mm-1021 === Subject: Re: Sets That Resemble Derivatives Somewhat by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1JDEKP11801; >Define: >1) (A-->B) = (AB) = A U B >for any two sets A, B, where Ô is complement and Adjacent letters >are intersected. This is the set analog of logical a-->b for a, >b propositions, the latter defined as ~(a ^ ~b) = ~a V b with Before replying to the other comments on my previous posting, which I prefer to do tomorrow morning (night debates arent my favorite way of going to sleep at age 65, whether positive or negative), I would like to point out some things about complements. 1) A U A = Universe This says intuitively that a set (A) and/or its complement (the part of the Universe outside the set) comprises the Universe. More specifically, it comprises the Mathematical Universe, not necessar- ily the Physical Universe. If we translated the idea behind this into Real or Complex Analysis, or even Tensor Analysis and so on, we would get something like: 1) A variable and/or a function and its change constitute Universe of study or Knowledge about the variable/function. This is what Garrett Birkhoff of Harvard meant by saying that Differential Equations contain causation/causality. Of course, in causation/causality the change is usually with respect to time, but even in spatial change (partial derivatives in general) it takes time in a sense to scan different spatial regions - even on a light wave, where time is supposed to be instantaneous, some temporal notion of the order of attention to different places can be introduced, since for one thing simultaneous attention to many different places contains some paradoxes depending on how intense the attention is. Ill mention one more aspect: 2) (A) = A This doesnt correspond to differentiation, but I \ dont think it is the most critical feature of differentiation. It is more like d^2 in exterior algebra. Intuitively speaking, when A is a set and Ô is complement, we could translate this as: 2) The change in a change of A is the change back to A. We would, I admit, sacrifice second order derivatives and higher in this scenario, but we already know that higher order equations can often be expressed as systems of first order equations. I think that the advantages outweigh the orders so to speak, and we may eventually find a generalization of the usual \ derivative order in some other ideas. I especially like the fact that the Riccati Differential Equation (with of course its Algebraic version) which underlies Growth-Expansion-Contraction processes and events (e.g., in biology - see the Logistic and Simple Positive or Neg- ative Exponential Differential Equations which are special cases of the Riccati Equation, and the generalization to a Bernoulli equation) is first order. Quantitative Biology is a new and very valuable branch of Arxiv.org. Ill try to discuss the differences between Curvilinear One-Direction-At-A-Time motion and Simultaneous Many-Directions-At-A-Time Growth-Expansion-Contraction Motion (including radiation, expansion of the Physical Universe as a whole, etc. - possibly with different rates and amounts in different directions, relating to (optimal) control also) later. Osher Doctorow === Subject: Re: errors in an argument by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1JDELd11866; >> As far as I know the first ßying creatures they talk \ about are >> dinosaurs, no less. And quite big ones as well. >Thats a comment on your knowledge. Some dinosaurs were small; a quick >search turned up some with a mass below 10 kg. And all the references I >found talk about small dinosaurs as the starting point for the evolution >of wings. >-- >Daniel W. Johnson >panoptes@iquest.net >http:// members.iquest.net/~pano ptes/> The generator matrix of a systematic linear block code has the >> form G = [Ik : P]. How can it be shown that the parity check >> matrix is of the form H = [-P^T : In-k]? > With great ease. > If a vector (a : b) is supposed orthogonal to the code generated > by G then (a : b)G^t = 0 so a + b P^t = 0 or a = -b P^t etc. > Are you in the IEEE? Yes. Why do you ask? -- % Randy Yates % Ticket to the moon, ßight leaves here today %% Fuquay-Varina, NC % from Satellite 2 %%% 919-577-9882 % ÔTicket To The Moon %%%% % *Time*, Electric Light Orchestra