mm-172 === Subject: Recursion and the Quadratic FunctionIf Recursion is related to Summation then If Summation is a Quadratic Function then Perhaps recursion routines can be solved by 'nding the correct Quadratic Function. Dick Saunders Jr. End IfEnd If === Subject: Re: branch of log z>... stop throwing stones, weÍre both in the glass house.> I dont want to throw stones, i donÍt even know, am i inside or> outside. It took me 25 years to 'nd out, what is behind this veil of:> imaginary axis, argand graph, complex plane, imaginary part. Caspar> Wessel and Hamilton didnÍt use or invented these, as i know now. These> words are still used in teaching people, who are interested in math:> Ask Dr.math > Mathworld.wolfram.com > Complex plane> These two donÍt mention, that they are talking in their websites about> the R2, just the simple R2 with + and *.> At another level - You mentioned Rudin,Marsden, Royden (books for> 100$): i only found some lecture notes to Rudin. Following these, he> starts with C=RxR, but then for example calling a rectangular> coordinate system an Argand Graph making this special. But you are> not told, what is the difference - it canÍt be the multiplication, can> it?IÍm sure you could 'nd any of these at your local library.> So really i am thankfull, that You told me,that i got this - no> differennce between i-axis and y-axis - right. Just one similar> opinion was recently written to sci.math from Lynn Kurtz. It seems to> me, that other people try to avoid a clear statement. So many> mathematicians are reading this and no con'rmation or opposition from> them can be found here in sci.math about it.> In R3 you can write a position or an arrow with (x,y,z) and sometimes> itÍs written as x*i+y*j+z*k or as x*e1+y*e2+z*e3. (linear combination> of three basis-vectors). Different notations for the same thing (and > multiplication of vectors is not a necessary condition for it).Good point. I was a bit hasty in suggesting that this would beimpractical.> When you introduce the dot-product in R3 you call it euclidian> vectorspace. But you wonÍt say itÍs isomorph to the 'rst, the dot> ignored. It stays the same, with extra dot-multiplication.> So i think (a,b) =a+i*b. a is called the Real part and b the> Imaginary part. a is the 'rst component, the 'rst coordinate and> you ommit 1 or k or e3, as R is embedded ( In R4 they do it too) .And> b is the second component, second coordinate.> Are You going here a step backwards : R^2 and the Complex plane are> isomorphic (if..... ) ? The only strict de'nition of C is> (R^2,+,*) , as far as i could 'nd. Do You have another one?Well, not of C, but of R^2 as a vector space. To be a bit moreprecise, R^2 is a vector space with + and scalar * and C is a 'eldwith elements from R^2 and the operations +, scalar * and vector*. In the sense you described, I am going backwards.> looks like: complex analysis and R2 analysis -one must be part of the> other. Is this so?Yes, complex analysis is a part of real analysis. The problem withthinking that R^2 analysis and complex analysis are the same is thatR^2 analysis usually works on the vector space R^2, whereas if oneworks with C, one usually works with the complex 'eld. Thatmultiplication issue is very important, since it puts constraints onwhich functions are differentiable (for instance). Have you heard ofthe Cauchy Riemann equations? Let f(z) be written as f(x,y) = u(x,y)+ iv(x,y). f is complex differentiable iff it is R^2 differentiable(that is, component-wise) and the following equations hold: du/dx =dv/dy and du/dy = -dv/dx. Check outhttp://mathworld.wolfram.com/Cauchy-RiemannEquations.html for a fullexposition.ItÍs a very elegant subject--my favorite thus far.> Feeling like a snail, unable to throw stones, creeping up an parking> ramp to an understanding of branches of log,> Hero === V|9[5IY]/OB15+}/e : m!6.qwÍKzn@$#CJQl/UvR,&G?sNO{#LEEnS3lQRnT#Ql#' 3ySQ80}u7Q<`+B@; Y;81AnKÍ_PD^fm.3T/<-i~YdZ%Q}b?:-]{Lr#s6.N,r(}Hi Bilge, Why do spend so much time in alt.morons ?The fact that you ignore the way Usenet actually works, while blabbering on about useless documentation, is very revealing.You canÍt ever win, can you Bilge ? === Subject: Re: . Very revealing . Jeff Relf: >Hi Bilge, Why do spend so much time in alt.morons ? > You must be even stupider than you appear from your posts. === Subject: Re: Lebesgue decomposition> U,v 'nite measures on (X,A). Suppose v has Lebesgue decomposition (d,N) with> respect to a measure u. How do you determine the Lebesgue decomposition of u> with respect to v?I am not inclined to check out the details. But suppose that the absolutely continuous part has Radon-Nikodym derivative f w.r.t. u. Then I would say that the singular part of u with respect to v is u restricted to the set {x:f(x)=0}.Well I am guessing that this might be a homework problem, so I am leaving you with the problem of (a) checking whether I am correct and (b) providing the details if I am correct. Can anyone help me to 'nd a paper about the application of broken?http://madeira.cc.hokudai.ac.jp/RD/takai/ automa.htmlhttp://cafaq.com/apps/index.php?open=complexityhttp ://www.ericweisstein.com/encyclopedias/books/ CellularAutomata.htmlhttp://mathforum.org/library/topics/ cellular_auto/http://ieeexplore.ieee.org/xpl/abs_free.jsp? arNumber=338094http://www.stephenwolfram.com/publications/ books/ca-wspc.htmlhttp://portal.acm.org/citation.cfm?id= 966317.966338&coll=GUIDE&dl=GUIDE[ comp.ai is moderated. To submit, just post and be patient, or if ][ ask your news administrator to 'x the problems with your system. ] === Subject: Re: Metamath Axiom of Choice> poohonlsd@yahoo.com because, as *everyone* parsed your> post, you are incapable of reading plain English but you pretend to> have the expertise to respond to questions anyway?> Well, I guess you must be right, then. Since *everyone* says so. By> the way, you should note that what I said was _accurate_. I only> responded to the wrong question. I really donÍt see why my gaffe has> inspired you to post two insulting messages about me.Correction: One insulting message. Your 'rst post was quitecordial--and correct, not the ascerbic dribble you posted later.> It shouldnÍt seem patronizing after all, since as he (and other> respondents) read your post, it was an answer an ignoramus might> write.> IÍm not exactly sure why youÍd want to get into a pissing contest with> me. But your reply is completely pointless, since it is in response> to a post that wasnÍt addressed to you.> ïcid ïooh> PS. Chill out. This has nothing to do with you. === V|9[5IY]/OB15+}/e : m!6.qwÍKzn@$#CJQl/UvR,&G?sNO{#LEEnS3lQRnT#Ql#' 3ySQ80}u7Q<`+B@; Y;81AnKÍ_PD^fm.3T/<-i~YdZ%Q}b?:-]{Lr#s6.N,r(}Hi Ha Ha Hanson, You incorrectly made the following equation, hanson@quick.net = !ed99b2be ,While hanson@quick.net is sure to be invalid, identify you every time you use Earthlink to post.I was just using your headers in a vain attempt to explain something to Bilge, IÍm not verifying you. === Subject: Re: help with solutions for three questions from the past contest> IÍd greatly appreciate your help on the solutions for these questions.> 1. Al and Bob are at opposite ends of a diameter of a silo in the> shape of a tall right circular cylinder with radius 150 ft. al is due> west of Bob. Al begins walking along the edge of the silo at 6 ft. per> second at the same moment that Bob begins to walk due east at the same> speed. The value closest to the time in seconds when Al 'rst can see> Bob is what? answer: 48> 2. if a, b, c, and d are nonzero numbers such that c and d are> solutions of x^2+ax+b=0 and a and b are solutions of x^2+cx+d, 'nd> a+b+c+d. ans: -2> 3. A boat with an ill passenger is 7.5 mi north of a straight> coastline which runs east and west. A hospital on the coast is 60> miles from the point on shore south of the boat. If the boat starts> toward shore at 15 mph at the same time an ambulance leaves the> hospital at 60 mph and meets the ambulance, what is the total distance> (to the nearest 0.5 mile) traveled by the solution for no. 2 so only help needed for no.1and no. 3. === Subject: Re: help with solutions for three questions from the past contest> 3. A boat with an ill passenger is 7.5 mi north of a straight> coastline which runs east and west. A hospital on the coast is 60> miles from the point on shore south of the boat. If the boat starts> toward shore at 15 mph at the same time an ambulance leaves the> hospital at 60 mph and meets the ambulance, what is the total distance> (to the nearest 0.5 mile) traveled by the boatand |___________________________________.Hospital x 60-xHours along distance z at 15 mph must equal hours along distance60-x at 60 mph.And at constant speeds, distance = speed * time. === Subject: Re: help with solutions for three questions message > 3. A boat with an ill passenger is 7.5 mi north of a straight> coastline which runs east and west. A hospital on the coast is 60> miles from the point on shore south of the boat. If the boat starts> toward shore at 15 mph at the same time an ambulance leaves the> hospital at 60 mph and meets the ambulance, what is the total distance> (to the nearest 0.5 mile) traveled by the boatand the ambulance? ans:> 62.5> Choogu> Boat .> |> | > 7.5| z> | > Shore |___________________________________.Hospital> x 60-x Hours along distance z at 15 mph must equal hours along distance> 60-x at 60 mph. And at constant speeds, distance = speed * time.IÍll be curious if thatÍs the best solution. (It may be the one expected.)A better solution would be to minimize the time to the hospital. ThatÍs astandard calculus question (usually phrased in terms of a swimming pool),with the additional proviso that if the ambulance wouldnÍt be to the meetingpoint at the time the boat gets there, then this solution should be used.Of course, I wouldnÍt be a proper kibbitzer if I actually worked out thedetails before posting this criticism.Dare we point out that, if itÍs a true emergency, the coast guard sends ahelicopter?Jon Miller === Subject: Re: help with solutions for three questions message > 3. A boat with an ill passenger is 7.5 mi north of a straight> coastline which runs east and west. A hospital on the coast is 60> miles from the point on shore south of the boat. If the boat starts> toward shore at 15 mph at the same time an ambulance leaves the> hospital at 60 mph and meets the ambulance, what is the total distance> (to the nearest 0.5 mile) traveled by the boatand the ambulance? ans:> 62.5> Choogu> Boat .> |> | > 7.5| z> | > Shore |___________________________________.Hospital> x 60-x Hours along distance z at 15 mph must equal hours along distance> 60-x at 60 mph. And at constant speeds, distance = speed * time.> IÍll be curious if thatÍs the best solution. (It may be the one expected.)> A better solution would be to minimize the time to the hospital. This does minimize the time to the hospital! === Subject: Re: help with solutions for three get no.2 and no. 3 'ne. but no luck on this one.any help will be appreciated.Choogu1. Al and Bob are at opposite ends of a diameter of a silo in theshape of a tall right circular cylinder with radius 150 ft. al is duewest of Bob. Al begins walking along the edge of the silo at 6 ft. persecond at the same moment that Bob begins to walk due east at the samespeed. The value closest to the time in seconds when Al 'rst can seeBob is what? answer: 48 === Subject: Need Code for Fractional Brownian Motion Analysis and SimulationI have looked everywhere for Matlab code for fractional Brownianmotion analysis and simulation. The only thing I found aftersearching with both Yahoo and Googe was a dead link. IÍm postingbecause I have no other recourse.Please sent me any info you can. Or post it to === Subject: covering compact set w/squaresGiven a compact set K in the plane s.t. each pt x is the center of a squareQ_x, prove that you can 'nd a subsequence Q_x_i of squares s.t. K iscovered by the Q_x_i and the sum of the areas of each of the Q_x_i is nomore than 4 times the area of the union of the Q_x_i.I think this is a totally geometric thing, and I know it suf'ces to showthat in such a minimal cover no point is contained in more than 4 squares,but how do i show this? I know how to do this with intervals on the realline (there the answer is 2 times the area....), but i get lost when goingup a dimension. === Subject: Re: covering compact set w/squaresX-DMCA-Noti'cations: 22:54:35 -0500, melnick Q_x, prove that you can 'nd a subsequence Q_x_i of squares s.t. K is>covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no>more than 4 times the area of the union of the Q_x_i.Aargh. I think IÍll just spend the rest of my life posting replies tothis message. I have to say that the two counterexamples IÍvegiven were wrong (I was reading the statement of the problemas no more than 4 times the area of K showing that itÍs not true thatin a minimal cover no point is in more than four squares wascorrect. (IÍm still not sure whether the result itself is true andI still wonder whether itÍs really what you wanted to prove.)>I think this is a totally geometric thing, and I know it suf'ces to show>that in such a minimal cover no point is contained in more than 4 squares,>but how do i show this? I know how to do this with intervals on the real>line (there the answer is 2 times the area....), but i get lost when going>up a dimension.>************************David C. Ullrich === Subject: Re: covering compact set w/squaresOh, I think I did misread. How about this:Given a compact set K in the plane s.t. each pt x is the center of a squareQ_x, prove that you can 'nd a subsequence Q_x_i of squares s.t. K iscovered by the Q_x_i andsum(over i) Char(Q_x_i) <= 4 Char (union(over i) Q_x_i),where Char(X) is the characteristic (indicator) function of X.David C. Ullrich Given a compact set K in the plane s.t. each pt x is the center of asquare>Q_x, prove that you can 'nd a subsequence Q_x_i of squares s.t. K is>covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no>more than 4 times the area of the union of the Q_x_i. Aargh. I think IÍll just spend the rest of my life posting replies to> this message. I have to say that the two counterexamples IÍve> given were wrong (I was reading the statement of the problem> as no more than 4 times the area of K instead of what you true that> in a minimal cover no point is in more than four squares was> correct. (IÍm still not sure whether the result itself is true and> I still wonder whether itÍs really what you wanted to prove.)I think this is a totally geometric thing, and I know it suf'ces to show>that in such a minimal cover no point is contained in more than 4squares,>but how do i show this? I know how to do this with intervals on the real>line (there the answer is 2 times the area....), but i get lost whengoing>up a dimension. ************************ David C. Ullrich === Subject: Re: covering compact set w/squaresX-DMCA-Noti'cations: 09:59:03 -0500, melnick Q_x, prove that you can 'nd a subsequence Q_x_i of squares s.t. K is>covered by the Q_x_i andsum(over i) Char(Q_x_i) <= 4 Char (union(over i) Q_x_i),where Char(X) is the characteristic (indicator) function of X.Not sure. That inequality is just another way to say that no pointis contained in more than 4 of the Q_x_i; IÍm not sure whetherone can get that, but we know it doesnÍt follow just from takinga minimial cover.When you say Oh, I think I did misread. How about this: itsounds like youÍre still not certain what the statement is.This might be a little simpler if youÍd get the statementstaight to begin with.(Um: IÍve been assuming that this is like a homeworkproblem. If itÍs not, something youÍre trying to do forsome other reason, I can tell you a standard result thatÍsvery much like what youÍre trying to prove, and which youmight 'nd equally Given a compact set K in the plane s.t. each pt x is the center of a>square>>Q_x, prove that you can 'nd a subsequence Q_x_i of squares s.t. K is>>covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no>>more than 4 times the area of the union of the Q_x_i.>> Aargh. I think IÍll just spend the rest of my life posting replies to>> this message. I have to say that the two counterexamples IÍve>> given were wrong (I was reading the statement of the problem>> as no more than 4 times the area of showing that itÍs not true that>> in a minimal cover no point is in more than four squares was>> correct. (IÍm still not sure whether the result itself is true and>> I still wonder whether itÍs really what you wanted to prove.)>>I think this is a totally geometric thing, and I know it suf'ces to show>>that in such a minimal cover no point is contained in more than 4>squares,>>but how do i show this? I know how to do this with intervals on the real>>line (there the answer is 2 times the area....), but i get lost when>going>>up a dimension.>> ************************>> David C. Ullrich>************************David C. Ullrich === Subject: Re: covering compact set w/squaresNo, itÍs not HW...it was an aside in a measure theory course that seemedkind of curious but that I couldnÍt immediately (and at this point justplain couldnÍt) verify.David C. Ullrich Oh, I think I did misread. How about this:Given a compact set K in the plane s.t. each pt x is the center of asquare>Q_x, prove that you can 'nd a subsequence Q_x_i of squares s.t. K is>covered by the Q_x_i andsum(over i) Char(Q_x_i) <= 4 Char (union(over i) Q_x_i),where Char(X) is the characteristic (indicator) function of X. Not sure. That inequality is just another way to say that no point> is contained in more than 4 of the Q_x_i; IÍm not sure whether> one can get that, but we know it doesnÍt follow just from taking> a minimial cover. When you say Oh, I think I did misread. How about this: it> sounds like youÍre still not certain what the statement is.> This might be a little simpler if youÍd get the statement> staight to begin with. (Um: IÍve been assuming that this is like a homework> problem. If itÍs not, something youÍre trying to do for> some other reason, I can tell you a standard result thatÍs> very much like what youÍre trying to prove, and which you> might 'nd equally Given a compact set K in the plane s.t. each pt x is the center of a>square>>Q_x, prove that you can 'nd a subsequence Q_x_i of squares s.t. K is>>covered by the Q_x_i and the sum of the areas of each of the Q_x_i isno>>more than 4 times the area of the union of the Q_x_i.>> Aargh. I think IÍll just spend the rest of my life posting replies to>> this message. I have to say that the two counterexamples IÍve>> given were wrong (I was reading the statement of the problem>> as no more than 4 times the area of K instead of what you not true that>> in a minimal cover no point is in more than four squares was>> correct. (IÍm still not sure whether the result itself is true and>> I still wonder whether itÍs really what you wanted to prove.)>>I think this is a totally geometric thing, and I know it suf'ces toshow>>that in such a minimal cover no point is contained in more than 4>squares,>>but how do i show this? I know how to do this with intervals on thereal>>line (there the answer is 2 times the area....), but i get lost when>going>>up a dimension.>> ************************>> David C. Ullrich ************************ David C. Ullrich === Subject: Re: covering compact set w/squaresX-DMCA-Noti'cations: 22:54:35 -0500, melnick Q_x, prove that you can 'nd a subsequence Q_x_i of squares s.t. K is>covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no>more than 4 times the area of the union of the Q_x_i.Sorry to reply three times, but the counterexample I gave wasutterly stupid, since it can be hugely simpl'ed: Let K be anycompact set with area = 1, and de'ne each Q_x to have sidelength 2. Duh.>I think this is a totally geometric thing, and I know it suf'ces to show>that in such a minimal cover no point is contained in more than 4 squares,>but how do i show this? I know how to do this with intervals on the real>line (there the answer is 2 times the area....), but i get lost when going>up a dimension.>************************David C. Ullrich === Subject: Re: covering compact set w/squaresX-DMCA-Noti'cations: 22:54:35 -0500, melnick Q_x, prove that you can 'nd a subsequence Q_x_i of squares s.t. K is>covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no>more than 4 times the area of the union of the Q_x_i.Ah. This is not true (hence the question: what you were really askedto show?). A minute ago I gave a counterexample to the statementbelow; a slight modi'cation gives a counterexample to the theorem:Let Q^1, ... Q^5 be squares of side length 1, such that the lowerleft corner of Q^j is the point (j/100, j/100). Let K be the union ofQ^1, ... Q^5.Now if x is in K and x happens to be the center of one of the Q^jde'ne Q_x = Q^j. If x is in K and x is not the center of one of theQ^j let Q_x be the square with center x and side length 1,000.ItÍs easy to see that there is no subcollection of the Q_x thatcovers K and has sum of the areas less than four times thearea of K: We certainly canÍt include any of the Q_x with sidelength 1000, but if we donÍt use any of them we need touse all 've of the Q^j to cover K, and the sum of the areasis almost 've times the area of K.>I think this is a totally geometric thing, and I know it suf'ces to show>that in such a minimal cover no point is contained in more than 4 squares,>but how do i show this? I know how to do this with intervals on the real>line (there the answer is 2 times the area....), but i get lost when going>up a dimension.>************************David C. Ullrich === Subject: Re: covering compact set w/squaresX-DMCA-Noti'cations: 22:54:35 -0500, melnick Q_x, prove that you can 'nd a subsequence Q_x_i of squares s.t. K is>covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no>more than 4 times the area of the union of the Q_x_i.Are you sure that this is exactly what youÍre supposed to show? Thereason I ask is that itÍs not entirely clear to me that this statementis true, but I know a true fact that I can imagine one could possiblymisread as the statement above.>I think this is a totally geometric thing, and I know it suf'ces to show>that in such a minimal cover no point is contained in more than 4 squares,>but how do i show this? IÍm not sure whether what you say youÍre trying to prove is true,but I can tell you that itÍs _not_ true that in a minimal cover bysquares no point is in more than four squares!For example, let Q_1, ... Q_5 be squares of side length 1, suchthat the lower left corner of Q_j is at the point (j/100, j/100),and let K be the union of the Q_j. Then the Q_j form a minimalcover of K, but there are points of K contained in all 've ofthe Q_j.>I know how to do this with intervals on the real>line (there the answer is 2 times the area....), but i get lost when going>up a dimension.Re-read the statement of what youÍre supposed to prove and letus know whether itÍs really exactly the above.************************David C. Ullrich === Subject: How to choose a matrix PNow I am thinking of this question.Suppose Y1=D - PÍEP Y2=D-E where D,E,P : nxn matrix||P||=1How to choose P, so that rank Y1 >= rank Y2 and === Subject: Re: How to choose a matrix P>Now I am thinking of this question.>Suppose >Y1=D - PÍEP >Y2=D-E >where D,E,P : nxn matrix>||P||=1>How to choose P, so that >rank Y1 >= rank Y2 >and >||Y1||>= ||Y2|| The identity matrix would seem to work...Maybe you have some more requirements youÍre not telling us.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: How to choose a matrix PNo D!=Eso P canÍt be an identity matrix>Now I am thinking of this question.>Suppose >Y1=D - PÍEP >Y2=D-E where D,E,P : nxn matrix>||P||=1>How to choose P, so that >rank Y1 >= rank Y2 >and >||Y1||>= ||Y2|| > The identity matrix would seem to work...> Maybe you have some more requirements youÍre not telling us.> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 === Subject: Re: How to choose a matrix P>No D!=E>so P canÍt be an identity matrixBut P = I works no matter what D and E are. Maybe you meanY1 != am thinking of this question.>>Suppose >>Y1=D - PÍEP >>Y2=D-E >>where D,E,P : nxn matrix>>||P||=1 >>How to choose P, so that >>rank Y1 >= rank Y2 >>and >>||Y1||>= ||Y2|| >> The identity matrix would seem to work...>> Maybe you have some more requirements youÍre not telling us. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: proving integrability of binomial numbers> En el Larrosa Ca.96estro escribi.97:> En el escribi.97:>> I know this is a well known integrability. But what would a formal>> proof of it look like ? Viewing that Comb(n, k) is the number, necessary integer, of ways to> choose k elemnts from n, without order. But if you insist in a arthmetic prove, show that Comb(n, 0) => Comb(n, 1) for n >= 0 and Comb(n, k) = Comb(n-1, k - 1) + Comb(n - 1,> k) (TartagliaÍs or PascalÍs triangle)> It must be:> But if you insist in a arthmetic prove, show that Comb(n, 0) = Comb(n, n) => 1 for n >= 0> and Comb(n, k) = Comb(n-1, k - 1) + Comb(n - 1, k) (TartagliaÍs or PascalÍs> triangle) Of course the proof using recursion is the standard proof and theeasiest proof. There is another arithmetic proof in GaussÍsDisquisitionae Arithmeticae, although I canÍt remember where it is orget my hands on my packed-away copy of it. He actually considered theprime factors of the numerator and denominator. I suppose you coulduse the formula for the number of powers of a prime p that divide n!,namely [n/p]+[n/p^2]+[n/p^3]+..., ([x] is the greatest integerfunction), where this is really a 'nite sum since all terms are 0once p^a > n. A small language note. The condition of something being a wholenumber is its integrality, not its integrability. We talk about theintegrability of a function to ask if there is an integral as inf(x)dx of this function. Must come from the same root, but the use PascalÍs triangle referred to asTartagliaÍs triangle. I only know of Tartaglia as a contributor tosolving the general cubic. Is this because I mostly have access toAmerican and other English language sources? === Subject: Re: proving integrability of binomial numbersEn el Nakhash, the Loving Snake escribi.97:> Ignacio Larrosa Ca.96estro have never heard of PascalÍs triangle referred to as> TartagliaÍs triangle. I only know of Tartaglia as a contributor to> solving the general cubic. Is this because I mostly have access to> American and other English language sources?Google [Tartaglia triangle] or [Tartaglia triangulo], or of course,[Tartaglia triangolo] ...At secondary school, I always saw TaratagliaÍs triangle (well, actuallyÍTri.87ngulo Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: How to choose 3 unity vectors from 1000 unity vectors, the volume of the parallelepiped formed by the 3 unity vectors are the support1.mathforum.org (8.11.6/8.11.6/The Math Forum, three dimensional unity vectors, one can form aparallelepiped using arbitrary 3 of them.I want to choose the 3 unity vectors from the 1000 unity vectors, thevolume of the parallelepiped spanned by these 3 unity vectors shouldbe the largest amongst all the parallelepiped formed by any possible 3 unity vectors in the 1000 unity vectors.I cannot choose the speci'ed 3 unity vectors by calculate the volumes of all parallelepiped formed by all possible combination of 3 unity vectors from the 1000 unity vectors, because the combination number is too large: about 1.6e8.Is there any easier way to choose the speci'ed 3 Polytechnic University. === Subject: Is the Div Theorem well behaved if the surface isnÍt closed?HelloThe subject says it all? Every reference I have found describing theDivergence Theorem always considers a closedsurface for the surface of integration. What happens if the surface === Subject: Re: Is the Div Theorem well behaved if the surface isnÍt closed?> Hello> The subject says it all? Every reference I have found describing the> Divergence Theorem always considers a closed> surface for the surface of integration. What happens if the surface isnt> then it doesnÍt really bound a region,then, does it?However, if the two surfaces S1 & S2 share the same boundary, thenthe Divergence Theorem relates the integrals of V.dN over S1 and S2,via the integral of div(V) over the volume bounded by the union ofS1 and S2.Dale. === Subject: Re: Math Too Advanced For Mainstream EconomistsThe pitiful child who posts below is teaches economics at a leadingsecond-tier department of economics in the U.S.A. Hard to believe.> YouÍve done enough. Have you no sense of decency, sir? At long last,> have you left no sense of decency?> -- Joseph Welch to Sen. Joseph McCarthy, 1954. Army-McCarthy> Hearings.> also> -- Mark the sci.math newsgroup for no reason. Stop this.> ThatÍs the longest answer I ever heard to a yes or no question.> -- John Presidential Debate> also> -- Mark Witte to Robert Vienneau, supply-and-demand is ever valid.> Samuel Johnson observed that that patriotism is the last refuge of> scoundrels, and evidently aggressive editing is the last refuge of> those losing newsgroup slanging matches. Anyone who is interested in> a strange Sraffarian modus operandi is invited to read up on this sad> thread.> =rvien-0D64F> 26lr%3D%26ie> %3DUTF-8%26group%3Dsci.econ> Robert Vienneau are determined > by the intersection of well-behaved supply and demand curves in > the labor market. > and> Poor Mark Witte seems to believe that the supply and demand model> is sometimes reasonable.> I tired to explain that his linked page doesnÍt really critique the> competitive labor market supply and demand model, where quantities are> related to prices, and other factors shift said curves. It evidently> didnÍt take, probably owing to confusion on Mr. VienneauÍs part on the> difference between partial and general equilbrium. ItÍs a common> intermediate mistake.> I then asked whether Robert had the courage of his convictions to say> whether he felt that supply and demand was ever a reasonable framework> for explaining our world. I made it easy, but in trying to bring the> mountain to Mohammed somehow only led Mohammed to run away.> RobertÍs citing of empirically based papers on other topics in support> of a model that endeavors to address labor market issues is shameful,> and he does it here yet again.> There are many models of labor market behavior, supply-and-demand> being a very useful on among them. In the tradition of the following> book,> http://www.amazon.com/exec/obidos/tg/detail/-/0226458083/qid= 1076997593//r> ef=sr_8_xs_ap_i1_xgl14/103-6894108-3696656?v=glance&s=books&n= 507846> this thread will gain my reply when the approach Robert Vienneau> advances is shown to explain something, anything, observable about> labor markets that other approaches donÍt.> To beat a model, one must either show it is invalid, or that another> model is superior empirically. If supply and demand is not invalid,> then for some other model to be preferred, that model must dominate> supply and demand on some empirical dimension. ItÍs that simple, and> many models meet this criterion. RobertÍs evidently does not.> Until some real evidence is offered in support of the attempted labor> model under discussion, this thread will just be a weird polemic, the> sound of one hand clapping.-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question 't perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Math Too Advanced For Mainstream EconomistsRobert Vienneau threatening to hold his breath until he dies once again,> The pitiful child who posts below is teaches economics at a leading> second-tier department of economics in the U.S.A. Hard to believe.Maybe you can get your big brother to beat him up. === Subject: any good book on general am studying general optimization, particularly continuous multivariateoptimization, something like First Order Nessecery Condition, Second OrderNessecary Condition, Second Order Suf'cient Condition for localminimizers... etc. I got really confused. The reasons are1) Multivariate derivatives, gradient, etc. are what I am not very familiarwith, so I got confused;2) Some boundary condition and degenerate cases got me confused: forexample, it is easy to have FONC, SONC, SOSC for interior points, thatÍseasy, but what if the FONC, SONC, SOSC on boundary point, involving thedirectional derivative vector d, etc. And some degenerate cases such as theHessian is inde'nite... etc. And the relationship between FONC, SONC, SOSC,etc. these all got me confused...Please point me to some good books/online lecture notes, etc. that can helpme clearify === Subject: Re: any good book on general optimization, continuous multivariate optimization, particularly continuous multivariate> optimization, something like First Order Nessecery Condition, Second Order> Nessecary Condition, Second Order Suf'cient Condition for local> minimizers... etc. I got really confused. The reasons are> 1) Multivariate derivatives, gradient, etc. are what I am not very familiar> with, so I got confused;> 2) Some boundary condition and degenerate cases got me confused: for> example, it is easy to have FONC, SONC, SOSC for interior points, thatÍs> easy, but what if the FONC, SONC, SOSC on boundary point, involving the> directional derivative vector d, etc. And some degenerate cases such as the> Hessian is inde'nite... etc. And the relationship between FONC, SONC, SOSC,> etc. these all got me confused...> Please point me to some good books/online lecture notes, etc. that can help> me clearify these concepts?I recommend you begin by getting an overview using the excellent book byGill, Murray and Wright. Itmisses some recent stuff that you probably donÍtneed anyway but otherwise explains everything very well, presenting themodern point of view (Luenberger is now quite dated) without highmathematical sophistication. If you donÍt understand the math in thereyouÍll need to look at a suitable math text. One cannot get a goodunderstanding of optimization without becoming familiar with gradientsand Hessians.A good account of necessary/suf'cient conditions is in FletcherÍsbook. (A comprehensive account is in MangasarianÍs book, but this iswritten for experts.)Arnold Neumaier === Subject: Re: any good book on general optimization, continuous multivariate optimization? walala general optimization, particularly continuous multivariate >optimization, something like First Order Nessecery Condition, Second Order >Nessecary Condition, Second Order Suf'cient Condition for local >minimizers... etc. I got really confused. The reasons are >1) Multivariate derivatives, gradient, etc. are what I am not very familiar >with, so I got confused; 2) Some boundary condition and degenerate cases got me confused: for >example, it is easy to have FONC, SONC, SOSC for interior points, thatÍs >easy, but what if the FONC, SONC, SOSC on boundary point, involving the >directional derivative vector d, etc. And some degenerate cases such as the >Hessian is inde'nite... etc. And the relationship between FONC, SONC, SOSC, >etc. these all got me confused... >Please point me to some good books/online lecture notes, etc. that can help >me >http://plato.la.asu.edu/topics/tutorials.htmlhthpeter === Subject: Is there a Second Order Suf'cient Condtion(SOSC) for boundary and non-interior points?Hi all,I only learned SOSC for interior points, but I 'nd no where about SOSC forboundary and all(boundary, noninterior) points... and how about First lot,-WAlala === Subject: Re: Is there a Second Order Suf'cient Condtion(SOSC) for boundary and non-interior points? walala for interior points, but I 'nd no where about SOSC for >boundary and all(boundary, noninterior) points... and how about First Order >Suf'cient Condition for local minimizer?'rst order suf'cient? -> convex case plus slaters constraint regular constraint point it is easy:let the constraints be h(x)=0 h has p componentsandg(x) <=0 g has m componentsassume that the gradients of the h_j and those g_i with g_i(x*)=0are linearly independent . assume there are a components gthen you can elimininate from the systemh(x)=0g_i(x)=0 for i such that g_i(x*)=0 p+a variables , with n-p-a remaining free. (if you have problems with the implicit function theorem which covers this: think of those equations as beinglinear... ) then you can express the objective function as one depending only on the free variables. apply the criteria for the unconstrained case to thisso called reduced function. write down the result. then translate everythingback in terms of the original functions. you obtain the 'rst order necessary,the second order necessary and the second order suf'cient conditions this way considering the active inequality constraints as equality constraints. it remmainsto consider strictly feasible directions with respect to the active inequalityconstraints. if you assume strict complementarity, then applying taylors theormonce more you see that in the case considered here these apply also to the inequality constrained case.hthpeter === Subject: how to understand this statement of Second order Necessary minimizer as follows:Let x* a local minimizer of function f over some constraint set O, and d afeasible direction at this point x*. If dÍ*gradient=0, then dÍ*H*d>=0. where dÍ is the transpose of d column vector...I got confused here because this seems to me is a half statement:If dÍ*gradient=0, then dÍ*H*d>=0.but what if dÍ*gradient>0 but not =0 at that point x*?Is this a possible case in the SONC statment? Is that true that whendÍ*gradient>0 but not =0, there is no statement about you,-Walalal === Subject: Re: how to understand this statement of Second order Necessary Condition?> I got confused here because this seems to me is a half statement:> If dÍ*gradient=0, then dÍ*H*d>=0.> but what if dÍ*gradient>0 but not =0 at that point x*?In mathematics, if the assumptions are not satis'ed, nothing can beconcluded.> Is this a possible case in the SONC statment? Is that true that when> dÍ*gradient>0 but not =0, there is no statement about the SONC?Yes. Yes.Arnold Neumaier === Subject: Re: how to understand this statement of Second order Necessary Condition?X-AUTHid: the statement of SONC for local minimizer as follows:> Let x* a local minimizer of function f over some constraint set O,> and d a feasible direction at this point x*. If dÍ*gradient=0, then> dÍ*H*d>=0. > where dÍ is the transpose of d column vector...> I got confused here because this seems to me is a half statement:> If dÍ*gradient=0, then dÍ*H*d>=0.> but what if dÍ*gradient>0 but not =0 at that point x*?> Is this a possible case in the SONC statment? Is that true that when> dÍ*gradient>0 but not =0, there is no statement about the SONC?> Please help me understand this positive inner-product with the gradient is not a problem, since x* is a local minimizer, not a maximizer. I suspect youÍre really interested in the case dÍ*gradient < 0. That cannot happen at a local minimizer when d is a feasible direction, since for some small t > 0 x*+td would be feasible and f(x*+td) would be less than f(x*) (by virtue of the gradient condition), contradicting x* being a minimizer.-- Paul********************************************************** ***************Paul A. Rubin Phone: (517) 432-3509Department of Management Fax: (517) 432-1111The Eli Broad Graduate School of Management E-mail: rubin@msu.eduMichigan State University http://www.msu.edu/~rubin/East Lansing, MI 48824-1122 (USA)********************************************************* ****************Mathematicians are like Frenchmen: whenever you say something to them,they translate it into their own language, and at once it is somethingentirely different. J. W. v. GOETHE === Subject: Re: how to understand this statement of Second order Necessary Condition? walala minimizer as follows: >Let x* a local minimizer of function f over some constraint set O, and d a >feasible direction at this point x*. If dÍ*gradient=0, then dÍ*H*d>=0. >where dÍ is the transpose of d column vector... >I got confused here because this seems to me is a half statement: >If dÍ*gradient=0, then dÍ*H*d>=0. >but what if dÍ*gradient>0 but not =0 at that point x*? >Is this a possible case in the SONC statment? Is that true that when > dÍ*gradient>0 but not =0, there is no statement about the SONC? > d is a feasible direction if for some t0 and 0<= t <= t0 x*+t*d is feasible. now consider f(x*)<= f(x*+t*d) (by assumption) = f(x*)+gÍ*d*t + dÍ*H*d*t^2 + o(t^2) by Taylors theorem for two times continuously differentiable t.let t>0, subtract f(x*) from both sides, divide by t and let t tend to zero...donehthpeter === Subject: Collatz Conjecture : Symmetry question.Every time I post on this problem I get beat up so please be kind. I have found many properties in the landscape of this problem and theone that has most of my interest the gross amount of non-regularitytightly coupled to large a domain of symmetry. I have noticed 2 formsof symmetry in this problem which have me thoroughly intrigued, but donot know where to go with these notions at this point.Symmetry 1;I have noticed that given any tree pattern of any size, it is exactlyduplicated in'nitely many times. For example, take the entire treestarting at A and extend out all branches to a depth of any N. Thenthat pattern taken generically is repeated at an offset of A +n*(constant value) with (n = 1, 2, 3, ... in'nity), with the constantvalue is a function of N.Symmetry 2;I have also noticed that all values at N (here I include allrationales and integers by taking all possible paths from A to N,keeping NON_INTEGER branches in this process) produces NÍ sets of acollective 2^(N-1) number of values where the number of items in eachset follows the Binomial Coef'cients or PascalÍs Triangle (which everyou normally call it). If you take each of these NÍ sets of values andcalculate the ordered set difference (NÍÍ) that the difference sets(NÍÍ) are constant and equal for all N independent of A.This means that using only the NÍÍ sets, all sets at N in groups ofNÍ, thus all values at N, for any A, can be calculated withoutextending the tree from A to N through all the intermediate NÍs.In other words, all values of absolute stopping distance of N, for anyA, can be calculated directly with out traversing the tree to 'ndeach of those values.Is this interesting to anyone? Or better yet, Is this anything o'nterest for this problem. === Subject: Re: Collatz Conjecture : Symmetry question.> Every time I post on this problem I get beat up so please be kind. I hope that wasnÍt directed at me. I never beat anyone. It may seemthat way to Ernst Berg, but I was just trying to help him by offeringconstructive criticism. Apparently some people canÍt get past the criticism part and get all bent out of shape to the point where theyignore the constructive part. But thatÍs ErnstÍs problem.> I have found many properties in the landscape of this problem and the> one that has most of my interest the gross amount of non-regularity> tightly coupled to large a domain of symmetry. I have noticed 2 forms> of symmetry in this problem which have me thoroughly intrigued, but do> not know where to go with these notions at this point.> Symmetry 1;> I have noticed that given any tree pattern of any size, it is exactly> duplicated in'nitely many times. For example, take the entire tree> starting at A and extend out all branches to a depth of any N. Then> that pattern taken generically is repeated at an offset of A +> n*(constant value) with (n = 1, 2, 3, ... in'nity), with the constant> value is a function of N.That seems reasonable since any sequence vector evaluates to a simpleformulaaÍ = (X*a - Z)/Ywhich means there are always an in'nte set of integer solutions to everysequence vector.> Symmetry 2;> I have also noticed that all values at N (here I include all> rationales and integers by taking all possible paths from A to N,> keeping NON_INTEGER branches in this process) produces NÍ sets of a> collective 2^(N-1) number of values where the number of items in each> set follows the Binomial Coef'cients or PascalÍs Triangle (which ever> you normally call it). If you take each of these NÍ sets of values and> calculate the ordered set difference (NÍÍ) that the difference sets> (NÍÍ) are constant and equal for all N independent of A.IÍm not sure I follow this, but since there are only two operations (3x+1and x/2) every number has two possible ancestors, so carried to N levels(and ignoring whether the ancestors are integers) there would certainlybe 2^N numbers at distance N.> This means that using only the NÍÍ sets, all sets at N in groups of> NÍ, thus all values at N, for any A, can be calculated without> extending the tree from A to N through all the intermediate NÍs.> In other words, all values of absolute stopping distance of N, for any> A, can be calculated directly with out traversing the tree to 'nd> each of those values.But there are vastly more total numbers at a given level than there areintegers, so although this is true, IÍm not sure it leads anywhere.For example, I recently found all the integers that are at level 84(from 1):179,441,377If I included all the potential non-integer ancestors, I would have had tosift through 2^84 or19,342,813,113,834,066,795,298,816numbers to 'nd which ones are integers. In this case, it is much quicker to traverse the tree from level 1 up to level 84. I have a program that canbuild each level from the previous one. Since it only generates integerresults, it is much more ef'cient than testing each binary permutaionof the two operations (provided you record the integers you collected from the previous level).It would be nice to be able to just start at an arbitrary level and 'ndall the integers. So although it can be done in theory, it canÍt be donein practice for high levels.Unless you have some means of determining which of the 2^84 numbers at level84 are integers without having to iterate through them all.> Is this interesting to anyone? Yes.> Or better yet, Is this anything of interest for this problem.IÍm not an expert, so I canÍt say. === Subject: Re: Collatz Conjecture : Symmetry question.> But there are vastly more total numbers at a given level than there are> integers, so although this is true, IÍm not sure it leads anywhere.> For example, I recently found all the integers that are at level 84> (from 1):> 179,441,377> If I included all the potential non-integer ancestors, I would have had to> sift through 2^84 or> 19,342,813,113,834,066,795,298,816Yes, this is true. But, taking the fact that within the19,342,813,113,834,066,795,298,816 values - if looked at as an orderedlist in magnitude, then the location of integers is static for allsymmetric trees. Hence, a set of 179,441,377 difference values can beused to calculate all integer values directly (I am leaving out a biton how to do this but is is quite easy). However, the continuity ofconnectivity goes out the window as you start jumping around amoungthe different symmetric trees. Bummer. It seems there is no end to thewalls between us and a solution. === Subject: Re: In'nite polynomial recently came across the following in'nite product:(1-x)(1-x^2)(1-x^3)(1-x^4)...,where abs(x) < 1.> Not elementary, but itÍs related to the Dedekind eta function and> EulerÍs Pentagonal Number Theorem.> See e.g.> .