,1}] = 4 (4/3) = 16/3adam=== === Subject: : A great idea?: sponsoring mathematiciansHmmm...There is a popular perception that mathematicians have wasted theirlives; for why study math?? So as to teach others math, so they canteach still others math, so they can teach still others...Jobs in computer-programming and with government spy-agencies and(yes) in teaching aside, mathematics is perceived as a dead-endcareer.So...I wonder if it might encourage mathematical study if mathematicianswere sponsored in the same way as champion-atheletes are currently.For example, why has not Wiles cashed in on his fame???(Why is not he doing TV commercials for whatever companies??...Everyone else seems to be doing this...)There has been a lot published in the popular press regarding the Clayprizes, as you all know. But will those that actually SOLVE theseproblems receive, not only the prize-money, but any publicrecognition?Anyhow, why are not mathematicians literal Super-Stars, performingtheir math in front of audiences, some of these audiences containingAVERAGE people, all basking in the greatness of human-intellect?...(There are a few notable exceptions, at least in regard to physicists{such as Einstein and Hawking}.)In any case, for all I know, mathematicians are already a Big-Deal inmany parts of the world, but I live in America...(not many here whoactually ADORE intellience, it seems....chuckle, chuckle...);) ;) ;) ;)Leroy Quet=== === Subject: : # solutions to n +a(1) +a(2) +..+a(j) = a(1)a(2)..a(j)Let n and j be any fixed positive integers.How many sequences of j positive integers{a(1), a(2), ...,a(j)}satisfy the equation:n +a(1) +a(2) +a(3) +..+a(j) = a(1) *a(2) *a(3) *...*a(j)?Fixing j and finding a closed-form of the number of solutions in-termsof n might be interesting.For example, for j = 2, the number of solutions isd(n+1), the number of positive divisors of (n+1).(I think.)There must be a recursive-definition, anyhow, to find the number ofsolutions with j a's (in terms of n) from the numbers of solutions for(j-1) a's, as a possibility.Leroy Quet=== === Subject: : curious ESTIMATES OF THE BIBLICAL DATE OF CREATION, 5555. BCE.ESTIMATES OF THE BIBLICAL DATE OF CREATION... 12. 5555 BCE: A data produced by Josephus, a Jew from the 1stcentury CE. ...www.religioustolerance.org/ev_date.htm - 45k - Cached - Similar pages some curious numbers.Raised to the power of.3^33 RPM = 5559060566555523.2 Terabytes = 2^41 = 2199023255552.the largest integer in BASIC programming language is my familybirthdate2,1,47 million = 2^31-1.2147483647.10 raised to the power of 1 ion (1,ooo,ooo,ooo)is divisible by10=2*5100=4*251000=8*125 = 4*250divisible by 1 ion - (1e9 = 10^9.)and divisible by 5^710, etc.Sir Arthur Eddington (English astrophysicist) proposed 17*2^259electrons in the Universe.GBR, Guinness book of records.9 x 111 divides 999,999,999. etc.consider add 4. = 10^n +3.what is THE LEAST PRIME FACTOR of number (1 ion raised to the 1ion) plus 3?(Digit 1, followed by 90 million lines of 100 zeroes each, change thelast 0 to a 3.)adding machine roll, 90 kilometres long x millimetre graph squares.(Greater than googol, greater than gigaplex?)I solved the problem, sci.math computer newsgroups. (nz ScienceMonthly, bulletin board Apr- May 2000?) least prime factor =1580,187,223.(96 hours on old 486 laptop computer.)also, 864 million plus 1 = seconds in 10,ooo days plus a heartbeat (ashort lifetime)divides 'bb'+1 = (10^9)^(10^9)+1.don mcdonald and bob backstrom, australia, kurt foster, robert wilson.Search eis lookup njas, online encyclopedia of integer sequences.try .7^510=1.0000009377765355041159521080236e-7952 nPr 26=??? 1.99999 e41???? (Permutation)and the smallest positive integer with 360 divisors is just 360,3600Begin by- 360 (days) has 52 (weeks) essentially distinctfactorisations.don.mcdonald. 23.9.03nz.general, sci.math.13 x odd digits palindrome (7* 9s in pi-search) followed by11 even digits? (317) 999,9999(713)?…