> and the many references given at the last URL.ThereÍs a proof of the Pentagonal Number Theorem onhttp://planetmath.org/encyclopedia/ PentagonalNumbersTheorem.htmlLH === Subject: Re: JSH: Not a solution>But hey, weÍre mostly Americans here! Surprise. Surprise.Not true. YouÍve kept me amused here in Oz for the last three years,and the best part of 've years before that in === Subject: Re: JSH: Not a solutionLap mostly Americans here! Surprise. Surprise. Not true. YouÍve kept me amused here in Oz for the last three years,> and the best part of 've years before that in the UK.He was referring to his multiple personalities, most of whom are American.Doug === Subject: Re: countable sets>> Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Prove>> that union of Ek as k = 1 to oo is countable. I do not need help proving>> this. I need someone to explain exactly what is being asked. In Let E1, E2,>> E3,... be a sequence of pairwise disjoint countable sets is that saying>> that the numbers of EiÍs are countable or is it saying that each Ei has a>> countable number of elements??> Others have explained what is being asked, but I was intrigued by your> statement that you donÍt need help proving it. Now that the question has> been clari'ed, you might look again at proving it.> Hint: If your proof does not invoke the axiom of choice, then your proof> is wrong.Not so. The countable union of disjoint countable setsis in 1-1 correspondence with NxN -- the pair (n, k) maps tothe the k-th element of the n-th set (the set of sets and eachset is alreadyknown to be in 1-1- correspodence with N). Thisis clearly surjctive, and the disjointness implies that itÍsinjective.And it does not require AC to show that NxN is countable. === Subject: Re: countable sets> Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Prove> that union of Ek as k = 1 to oo is countable. I do not need help proving> this. I need someone to explain exactly what is being asked. In Let E1, E2,> E3,... be a sequence of pairwise disjoint countable sets is that saying> that the numbers of EiÍs are countable or is it saying that each Ei has a> countable number of elements??> Others have explained what is being asked, but I was intrigued by your>> statement that you donÍt need help proving it. Now that the question has>> been clari'ed, you might look again at proving it.> Hint: If your proof does not invoke the axiom of choice, then your proof>> is wrong.> Not so. The countable union of disjoint countable sets> is in 1-1 correspondence with NxN -- the pair (n, k) maps to> the the k-th element of the n-th set (the set of sets and each> set is alreadyknown to be in 1-1- correspodence with N). This> is clearly surjctive, and the disjointness implies that itÍs> injective.See my other post. You used AC when you selected for each i a particularbijection between E_i and N.> And it does not require AC to show that NxN is countable.Agreed. The problem lies elsewhere.-- Dave SeamanJudge YohnÍs mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: countable sets >Let E1, E2, E3, ... be a sequence of pairwise disjoint countable sets. Prove>that union of Ek as k = 1 to oo is countable. I do not need help proving>this. I need someone to explain exactly what is being asked. In Let E1, E2,>E3,... be a sequence of pairwise disjoint countable sets is that saying>that the numbers of EiÍs are countable or is it saying that each Ei has a>countable number of elements??>Others have explained what is being asked, but I was intrigued by your>>statement that you donÍt need help proving it. Now that the question has>>been clari'ed, you might look again at proving it.>>Hint: If your proof does not invoke the axiom of choice, then your proof>>is wrong.>Not so. The countable union of disjoint countable sets>is in 1-1 correspondence with NxN -- the pair (n, k) maps to>the the k-th element of the n-th set (the set of sets and each>set is alreadyknown to be in 1-1- correspodence with N). This>is clearly surjctive, and the disjointness implies that itÍs>injective.And it does not require AC to show that NxN is countable.>But you *have* used the axiom of choice in selecting in'nitely many enumerations (one for each E_i).-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: countable countable. But you *have* used the axiom of choice in selecting in'nitely many> enumerations (one for each E_i).But if memory does not betray me, countable choiceis a lot weaker === Subject: Re: countable sets>> Let E1, E2, E3, ... be a sequence of pairwise disjoint countable> sets. Prove that union of Ek as k = 1 to oo is countable. I do not> need help proving this. I need someone to explain exactly what is> being asked. In Let E1, E2, E3,... be a sequence of pairwise> disjoint countable sets is that saying that the numbers of EiÍs> are countable or is it saying that each Ei has a countable number> of elements??>> Others have explained what is being asked, but I was intrigued by>> your statement that you donÍt need help proving it. Now that the>> question has been clari'ed, you might look again at proving it.>> Hint: If your proof does not invoke the axiom of choice, then your>> proof is wrong. Not so. The countable union of disjoint countable sets> is in 1-1 correspondence with NxN -- the pair (n, k) maps to> the the k-th element of the n-th set (the set of sets and each> set is alreadyknown to be in 1-1- correspodence with N).AC is quite tricky. Think again: all you know is that there exist abijection between the n-th set and IN; the k-th object of the n-th set iszill-de'ned, since you donÍt know explicitly the n-th bijection ;-(This> is clearly surjctive, and the disjointness implies that itÍs> injective. And it does not require AC to show that NxN is countable. === Subject: two other sides of triangleHow we can 'nd two sides of triangle if the third side, the radius o'nscribed circle and the angle opposite to the third side is given? === Subject: Re: two other sides of triangle> How we can 'nd two sides of triangle if the third side, the radius of> inscribed circle and the angle opposite to the third side is given?Let c be the known side, C the corresponding angle and r the inradius.Then youhave the equations (a, b being the unknown sides) :ab sinC = r (a + b + c) --> a + b = ab/r sin C - c (1) (Formula forinradius)c^2 = (a + b)^2 - (2 + 2cosC) ab (2) (Cosine rule)Substitute in (2) the formula for a+b given in (1) and you have a(remarkably simple !) quadratic in ab. Solve, evaluate a+b using (1)and then solve x^2 - (a+b)x + ab = 0 whose two roots will give you thesides.Hope this helps,Dimitris === Subject: Re: two other sides of triangle>How we can 'nd two sides of triangle if the third side, the radius of>inscribed circle and the angle opposite to the third side is given?>In case this is homework, I give only a brief outline:Draw the angle bisectors from the endpoints of the known side. Drop a perpendicular to that side. Let x and y be the lengths of the two subsegments of the side so determined. We can now use trigonometry to determine the tangents of half of each of the unknown angles in terms of x and y. Use the tangent addition formula to get an equation expressing the tangent of the known angle in terms of x and y. Combine this with the fact that the length of the known side is x+y sines of the unknown angles.If this is a school assignment, please cite assistance received in submitted work. If not and you would like more details, ask again and I will gladly provide them.By the way, I am not sure this is the most elegant method.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: IÍd like to join below really prove that 2=1 ?? Y N > 1) X=Y ; Given> 2) X^2=XY ; Multiply both sides by X> 3) X^2-Y^2=XY-Y^2 ; Subtract Y^2 from both sides> 4) (X+Y)(X-Y)=Y(X-Y) ; Factor> 5) X+Y=Y ; Cancel out (X-Y) term> 6) 2Y=Y ; Substitute X for Y, by equation 1> 7) 2=1 ; Divide both sides by YGo back to step 4. What it really proves is 2x0=1x0. Now try this:1) I am nothing without her. (Shakespeare)2) ME - SHE = 0 (Algebraic representation)3) ME = SHE (Added SHE to both sides)4) M = SH (Divided by E)5) MIT = (Multiplied by IT)So there you have the mathematical proof of what Shakespeare thought of the future American techincal education. :-) === Subject: Re: IÍd like to join this group!>> Does the Equation below really prove that 2=1 ?? Y N >> 1) X=Y ; Given>> 2) X^2=XY ; Multiply both sides by X>> 3) X^2-Y^2=XY-Y^2 ; Subtract Y^2 from both sides>> 4) (X+Y)(X-Y)=Y(X-Y) ; Factor>> 5) X+Y=Y ; Cancel out (X-Y) term>> 6) 2Y=Y ; Substitute X for Y, by equation 1>> 7) 2=1 ; Divide both sides by Y> Go back to step 4. What it really proves is 2x0=1x0. > Now try this:> 1) I am nothing without her. (Shakespeare)> 2) ME - SHE = 0 (Algebraic representation)> 3) ME = SHE (Added SHE to both sides)> 4) M = SH (Divided by E)> 5) MIT = (Multiplied by IT)> So there you have the mathematical proof of what Shakespeare > thought of the future American techincal education. :-)You Guys are too sharp.That was actually printed in OMNI magazine Nov 79 , as proof that 2=1 . ;-) === Subject: Re: IÍd like to join this Nov 79 , as proof that 2=1 .;-)It was also printed in the Wall Street Journal to explain how companyearnings are up. ;-)-- RichGrowing old is mandatory.Growing up is optional. === Subject: Re: IÍd like to join this printed in OMNI magazine Nov 79 , as proof that 2=1 .> ;-)> It was also printed in the Wall Street Journal to explain how company> earnings are up. ;-)I am sure GW has used it in his calculatations. === Subject: Re: looking for an eltry soln to an old problem>A long time ago, J.J. Sylvester posed the problem:>if I have arbitrarily many 5 cent or 17 cent stamps,>what is the largest denomination I cannot make?>In general, if we have p and q cent stamps, it turns>out the answer is pq-p-q (granted p and q are coprime).>I have derived a solution>to the problem, but IÍd like to teach this to my>undergraduates, some of whom have a limited background.>So my question is: is there a very nice & friendly>proof of this fact? (For example, which avoids any>nonobvious facts from number theory.)You need a couple of not-quite-obvious facts.Fact 1: if x = a p + b q, then x = (a + n q) p + (b - n p) q forany integer n, and these are all the ways of writing x as amultiple of p plus a multiple of q.Fact 2: if p and q are coprime, every integer x can be written as x = a p + b q for some (not necessarily positive) integers a,b.This can be done using the Euclidean algorithm; or the fact thatt -> t p mod q is one-to-one on the integers in {1,...,q-1} coprimeto q, plus the pigeonhole principle.Suppose for some x it canÍt be done with a, b >= 0. Ifx = a p + b q is one representation with integers a,b, then for each n we have either a + n q < 0 or b - n p < 0. Take aÍ to be the largest a + n q < 0, so -q <= aÍ < 0 and x = aÍ p + bÍ q with bÍ - p < 0.But then x <= -p + (p-1) q = pq - p - q.On the other hand, pq - p - q = ap + bq with a=q-1 and b=-1; a+nq < 0 if n < 0 and b-np < 0 if n >=0.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: looking for an eltry soln to an old problemTyler Neylon> A long time ago, J.J. Sylvester posed the problem:> if I have arbitrarily many 5 cent or 17 cent stamps,> what is the largest denomination I cannot make? In general, if we have p and q cent stamps, it turns> out the answer is pq-p-q (granted p and q are coprime).> I have derived a solution> to the problem, but IÍd like to teach this to my> undergraduates, some of whom have a limited background.> So my question is: is there a very nice & friendly> proof of this fact? (For example, which avoids any> nonobvious facts from number theory.)1. If an amount x in [1,pq-1] is not doable, then x is not a multiple ofp or q.2. If y is any element of [1,pq-1] other than a multiple of p or q, then yor pq-y is doable, but not both.3. So, since p+q is the smallest doable number not a multiple of p or q, thelargest non-doable number is pq-p-q.But how to show (2) in a suf'ciently digestible fashion?LH === Subject: Re: Number property>I am searching for the most general set of numbers {a_i} (i=1...n) >, 0<=a_i<=2pi , such that :> Better to say < 2 pi, because 0 and 2 pi are equivalent in respect> to addition mod (2 pi). >a_i + a_j = a_t mod(2pi) for every i,j=1...n>and such that the set of all {a_t} obtained in this way is equal to>the set {a_i} we started with.> In other words, you want a 'nite sub-semigroup of T = R/(2 pi Z)> under addition. > Any 'nite sub-semigroup of a group is a subgroup (in fact any> 'nite semigroup such that t->a*t and t->t*a are one-to-one for all > a in the semigroup is a group).> So youÍre looking for the 'nite subgroups of T. >And these are all>of the form {2pi k/n: k=0...n-1} for positive demonstration of :And these are all of the form {2pi k/n: === Subject: Re: No Set Contains Every Computable Natural Discussion, that a TM canÍt decide if a string is in'nite.Duh.-- Run mathematicians, RUN!!! IÍm coming for you. It may take a fewmonths, but IÍll get [computer veri'cation of my proof] and then yourlives will be ended as you previously knew it. -- JSH meets PVS === Subject: Re: No Set Contains Every Computable admission that a TM canÍt decide if a string is in'nite. Duh.But a human can, right? === Subject: Re: No Set Contains Every Computable Natural <871xosptgg.fsf@phiwumbda.org> Discussion, TM canÍt decide if a string is in'nite.>> Duh. But a human can, right?I canÍt imagine how. Tell you what, though. LetÍs test it. Why donÍt you create a fewinput tapes, some of which contain an in'nite number of 1Ís? Thentraipse down to your local research institute and ask the good fellasif they can determine which of your tapes have an in'nite number of1Ís and which do not.Clearly, whether a string is in'nite is easily refutable but notdecidable. Also clearly, all this talk about in'nite strings is mereidealization.-- Sure, [my Usenet presence is] like Shaq playing against you in yourbackyard, but that has its perks, as I 'nd ways to have my fun *and*I can send messages to certain people in the United States Governmentwithout concern that the rest of you understand them. -- James Harris === Subject: Re: No Set Contains Every Computable Natural <3sRWb.24383$cM.3142@newssvr25.news.prodigy.com> You still need a human to decide if a representation is 'nite.> No TM can make this determination. Same comment as above.> A blank is just another symbol.> Why would we allow input tapes with an in'nite string of blanks> and not allow inputs with an in'nite number of 1Ís. Because then the input is not *'nite-sized* . A TM is essentially an algorithm. The de'nition of an algorithm requires that it terminates in a 'nite number of steps. If a TM (an algorithm) is required to read its input (which is a usual requirement) then it will never read its input in a 'nite number of steps if the input is in'nite. This is a good reason why such inputs are not allowed.> I can easily represent natural numbers in unary using blanks.> The input tape would be a string of blanks followed by a 0.> b0=1, bb0=2, bbb0=3, etc.> A TM that decided this language would not halt on a blank tape. You canÍt ask a TM to decide whether an in'nitely long string has any properties. You have to ask it something like does this tape have the 'rst 100 numbers? the 'rst 1000000? the 'rst k?> Only if we assume that a human operator decided the iinput string was> 'nite. Again, if the human is not working with a 'nite string, then the human should not be feeding it to a TM.> And, yes, if you include an in'nitely-long string of 1s, then no TM can> decide it (in a 'nite amount of time, which is a crucial requirement.)> This is why in'nitely-long strings are not usually considered in> languages.> Then, no decidable language can contain an irrational number. Depends on how you represent them. Here is a set of all possible solutions to quadratic equations with integer coef'cients: { (a,b,c,d) such that a,b,c,d are integers }An element of this set (a,b,c,d) should be interpreted to mean (a+b*sqrt(c))/d . > Actually, it is quite simple to decide if M is a valid representation.> Not for a TM. It is impossible for a TM to decide the representation> is 'nite. I donÍt think it is that simple for human operators, either. Again, this depends on the representation. I can make a TM that requires an input to be (q_0, Q, T, F, S) where Q,T,F,S are speci'ed by a list of their elements. Then any 'nite input will have to be describing 'nite sets. If the data structure, given in this speci'ed way, is not 'nite, then you canÍt make it an input to a TM in the 'rst place.> It is easy to check, given a data structure, whether the input is a valid TM or not.> Only if the human operator lives forever. Again, it depends on the input format. If I ask for a list of all the elements, then I can be sure it is 'nite. If you want to specify a set of states having a state for every even number representable as a sum of two prime numbers, then no one knows if it is 'nite. The point, again, is that such claims are representation-dependent.> Turing gives an example of such a TM.> He doesnÍt explicitely claim this TM produces every natural number.> But, why would he give it as an example unless he was trying to show> a TM CAN produce a tape with every natural number? I donÍt know what TuringÍs intentions were and I donÍt know the example you are referring to.> Taking about withing a reasonable amount of time has very little to do> with decidable sets or recursively enumerable ones.> I think it has a lot to do with decidable sets. No, time and number-of-steps have nothing to do with decidable sets (besides, of course, the fact that time and number-of-steps are supposed to be 'nite.) Talking about time and related properties of TMs (or algorithms) essentially leads to algorithm runtime, NP-completeness, and complexity theory on that sense, but computability theory is independent of time. > These sets do not say anything about the time required to accomplish> any computation; they are merely telling you what can and can not be> computed.> Implicit in these de'nitions is the assumption that an in'nite number> of computations can be done in a 'nite amount of time. An in'nite number of computations is never done in 'nite time. Yes, it might be considered implicit or hidden assumptions to you, but 'niteness of algorithms, turing machine sizes, and input sizes are all *explicitly* de'ned for all these concepts.J === Subject: Re: No Set Contains Every Computable Finite length must be decided by a human operator. The point here is that a TM can never be given an in'nitely sized> input.A TM canÍt be given an in'nitely long blank tape as input.> You still need a human to decide if a representation is 'nite.> No TM can make this determination. Same comment as above.See mine.> A blank is just another symbol.> Why would we allow input tapes with an in'nite string of blanks> and not allow inputs with an in'nite number of 1Ís. Because then the input is not *'nite-sized* .> A TM is essentially an algorithm. The de'nition of an algorithm> requires that it terminates in a 'nite number of steps. If a TM (an> algorithm) is required to read its input (which is a usual requirement)> then it will never read its input in a 'nite number of steps if the input> is in'nite. This is a good reason why such inputs are not allowed.In'nitely long blank tapes are part of the de'nition of a TM.> I can easily represent natural numbers in unary using blanks.> The input tape would be a string of blanks followed by a 0.> b0=1, bb0=2, bbb0=3, etc.> A TM that decided this language would not halt on a blank tape. You canÍt ask a TM to decide whether an in'nitely long string has any> properties. You have to ask it something like does this tape have the> 'rst 100 numbers? the 'rst 1000000? the 'rst k? Only if we assume that a human operator decided the iinput string was> 'nite. Again, if the human is not working with a 'nite string, then the human> should not be feeding it to a TM.A language is decidable because a human decided the input is 'nite.> And, yes, if you include an in'nitely-long string of 1s, then no TMcan> decide it (in a 'nite amount of time, which is a crucialrequirement.)> This is why in'nitely-long strings are not usually considered in> languages.Instead, we rely on a human operator to decide in advancewhether all of the elements of the language are 'nite.> Then, no decidable language can contain an irrational number. Depends on how you represent them. Here is a set of all possible> solutions to quadratic equations with integer coef'cients: { (a,b,c,d) such that a,b,c,d are integers } An element of this set (a,b,c,d) should be interpreted to mean> (a+b*sqrt(c))/d .Is this supposed to represent a decidable language?> Actually, it is quite simple to decide if M is a validrepresentation. Not for a TM. It is impossible for a TM to decide the representation> is 'nite. I donÍt think it is that simple for human operators, either. Again, this depends on the representation. I can make a TM that requires> an input to be (q_0, Q, T, F, S) where Q,T,F,S are speci'ed by a list of> their elements. Then any 'nite input will have to be describing 'nite> sets. If the data structure, given in this speci'ed way, is not 'nite,> then you canÍt make it an input to a TM in the 'rst place.A human operator will decide in advance if the data structure is 'nite?> It is easy to check, given a data structure, whether the input isa> valid TM or not. Only if the human operator lives forever. Again, it depends on the input format. If I ask for a list of all the> elements, then I can be sure it is 'nite.You will examine every element yourself?I think it will require your children and your childrenÍs children.Even then, I doubt they can examine every 'nite element.> If you want to specify a set of> states having a state for every even number representable as a sum of two> prime numbers, then no one knows if it is 'nite. The point, again, is> that such claims are representation-dependent.Whether or not an arbitrary irrational has a 'nite number of digits dependson the representation? Why not just assign a single symbol to every realnumber?> Turing gives an example of such a TM.> He doesnÍt explicitely claim this TM produces every natural number.> But, why would he give it as an example unless he was trying to show> a TM CAN produce a tape with every natural number? I donÍt know what TuringÍs intentions were and I donÍt know the example> you are referring to. > Taking about withing a reasonable amount of time has very littleto do> with decidable sets or recursively enumerable ones. I think it has a lot to do with decidable sets. No, time and number-of-steps have nothing to do with decidable sets> (besides, of course, the fact that time and number-of-steps are supposed> to be 'nite.)Except for the fact that a human has decide if the decidable setis actually decidable. There is no automated way to make thisdetermination.> Talking about time and related properties of TMs (or> algorithms) essentially leads to algorithm runtime, NP-completeness, and> complexity theory on that sense, but computability theory is independent> of time.No, it is not. Maybe people wish it was.> These sets do not say anything about the time required to accomplish> any computation; they are merely telling you what can and can not be> computed. Implicit in these de'nitions is the assumption that an in'nite number> of computations can be done in a 'nite amount of time. An in'nite number of computations is never done in 'nite time. Yes, it> might be considered implicit or hidden assumptions to you, but> 'niteness of algorithms, turing machine sizes, and input sizes are all> *explicitly* de'ned for all these concepts.The de'nition explicitly states that a language is decidable if thereexists a TM that can decide if a string, x, is a member of the language.Now, it appears that a human must make this decision in advancebefore the string can be given to the TM.Russell- 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural> length must be decided by a human operator. The point here is that a TM can never be given an in'nitely sized> input.> A TM canÍt be given an in'nitely long blank tape as de'nition of a TM.And is black the same as white? today ?? Russell> - 2 nany 2 count> === Subject: Re: No Set Contains message> There is no TM that can decide if the string has an> in'nite string of 1Ís. Has anyone claimed otherwise?People have claimed the set of all natural numbersis a recusive set. This is equivalent to claiming thereis a TM that can decide, in a 'nite number of steps,that an in'nite string of 1Ís does not represent a natural number.> TuringÍs TM doesnÍt produce a unary representation of every natural? Which ones does it leave out?I describe a three state TM that calculates which one.Are you sure you want me to post it?How fast is your modem?> I can de'ne a three state TM that will 'nd a natural number> that is not on this tape:> 1) Read right until there is a 0.> 2) Read right until a second 0 is found.> 3) Backup and write a 1 on the previous 0.> Repeat steps (1) through (3).> This TM will always produce a tape that has exactly one 0.> This 0 will be at a 'nite position on the tape and the string of1Ís> that preceds this 0 represents a natural number that was not on> the original input tape. This only holds if the initial tape is 'nite and your last zero comes> before the end of the tape.This certainly applies if the input tape is 'nite.It also applies if the input tape is in'nitely long.Russell- 2 many 2 count === Subject: Re: No Set Contains Every Computable NaturalRussell Easterly says...>Virgil decide if the string has an>> in'nite string of 1Ís.>> Has anyone claimed otherwise?People have claimed the set of all natural numbers>is a recusive set.The usual notion of recursive set is that a set S isrecursive as a *subset* of the naturals if there is aTuring machine T such that for any *natural* number n(in unary notation, to be speci'c) T(n) halts and outputs 1 <-> n is an element of S T(n) halts and outputs 0 <-> n is not an element of SIf x is some input that is not a representation of a naturalnumber, then there is no constraint on what T(n) does.--Daryl McCulloughIthaca, NY === Subject: Re: No Set Contains Every Computable message> Russell Easterly says...>Virgil has an>> in'nite string of 1Ís.>> Has anyone claimed otherwise?People have claimed the set of all natural numbers>is a recusive set. The usual notion of recursive set is that a set S is> recursive as a *subset* of the naturals if there is a> Turing machine T such that for any *natural* number n> (in unary notation, to be speci'c)The set of all natural numbers is not a proper subset ofthe set of all natural numbers.> T(n) halts and outputs 1 <-> n is an element of S> T(n) halts and outputs 0 <-> n is not an element of S If x is some input that is not a representation of a natural> number, then there is no constraint on what T(n) does.>According to this de'nition, the set of all natural numbersis recursive because there exists a TM that doesnÍt evenread the input and always outputs a 1.This is a one state TM that decides the set of all natural numbers:1) Write a 1.Russell- 2 many 2 count === Subject: Re: No Set Contains Every Computable message> Russell Easterly says...>Virgil can decide if the string has an>> in'nite string of 1Ís.> Has anyone claimed otherwise?>People have claimed the set of all natural numbers>is a recusive set. The usual notion of recursive set is that a set S is> recursive as a *subset* of the naturals if there is a> Turing machine T such that for any *natural* number n> (in unary notation, to be speci'c) The set of all natural numbers is not a proper subset of> the set of all natural numbers. T(n) halts and outputs 1 <-> n is an element of S> T(n) halts and outputs 0 <-> n is not an element of S If x is some input that is not a representation of a natural> number, then there is no constraint on what T(n) does.> According to this de'nition, the set of all natural numbers> is recursive because there exists a TM that doesnÍt even> read the input and always outputs a 1.>I think DarrylÍs de'nition has confused you. The use of integers inde'ning languages is not as popular as it once was. Modern books are morelikely to de'ne a language as a subset of E*, where E is a 'nite set ofsymbols. In DarrylÍs de'nition, every natural n is equivalent to a singleunique string s from E*. A TM outputting 1 after reading n = A TM acceptings. A TM outputting 0 after reading n = A TM rejecting s.In other words, we may have an alphbabet {0,1}, where (letting e = the emptystring):n -> s table for N -> {0,1}*------------------------------1 -> e ; 2 -> 0 ; 3 -> 1 ; 4 -> 00 ; 5 -> 01 ; 6 -> 10 ; 7 -> 11 ; 8 -> 000; 9 -> 001 ;10->010 ; 11 -> 011 ; 12 -> 100 ; 13 -> 101 ; 14 -> 110 ; 15 -> 111 ;.......etc....So DarrylÍs version of {1^n | n in N}, would contain {3, 7, 15, ....}. Itwould not neccessarily be every single natural number.l8r, Mike N. Christoff === Subject: Re: No Set Contains Every Computable NaturalRussell Easterly says...>> The usual notion of recursive set is that a set S is>> recursive as a *subset* of the naturals if there is a>> Turing machine T such that for any *natural* number n>> (in unary notation, to be speci'c)The set of all natural numbers is not a proper subset of>the set of all natural numbers.But it *is* a subset. A recursive subset.>> T(n) halts and outputs 1 <-> n is an element of S>> T(n) halts and outputs 0 <-> n is not an element of S>> If x is some input that is not a representation of a natural>> number, then there is no constraint on what T(n) does.>>According to this de'nition, the set of all natural numbers>is recursive because there exists a TM that doesnÍt even>read the input and always outputs a 1.Right.--Daryl McCulloughIthaca, NY === Subject: Re: No Set Contains Every Computable NaturalDaryl McCullough says...> The usual notion of recursive set is that a set S is>> recursive as a *subset* of the naturals if there is a>> Turing machine T such that for any *natural* number n>> (in unary notation, to be speci'c)The set of all natural numbers is not a proper subset of>the set of all natural numbers. But it *is* a subset. A recursive subset.> T(n) halts and outputs 1 <-> n is an element of S>> T(n) halts and outputs 0 <-> n is not an element of S>> If x is some input that is not a representation of a natural>> number, then there is no constraint on what T(n) does.>>According to this de'nition, the set of all natural numbers>is recursive because there exists a TM that doesnÍt even>read the input and always outputs a 1. Right.The set of all natural numbers is recursive by 'at.Why bother to include computability in the de'nition?A human must decide if x is a member of this setbefore x can be given to the TM.Russell- 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural> Daryl McCullough says... >> The usual notion of recursive set is that a set S is>> recursive as a *subset* of the naturals if there is a> Turing machine T such that for any *natural* number n>> (in unary notation, to be speci'c)>The set of all natural numbers is not a proper subset of>the set of all natural numbers. But it *is* a subset. A recursive subset. >> T(n) halts and outputs 1 <-> n is an element of S>> T(n) halts and outputs 0 <-> n is not an element of S>> If x is some input that is not a representation of a natural>> number, then there is no constraint on what T(n) does.>According to this de'nition, the set of all natural numbers>is recursive because there exists a TM that doesnÍt even>read the input and always outputs a 1. Right.> The set of all natural numbers is recursive by 'at.No, not by 'at. It is a trivial consequence of a general de'nition--the de'nition that Daryl McCullough has reproduced for you.> Why bother to include computability in the de'nition?There is no separate de'nition for the case of all natural numbers.There is one de'nition that applies to all subsets of the set of naturalnumbers. The set of all natural numbers (henceforth N) is a subset ofN. We apply the de'ntition to N and 'nd out trivially that it isrecursive.It is not a trivial de'nition. It is quite profound. But it istrivial that N is recursive.> A human must decide if x is a member of this set> before x can be given to the TM.You are seeking to create a new de'nition and theory of computability.Good. You want to consider a larger context of inputs other than just natural numbers. For example, you want to include in'nite stings as acceptable inputs. Fine. I would agree with you thatif we enlarged the universe of acceptable inputs from just N to a largerset containing in'nite strings, then N would probably not be recursive in any reasonable new de'nition of the word. YouÍre right,a TM would not be able to conclude that an in'nite string of 1Ís isnot a natural number. I would say that in this enlarged context, thatN would be recursively enumerable but not recursive.But your problem is that you are using terms with agreed upon de'nitionsby the mathematical community (like recursive) incorrectly. You haveyour own personal de'nition that you have failed to state. This is whymathematicians are arguing with you and pointing out your obvious errors. If you would like to re-de'ne recursive by all means do so.But make your new de'nition known. It would also be good to giveit another name to avoid confusion, like Easterly-Recursive.Until you do this, you will probably continue to make false statementsthat are easily refuted.Remember, mathematics is precise.By the way there are already in existence more generalized recursion theories,which deal with recursion on other domains besides N.-Leonard Blackburn> Russell> - 2 many 2 count === Subject: Re: No Set Contains Every Computable NaturalLeonard Blackburn says...>> The usual notion of recursive set is that a set S is>> recursive as a *subset* of the naturals if there is a> Turing machine T such that for any *natural* number n>> (in unary notation, to be speci'c)>The set of all natural numbers is not a proper subset of>the set of all natural numbers.> But it *is* a subset. A recursive subset.>> T(n) halts and outputs 1 <-> n is an element of S>> T(n) halts and outputs 0 <-> n is not an element of S>> If x is some input that is not a representation of a natural>> number, then there is no constraint on what T(n) does.>According to this de'nition, the set of all natural numbers>is recursive because there exists a TM that doesnÍt even>read the input and always outputs a 1.> Right. The set of all natural numbers is recursive by 'at. No, not by 'at. It is a trivial consequence of a general de'nition--> the de'nition that Daryl McCullough has reproduced for you.Using Daryl McCulloughÍs de'nition, I can claim the set ofall natural numbers is (1,2,3). Since the string 1111 doesnot represent a natural number in my system, it is too largeto be the input to a Turing machine.> Why bother to include computability in the de'nition? There is no separate de'nition for the case of all natural numbers.> There is one de'nition that applies to all subsets of the set of natural> numbers. The set of all natural numbers (henceforth N) is a subset of> N. We apply the de'ntition to N and 'nd out trivially that it is> recursive. It is not a trivial de'nition. It is quite profound. But it is> trivial that N is recursive.I 'nd the idea that N=(1,2,3) quite profound.> A human must decide if x is a member of this set> before x can be given to the TM. You are seeking to create a new de'nition and theory of computability.> Good. You want to consider a larger context of inputs other than> just natural numbers. For example, you want to include in'nite> stings as acceptable inputs.Who decides the string is 'nite?A TM will always think the input is 'nite.I have given proofs of this.> Fine. I would agree with you that> if we enlarged the universe of acceptable inputs from just N to a larger> set containing in'nite strings, then N would probably not be> recursive in any reasonable new de'nition of the word. YouÍre right,> a TM would not be able to conclude that an in'nite string of 1Ís is> not a natural number.A TM will say that any string of 1Ís represents a natural number.> I would say that in this enlarged context, that> N would be recursively enumerable but not recursive. But your problem is that you are using terms with agreed upon de'nitions> by the mathematical community (like recursive) incorrectly. You have> your own personal de'nition that you have failed to state.I used a standard de'nition of decidable language.The standard de'nition states that a human must decidethat a string is 'nite before giving it to the TM.Ultimately, a human being must decide if a string, x,represents a natural number.Why pretend there is an algorithm that can do this?> This is why> mathematicians are arguing with you and pointing out your obvious> errors. If you would like to re-de'ne recursive by all means do so.I am not rede'ning recursive.I am pointing out that if the de'nition of decidable languangesays there is a TM that can decide if string, x, is a memberof the language, then the language of all unary representationsof the natural numbers is NOT a decidable language.If we include the requirement that a human being decideswhether x is a 'nite string, then I agree N is recursive.This additional requirement allows me to choose N to beany set I want it to be.> But make your new de'nition known. It would also be good to give> it another name to avoid confusion, like Easterly-Recursive.> Until you do this, you will probably continue to make false statements> that are easily refuted.Are you refuting my claim that N=(1,2,3)?> Remember, mathematics is precise.N is precisely equal to (1,2,3).> By the way there are already in existence more generalized recursiontheories,> which deal with recursion on other domains besides N.Can you give references?Obviously, I need to learn more about this stuff.Russell- 2 many 2 count === Subject: Re: No Set Contains Every Computable NaturalRussell Easterly says...>Using Daryl McCulloughÍs de'nition, I can claim the set of>all natural numbers is (1,2,3). Since the string 1111 does>not represent a natural number in my system, it is too large>to be the input to a Turing machine.How does my de'nition imply that the set of all naturalnumbers is {1,2,3}?If you want to be more general than the usual treatment,you can do the following: Let A be some 'nite alphabet (containing 0 and 1, at least), and let A^omega be the set of in'nite sequences of elements of A. Every such in'nite sequence is a possible state for the in'nite tape of a Turing machine. Now, let S be any subset of A^omega, and let R be any subset of S. Then we can say that R is recursive relative to S (my own terminology, not any kind of standard) if there is a Turing machine T such that for any tape state x: 1. If x is an element of R, then T(x) halts and outputs 1. 2. If x is an element of S, but not an element of R, then T(x) halts and outputs 0. 3. If x is not an element of S, then there is no restriction on what T(x) does. We can also de'ne a notion of being recursively recognizable relative to S: R is recursively recognizable relative to S if R is a subset of S and if there is a Turing machine T such that for any tape state x: 1. If x is an element of R, then T(x) halts and outputs 1. 2. If x is an element of S, but not an element of R, then T(x) either halts and outputs 0, or never halts. 3. Again, if x is not an element of S, then there is no restriction on what T(x) does.Of particular interest is the case when S = the set of unaryrepresentations of natural numbers (that is, each tape statein S consists of a 'nite number of consecutive 1s followedby an in'nite number of consecutive 0s). In that case, ifR is a subset of S, then R is recursive relative to S if andonly if R is the set of unary representations of a recursiveset of natural numbers. R is recursively recognizable relativeto S if and only if R is the set of unary representation ofan r.e. set of natural numbers.As you point out, the set of unary representations of naturalsis not recursive (or recursively recognizable) relative to theset A^omega of all tape states.--Daryl McCulloughIthaca, NY === Subject: Re: No Set Contains Every Computable NaturalDaryl McCullough McCulloughÍs de'nition, I can claim the set of>all natural numbers is (1,2,3). Since the string 1111 does>not represent a natural number in my system, it is too large>to be the input to a Turing machine. How does my de'nition imply that the set of all natural> numbers is {1,2,3}?We have already agreed there is no way a TM can decideif the number of symbols on a tape is 'nite(in a 'nite number of steps).If I get to decide what is 'nite, then I can say anythingwith three or less symbols is 'nite and anything withmore than three symbols is in'nite.Of course, most people would use the standardde'nition of N we all know and love.> If you want to be more general than the usual treatment,> you can do the following: Let A be some 'nite alphabet (containing 0 and 1, at least),> and let A^omega be the set of in'nite sequences of elements of A.> Every such in'nite sequence is a possible state> for the in'nite tape of a Turing machine. Now, let S be any subset of A^omega, and> let R be any subset of S. Then we can say> that R is recursive relative to S (my own> terminology, not any kind of standard) if there is a> Turing machine T such that for any tape state x: 1. If x is an element of R, then T(x) halts> and outputs 1. 2. If x is an element of S, but not an element of R,> then T(x) halts and outputs 0. 3. If x is not an element of S, then there is no> restriction on what T(x) does. We can also de'ne a notion of being recursively> recognizable relative to S: R is recursively recognizable> relative to S if R is a subset of S and if> there is a Turing machine T such that> for any tape state x: 1. If x is an element of R, then T(x) halts and outputs 1.> 2. If x is an element of S, but not an element of R, then> T(x) either halts and outputs 0, or never halts.> 3. Again, if x is not an element of S, then there is no> restriction on what T(x) does. Of particular interest is the case when S = the set of unary> representations of natural numbers (that is, each tape state> in S consists of a 'nite number of consecutive 1s followed> by an in'nite number of consecutive 0s).There is that word, 'nite, again.Can I assume that it is never the case there is a stringwith an in'nite number of 1s followed by an in'nitenumber of 0Ís?This would mean there is exactly one string in A^omegathat contains an in'nite number of 1Ís with nopreceding 0Ís.> In that case, if> R is a subset of S, then R is recursive relative to S if and> only if R is the set of unary representations of a recursive> set of natural numbers. R is recursively recognizable relative> to S if and only if R is the set of unary representation of> an r.e. set of natural numbers. As you point out, the set of unary representations of naturals> is not recursive (or recursively recognizable) relative to the> set A^omega of all tape states.>I guess I should be happy you agree with me.Of course, that would be too simple.I really donÍt understand why R is not recursively recognizable.I warned you that I sometimes contradict myself.I think I can prove that a TM will say that anystring of consecutive 1s is 'nite.This proof also shows there is no such thing as anin'nite string of consecutive 1Ís, at least asfar as a TM is concerned.Before I give a complicated proof, I should givea simple explaination. Assume you want to countto in'nity. You start with 1, 2, 3, etc.Will you ever get to in'nity? Of course not.You can never add 1 to a natural number and getin'nity. A TM can add, subtract, multiply, and,maybe divide (division is kind of iffy). You canÍtcompute in'nity using these operations. A TMwill never say a string is in'nite because it isincapable of doing so.Now, the complicatd proof:I de'ne two TMs, A and B.TM A1) write 12) move rightrepeat steps (1) and (2)We might think this TM will write an in'nite string of 1s,but it doesnÍt.TM B1) Find a blank2) Find a second blank3) Backup and write a 1 on the previous blankrepeat steps (1) through (3)TM B will write a 'nite number of 1s followedby a blank. The number of 1s written by TM A isa multiple of the number written by TM B.1) write 1 read blank2) move right move right3) write 1 read blank4) move right move left5) write 1 for every 1 written by TM B.Either both TMs write a 'nite number of 1sor both TMs write an in'nite number of 1s.Either way, the output tape will still containan in'nite string of trailing blanks.(Of course, we can replace the blanks with 0).Russell- 2 many 2 count === Subject: Re: No Set Contains Every Computable NaturalRussell Easterly says...>The set of all natural numbers is recursive by 'at.>Why bother to include computability in the de'nition?>A human must decide if x is a member of this set>before x can be given to the TM.Right. The notion of recursive set is only nontrivialfor proper subsets of the naturals. For the complete set,it is recursive by de'nition.--Daryl McCulloughIthaca, NY === Subject: Re: No Set Contains Every Computable Natural <3_OdnTxoJNyqVK_dRVn-hA@comcast.com> Discussion, natural numbers is recursive by 'at.> Why bother to include computability in the de'nition?> A human must decide if x is a member of this set> before x can be given to the TM.Really?And if we give a tape with an in'nite number of 1Ís on it, how doesthe human decide whether thereÍs a 1 on every square (to the right ofthe start square) or just very many 1Ís to the right? What wonderfulfaculty have humans for determining the contents of an in'nitely longtape? -- All intelligent men are cowards. The Chinese are the worldÍs worst'ghters because they are an intelligent race[...] An average Chinesechild knows what the European gray-haired statesmen do not know, thatby 'ghting one gets killed or maimed. -- Lin Yutang === Subject: Re: No Set Contains Every Computable NaturalJesse F. Hughes is recursive by 'at.> Why bother to include computability in the de'nition?> A human must decide if x is a member of this set> before x can be given to the TM. Really? And if we give a tape with an in'nite number of 1Ís on it, how does> the human decide whether thereÍs a 1 on every square (to the right of> the start square) or just very many 1Ís to the right? What wonderful> faculty have humans for determining the contents of an in'nitely long> tape?Hey, thatÍs my argument.Using the logic OP have presented, I can choose the set ofall natural numbers to be any set I want.(1,2,3) is the recursive set of all natural numbers.Proof:This is the TM that decides if x is a member of (1,2,3).1) Print 12) HaltYou might think 1111 represents a natural number,but you are wrong. 1111 is too large to be theinput for a TM.Russell- Solution to halting problem: Ctrl Alt Del === Subject: Re: No Set Contains Every Computable Natural <3_OdnTxoJNyqVK_dRVn-hA@comcast.com> <8765e4q2fe.fsf@phiwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> Discussion, numbers is recursive by 'at.>> Why bother to include computability in the de'nition?>> A human must decide if x is a member of this set>> before x can be given to the TM.>> Really?>> And if we give a tape with an in'nite number of 1Ís on it, how does>> the human decide whether thereÍs a 1 on every square (to the right of>> the start square) or just very many 1Ís to the right? What wonderful>> faculty have humans for determining the contents of an in'nitely long>> tape? Hey, thatÍs my argument. Using the logic OP have presented, I can choose the set of> all natural numbers to be any set I want.> (1,2,3) is the recursive set of all natural numbers.No, itÍs not your argument.We have a well-de'ned notion of N, and also of functions N -> N. Wehave a well-de'ned notion of algorithms (in the form of Turingmachines). We de'ne what it means for an algorithm to represent aTuring machine (by de'ning which tapes represent which sequences ofnatural numbers) and hence derive a notion of computability, recursivesets, etc. These notions as stated make the claim that N is recursivetrivial. ThatÍs all resembles your errantnonsense. It does not.> Proof: This is the TM that decides if x is a member of (1,2,3). 1) Print 1> 2) Halt You might think 1111 represents a natural number,> but you are wrong. 1111 is too large to be the> input for a TM.This is just more errant nonsense.-- These mathematicians are worse than communists, as how do you explaintheir behavior? I *am* the American Dream, 'ghting for what should bemine, having to get past weak-minded academics who are 'ghting toblock my success. But I shall prevail!!! -- James S. Harris === Subject: Re: No Set Contains Every Computable NaturalJesse F. Hughes numbers is recursive by 'at.>> Why bother to include computability in the de'nition?>> A human must decide if x is a member of this set>> before x can be given to the TM.>> Really?>> And if we give a tape with an in'nite number of 1Ís on it, how does>> the human decide whether thereÍs a 1 on every square (to the right of>> the start square) or just very many 1Ís to the right? What wonderful>> faculty have humans for determining the contents of an in'nitely long>> tape? Hey, thatÍs my argument. Using the logic OP have presented, I can choose the set of> all natural numbers to be any set I want.> (1,2,3) is the recursive set of all natural numbers. No, itÍs not your argument. We have a well-de'ned notion of N, and also of functions N -> N.Then why pretend that there is an algorithm that can decide them?> We> have a well-de'ned notion of algorithms (in the form of Turing> machines). We de'ne what it means for an algorithm to represent a> Turing machine (by de'ning which tapes represent which sequences of> natural symbols.Suddenly, TMÍs only write sequences of natural nuumbers.> and hence derive a notion of computability, recursive> sets, etc. These notions as stated make the claim that N is recursive> trivial.N is a recursive set by 'at.Why bother to pretend N is computable?Obviously, a human has to decide if a symbol representsa natural number. No TM can do this.A TM doesnÍt have a well-de'ned notion of N.Russell- 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural <3_OdnTxoJNyqVK_dRVn-hA@comcast.com> <8765e4q2fe.fsf@phiwumbda.org> <8JCdncK7SMPAiK7dRVn-vw@comcast.com> <87hdxovh3h.fsf@phiwumbda.org> Discussion, argument.>> We have a well-de'ned notion of N, and also of functions N -> N. Then why pretend that there is an algorithm that can decide them?> We>> have a well-de'ned notion of algorithms (in the form of Turing>> machines). We de'ne what it means for an algorithm to represent a>> Turing machine (by de'ning which tapes represent which sequences of>> natural symbols.> Suddenly, TMÍs only write sequences of natural nuumbers.Who the said that? Read, son, read. We adopt a convention that certain of the tapes produced representnatural numbers. Other tapes do not represent natural numbers. If wewant to talk about representations of functions N -> N, then we ignorethose outputs which donÍt represent natural numbers (typically, weadopt the convention that function de'ned by a Turing machineconverges iff the machine halts and the tape thus produced representsa natural number).> and hence derive a notion of computability, recursive>> sets, etc. These notions as stated make the claim that N is recursive>> trivial. N is a recursive set by 'at.I think thatÍs a fair assessment. We have the mathematical object N. We have the (also mathematical)notion of Turing machines which produce marks on tapes. If we want toreason about computability of functions involving the former using thelatter, then we must choose a representation of natural numbers bytapes. If we ask whether N is recursive, then it depends on what context wemean. We may mean whether there is a computable function N -> N whichreturns 1 iff its input is in N and 0 else. To prove that there issuch a function, we need to reason about Turing machines, but onlyabout their behavior on input tapes which represent natural numbers.On the other hand, we might ask whether there is a total functionwhich, on *any* input tape whatsoever, returns 1 iff the taperepresents a natural number and 0 else. Now, if any includes tapeswhich have in'nitely many ones at the start, then clearly the answeris no. In *this* sense, N is not recursive.Again, there is a difference between the following two questions:(1) Does there exist a TM t such that, for every input tape i whichrepresents some natural number n, t(i) = 1?(2) Does there exist a TM t such that, on every input tape (includingthose with an in'nite number of ones) i, t halts and outputs 1 iff irepresents some natural number?> Why bother to pretend N is computable?I donÍt know what it means for N to be computable. I think itÍspretty clear what I mean when I say N is a recursive set. Evidently,itÍs not clear to you, but maybe the above helps.> Obviously, a human has to decide if a symbol represents> a natural number. No TM can do this.No TM can devise a convention, if thatÍs what you mean. At least, noTM can devise a convention in the sense that matters here.> A TM doesnÍt have a well-de'ned notion of N.I wouldnÍt want to claim that a TM has any notions at all.-- But he himself was not to blame for his vices. They grew out of a personaldefect in his mother. She did her best in the way of §ogging him while aninfant... but, poor woman! she had the misfortune to be left-handed, and achild §ogged left-handedly had better be left un§ogged. -- E.A. Poe === Subject: Re: No Set Contains Every Computable derive a notion of computability, recursive>> sets, etc. These notions as stated make the claim that N is recursive>> trivial.Not completely trivial. We are letting a human decide what is computable.> N is a recursive set by 'at. I think thatÍs a fair assessment. We have the mathematical object N. We have the (also mathematical)> notion of Turing machines which produce marks on tapes. If we want to> reason about computability of functions involving the former using the> latter, then we must choose a representation of natural numbers by> tapes. If we ask whether N is recursive, then it depends on what context we> mean. We may mean whether there is a computable function N -> N which> returns 1 iff its input is in N and 0 else. To prove that there is> such a function, we need to reason about Turing machines, but only> about their behavior on input tapes which represent natural numbers.Ultimately, a human decides what represents a natural number.Computability has nothing to do with it.> On the other hand, we might ask whether there is a total function> which, on *any* input tape whatsoever, returns 1 iff the tape> represents a natural number and 0 else. Now, if any includes tapes> which have in'nitely many ones at the start, then clearly the answer> is no. In *this* sense, N is not recursive.This is what I was trying to say.> Again, there is a difference between the following two questions: (1) Does there exist a TM t such that, for every input tape i which> represents some natural number n, t(i) = 1?Trivially, yes.1) Print 12) Halt> (2) Does there exist a TM t such that, on every input tape (including> those with an in'nite number of ones) i, t halts and outputs 1 iff i> represents some natural number?No TM can do this and halt after a 'nite number of steps.> Why bother to pretend N is computable? I donÍt know what it means for N to be computable. I think itÍs> pretty clear what I mean when I say N is a recursive set. Evidently,> itÍs not clear to you, but maybe the above helps.Given any set, S, of representations of natural numbers,I can write a TM that will 'nd a representation of a natural numberthat is not in S. (No, this TM does not halt after a 'nite numberof steps.)> Obviously, a human has to decide if a symbol represents> a natural number. No TM can do this. No TM can devise a convention, if thatÍs what you mean. At least, no> TM can devise a convention in the sense that matters here.But, a TM can extend the convention in such a way that no setcontains every symbol that represents a natural number.> A TM doesnÍt have a well-de'ned notion of N. I wouldnÍt want to claim that a TM has any notions at all.Exactly.Computers do what you tell them to do,not what you want them to do.Russell- 2 many 2 count === Subject: Re: No Set Contains Every Computable am using the de'nition of decidable language given at wikipedia: A decidable or recursive language is a formal language that is arecursive> set, i.e., for which there exists an algorithm to solve the following> decision problem: Given string w, does w belong to the language? The> algorithm is not allowed to run into an in'nite loop and has to producea> YES/NO answer for any input string after a 'nite amount of time. To> formalize the rather vague term algorithm, one usually employs Turing> machines, but several other equivalent approaches are possible. Note that it says a 'nite amount of time.I ponted out in an earlier post that this should say in a 'nite number ofsteps.Hypercomputers can perform an in'nite number of operationsin a 'nite amount of time.> All regular, context-free and context-sensitive languages are recursive,> but there exist recursively enumerable languages that are not recursive;> one example is given by the halting problem. Did you realize that the natural numbers in unary (or binary, or> any standard base) is a *regular* language, and hence context-free, and> recursive (i.e. decidable.).Recursively enumerable, but not recursive.> You reference this site, and then you try to make claims contradicting> them.I contradict lots of people.I even contradict myself, sometimes.> I have shown why a Turing machine can not decide if a string, w,> represents a natural number. ThatÍs because you think the string of an in'nite number of 1s is a> natural number. Once again, you have to understand that every natural> number is 'nite, even in unary representation.I never said an in'nite string of 1Ís represents a natural number.I said such a string does NOT represent a natural number.No TM can decide that an in'nite string of 1Ís does not represent anatural number in a 'nite number of steps.> The set of all natural numbers is not a recursive set if our de'nition> of recursive set requires there exist a TM that can decide membership> in the set. MathworldÍs de'nition indicates there are differences> between de'nitions of general recursive function.> This appears to be one of those differences. Did you even notice that your de'nitions of these sets are based on> *formal languages* ? Did you even look up what a formal language is?> Here you go, from Wikipedia: > In mathematics, logic and computer science, a formal language is a> set of 'nite-length words (or strings) over some 'nite alphabet. Note that a formal language is a set of FINITE-LENGTHED strings.No TM can decide whether an arbitrary set of strings is a formal language.Russell- 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural I ponted out in an earlier post that this should say in a 'nite number of> steps.> Hypercomputers can perform an in'nite number of operations> in a 'nite amount of time. Time for a turing machine is the number of steps. We are talking about turing machines here, not ïhypercomputersÍ or other constructs.> Did you realize that the natural numbers in unary (or binary, or> any standard base) is a *regular* language, and hence context-free, and> recursive (i.e. decidable.).> Recursively enumerable, but not recursive. Any regular set is recursive. Any context-free set is recursive. Perhaps you should learn about these two classes of sets before moving up on the Chomsky Hierarchy.> You reference this site, and then you try to make claims contradicting> them.> I contradict lots of people.> I even contradict myself, sometimes. And everyone knows one can deduce anything from a contradiction. I suppose thatÍs how you prove a lot of your statements.> ThatÍs because you think the string of an in'nite number of 1s is a> natural number. Once again, you have to understand that every natural> number is 'nite, even in unary representation.> I never said an in'nite string of 1Ís represents a natural number.> I said such a string does NOT represent a natural number.> No TM can decide that an in'nite string of 1Ís does not represent a> natural number in a 'nite number of steps. And this is essentially why an input to a turing machine MUST be 'nite. The turing machine must be guaranteed that it will eventually hit the end of input if it continues to read along its tape. You need to start accepting the fact that input sizes are 'nite-sized, and a lot of these issues of yours will disappear.> In mathematics, logic and computer science, a formal language is a> set of 'nite-length words (or strings) over some 'nite alphabet. Note that a formal language is a set of FINITE-LENGTHED strings.> No TM can decide whether an arbitrary set of strings is a formal language. TMs are not supposed to reason about anything that is NOT (as a precondition) a formal language. DonÍt expect any further responses from me in this thread because this is essentially arguing something similar to the existence of God. You seem to be accepting a different set of axioms and de'nitions from standard ones.J === Subject: Re: Bound of a genius, but I need to 'nd some way to bound>>k-1 n>sum sum 2/(j-i+1)>i=1 j=k+1>>1 <= k<= n.>>Any upper bound that is some constant by n is OK.>Write the sum as >S = sum_{r=3}^n f(r,k)/r>where f(r,k) = min(k-1,n-r+1) - max(0,k-r+1).Oops, somehow I lost the 2. So my bound would be 2n.I donÍt know if if matters to the original poster, but I think I canimprove the constant to log(4).Let us 'rst reindex and then count how many times 2/j appears for eachvalue of j to get k-1 n --- --- 2 > ----- --- --- j-i+1 i=1 j=k+1 k-2 n-i --- --- 2 = > - --- --- j i=0 j=k-i+1 n --- 2 = > - (min(n-j,k-2) - max(k-j+1,0) + 1) --- j j=3 n --- n-1 - |n-j-k+2| - |k-j+1| = > ------------------------- [1] --- j j=3When j is near 3, the numerator of the summand in [1] is 2(j-2). Whenj is near n, the numerator of the summand is 2(n-j+1). The numeratorreaches a plateau of 2 min(k-1,n-k). ____________ <--- 2 min(k-1,n-k) / / / / <--- 2 3 n Fig. 1: Numerator of the summand in [1]Therefore, when k = [(n+1)/2], the summation in [1] is maximum for thegiven n. For that k, when j <= k+1 the numerator is 2(j-2) and whenj >= k+2, the numerator is 2(n-j+1).Computing the sum for k = [(n+1)/2], we get [1] k+1 n --- 2(j-2) --- 2(n-j+1) = > ------ + > -------- --- j --- j j=3 j=k+2 = n log(4) - 4 log(n) + 6 log(2) - 4 gamma + 1 - (16+(-1)^n)/2n + O(1/n^2) [2]Although a O(1/n) estimate would have probably been enough, I computedthe 1/n term since I had expected the even n case to be about 1/n lessthan the odd n case since the triangle in Fig. 1 is truncated for evenn. Estimate [2] veri'es this.In any case, both n log(4) - 4 log(n) + 6 log(2) - 4 gamma + 1 andn log(4) are overestimates for the sum when n >= 3.Rob Johnson take out the trash before replying === Subject: Re: Bound of a sum> I donÍt know if if matters to the original poster, but I think I can> improve the constant to log(4).ThatÍs the bound I got earlier. See === Subject: Re: . 53ab2750 is an anonymous user .hanson laugh like you ... when she was 2 years old === Subject: Re: . Produced By Microsoft MimeOLE V6.00.2800.1165Gauge AHahahahaha....AHAHAhahahaha...ahahaha....> ..... you too get three attaboys for your reply !> LifeÍs a bowl of cherries, pits and nuts and all....ahahahahaha.......ahahahahanson Yeah. My niece used to laugh like you ... when she was 2 years old>ehhhhh, hmmmmmm.........ahahahaha......Your retort has a pH 12+ for some reason. Is it due to an after effect from ~ ....after chemo, I, Peter Brown was never the same?Cheer up, Peter, *laugh*, thatÍs all thatÍs there, thatÍs all you got left!LifeÍs a bowl of cherries, pits and nuts & all....Do you git it? git it git it?hanson, bonvivant and commentator on all walks of life...AHAhhahaha..Ahahahahahaha.......... AHAHAHHAHHAAHAHAHHAHa........ahahahahaha.. === Subject: Re: How fast is the Con'nued Fraction factorization algorithm?>>My impression, based on not very scienti'c>>(or knowledgeable) experiment>>is that the Quadratic Sieve (with many polynomials)>>is unlikely to factorize number with more than 80 digits>>in a reasonable time on a reasonable computer>>(say 1 day on my 667MHz laptop).> What do you mean by unlikely? It makes no sense in this> context. QS succeeds with virtual certainty in time that depends> only on the size of n (with some small statistical variation).ThatÍs just not true in my experience.And my comment makes perfect sense.The quadratic sieve completes in one or two minuteswith some 70-digit numbers, and takes hours with others.Which is exactly what I would expect.What reason do you have to suppose the time has onlysmall statistical variation?Have you actually tried it?What if the number is prime?(I gave random 60-digit numbers for factorisationas an assignment to a class of 30.One was prime, and there was a huge variationin the time taken - using quadratic sieve -to factorise the others.)-- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: How fast is the Con'nued Fraction factorization algorithm?>My impression, based on not very scienti'c>(or knowledgeable) experiment>is that the Quadratic Sieve (with many polynomials)>is unlikely to factorize number with more than 80 digits>in a reasonable time on a reasonable computer>(say 1 day on my 667MHz laptop).>>What do you mean by unlikely? It makes no sense in this>>context. QS succeeds with virtual certainty in time that depends>>only on the size of n (with some small statistical variation).> ThatÍs just not true in my experience.> And my comment makes perfect sense.> The quadratic sieve completes in one or two minutes> with some 70-digit numbers, and takes hours with others.> Which is exactly what I would expect.> What reason do you have to suppose the time has only> small statistical variation?> Have you actually tried it?> What if the number is prime?> (I gave random 60-digit numbers for factorisation> as an assignment to a class of 30.> One was prime, and there was a huge variation> in the time taken - using quadratic sieve -> to factorise the others.)> This is probably because your implementation runs trial division, the p+-1 and/or a couple of runs of ECM before running the quadratic seive. You can probably test this by factoring a few hundred large numbers and looking a a histogram of the run times. It will be extremely bimodal. The p+-1 and ECM are extremely fast, but probabilistic algoritms (they also have expected run times based on the size of the factors, not on the size of the original number.)Theoriginal quadratic seive (if I recall correctly) worked like this:To factor n1) Choose a factor base of small primes.2) Choose a random number r between 1 and n-13) If the least positive residue of r^2 mod n factors completely using only the primes in the factor base save r and a list of the primes occuring to odd powers.4) repeat steps 2 and 3 till yo have about as many values of r as there are primes in the factor base.5) You now have almost enough info to 'nd a pair of numbers such that x^2=y^2 mod n.Improvements have increased the probability of 'nding numbers that factor completely, but the philosopy is the same. The run time will beproportional to the number of trials it takes to 'nd enough numbers that completely factor. === Subject: Re: How fast is the Con'nued Fraction factorization algorithm?Timothy Murphy on not very scienti'c>>(or knowledgeable) experiment>>is that the Quadratic Sieve (with many polynomials)>>is unlikely to factorize number with more than 80 digits>>in a reasonable time on a reasonable computer>>(say 1 day on my 667MHz laptop).> What do you mean by unlikely? It makes no sense in this> context. QS succeeds with virtual certainty in time that depends> only on the size of n (with some small statistical variation).> ThatÍs just not true in my experience.> And my comment makes perfect sense.> The quadratic sieve completes in one or two minutes> with some 70-digit numbers, and takes hours with others.> Which is exactly what I would expect.> What reason do you have to suppose the time has only> small statistical variation?> Have you actually tried it?> What if the number is prime?> (I gave random 60-digit numbers for factorisation> as an assignment to a class of 30.> One was prime, and there was a huge variation> in the time taken - using quadratic sieve -> to factorise the others.)My experience is mainly with 50-70 digit numbers, and I 'nd the variation to be minimal. A factor of less than 2 differentiates the slow ones from that fast ones, absolute max. Never a factor of over 30 like you suggest. And certainly I see no basis for you to /expect/ such variation. On what are your expectations founded?It just so happens that just now my Lucas project has dumped on me 2 similar sized C65s, that I wasnÍt going to bother factoring, but just for the sake of this thread I can crack them as anecdotal evidence:<<CPU time: 3m45s<<CPU time: 2m38s(Satoshi TomabechiÍs PPSIQS)ThatÍs pretty close, and about what IÍve come to expect for C65s.However, donÍt just take my word for it - my view is entirely suported by data from other factoring projects: http://www.asahi-net.or.jp/~KC2H-MSM/mathland/part/ ppsiqs.htmThe times in each column have no more than a factor of 2 spread, and almost always the highest time will correspond to a number a couple of bits larger than the smallest time.And primes are a red herring, it takes a fraction of a second to do a PSW test on numbers that size, or even an APRCL test.Phil-- Unpatched IE vulnerability: NavigateAndFind protocol historyDescription: cross-domain scripting, cookie/data/identity theft, command executionReference: http://safecenter.net/liudieyu/NAFjpuInHistory/ NAFjpuInHistory-Content.HTMExploit: http://safecenter.net/liudieyu/NAFjpuInHistory/ NAFjpuInHistory-MyPage.HTM === Subject: Re: the anticlassicalist }{ i: linguistic negation> -=-=-=-=-= linguistic negation =-=-=-=-=-=- The logic of natural language has been studied by > That they have one and all avoided coming to any unique conclusion> emphasizes the bankruptcy of the Liberal Arts. ItÍs a hen party.> Anybody can gossip and all are appreciated. ItÍs bull. How would> you like to be doing 80 mph in a car that was built by mob rule> instead of cold, hard, dry engineering?> Oh, but that example is different! It certainly is. Liberal Arts are> bull and engineering is real world.> Everything in the Liberal Arts and Social Sciences is both right and> wrong. No conclusion at all can be drawn except the need for more> funding. Say what you want - it makes no difference at all (except> for grant funding - then you must agree with the mob or be destroyed).> Klingon was speci'cally created to be the worst language possible by> folks who knew linguistics. It was enthusiastically embraced by the> mob and it is as good as Korean or Chinese for transferring content.Well, actually, Uncle Al, youÍre wrong about this. But then IÍm just alinguist so what the would I know? And youÍre a ing chemist orsome sucha whizpoop hot-lead scientologician, which means that youropinions about linguistics are worth-- what, 3.978654549 kazillion golddollars to the peso? -- over anything that somebody who knows somethingabout the subject might say.Okrand says he violated a few human language universals in inventingKlingon, since of course it wasnÍt meant to be a human language. Nothingabout worst language possible. Studies of how Klingon-users actuallyspeak might be interesting. Do you know of any? And anyhow what was thismeant to prove, even if it were all true? Sure as hell wouldnÍt provewhatever you were salivating about in the previous paragraph anyway.Ross Clark === Subject: Re: the anticlassicalist }{ i: linguistic negation> Klingon was speci'cally created to be the worst language possible by> folks who knew linguistics. It was enthusiastically embraced by the> mob and it is as good as Korean or Chinese for transferring content.> Okrand says he violated a few human language universals in inventing> Klingon, since of course it wasnÍt meant to be a human language. Nothing> about worst language possible. Studies of how Klingon-users actually> speak might be interesting. Do you know of any? And anyhow what was this> meant to prove, even if it were all true? Sure as hell wouldnÍt prove> whatever you were salivating about in the previous paragraph anyway.You can 'nd a signi'cant community of people who speak Klingon atthe Klingon Language Institute. http://www.kli.org/The archives of the KLIÍs email discussion forum might be useful ifyouÍre serious about studying how Klingon is spoken. http://www.kli.org/tlhIngan-Hol/As for worst language possible, that would be INTERCAL. Klingon isactually a very easy language. Compared to natural languages, itÍsessentially === a toy, but itÍs a very powerful and interesting toy.Subject: Re: the anticlassicalist }{ i: linguistic negation > As for worst language possible, that would be INTERCAL.I found this:http://www.antipope.org/charlie/journo/ === anti-universals etc. (Re: the anticlassicalist }{ i: linguistic negation) > Okrand says he violated a few human language universals in inventing> Klingon, since of course it wasnÍt meant to be a human language. IÍve got the Klingon dictionary (which is actually a handbook),IÍve got the tapes, IÍve got the later Klingon Berlitz Guide(not its true title, itÍs the Klingon dictionary again,adapted for galactic hitchhikers, canÍt remember its realtitle), IÍve been through the lot, I havenÍt come across anything even remotely weird. If you want something reallyweird, try Lojban instead.(No apologies for cross-posting to alt.phil, sci.m, sci.l,sci.p, someone out there started this , not me) === Subject: Re: the anticlassicalist }{ i: linguistic negation <103513r3gafocff@corp.supernews.com> <1035esdt30h4d38@corp.supernews.com> In message <1035esdt30h4d38@corp.supernews.com>, galathaea the references to math and physics,Did I blink? I havenÍt seen any physics yet.-- Richard Herring === Subject: Re: the anticlassicalist }{ i: linguistic negation > They donÍt know nothing!>How bland. LetÍs really mean it: They donÍt know nothing about nothing.> The not-house will be built>Is this to mean the not-house wonÍt be not-built?Do even number of negations af'rm and odd number of negations deny? If he didnÍt not double negate, didnÍt he af'rm? If he didnÍt not double negate, didnÍt he not af'rm?How about double af'rmatives implying negation?ArenÍt my ideas the greatest?Oh yea, === yea.Subject: Re: the anticlassicalist }{ i: linguistic negation > The not-house will be built> Is this to mean the not-house wonÍt be not-built?No, itÍs a solemn undertaking to build a trampoline.Some argue no, not a trampoline, but a bridge,which is not a house either. I counter thattrolls are known to dwell under bridges (LordBritish, pers.com.), and therefore a bridge is ahouse since it houses trolls. So a trampoline it must be. === Subject: [PhD-position] Statistics. +0100 == of insurance portfolios University of Twente, The Netherlands =The Department of Applied Mathematics of the Faculty of ElectricalEngineering, Mathematics and Computer Science at the University ofTwente, Enschede, The Netherlands has available a PhD-positionFor information contact:prof.dr. W. Albers (w.albers@math.utwente.nl, tel. 053-4893816/3434).Title:Statistical analysis of dependence effects on insurance portfoliosResearch Group:The project will be performed in the chair Statistics and Probability.It 'ts in the research programme Financial Engineering from theCentre for Telematics and Information Technology (CTIT), one of thekey research institutes of the University of Twente. The FinancialEngineering Laboratory (FELab) is a joint initiative of researchersin this 'eld within the School of Mathematical Sciences and theDepartment of Finance and Accounting (FMBE) at the School ofTechnology & Management. At present, 13 full- and 6 part-time facultymembers, as well as 6 PhD-students, participate in this laboratory.Supervision:prof.dr. W. Albers (UT; email: w.albers@math.utwente.nl)dr. W.C.M. Kallenberg (UT; email: w.c.m.kallenberg@math.utwente.nl)Funding source:This position is funded by the Technology Foundation STW, appliedscience division of NWO and the technology programme of the Ministryof Economic Affairs.Period:The project can start immediately; it will take 4 years.Required:The candidate should have a MSc in mathematical statistics, probabilityor econometrics, and a keen interest in applicationsDescription:Quantities of interest for insurance portfolios are among others the sumS of the claims of the individual risks during a reference period andthe stop-loss premiums E{max(0,(S-a))} for various retentions a.Typically it is assumed that the summands in S are independent, althoughit is clear that in practice dependencies will occur. Implicit in thisassumption apparently is the hope that the effects of such dependencieswill be negligible, or at most small for practical purposes. Unfortunately,however, this is often not the case: introducing a small dose ofdependence may already lead to gross deviations of the stop-loss premiums. To deal with the problem, families of models incorporating dependenciesneed to be proposed and studied, with an emphasis on practical usefulness.Typically, a balance will have to be found between degree of complicationand realistic content. Next, techniques have to be found to obtainadequate approximations to the above mentioned quantities of interest.This is also a nontrivial part, as already in the independent case exactresults are generally intractable. Finally, estimation methods have to bedeveloped to 't the models obtained to data occurring in practice. As concerns the 'rst two of these three stages in the above program,some progress has already been made in [1]. Here a simple stochasticmixture model is proposed, which is demonstrated to incorporate to 'rstorder many models in common practical use Second order Edgeworth expansionsturn out to be quite successful for obtaining adequate approximations tostop-loss premiums, as long as normally distributed claims are used.In [2], the scope is widened considerably by including various otherapproximations from the literature. For a wide variety of claim sizedistributions and retention levels, such approximations are compared in thispaper to each other, as well as to a quantitative criterion. The third step,'tting such a model to real data, still has to be taken. It is expectedthatthis will generate a new iteration of the program by suggesting directionsinto which the basic model should be altered or generalized.[1] Albers, W. (1999). Stop-loss premiums under dependence. Insurance: Math. & Econ. 24, 173-185.[2] Reijnen, R., Albers, W. and Kallenberg, W.C.M. TW-report 1695. == === ==Subject: Sets That Resemble Derivatives support1.mathforum.org (8.11.6/8.11.6/The Math Forum, (ABÍ)' = AÍ U Bfor any two sets A, B, where ï is complement and Adjacent lettersare intersected. This is the set analog of logical a-->b for a,b propositions, the latter de'ned as ~(a ^ ~b) = ~a V b with~ being not (negation), ^ and (conjunction), V or (disjunctionincluding the possibility of conjunction). Then it is fairly easy to prove that:2) [(A-->B)(AÍ-->BÍ)]' = ABÍ U AÍB3) [(AÍ-->B)(A-->BÍ)]' = AÍBÍ U ABEquation (2) resembles the derivative or derivation if ï (comple-ment) takes the place of derivative and the set on the lefthand side inside brackets replaces AB, somewhat like (fg)' =fgÍ + fÍg for differentiable functions f, g. Equation (3) putsan even number of primes on expressions that look like AB onthe right hand side, that is to say even from 0 to 2. Thelatter generalizes one idea of the derivative. Derivations are a topic in the literature on Lie Algebras and otherstructures, but (2) and (3) have a rather different §avor.Notice on the left-hand-side inside the brackets there are 0 or2 primes in one pair of parentheses in (2), while in (3) thereis one prime in one pair of parentheses, a sort of reverse of theright hand side situation for (2) and (3).Could the complement of a set be a generalized derivative? Well,f(x + h) and f(x - h) get compared with f(x) in a derivative forreal functions, and the complement of a set is certainly outsidethe set in a sense similar to the way x + h and x - h are outsidex (to left and right). On the other hand, since the complementof a set is the universe minus the set, we donÍt really have tocompare the complement with the original set since in a sense thelatter is determined once the former is, so we can just look atf(x + h) and not f(x) 'guratively speaking. To see this moreexplicitly, notice that:4) ABÍ U AÍB U AÍBÍ U AB = universewhich you can visualize rapidly by using Venn diagrams or prove byelementary set theory, for any two different sets A, B.Osher Doctorow === Subject: Re: Sets That Resemble Derivatives SomewhatthereÍs a general theme that some people have explored that says thatthe fact that certain sorts of differential operators obeyleibniz-like identities is conceptually subordinate to the fact thatcertain sorts of boundary operators in geometry and topology obeyleibniz-like identities, roughly (or sometimes precisely, depending onthe particular concept of boundary being used) boundary(x X y) =(boundary(x) X y) + (x X boundary(y)).-- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Sets That Resemble Derivatives Somewhat> 2) [(A-->B)(AÍ-->BÍ)]' = ABÍ U AÍB> 3) [(AÍ-->B)(A-->BÍ)]' = AÍBÍ U AB>2) is just symmetric difference, assuming the existence of some universe === Subject: Re: Sets That Resemble Derivatives Somewhat> De'ne:> 1) (A-->B) = (ABÍ)' = AÍ U B> for any two sets A, B, where ï is complementin what?> and Adjacent letters> are intersected. This is the set analog of logical a-->b for a,> b propositions, the latter de'ned as ~(a ^ ~b) = ~a V b with> ~ being not (negation), ^ and (conjunction), V or (disjunction> including the possibility of conjunction).> Then it is fairly easy to prove that:> 2) [(A-->B)(AÍ-->BÍ)]' = ABÍ U AÍB> 3) [(AÍ-->B)(A-->BÍ)]' = AÍBÍ U AB> Equation (2) resembles the derivative or derivation if ï (comple-> ment) takes the place of derivative and the set on the leftNo. The right side looks similar, ïcos of your denoting complementby the same notation as derivative, but the right side is completelydifferent!> hand side inside brackets replaces AB, somewhat like (fg)' => fgÍ + fÍg for differentiable functions f, g. Equation (3) puts> an even number of primes on expressions that look like AB on> the right hand side, that is to say even from 0 to 2. The> latter generalizes one idea of the derivative.One idea? Not the idea of it being a limit of differencequotients methinks.> Derivations are a topic in the literature on Lie Algebras and other> structures, but (2) and (3) have a rather different §avor.I.e., by not resembling derivation at all?> Notice on the left-hand-side inside the brackets there are 0 or> 2 primes in one pair of parentheses in (2), while in (3) there> is one prime in one pair of parentheses, a sort of reverse of the> right hand side situation for (2) and (3).applause!> Could the complement of a set be a generalized derivative? Well,> f(x + h) and f(x - h) get compared with f(x) in a derivative for> real functions, and the complement of a set is certainly outside> the set in a sense similar to the way x + h and x - h are outside> x (to left and right). Like er yeah! I mean yeah if x + h is different to x then er yeahthen x + h is outside x. Wow, groovy!> On the other hand, since the complement> of a set is the universe minus the set, we donÍt really have to> compare the complement with the original set since in a sense the> latter is determined once the former is, so we can just look at> f(x + h) and not f(x) 'guratively speaking. To see this more> explicitly, notice that:> 4) ABÍ U AÍB U AÍBÍ U AB = universeAha! The universe itslef. Next youÍll be giving us the answerto life the universe and everything.> which you can visualize rapidly by using Venn diagrams or prove by> elementary set theory, for any two different sets A, B.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Perplexing by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, follow these messages about perplex patterns but I am a little lost.Could someone summarize what has already been done about this subject, that is how many patterns have been found for n from 1 to 20 (or something like that) and also how many one can expect to discover.Joaquim Nogueira === Subject: Re: e is transcendental (was: classes of transcendental numbers support1.mathforum.org (8.11.6/8.11.6/The Math Forum, Ing Panagiotis Stefanides) exp(i pi) is cos(pi), and its imaginary part>is sin(pi), so all you are saying is that cos(pi) = -1 and >sin(pi) = 0, and we were already aware of these facts. There is >NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect.>> Panagiotis Stefanides>>Yes, but you DONÍT tell us what your reason is. You canÍt expect us to >accept your claims without giving support for those claims. So what >possible reason could you have for expecting us to agree with your claim >that exp(i pi) = 0?>The reason is simply that exp(ipi0=-1 should be accompanied>>by the statement that this is the real part solution.>>Is it fair?>>No, that is not a fair comment. exp(i pi), the complex number, is>equal to -1, the complex number. There is no need to appeal to the >real part. Also, what does exp(i pi) = -1 is the real part solution>mean? You look like you are using terminology in a manner not >recognized in mathematics.>>David McAnally>>-------------->>e^[i*pi] ,as accepted ,is a phasor. >>In most of the relevant 'elds of mathematics, e^[i pi] is a complex >number.>It is only fair to state that its >>polar representation is :>e^[i*pi] = MOD 1 , ARG 180 .>>That is Arg 180 degrees, not just Arg 180. And so what? That does >not lead to your assertion that exp[i pi] = 0, a result for which >you have given absolutely no support. Why donÍt you just give up?>>David McAnally>> At the moment, they (the Time Lords) are far from being all-powerful.> ThatÍs why itÍs been left up to me and me and me.> quote by: Patrick Troughton in The Three Doctors>>------->I, have made myself very explicit.>My original question of the implication>>of the imaginary component: e^[ipi]=j*0 (to the related proof)Complex Notation, chapter 12 ,Electrical Technology 3RD ED.Edward Hughes Longmans ,page 338:Stares: OA*=OB+jOC=OA(COStheta+jSINtheta)* Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,..Here is very clear the difference between PHASOR and MAGNITUDES[which(MAGNITUDES) I, referred to as REAL PART or IMMAGINAR PART, I should have stated COMPONENTS ].>Perhaps you mean that (the imaginary part of e^[i pi]) = 0, in which>case you are correct, but you have had a lot of problem expressing >yourself, especially in view of the way that you initially made>the claim by stating that e^[i pi] = 0, which you described as the >imaginary part solution, using a terminology that nobody but you >knows. e[i*theta]=COStheta+iSINthetathetas could be given and calculations could be performedfor numerical evaluations.In books is stated that it is FORMULAand also terms such as evaluate:(-1+i*sqrt[3])^10Are these not solutions to problems?>Nobody knows what you mean when you make a statement like exp[i pi] = -1>is the real part solution of exp[i pi] = -1, or that exp[i pi] = 0>is the imaginary part solution of exp[i pi] = -1. I asked you to >explain your terminology but you havenÍt bothered.I, doubt it but ,still I, exlpained it anyway.>As I, exlained earlier, but I, am not the kind of not being bothered>When I take the imaginary part of the equation exp[i pi] = -1, I get>sin(pi) = 0, a fact which is already known to us.Have you thought also of the fact that if you have exp[i pi] = -1 (your >real part solution) and exp[i pi] = 0 (your imaginary part solution),>then you could conclude that 0 = exp[i pi] = -1, and get a contradiction?Well tthere is always the possibility of an answer such as MULTIVALUED , examplum gratias: e^[2i*pi]=1 ln[1]=0=2i*piI, make use of the comlex notation , but still ,I, have my natural questions ,and do not accept everything for granted.I, give an example in the form of question:It is required that COS[-i]+i*COS[-i] be evaluatedso that is its Modulus and Argumentbe evaluated Stefanideshttp://www.stefanides.gr/why_logarithm.htm>>was not answered and still it is open.Because under no circumstances can you ever claim that exp[i pi] = 0.David McAnally === Subject: Re: e is transcendental (was: classes of transcendental numbers ?Why do I even try? Are you even TRYING to learn? Or are you so closed-minded that you canÍt be bothered?panamars@otenet.gr (Eur Ing Panagiotis Stefanides) Panagiotis Stefanides) part of exp(i pi) is cos(pi), and its imaginary part>>is sin(pi), so all you are saying is that cos(pi) = -1 and >>sin(pi) = 0, and we were already aware of these facts. There is >>NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect.> Panagiotis Stefanides>>Yes, but you DONÍT tell us what your reason is. You canÍt expect us to >>accept your claims without giving support for those claims. So what >>possible reason could you have for expecting us to agree with your claim >>that exp(i pi) = 0?>The reason is simply that exp(ipi0=-1 should be accompanied>by the statement that this is the real part solution.>Is it fair?>>No, that is not a fair comment. exp(i pi), the complex number, is>>equal to -1, the complex number. There is no need to appeal to the >>real part. Also, what does exp(i pi) = -1 is the real part solution>>mean? You look like you are using terminology in a manner not >>recognized in mathematics.>>David McAnally>>-------------->e^[i*pi] ,as accepted ,is a phasor. >>In most of the relevant 'elds of mathematics, e^[i pi] is a complex >>number.>It is only fair to state that its >polar representation is :>e^[i*pi] = MOD 1 , ARG 180 .>>That is Arg 180 degrees, not just Arg 180. And so what? That does >>not lead to your assertion that exp[i pi] = 0, a result for which >>you have given absolutely no support. Why donÍt you just give up?>>David McAnally>> At the moment, they (the Time Lords) are far from being all-powerful.>> ThatÍs why itÍs been left up to me and me and me.>> quote by: Patrick Troughton in The Three Doctors>>------->I, have made myself very explicit.>My original question of the implication>of the imaginary component: e^[ipi]=j*0 (to the related proof)>Complex Notation, chapter 12 ,Electrical Technology 3RD ED.>Edward Hughes Longmans ,page 338:>Stares:> OA*=OB+jOC=OA(COStheta+jSINtheta)>* Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,..>Here is very clear the difference between PHASOR and MAGNITUDESI know the difference between a complex number and its modulus (or magnitude). You donÍt have to explain the difference tome.>[which(MAGNITUDES) I, referred to as REAL PART or IMMAGINAR PART,The magnitude of a complex number is generally not equal to eitherits real or imaginary part. How could you claim that it is?> I should have stated COMPONENTS ].>>Perhaps you mean that (the imaginary part of e^[i pi]) = 0, in which>>case you are correct, but you have had a lot of problem expressing >>yourself, especially in view of the way that you initially made>>the claim by stating that e^[i pi] = 0, which you described as the >>imaginary part solution, using a terminology that nobody but you >>knows.> e[i*theta]=COStheta+iSINthetaI know that it is true that exp[i theta] = cos(theta)+i sin(theta). It follows that exp[i pi] = -1. Incidentally, e[i theta] = i e theta, unlike what you have written.>thetas could be given and calculations could be performed>for numerical evaluations.And for other results as well.>In books is stated that it is FORMULA>and also terms such as evaluate:(-1+i*sqrt[3])^10>Are these not solutions to problems?No. Evaluate (-1+i sqrt(3))^10 is a problem for which you canget a solution using exp(i theta) = cos(theta)+i sin(theta). Thisdoes not mean that exp(i theta) = cos(theta)+i sin(theta) is itselfa solution. You need a problem before you can describe anything asa solution.>>Nobody knows what you mean when you make a statement like exp[i pi] = -1>>is the real part solution of exp[i pi] = -1, or that exp[i pi] = 0>>is the imaginary part solution of exp[i pi] = -1. I asked you to >>explain your terminology but you havenÍt bothered.>I, doubt it but ,still I, exlpained it anyway.I did ask you. And you did not explain your bizarre terminology. Therewas no problem, hence there is no solution, whether real part orimaginary part.>>As I, exlained earlier, but I, am not the kind of not >being botheredThis was not a quote from me. Why are you pretending that it was?>>When I take the imaginary part of the equation exp[i pi] = -1, I get>>sin(pi) = 0, a fact which is already known to us.>>Have you thought also of the fact that if you have exp[i pi] = -1 (your >>real part solution) and exp[i pi] = 0 (your imaginary part solution),>>then you could conclude that 0 = exp[i pi] = -1, and get a contradiction?>Well tthere is always the possibility of an answer such as >MULTIVALUED , examplum gratias:It is known that the exponential function is single-valued (in fact,it is known that the exponential function is entire).> e^[2i*pi]=1 > ln[1]=0=2i*piThe logarithm has a branch point at 0. The exponential function doesnot have a branch point. The analogy is invalid. Because the exponential function is single-valued, my objection here still stands.>I, make use of the comlex notation , but still ,I, have my >natural questions ,and do not accept everything for granted.But you shouldnÍt start making up mathematics to suit yourself.>I, give an example in the form of question:>It is required that COS[-i]+i*COS[-i] be evaluated>so that is its Modulus and Argument>be evaluated (NUMERICALLY].cos(-i) = cosh(1), so cos(-i)+i cos(-i) = sqrt(2) cosh(1) exp(i p/4).The magnitude is sqrt(2) cosh(1), which is approximately 2.18225,and its argument is pi/4, which is approximately 0.7854. Alternatively, the argument is exactly 45 degrees.David McAnally-------------- === Subject: Re: We come from your future> The Question is: What is The Question? Wheeler43 === Subject: Help needed - primes programHallo,I need help with the following :I made a program which gives all the primes from selected area. I made it inPascal. The only problem is that it hasnÍt a very good algorithm which Ineed.Here is the code:--------------------------------------------------------- ----------------------Program Primes;Var x,y:word; f:word; i,n:word; ok:boolean;Begin Write(ïEnter X and Y -> ï); Readln(x,y); For i:=x to y do Begin f:=2; ok:=true; While f<=i-1 do Begin If i mod f = 0 then Begin f:=i; ok:=false; End else f:=f+1; End; If ok=true then Write(i,' is a prime. ï); End; Readln;End.--------------------------------------------------- ----------------------------I wanted to give you attachment with the *.EXE 'le but I canÍt so if ypuneed to compile this code download Free Pascal (current version is 1.0.10)form www.freepascal.org .Any suggestion and similar is === Subject: Re: Help needed - primes programWell how about this:--------------------------------------------------------- ----------------------Program PB;Var x,y:word; f:word; i,n:word; ok:boolean;Begin Write(Enter X and Y -> ï); Readln(x,y); For i:=x to y do Begin f:=3; ok:=true; If (i=1) or (i mod 2 = 0) then Begin f:=i+1; ok:=false; End; While f<=sqrt(i) do Begin If i mod f = 0 then Begin f:=i; ok:=false; End else f:=f+2; End; If ok=true then Write(i,' is a prime. ï); End; Readln;End.--------------------------------------------------- ----------------------------It ignores all number with form k*2 , k{2,3,4,...} and it goes till sqrt(i)insted of i.Is this faste or slower than sieve of eEatosthenes? Why? === Subject: Re: Help needed - primes program| Well how about this:| -------------------------------------------------------------- --------------| Program PB;|| Var| x,y:word;| f:word;| i,n:word;| ok:boolean;|| Begin| Write(Enter X and Y -> ï);| Readln(x,y);|| For i:=x to y do| Begin| f:=3;| ok:=true;| If (i=1) or (i mod 2 = 0) then| Begin| f:=i+1;| ok:=false;| End;|| While f<=sqrt(i) do| Begin| If i mod f = 0 then| Begin| f:=i;| ok:=false;| End| else| f:=f+2;| End;|| If ok=true then Write(i,' is a prime. ï);| End;|| Readln;| End.| -------------------------------------------------------------- --------------| It ignores all number with form k*2 , k{2,3,4,...} and it goes till sqrt(i)| insted of i.|| Is this faste or slower than sieve of eEatosthenes? Why?The above program doesnÍt show two (2) as a prime. __________________Gerard S. === Subject: Re: Help needed - primes program| Well how about this:| -------------------------------------------------------------- --------------| Program PB;|| Var| x,y:word;| f:word;| i,n:word;| ok:boolean;|| Begin| Write(Enter X and Y -> ï);| Readln(x,y);|| For i:=x to y do| Begin| f:=3;| ok:=true;| If (i=1) or (i mod 2 = 0) then| Begin| f:=i+1;| ok:=false;| End;|| While f<=sqrt(i) do| Begin| If i mod f = 0 then| Begin| f:=i;| ok:=false;| End| else| f:=f+2;| End;|| If ok=true then Write(i,' is a prime. ï);| End;|| Readln;| End.| -------------------------------------------------------------- --------------| It ignores all number with form k*2 , k{2,3,4,...} and it goes till sqrt(i)| insted of i.|| Is this faste or slower than sieve of eEatosthenes? Why?No, itÍs slower because you are dividing the numbers by all odd numbers, thebest you could do (here) is just divide the numbers by all odd primes(handling the even numbers as a it to the original program(also rewritten in REXX) and sped it up by a factor of three:(orginal algorithm in REXX):________________________________________________________ _____________________/**/ arg x y .; if x==ÍÍ then x=1; if y==ÍÍ then y=12345; tell=y<0; y=abs(y)p=0; call time ïRÍ do j=x to y f=3; ok=1 if (j=1) | (j//2==0) then do; f=j+1; ok=0; end isq=$isq(j) do while f<=isq If j//f==0 then do; f=j; ok=0; end else f=f+2 end if ok then p=p+1 if tell then say j ïis prime.Í endt=format(time(ïEÍ),,2)say p ïprimes found betweenÍ x ïandÍ y ïand tookÍ t ïsecondsÍexit/*-----------------interger squart root subroutine-----------*/$isq: procedure; parse arg x; x=trunc(x); r=0; q=1 do while q<=x; q=q*4; end do while q>1; q=q%4; h=r+q; r=r%2; _=x-h if _>=0 then do; x=_; r=r+q; end endreturn r_____________________________________________________________ returns the integersquart root, which can be de'nes as §oor(sqrt(x))Note that REXX (the language I used) has no SQRT function anyway, so thiswould make it faster than if I used a regular SQRT function.Note that the orginal version has an error in it that doesnÍt show the (only)even prime (that is, 2).Here is the faster version (and correct) version of your algorithm:____________________________________________________ ___________________________/**/ arg x y .; if x==ÍÍ then x=1; if y==ÍÍ then y=12345tell=y<0 /*show primes if y is neg. */y=abs(y) /*use postive value of Y. */p=0 /*number of primes found. */ox=x /*save original x speci'ed.*/call time ïRÍif 2>=x then do /*special case of even prime*/ p=p+1 /*add 1 to the num of primes*/ if tell then say ï2 is prime.Í endx=max(x,3) /*start with 3 as a minimum.*/x=x+(x//2==0) /*if x is even, add 1 to it.*/ do j=x to y by 2 /*now, step trough odd nums.*/ do f=3 to $isq(j) by 2 /*divide j by all odd numbs.*/ if j//f==0 then iterate j /*not prime, try another num*/ end p=p+1 /*add 1 to the num of primes*/ if tell then say j ïis prime.Í endt=format(time(ïEÍ),,2)say p ïprimes found betweenÍ ox and y ïand tookÍ t seconds.exit/*-----------------interger squart root subroutine-----------*/$isq: procedure; parse arg x; x=trunc(x); r=0; q=1 do while q<=x; q=q*4; end do while q>1; q=q%4; h=r+q; r=r%2; _=x-h if _>=0 then do; x=_; r=r+q; end endreturn r_____________________________________________________________ _________________The next step in the optimization of the program is to only divide by theodd primes found so far (up to the ISQRT of J), instead of all the odd numbers.______________________________________________________ _________________Gerard S. === Subject: Re: Help needed - primes help with the following :> I made a program which gives all the primes from selected area. I made it in> Pascal. The only problem is that it hasnÍt a very good algorithm which I> need.Use the Sieve of Eratosthenes.Google will 'nd several versions and descriptions of it, as will the prime links at http://primepages.org/Phil-- Unpatched IE vulnerability: XSS in Unparsable XML FilesDescription: Cross-Site Scripting on any site hosting 'les that can be misrendered in MSXMLReference: http://sec.greymagic.com/adv/gm013-ie/Exploit: http://sec.greymagic.com/adv/gm013-ie/ === Subject: Re: Help needed - primes program>Hallo,I need help with the following :I made a program which gives all the primes from selected area. I made it in>Pascal. The only problem is that it hasnÍt a very good algorithm which I>need.Here is the code:>-------------------------------------------------------- -------------------->--->Program Primes;Var> x,y:word;> f:word;> i,n:word;> ok:boolean;Begin> Write(ïEnter X and Y -> ï);> Readln(x,y); For i:=x to y do> Begin> f:=2;> ok:=true; While f<=i-1 do> Begin> If i mod f = 0 then> Begin> f:=i;> ok:=false;> End> else> f:=f+1;> End; If ok=true then Write(i,' is a prime. ï);> End; Readln;>End.>------------------------------------------------- --------------------------->---I wanted to give you attachment with the *.EXE 'le but I canÍt so if ypu>need to compile this code download Free Pascal (current version is 1.0.10)>form www.freepascal.org .Any suggestion and similar is numbers in the loops, itwould double the speed.You can also likehttp://www.nist.gov/dads/HTML/sieve.html including implementations inpascal. === Subject: Re: Help needed - primes program> Hallo, I need help with the following : I made a program which gives all the primes from selected area. I made it in> Pascal. The only problem is that it hasnÍt a very good algorithm which I> need.I am not going to do your homework for you, but I marked belowsome obvious areas where you can improve your algorithm.> Here is the code:> -------------------------------------------------------------- --------------> ---> Program Primes; Var> x,y:word;> f:word;> i,n:word;> ok:boolean; Begin> Write(ïEnter X and Y -> ï);> Readln(x,y); For i:=x to y do> Begin> f:=2;> ok:=true; While f<=i-1 doYou can break off the search much before you get to i-1.> Begin> If i mod f = 0 then> Begin> f:=i;> ok:=false;Once you get here, why do you continue to iterate? End> else> f:=f+1;Why do you bother with even numbers and multiples of 3 at all?> End; If ok=true then Write(i,' is a prime. ï);> End; Readln;> End. === Subject: Re: How fast is the Con'nued Fraction support1.mathforum.org (8.11.6/8.11.6/The Math Forum, on not very scienti'c>(or knowledgeable) experiment>is that the Quadratic Sieve (with many polynomials)>is unlikely to factorize number with more than 80 digits>in a reasonable time on a reasonable computer>(say 1 day on my 667MHz laptop).> What do you mean by unlikely? It makes no sense in this>> context. QS succeeds with virtual certainty in time that depends>> only on the size of n (with some small statistical variation).ThatÍs just not true in my experience.And how much is that?>And my comment makes perfect sense.Sorry, but it makes ZERO sense.>The quadratic sieve completes in one or two minutes>with some 70-digit numbers, and takes hours with others.This is grossly false unless there is a major problem with yourimplementation. While there is some variation, depending on thequality of the multiplier, that variation is at worst a factor of 2 to 3. On a 1 GHz Pentium, I expect 70 digits to take about 20 minutesto an hour. Your claim of 1-2 minutes *seems* ludicrous, as doesthe claim of hours. Are you sure you are not talking about ECM?Such variation with ECM would not surprise me at all. In fact, I wouldexpect it.>Which is exactly what I would expect.Huh? QS runs in time that depends on the size of the number beingfactored. Why would you expect such a large variation?>What reason do you have to suppose the time has only>small statistical variation?>Have you actually tried it?ROTFL.I have factored more numbers with QS than anyone else in the world to date. I have more experience with the method than anyone else. I suggest you read my paperR. SilvermanThe Multiple Polynomial Quadratic Sieve Math Comp 1987In particular, read the section discussing the Knuth-Schroeppel function and its meaning/implications.>What if the number is prime?Then the 'nal congruences that get constructed, i.e.x^2 = y^2 mod Nwill ALWAYS have x = y or x = -y mod N. When N is prime, there areonly two solutions to x^2 = a mod N for any a. If N =pq there willbe 4 solutions. Two of these will be x = y or x=-y mod N. The othertwo yield factors. [hint: think about the Chinese Remainder Theorem]>(I gave random 60-digit numbers for factorisation>as an assignment to a class of 30.>One was prime, and there was a huge variation>in the time taken - using quadratic sieve ->to factorise the others.)Which only shows that there was wide variation in the quality of the implementations. QS does have some variation, depending on the value of the Knuth-Schroeppel function. But it does not vary by morethan a factor of 3 for numbers of 'xed size. === Subject: Who can tell me the recent develepoment on the GCH?W. Huge Woodin said he gave a solution that yield CH is false. Whatwas it gonging on? === Subject: Re: Simple numbersHallo;Well, p(n) is a n-th prime but if n=123456 how do I know what is the123456-th prime? I need the correct number (for exampel 1-st prime is - I know that every simple number (beside 2 and 3) has its one formula :> S.N => 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1> gives> the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple ,> 5,7|35) By simple number I think you must mean prime. > so my question is: > If n={1,2,3,...} what is the formula which gives the simple numbers> (beside> 2 and 3) for any n? Evidently you donÍt consider p(n) is the nÍth prime to be a formula.> Any> particular reason? Computational complexity? Jon Miller === Subject: Re: Simple numbers> Hallo;> Well, p(n) is a n-th prime but if n=123456 how do I know what is the> 123456-th prime? I need the correct number (for exampel 1-st prime is -> 2)!> Jonathan Miller message>> Hallo,>> I know that every simple number (beside 2 and 3) has its one formula :>> S.N =>> 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1>> gives> the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple ,>> 5,7|35)>> By simple number I think you must mean prime.>> so my question is:>> If n={1,2,3,...} what is the formula which gives the simple numbers>> (beside>> 2 and 3) for any n?>> Evidently you donÍt consider p(n) is the nÍth prime to be a formula.>> Any>> particular reason?>> Computational complexity?>> Jon MillerGiven any n, there is a simple algorithm that computes p(n). You could spend alifetime trying to make the algorithm more ef'cient, but the basic algorithmis quite simple. === Subject: Re: Simple numbersWell what is that algorithm?Dave Seaman n=123456 how do I know what is the> 123456-th prime? I need the correct number (for exampel 1-st prime is ->2)! Jonathan I know that every simple number (beside 2 and 3) has its oneformula :>> S.N =>> 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n+/- 1>> gives>> the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple,>> 5,7|35)>> By simple number I think you must mean prime.>> so my question is:>> If n={1,2,3,...} what is the formula which gives the simple numbers>> (beside>> 2 and 3) for any n?>> Evidently you donÍt consider p(n) is the nÍth prime to be aformula.>> Any>> particular reason?>> Computational complexity?>> Jon Miller Given any n, there is a simple algorithm that computes p(n). You couldspend a> lifetime trying to make the algorithm more ef'cient, but the === basicalgorithm> is quite simple.Subject: Re: Simple numbers what is that algorithm?Checkhttp://www.fpx.de/fp/Software/Sieve.htmlhttp:// www.math.utah.edu/~alfeld/Eratosthenes.htmlhttp:// www.math.utah.edu/history/eratosthenes.htmlhttp://www.utm.edu/ research/primes/prove/prove2_1.htmlhttp://math.nmsu.edu/~ pmorandi/CourseMaterials/FindingPrimes.htmlhttp:// primes.utm.edu/glossary/page.php?sort=SieveOfEratosthenes === Subject: Re: Simple numbers> Well what is that algorithm?ItÍs called the Sieve of Eratosthenes.-- Dave SeamanJudge YohnÍs mistakes revealed in Mumia Abu-Jamal ruling.Subject: re:How big can a manifold IÍm mainly interested in 1-dimentional manifolds anyway. I donÍt seehow WhitneyÍs Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: How big can a manifold be?well its === emmbeddable in R^3 its obvious of cardinality c, no?Subject: $$f_w91tn_g Only works for aleph1. It gives an example of a really big manifold,but still of cardinality 2^aleph0. My conjecture is that itÍs as bigas you get, but I donÍt know how Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: How big can a manifold be?what about using whitneyÍs embedding theorem? a jumpin dimension will have no effect on the cardinality. === Subject: Re: Goldberg dualRe:http://www.geocities.com/nicholasshea1/index.html and the radius for any number of points? on whether the convex hull is allowed to have square faces orwhether> those faces are triangulated.There is a broader generalization to be made about this. Theratio between an average edge length of a tesselated sphereand the radius of the sphere, call it beta, will vary between amaximum value and a minimum values. The maximum valuewill be come from the case of a completely triangulatedarrangement of edges such as in a typical geodesic sphere. Theminimum value for beta will be for a tesselation with all 3-wayvertexes such as in a Goldberg-like structure.The 'rst case has the greatest edge density(edges/area) andthe least volume. The second case has the least edgedensity(edges/area) with the greatest volume. Beta is alwayssomewhere between these extremes.For any number of points distribute on a sphere, beta will bewithin a 'nite range. This range depends on the arrangement ofthe edge.Dick === Subject: Re: Why am I so important?James Harris posts, one of which had an obvious sign error. Now then, probably quite a few posters will gleefully leap upon those> posts, as if itÍs such a big deal. Why? Why do I see webpages scattered across the Internet dedicated to> pushing negatives about me? Why am I so important to these people? Given the fact that IÍve been posting for years, and making mistakes> for years, it stands to reason that posters who so happily reply to me> as if it matters are strangely deluded. What changes? ItÍs a bizarre drama, where IÍm this guy, who posts a lot and has a> blog. For some odd reason, there are all these other people, like Dik> Winter, Erik Max Francis and Rick Decker who have these webpages,> which always 'nd some way to insult me. Now sure, maybe they get their jollies doing that, but why keep at it> for YEARS. Look at DeckerÍs webpage, and notice the date. Look at Dik WinterÍs webpages and notice the dates to which heÍs> referring. Clearly IÍm VERY IMPORTANT to these people!!! Oh well, IÍll just keep posting, and I guess these people will keep> replying.> James HarrisJames, QUIT COMPLAINING. If you donÍt want replies from certain people,donÍt post. You act like you own usenet and these threads. NEWSFLASH: youdonÍt.David Moran === Subject: strain for the last 20 years inthe 'eld as a result of which my math is somewhat basic. I have agood handle of linear systems, but am taking my 'rst baby stepswith non-lineear systems.I have read up a few books on non-linear systems looking for acanned solution for a problem I am trying to solve. But while I seea lot of stuff on the Duf'ng equation I donÍt see anything thatresembles the equation I am trying to solve - a mass-spring-dampersystem where the spring softens exponentially.i.e. the equation I am trying to solve is as follows:yÍÍ + cyÍ + [ko*e^(-alpha*t)]y = 0Any help would be most appreciated.sincerelyPaul Joseph(pjoseph@excite.com) === Subject: Re: strain softening engineer who has been working for the last 20 years in> the 'eld as a result of which my math is somewhat basic. I have a> good handle of linear systems, but am taking my 'rst baby steps> with non-lineear systems. I have read up a few books on non-linear systems looking for a> canned solution for a problem I am trying to solve. But while I see> a lot of stuff on the Duf'ng equation I donÍt see anything that> resembles the equation I am trying to solve - a mass-spring-damper> system where the spring softens exponentially. i.e. the equation I am trying to solve is as follows: yÍÍ + cyÍ + [ko*e^(-alpha*t)]y = 0 Any help would be most appreciated.Well, if you want an intuitive (albeit approximate) solution, you shouldstart by considering the relative size of the parameters. Speci'cally, youhave ordinary damping due to the ïcÍ parameter, and then you have theexponential softening, i.e. the ïalphaÍ parameter. These both have units o'nverse time and you should start by determining which range of values youare interested in, e.g. c << alpha and/or c >> alpha. In either of thesecases, the time scales are very different, and you can therefore make someassumptions. For instance, if alpha is very small, then the softeninghappens on a very slow time scale and therefore the spring constant isalmost, well, constant. You can therefore assume that the solution is anordinary damped oscillation and simply replace k0 in the solution with k0 *e^(-alpha*t).Another line of approach is to try to eliminate the damping by substitutingy(t) = z(t) * e^(-beta*t) for some suitable chosen value of beta.A third idea is to make a vector plot. Introduce z=yÍ and for each point(y,z) draw a vector whose direction is (yÍ,zÍ). Try it 'rst with alpha = 0,because then the vectors are independent of time. You will see spirals thatlead to the point (0,0). When alpha is nonzero, the vector plot istime-dependent, but you might be able to say something qualitative aboutthem anyway, e.g. whether they spiral faster or slower towards (0,0).-Michael. === Subject: Re: strain as follows:yÍÍ + cyÍ + [ko*e^(-alpha*t)]y = 0Any help would be most appreciated.sincerely>Paul Joseph>(pjoseph@excite.com)Maple gives a solution in terms of Bessel functions. I posted a 'lecalled DE.pdf showing the solution at:http://math.asu.edu/~kurtz/de/--Lynn === Subject: Re: puzzle: GCDs of In'nite Set of Integer Pairs>Playing the physics card isnÍt helpful here. IÍm claiming that itÍs>*mathematically* impossible to randomly choose a real from [0,1]. Not physically>impossible. IÍm asking you to convince me otherwise, use a thought experiment if>you want, or not. >>The canonical example is a spinner which can take on a random angle.>>This gets you [0,2pi)I think we had this one last time we explored this concept. At some point the>quantum state of the universe takes over. There are a countable number of angles>your spinner can indicate. Not surprising that they all have non-zero>probability, and that one actually happened.Angular momentum is quantized, but angle is not (so far, anyway). Inany case, bringing quantum physics to a thought experiment is farless valid than bringing Lebesgue measure to a question aboutprobability.-- Matthew T. Russotto mrussotto@speakeasy.netExtremism in defense of liberty is no vice, and moderation in pursuitof justice is no virtue. But extreme restriction of liberty in pursuit of a modicum of === security is a very expensive vice.Subject: re:puzzle: GCDs of $$f_w91tn_g My initial guess is 1 - pi*sin(sqrt(6)) / Newsfeed.Com - Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: puzzle: GCDs of In'nite Set of Integer Pairs> ie, generate a random number by §ipping a coin to determine whether itÍs>> bigger than 0.5, the again to determine whether itÍs bigger than 0.25 (or>> 0.75), repeat forever.> ThereÍs the kicker - forever. When forever comes around, IÍll let you know how>> you went wrong. Or you can say I told you so. Either way, itÍs not going to>> happen. You never chose your number, youÍre still choosing it.That argument leads to a denial of the existence of most real numbers.mmmm... in a sense it does. But then again in a sense, most real numbers donÍthave a proper existence. In terms of proportions, 100% of real numbers are notthought of, used, speci'ed, and donÍt occur naturally as a measure ofsomething. IÍm excluding extremely vague usage such as you specify ïall realslarger than xÍ.-- Patrick Hamlyn posting from Perth, Western AustraliaWindsur'ng capital of the Southern HemisphereModerator: polyforms group (polyforms-subscribe@egroups.com) === Subject: Re: puzzle: GCDs of In'nite Set of Integer Pairs> mmmm... in a sense it does. But then again in a sense, most real numbers donÍt> have a proper existence. In terms of proportions, 100% of real numbers are not> thought of, used, speci'ed, and donÍt occur naturally as a measure of> something. IÍm excluding extremely vague usage such as you specify ïall reals> larger than xÍ.The set of real numbers that have a 'nite description is countable;this is a superset of the algebraic reals. If your proper existenceis equivalent to 'nite description, your assertion is trivial.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: puzzle: GCDs of In'nite Set of Integer Pairs> ie, generate a random number by §ipping a coin to determine whether itÍs> bigger than 0.5, the again to determine whether itÍs bigger than 0.25 (or> 0.75), repeat forever.> ThereÍs the kicker - forever. When forever comes around, IÍll let you know how> you went wrong. Or you can say I told you so. Either way, itÍs not going to> happen. You never chose your number, youÍre still choosing it.>>That argument leads to a denial of the existence of most real numbers.> mmmm... in a sense it does. But then again in a sense, most real numbers donÍt> have a proper existence. In terms of proportions, 100% of real numbers are not> thought of, used, speci'ed, and donÍt occur naturally as a measure of> something. IÍm excluding extremely vague usage such as you specify ïall reals> larger than xÍ.Do you agree that the intermediate value property holds?That is, if f: [a,b] -> R is continuous, f(a) < 0, and f(b) > 0, then doyou agree that there is a real number x in [a,b] such that f(x) = 0?-- Dave SeamanJudge YohnÍs mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: puzzle: GCDs of In'nite Set of Integer Pairs>The canonical example is a spinner which can take on a random angle.>>This gets you [0,2pi)I think we had this one last time we explored this concept. At some point the>quantum state of the universe takes over. There are a countable number of angles>your spinner can indicate. Not surprising that they all have non-zero>probability, and that one actually happened.Pray tell, what is the non-zero real number that sums countably to 1?And what does the quantum state of the universe have to do withrigorous mathematical de'nitions? At 'rst I thought you were a 'nitist but now you turn out to be justanother crank. Oh well.-- IÍm not interested in mathematics that might have anythingto do with reality. -- Russell Easterly, in sci.math === Subject: Re: puzzle: GCDs of In'nite Set of Integer PairsOriginator: tchow@markov.mit.edu.mit.edu (Timothy Chow)>ThereÍs the kicker - forever. When forever comes around, IÍll let you know>how you went wrong. Or you can say I told you so. Either way, itÍs not>going to happen. You never chose your number, youÍre still choosing it.Forever isnÍt going to come around physically, but how do you know that itis impossible for it to come around mathematically? Suppose I choose the'rst number at time t = 1/2, the second number at time t = 3/4, the thirdnumber at time t = 7/8, and so on. (There are physical problems withaccelerating the choosing process without limit, but weÍre not talkingphysics.) At time t = 1, what are you going to tell me went wrong?-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: puzzle: GCDs of In'nite Set of Integer Pairs>>The probability space is simply ([0,1],L,m), where L is the set of all>>Lebesgue-measurable subsets of [0,1], and m is Lebesgue measure on [0,1].>>See .>>For each x in [0,1], the probability of the event X element {x} is the>>Lebesgue measure of the set {x}, which is zero.>>Done.> I trawled up and down those de'nitions, nothing seemed to jump out as> supporting the ïzero-probability-event actually happenedÍ scenario. Not that I> pretend to understand very much of those pages.I donÍt even know what it means to say that an event actually happened.I know what it means to say that an event is nonempty. An event isreally a set, and the probability of an event is its measure.Look at and noticewhere it says that the Cantor set (an uncountable set) has measure zero.Thus, if we transfer that language to the probability space we have beendiscussing, then the probability of the event (X in Cantor Set) is 0,despite the fact that the Cantor set contains uncountably many points.ItÍs a very long way from being empty.> And the fact that you canÍt specify which real you chose is not the only reason> you canÍt choose one. If you 'nd some way of specifying which one you chose,> IÍll 'nd a way to show you where you failed to do what you set out to do. >>And I suppose you also object to things like geometric constructions, on>>the grounds that they canÍt be carried out exactly in the real world.>>Yawn.> You can specify your real however you want. Geometric constructions translate> just 'ne into the real world. But what did you have in mind? Choose an angle> randomly from 0 to 90 degrees, and then construct a line from the 90 degree> vertex of a triangle to the opposite side (which has length 1), at that angle> from one of the other sides?> YouÍre just translating the domain, choosing the angle is effectively the same> as choosing the real directly.I didnÍt say anything about angles. I didnÍt say anything aboutchoosing a number at all. I gave you the de'nition of a particularprobability space and I pointed out that in that space, there arenonempty events that have probability zero.> Once again, you canÍt hide behind the ïreality cardÍ. You can say IÍll pick an> integer randomly from a uniform distribution across the whole set but thatÍs> garbage mathematically speaking, and IÍm claiming that choosing a real from> [0,1] is roughly the same §avour of garbage.>>No, there is an enormous difference between the two. There is no such>>thing as a uniform probability measure on a countably in'nite set, but>>uncountable sets are quite a different matter, as the Lebesgue measure>>example demonstrates.>>The example I gave is correct. > ItÍs so ïcos I said itÍs so doesnÍt wash with me.>>ItÍs not my fault if you donÍt know the de'nition of a probability>>space.> No. But itÍs disappointing (to me at least) that my error canÍt be explained to> me in terms I can understand. It seems such a simple concept, and we have to> resort to opaque de'nitions like Probability Spaces de'ned in terms of a> Lebesgue Measure of Lebesgue Measurable subsets.What is the length of a point? Zero, right?What is the length of the unit interval [0,1]? ItÍs 1, right?How many points are there in [0,1]? Uncountably many.This shows that you canÍt 'nd the length of the interval by adding upthe lengths of all the consitituent points. There are too many of them.Probability (like measure in general) is countably additive, which meansthat if you have a *sequence* of pairwise disjoint events, then theprobability of the union is the sum of the probabilities. That propertydoesnÍt hold for the [0,1] model because there are too many points to becontained in a sequence.The basic problem is that there are exactly the same number of points inthe interval [0,1/2] as there are in [0,1]. If you could 'nd the totalprobability by just adding up the constituent parts, then everynondegenerate interval would have the same probability, which doesnÍt 'tvery well with our notion of how probability ought to behave.><...>Look at and read the>>third paragraph down, where the concept of a probability function is>>mentioned. Notice that the function f: [0,1] -> R given by f(x) = 1 is>>an example of a probability function or probability density function>>as described there. The fact that the integral of f over [0,1] = 1 is>>what makes the function properly normalized to de'ne a probability>>measure, which turns out to be ordinary Lebesgue measure on [0,1].>>Look at the probability axioms that are stated farther down on that page.>>A probability measure is something that satis'es those axioms. Notice>>that the description of probability says nothing about what you can or>>cannot do in the real world; itÍs an abstract mathematical concept.> So... youÍre saying that if we accept some set of de'nitions and constructs> which donÍt relate to the real world, we can then say ïa zero-probability-event> can happenÍ in terms of those de'nitions.> Well maybe you can, but then youÍre not saying what I understand by the phrase.I am saying that events are sets, and a zero-probability event may be anonempty set. If an event is represented by a nonempty subset of the samplespace, then we can say that it represents a possible outcome.-- Dave SeamanJudge YohnÍs mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: tensors for tots> I am also familiar with the idea of transforming matrices themselves> to a new basis. If B is the transformation matrix (in my sense),> then the prescription for transforming a matrix A is:> AÍ = BAB^(-1)> Observe that this is the Transformation rule for a (1,1)-Tensor. > IÍm uncertain how we wish to transform v*Ok, let there be a 'xed dual vector, say (v*), which is a linearmapping (v*) : V -> R. (V a R-vector space)The value of (v*) at any vector (v) does clearly not depend on thebasis, let it be x := <(v*), (v)>(The value of (v*) at (v))Now we introduce a basis and have coordinate-vectors of (v*) and (v),say v* and v. We write v* and v as a column. We get thevalue of (v*) at (v) by matrix multiplication of theses two vectors:x = <(v*), (v)> = v*^t v.Now we change the basis. We know the new coordinates of (v), these arevÍ = Bv (your notation :-))Let v*' be the new coordinates of (v*).These are given byv*' = Av* with some Matrix AWhat is A? Look atx = v*^t v = v*'^t vÍ = (Av*)^t B v = v*^t A^t B vwhich implies A^t B = identity matrix,A = B^(-1)^t.Here we are. > I have a hunch this has to do with the metric tensorA normal matrix, so to say, is a (1,1)-Tensor.It can be multiplied by a row from the left and by a column fromthe right, that is by a covariant and a contravariant vector.The metric tensor eats two contravariant vectors, his matrix Mrepresents a different object. It stands for the mapping(v,w) -> v^t M w(here v and w are two columns)The transposing of v is built in.This implies a different behaviour under change of coordinates. === Subject: Re: tensors for tots> Anyway, to take one more baby step, if I have not exhausted my> credits, am I correct in thinking that a reasonable way to introduce a> metric tensor into babyland would be to de'ne the inner product of> a row vector and a column vector, in a particular coordinate system,> to be a bilinear form with a matrix sandwiched in the middle, that> matrix our metric?Urg, itÍs been a while since IÍve done this. This sounds about right,but you should de'nitely check out Michael SpivakÍs book onDifferential Geometry. . Most of the work he does is in terms o§inear algebra and differential forms. Granted, the latter are atensor analysis concept, but if you understand the Generalized StokeÍsTheorem, you probably know enough about them to get a good start.Ícid ïooh === Subject: Re: x^2 + y^4 = z^4> + y^4 = z^4 has no positive-integer solutions. Is the proof of this > result short enough for some kind soul to post it, or need I make a trip > to the library? (I have citations.)An elementary proof (no elliptic curves) is based on descent. Thatis, assume a solution exists in positive x, y, z, and show thatanother one can be found with smaller z. The proof is notparticularly dif'cult but too long to type out here (one single-spacetextbook page). It can be found in many basic number theory books,e.g. Niven and Zuckerman in the chapter on Diophantine equations, p.107 in the 3rd edition. === Subject: Re: x^2 + y^4 = z^4> + y^4 = z^4 has no positive-integer solutions. Is the proof of this> result short enough for some kind soul to post it, or need I make a trip> to the library? (I have citations.)ItÍs relatively easy to prove there are no solutions using elliptic curves.Firstly, the given equation has no non-trivial integer solutionsif and only if the equationy^2 = x^4 - 1has no non-trivial rational solutions.The transformationu = 1/(x-1), v = y/(x-1)^2brings this to standard elliptic formv^2 = cubic in u,and a little manipulation brings this toy^2 = x^3 + 2x^2 + 4x.The standard method (as described in Silverman & Tate)with the associated curvey^2 = x^3 - 4x^2 - 12xshows that this curve has rank 0,and it is easy to verify that there are no points of 'nite orderexcept for the point (0,0) of order 2 and the zero point.-- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: x^2 + y^4 = z^4> it is known that the equation x^2 > + y^4 = z^4 has no positive-integer solutions. Is the proof of this > result short enough for some kind soul to post it, or need I make a trip > to the library?Depends on what youÍre willing to accept as proved.(1) (z^4 - y^4)(y^2)(z^2) = (x^2)(y^2)(z^2).But we also (always) have(2) (2(y^2)(z^2))^2 + (z^4 - y^4)^2 = (z^4 + y^4)^2.So, by (1), the Pythagorean triangle with sides 2(y^2)(z^2) and z^4 - y^4 wouldhave an area that is a square. Since 1 is not a congruent number, this isimpossible.John Robertson === Subject: Re: x^2 + y^4 = z^4 === Subject: Re: x^2 + y^4 = z^4 >it is known that the equation x^2 + y^4 = z^4 has no >positive-integer solutions. >Depends on what youÍre willing to accept as proved. >(1) (z^4 - y^4)(y^2)(z^2) = (x^2)(y^2)(z^2). >But we also (always) have >(2) (2(y^2)(z^2))^2 + (z^4 - y^4)^2 = (z^4 + y^4)^2. >So, by (1), the Pythagorean triangle with sides 2(y^2)(z^2) and >z^4 - y^4 would have an area that is a square. Since 1 is not a >congruent number, this is impossible.Would you explain why?---- === Subject: Re: Mean Value Theorem> Two places IÍve seen the equivalent statement: If f is continuously> differentiable on [0,1], then |f(x)-f(y)|<=sup|fÍ(u)||x-y| where the> sup is extended over 0<=u<=1.> I canÍt see why we need fÍ to be continuous. It seems by the MVT we> have> f(x)-f(y)=fÍ(c)(x-y) for some c in [0,1],> so that > |f(x)-f(y)| = |fÍ(c)||x-y|, for some c in [0,1],> <= missing something subtle. However, it turns out, later in the discussion, they start talkingabout rates of convergence, so that there is in fact a need for sup|fÍ(u)| to be 'nite--itÍs not just a super§uous nicety. Does anyone know of a case where a textbook purposely included super§uous infomation as part of an exercise? Or how about on a test question? It seems that it would make the problems somewhat more realistic, albeit esthetically unpleasing. It may make studentsthink more about the problem. Students seem to develop the unhealthy assumption that if it is given, then is must be used. === Subject: Super§uous information; was: Re: Mean Value Theorem>Does anyone know of a case where a textbook purposely included >super§uous infomation as part of an exercise? Or how about on a >test question? It seems that it would make the problems somewhat >more realistic, albeit esthetically unpleasing. It may make students>think more about the problem. Students seem to develop the unhealthy >assumption that if it is given, then is must be used.>I have often put super§uous information in homework and exams (mainly in operations research and statistics) for just this reason.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Mean Value <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_> Two places IÍve seen the equivalent statement: If f is continuously> differentiable on [0,1], then |f(x)-f(y)|<=sup|fÍ(u)||x-y| where the> sup is extended over 0<=u<=1.> I canÍt see why we need fÍ to be continuous. It seems by the MVT we> have> f(x)-f(y)=fÍ(c)(x-y) for some c in [0,1],> so that > |f(x)-f(y)| = |fÍ(c)||x-y|, for some c in [0,1],> <= sup_u|fÍ(u)||x-y|.Maybe they intended f have values in R^n, so MVT is not available.Assuming fÍ exists everywhere and fÍ is integrable, we can stillintegrate fÍ on [x,y] to get the conclusion. And fÍ bounded willimply fÍ is Riemann integrable, so adding the assumptionfÍ is integrable is not needed. So I agree: even in thiscase, continuity of fÍ is not requried.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Mean Value TheoremX-DMCA-Noti'cations: 09:52:45 -0500, G. A. Edgar> Two places IÍve seen the equivalent statement: If f is continuously>> differentiable on [0,1], then |f(x)-f(y)|<=sup|fÍ(u)||x-y| where the>> sup is extended over 0<=u<=1.> I canÍt see why we need fÍ to be continuous. It seems by the MVT we>> have> f(x)-f(y)=fÍ(c)(x-y) for some c in [0,1],> so that > |f(x)-f(y)| = |fÍ(c)||x-y|, for some c in [0,1],>> <= sup_u|fÍ(u)||x-y|.Maybe they intended f have values in R^n, so MVT is not available.Assuming fÍ exists everywhere and fÍ is integrable, we can still>integrate fÍ on [x,y] to get the conclusion. And fÍ bounded will>imply fÍ is Riemann integrable, so adding the assumption>fÍ is integrable is not needed. So I agree: even in this>case, continuity of fÍ is not requried.Or you can just say this:If f : R -> R^n is differentiable then |f(x)-f(y)| <= sup_u|fÍ(u)||x-y|.Pf: Choose p in R^n with |p| = 1 and = |f(x) - f(y)|. Let g = andapply the MVT to g... QED.DoesnÍt require the fact that fÍ bounded implies thatfÍ is Riemann integrable (the proof of which is notobvious to me this second...)************************David C. Ullrich === Subject: Re: Mean Value TheoremX-DMCA-Noti'cations: -0800, stationaryset@yahoo.com (Jerome Davies)>Two places IÍve seen the equivalent statement: If f is continuously>differentiable on [0,1], then |f(x)-f(y)|<=sup|fÍ(u)||x-y| where the>sup is extended over 0<=u<=1.I canÍt see why we need fÍ to be continuous. It seems by the MVT we>have> f(x)-f(y)=fÍ(c)(x-y) for some c in [0,1],>so that > |f(x)-f(y)| = |fÍ(c)||x-y|, for some c in [0,1],> <= sup_u|fÍ(u)||x-y|.ThatÍs correct (noting that if fÍ is not continuous theexpression on the right may be in'nite; the reasonpeople state the thing the way they do is so theydonÍt have to consider the question of what the supof an unbounded function is, also because sayingthat something is <= in'nity doesnÍt seem all thatinteresting.)************************David C. Ullrich === Subject: Re: Mean Value TheoremGon.8dalo Rodrigues 16:11:09 -0800, stationaryset@yahoo.com (Jerome Davies)Two places IÍve seen the equivalent statement: If f is continuously>differentiable on [0,1], then |f(x)-f(y)|<=sup|fÍ(u)||x-y| where the>sup is extended over 0<=u<=1.I canÍt see why we need fÍ to be continuous. It seems by the MVT we>have f(x)-f(y)=fÍ(c)(x-y) for some c in [0,1],> Right.so that |f(x)-f(y)| = |fÍ(c)||x-y|, for some c in [0,1],> <= sup_u|fÍ(u)||x-y|. But without continuity (or something extra beyond differentiability)> how can you conclude that the derivative fÍ is bounded on [0, 1]? You> have to have boundedness in order to justify your last assign the value In'nity to sup_u|fÍ(u)|and then the relation still holds.So it seems that the differentiability condition is there purely fortechnical reasons (read: rigor), to avoid having to deal with In'nitybecause it does not belong to Rodrigues === 07:21:34 +0100, Michael JrgensenGon.8dalo Rodrigues 16:11:09 -0800, stationaryset@yahoo.com (Jerome Davies)>>Two places IÍve seen the equivalent statement: If f is continuously>>differentiable on [0,1], then |f(x)-f(y)|<=sup|fÍ(u)||x-y| where the>>sup is extended over 0<=u<=1.>>I canÍt see why we need fÍ to be continuous. It seems by the MVT we>>have>> f(x)-f(y)=fÍ(c)(x-y) for some c in [0,1],>> Right.>>so that>> |f(x)-f(y)| = |fÍ(c)||x-y|, for some c in [0,1],>> <= sup_u|fÍ(u)||x-y|.>> But without continuity (or something extra beyond differentiability)>> how can you conclude that the derivative fÍ is bounded on [0, 1]? You>> have to have boundedness in order to justify your last assign the value In'nity to sup_u|fÍ(u)|>and then the relation still holds.So it seems that the differentiability condition is there purely for>technical reasons (read: rigor), to avoid having to deal with In'nity>because it does not belong to R.>[Continuing the rambling]Well we can always, compactify R by adjoining -infty and infty, sothat should not be a problem. But what if x = y? We get0 <= infty 0What to make of === Subject: Re: Mean Value Theorem> On Davies)>>Two places IÍve seen the equivalent statement: If f is continuously>>differentiable on [0,1], then |f(x)-f(y)|<=sup|fÍ(u)||x-y| where the>>sup is extended over 0<=u<=1.>>I canÍt see why we need fÍ to be continuous. It seems by the MVT we>>have>> f(x)-f(y)=fÍ(c)(x-y) for some c in [0,1],>> Right.>>so that>> |f(x)-f(y)| = |fÍ(c)||x-y|, for some c in [0,1],>> <= sup_u|fÍ(u)||x-y|.>> But without continuity (or something extra beyond differentiability)>> how can you conclude that the derivative fÍ is bounded on [0, 1]? You>> have to have boundedness in order to justify your last assign the value In'nity to sup_u|fÍ(u)|>and then the relation still holds.So it seems that the differentiability condition is there purely for>technical reasons (read: rigor), to avoid having to deal with In'nity>because it does not belong to R. > [Continuing the rambling]> Well we can always, compactify R by adjoining -infty and infty, so> that should not be a problem. But what if x = y? We get> 0 <= infty 0> What to perhaps we can slip in an epsilon ad hoc. === Subject: Re: support1.mathforum.org (8.11.6/8.11.6/The Math Forum, posts from Hamilton College:You consider mathematicians *incapable* ?Do you or do you not consider mathematicians === Subject: intercept length when a random line intersect with ellipsoid shell or parallelepiped shell with 'xed thicknessIf a random line intersects with a ellipsoid shell or parellelepipedshell with a 'xed thickness, === how much is the statistical interceptlength?Subject: re:intercept length when a random line intersect with =http://www.newsgroups.com>Visit In order to answer your question, you need to be more speci'c aboutrandom for the line. A line in space can be speci'ed as all pointP satisfying P=X+tV, where X is a random point in space (needing aprobability distribution) and V is a random unit vector (isotropic?distribution), while t ranges over the entire real line.Once these distributions (for X and V) and the parameters of the shellare de'ned, you will have a tractable porblem. Services------------------------------------------------------ ---- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **---------------------------------------------------------- http://www.usenet.com === Subject: Re: intercept length when a random line intersect with ellipsoid shell or parallelepiped shell with 'xed thicknessEn el escribi.97:> If a random line intersects with a ellipsoid shell or parellelepiped> shell with a 'xed thickness, how much is the statistical intercept> length?You must de'ne in a precisse way how the random line is choosed. The answerdepend on it and it can be very different.-- (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: site. itÍs not real. Why donÍt you> actually read the whole site before you start condemning it. Moron.> Uh..I think heÍs promoting it. You might want to think hard about that> last word you used. ;^)Uh...doesnÍt READ like a promotion. I think IÍd rather play dumb.- Nick === Subject: Full Beal Conjecture (revised)Sorry for the mistake. My proof can be obtained === Subject: Re: Full Beal Conjecture (revised)> Sorry for the mistake. My proof can be obtained at> geocities.com/kerrymerry2000. www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Quadratics and reasonably easily get a generalidea of the type(s) of transformations that have been applied to x^2just by looking at the 'nished quadratic. That is, if I have 3x^2 + 7x + 2, can I easily relate the coef'cientsand constant to the dilations, re§ections and translations that thisfunction represents compared to x^2 ? I have tried thinking about it as a combined function, that is,f(x)=x^2, g[f(x)]=3x^2 + 7x + 2 and trying to 'nd g(x). Well, I havenÍttried with this example but one I tried earlier turned into a horriblemess and didnÍt give me what I wanted - I assumed that g(x) is simplythe inverse of g[f(x)] so maybe thatÍs why it got so messy ? It also seemed to be a lot of work!IÍm aware that I could determine the turning point and work out thetranslations and that the roots would indicate what horizontal dilationhas been applied. I expect that if I calculate two other points I wouldalso be able to determine re§ection and vertical dilation. Of course Icould always graph it but IÍm hoping to achieve what I want frominspection of the description of the function (is that the rightterminology?). Any and all help appreciated.Ivan. === Subject: Re: Quadratics and transformationIvan McDonagh seems to me that I should be able to reasonably easily get a general> idea of the type(s) of transformations that have been applied to x^2> just by looking at the 'nished quadratic. That is, if I have 3x^2 + 7x + 2, can I easily relate the coef'cients> and constant to the dilations, re§ections and translations that this> function represents compared to x^2 ?Yes. 3(x^2 + 7/3*x) + 2, complete the square inside the parentheses. Soyou end up with a(x-b)^2 + c. b is translation along the x-axis, c alongthe y-axis, a is the dilation factor. === If a is negative, thatÍs are§ection.Jon MillerSubject: re:JSH: I see what heÍs up to, weÍll 'll this thread with posts and heÍll goto somewhere else and say that heÍs important and will have thisthread to back him up, and the chain goes on forever. Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: JSH: Why am I so important?> Why am I so important to these people?You are important for the same reason Philip J. Fry is important.-- --Tim Smith === Subject: Re: JSH: Why am I so important? ... earth-shatteringly incisive analysis deleted ...> Clearly IÍm VERY IMPORTANT to these people!!!> Oh well, IÍll just keep posting, and I guess these people will keep> replying.> James HarrisYou mispelled impotent. I know that those words are pronounced (poe-nownst) the same way in Georgia (Joe-ja). People of yoursensitive disposition use the abbreviation ED (for ErectileDysfunction). Impotent seems, so, er, direct (pardon the implicitrhyme with erect).HereÍs the URL you want: === http://www.viagra.com/yer pal,Dale.Subject: re:Handling the How I hate it when wanna-be mathematicians write out cute identitiesand call it ïresearchÍ and then make wars about the philosophicalimplications of some useless notions. Then they have a go about theirmoral superiority and stuff... Stick to mainstream math and stop f*cking peopleÍs brains. Amegalomaniac cat is no way a ïbetter personÍ than the humble Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: Handling the truth Decker and a guy named Dik Winter say that w_1(x) w_2(x) = 7, and> these are *varying* functions that are the factors of (5a_1(x) + 7)> and (5a_2(x) + 7), in the ring of algebraic integers.> Can they derive w_1(x) and w_2(x)?> No.Yes. HereÍs a repost, in case you missed it.YouÍve seen what happens when x=1 and x=2. LetÍs try adifferent one, where things are simple enough to verify byhand.Suppose we take x = -3. Then we have (5a_1(-3) + 7)(5a_2(-3) + 7) = 7(25(9) + 30(-3) + 2) = 7(137)where the aÍs satisfy a^2 + 4a + 7((-3)^2 + (-3)) = a^2 + 4a + 42.We 'nd that a_1 = -2 + sqrt(-38) and a_2 = -2 - sqrt(-38)ItÍs easy enough to verify that the aÍs arenÍt divisible by7 or by sqrt(7).However, a_1 does share a factor in common with 7,namely w_1 = (1 + 3sqrt(-38))^{1/3}We can verify that1. w_1^3 divides 7^3, since (1 + 3sqrt(-38))(1 - 3sqrt(-38)) = 343 = 7^3 So w_1 = (1 + 3sqrt(-38))^{1/3} divides 7.2. w_1^3 divides a_1^3 = 220 - 26sqrt(-38), since (1 + sqrt(-38)(-8 - 2sqrt(-38)) = 220 - 26 sqrt(-38) So w_1 = (1 + 3sqrt(-38))^{1/3} divides a_1Thus w_1 is a common divisor of a_1 and 7. (In fact,itÍs a gcd of a_1 and 7.)In a similar way, we 'nd that w_2 = (1 - 3sqrt(-38))^{1/3}is a common divisor of a_2 and 7.In this case, with (5a_1 + 7)(5a_2 + 7) = 7(137)We see that the factor 7 on the right splits as w_1 * w_2 = 7and that f_1 = (5a_1 + 7)/w_1 = 5(-8 - 2sqrt(-38))^{1/3} + (1 - 3sqrt(-38))^{1/3}and f_2 = (5a_1 + 7)/w_1 = 5(-8 + 2sqrt(-38))^{1/3} + (1 + 3sqrt(-38))^{1/3} f_1 * f_2 = 137exactly as expected. Likewise, itÍs easy to show thatw_1, w_2, f_1, and f_2 are all algebraic integersand none of them are units.I hasten to add that this behavior is what happens foralmost all values of x, namely that in (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)with the aÍs satisfying a^2 -(x - 1)a + 7(x^2 + x)we will be able to 'nd algebraic integers w_1(x) and w_2(x)with w_1(x) * w_2(x) = 7 and w_i(x) dividing a_i(x) so thatwith f_i(x) = (5a_1(x) + 7)/w_i(x)we will have f_1(x) * f_2(x) = 25x^2 + 30x + 2as long as 7 doesnÍt divide 25x^2 + 30x + 2.With more or less dif'culty, one could do the sameconstruction for most integers x, just as I did for x=1and Keith did for x=2. In most cases weÍll 'nd that 7splits into two nonunit factors, distributed between(5a_1+7) and (5a_2+7). In general, as with w_i(-3) = (1 +/- 3sqrt(-38))^{1/3}it wonÍt be immediately obvious that w_1(x) and w_2(x)are divisors of 7 and of a_i(x), in the sense thatyou canÍt look at them and immediately notice thatthey divide 7 or a_i(x).> Instead you have one of their cohorts Keith Ramsay throwing out a> degree 22 polynomial claiming to have found it some kind of way, yet> if you look at> w_1(x) w_2(x) = 7> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > and> a^2 - (x - 1)a + 7(x^2 + x)> there is no way to derive the wÍs as varying functions of x.> The only thing that actually makes sense is something like> w_1(x) = 7, w_2(x) = 1.>No. See above. > I can point out the obvious, like why 7? Why not 13? Why canÍt you> just have something like> w_1(x) w_2(x) = 13> (5a_1(x) + 13)(5a_2(x) + 13) = 13(25x^2 + 30x + 2) > and> a^2 - (x - 1)a + 13(x^2 + x)?> Mathematically there has to be a reason, right?But you can. Just use (5a_1(x) + 13)(5a_2(x) + 13) = 13(25x^2 + 30x + 3)where the aÍs satisfy a^2 - (x - 2)a + 13(x^2 + x)Then an analogue to what I did above will allow you to'nd w_1(x) and w_2(x) for most integers x with thewÍs being algebraic integer divisors of 13 and theappropriate a, such that w_1(x) * w_2(x) === Subject: Re: message> Constant factors of polynomials used to be irrelevant multiples that> you just got rid of, until I came along. Now suddenly, youÍre told that constants are really products of> functions of x, and must be treated as such with polynomials. So what> changed? ItÍs as easy as looking at that example from Rick Decker, where heÍs> actually *supporting* the idea of a constant--7--being the product of> functions of x. Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his aÍs are roots of a^2 - (x - 1)a + 7(x^2 + x). If you check, it, yes, it all works out, but what about that 7. If> youÍre trained at all in mathematics then you should have learned to> discard it and focus on whether or not 25x^2 + 30x + 2 is reducible> over rationals, or on how to get its roots, but I changed the focus> with my research. So why decide that 7 is now a product of functions of x? Because the> truth hurts. The truth is that mathematicians have a de'nition that doesnÍt do> what they think it does, and you can either accept the truth, or you> can 'ght for what is now a religion. Decker and a guy named Dik Winter say that w_1(x) w_2(x) = 7, and> these are *varying* functions that are the factors of (5a_1(x) + 7)> and (5a_2(x) + 7), in the ring of algebraic integers. TheyÍre 'ghting for functions so that they can stay in the ring of> algebraic integers. Trouble is that I can show in multiple ways that what theyÍre saying> is false, besides the fact that itÍs silly. YouÍre *taught* to discard constant factors of polynomials like 7> *because* they are independent of x. Dik Winter and Rick Decker, and> people like them donÍt care about mathematics. They donÍt care how> many ways I prove them wrong. They just keep coming back with their silly claims. So why w_1(x) and w_2(x)? They canÍt tell you, except to say because,> because otherwise their beliefs about mathematicians and the ring of> algebraic integers are wrong, because then theyÍd have to admit that> IÍm right, because then theyÍd have to accept mathematical truth. Can they derive w_1(x) and w_2(x)? No. Instead you have one of their cohorts Keith Ramsay throwing out a> degree 22 polynomial claiming to have found it some kind of way, yet> if you look at w_1(x) w_2(x) = 7 (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) and a^2 - (x - 1)a + 7(x^2 + x) there is no way to derive the wÍs as varying functions of x. The only thing that actually makes sense is something like w_1(x) = 7, w_2(x) = 1. I can point out the obvious, like why 7? Why not 13? Why canÍt you> just have something like w_1(x) w_2(x) = 13 (5a_1(x) + 13)(5a_2(x) + 13) = 13(25x^2 + 30x + 2) and a^2 - (x - 1)a + 13(x^2 + x)? Mathematically there has to be a reason, right? But not if youÍre Dik Winter or Rick Decker promoting what is really a> *religion* and not mathematics. But most of you go along with them, you have to be going along as the> story is too big, too much important mathematics here for> mathematicians to try and claim that constants like 7 are no longer so> simple. You have to be complicit in their deceit, and an active> participant in betraying mathematics, or you would speak up. ThatÍs why IÍm better than you are. I can handle the truth. Years ago when I was working out various ideas that I thought proved> FermatÍs Last Theorem, dealing with a lot of hostile posters, calling> me names, insulting me, I found out I was wrong with an argument that> IÍd held on to for months. I went from thinking I was hero to feeling> like a zero. But I came out and admitted I was wrong because I realized that in> mathematics there is no gain from holding on to false beliefs. ThatÍs why IÍm a better person than you. If you donÍt learn that> lesson IÍm a better person than you can ever be. Not because I made some math discoveries, not because of my personal> characteristics, like I curse and get absusive myself, and hey, IÍm> sel'sh, but I can handle the truth, so no matter who you are, what> youÍve done, or what other people tell you about yourself or what you> tell yourself about yourself, IÍm a better person than you are while> you canÍt. IÍm better than you because I actually believe in mathematics, in> mathematical truth, even when it hurts. So you keep 'ghting along side Dik Winter and Rick Decker for what is> now your mathematical religion, 'ghing for lies, 'ghting for that 7> thatÍs now a product of functions! But no matter how many people stand next to you, and congratulate you,> or just silently support you, you will still be wrong because> mathematics is supposed to be more than a vote of con'dence. Mathematics is about the truth. Can you ever handle truth? Will any of you start growing as people?> Will any of you learn to better than you were?> James Harris Decker Quadratic Source Information> ---------------------> Recently Rick Decker, a professor at Hamilton College, apparently> trying to refute my research came up with a quadratic example, which I> like because itÍs a quadratic, and easier to manipulate than the> cubics IÍve used before. If you wish to see his original post here are some headers which also> show that he posts from Hamilton College: === Subject: Re: Mathematical consistency, courage Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his aÍs are roots of a^2 - (x - 1)a + 7(x^2 + x).James, shut up. Quit putting people down and STICK TO THE MATH. You are nobetter than us and vice versa.David Moran === Subject: Re: HELL :-) (Was: Re: the anticlassicalist }{ ii: the spectre continues)> *MATHEMATICS IS FULL OF TRICKS*French even allows you the conjugation :ma th.8ematiqueta th.8ematiquesa th.8ematiquenotre th.8fme .88 ticsvotre th.8fme attiqueleurre, tÍes mat, hic !> *REDUCING TOPICS TO THE SCIENCE OF NUMBER IS NUMEROLOGY*IÍd call that accounting. === Subject: Re: HELL :-) (Was: Re: the anticlassicalist }{ ii: the spectre continues) *MATHEMATICS IS FULL OF TRICKS* French even allows you the conjugation : ma th.8ematique> ta th.8ematique> sa th.8ematique> notre th.8fme .88 tics> votre th.8fme attique> leurre, tÍes mat, hic !> *REDUCING TOPICS TO THE SCIENCE OF NUMBER IS NUMEROLOGY* IÍd call that accounting.lol:-)mitch === Subject: Re: pentiamond rep-tiles, continued That leaves these two:> ____ __/> /____ /___/> I see no reason that these should not be rep-tilable. Neither do I see> a rep-tiling of either! Does anybody?I now see a reason that the pentiamond on the right should not berep-tilable. Suppose you have such a rep-tiling. Look at the littlepentiamonds along the edge of the big one, starting at the top vertexand following the edges on the right and bottom. You will 'nd thatall the boundaries between the little pentiamonds have the same orientationrelative to the edge of the big one: _______________You cannot escape from this pattern, even when you pass a 120-degreeangle. When you reach the lower left vertex, youÍre stuck.No solution yet on the I-, or straight, pentiamond. It canÍt be doneon a scale of 9x9 or smaller.-:- To what do I owe the honor of this unexpected visit, Lord Ruthven . . . alias Lyford Pemberton! H. C. Artmann, Tom Parker, International Detective-- Col. G. L. Sichermanhome: colonel@mail.monmouth.comwork: sicherman@att.comweb: === Subject: Re: errors in an argument>LetÍs just look at the probability of a protien made up of>100 amino acids forming by chance. All the amino acids>have to line up just right, and the probability of this>happening in any given try is 1/100! or 10^(-158).> There are only 20 different amino acids.> The more subtle error is that nobody says that life required> exactly this protein to be produced by a random permutation. > There are probably a huge number of different proteins that> could do more or less whatever this one does. And its production> would not be a matter of random permutations, but rather of> the gradual accumulation of small improvements. You might> look up genetic algorithms: this is not just a theoretical> construct but a method which is actually used to obtain a > nearly-optimal solution to a dif'cult problem by simulating > the process of evolution.> That said, I think itÍs fair to add: my impression, as an> outsider in this 'eld, is that while evolution itself (for > organisms that already have the usual genetic machinery) has pretty> good mathematical models, we are still far from understanding> at a quantitative level how that genetic machinery might have > arisen in the 'rst place.> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2> Robert, of course, mentioned exactly the errors I was thinking of,with the second being that we donÍt know how many possibleproteins there are with 100 amino acids in them. To say that foreach one, the probability is essentially 0, and so the collectiveprobability for the set of proteins is 0 is quite a mistake.If I were trying to explain the error to someone, I would probablyuse the analogy of rolling a 20 sided dice 100 times and recordingthe result. By the proof above, one canÍt do this, since theprobability of any particular sequence is (essentially) 0.I guess what I would really like to know is how poorly I shouldthink of the person who made the argument. Here are somequali'ers: He boasts of his degree in mathematics. Also, in histalk, he indicated that he would try to teach us how to thinkcritically about what scientists say. On the other hand, he isnow in his mid 60Ís, so he is far removed from his schooling.Finally, almost certainly (this is a guess of mine) he did notformulate the proof himself, but it had been presented to him.He liked the argument and missed the §aws.Should I not be too critical of his wishful thinking that thisproof actually means something?By the way, I certainly agree completely with RobertÍslast paragraph.John === Subject: Re: errors in an argument> If I were trying to explain the error to someone, I would probably> use the analogy of rolling a 20 sided dice 100 times and recording> the result. By the proof above, one canÍt do this, since the> probability of any particular sequence is (essentially) 0.What about the probability of any particular sperm actually being the oneto fertilize an egg? That would seem to preclude the possibility of sexualreproduction altogether.-- Joe Bramblett, KD5NRH === Subject: Re: Big Rip NY Science TimesIs it my imagination, or are this idiotÍs ramblings becomingeven *more* delusional and illucid than before?-E === Subject: Big Ripto appear in the book Super Cosmos on Amazon et-al ~ Summer additions and physics correctionsTwo big stories from the world of physics may portend the arrival of new weapons of mass destruction far more powerful and compact than atomic bombs. In recent years it has been discovered that our universe is being blown apart by a mysterious anti-gravity effect called dark energy. Mainstream physicists are scrambling to explain this mysterious acceleration in the expansion of the universe. Some physicists even believe that the expansion will lead to The Big Rip when all of the matter in the universe is torn asunder - from clusters of galaxies in appears to be made of two unknowns - roughly 23% is dark matter, an invisible source of gravity, and roughly 73% is dark energy, an invisible anti-gravity force. Ordinary matter constitutes perhaps 4 percent of the universe. Recently the British science news journal New Scientist revealed that the American military is pursuing new types of exotic bombs - including a new class of isomeric gamma ray weapons.That was an original idea of mine in 1963 at Cornell and I discussed it with Hans Bethe. That is one of the reasons Ron Bullough invited me to Harwell in 1966. No doubt others thought of it but probably later. I thought of it while at Tech/Ops in Lexington, Mass working for George Parrant Jr.Unlike conventional atomic and hydrogen bombs, the new weapons would trigger the release energy by absorbing radiation, and respond by re-emitting a far more powerful radiation. In this new category of gamma-ray weapons, a nuclear isomer absorbs x-rays and re-emits higher frequency gamma rays. The emitted gamma radiation has been reported to release 60 times the energy of the x-rays that trigger the effect.Gamma-ray weapons could trigger next arms race 19:00 13 August 03Exclusive from New Scientist Print Edition. Subscribe and get 4 free issues.An exotic kind of nuclear explosive being developed by the US Department of Defense could blur the critical distinction between conventional and nuclear weapons. The work has also raised fears that weapons based on this technology could trigger the next arms race. The explosive works by stimulating the release of energy from the nuclei of certain elements but does not involve nuclear 'ssion or fusion. The energy, emitted as gamma radiation, is thousands of times greater than that from conventional chemical explosives.The technology has already been included in the Department of DefenseÍs Militarily Critical Technologies List, which says: Such extraordinary energy density has the potential to revolutionise all aspects of warfare. Scientists have known for many years that the nuclei of some elements, such as hafnium, can exist in a high-energy state, or nuclear isomer, that slowly decays to a low-energy state by emitting gamma rays. For example, hafnium-178m2, the excited, isomeric form of hafnium-178, has a half-life of 31 years. The possibility that this process could be explosive was discovered when Carl Collins and colleagues at the University of Texas at Dallas demonstrated that they could arti'cially trigger the decay of the hafnium isomer by bombarding it with low-energy X-rays (New Scientist print edition, 3 July 1999). The experiment released 60 times as much energy as was put in, and in theory a much greater energy release could be achieved.http://www.newscientist.com/news/news.jsp?id= ns99994049I was thinking in 1963 of a gamma ray laser pumping a nuclear isomeric transition. Bethe at the time said it wouldnÍt work and basically discouraged me working on it.Bekkum continued:In the summer of 2000 I contacted Nick Cook, the former aviation editor and aerospace consultant to JaneÍs Defence Weekly, the international military affairs journal. Cook had been investigating black budget super-secret research into exotic physics for advanced propulsion technologies.Uh Oh :)I had been monitoring electronic discussions between various American and Russian scientists theorizing about rectifying the quantum vacuum for advanced space drive. Several groups of scientists, partitioned into various research organizations, were exploring what NASA calls Breakthrough Propulsion Physics - exotic technologies for advanced space travel to traverse the vast distances between stars. Partly inspired by the pulp science 'ction stories of their youth, and partly by recent reports of multiple radar tracking tapes of unidenti'ed objects performing impossible maneuvers in the sky, these scientists were on a quest to uncover the most likely new physics for star travel. The NASA program was run by Marc Millis, under the Advanced Space Transportation Program Of'ce (ASTP). Joe Firmage, a Silicon Valley entrepreneur, who at age 28 had found risen to CEO of a three billion dollar internet 'rm, began to fund research in parallel with NASA. He hired a NASA Ames nano-technology scientist, Creon Levit, to run the International Space Sciences Organization,Joe did that because I suggested it. I introduced Creon to Joe.Cook was intrigued by the apparent connections between various private investors, defense contractors, NASA, INSCOM (American military intelligence), and the CIA. While researching exotic propulsion technologies Cook had heard rumors of a new kind of weapon, a sub-quantum atomic bomb, being whispered about in the dark halls of defense research.I think that must have come from me regarding J. P. VigierÍs tight atomic states with experiments in Beograd by Z. Maric and G. Dragic. But how did Cook hear about that? We brought Vigier to ISSO in San Francisco several times along with physicist Gennady Shipov from Moscow. That story with photographs of Vigier and the group is in my autobiography Destiny Matrix. Dragic A, Maric Z, Vigier JP; Phys. Lett. A 265 (2000) 163. New quantum mechanical tight bound states and ïcold fusionÍ. Creon Levit and Vigier met with Maric in Budapest in 2000.Bekkum who is one of my on line students continued:Sub-quantum physics is a controversial re-interpretation of quantum theory, based on so-called pilot wave theories, where an information that the predictions of ordinary quantum mechanics could be recast into a pilot wave information theory. Recently Anthony Valentini of the Perimeter Institute has suggested that ordinary quantum theory may be a special case of pilot wave theories, leaving open the possibility of new and exotic non-quantum technologies. Even thought rumors of a sub-quantum bomb may be purely fantasy ItÍs not fantasy. It might not work, Maric and Dragic in Beograd, while not ostensibly trying to make a weapon by any means, were trying to test VigierÍs basic theory of the spatially extended electron, which I think is basically a correct idea and 'ts my own ideas including why the electron appears to shrink to less than 10-16 cm under high resolution imaging (i.e. scattering) and how the electric charge distribution is contained by the strongly attractive zero point energy exotic vacuum dark matter core (Abraham-Becker-Lorentz-Poincare stress problem of 100 years ago). This only works in the Bohm hidden or extra variable interpretation. That is, a classical spatially extended electric charge distribution is unstable. It explodes under its own self-repulsion. This is why physicists had to postulate a point electron because they did not understand that the strong gravity attraction of the positive zero point pressure in a possible state of exotic vacuum would hold the charge together. As Herbert Frohlich told me at UCSD in La Jolla in 1966 the basic thing wrong with physics is the idea of the point electron. The bad idea of the point electron gives the in'nite energy in quantum electrodynamics. Richard Feynman told me in his of'ce at Cal Tech in 1968 that in'nite renormalization is a shell game, and it is a scandal in physics that no one could do better than what he had done. They did not know 100 years ago that 1/3 or so of the universe was this kind of exotic vacuum. For example, there is a huge sphere of exotic vacuum of w = -1 positive pressure that holds our galaxy together preventing our solar system from going off into space on its own. This sphere looks like w = 0 cold dark matter from our vantage point. What works on this large scale also works on the small scale of the single electron (and all the charged lepto-quarks). A neutrino has some mass and is simply a micro-geon of pure zero point energy with positive pressure. there is no question that physicists seriously contemplate a phase transition in the quantum vacuum as a real possibility. The quantum vacuum de'es common sense, because empty space in quantum 'eld theory appear and disappear far too quickly to be detected directly, but their existence has been con'rmed by experiments that demonstrate their in§uence on ordinary matter.A major component of the physical quantum vacuum consists of virtual electrons frothing and bubbling at the Fermi surface edge of the Dirac negative energy sea. This is because of the Pauli exclusion principle that only none or one electron per quantum state. A virtual electron pops out of the vacuums Fermi surface leaving a hole behind. The hole is the virtual positron. The result is a virtual electron-positron pair. However, the virtual electron and the virtual positron attract because they have opposite charges and they are exchanging virtual photons. Therefore, some of them form a more stable bound state. An enormous number of these virtual pairs Bose-Einstein condense into the same center of mass quantum wave packet forming the Vacuum Coherence Field (AKA In§ation Field). This is a dynamic steady state of detailed balance in which there is a continual in§ow and out§ow of virtual pairs into and out of this giant quantum or macro-quantum super§uid. Essentially this is a vacuum phase transition, similar to the BCS transition from a normal metal to an electrical superconductor, from the globally §at micro-quantum electrodynamic vacuum without any gravity at all to the curved macro-quantum electrodynamic vacuum with emergent gravity. Einsteins 'eld equation of general relativity can be derived from the phase wiggles and ripples in the robust stable macroscopically occupied center of mass quantum wave packet of the bound state of the virtual electron-positron pair. The exotic vacuum dark energy and dark matter are simply the amplitude wiggles and ripples of this same virtual pair quantum wave packet. The wave packet spreads over the entire 3D space of the post-in§ationary bubble on which our Hubble-horizoned universe is located along with an in'nity of parallel universes next door as Max Tegmark explains in May correct, take the surface area of the expanding Hubble sphere that is the causal retarded boundary of 3D space of our past light cone at Earth and divide it by the quantum of area. That gives us the number of Bekenstein-Shannon c-bits and explains the arrow of time (AKA Second Law of Thermodynamics) of increasing thermodynamic entropy in terms of the dynamical expansion of the 3D space of the universe. Lenny Susskind calls this DeSitter Space.Such research should be forbidden!Too late. Pandoras Box is open. Schrodingers Cat has jumped out of it.In the early 1970Ís Soviet physicists were concerned that the vacuum of our universe was in fact only one possible state of empty space. The fundamental state of empty space is called the true vacuum. Our universe was considered to reside in a false vacuum, protected from the true vacuum by the wall of our world. A change from one vacuum state to another is known as a phase transition. This is analogous to the transition between frozen and liquid water. Lev Okun, a Russian physicist and historian recalls Andrei Sakharov, the father of the Soviet hydrogen bomb, expressing his concern about research into the phase transitions of the vacuum. If the wall between the vacuum states was to be breached, calculations showed that an unstoppable expanding bubble would continue to grow until it destroyed our entire universe! Sakharov declared Such research should be forbidden! since there was always the possibility that an experiment might accidentally trigger a vacuum phase transition.British Astronomer Royal, Sir Martin Rees, Master of Trinity College, and Director of the Cambridge University Institute of Theoretical Astronomy on Madingley Road where Stephen Hawking works discusses all this in Chapter 9 of his important book Our Final Hour.Could the wall of our universe be breached from within? The amount of energy required to punch a hole through the wall appeared to be enormous, and no known natural physical phenomena, even the most energetic, had punched through either. A recent report commissioned to examine potential dangers at the Large Hadron Collider, one of the next the best of our existing knowledge. Others are not so certain, however. At least one of the Russian physicists I had corresponded with was said to have been a former associate of Andrei Sakharov. He strongly hinted at new theories the Russians had developed which allow for the manipulation of the fundamental constants of nature, but he never revealed more than a sketch of his ideas. He claimed that a breakthrough was within reach, perhaps within 've years Who was that? Not George Ryazanov?Recent theoretical explorations may suggest another approach to the physics of the vacuum. The invisible gravitating dark matter could be the other side of the invisible dark energy coin, and that suggests the possibility of manipulating the vacuum for energy release.Now this is my original idea that you got from our communications over the past few years. I am the only physicist in the world today, as far as I know who has suggested this and has already published it in my two books of 2002 so its in the of'cial record at the Library of Congress.If a controllable parameter could be found to mediate the balance between the invisible dark forces, the result would unleash the vacuum energy of creation in all of its awful power and majesty. If it were possible to control the dark sides of the force then spacetime, the arena where everything we know takes place, could be bent and twisted with in'nitely greater ease than was ever suspected. This would open PandoraÍs box to everything from vacuum energy weapons of mass destruction (capable of destroying the universe!) to spacetime warp drives and time machines.Exactly, the above is the thesis of all my books since 2002 at least.A quick survey of the international electronic archive of physics papers at www.arXiv.org shows that research into the vacuum of spacetime for energy production is alive and well.I do not think that is true. You need to cite speci'cs here. There are lots of §akey new age papers on free energy on the Web written by people without any real credentials but they are not on www.arXiv.org which is not even allowing competent fringe papers in controversial topics like cold fusion. So what exactly are you thinking of here? Indeed, Paul Ginsparg, who controls the archive, does not even allow Carlos Castro to publish conservative competent papers on Clifford Algebras which are not fringe at all!Most authors are independent researchers struggling with limited funding and resources, yet their theoretical results suggest that somewhere in Nick CookÍs black world, a major breakthrough has already taken place. Most likely the United States and Russia are in the lead, but China, France, Ukraine, Iran, India and Saudi Arabia all have scientists actively pursuing the fundamental physics that determine the fabric of our reality, and are seeking the theory and the means to access the enormous energies locked inside of the vacuum since the creation of the universe. Even if the black budget world has yet to unleash the enormous potential of vacuum energy, there are signs that those in power may have begun to take notice. Dr. Harold Puthoff, a scientist with strong government connections, who has previously worked on classi'ed projects for the CIA, is a major proponent of vacuum energy physics. Nick CookÍs book, The Hunt for Zero Point, and his recent stories on zero point energy in JaneÍs Defence Weekly have also brought attention to the dangers and military potential of vacuum research. The American intelligence community 'nanced so-called psychic spies for over twenty years and through four presidential administrations. It is highly unlikely that they would ignore the potential of the quantum vacuum. Dr. George Chapline, of the Lawrence Livermore National Laboratory, and Dr. Jack Sarfatti in San Francisco, knew each other in the sixties in La Jolla, have independently proposed that the quantum vacuum may unstable to the formation of coherent virtual processes. Sarfatti suggests that gravity is an emergent property determined by the physics of the vacuum. His idea is to 'nd a means of directly interacting with the physics of the vacuum that controls the shape of spacetime. Such a possibility would be consistent with the reported success of EvgenyPodkletnov, the Russian scientist who is experimenting with spinning superconducting disks. PodkletnovÍs most recent papers report the appearance of a mysterious coherent beam of gravity like radiation with a measured force of 1000 G. In an interview on BBC radio, Nick Cook pointed out one immediate application of the Podkletnov beam - the destruction of missiles and satellites in §ight or in orbit around the earth. Cook showed the BBC internal documents from Boeing, the American aerospace contractor, proving their interest in PodkletnovÍs research.This beam stuff I am suspicious of. Of course if the experiment is good, I have to think more about it. I am not so sure if Podkletnovs experiment is any good and has been replicated.The connections between PodkletnovÍs results, and the kind of vacuum research explored by Sarfatti, beginning in 1999 at the International Space Sciences Organization are the latest threads in a trail that most likely originates in cold war disinformation, a game played by East and West against each other. Glasnost has shifted the balance of partnerships and the positions of the players, but not the stakes of an outcome that would leave the world with even more proli'c and powerful weapons of mass destruction.That is true, as shown in Destiny Matrix, however you leave out the most important intrigues me in this whole business are the connections. Although Nick Cook never revealed the identity of his deep throat contact called Dr. Dan Marckus in the book The Hunt for Zero Point, there was no question that the Podkletnov results had played a major part in 'tting together the pieces of the puzzle. The amount of interest was in PodkletnovÍs reports by NASA, Boeing, and others in the international arena of aerospace and military research communities was evidence that there was more here to explore than the latest musings of the intellectual elite. The truth is that a fundamental theory of gravity at the scales of subatomic nuclear physics does not exist. The fact is that no one understands the nature of the gravitational 'eld at very small scales. In fact gravity has barely been probed much below one millimeter. Every attempt to unify the physical theories of gravity with the well-known standard model physics of electromagnetism, and the strong and weak nuclear forces, has failed. More importantly there has been recent progress in the exotic areas of mainstream research, such as superstring theory, which suggest new kinds of physics, which might support explanations for PodkletnovÍs impulse gravity effect. One of the current fads in theoretical physics involves large extra dimensions of space that allow a much stronger version of gravity to leak off the membrane world of our ordinary three dimensions. The large dimensional picture allows for the well known forces of electromagnetism, and the strong and weak nuclear forces, to be con'ned to a three dimensional brane-world §oating in a higher dimensional space. Gravitons, the are able to slip off of our brane-world, which explains why the gravitational force is so much weaker than the other forces that hold matter together. Gravitons could be exchanged between our brane-world and another brane §oating nearby in the same higher dimensions.The Sarfatti picture offers a more direct interaction with the new physics than the brane world ideas. SarfattiÍs vision is to 'nd a means of using electromagnetic 'elds in the Josephson effect to couple to the virtual electron-positron pair giant coherent condensate in§ation 'eld inside the vacuum that controls the shape of spacetime to the real electron pair giant coherent condensate of a control high temperature superconductor. UCBs Ray Chiao has a similar idea using a superconductor to transduce electromagnetic far 'eld waves to gravity waves with high ef'ciency conversion. Sarfatti wants to do the same thing with non-propagating electromagnetic and gravity near 'elds. One wonders if the black budget world may have already produced some of the technology needed to explore and test these new realms.Not a chance. They are clueless about the theory. They are still stuck in HalÍs PV model and Bernie HaischÍs zero point ideas, which will never, in my opinion §y. They are not asking the right questions and do not have the right idea in their minds. It is my belief, until I see evidence to the contrary, that I am the only physicist on the planet today who is doing real theoretical work directly relevant to the achievement of practical metric engineering anchored in the now observed reality of dark energy. All my work is public. I would love to be proved wrong on this especially by Hal Puthoff, but I am not holding my breath.;-) Extraordinary claims require extraordinary proof. Everything else I have seen is either on the wrong track asking the wrong questions like the work of Puthoff, Haisch, Ibison & Rueda for example, which at least is real physics that has proved itself wrong in IbisonÍs PV cosmology paper, or else the claims are patently obvious nonsense that Feynman called Cargo Cult Science. There is also the Russian torsion work of Akimov and Shipov and I am not prepared to make a de'nitive statement on that, as the issue is not simple because of several factors some political. One must be careful there to separate ShipovÍs theoretical work from claims made about practical devices including weapons applications. I do, however, agree with you that there is a real issue here as de'ned in Ch. 9 of Martin ReesÍs Our Final Hour.Hal Puthoff coined the term metric engineering for §ying saucer technology. Hal has been working on this problem for many decades and has held high USG security clearances and has been privy to reliable information that the saucers are real and are alien. Otherwise he would not be working on the problem. However, HalÍs theories, both of the zero point energy and of the gravity 'eld will not solve the problem because they are too naively based and do not ask the right questions. The basic physics required for this task is way beyond the depth of Puthoffs self-described engineering approach and can be found in RovelliÍs new book on quantum gravity. === Subject: Re: Understanding taylor expansion for Christensen--Jonathan Christensen taylor exapnsion for the sine> function, sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ... IÍve been told that it uses the fact that e^(i*x) = cos(x) + i*sin(x),> and I can easily work out the taylor series for e^x, but IÍm not sure> how to use these two pieces to 'nd the expansion for sin(x). Any help would be certi'ed Virus Free.Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: e is transcendental approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id sci.math>Why do I even try? Are you even TRYING to learn? Or are you >so closed-minded that you canÍt be bothered?panamars@otenet.gr (Eur Ing Panagiotis Stefanides) (Eur Ing Panagiotis Stefanides) real part of exp(i pi) is cos(pi), and its imaginary part>is sin(pi), so all you are saying is that cos(pi) = -1 and >sin(pi) = 0, and we were already aware of these facts. There is >NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect.>> Panagiotis Stefanides>>Yes, but you DONÍT tell us what your reason is. You canÍt expect us to >accept your claims without giving support for those claims. So what >possible reason could you have for expecting us to agree with your claim >that exp(i pi) = 0?>The reason is simply that exp(ipi0=-1 should be accompanied>>by the statement that this is the real part solution.>>Is it fair?>>No, that is not a fair comment. exp(i pi), the complex number, is>equal to -1, the complex number. There is no need to appeal to the >real part. Also, what does exp(i pi) = -1 is the real part solution>mean? You look like you are using terminology in a manner not >recognized in mathematics.>>David McAnally>>-------------->>e^[i*pi] ,as accepted ,is a phasor. >>In most of the relevant 'elds of mathematics, e^[i pi] is a complex >number.>It is only fair to state that its >>polar representation is :>e^[i*pi] = MOD 1 , ARG 180 .>>That is Arg 180 degrees, not just Arg 180. And so what? That does >not lead to your assertion that exp[i pi] = 0, a result for which >you have given absolutely no support. Why donÍt you just give up?>>David McAnally>> At the moment, they (the Time Lords) are far from being all-powerful.> ThatÍs why itÍs been left up to me and me and me.> quote by: Patrick Troughton in The Three Doctors>>------->I, have made myself very explicit.>My original question of the implication>>of the imaginary component: e^[ipi]=j*0 (to the related proof)>>Complex Notation, chapter 12 ,Electrical Technology 3RD ED.>>Edward Hughes Longmans ,page 338:>>Stares:>> OA*=OB+jOC=OA(COStheta+jSINtheta)>>* Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,..>Here is very clear the difference between PHASOR and MAGNITUDESI know the difference between a complex number and its modulus >(or magnitude). You donÍt have to explain the difference to>me.>[which(MAGNITUDES) I, referred to as REAL PART or IMMAGINAR PART,The magnitude of a complex number is generally not equal to either>its real or imaginary part. How could you claim that it is?I, gave a reference:Complex Notation, chapter 12 ,Electrical Technology 3RD ED.Edward Hughes Longmans ,page 338:States: OA*=OB+jOC=OA(COStheta+jSINtheta)* Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,.. What are these magnitudes then ?>> I should have stated COMPONENTS ].>>Perhaps you mean that (the imaginary part of e^[i pi]) = 0, in which>case you are correct, but you have had a lot of problem expressing >yourself, especially in view of the way that you initially made>the claim by stating that e^[i pi] = 0, which you described as the >imaginary part solution, using a terminology that nobody but you >knows.> e[i*theta]=COStheta+iSINthetaI know that it is true that exp[i theta] = cos(theta)+i sin(theta). >It follows that exp[i pi] = -1. Incidentally, e[i theta] = i e theta, >unlike what you have written.>thetas could be given and calculations could be performed>>for numerical evaluations.And for other results as well.>In books is stated that it is FORMULA>>and also terms such as evaluate:(-1+i*sqrt[3])^10>Are these not solutions to problems?No. Evaluate (-1+i sqrt(3))^10 is a problem for which you can>get a solution using exp(i theta) = cos(theta)+i sin(theta). This>does not mean that exp(i theta) = cos(theta)+i sin(theta) is itself>a solution. You need a problem before you can describe anything as>a solution.Of course ,this crops up when theta is substituted by a given angle.>>Nobody knows what you mean when you make a statement like exp[i pi] = -1>is the real part solution of exp[i pi] = -1, or that exp[i pi] = 0>>is the imaginary part solution of exp[i pi] = -1. I asked you to >explain your terminology but you havenÍt bothered.>>I, doubt it but ,still I, exlpained it anyway.I did ask you. And you did not explain hence there is no solution, whether real part or>imaginary part.How else would do You call them,I, said:I should have stated COMPONENTS>As I, exlained earlier, but I, am not the kind of not >>being botheredThis was not a quote from me. Why are you pretending that it was?You are correct, this was my quote.>>When I take the imaginary part of the equation exp[i pi] = -1, I get>sin(pi) = 0, a fact which is already known to us.>>Have you thought also of the fact that if you have exp[i pi] = -1 (your >real part solution) and exp[i pi] = 0 (your imaginary part solution),>then you could conclude that 0 = exp[i pi] = -1, and get a contradiction?>Well tthere is always the possibility of an answer such as >>MULTIVALUED , examplum gratias:It is known that the exponential function is single-valued (in fact,>it is known that the exponential function is entire).> e^[2i*pi]=1 > ln[1]=0=2i*piThe logarithm has a branch point at 0. The exponential function does>not have a branch point. The analogy is invalid. Because the >exponential function is single-valued, my objection here still stands.>I, make use of the comlex notation , but still ,I, have my >>natural questions ,and do not accept everything for granted.But you shouldnÍt start making up mathematics to suit yourself. This is not my intention,believe me.>I, give an example in the form of question:>It is required that COS[-i]+i*COS[-i] be evaluated>>so that is its Modulus and Argument>>be evaluated (NUMERICALLY].cos(-i) = cosh(1), so cos(-i)+i cos(-i) = sqrt(2) cosh(1) exp(i p/4).The magnitude is sqrt(2) cosh(1), which is approximately 2.18225,>and its argument is pi/4, which is approximately 0.7854. >Alternatively, the argument is exactly 45 degrees.I, thank McAnally-------------- === Subject: Teaching philosophyI need to write a description of my teaching philosophy. In order todo so accurately, I think I need to write at greater length and ingreater detail on this topic than any hiring committee will want to read.Moreover, I need to present my views accurately but, somehow, in sucha manner as not to vitiate the consideration of my application.That being the case, can someone please tell me what my teachingphilosophy is?Ignorantly,Allan Adlerara@zurich.ai.mit.edu************************************ ***************************************** ** Disclaimer: I am a guest and *not* a member of the MIT Arti'cial ** Intelligence Lab. My actions and comments do not re§ect ** in any way on MIT. Moreover, I am nowhere near the Boston ** metropolitan area. ** ************************************************************** *************** === Subject: Re: Teaching philosophy> I need to write a description of my teaching philosophy. In order to> do so accurately, I think I need to write at greater length and in> greater detail on this topic than any hiring committee will want to read.Instead, go for clarity and brevity. === 12:44:57 -0500, Allan Adler I need to write a description of my teaching philosophy. In order to>do so accurately, I think I need to write at greater length and in>greater detail on this topic than any hiring committee will want to read.>Moreover, I need to present my views accurately but, somehow, in such>a manner as not to vitiate the consideration of my application.That being the case, can someone please tell me what my teaching>philosophy is?Ignorantly,>Allan AdlerA more serious answer:I promise to cheerfully accept students who donÍt have theprerequisites and canÍt add fractions without a calculator or solvelinear equations. I will have them work in groups so, hopefully,someone in each group can do the work. I will be their best friend sothey donÍt think of me as a teacher. We will have nice pleasantclasses where we all discover things together so nobody feels left outor has their self-esteem bruised. Before each exam I will give out apractice exam ****very**** similar to the real exam so they will knowwhat to expect. I will be very generous with part credit for thosethat still donÍt get it, even giving re-takes if necessary. I**guarantee** a very high success rate in my classes. There will be somany AÍs and BÍs that I will get awesome student evaluations. Studentswill §ock to my classes because I am such a great teacher. The Deanwill notice the improvement in my departmentÍs success rate and willprobably increase itÍs funding. Over time, as my methods catch on,even the University administration and maybe even the Regents willhold our department up as a shining example of what is being done toimprove higher education.--Lynn === Subject: Re: Teaching do so accurately, I think I need to write at greater length and in>greater detail on this topic than any hiring committee will want to read.>Moreover, I need to present my views accurately but, somehow, in such>a manner as not to vitiate the consideration of my application.That being the case, can someone please tell me what my teaching>philosophy is?Ignorantly,>Allan Adler> A more serious answer:> I promise to cheerfully accept students who donÍt have the> prerequisites and canÍt add fractions without a calculator or solve> linear equations. I will have them work in groups so, hopefully,> someone in each group can do the work. I will be their best friend so> they donÍt think of me as a teacher. We will have nice pleasant> classes where we all discover things together so nobody feels left out> or has their self-esteem bruised. Before each exam I will give out a> practice exam ****very**** similar to the real exam so they will know> what to expect. I will be very generous with part credit for those> that still donÍt get it, even giving re-takes if necessary. I> **guarantee** a very high success rate in my classes. There will be so> many AÍs and BÍs that I will get awesome student evaluations. Students> will §ock to my classes because I am such a great teacher. The Dean> will notice the improvement in my departmentÍs success rate and will> probably increase itÍs funding. Over time, as my methods catch on,> even the University administration and maybe even the Regents will> hold our department up as a shining example of what is being done to> improve higher education.> --LynnOh, sod the itÍs its confusion, this is the funniest and most heartwarming thing IÍve ever read in this group.I had started a reply based on my experiences there, but thought ittoo much like sour grapes. === Subject: Re: Teaching philosophy> I promise to cheerfully accept students who donÍt have the> prerequisites and canÍt add fractions without a calculator or solve> linear equations. I will have them work in groups so, hopefully,> someone in each group can do the work. I will be their best friend so> they donÍt think of me as a teacher. We will have nice pleasant> classes where we all discover things together so nobody feels left out> or has their self-esteem bruised. Before each exam I will give out a> practice exam ****very**** similar to the real exam so they will know> what to expect. I will be very generous with part credit for those> that still donÍt get it, even giving re-takes if necessary. I> **guarantee** a very high success rate in my classes. There will be so> many AÍs and BÍs that I will get awesome student evaluations. Students> will §ock to my classes because I am such a great teacher. The Dean> will notice the improvement in my departmentÍs success rate and will> probably increase itÍs funding. Over time, as my methods catch on,> even the University administration and maybe even the Regents will> hold our department up as a shining example of what is being done to> improve higher education.Hah, pretty good! === Subject: Re: Teaching philosophy> I promise to cheerfully accept students who donÍt have the> prerequisites and canÍt add fractions without a calculator or solve> linear equations. I will have them work in groups so, hopefully,> someone in each group can do the work. I will be their best friend so> they donÍt think of me as a teacher. We will have nice pleasant> classes where we all discover things together so nobody feels left out> or has their self-esteem bruised. Before each exam I will give out a> practice exam ****very**** similar to the real exam so they will know> what to expect. I will be very generous with part credit for those> that still donÍt get it, even giving re-takes if necessary. I> **guarantee** a very high success rate in my classes. There will be so> many AÍs and BÍs that I will get awesome student evaluations. Students> will §ock to my classes because I am such a great teacher. The Dean> will notice the improvement in my departmentÍs success rate and will> probably increase itÍs funding. Over time, as my methods catch on,> even the University administration and maybe even the Regents will> hold our department up as a shining example of what is being done to> improve higher education.> Hah, pretty good!Yes, thereby showing he stands with the students & not the academic snobs.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === 09:13:21 +1100, Gerry Myerson>Yes, especially the part where with the students & not the academic snobs.Yeah, I 'gured someone would catch that. Is & a word? :-)--Lynn === Subject: Re: Teaching philosophy>Yeah, I 'gured someone would catch that. Is & a word? &ante Melancholico To &rea (names have been changed to protect the guilty) & after all is said & done My life has been a lonely one, & when I sometimes think of you I underst& your point of view. & often whilst I contemplate, The at www.herring.pwp.blueyonder.co.uk/poetry/andante.htm ] === -0500, Allan Adler I need to write a description of my teaching philosophy. In order to>do so accurately, I think I need to write at greater length and in>greater detail on this topic than any hiring committee will want to read.>Moreover, I need to present my views accurately but, somehow, in such>a manner as not to vitiate the consideration of my application.That being the case, can someone please tell me what my teaching>philosophy is?Ignorantly,>Allan Adler>ara@zurich.ai.mit.edu>I would be happy to do so, but I only know how to take multiple choicetests. What are the choices?:-)--Lynn === Subject: Re: Teaching philosophy> I need to write a description of my teaching philosophy. In order to> do so accurately, I think I need to write at greater length and in> greater detail on this topic than any hiring committee will want to read.> Moreover, I need to present my views accurately but, somehow, in such> a manner as not to vitiate the consideration of my application. That being the case, can someone please tell me what my teaching> philosophy is?What kind of philosophy? Metaphysics? Epistemology? Ethics? Aesthetics? Kantian? Pragmatic? Platonic? Aristotelean?Bob Kolker === Subject: Re: question support1.mathforum.org (8.11.6/8.11.6/The Math Forum, sci.math>If you have two periodic functions f(t) and g(t) with periods a and b,>then the function f(t) + g(t) is periodic with period c where c is the>least-common-multiple of a and b. Here the 'rst period is 1/.018=500/9find>the second period is 1/.02=50. So the least-common-multiple is 500, i.e.>1/.002. That should be the period of the combined function.How to prove the theorem? Any hint is welcome. TIA. === Subject: Re: question about periodic function>>If you have two periodic functions f(t) and g(t) with periods a and b,>>then the function f(t) + g(t) is periodic with period c where c is the>>least-common-multiple of a and b. Here the 'rst period is 1/.018=500/9find>>the second period is 1/.02=50. So the least-common-multiple is 500, i.e.>>1/.002. That should be the period of the combined function.>How to prove the theorem? Any hint is welcome. TIA.DonÍt read too much into this. ItÍs a period, not the period, thatis lcm(a,b): there might be a smaller period. All thatÍs involved hereis the rather obvious fact that any common multiple of a and b is a period of both f(t) and g(t), and thus of f(t)+g(t). For example, f(t) = cos(t/2) + cos(t/3) has period 12 pi andg(t) = cos(t/5) - cos(t/2) has period 20 pi. So the lcm of 12 pi and20 pi, namely 60 pi, is a period of f(t) + g(t). But f(t) + g(t)= cos(t/5) + cos(t/3) also has period 30 pi.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: question about periodic function>If you have two periodic functions f(t) and g(t) with periods a and b,>then the function f(t) + g(t) is periodic with period c where c is the>least-common-multiple of a and b. Here the 'rst period is 1/.018=500/9find>the second period is 1/.02=50. So the least-common-multiple is 500, i.e.>1/.002. That should be the period of the combined function.> How to prove the theorem? Any hint is welcome. TIA.> The sum will only be periodic if the ratio of periods is rational.If that ratio is irrational f(t) + g(t) is not periodic. === Subject: Re: question about periodic function>>If you have two periodic functions f(t) and g(t) with periods a and b,>>then the function f(t) + g(t) is periodic with period c where c is the>>least-common-multiple of a and b. Here the 'rst period is 1/.018=500/9find>>the second period is 1/.02=50. So the least-common-multiple is 500, i.e.>>1/.002. That should be the period of the combined function.How to prove the theorem? Any hint is welcome. TIA.If f(t) has period a then f(t + na) = f(t) for any integer n, right?Same idea for g: g(t + mb) = g(t) for any integer m.Hmmmm....how could I 'nd an integer that works for both.....?--Lynn === Subject: Re: CAN The ïcaps lockÍ key is on the left-hand side of the keyboard, above> ïshiftÍ and below ïtabÍ. It should look like this:> +--------+> | |<--- | <-- tab key> | --->| |> +--------+-+> | Caps | <-- caps lock key> | Lock |> +--------+-+> | / |> | || | <-- shift key> +--------+ Over on the right-hand side of the keyboard, above the numeric keypad,> should be three LEDs, labelled (from left to right) ïNum LockÍ, ïCaps> LockÍ and ïScroll LockÍ Press your ïcaps lockÍ key until the LED marked ïCaps LockÍ is no longer> lit. Now you are ready to post on usenet. 1.DETERMINE DOMAIN OF THE FUNCTION f(X)=X3/X2+2X+1> What happens when x^2 + 2x + 1 = 0?Excellent! I (along with Baghdad Bob and most readers) assumed he meant(x^3/x^2) + 2x + 1 (following the standard rules for order of operations),but you ferreted out the probable real meaning, which for OPÍs clari'cationshould have been written x^3/(x^2 + 2x + 1). FIND EXTREME FUNCTIONS f AND DRAW ITS GRAPHextreme points? extreme points of the function? 2.BY NEWTONS METHOD SOLVE EQUATION: X3-X-2=0Your book or your notes have NewtonÍs method (or Newton-Raphson iteration).Google also gets about 37,000 hits (only 4,780 for Newton-Raphson), but someof these are advanced treatments. You can still get good info, but you haveto look for elementary treatments.Jon Miller === Subject: Re: the anticlassicalist }{ iv: from maps to logic> -=-=-=-=-= from maps to logic =-=-=-=-=-=-> Now let us look at interpretting this arrow symbol, such as in process,> categorial inclusion, rule re'nement, or progressions of state. We want> to be able to evaluate logical relationships between objects. So we look to> de'ne on a recognition map new objects de'ned by any two objects which> allows us to collectively compare two objects logically. The 'rst one of> these, which can be called the least upper bound or disjunction of two> states ïaÍ and ïbÍ, written (a / b) which obeys the rules that> a -> (a / b)> b -> (a / b)> forall x in the lattice with a -> x and b -> x, we have (a / b) -> x> where it is important to differentiate when these are derivationally found> on a lattice (objects already de'ned through previous axioms with the> appropriate universal property derivable) and when the existence of new> objects is being de'ned (for each new de'nition requires new consistency /> completeness / etc. checks).You mean it provides a natural framework for conjectures ?> dualityDoesnÍt tell much about that. Duality is negation in boolean algebras, isnÍt it ? === Subject: Re: How many different resistances with n resistors?Originator: tchow@lagrange.mit.edu.mit.edu Corner in MITÍs magazine Technology Review raised the>> question of the number of distinct resistances achievable by connecting>> ten unit resistors.Tony Bartoletti (seq. A048211) considers only cases that can be reduced by>applying the formula for parallel/series. If you have 6 resistors there is 1>con'guration where this is not possible. I suppose we get an additional>resistance value.WhatÍs the 6-resistor con'guration you have in mind?And has anyone calculated the answer to the problem if we donÍt assumeseries/parallel circuits?-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: How many different E2viu0ZC8eLZuWnmdUrIGOhhK5Qw9k9v9kIYQ8-4TI9+4-xlZWn+kbtchow@ Puzzle Corner in MITÍs magazine Technology Review raised the>> question of the number of distinct resistances achievable by connecting>> ten unit resistors.Tony Bartoletti (seq. A048211) considers only cases that can be reduced by>applying the formula for parallel/series. If you have 6 resistors there is 1>con'guration where this is not possible. I suppose we get an additional>resistance value.> WhatÍs the 6-resistor con'guration you have in mind?An H with asymmetric legs. I thought that 6 is the smallest number with thisproperty, but obviously a H with 5 resistors gives us r=1 which seems impossiblewith other combinations of 5 resistors. I did not yet check, whether the 6-H hasa resistance not occuring among the (other) 53 possibilities. I will play withthis a little tomorrow. I think we would 'nd a reference in encyclopedia ifsomeone had solved the general case.Klaus> And has anyone calculated the answer to the problem if we donÍt assume> series/parallel circuits?> --> Tim Chow tchow-at-alum-dot-mit-dot-edu> The range of our projectiles---even ... the artillery---however great, will> never exceed four of those miles of which as many thousand separate us from> the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: errors in support1.mathforum.org (8.11.6/8.11.6/The Math Forum, critical of his wishful thinking that this>proof actually means something?>His argument is an overly sophisticated version of one of the most basic arguments against evolution, one that I canÍt rule out. How does super evolved, complex, 'ne tuned to the millimeter biological structures appear when their evolutionary bene't seems to be absolutely nothing even with 90% or 95% of them installed. People used to point out that eyes are one of these mechanisms from the very beginning,but that, I think, is a bad example because any ability to sense this way, even on a veryprimitive level, must be a huge advantage.My favorite is birds : How can one evolve into §ying ? So many things have to happen inorder to create functional wings it seems impossible any awkward animal developing inthis direction would survive itÍs cost, or any failing §ying attempts. The evolutionistarguments sound totally absurd : they ran fast or jumped from tree to tree and got anadvantage slowly developing wings. Sounds like absolute voodoo to me. === Subject: Re: errors in an argument> My favorite is birds : How can one evolve into §ying ? So many things> have to happen in order to create functional wings it seems impossible any> awkward animal developing in this direction would survive itÍs cost, or> any failing §ying attempts. The evolutionist arguments sound totally> absurd : they ran fast or jumped from tree to tree and got an advantage> slowly developing wings. Sounds like absolute voodoo to me.That depends on the size of the animal. If the animal is small enough,the terminal velocity can result in a non-lethal impact with theground. (I believe the threshold is around the size of a mouse or acat.) Incremental levels of control over the landing point then have anobvious bene't. -- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: the anticlassicalist }{ v: universal truths> And on these lattices we can make more explicit our theories of negation.> On sci.logic, mitch recently gave a very interesting post on Stone algebras.> When we want to model uses of various de'nitions, we need to show them as> an axiom set. Stone algebras study the axiom set:> ~(a) / ~(~(a)) <--> TBtw, can complex numbers model it ? LetÍs see~ : r exp(i s) --> r/2 exp(i s/2)a = 1~(a) = -1/2~(~(a)) = i/4letÍs say T is 0...Well, then I am sure a Lorentz-Moebius transform can save the situation, I mean, provide a natural interpretation to the same formula graphics. The Moebius group is strictly 3-transitive, and z / w <---> ucan pass as denoting -the- moebius transform that swaps z and w while keeping u invariant. Now idea if the thing can mimick a lattice any further, though.IsnÍt the symmetric group attached to the boolean algebra in such a manner that extremely transitive (but non-symmetric) groups shine as interesting analogues ? === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational I will use this to show that > if cos x = (1 + sqrt(5))/2 then x is transcendental. (Hence, it is > irrational.)> But that wasnÍt the original question. The original question > was whether x / pi is irrational. A whole nother kettle of 'sh.> To make this question actually look interesting, note that if > cos(x)=( (1+sqrt(5)) / 4 ), then x/pi is equal to 1/5.> (Also, the original question has sqrt(5)-1 instead of sqrt(5)+1 .> Again, cos(x) = ( (sqrt(5)-1)/4 ) implies x/pi = 2/5 . The question is > to show that this 2/5 value turns transcendental when that ï4Í on the > denominator changes to a ï2.')> J> So sorry for the bleeps in my non-solution.> Yes, the original question did involve (sqrt(5)-1)/(2 *p)i not> sqrt(5)+1 and LindemannÍs lemma and> my proof are only true if x ne 0. Of course e^0 = 1.> I will go back to the drawing board and see if my original proof can> be 'xed.> Sorry for the whole nother kettle of solution(hopefully).I donÍt have time to post all the details right now, but here is abrief outline of the steps:1). Use cos(2*pi/5)= (-1 + sqrt(5))/4 to 'nd all the roots of unityin Q(exp(2*pi*i/5)).(There are 10 of them.)2). Now consider cos(pi*r)= (-1 + sqrt(5))/2.Then, by the usual formulas of trigonometry, we getcos(2*pi*r)= 2-sqrt(5)andsin(2*pi*r)= 2*sqrt( -2+sqrt(5)).If r were rational, cos(2*pi*r) + i*sin(2*pi*r) (*)would be a root of unity in Q(exp(2*pi*i/5)).3). Since (*) doesnÍt match any of the answers of step 1, r cannot berational.Hence, r is Steiner === Subject: A newbieÍs question -- about real numberTwo real numbers r1 = 0.89, r2 = 0.889999999..... (with in'nite === Subject: Re: A newbieÍs question -- about real number> Two real numbers r1 = 0.89, r2 = 0.889999999..... (with in'nite 9s), > can we say r1 = r2 ? = .9999...thus illustrating the mathematicianÍs favorite pastime of reducing this problem to another problem weÍve seen before . . . === Subject: Re: A newbieÍs question -- easier numbers r1 = 0.89, r2 = 0.889999999..... (with in'nite 9s),> show the equality is as follows:r1 = .89 = 8/10 + 9/100r2 = .889999... = 8/10 + 8/100 + 9/900r1 - r2 = 9/100 - 8/100 - 9/900 = 9/100 - 81/900 = 0.0Skip === Subject: Re: A newbieÍs question -- about real number> Two real numbers r1 = 0.89, r2 = 0.889999999..... (with in'nite 9s), > can we say r1 = r2 ?Yes. The problem is that we present in'nite decimal expansions in school before their meaning can be properly understood. Once this meaning is made clear, proving r1 = r2 is a very simple exercise. === Subject: Re: A newbieÍs question -- about real number Adjunct Assistant Professor at the University of Montana.>Two real numbers r1 = 0.89, r2 = 0.889999999..... (with in'nite 9s), >can we say r1 = r2 ? number. The'rst is the real number 8/10 + 9/100.The second is taken to represent the limit of the in'nity series8/10 + 8/100 + (9/10^3 + 9/10^4 + 9/10^5 + ....)Since Sum_{i=3}^{in'nity} 9/10^i = 9/10^3*(Sum_{i=0}^{in'nity} 1/10^i),and the series (1/1 + 1/10 + 1/100 + 1/1000 + ...) converges to1/(1-(1/10)) = 1/(9/10) = 10/9, we have that the original in'niteseries8/10 + 8/100 + (9/10^3 + 9/10^4 + 9/10^5 + ....)converges to8/10 + 8/100 + 9/10^3*(10/9) = 8/10 + 8/100 + 1/100 = 8/10 + 9/100,which is equal to r1.-- =ItÍs not denial. IÍm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: A newbieÍs question -- about real numberlisong@iastate.edu asks this about real numbers:>Two real numbers r1 = 0.89, r2 = 0.889999999..... (with in'nite 9s), >can we say r1 = r2 ? calculations in which only two signi'cant'gures are reasonable, then r1=r2. The context must be appropriate. G C === Subject: Re: A newbieÍs question -- about real numberX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>lisong@iastate.edu asks this about real numbers:>Two real numbers r1 = 0.89, r2 = 0.889999999..... (with in'nite 9s), >>can we say r1 = r2 ? life measurement or result calculations in which only two signi'cant>'gures are reasonable, then r1=r2. What does that have to do with _real numbers_?(Heh, I just realized that you might be fooled by the wordreal. In fact the real numbers are not real in a usualsense of the word, theyÍre mathematical abstractions.)>The context must be appropriate. G C************************David C. Ullrich === Subject: Re: A newbieÍs question -- about real number> Theoretically NO.> In real life measurement or result calculations in which only two signi'cant> 'gures are reasonable, then r1=r2. The context must be appropriate.Theoretically YES, and given that we discuss mathematics here, that is the only meaningful answer. === Subject: Re: A newbieÍs question -- about real numberlisong@iastate.edu asks this about real numbers:>Two real numbers r1 = 0.89, r2 = 0.889999999..... (with in'nite the rationals are uncountable! Rich === Subject: Re: A newbieÍs question -- about real number> Hence the rationals are uncountable! Nice try, but no cigar. The set of decimal sequences that terminate in all 0Ís or all 9Ís is countable. === Subject: Re: A newbieÍs question -- about real number> Hence the rationals are uncountable! > Nice try, but no cigar. The set of decimal sequences that terminate in all > 0Ís or all 9Ís is countable.I think you missed the point, which I took to be that a false staement implies any statement. === Subject: Re: A newbieÍs question -- about real number>> Hence the rationals are uncountable! Nice try, but no cigar. The set of decimal sequences that terminate in all >0Ís or all 9Ís is countable.>No cigar? Bummer. But I *can* 'nd an uncountable number of non-terminatingrepresentations of 89/100 and 'gured that since r1<>r2 the conclusionfollowed. If that is not true then clearly 89/100 is irrational! Even === Subject: Re: A newbieÍs question -- about real number>> Hence the rationals are uncountable! Nice try, but no cigar. The set of decimal sequences that terminate in all >0Ís or all 9Ís is countable. > No cigar? Bummer. But I *can* 'nd an uncountable number of non-terminating> representations of 89/100 Not in any standard form of basal notation.> and 'gured that since r1<>r2 the conclusion followed.If r1 = 0.89 and r2 = 0.88999... = 0.88 + .001*sum[i=0..oo, 9/10^-i],then r1 = r2 If that is not true then clearly 89/100 is irrational! Or you Are. === Subject: Re: A newbieÍs question -- about real number> But I *can* 'nd an uncountable number of non-terminating> representations of 89/100You canÍt count to 1?-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: A newbieÍs question 'nd an uncountable number of non-terminating>> representations of 89/100You canÍt count to 1?Of course not! Given r1<>r2 itÍs clear that 1, like 89/100, is irrational(provided the rationals are not uncountable).Rich === Subject: De'nition of Separable Space (basic topology question)Let A be a subset of (X,T). Then A is dense in X iff for every non-emptyopen subset U of X, A / U != {}.I have seen two de'nitions of ïseparable topological spaceÍ:a) (X,T) is separable if there exists A X, where A is countable anddense in Xb) (X,T) is separable if there exists A X, where A iscountable and dense in XGiven def (a), its simple to prove that all countable spaces X areautomatically separable since the subset X is countable and non-triviallyintersects every non-empty open subset. However, this wonÍt work given def(b). I 'rst became suspicious when I saw a proof that all countable spaceswere separable that seemed ridiculously complex in comparison to theseemingly obvious 1 step proof above. I later found de'nitions ofseparable like def (b).l8r, Mike N. Christoff === Subject: Re: De'nition of Separable Space (basic <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_> Let A be a subset of (X,T). Then A is dense in X iff for every non-empty> open subset U of X, A / U != {}.> I have seen two de'nitions of ïseparable topological spaceÍ:> a) (X,T) is separable if there exists A X, where A is countable and> dense in X> b) (X,T) is separable if there exists A X, where A is> countable and dense in X> Given def (a), its simple to prove that all countable spaces X are> automatically separable since the subset X is countable and non-trivially> intersects every non-empty open subset. However, this wonÍt work given def> (b). I 'rst became suspicious when I saw a proof that all countable spaces> were separable that seemed ridiculously complex in comparison to the> seemingly obvious 1 step proof above. I later found de'nitions of> separable like def (b).> l8r, Mike N. ChristoffI have never seen de'nition (b). I have seen the symbol $subset$used to mean subset, not proper subset, so maybe that is what you saw.Certainly a one-point space with the discrete topology is to beconsidered separable, even though it has no dense proper subsets.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: CauchyÍs Interlacing TheoremHi folks,IÍm looking for a proof of the following...Let A = [a0 a^h ; a A2] >= 0 be a NxN hermitian matrixwhere a0 is a scalar > 0 and A2 is the lower-right N-1xN-1(principal) submatrix of A.According to CauchyÍs interlacing theorem (1829 I guess),the N Eigenvalues gamma_i of A and the N-1 Eigenvalues alpha_iof A2 interlace, i.e., gamma_1 >= alpha_1 >= gamma_2 >= ...Interestingly, having played a little bit with MATLAB,also the Eigenvalues gamma_i of A and the Eigenvalues beta_iof the N-1xN-1 matrix A2-a*a^h/a0 seem to interlace. However,this seems to be true only for nonnegative-de'nite A.Does anyone have a proof, or reference to a proof for that??Initially, I wanted to show that if A >= 0 with rank A >= 2,then also A2-a*a^h/a0 >= 0; but, then I discovered CauchyÍstheorem... ;)(If rank A = 1, then A2-a*a^h/a0 yields the N-1xN-1 zero matrix,and if rank A >= 2, then rank A2-a*a^h/a0 >= 1, at least thisis easily seen).Once the above assumption had been proven, A2-a*a^h/a0 >= 0would directly follow as beta_N-1 >= gamma_N >= 0(comparing the smallest === Subject: Graph Theory participating in anREU for discrete math and combinatorics. I am looking for a goodgraph theory textbook to learn the basics from. I have seen the booksthat Dover (which of course are very cheap) has to offer on thesubject, but unfortunately they seem too elementary. Is the SpringerGTM Graph Theory by Reinhard Diestel any good? Any suggestions arehighly === Subject: Re: Graph Theory IÍm an undergraduate, and this summer I will be participating in an> REU for discrete math and combinatorics. I am looking for a good> graph theory textbook to learn the basics from. I have seen the books> that Dover (which of course are very cheap) has to offer on the> subject, but unfortunately they seem too elementary. Is the Springer> GTM Graph Theory by Reinhard Diestel any good? Any suggestions are> highly Introduction to Graph Theory by Robin Wilson. ItÍs runs about $40-50 used, so itÍs not the cheapest book, especially for paperback. However, I thought that the exercises were very appropriate. They were some that required modern algebra and a few that required beginning combinatorics. Some questiosn were simple verify problmes and there were a few proofs, but nothing way over the top. Answers to selected exercises were in the back, including a few of the proof answers. The proofs that were covered in the text were, for the most part, standard proofs for the topics. The book also did a good job de'ning and explaining terms. It also has a chapter covering matroids. For undergrads, you can probably skip that chapter and not lose anything. However, if you do cover that chapter, I suggest looking at another book for a more complete coverage.Oh, the class I took was an upper-level undergrad, lower-level graduate course. - Tim-- Timothy M. BrauchGraduate === StudentDepartment of MathematicsWake Forest UniversitySubject: Re: strange identi'cation spaces argh! I meant one unit in *height*. sorry for the multiple posts.So to reiterate: the rectangle is one unit in height, and four units in length. === Subject: Anyone knows a formula for this? (combinatorics)HiLet n be the number of cards drawn from an ordinary deck (52 cards).For example according to the book, if n=5i.e if we pick 5 cards randomly from a deck of 52 cardsThe probability of getting a) A full house is:[C(4,2)*C(4,3)*2*C(13,2)]/C(52,5)b) Exactly two pairs is:[C(4,2)*C(4,2)*C(44,1)*C(13,2)]/C(52,5)So i was wondering if thereÍs a general result (a formula)? for a) and b)where n === Subject: Re: Anyone knows a formula for this? (combinatorics) Adjunct Assistant Professor at the University of Montana.>Hi>Let n be the number of cards drawn from an ordinary deck (52 cards).>For example according to the book, if n=5>i.e if we pick 5 cards randomly from a deck of 52 cards>The probability of getting a) A full house is:>[C(4,2)*C(4,3)*2*C(13,2)]/C(52,5)b) Exactly two pairs is:>[C(4,2)*C(4,2)*C(44,1)*C(13,2)]/C(52,5)So i was wondering if thereÍs a general result (a formula)? for a) and b)>where n is the number of cards drawn.You should try to understand what the numbers mean.For example, with full house, you have C(52,5): thatÍs the number ofdifferent ways in which you can randomly pick 5 cards from a deck of52 cards. It is the total space. If you are randomly pickin k cardsfrom a deck of 52, then that number should obviously be replaced byC(52,k).What do the other numbers mean? I.e., C(4,2)*C(4,3)*2*C(13,2)should be the number of different 5 card hands which are a fullhouse. How is this accomplished? Why does this count that?Well, a full house is determined by stating the following things: (1) The rank of the 3-of-a-kind (i.e., is it 3 aces, 3 jacks, 3 tens, etc); and (2) The rank of the pair; and (3) The suits involved in the 3-of-a-kind; and (4) The suits involved in the pair.That completely determines the hand, since there are no cards leftoverin a 5-card hand.Now, C(4,2) means how many ways to choose 2 cards out of 4; thatrepresents (4), how to pick which suits the pair will involve.C(4,3) is how you pick the suits for the 3-of-a-kind.And C(13,2) picks the two ranks that will be involved; you multiply by2 because it matters which goes 'rst (i.e., is it tens over fours, orfours over tens)? So, say you are drawing k cards instead, k>5, and you want to know howmany ways in which you can have exactly a full house. Obviously, somevalues of k will not allow it, so it becomes more complicated (if youdraw 50 cards, say, then you will surely have a better hand than afull house to pick out of there). But you determine the full house thesame way. What remains is to ensure that the left-over cards in yourhand do not add more to your hand. So, for example, you would need tochoose (k-5) more cards, making sure that none of them are in the samerank as the pair or as the 3-of-a-kind; you would also have to ensurethat they are not pairs among themselves, say, that you do not drawenough to get a 5-card §ush, etc.Some easy ones, though: say we have k=7; thereÍs no danger of a §ushor a straight, so all you have to do is make sure you do not draw fromthe same rank as the pair or the 3-of-a-kind, and that the two extracards are not a pair. So you need to pick 2 cards of different ranks;there are 11 ranks left, so you multiply by C(11,2). And you need toselect their suits. Pick one suit for the higher card, C(4,1), andanother suit for the other card, C(4,1). So we get that, drawing 7cards, the number of hands that have exactly a full house and nobetter (and do not have a 3-of-a-kind and two pairs) are:C(4,2)*C(4,3)*2*C(13,2)*C(11,2)*C(4,1)*C(4,1)out of a total of C(52,7) possible hands.So the probability would beC(4,2)*C(4,3)*2*C(13,2)*C(11,2)*C(4,1)*C(4,1)/ C(52,7).Higher number of cards complicates matters, since you would have todiscard hands that get a straight, a §ush, etc.-- ItÍs not denial. IÍm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu === Subject: RFI:WanlessÍ Fourth Conjecture/DirichletÍs Geometric TheoremThe series a^n+b [hcf(a,b)=1] includes an in'nite number of primes [overall n] === Subject: Re: RFI:WanlessÍ Fourth Conjecture/DirichletÍs Geometric Theorem_If_ itÍs true - can one prove it from DirichletÍs [Arithmetic ie message> The series a^n+b [hcf(a,b)=1] includes an in'nite number of primes [over> all n] === Subject: Re: RFI:WanlessÍ Fourth Conjecture/DirichletÍs Geometric Theorem> _If_ itÍs true - can one prove it from DirichletÍs [Arithmetic ie standard]> Theorem?A) Regardless of its truth or falsity, I am sure that you can prove it from DirichletÍs Theorem.B) Consider a=234578273658273645827341 b=2348798769847962341> J> James a^n+b [hcf(a,b)=1] includes an in'nite number of primes [over>>all n]> === Subject: Re: RFI:WanlessÍ Fourth Conjecture/DirichletÍs Geometric TheoremImmediate proviso:a>1,but apart from that...?James Wanless [hcf(a,b)=1] includes an in'nite number of primes [over> all === n]Subject: Re: :: towards a constructive education :: (news server friendly) And stop grinning, until you put up a decent argument. ItÍs just sad.What is it with imbeciles like you? decent argument ???Look Daddy, I got on the debating === team!!!ad nauseum:-)mitchSubject: Re: Combinatoric:4 groups of 4 Look up references on 'nite geometries or combinatorial designs, andaf'ne planes in particular.> This is a planning problem for scheduling players in teams of 4.> Not sure if it is feasible.> 16 players, 5 days of play, 4 teams each day.> Each player teamed with each other player only once over the 5 days....> Any suggestions or pointers to places of study are much appreciated. === Subject: in'nitely many palindromic Theory of Numbers_(Dover, 1964) one 'nds the raw assertion that there are in'nitelymany palindromic primes, primes whose base 10 representation reads thesame forwards and backwards.The result is certainly believable, e.g., as a standard heuristicpredicts in'nitely many primes consisting solely of 1Ís, but I wassurprised to see it claimed as a theorem. Is this really the case? Anold (1984) sci.math post from Bob Silverman suggests that it is andthat the proof is connected with results on automorphs of binaryquadratic forms. (The post in question discusses classes ofautomorphic binary quadratic forms, but the references to Dicksonseem to be to theorems about automorphs of binary forms and theirconnection with the Pell equation.) Can anyone elucidate for me much,Paul === Subject: Metric Enginer The Big Rip?This is expanded versionNew 269 pageSuper Cosmosnow athttp://qedcorp.com/destiny/SuperCosmos.pdfThe Hunt for the Zero Point Farcehttp://www.salon.com/books/review/2002/08/05/zero_gravity /May The Farce be not with you.FYI draft in progress with more re: ISSO/Sarfatti etc.DARK MATTERS SURROUND DARK ENERGYTwo big stories from the world of physics may portend the arrival of new weapons of mass destruction far more powerful and compact than atomic bombs. In recent years it has been discovered that our universe is being blown apart by a mysterious anti-gravity effect called dark energy. Mainstream physicists are scrambling to explain this mysterious acceleration in the expansion of the universe. Some physicists even believe that the expansion will lead to The Big Rip when all of the matter in the universe is torn asunder - from clusters of galaxies in appears to be made of two unknowns - roughly 23% is dark matter, an invisible source of gravity, and roughly 73% is dark energy, an invisible anti-gravity force. Ordinary matter constitutes perhaps 4 percent of the universe. Recently the British science news journal New Scientist revealed that the American military is pursuing new types of exotic bombs - including a new class of isomeric gamma ray weapons.That was an original idea of mine in 1963 at Cornell and I discussed it with Hans Bethe. That is one of the reasons Ron Bullough invited me to Harwell in 1966. No doubt others thought of it but probably later. I thought of it while at Tech/Ops in Lexington, Mass working for George Parrant Jr.Unlike conventional atomic and hydrogen bombs, the new weapons would trigger the release energy by absorbing radiation, and respond by re-emitting a far more powerful radiation. In this new category of gamma-ray weapons, a nuclear isomer absorbs x-rays and re-emits higher frequency gamma rays. The emitted gamma radiation has been reported to release 60 times the energy of the x-rays that trigger the effect.Gamma-ray weapons could trigger next arms race 19:00 13 August 03Exclusive from New Scientist Print Edition. Subscribe and get 4 free issues.An exotic kind of nuclear explosive being developed by the US Department of Defense could blur the critical distinction between conventional and nuclear weapons. The work has also raised fears that weapons based on this technology could trigger the next arms race. The explosive works by stimulating the release of energy from the nuclei of certain elements but does not involve nuclear 'ssion or fusion. The energy, emitted as gamma radiation, is thousands of times greater than that from conventional chemical explosives. The technology has already been included in the Department of DefenseÍs Militarily Critical Technologies List, which says: Such extraordinary energy density has the potential to revolutionise all aspects of warfare. Scientists have known for many years that the nuclei of some elements, such as hafnium, can exist in a high-energy state, or nuclear isomer, that slowly decays to a low-energy state by emitting gamma rays. For example, hafnium-178m2, the excited, isomeric form of hafnium-178, has a half-life of 31 years. The possibility that this process could be explosive was discovered when Carl Collins and colleagues at the University of Texas at Dallas demonstrated that they could arti'cially trigger the decay of the hafnium isomer by bombarding it with low-energy X-rays (New Scientist print edition, 3 July 1999). The experiment released 60 times as much energy as was put in, and in theory a much greater energy release could be achieved.http://www.newscientist.com/news/news.jsp?id= ns99994049I was thinking in 1963 of a gamma ray laser pumping a nuclear isomeric transition. Bethe at the time said it wouldnÍt work and basically discouraged me working on it.Bekkum continued:In the summer of 2000 I contacted Nick Cook, the former aviation editor and aerospace consultant to JaneÍs Defence Weekly, the international military affairs journal. Cook had been investigating black budget super-secret research into exotic physics for advanced propulsion technologies.Uh Oh :)I had been monitoring electronic discussions between various American and Russian scientists theorizing about rectifying the quantum vacuum for advanced space drive. Several groups of scientists, partitioned into various research organizations, were exploring what NASA calls Breakthrough Propulsion Physics - exotic technologies for advanced space travel to traverse the vast distances between stars. Partly inspired by the pulp science 'ction stories of their youth, and partly by recent reports of multiple radar tracking tapes of unidenti'ed objects performing impossible maneuvers in the sky, these scientists were on a quest to uncover the most likely new physics for star travel. The NASA program was run by Marc Millis, under the Advanced Space Transportation Program Of'ce (ASTP). Joe Firmage, a Silicon Valley entrepreneur, who at age 28 had found risen to CEO of a three billion dollar internet 'rm, began to fund research in parallel with NASA. He hired a NASA Ames nanotechnology scientist, Creon Levit, to run the International Space Sciences Organization,Joe did that because I suggested it. I introduced Creon to Joe.Cook was intrigued by the apparent connections between various private investors, defense contractors, NASA, INSCOM (American military intelligence), and the CIA. While researching exotic propulsion technologies Cook had heard rumors of a new kind of weapon, a sub-quantum atomic bomb, being whispered about in the dark halls of defense research.I think that must have come from me regarding J. P. VigierÍs tight atomic states with experiments in Beograd, Serbia by Z. Maric (same name as Einsteins wife, Mileva Maric) and G. Dragic. But how did Cook hear about that? We brought Vigier to ISSO in San Francisco several times along with physicist Gennady Shipov from Moscow. That story with photographs of Vigier and the group is in my autobiography Destiny Matrix. Dragic A, Maric Z, Vigier JP; Phys. Lett. A 265 (2000) 163. New quantum mechanical tight bound states and ïcold fusionÍ. Creon Levit and Vigier met with Maric in Budapest, Hungary in 2000.Bekkum who is one of my ISEP on line virtual students continued:Sub-quantum physics is a controversial re-interpretation of quantum theory, based on so-called pilot wave theories, where an information that the predictions of ordinary quantum mechanics could be recast into a pilot wave information theory. Recently Anthony Valentini of the Perimeter Institute has suggested that ordinary quantum theory may be a special case of pilot wave theories, leaving open the possibility of new and exotic post-quantum technologies. Even thought rumors of a sub-quantum bomb may be purely fantasy ItÍs not fantasy. It might not work, Maric and Dragic in Beograd, while not ostensibly trying to make a weapon by any means, were trying to test VigierÍs basic theory of the spatially extended electron, which I think is basically a correct idea and 'ts my own ideas including why the electron appears to shrink to less than 10-16 cm under high resolution imaging (i.e. scattering) and how the electric charge distribution is contained by the strongly attractive short-range zero point stress-energy density w = -1 exotic vacuum tiny dark matter core only roughly one Fermi (10-13 cm) across. This solves the Abraham-Becker-Lorentz-Poincare compensating stress problem of 100 years ago. My new explanation of this long-standing mystery that stumped even Richard Feynman, only works in the Bohm hidden or extra variable interpretation. That is, a classical spatially extended electric charge distribution is unstable. It explodes under its own self-repulsion. This is why physicists had to postulate a point electron because they did not understand that the strong short-range gravity attraction of the positive zero point pressure in the dark matter phase of exotic vacuum would hold the charge together. As Herbert Frohlich told me at UCSD in La Jolla in 1966 the basic thing wrong with physics is the idea of the point electron. The bad idea of the point electron gives the in'nite energy in quantum electrodynamics. Richard Feynman told me in his of'ce at Cal Tech in 1968 that in'nite renormalization is a bait-and-switch shell game like shoddy used car salesmen use, and like the lame WMD excuse to liberate Iraq, and it is a scandal in physics that no one could do better than what he had done. No wonder all our top theoretical physicists did not balk at sharing Werner Erhards 'ne Cuban cigars and liquour in his attic at The Brown House on Franklin Street in San Francisco. (Grin) Physicists did not know 100 years ago that 1/3 or so of the universe was this kind of exotic vacuum. For example, there is a huge sphere (AKA galactic halo) of exotic vacuum of w = -1 positive pressure that holds our galaxy together preventing our solar system from going off into space on its own. This sphere looks like w = 0 cold dark matter from our vantage point. What works on this large scale also works on the small scale of the single electron (and all the charged lepto-quarks). A neutrino has some mass and is simply a micro-geon of Mass without mass pure zero point energy with positive pressure. One can plausibly metric engineer exotic vacua into arti'cial black holes and traversable wormhole star gates as well as weightless warp drives as we allegedly see in the seemingly impossible Wheeler wormhole with at least two mouths (like 2D spherical surfaces in 3D space) and quantized electro-weak-strong Faraday lines of gauge force forming closed loops with no free ends threading the mouths. The fact that there is very little naturally occurring anti-matter in our universe is direct evidence for at least one parallel universe next door where all the twin wormhole mouths are attached to. This is consistent with Jacques Vallees Fastwalker. Furthermore, none of the dark matter detector experiments now planned or underway will ever click with the right stuff if my ideas here are correct. This is a matter of fundamental principle. there is no question that physicists seriously contemplate a phase transition in the quantum vacuum as a real possibility. The quantum vacuum de'es common sense, because empty space in quantum 'eld theory appear and disappear far too quickly to be detected directly, but their existence has been con'rmed by experiments that demonstrate their in§uence on ordinary matter.A major component of the physical quantum vacuum consists of virtual electrons frothing and bubbling, like The Brew of the Three Witches in Shakespeares Macbeth, at the foamy Fermi surface edge of the Dirac negative energy sea. This is because of the Pauli exclusion principle that only none or one electron per quantum state. A virtual electron pops out of the vacuums Fermi surface leaving a hole behind. The hole is the virtual positron. The result is a virtual electron-positron pair. However, the virtual electron and the virtual positron attract because they have opposite charges and they are exchanging virtual photons. Therefore, some of them form a more stable bound state. An enormous number of these virtual pairs Bose-Einstein condense into the same center of mass quantum wave packet of the single pair forming the Vacuum Coherence Field (AKA In§ation Field). This is a dynamic steady state non-equilibrium dissipative structure of detailed balance in which there is a continual in§ow and out§ow of virtual pairs into and out of this giant quantum or macro-quantum super§uid. That is, virtual pair bound states are continually created and destroyed from and into the electrically neutral ionized plasma of broken virtual pairs which is the zero point dark energy/matter normal §uid component of the exotic vacuum that sustains the robust generalized phase rigidity of the macro-quantum vacuum coherence 'eld. This P.W. Anderson phase rigidity explains Andrei Sakharovs metric elasticity or space-time stiffness or string tension. Essentially this is a vacuum phase transition, similar to the BCS phase transition from a normal metal to an electrical superconductor, from the globally §at micro-quantum electrodynamic vacuum without any gravity at all to the curved macro-quantum electrodynamic vacuum with emergent gravity. Einsteins 'eld equation of general relativity can be derived from the phase wiggles and ripples in the robust stable macroscopically occupied center of mass quantum wave packet of the bound state of the virtual electron-positron pair. The exotic vacuum dark energy and dark matter are simply the amplitude wiggles and ripples of this same virtual pair quantum wave packet. The wave packet spreads over the entire 3D space of the post-in§ationary bubble on which our Hubble-horizoned universe is located along with an in'nity of parallel universes next door as Max Tegmark explains in May idea is correct, take the surface area of the expanding Hubble sphere that is the causal retarded boundary of 3D space of our past light cone at Earth and divide it by the quantum of area. That gives us the number of Bekenstein-Shannon c-bits and explains the irreversible arrow of time (AKA Second Law of Thermodynamics) of increasing thermodynamic entropy in terms of the dynamical expansion of the 3D space of the universe. Lenny Susskind calls this kind of idea DeSitter Space, which is a large-scale homogeneous isotropic limiting case of what I am pointing to on all scales.As Above So BelowSuch research should be forbidden!Sakharov to Lev OkunTiger TigerIts already too late. Pandoras Box is open. Schrodingers Cat has jumped out of it. This Kitty has grown up to William Blakes Tyger, Tyger shining bright in the darkness of the night and I am holding it by the tail. Hold on to your hats for a bumpy ride until we switch on zero g-force weightless warp drive!Tiger, tiger, burning brightIn the forests of the night,What immortal hand or eyeCould frame thy fearful symmetry?In what distant deeps or skiesBurnt the 're of thine eyes?On what wings dare he aspire?What the hand dare seize the 're?And what shoulder and what artCould twist the sinews of thy heart?And, when thy heart began to beat,What dread hand and what dread feet?What the hammer? What the chain?In what furnace was thy brain?What the anvil? What dread graspDare its deadly terrors clasp?When the stars threw down their spears,And water`d heaven with their tears,Did He smile His work to see?Did He who made the lamb make thee?Tiger, tiger, burning bright,In the forests of the night,What immortal hand or eyeDare frame thy fearful symmetry?In the early 1970Ís Soviet physicists were concerned that the vacuum of our universe was in fact only one possible state of empty space. The fundamental state of empty space is called the true vacuum. Our universe was considered to reside in a false vacuum, protected from the true vacuum by the wall of our world. A change from one vacuum state to another is known as a phase transition. This is analogous to the transition between frozen and liquid water. Lev Okun, a Russian physicist and historian recalls Andrei Sakharov, the father of the Soviet hydrogen bomb, expressing his concern about research into the phase transitions of the vacuum. If the wall between the vacuum states was to be breached, calculations showed that an unstoppable expanding vacuum bubble would continue to grow until it destroyed our entire universe! Sakharov declared Such research should be forbidden! since there was always the possibility that an experiment might accidentally trigger a vacuum phase transition.British Astronomer Royal, Sir Martin Rees, Master of Trinity College, and Director of the Cambridge University Institute of Theoretical Astronomy on Madingley Road where Stephen Hawking works discusses all this in Chapter 9 of his important book Our Final Hour.Could the wall of our universe be breached from within? The amount of energy required to punch a hole through the wall appeared to be enormous, and no known natural physical phenomena, even the most energetic, had punched through either. A recent report commissioned to examine potential dangers at the Large Hadron Collider, one of the next the best of our existing knowledge. Others are not so certain, however. At least one of the Russian physicists I had corresponded with was said to have been a former associate of Andrei Sakharov. He strongly hinted at new theories the Russians had developed which allow for the manipulation of the fundamental constants of nature, but he never revealed more than a sketch of his ideas. He claimed that a breakthrough was within reach, perhaps within 've years Who was that? Not George Ryazanov?Recent theoretical explorations may suggest another approach to the physics of the vacuum. The invisible gravitating dark matter could be the other side of the invisible dark energy coin, and that suggests the possibility of manipulating the vacuum for energy release.Now this is my original idea that you got from our communications over the past few years. I am the only physicist in the world today, as far as I know who has suggested this and has already published it in my two books of 2002 so its in the of'cial record at the Library of Congress.If a controllable parameter could be found to mediate the balance between the invisible dark forces, the result would unleash the vacuum energy of creation in all of its awful power and majesty. If it were possible to control the dark sides of the force then spacetime, the arena where everything we know takes place, could be bent and twisted with in'nitely greater ease than was ever suspected. This would open PandoraÍs box to everything from vacuum energy weapons of mass destruction (capable of destroying the universe!) to spacetime warp drives and time machines.Exactly, the above is the thesis of all my books since 2002 at least.A quick survey of the international electronic archive of physics papers at www.arXiv.org shows that research into the vacuum of spacetime for energy production is alive and well.I do not think that is true. You need to cite speci'cs here. There are lots of §aky new age papers on free energy on the Web written by people without any real credentials but they are not on www.arXiv.org which is not even allowing competent fringe papers in controversial topics like cold fusion. So what exactly are you thinking of here? Indeed, Paul Ginsparg, who controls the archive, does not even allow Carlos Castro to publish conservative competent papers on Clifford Algebras which are not fringe at all!Most authors are independent researchers struggling with limited funding and resources, yet their theoretical results suggest that somewhere in Nick CookÍs black world, a major breakthrough has already taken place. Most likely Israel, UK, the United States and Russia are in the lead, but China, France, Serbia, Ukraine, Iran, India and Saudi Arabia all have scientists actively pursuing the fundamental physics that determine the fabric of our reality, and are seeking the theory and the means to access the enormous energies locked inside of the vacuum since the creation of the universe. Even if the black budget world has yet to unleash the enormous potential of vacuum energy, there are signs that those in power may have begun to take notice. Dr. Harold Puthoff, a scientist with strong government connections, who has previously worked on classi'ed projects for the CIA, is a major proponent of vacuum energy physics. Nick CookÍs book, The Hunt for Zero Point, and his recent stories on zero point energy in JaneÍs Defence Weekly have also brought attention to the dangers and military potential of exotic vacuum research. The American intelligence community 'nanced so-called psychic spies for over twenty years and through four presidential administrations. It is highly unlikely that they would ignore the potential of the quantum vacuum. Dr. George Chapline, of the Lawrence Livermore National Laboratory, and Dr. Jack Sarfatti in San Francisco, knew each other in the sixties in La Jolla, have independently proposed that the quantum vacuum may unstable to the formation of coherent virtual processes. Sarfatti suggests that gravity is an emergent property determined by the physics of the vacuum. His idea is to 'nd a means of directly interacting with the physics of the vacuum that controls the shape of spacetime. Such a possibility would be consistent with the reported success of Evgeny Podkletnov, the Russian scientist who is experimenting with spinning superconducting disks. PodkletnovÍs most recent papers report the appearance of a mysterious coherent beam of gravity like radiation with a measured force of 1000 G. In an interview on BBC radio, Nick Cook pointed out one immediate application of the Podkletnov beam - the destruction of missiles and satellites in §ight or in orbit around the earth. Cook showed the BBC internal documents from Boeing, the American aerospace contractor, proving their interest in PodkletnovÍs research.This beam stuff I am suspicious of. Of course if the experiment is good, I have to think more about it. I am not so sure if Podkletnovs experiment is any good and has been replicated.The connections between PodkletnovÍs results, and the kind of vacuum research explored by Sarfatti, beginning in 1999 at the International Space Sciences Organization are the latest threads in a trail that most likely originates in cold war disinformation, a game played by East and West against each other. Glasnost has shifted the balance of partnerships and the positions of the players, but not the stakes of an outcome that would leave the world with even more proli'c and powerful weapons of mass destruction.That is true, as shown in Destiny Matrix, however you leave out the most important evidence -UFOs!When I 'rst business are the connections. Although Nick Cook never revealed the identity of his deep throat contact called Dr. Dan Marckus in the book The Hunt for Zero Point, there was no question that the Podkletnov results had played a major part in 'tting together the pieces of the puzzle. The amount of interest was in PodkletnovÍs reports by NASA, Boeing, and others in the international arena of aerospace and military research communities was evidence that there was more here to explore than the latest musings of the intellectual elite. The truth is that a fundamental theory of gravity at the scales of subatomic nuclear physics does not exist. The fact is that no one understands the nature of the gravitational 'eld at very small scales. In fact gravity has barely been probed much below one millimeter. Every attempt to unify the physical theories of gravity with the well-known standard model physics of electromagnetism, and the strong and weak nuclear forces, has failed. More importantly there has been recent progress in the exotic areas of mainstream research, such as superstring theory, which suggest new kinds of physics, which might support explanations for PodkletnovÍs impulse gravity effect. One of the current fads in theoretical physics involves large extra dimensions of space that allow a much stronger version of gravity to leak off the membrane world of our ordinary three dimensions. The large dimensional picture allows for the well known forces of electromagnetism, and the strong and weak nuclear forces, to be con'ned to a three dimensional brane-world §oating in a higher dimensional hyperspace. Gravitons, the quanta of the gravitational force, are viewed as closed strings, and are able to slip off of our brane-world, which explains why the gravitational force is so much weaker than the other forces that hold matter together. Gravitons could be exchanged between our brane-world and another brane §oating nearby in the same higher dimensions.The Sarfatti picture offers a more direct interaction with the new physics than the brane world ideas. SarfattiÍs vision is to 'nd a means of using electromagnetic 'elds in the Josephson effect to couple to the virtual electron-positron pair giant coherent condensate in§ation 'eld inside the vacuum that controls the shape of spacetime to the real electron pair giant coherent condensate of a control high temperature superconductor. UCBs Ray Chiao has a similar idea using a superconductor to transduce electromagnetic far 'eld waves to gravity waves with high ef'ciency conversion. Sarfatti wants to do the same thing with non-propagating electromagnetic and gravity near 'elds. One wonders if the black budget world may have already produced some of the technology needed to explore and test these new realms.Not a chance. They are clueless about the theory. They are still stuck in HalÍs PV model and Bernie HaischÍs zero point ideas, which will never, in my opinion §y. They are not asking the right questions and do not have the right idea in their minds. It is my belief, until I see evidence to the contrary, that I am the only physicist on the planet today who is doing real theoretical work directly relevant to the achievement of practical metric engineering anchored in the now observed reality of dark energy. All my work is public. I would love to be proved wrong on this especially by Hal Puthoff, but I am not holding my breath.;-) Extraordinary claims require extraordinary proof. Everything else I have seen is either on the wrong track asking the wrong questions like the work of Puthoff, Haisch, Ibison & Rueda for example, which at least is real physics that has proved itself wrong in IbisonÍs PV cosmology paper, or else the claims are patently obvious nonsense that Feynman called Cargo Cult Science. There is also the Russian torsion work of Akimov and Shipov and I am not prepared to make a de'nitive statement on that, as the issue is not simple because of several factors some political. One must be careful there to separate ShipovÍs theoretical work from claims made about practical devices including weapons applications. I do, however, agree with you that there is a real issue here as de'ned in Ch. 9 of Martin ReesÍs Our Final Hour.Hal Puthoff coined the term metric engineering for §ying saucer technology. Hal has been working on this problem for many decades and has held high USG security clearances and has been privy to reliable information that the saucers are real and are alien. Otherwise he would not be working on the problem. However, HalÍs theories, both of the zero point energy and of the gravity 'eld will not solve the problem because they are too naively based and do not ask the right questions. The basic physics required for this task is way beyond the depth of Puthoffs self-described engineering approach and can be found in RovelliÍs new book on quantum gravity.http://www.cpt.univ-mrs.fr/~rovelli/book.pdf Subject : Re: How to diagonalize a Hermitian matrix> Hello group, Could somebody please tell me a numerical algorithm to> diagonalize a Hermitian matrix H, so that I end up not only with the> eigenvalues perched along the diagonal, but also with the unitary> matrix U that conjugates with H to diagonalize it.> I should probably say what I really want. I have a> positive-de'nite Hermitian matrix H, and I need its matrix square> root. The above is simply the approach that advance. Achava> The group sci.math.num-analysis is a better forum for your question.And: get hold of Matrix Computations by Gene. G. Golub and Charles F. Van Loan(newest edition);it has both the diagonalization and square root algorithms, 'nely tunedand analyzed.For ready-made routines for matrices of moderate size, use MATLAB or itsfree variations === Subject: Re: How to diagonalize a Hermitian matrix>Hello group,> Could somebody please tell me a numerical algorithm to>diagonalize a Hermitian matrix H, so that I end up not only with the>eigenvalues perched along the diagonal, but also with the unitary>matrix U that conjugates with H to diagonalize it.> I should probably say what I really want. I have a>positive-de'nite Hermitian matrix H, and I need its matrix square>root. The above is simply the approach that occurred to available algorithms to do this diagonalization.Any good library should have such.I am assuming that by square root you mean the Hermitiansquare root. If you just want a matrix A with H=A*A~, theconjugate transpose of A, the Cholesky decomposition is thefast way to do it.One can use NewtonÍs method to obtain the square root of amatrix; the rate of convergence is not bad. There are otherways as well. I suggest you ask the question insci.math.num-analysis.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: How to diagonalize a Hermitian matrix>Hello group,> Could somebody please tell me a numerical algorithm to>diagonalize a Hermitian matrix H, so that I end up not only with the>eigenvalues perched along the diagonal, but also with the unitary>matrix U that conjugates with H to diagonalize it.> I should probably say what I really want. I have a>positive-de'nite Hermitian matrix H, and I need its matrix square>root. The above is simply the approach that advance.> There are available algorithms to do this diagonalization.> Any good library should have such.> I am assuming that by square root you mean the Hermitian> square root. If you just want a matrix A with H=A*A~, the> conjugate transpose of A, the Cholesky decomposition is the> fast way to do it.> One can use NewtonÍs method to obtain the square root of a> matrix; the rate of convergence is not bad. There are other> ways as well. I suggest you ask the question in> with A^2 = H. At least Ithink that is what I am looking for. I === Subject: Re: Dik WinterÍs claims revisited, dependency issue solution for the wÍs, but how did he pick wÍs from in'nity?> Given that his polynomial is of degree 22, itÍs quite possible that he> just chose a really big polynomial!> Well I read over what Keith Ramsay has and it looks ok to me, though IdidnÍt check all his numbers.So I was wrong here, no big deal.James Harris === Subject: Re: Dik WinterÍs claims revisited, wrong isnÍt a big deal. But consistently accusing those who discover your errors of being liars *is*a big deal. It demonstrates clearly that you have NO integrity.> James Often in error, but never in doubt. Harris--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: Dik WinterÍs claims revisited, dependency issue> jstevh@msn.com (James that Keith Ramsay even claimed to have posted a> solution for the wÍs, but how did he pick wÍs from in'nity?> Given that his polynomial is of degree 22, itÍs quite possible that he> just chose a really big polynomial!> Well I read over what Keith Ramsay has and it looks ok to me, though I> didnÍt check all his numbers.> So I was wrong here, no big deal.> James HarrisThat makes a perfect score of none right in all these protracted disputes and a humongous number of times wromg.DoesnÍt this record indicate that your forte is not in mathematical speculations? === Subject: Re: Dik WinterÍs claims revisited, dependency issue Decker put forward the quadratic> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his aÍs are roots of > a^2 - (x - 1)a + 7(x^2 + x). Dik Winter has repeatedly asserted that there exists varying algebraic> integer functions w_1(x) and w_2(x), such that> w_1(x) w_2(x) = 7> and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) and> w_2(x) as factors, respectively.> Right. YouÍve seen what happens when x=1 and x=2. LetÍs try adifferent one, where things are simple enough to verify byhand.Suppose we take x = -3. Then we have (5a_1(-3) + 7)(5a_2(-3) + 7) = 7(25(9) + 30(-3) + 2) = 7(137)where the aÍs satisfy a^2 + 4a + 7((-3)^2 + (-3)) = a^2 + 4a + 42.We 'nd that a_1 = -2 + sqrt(-38) and a_2 = -2 - sqrt(-38)ItÍs easy enough to verify that the aÍs arenÍt divisible by7 or by sqrt(7).However, a_1 does share a factor in common with 7,namely w_1 = (1 + 3sqrt(-38))^{1/3}We can verify that1. w_1^3 divides 7^3, since (1 + 3sqrt(-38))(1 - 3sqrt(-38)) = 343 = 7^3 So w_1 = (1 + 3sqrt(-38))^{1/3} divides 7.2. w_1^3 divides a_1^3 = 220 - 26sqrt(-38), since (1 + sqrt(-38)(-8 - 2sqrt(-38)) = 220 - 26 sqrt(-38) So w_1 = (1 + 3sqrt(-38))^{1/3} divides a_1Thus w_1 is a common divisor of a_1 and 7. (In fact,itÍs a gcd of a_1 and 7.)In a similar way, we 'nd that w_2 = (1 - 3sqrt(-38))^{1/3}is a common divisor of a_2 and 7.In this case, with (5a_1 + 7)(5a_2 + 7) = 7(137)We see that the factor 7 on the right splits as w_1 * w_2 = 7and that f_1 = (5a_1 + 7)/w_1 = 5(-8 - 2sqrt(-38))^{1/3} + (1 - 3sqrt(-38))^{1/3}and f_2 = (5a_1 + 7)/w_1 = 5(-8 + 2sqrt(-38))^{1/3} + (1 + 3sqrt(-38))^{1/3} f_1 * f_2 = 137exactly as expected. Likewise, itÍs easy to show thatw_1, w_2, f_1, and f_2 are all algebraic integersand none of them are units.I hasten to add that this behavior is what happens foralmost all values of x, namely that in (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)with the aÍs satisfying a^2 -(x - 1)a + 7(x^2 + x)we will be able to 'nd algebraic integers w_1(x) and w_2(x)with w_1(x) * w_2(x) = 7 and w_i(x) dividing a_i(x) so thatwith f_i(x) = (5a_1(x) + 7)/w_i(x)we will have f_1(x) * f_2(x) = 25x^2 + 30x + 2as long as 7 doesnÍt divide 25x^2 + 30x + 2.With more or less dif'culty, one could do the sameconstruction for most integers x, just as I did for x=1and Keith did for x=2. In most cases weÍll 'nd that 7splits into two nonunit factors, distributed between(5a_1+7) and (5a_2+7). In general, as with w_i(-3) = (1 +/- 3sqrt(-38))^{1/3}it wonÍt be immediately obvious that w_1(x) and w_2(x)are divisors of 7 and of a_i(x), in the sense thatyou canÍt look at them and immediately notice thatthey === Subject: Re: Dik WinterÍs claims revisited, dependency issueX-DMCA-Noti'cations: -0800, norabaron@hotmail.com (Nora Baron)>>[...]>> IÍm still pissed. > Too bad. In fact, seeing the number of threads you have started today>on this, and their content, it looks to me like you are frantic,>perhaps even dangerously upset. You need to calm down about >this.DonÍt tell him that. HeÍs a lot more fun to read when heÍs inraving lunatic mode.Oops, I guess that wasnÍt a very nice thing to say. I for onehave decided that the idea that weÍre supposed to be niceto everyone, regardless of how they behave, is just silly.> We are not evil conspirators trying to rob you of your>rightful credit. We are just ordinary people trying to show you>where you are making mistakes. Nora B.************************David C. Ullrich === Subject: Re: Dik WinterÍs claims revisited, norabaron@hotmail.com (Nora Baron)>>[...]>> IÍm still pissed. > Too bad. In fact, seeing the number of threads you have started today>on this, and their content, it looks to me like you are frantic,>perhaps even dangerously upset. You need to calm down about >this.> DonÍt tell him that. HeÍs a lot more fun to read when heÍs in> raving lunatic mode.> Oops, I guess that wasnÍt a very nice thing to say. I for one> have decided that the idea that weÍre supposed to be nice> to everyone, regardless of how they behave, is just silly.> Some professor you turned out to be Ullrich.YouÍre a joke and a disgrace to your profession.Oh yeah, off.James Harris === Subject: Re: Dik WinterÍs claims revisited, dependency issueX-DMCA-Noti'cations: 16:44:40 -0800, norabaron@hotmail.com (Nora Baron)>[...]> IÍm still pissed. >> Too bad. >> In fact, seeing the number of threads you have started today>>on this, and their content, it looks to me like you are frantic,>>perhaps even dangerously upset. You need to calm down about >>this.>> DonÍt tell him that. HeÍs a lot more fun to read when heÍs in>> raving lunatic mode.> Oops, I guess that wasnÍt a very nice thing to say. I for one>> have decided that the idea that weÍre supposed to be nice>> to everyone, regardless of how they behave, is just silly.>> Some professor you turned out to be Ullrich.YouÍre a joke and a disgrace to your profession.Actually being polite to people who behave like raving lunaticsbut insist they understand mathematics much better than_every_ professional mathematician on the planet is notpart of the job description.If people had been refusing to help you learn math youÍdhave a valid complaint about what sort of professors theywere. But youÍve shown over and over for years that youhave no interest in learning any math.>Oh yeah, off.>James Harris************************David C. Ullrich === Subject: Re: Dik WinterÍs claims revisited, >>[...]>> IÍm still pissed.> Too bad.> In fact, seeing the number of threads you have started today>on this, and their content, it looks to me like you are frantic,>perhaps even dangerously upset. You need to calm down about>this. DonÍt tell him that. HeÍs a lot more fun to read when heÍs in> raving lunatic mode. Oops, I guess that wasnÍt a very nice thing to say. I for one> have decided that the idea that weÍre supposed to be nice> to everyone, regardless of how they behave, is just silly.> Some professor you turned out to be Ullrich. YouÍre a joke and a disgrace to your profession. Oh yeah, off.> James HarrisSome man you are. You should either be ashamed, embarrassed, or both. IÍvenever seen a man as immature as you are, James.David Moran === Subject: Re: Dik WinterÍs claims revisited, dependency issueJames Harris:|> TheyÍve gone so far that Keith Ramsay even claimed to have posted a|> solution for the wÍs, but how did he pick wÍs from in'nity?IÍm not sure what you mean here by pick wÍs from didnÍt. There is a well-de'ned algorithm for computing|the wÍs, starting with the number (1 + sqrt(-167))/2. If you|had replaced 7 with 17, the starting number would have been|(-1 + sqrt(-407))/2 and the result would be different.LetÍs recall what I posted.|It appears that Q(sqrt(-167)) has a class group of order 11.|Let r stand for (1+sqrt(-167))/2 (which is an algebraic integer,|in spite of the 2 in the denominator).||r^11 = (-592764018-86559857*r)| = (44555-222*r) (-12882-2017*r)||and||7^11 = (44555-222*r)(44555+222*r).As an intermediate step, I found the GCD of ((1+sqrt(-167))/2)^11and 7^11, which is an 11-th power of the GCD of (1+sqrt(-167))/2and for JamesHarris to check the result using a calculator or something like that.|Also (-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which is|relatively prime to 7^11.||The factor 44555-222*r = 44444 - 111 sqrt(-167) and its conjugate|44444 + 111 sqrt(-167) are roots of x^2-88888*x+7^11=0. One GCD of|r with 7 is then (44444-111 sqrt(-167))^(1/11), which is a root of|x^22 - 88888 * x^11 + 7^11 = 0.The only point of giving the degree 22 polynomial was to show thatthe GCD (44444-111*sqrt(-167))^(1/11) actually is an algebraic integer,by giving a monic polynomial with integer coef'cients of which itÍsa root.James Harris:|> Given that his polynomial is of degree 22, itÍs quite possible that he|> just chose a really big polynomial!ThatÍs just silly. ItÍs easy to 'nd a quadratic that 44444-111*sqrt(-167)is a root of. Getting the polynomial of degree 22 is just a matter ofsubstituting x^11 for x. ItÍs a very simple polynomial.Nora Baron:| Actually I think it had degree 11. I think he used a program|to do the computation, and I think it 'nds the lowest-degree|polynomial that works. I believe it computes powers of an|ideal, ^M, until it 'nds an M where that ideal|is principal. It is not a matter of human choice, just a |programmable algorithm.The package Pari-GP has a way of representing ideals in the ringof integers of a number 'eld. There are functions for multiplying ortaking powers of ideals, as well as one for determining whether ornot an ideal is principal. It also computes the class group of thenumber 'eld. The class group of Q(sqrt(-167)) has order 11, andthe ideal generated by 7 and (1+sqrt(-167))/2 isnÍt principal. So itmust be that the 11-th power of the ideal is principal. I then usedthe ideal power function and the one for 'nding the principal generatorof an ideal to get the 11-th power of the GCD.I donÍt know whether you can multiply the GCD by a unit to get onewhose minimal polynomial has degree less than 22. That might bea cute exercise for someone. I would guess not, but I havenÍt givenit much thought.Keith Ramsay === Subject: Re: Dik WinterÍs claims revisited, dependency issue> James Harris:> |> TheyÍve gone so far that Keith Ramsay even claimed to have posted a> |> solution for the wÍs, but how did he pick wÍs from in'nity?> IÍm not sure what you mean here by pick wÍs from in'nity.> There is a well-de'ned algorithm for computing> |the wÍs, starting with the number (1 + sqrt(-167))/2. If you> |had replaced 7 with 17, the starting number would have been> |(-1 + sqrt(-407))/2 and the result would be different.> LetÍs recall what I posted.> |It appears that Q(sqrt(-167)) has a class group of order 11.> |Let r stand for (1+sqrt(-167))/2 (which is an algebraic integer,> |in spite of the 2 in the denominator).> |> |r^11 = (-592764018-86559857*r)> | = (44555-222*r) (-12882-2017*r)> |> |and> |> |7^11 = (44555-222*r)(44555+222*r).> As an intermediate step, I found the GCD of ((1+sqrt(-167))/2)^11> and 7^11, which is an out the products this way partly to make it easier for James> Harris to check the result using a calculator or something like that.> |Also (-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which is> |relatively prime to 7^11.> |> |The factor 44555-222*r = 44444 - 111 sqrt(-167) and its conjugate> |44444 + 111 sqrt(-167) are roots of x^2-88888*x+7^11=0. One GCD of> |r with 7 is then (44444-111 sqrt(-167))^(1/11), which is a root of> |x^22 - 88888 * x^11 + 7^11 = 0.> Well I 'gure youÍre not lying here Keith, so IÍll go with youranalysis.So I was wrong about being able to 'nd a factor in the ring ofalgebraic === Subject: Re: Dik wrong about being able to 'nd a factor in the ring of> algebraic integers. Interesting.Now go back and count how many times this was proved to you, and count how many times you accusedother posters of being liars, co-conspirators, etc. for many times do you have to be smacked with the 2x4-of-truth before it leaves a dent?> James Often in error, but never in doubt. HarrisWacky, isnÍt it? But, hey, itÍs just math. Yup, yup, yup.--Mixing medications can be dangerous. Never take a strong laxative and a sleeping pill at the sametime..--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: Dik WinterÍs claims revisited, dependency issue> James Harris:> |> TheyÍve gone so far that Keith Ramsay even claimed to have posted a> |> solution for the wÍs, but how did he pick wÍs from in'nity?> IÍm not sure what you mean here by pick wÍs from didnÍt. There is a well-de'ned algorithm for computing> |the wÍs, starting with the number (1 + sqrt(-167))/2. If you> |had replaced 7 with 17, the starting number would have been> |(-1 + sqrt(-407))/2 and the result would be different.> LetÍs recall what I posted.> |It appears that Q(sqrt(-167)) has a class group of order 11.> |Let r stand for (1+sqrt(-167))/2 (which is an algebraic integer,> |in spite of the 2 in the denominator).> |> |r^11 = (-592764018-86559857*r)> | = (44555-222*r) (-12882-2017*r)> |> |and> |> |7^11 = (44555-222*r)(44555+222*r).> As an intermediate step, I found the GCD of ((1+sqrt(-167))/2)^11> and 7^11, which is an out the products this way partly to make it easier for James> Harris to check the result using a calculator or something like that.> |Also (-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which is> |relatively prime to 7^11.> |> |The factor 44555-222*r = 44444 - 111 sqrt(-167) and its conjugate> |44444 + 111 sqrt(-167) are roots of x^2-88888*x+7^11=0. One GCD of> |r with 7 is then (44444-111 sqrt(-167))^(1/11), which is a root of> |x^22 - 88888 * x^11 + 7^11 = 0.> OOPS! IÍm re-thinking my previous post. IÍll consider Keith RamsayÍspoints here some more.For all I know IÍm wrong, but then again, maybe IÍm still just missingsomething.James Harris === Subject: Re: Dik WinterÍs claims revisited, dependency issue> |Also (-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which is> |relatively prime to 7^11.> |> |The factor 44555-222*r = 44444 - 111 sqrt(-167) and its conjugate> |44444 + 111 sqrt(-167) are roots of x^2-88888*x+7^11=0. One GCD of> |r with 7 is then (44444-111 sqrt(-167))^(1/11), which is a root of> |x^22 - 88888 * x^11 + 7^11 = 0.> OOPS! IÍm re-thinking my previous post. IÍll consider Keith RamsayÍs> points here some more.> For all I know IÍm wrong, but then again, maybe IÍm still just missing> something.Both! === Subject: Re: Dik WinterÍs claims revisited, dependency issue> James Harris:> |> TheyÍve gone so far that Keith Ramsay even claimed to have posted a> |> solution for the wÍs, but how did he pick wÍs from in'nity?> IÍm not sure what you mean here by pick wÍs from in'nity.> There is a well-de'ned algorithm for computing> |the wÍs, starting with the number (1 + sqrt(-167))/2. If you> |had replaced 7 with 17, the starting number would have been> |(-1 + sqrt(-407))/2 and the result would be different.> LetÍs recall what I posted.> |It appears that Q(sqrt(-167)) has a class group of order 11.> |Let r stand for (1+sqrt(-167))/2 (which is an algebraic integer,> |in spite of the 2 in the denominator).> |> |r^11 = (-592764018-86559857*r)> | = (44555-222*r) (-12882-2017*r)> |> |and> |> |7^11 = (44555-222*r)(44555+222*r).> As an intermediate step, I found the GCD of ((1+sqrt(-167))/2)^11> and 7^11, which is an out the products this way partly to make it easier for James> Harris to check the result using a calculator or something like that.> |Also (-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which is> |relatively prime to 7^11.> |> |The factor 44555-222*r = 44444 - 111 sqrt(-167) and its conjugate> |44444 + 111 sqrt(-167) are roots of x^2-88888*x+7^11=0. One GCD of> |r with 7 is then (44444-111 sqrt(-167))^(1/11), which is a root of> |x^22 - 88888 * x^11 + 7^11 = 0.> All those numbers are hard to parse through, so I just went back toconsidering thata_1(x) + a_2(x) = (x-1)so assuming factors f_1(x), and f_2(x), with yourw_1(2) = (44444-111 sqrt(-167))^(1/11), andw_2(x) = (44444+111 sqrt(-167))^(1/11), I have w_1(2) f_1(2) + w_2(2) f_2(2) = 1and assuming the wÍs are symmetric by sign, I have that f_1(2) mustequal f_2(2).But f_1(2)f_2(2) = 6as the problem I see is that youÍre stuck, if the wÍs are symmetric bysign, so are you saying they are not?James Harris === Subject: Re: Dik WinterÍs claims revisited, dependency issue Consider the Decker quadratic example (reference at bottom).> Decker put forward the quadratic> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his aÍs are roots of > a^2 - (x - 1)a + 7(x^2 + x).> Dik Winter has repeatedly asserted that there exists varying algebraic> integer functions w_1(x) and w_2(x), such that> w_1(x) w_2(x) = 7> and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) and> w_2(x) as factors, respectively.> Now then, introducing f_1(x) and f_2(x) as the other factors of the> aÍs, you have> w_1(x) f_1(x) = 5a_1(x) + 7, and w_2(x) f_2(x) = 5a_2(x) + 7, > so> a_1(x) = (w_1(x) f_1(x) - 7)/5, and a_2(x) = (w_2(x) f_2(x) - 7)/5.> Now it matters to use the fact that the aÍs are roots of> a^2 - (x - 1)a + 7(x^2 + x),> as solving that quadratic, and picking a_1(x) for the positive sign> root gives> a_1(x) = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, soTypo (or thinko) here and below. ... + 28(x^2 + x)))/2 should be... - 28(x^2 + === Subject: Re: Dik WinterÍs claims revisited, dependency issue> ... stuff deleted ...>>Given that his polynomial is of degree 22, itÍs quite possible that he>>just chose a really big polynomial!> Actually I think it had degree 11. I think he used a program> to do the computation, and I think it 'nds the lowest-degree> polynomial that works. I believe it computes powers of an> ideal, ^M, until it 'nds an M where that ideal> is principal. It is not a matter of human choice, just a > programmable algorithm.> I thought the gcd was of degree 11 over the 'eld Q(sqrt(-167)), butwhen you put it in terms of rational integers, the degree had to be 22.It turned out that way in KASH, which I used. The class number of thesplitting 'eld of (whatever polynomial it was) was 11, so it tookraising the ideal to the 11th power to force it to beprincipal, as you pointed out. However, this is still relative tothe (integers of the) splitting 'eld of that quadratic. ... the rest deleted ...Except for this:> In fact, seeing the number of threads you have started today> on this, and their content, it looks to me like you are frantic,> perhaps even dangerously upset. You need to calm down about > this. We are not evil conspirators trying to rob you of your> rightful credit. We are just ordinary people trying to show you> where you are making mistakes.> Nora B.> !labaC lacitamehtaM livE terceS eht tuoba no tel ot ton tseBOr, ixnay on the onspiracycay!>>But thereÍs not a lot I can do when people like Dik>>Winter can so easily get away with basic problems in their claims, on>>a newsgroup that doesnÍt seem to give a damn about mathematical truth.>>As Dan Quayle said, How terrible it is to lose your mind. Or not to have a mind at all. How true that is.>>James Harris> Dale. === Subject: Re: Dik WinterÍs claims revisited, dependency issueNntp-Posting-Host: apps.cwi.nl > Decker put forward the quadratic > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his aÍs are roots of > a^2 - (x - 1)a + 7(x^2 + x). > Dik Winter has repeatedly asserted that there exists varying algebraic > integer functions w_1(x) and w_2(x), such that > w_1(x) w_2(x) = 7Yup, right, they were de'ned in terms of the gcd function on algebraicintegers. > and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) and > w_2(x) as factors, respectively.It is a property of the gcd function that gcd(5a_1(x) + 7, 7) is afactor of both (5a_1(x) + 7) and of 7. I constructed such w_1(x)and w_2(x) such that also w_1(x)w_2(x) = 7. You never pointed toan error in those de'nitions. > Now then, introducing f_1(x) and f_2(x) as the other factors of the > aÍs, you have > w_1(x) f_1(x) = 5a_1(x) + 7, and w_2(x) f_2(x) = 5a_2(x) + 7, Indeed. > so > a_1(x) = (w_1(x) f_1(x) - 7)/5, and a_2(x) = (w_2(x) f_2(x) - 7)/5. > Now it matters to use the fact that the aÍs are roots of > a^2 - (x - 1)a + 7(x^2 + x), > as solving that quadratic, and picking a_1(x) for the positive sign > root gives > a_1(x) = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, so > (w_1(x) f_1(x) - 7)/5 = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 > so > w_1(x) f_1(x) - 14 = 5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, soGetting sloppy again? 14 should be 7. > f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x), > which implies a relationship between f_1(x) and w_1(x), but > f_1(x) f_2(x) = 25x^2 + 30x + 2But there *is* such a relation: w_1(x)f_1(x) = (5a_1(x) + 7), > and it is arbitrary that 7 was the multiple before as it could have > been 11 or 13 or any of an in'nity of other numbers, and w_1(x) and > w_2(x) are themselves not determined, so itÍs a spurious appearance of > dependency.I do not understand what you are writing here at all. If you meanthat w_1(x).w_2(x) = 7 could just as easily have been 11 or 13, youare talking nonsense. They were *de'ned* such that their productis 7. And if I provide de'nitions in full detail, I do not expectthat somebody says they are not determined. Go over the de'nitionsand play state which part is incomplete, does not de'ne an algebraicinteger, or what. > ThatÍs the little detail that Dik Winter never bothered to address.There is one little detail that you never bothered to address. Thereare complete de'nitions of w_1(x) and w_2(x). > Now then, if f_1(x) is in fact independent of w_1(x), then how do you > account for the appearance of a dependency in > f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x)?f_1(x) is *not* independent of w_1(x), you have just de'ned it suchthat w_1(x)f_1(x) = (5a_1(x) + 7). So where do you 'nd theindenpendence? And when you replace 14 by 7 above, it just statesthe same. > If you try to push the issue that it isnÍt so required consider what > happens if you try to divide through by w_1(x), as then you have > f_1(x) = > (5((x-1)/w_1(x)+sqrt((x-1)^2 + 28(x^2 + x)))/(2w_1(x) + 2w_2(x))You have lost me here. > and the problem is that all of the wÍs need to go away. So assuming > that a_1(x) has w_1(x) as a factor means, introducing g(x), that you > haveA strong assumption you make here, and it is not valid. a_1(x) does*not* have w_1(x) as a factor, it is (5a_1(x) + 7) that has w_1(x)as a factor. > which indicates that f_1(x) is dependent on w_2(x).Whacky, but indeed, it is so, because w_1(x) is dependent on w_2(x)and the reverse. They are connected through the relation w_1(x)w_2(x) = 7. > However, as I pointed out w_1(x) w_2(x) = 7 does NOT determine them as > an in'nity of functions will work, while f_1(x) is independent of > w_1(x) and w_2(x) since > f_1(x) f_2(x) = 25x^2 + 30x + 2.This does not make them independent on w_1(x) and w_2(x). You have*de'ned* f_1(x) as the function such that w_1(x)f_1(x) = (5a_1(x) + 7)so you yourself has made it dependent on w_1(x). Similar withf_2(x) and w_2(x). > The point is you can multiply some polynomial like 25x^2 + 30x + 2, by > anything you choose. ThereÍs no way that itÍs locking into it that > functions that are factors of 7 are required.*You* required it by having the factorisation (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2). > TheyÍve gone so far that Keith Ramsay even claimed to have posted a > solution for the wÍs, but how did he pick wÍs from in'nity? > Given that his polynomial is of degree 22, itÍs quite possible that he > just chose a really big polynomial!Ah, I understand, you did not check what he you have two > independent equations: > w_1(x) w_2(x) = 7 and > f_1(x) f_2(x) = 25x^2 + 30x + 2 > To try and dispute my results, posters like Dik Winter or Nora Baron > simply skip past mathematical consequences of their claims, like the > independence between those equations.Try to read my de'nitions of w_1(x) and w_2(x) and see that they force w_1(x)f_1(x)w_2(x)f_2(x) = (5a_1(x) + 7)(5a_2(x) + 7) = = 7(25x^2 + 30x + 2).And as w_1(x), f_1(x), w_2(x) and f_2(x) are (with my de'nitions) allalgebraic integer functions I see no problem. > IÍm still pissed. But thereÍs not a lot I can do when people like Dik > Winter can so easily get away with basic problems in their claims, on > a newsgroup that doesnÍt seem to give a damn about mathematical truth.It is clear through your remark on Keith RamsayÍs solution that youreally do not read is correct or not, you just assume.Neither (I assume) did you check my de'nitions of w_1(x) and w_2(x).-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Dik WinterÍs claims revisited, dependency issue Adjunct Assistant Professor at the University of Montana. [.snip.]>A strong assumption you make here, and it is not valid. a_1(x) does>*not* have w_1(x) as a factor, it is (5a_1(x) + 7) that has w_1(x)>as a factor.Actually, in this context it is valid: w_1(x) divides 7, and divides5a_1(x)+7, so it divides 5a_1(x); since w_1(x) divides 7, it iscoprime to 5, and therefore divides a_1(x).-- =ItÍs not denial. IÍm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: Dik WinterÍs claims revisited, dependency issueThere was a small mistake in my previous reply to this:where I said a^2 - (x - 1)*a + 17it should have been a^2 - (x - 1)*a + 17*(x^2 + x).Nora B.[rest deleted] === Subject: JSH: Apology to Ramsay, why I postItÍs simpler to just post to all the newsgroups that I posted beforean apology for questioning Keith RamsayÍs honesty. It seems he did infact post a non-unit algebraic integer factor of (1+sqrt(-167))/2 and7, so I was wrong.So I apologize to Keith Ramsay for questioning his honesty here.As for the rest of the obsessive repliers, no apologies to thatdogmatic crew!TheyÍre still at it, replying, and replying and replying.Still now you can see why Usenet IS useful to me, as I can have peoplecheck things for me, like Keith Ramsay did.James Harris === Subject: Re: JSH: Apology to Ramsay, why I post > ItÍs simpler to just post to all the newsgroups that I posted before > an apology for questioning Keith RamsayÍs honesty. It seems he did in > fact post a non-unit algebraic integer factor of (1+sqrt(-167))/2 and > 7, so I was wrong.Yup. > So I apologize to Keith Ramsay for questioning his honesty here.Why not apologise to those who did repost that result for your perusal? > As for the rest of the obsessive repliers, no apologies to that > dogmatic crew! > TheyÍre still at it, replying, and replying and replying. > Still now you can see why Usenet IS useful to me, as I can have people > check things for me, like Keith Ramsay did.But if people check things they are abused...What happens is that if somebody posts a counterexample or something,you just state it is nonsense, without ever checking. This example hasbeen posted in total 10 times by 4 different posters (Keith Ramsay,Nora Baron, Rick Decker and me) in a period of 6 weeks (the 'rst oneon 7 January). And 'nally after 6 weeks it took you only one falsestart to check that the result was true indeed... If you had doneyour checking 6 weeks ago Usenet would have been useful for you andit would have spared us a large number of threads.Now when are you going to check my de'nitions of w_1(x) and w_2(x)?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Apology to Ramsay, why simpler to just post to all the newsgroups that I posted before> an apology for questioning Keith RamsayÍs honesty. It seems he did in> fact post a non-unit algebraic integer factor of (1+sqrt(-167))/2 and> 7, so I was wrong.Why is the non-unit algebraic integer factor of (1 + sqrt(-167))/2 and 7important ?Likewise, is there any signi'cance to the non-unit irrational advance.lt === Subject: Re: Apology to Ramsay, why I to all the newsgroups that I posted before> an apology for questioning Keith RamsayÍs honesty. It seems he did in> fact post a non-unit algebraic integer factor of (1+sqrt(-167))/2 and> 7, so I was wrong.> Why is the non-unit algebraic integer factor of (1 + sqrt(-167))/2 and 7> important ?It is not all that important in the great scheme of things, but it isimportant in the ongoing discussion of Mr. HarrisÍs ideas, because hehad previously stated that there was no such factor. > Likewise, is there any signi'cance to the non-unit irrational advance.> lt> === Subject: Re: JSH: Apology to Ramsay, why I honesty here. As for the rest of the obsessive repliers, no apologies to that> dogmatic crew! TheyÍre still at it, replying, and replying and replying.And youÍll still be at it, too, lying and lying and lying.> James Often in error, but never in doubt. Harris--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: JSH: Apology to Ramsay, why I postX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>ItÍs simpler to just post to all the newsgroups that I posted before>an apology for questioning Keith RamsayÍs honesty. It seems he did in>fact post a non-unit algebraic integer factor of (1+sqrt(-167))/2 and>7, so I was wrong.So I apologize to Keith Ramsay for questioning his honesty here.As for the rest of the obsessive repliers, no apologies to that>dogmatic crew!In particular none to the people that youÍve accused of various sorts of dishonesty when they pointed out that heÍd given thisexample? Huh.ing moronic asshole.>TheyÍre still at it, replying, and replying and replying.Still now you can see why Usenet IS useful to me, as I can have people>check things for me, like Keith Ramsay did.>James Harris************************David C. Ullrich === Subject: Re: JSH: Apology to Ramsay, why I postOriginator: a@shell3.shore.net (a)>Still now you can see why Usenet IS useful to me, as I can have people>check things for me, like Keith Ramsay did.Of course. ThatÍs a symptom of your Narcissistic Personality Disorder: 6. is interpersonally exploitative, i.e., takes advantage of others to achieve his or her own endshttp://www.mentalhealth.com/dis1/p21-pe07.html === Subject: Re: JSH: Apology to Ramsay, why I post> ItÍs simpler to just post to all the newsgroups that I posted before> an apology for questioning Keith RamsayÍs honesty. It seems he did in> fact post a non-unit algebraic integer factor of (1+sqrt(-167))/2 and> 7, so I was wrong.> So I apologize to Keith Ramsay for questioning his honesty here.> As for the rest of the obsessive repliers, no apologies to that> dogmatic crew!> TheyÍre still at it, replying, and replying and replying.> Still now you can see why Usenet IS useful to me, as I can have people> check things for me, like Keith Ramsay did.> James HarrisKeith Ramsay merely contributed futher proof of JSHÍs inanity.There is no evidence that JSh will learn anything from it. === Subject: branch of log zsci.math is whatÍs inside the glass-house, you were talking about.It really helps !Keep the glasses clean and un-veiled.Keep expressing, what You really think -there are not too many likeYou.Hero === (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id sci.math> My favorite is birds : How can one evolve into §ying ? So many things>> have to happen in order to create functional wings it seems impossible any>> awkward animal developing in this direction would survive itÍs cost, or>> any failing §ying attempts. The evolutionist arguments sound totally>> absurd : they ran fast or jumped from tree to tree and got an advantage>> slowly developing wings. Sounds like absolute voodoo to me.That depends on the size of the animal. If the animal is small enough,>the terminal velocity can result in a non-lethal impact with the>ground. (I believe the threshold is around the size of a mouse or a>cat.) Incremental levels of control over the landing point then have an>obvious bene't. >-- >Daniel W. Johnson>panoptes@iquest.net>http:// members.iquest.net/~panoptes/ As far as I know the 'rst §ying creatures they talk about are > dinosaurs, no less. And quite big ones as well.ThatÍs a comment on your knowledge. Some dinosaurs were small; a quicksearch turned up some with a mass below 10 kg. And all the references Ifound talk about small dinosaurs as the starting point for the evolutionof wings.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ abuse@teranews.com === Subject: Re: What is the Origin of Space and Time?>>What is the origin of space and time?> Where and when could space and time have originated, if there was no> space and time before they originated?What would where and when mean before they came into existence?-- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people abuse@teranews.com === Subject: Re: there is no such thing as in'nity>> Assembler? Bah!! Raw machine language burnt into a ROM is the only way>> to 'nd the largest number. In any case, I have the worldÍs largest>> number written on my of'ce blackboard, so you all might as well give up. Add one to it and see what the sum is.How about just raising it to something greater than 1? :)-- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people abuse@teranews.com === Subject: Re: there is no such thing as in'nity>> And can be written in exponentially more obfuscated a manner than even C>> or assembler... :)> Assembler? Bah!! Raw machine language burnt into a ROM is the only way> to 'nd the largest number. Ah, I remember when I 'rst started programming. All I had was a really hotneedle and had to write my code directly to the SIMMs. It was a major leapforward when I upgraded and could just type copy con 'le.exe...> In any case, I have the worldÍs largest> number written on my of'ce blackboard, so you all might as well give up. Actually, *I* have the largest number: itÍs written in 0.1pt type on theoutside of my 2-story house, covering all four walls, the ItÍs raining again! Gotta gorecalculate! -- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people think, Mr. Feynman? === Subject: Request for ideas on tricky little problemConsider a bus route with k stops. When the bus starts its journey, ppersons are on the bus. Person j wants to get off at stop s_j<=k. Alsos_j>=1. However the bus cant stop at more than z < k stops. So a personwho wants to get off at a stop s_j which is not among the z stops has to getoff at the nearest one and then walk. Let f(x_j) = (x_j)^3 be the distanceperson j has to walk to get to his destination, where x is the number ofstops he has to walk to get to his destination. Let T be the sum f(x_1) +f(x_2) + ... + f(x_p)The problem is now to construct a method(algorithm) where the bus stops atstops, so that the T is minimized. Any ideas on adecent strategy here ? === Subject: Re: Request for ideas on tricky little problem> Consider a bus route with k stops. When the bus starts its journey, p> persons are on the bus. Person j wants to get off at stop s_j<=k. Also> s_j>=1. However the bus cant stop at more than z < k stops. So a person> who wants to get off at a stop s_j which is not among the z stops has to get> off at the nearest one and then walk. Let f(x_j) = (x_j)^3 be the distance> person j has to walk to get to his destination, where x is the number of> stops he has to walk to get to his destination. Let T be the sum f(x_1) +> f(x_2) + ... + f(x_p)> The problem is now to construct a method(algorithm) where the bus stops at> stops, so that the T is minimized. Any ideas on a> decent strategy here ?If a choice of bus stops is optimal, then there can be no improvement, and especially no trivial improvement. So if you have an optimal choice of bus stops, and I take a look at lets say three consecutive stops, then I wonÍt be able to demonstrate an improvement based on these stops. This can be used as the basis of an algorithm: For i = 3, 4, ..., k, 'nd all sequences of i consecutive stops that donÍt allow a trivial improvement. Start with i = 3. Lets say I choose the 'rst stop at x, the last at z, and the middle one would be at y for some x < y < z. Try all possible choices of y, calculate the cost for all people who want to leave at one of the stops x to z for each choice of y, and you will 'nd all choices for y that cannot be trivially improved. Do this for all x and z, and you have all seqences of three stops that cannot be trivially improved.Continue with i = 4, so you look at all sequences of stops x, y, z, u. Pick x and u. Then choose x+2 <= z <= u-1. For each z, choose those values of y such that (x, y, z) cannot be improved and (y, z, u) cannot be improved. For each pair (y, z) calculate the total cost of all people exiting at stops x to u. Find all pairs (y, z) that minimise the cost. Do this for all x and u, and you have all sequences of four stops that cannot be improved trivially. Continue with i = 5, 6 etc. until i = number of stops that the bus can make. Only difference when i = number of stops is that you calculate the total cost. I havenÍt tried this, so I donÍt have any idea how the amount of data will grow, but I would hope that it isnÍt too bad. === Subject: Re: JSH: GOT IT!!! Algebraic integers check proof> Well I kept 'ddling at it, and now have on my blog the complete proof> that thereÍs a problem with the ring of algebraic integers by checking> what happens with the roots of x^2 - x + 42 and y^2 + by - 7, with an> algebraic integer b.> ItÍs all at> http://mathforpro't.blogspot.com/> Nope. I was wrong so I deleted off that blog entry.Sorry about the mistake there, but it happens.James Harris === Subject: Re: ïerfÍ function in C> If you de'ne> R(x) = cPhi(x)/phi(x)> then the function R is a well behaved function that has> a rapidly convergent Taylor expansion, an expansion> with terms easily determined by a two-step recursion.> (Try to derive the recursion yourself:> r[k+1](x)=x*r[k](x)+k*r[k-1](x),> where r[k](x) is the kth derivative of R(x).)> ...> double Phi(double x)> {long double s,t=0,a=1.2533141373155L,z=0,b=-1,pwr=1;> int i;> s=a+b*x;> for(i=2;s!=t;i+=2)> { a=(a+z*b)/i;> b=(b+z*a)/(i+1);> pwr=pwr*x*x;> t=s;> s+=pwr*(a+x*b);> }> return 1.-s*exp(-.5*x*x-.91893853320467274178L);> }It is not clear what role z is supposed to play here, since it isinitialized to 0 and never modi'ed. Removing z, and the termsinvolving z, corresponds to your de'nition of r[k](x):double Phi(double x){long double s,t=0,a=1.2533141373155L,b=-1,pwr=1; int i; s=a+b*x; for(i=2;s!=t;i+=2) { a/=i; b/=(i+1); pwr=pwr*x*x; t=s; s+=pwr*(a+x*b); } return 1.-s*exp(-.5*x*x-.91893853320467274178L);}A possible improvement is to note that the even part of R(x),(R(x)+R(-x))/2, is equal to 1/phi(x), thus only the odd part of R(x),(R(x)-R(-x))/2, needs to be computed:double Phi(double x){long double s,t=0,b=1,pwr=x; int i; s=x; for(i=2;s!=t;i+=2) { b/=(i+1); pwr=pwr*x*x; t=s; s+=pwr*b; } return .5+s*exp(-.5*x*x-.91893853320467274178L);}IÍve been using a similar algorithm, using R(x) = 1/(Phi(x)*Phi(-x)),which also has a nice Taylor series expansion with a couple of otherdesirable properties: it is an even function, thus only the terms inx^2i need to be computed; and all the coef'cients are positive, thuseliminating the loss of precision resulting from subtracting two largenumbers to get a small answer. This leads to (2*Phi(x)-1)^2 = 1 -4/R(x), thus Phi(x) = 1/2 +/- sqrt(1/4-1/R(x)), where the sign of thesqrt should be the same as the sign of x. This has the advantage thatwhen |x| is large, R(x) is also large, thus 1/R(x) is quite close to 0and the result is fairly close to the correct value in absolute terms. === Subject: Re: ïerfÍ function in implementations in Fortran, HP-41C, Pascal, Modula 2,> Clipper and recently Visual Basic of the Taylor Series expansion of> Phi(x) given in 26.2.11 in A&S since around 1980: Phi(x) = 1/2 + phi(x)*(x + x^3/3 + x^5/(3*5) + x^7/(3*5*7) + ...) Here are my VB results against your algorithm (also translated to> Visual Basic). It looks like the VB version loses about 1 digit in> precision against the C version (IÍm not surprised, but it is not too> bad either). x Marsaglia (my VB Version)> Marsaglia (C Version, from your table)> Klaey (VB)> Maple 20 places> 0.123 0.54894645101643700000> 0.5489464510164369> 0.54894645101643700000> 0.54894645101643675909 1.2 0.88493032977829200000> 0.8849303297782918> 0.88493032977829200000> 0.88493032977829173198 2.4 0.99180246407540400000> 0.9918024640754040> 0.99180246407540400000> 0.99180246407540387055 6.1 0.99999999946965800000> 0.9999999994696578> 0.99999999946965800000> 0.99999999946965767370 -6.1 0.00000000053034350051> 0.0000000005303427> 0.00000000053034221459> 0.00000000053034232638 -1.1 0.13566606094638300000> 0.1356660609463828> 0.13566606094638200000> 0.13566606094638267517 7.2 0.99999999999969900000> 0.9999999999996990> 0.99999999999969900000> 0.99999999999969893721 IÍm quite pleased how my algorithm still stands up today :-)> With kind suggestion. While it was not so in the 1960Ís when Ideveloped the method for Fortran via cPhi(x)/phi(x),the ordinary Taylor series about zero for the normalintegral,Phi(x)=.5+integral exp(-t^2/2)/sqrt(2*pi), x=0..t= .5+(x-x^3/6+x^5/40-x^7/336+...)/sqrt(2*pi)provides, with current double precisionarithmetic, an excellent method for providingnormal probabilities arithmetic for the rangeof xÍs one usually encounters, say to +/- fouror 've sigmas, letting the CPU do the workafter being given a few simple instructions.Here is a sample implementation:----------------------------------------------- #include #include double Phi(double x){ long double z,t=0,s=.3989422804014327L; int i; s*=x; z=s; t=0; for(i=3;s!=t;i+=2) {z*=-x*x*(i-2)/(i*i-i); t=s; s+=z;} return .5+s;}int main(){double x;while(1){printf(Enter your x value:);scanf(%lf,&x);printf(Normal Prob(X<%f)=%20.16fn,x,Phi(x)); } }---------------------------------------------Here are a few results, with the true valuesprovided by setting Digits:=30 in Maple: x Phi(x) .123 .5489464510164368 .5489464510164367590816... true value 2.34 .9903581300546417 .9903581300546416673759... -2.34 .0096418699453583 .0096418699453583326240... 3.45 .9997197067231839 .9997197067231838225884... 4.56 .9999974423189606 .9999974423189605484746... 6.6 .9999999999809187 .9999999999794421109060... -6.6 .0000000000190813 .0000000000205578890939... 6.8 .9999999999950689 .9999999999947690424558... -6.8 .0000000000049310 .0000000000052309575441... 7. 1.0000000000163602 .9999999999987201874561... -7 -.0000000000163602 .0000000000012798125438...Again, I hope to access 80-bit§oating point processors by usinglong doubleÍs. With plain doubleÍs,there is little difference for |x|<4;beyond that accuracy drops to 13 then12,11,10,... digits until cases suchas Phi(6.5), which returns a value >1.Those interested are invited to compareon other platforms the approach usingthe Taylor series for cPhi(x)/phi(x)with that of the Taylor series for plainPhi(x), using long doubleÍs then plain doubleÍs.Either approach may be preferable to thesometimes unavailable and/or mysterious erf.George Marsaglia === Subject: Re: ïerfÍ function in CX-Enigmail-Version: 0.76.1.0X-Enigmail-Supports: phi(t) be the normal density, phi(t) = exp(-t^2/2)/sqrt(2*Pi),> so that> Phi(x) = integral phi(t), t = -in'nity to x.> and> cPhi(x) = 1-Phi(x) = integral phi(t), t = x to in'nity.> If you de'ne> R(x) = cPhi(x)/phi(x)> then the function R is a well behaved function that has> a rapidly convergent Taylor expansion, an expansion> with terms easily determined by a two-step recursion.Very nice. But can someone explain why you need R = cPhi / phirather than just S = Phi/phi? The maths seems to work out just as well in the second case, and you donÍtneed to keep on writing (1-) during it. Thus PhiÍ = phi phiÍ = -x phitherefore SÍ = (phi PhiÍ - Phi phiÍ) / phi^2 = (phi^2 + x Phi phi) / phi^2 = 1 + x Sgiving you the recurrence in the terms of the power series for S. === Subject: Re: Got a speeding ticket and need to 'ght back> Good one, chief. I hear the WB network is hiring comedy writers.> WB is also looking for lawyers who can play sidewalk Santas for theirchristmas in july specials. Btw, WB is also looking for statistcianswho can prove that christmas is supposed to start in july.-suresh === Subject: Re: Got a speeding ticket and need to 'ght backSuresh __NoJunkMail kumar hear the WB network is hiring comedy writers. WB is also looking for lawyers who can play sidewalk Santas for their> christmas in july specials. Btw, WB is also looking for statistcians> who can prove that christmas is supposed to start in july.It took you six days to come up with *that*? Pathetic.Doug === Subject: Re: Got a speeding ticket and need to 'ght back> The Lord of Chaos (Suresh Devanathan)> A Prince, is supposed to be intelligent. duh!!!> I thought that when Euler said There is no prince way to mathematics he> was politely meaning the opposite? Or by Prince with capital P you do> refer to some unfamous singer. That one looks intelligent for sure!Hey! no wonder why most noble prize winners in physics areridiculously good looking and so are many famous mathematicians!!! === Subject: JSH: Making it personalIÍm concerned about all the histrionics that has darkened even my ownposts as itÍs all too personal at this point.Right now I *despise* posters like Dik Winter, Nora Baron, and DavidUllrich, which necessarily involves some fear: theyÍre potent at'ghting for their positions.They also just so happen to ultimately be wrong.Dik Winter keeps up webpages on some of my OLD failed attempts atproving FermatÍs Last Theorem.ThatÍs just bad behavior.Trotting out someone elseÍs mistakes made years ago, and pushing themon your own personal webpages is not something thatÍs defensible.ItÍs clearly making a statement.Dik Winter invites my contempt, and unfortunately, I have to fear theeffectiveness of such tactics.ItÍs sad really.IÍm going to see about dropping that fear as I think it gets in theway of research, but right now I do despise these people becausetheyÍre so contemptible, and so successful in screwing up thedialogue.James Harris === Subject: Re: JSH: Making it personal> IÍm concerned about all the histrionics that has darkened even my own> posts as itÍs all too personal at this point.You can control your side of it. Put up or shut up.> Right now I *despise* posters like Dik Winter, Nora Baron, and David> Ullrich, which necessarily involves some fear: theyÍre potent at> 'ghting for their positions.They have been 'ghting for the truth. WhatÍs *your* position?> They also just so happen to ultimately be wrong.Nope. You are wrong.> Dik Winter keeps up webpages on some of my OLD failed attempts at> proving FermatÍs Last Theorem.So what happens if he replaces them with your NEW failed attempts atproving FLT? Will you be happy then?> ThatÍs just bad behavior.Posting public records is not bad behavior. Posting false theories and§awed mathematics, such as you have consistently done, and launchingpersonal attacks on those who identify your errors *is* bad behavior.> Trotting out someone elseÍs mistakes made years ago, and pushing them> on your own personal webpages is not something thatÍs defensible.DonÍt post mistakes. In a public forum anything you post is subject toreview. And what if he deletes all the mistakes made years ago andreplaces them with mistakes made recently? Is that better?> ItÍs clearly making a statement.Duh!> Dik Winter invites my contempt, and unfortunately, I have to fear the> effectiveness of such tactics.If the tactic you refer to is making your ridiculous claims and stupidmistakes easily accessible, it is worthy of praise. It is *you* who invitecontempt.> ItÍs sad really.Well, IÍd certainly be sad if my ignorance were as easily discovered asyours is.> IÍm going to see about dropping that fear as I think it gets in the> way of research, but right now I do despise these people because> theyÍre so contemptible, and so successful in screwing up the> dialogue.You despise these people because they have identi'ed errors in yourpompous claims, and have refused to back down in the face of yourslanderous gutter language.> James Often in error, but never in doubt. Harris--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: JSH: Making it personalNntp-Posting-Host: apps.cwi.nl > Right now I *despise* posters like Dik Winter, Nora Baron, and David > Ullrich, which necessarily involves some fear: theyÍre potent at > 'ghting for their positions.Because we understand the mathematics perhaps? > They also just so happen to ultimately be wrong.Are we? > Dik Winter keeps up webpages on some of my OLD failed attempts at > proving FermatÍs Last Theorem. > ThatÍs just bad behavior.I have explicitly said *why* I keep them up. As your current way todiscuss is just exactly the same as is succinctly shown on thosepages I see no reason to pull them down. It is your discussions(or actually lack of discussions) that are shown. And your way todiscuss it just bad behaviour, I would say. > Trotting out someone elseÍs mistakes made years ago, and pushing them > on your own personal webpages is not something thatÍs defensible.Well, the kind of mistakes you made back than are similar to the kindof mistakes you make now. Moreover, the way you respond to peoplepointing out the mistakes has also remained the same (although currentlyyou are quite a bit more foul-mouthed). > ItÍs clearly making a statement.Indeed. You are inept at handling contradictions to what you think istrue. As has been shown in the last 6 weeks over a common factorbetween two numbers. I may note that the *existance* of such a commonfactor was already shown much earlier. At least October last year. > Dik Winter invites my contempt, and unfortunately, I have to fear the > effectiveness of such tactics.Oh, well. Do not fear, I will not hurt you. > ItÍs sad really.Yup. > IÍm going to see about dropping that fear as I think it gets in the > way of research, but right now I do despise these people because > theyÍre so contemptible, and so successful in screwing up the > dialogue.I thought you were the screwer. You never verify what is explicitlystated, you only say it is wrong.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: message> IÍm concerned about all the histrionics that has darkened even my own> posts as itÍs all too personal at this point. Right now I *despise* posters like Dik Winter, Nora Baron, and David> Ullrich, which necessarily involves some fear: theyÍre potent at> 'ghting for their positions. They also just so happen to ultimately be wrong. Dik Winter keeps up webpages on some of my OLD failed attempts at> proving FermatÍs Last Theorem. ThatÍs just bad behavior. Trotting out someone elseÍs mistakes made years ago, and pushing them> on your own personal webpages is not something thatÍs defensible. ItÍs clearly making a statement. Dik Winter invites my contempt, and unfortunately, I have to fear the> effectiveness of such tactics. ItÍs sad really. IÍm going to see about dropping that fear as I think it gets in the> way of research, but right now I do despise these people because> theyÍre so contemptible, and so successful in screwing up the> dialogue.> James HarrisIf you make it known youÍre pissed about DikÍs webpage, heÍs probably notgoing to change it. Besides, you exhibit the same bad behavior when you callus names and such.David Moran === about dropping that fear as I think it gets in the> way of research, but right now I do despise these people because> theyÍre so contemptible, and so successful in screwing up the> dialogue. James HarrisMr. Harris,Given your eight years of human-sewage on sci.math, can you please summarizeyour research results for us?Can you please site the number of papers accepted by journals and actuallysite those papers?Do you realize that you are wasting your life?You have called mathematicians names, scolded people who know more than youdo, lied, performed chest beating feats the likes of which no one has everseen and where has all of that led you?Are we to feel pity for you?Are we to feel sorry for you?Do you realize that you are in need of professional help?You have brought upon yourself any and all criticisms only by the path youhave === chosen.DonÍt cry now, eh?Subject: Klingon misconceptions (Re: the anticlassicalist }{ i: linguistic negation) > Klingon was speci'cally created to be the worst language possible by> folks who knew linguistics.Nope. It was created to 't the model of the archetypalKlingons when they were unsullied baddies (those youÍdzap with phasers and photon torpedoes on your universityÍsIBM-370). Klingon was designed to be the worst languageonly in:(1) acoustic phonetics--it had to sound foul(2) articulatory phonetics--handbooks of Klingon advise that spraying the audience with spittle is the right way to pronounce Klingon aspirated stops. Pity Ockrand didnÍt know about the interlabial trills of Malekula!Apart from that, Okrand had a bit of harmless fun picking a word order that few languages follow,and making up a set of fused pronominal verbal pre'xes. But Greenlandic does that on such agrand scale that, compared to it, Klingon is asEsperanto is to Sanskrit. Plus, without even trying, Okrand came up with phonotactics that madeautosegmentation dead easy (good for natural-languageprocessing, that). The authors of Lojban tried hard,and made a mess of it. To sum it up, Klingon is an ideal arti'cial language.It is much easier than... you kn*w (blush, whisper),whilst being just weird enough so that no-one enjoysa great advantage over anyone else when learning it. === Subject: Re: Peer ReviewAnother anecdote: Simon Singh in his book FermatÍs Last Theorem notes what may havebeen the 'rst instance of peer review injustice. Hippasus in theschool of Pythagoras realized that irrational numbers exist. This wasregarded as heresy and he was put to death for it. === Subject: Re: How many ways to put 5 balls into 500 ordered cups?>> Stars and bars proof: .. <> No fair doing peopleÍs homework for them :-) < You are given 500 numbered cups and 've identical balls. Any cup can> hold up to 've balls. How many ways can you put the 've balls into> the 500 cups?Three; you can use your left hand, your right hand, or both. Of course,you could use some other sort of mechanical aid, but that would be aquestion for the engineering groups.-- Joe Bramblett, KD5NRH === Subject: Re: How many ways to put 5 balls into 500 ordered cups?--CELKO-- identical balls. Any cup can> hold up to 've balls. How many ways can you put the 've balls into> the 500 cups?Lets make it more interesting: How about *ten* balls (but each cup can stillonly hold up to 've balls)?-Michael. === Subject: Re: How many ways to put 5 balls into 500 ordered cups?Michael numbered cups and 've identical balls. Any cup>> can hold up to 've balls. How many ways can you put the 've balls>> into the 500 cups?> Lets make it more interesting: How about *ten* balls (but each cup can> still only hold up to 've balls)?> -Michael.> Ah, generating functions. Now youÍve given me something to think about all day. - Tim-- Timothy M. BrauchGraduate StudentDepartment of MathematicsWake Forest University === Subject: Re: How many ways to put 5 balls into support1.mathforum.org (8.11.6/8.11.6/The Math Forum, numbered cups and 've identical balls. Any cup can>hold up to 've balls. How many ways can you put the 've balls into>the 500 cups?As a warm-up, letÍs try smaller sets of ordered cups:Cups Arrangements> =>1 1>2 6>3 21>4 56>5 126>6 252>7 462>8 792>9 1287>10 2002This is not EulerÍs partition function because the cups are ordered;>thus, the two-cup case is:(0, 5)>(5, 0)>(1, 4)>(4, 1)>(2, 3)>(3, 2)My 'rst thought was that if I throw the 'rst ball into the cups, it>can land #1, thru #500; likewise the second, third, fourth and 'fth>balls. So the answer would be (500^5) = 31, 250, 000, 000, 000 ways. >This is wrong, but it sounds good.What IÍd like to get is a function (generator, recurrence relation or>closed form) for this that I can explain to someone relatively easily.I have a brute force solution using an SQL query from hell (my>specialty), which is not quite the same thing as a formula and proof. > I cannot remember enough Combinatorics and Finite Differences to make>the next step. ARRRGH!It is bothering me enough that I will offer a copy of my next book>(TREES & HIERARCHIES IN SQL) as a prize for the best solution.Why not break it into cases?CASE 1: All 5 balls in 1 cup500-Choose-1 gives 500 waysCASE 2: 4 balls in 1 cup and 1 ball in another cupThere are 500-Choose-2 ways to choose which two cups are used.Then there are 2! ways to permute which cup has 4 balls and which cup has 1 ballSo you have 500-Choose-2 * 2!CASE 3: 3 balls in 1 cup and two balls in another cupThere are 500-Choose-2 ways to choose which two cups are used.Then there are 2! ways to permute which cup has 3 balls and which cup has 2 ballsSo you have 500-Choose-2 * 2!CASE 4: 3 balls in 1 cup, 1 ball in another cup, 1 ball in another cupThere are 500-Choose-3 ways to choose which three cups are used.Then there are 3!/2! ways to permute which cup has 3 balls and which 2 cups have 1 ball each.So you have 500-Choose-2 * 3!/2!CASE 5: 2 balls in 1 cup, 2 balls in another cup, 1 ball in another cupThere are 500-Choose-3 ways to choose which three cups are used.Then there are 3!/2! ways to permute which cup has 1 ball and which 2 cups have 2 balls each.So you have 500-Choose-3 * 3!/2!CASE 6: 2 balls in 1 cup, 1 ball in another cup, 1 ball in another cup, 1 ball in another cupThere are 500-Choose-4 ways to choose which four cups are used.Then there are 4!/3! ways to permute which cup has 2 ball and which 3 cups have 1 ball each.So you have 500-Choose-4 * 4!/3!CASE 7: 5 cups with 1 ball eachThere are 500-Choose-5 ways to choose which three cups are used.Then there is 5!/5! = 1 way to permute which cups have 1 ball each.So you have 500-Choose-5SUMMARY:(1) Find the ways to partition the number 5:54 13 23 1 12 2 12 1 1 11 1 1 1 1(2) For a given case, if there are n numbers that add to 5, then 500-Choose-n cups are used. Then there are n!/c! ways to permute which of the n cups have which number of balls, where c is the number of repititions of the same number of balls in a cup.Multiply these two values to get the total for the case.(3) Sum over partitioning 19 balls into 500 cups.One of the ways to partition 19 is4 3 3 3 2 2 1 1Here you are choosing 8 cups, so you have 500-Choose-8.There are 8!/(1! 3! 2! 2!) ways to permute the numbers of balls among the cups.So, for this case, you have 500-Choose-8 * 8!/(1! 3! 2! 2!). === Subject: Re: 15:58:04 -0800, qqquet@mindspring.com (Leroy Quet)>In this post, I write of a speci'c and altered case of the puzzle>mentioned in these previous threads: =3&prev= =4&prev= =1&prev=In the puzzle, we try to write, in order, the positive integers into a>grid, one integer per grid-square, such that:Each integer n is adjacent (above /left of /right of /below)>of the integer (n+1);And each adjacent pair of integers (above /left of /right of /below)>are coprime.And the goal is to completely 'll the grid.>But here I am asking about 'lling an in'nite grid which is bounded>along 2 perpendicular sides.ie. the grid is an entire quadrant of the Cartesian plane, bounded by>the x-axis and the y-axis.I think I found a simple procedure which *might* ensure a successful>'lling of the grid with coprime-adjacent integers.>(Sorry to those on rec.puzzles, but I will give my procedure below.>You can still post your own algorithm, however, or con'rm that mine>can really work for the entire grid without problems.)I illustrate with the 'rst 99 terms:>('gured by hand, so maybe in-error)99>98>97 96 95>54 55 94 93 92>53 56 57 58 91 90 89>52 51 50 59 60 61 88 87>21 22 49 48 47 62 63 86>20 23 24 25 46 45 64 85>19 18 17 26 27 44 65 84>06 07 16 15 28 43 66 83 82 81 80 79>05 08 09 14 29 42 67 68 69 70 71 78>04 03 10 13 30 41 40 39 38 37 72 77 76>01 02 11 12 31 32 33 34 35 36 73 74 75Basically, the path swings clockwise and counterclockwise, running>along the outside of the already-'lled section.>When it gets to either the x-axis or y-axis, it forms a ïpeninsulaÍ,>the length of which is the shortest needed to avoid uncoprime integers>being placed next to each other in the path-section which runs from>that peninsulaÍs axis to the other axis.Now, we do not want a situation where, following the algorith>precisely,>there is NO peninsula-length which would avoid uncoprime neighbors.I am not certain, but I believe this issue is not a problem.Fun perhaps: Show if my algorithm is foolproof...or just foolish. > Well, what I know about number theory and 50 cents will get me a cup> of coffee, but it seems to me that the way to attack this solution> is to show that once the peninsula length exceeds some number N,> you will always have a noncoprime pair when it loops its way back.> It seems to me that this ought to be true. > Of course, that stil wouldnÍt prove that your algorithm canÍt> work, because you might never have to buld a peninsula out that big,> but it would be a start.>(I know I do not give the peninsulaÍs length. Also fun perhaps: try to>determine the shortest length needed for each peninsula, given that>the algorithm never leads to a do not believe you are necessarily right about the 'lling-algorithmworking for all peninsula-lengths above N(k), for the k_th peninsula.For if we lengthen the peninsula by r, each integer following thepeninsula in the path is increased by 2r.But, for p = any ODD prime, and integer n in the path after thelengthened peninsula, n is originally congruent to m (mod p).And after adding (2r), n becomes congruent to (m+2r)(mod p),obviously.But if n was originally next to a multiple of p (so in the originalpath, m was coprime with p), then we get after adding 2r, n + 2r == m+2r (mod p), and(m+2r) is divisible by p for an in'nite number of rÍs, given that pis odd.Any progress by anyone out there, or anything more of interest toadd?...(Yes, yes, I know: It is not how long your Quet === Subject: Re: What is a number?/What is not a number? <4030da8f$12$fuzhry+tra$mr2ice@news.patriot.net> tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Grif'th X-Treme: C&C,DWSIn have three ways to proceed:More than that.>1) revamp the de'nition above to remove polynomials,>2) accept polynomials as numbers, or>3) de'ne a number set as any mathematical set at all.4) De'ne a number as any element of an algebraic structure5) DonÍt use the term number at all, except as a constituent of compound terms, e.g., real number.There are, of course, yet other options.My prefered option is 5.-- 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 === Subject: Re: What is a number?/What is not a number?Nntp-Posting-Host: apps.cwi.nl... > The question is, What is it we really mean by the label number?The answer is, there is no clear-cut de'nition in use. > There are two ways to de'ne the set of number sets: One way is to > simply list all number sets, the other is to give a rule (given by a > de'nition) that takes the place of the listing of number sets. My > question is, Is there a de'nition that provides a rule that creates > the same set of number sets as there are in the set of number sets > created as a listing by 'at?Again, the term number is erratic. Rarely do mathematicians use theterm number without quali'cation. And if they do it is (or shouldbe) in a well understood context. > So, we have the set of number sets S = {N, Z, Rationals, Reals, > Complex, Cayley, ...}. So, what de'nition of number gives us the > set S?All elements of sets obtained by the standard Cayley-Dickson doublingfrom the real numbers. This will also give quaternions, sedenions,etc.. But I would think that most would consider high order numbersin this sequence not really numbers. And it excludes the cardinalsand ordinals.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Integrals without explicit formulasI am in a differential calculus class. Just about all the problemsinvolve integrals, but I have been informed that some of the integralsin these problems have no explicit formulas. Can you give me any rulesof thumb so I know what kind of integrals to not bother trying tointegrate? === Subject: Re: Integrals without explicit formulas about all the problems> involve integrals, but I have been informed that some of the integrals> in these problems have no explicit formulas. Can you give me any rules> of thumb so I know what kind of integrals to not bother trying to> integrate?First of all, why are you integrating in a differential calculus class? Thatreally doesnÍt make sense to me.Whenever I integrate, I look at what IÍm working with. Polynomials, trigfunctions, log and exponential functions, and rational functions (though notalways) are usually pretty straight forward. Rational functions can getmessy, especially if youÍre working with partial fractions. Most functionsinvolving radicals will lead into trigonometric substitution which will leadto a trigonometric integral. Some functions (i.e. sin(x^2)) will requirepower series and for others that you canÍt integrate, you can use a RiemannSum. IÍd look up the Trapezoidal Rule, as well as breaking the area intosubintervals and then 'nd the area of each rectangular subinterval.David Moran === Subject: Squares Of Certain Areas: puzzleYou have a FIXED sequence of distinct positive integers, {k}, which isto be determined.For all kÍs <= EACH positive integer m, we can take a number n (n <= m) of (geometric) squares,each of *integer* side-length,where the j_th square is of the largest squarewith an integer area that is <= m/(k_j).But {k} is such that, if we lined the squares up vertically so thattheir horizontally-running edges coincide and they do not overlap,the length from the upper side of the top square to the lower side ofthe bottom square will ALWAYS be m.2 questions: 1) Which integers make up {k_j}?2) Give a formula for the total area of the (n number of) squares.Example:For m = 4,We have 1, 2, 3 in {k}, but not 4.So, we have the squares, as lined up: ------! ! ! ------ k=1, area = 4 <= 4/1! ! ! ==----! ! k=2, area = 1 <= 4/2 ==! ! k=3, area = 1 <= 4/3 --Height of the stack is m = 4. Area is === Subject: If We Replaced Each Prime With -1...Let c(k) = the sum of the exponents in the prime factorization of k.So, if we exchanged each prime in the prime factorization of k with(-1),we would get (-1)^c(k).What I am wondering, however, iswhat is C(m) =sum{k=1 to m} (-1)^c(k)asymptotical towards?C(m) also equals:sum{k=1 to m} §oor(sqrt(m/k)) *mu(k),where mu() is the Mobius function(or Moebius function).So, I 'rst thought that sqrt(m) had something to do with theasymptotics of C(m).But the difference in the sum before and after taking the §oor is toosigni'cant for me to claim anything.I might as well write the above equations in ascii-art mode for thoseexpecting that:;)C(m) = m--- c(k)/ (-1) = ---k=1 m--- _____ | / |/ | V m/k | *mu(k)--- === Subject: Coprime Grid / Increasingly-Sized StepsHere is a question which combines 2 earlier puzzles of mine into one.As =12&prev=we attempt to: Start with an in'nite (in every direction) rectangular grid/lattice.Put 1 in a square of the grid (or at a vertex of the lattice).Place 2, 3, 4, 5, ....in'nity, by some algorithm, into thegrid/lattice so that each integer (1+m) is exactly m squares (in onlydirections of either up,down,left,or right) from the square/vertexwith the integer m, for all mÍs.And only one, no more/ no fewer, integer per vertex/square.But...as =the integers are to be placed so that EVERY pairof adjacent squares contains two integers which are COPRIME with eachother...(Adjacent is as de'ned as immediately next to in the directions ofeither up, down, left, or right.)A few things:I am wondering if it can be proved it is possible to eventually placeall the integers, using the restictions above, into the in'niteunbounded grid with exactly one integer per grid-square.My instinct is that there are ways to 'll the in'nite gridsuccessfully.And if so, what would the ïmost compactÍ 'lling be?By ïmost compactÍ, I mean:which placement of integers would minimize:limit{n -> oo}(area of convex hull of 'rst n integersÍ positions)/n? (I could also ask this question about the less speci'c in'nitevariations of the 2 Quet === Subject: Advice on future with MathIÍm an upperclass math major that was/is planning on attempting a masters orphd in math. Here lately, IÍve wondered if this is a great idea. My 'rst question is this: How much should you study (reading the material,working problems) for, say, an abstract algebra or advanced calculusjunior/senior level class? I talked to a graduate student that said he studiedfor a couple of his 'rst year grad classes at least 4 hours a day on weekdaysand about 8 hours on Saturday and Sunday. Needless to say, I do nothhing likethat. Should I be doing that or is that just an insane amount of time to bespending on it? It seems that if youÍre good at something, it shouldnÍtrequire so much work.At times I have dif'culty understanding things that, after 'nding out where Iwent wrong, seem pretty simple. I assume others donÍt have this problembecause normally they can respond right away with what some math concept meant or, less often, help me where I am having a problem understanding part of aproof. Is it normal to not follow a proof or just a sign that IÍm just weakmathematically (weÍre again talking junior/senior level advanced calc/abstractalgebra here)?I have a bad tendency to be hot/cold when it comes to math. It seems like IÍllhave the highest/near highest score on an exam and then really screw up on thenext exam, getting something in the bottom half of the grades. I wish that Icould attribute it to something like when I 'rst started school: I would dono work until after I screwed up on the 'rst exam, and then work harder afterthat and do quite well. Now it seems that IÍll do well on the 'rst exam andthen screw up on the next one. I really donÍt think that I will have studiedmuch less for the second one but I guess it could be a possibility. When I dobad on the other exam, I begin to wonder if I just got lucky on the problemsthat were given on the good exam and would have done lousy if a few differentproblems were picked.Other times, I wonder if I am just not good at proofs, thus making me about thecrappiest grad school candidite out there. I sit and stare at the homeworkproblems and often canÍt get anywhere. After I see an answer, it doesnÍt looktoo bad but it just seems that theyÍre impossible at times. I had a bad proofsclass that was designed for just about anyone wanting to take it, so it wasreally easy. IÍm not sure if I can use that excuse at this point though. Mypre-calculus education was really bad since I was from a really poor school (Ioften have no clue what certain things that were assumed as ïknowledgeÍ fromhigh school) but again this doesnÍt seem to matter too much in the proof-basedclasses.Well, that was kind of long-winded. I hope that I got the point across. Anyadvice would be helpful. === suggestions to books on game theory. Not game theory in thesense of economics at all, but a book that deals with actual games, andconcrete mathematics. I hear there is a good book in the MAA library, but do not know the name of it.book level should be at the upper undergraduate / early graduate level in termsof dif'culty but should also be accessible to high school students with aproblem solving background.All === Subject: Re: Book on MATH game theoryI just happened to be on amazon when you asked and I tried MAA and gametheory and received:Game Theory and Strategyby Philip D. Jr Straf'nhttp://www.amazon.com/exec/obidos/tg/detail/-/ 0883856379/qid=1077161183/sr=1-2/ref=sr_1_2/102-3578915- 2424159?v=glance&s=booksIÍm not sure if thatÍs the dif'culty level youÍre looking for but it soundslike your description otherwise. === Kurtz for attempting to help me 'nd out what myteaching philosophy is. Admirable as LynnÍs composition is, Ihave two problems with it:(1) It is LynnÍs point of view (maybe) but not mine.(2) It is actually a contract of sorts, containing a number of promises, and a prognosis of sorts for the results of carrying out those promises. It is not philosophy.Of course, I really have no way of knowing whether, when a hiring committeeasks for philosophy, it really means philosophy. ThatÍs part of the question.It has also been pointed out that I have not stated in other terms whatmy philosophy of teaching is. And one can correctly conclude thatI am not submitting something here for revision. But IÍve been postingto sci.math for a long time, including on topics related to education,and I was hoping that people who might have some familiarity with mypoint of view might also be able to explain the interface between thepositions IÍve taken and what the anonymous hiring committees mightbe asking for.It has also been suggested (privately) that I describe my teachingexperiences or some of the innovative things I have tried. But I thinkthat falls under the category of gimmicks or technique, not philosophyper se.I have no lack of things to tell a hiring committee. I just have noidea what they are asking for, nor how much of it.Ignorantly,Allan Adlerara@zurich.ai.mit.edu************************************ ***************************************** ** Disclaimer: I am a guest and *not* a member of the MIT Arti'cial ** Intelligence Lab. My actions and comments do not re§ect ** in any way on MIT. Moreover, I am nowhere near the Boston ** metropolitan area. ** ************************************************************** *************** === F1vU=Jm9LchOj|&)y/yÍn33$K{,'|+j:PBvlXn]+u&LG@{ zY!w}Yp)i57ga,S] B3`Xj1LZGa)d5?ma~E_^>yu_xS(E:wqD.(nw&)SmG)LQK:H;^S;&*0!.uuW+ 59UwyslR=xGd RtodT9/H-B.I_>&f9CXI;dty3[...snipped...]> Of course, I really have no way of knowing whether, when a hiring committee> asks for philosophy, it really means philosophy. ThatÍs part of the question.> My impression is that teaching philosophy really means teachingstatement. Does that help?> It has also been pointed out that I have not stated in other terms what> my philosophy of teaching is. And one can correctly conclude that> I am not submitting something here for revision. But IÍve been posting> to sci.math for a long time, including on topics related to education,> and I was hoping that people who might have some familiarity with my> point of view might also be able to explain the interface between the> positions IÍve taken and what the anonymous hiring committees might> be asking for.> It has also been suggested (privately) that I describe my teaching> experiences or some of the innovative things I have tried. But I think> that falls under the category of gimmicks or technique, not philosophy> per se.> Well, thatÍs probably what they want. A good resource ishttp://www.ams.org/employment/academic-job-search.htmlIn it, Tom Rishel, a man whose advice you should take seriously, says,First, because itÍs easier to write a statement than a philosophy,letÍs start with I began teaching in 1993 rather than Teaching islove and caring. .....Start with concrete details of your teachingexperience. What courses have you TAÍd or taught and at what levels? How do you conduct a typical class? Do you have a particularassignment or class that pleased you and that you wish to describe? Was there a classroom situation that taught you something meaningfulabout the way you teach and the way students react to your methods?Have you checked this resource out before? If not, IÍm resources. === Subject: RudinThis is a pain-in-the-rear question as it involves looking somethingup in a book. In RudinÍs Principles 3rd. ed. IÍm having a problemunderstanding the inequality near the top of p 305 (section onLebesque Theory, proof that u*(A)= u(A) for all A in E, E being thefamily of all elementary subsets of R^p). Speci'cally I canÍt seethe reason forSum(n=1 to N) u(A_n) <= Sum (n=1 to oo) u(A_n) + epsilon. All help much appreciated. === Subject: Re: oil of bitter almonds: Woehler and LiebigIÍve made a little progress 'guring out what Woehler and Liebig aretalking about. First of all, they do a combustion analysis of two samplesof oil of bitter almonds, namelyI. 0.386 gramme = 1.109 carbonic acid, and 0.200 water.II. 0.341 = 0.982 0.175 which are fairly close to what I get from their data: I IICarbon ........... 79.438 .......... 79.603Hydrogen ......... 5.756 .......... 5.734Oxygen ......... 14.808 .......... 14.663carbon, hydrogen and oxygen or, what is the same, the ratio 14:12:2, asWoehler and Liebig do. How exactly they decided on 14 instead of 7 isnÍtclear to me, but given that they did, the next table: 14 atoms of carbon 1070.118 79.56 12 hydrogen 74.877 5.56 2 oxygen 200.000 14.88 -------- ------ 1344.995 100.00becomes perfectly clear. For example, if one uses BerzeliusÍ scale,wherein the weight of oxygen is taken to be 100, the weight of carbonis about 76.44 and that is close to what one gets by dividing 1070.118by 14. So, the column on the left is simply their calculation of theweights of 14 carbon atoms, of 12 hydrogen atoms and of 2 oxygen atoms.Throughout the rest of the paper, they use their own scale in which theweight of oxygen is taken to be 10, and whenever they have 14 carbonsafter an analysis, they give the weight as 107.0118, except for one placethat is probably a typo. So, that explains the column on the left.As for the column on the right, of one expresses these weights as percentagesof the total weight 1344.995, one gets the column on the right, more or less.So, now I have an overview of the kind of computations they are presenting.What IÍm still confused about are their conventions for working with theirnumbers and for dealing with questions of precision, etc. Note that theytake all their numbers to three decimal places, even after successivecomputations. Some of their results suggest that they simply truncatedtheir numbers, no matter what came afterwards (I think MFÍs lab instructorwould be pleased). But I donÍt have it all nailed down.If, instead of 1.109 for the weight of carbon dioxide, the value was really1.10891, and they used that instead of the rounded value 1.109 in theiractual computations, and if the weight of the oil of bitter almonds wasexactly 0.386, and if one used their value (according to BenfeyÍs footnote)of 7.644 for the atomic weight of carbon and 10 for that of oxygen, then onewould get their value of 79.438 for carbon in the 'rst column of thecombustion analysis. I havenÍt tried to 'nd out all possible modi'cationsbut note also that 1.1089 and 1.10892 wonÍt work to do that. So, if that iswhat they did, it would imply that they had a method of weighing thathad a precision of 5 decimal places. Did they?So, after all this math, I have a question about the chemical laboratoryof Woehler and Liebig: how exactly did they weigh things and what precisioncould they achieve with their apparatus?I am cross posting this to sci.math, since someone there might know somethingabout the history of numerical methods in the time of Woehler and Liebigand have some idea of what conventions they might have used in dealingwith numbers. If someone can explain how to get EXACTLY the numbersWoehler and Liebig claim to have derived from their initial combustionmeasurements that would be very helpful.Ignorantly,Allan Adlerara@zurich.ai.mit.edu************************************ ***************************************** ** Disclaimer: I am a guest and *not* a member of the MIT Arti'cial ** Intelligence Lab. My actions and comments do not re§ect ** in any way on MIT. Moreover, I am nowhere near the Boston ** metropolitan area. ** ************************************************************** ***************