http://home.earthlink.net/~yatescr === Subject: Are the derivatives of abs[(x-a)^3] different for x>a and xa and x Often, they skip > steps or say somethign is obvious when it isnt (unless \ you know som > obscure theorem). There is an old story about a lecturing professor who claimed It is obvious that ... and a student challenged him, saying, It is not obvious. Then the professor paced back and forth for ten minutes, thinking. Finally he announced, It IS obvious, and he resumed his lecture. There is a corresponding story in music. Nadia Boulanger assigned a harmonization exercise to a student, from a book. She looked at his result for several minutes, finally said, The alto part is not very interesting, and handed it back to the student. David Ames === Subject: Re: Graph Theory Textbook Is the Springer > GTM Graph Theory by Reinhard Diestel any good? Any suggestions are > highly welcomed. will help you) so that you can make up your mind by yourself. -- Julien Santini === Subject: Re: Teaching philosophy teaching philosophy. I think that gives me a good idea of the genre. What about the length of the statement? Teaching Philosophy: Les ßeurs qui pousse de la merde by Allan Adler Evil is the root of all money. Hence if funding has been found for a position, it necessarily cometh of evil. Yet, though mushrooms nourish themselves on excrement, they in turn nourish people and, in like manner, our fate as educators is to be a kind of mushroom. Let me therefore describe some recipes in which this mushroom has been used. [details omitted for the purpose of this posting] (I know, cut the first paragraph...) Allan Adler ara@zurich.ai.mit.edu ************************************************************* *************** * * * Disclaimer: I am a guest and *not* a member of the MIT Artificial * * Intelligence Lab. My actions and comments do not reßect * * in any way on MIT. Moreover, I am nowhere near the Boston * * metropolitan area. * * * ************************************************************* *************** === Subject: Re: Are the derivatives of abs[(x-a)^3] different for x>a and x given abs[(x-a)^3], where abs[ ] means taking the absolute value, and > a is a constant, are the first, second, third derivatives (with > respect to x) different for x>a and x I forgot how to write down the derivative of an absolute function in a > formal way, I mean, not using if else, but using something like > sign(x-a). The usual method is to define sgn(x) in terms of the heaviside function H(x) = { 0, x < 0; 1, x > 0; 0.5, x = 0 }, and write formally dH/dx = delta(x) where delta(x) = { 0, x != 0 } is the Dirac delta Ôfunction (integral of delta(x) over any \ interval containing 0 is 1 by definition, and 1/2 if 0 is an endpoint). sgn(f(x)) = 2H(f(x)) - 1 |f(x)| = f*sgn(f) d(sgn(f(x))/dx = 2*delta(f(x))*f(x) d(sgn(f)f)/dx = sgn(f)df/dx + 2*f*f*delta(f) Terms involving delta(f) vanish if one avoids points where |f| = 0. > I think the first derivative would look like: > 3(x-a)^2 * sign(x-a) Yes. > But then I dont know how to proceed to the second derivative. You have to exclude points where |f| = 0 from the domain to proceed further; d/dx(delta(x)) is not well-defined. D(sgn(f)Df) = sgn(f)(D^2)f + (Df)*(D(sgn(f))) (D = d/dx) = sgn(f) d^2f/dx^2 if |f| != 0. Similarly, D^n(sgn(f)f) = sgn(f)*(D^n)f if |f| != 0. -- P.A.C. Smith The vast majority of Iraqis want to live in a peaceful, free world. And we will find these people and we will bring them to justice. === Subject: Free Interactive Proof Writing Tutorial Students, My new proof writing program, DC Proof 1.0, can help you learn the basics of logic and mathematical proof. Included with the free download is a self-study tutorial with exercises (plus hints and full solutions). It serves as both an introduction to writing proofs and to all the main features of the DC Proof program itself. It is aimed at the non-specialist undergraduate and advanced high school student. You can also participate in a forum for DC Proof users, hosted by Yahoo Groups. And you can have your proofs published at the DC Proof Online website! No need to worry about making mistakes -- DC Proof checks every line of your proof as you enter it. Visit DC Proof Online now for your FREE download at: http://www.dcproof.com System requirements: Win95 or later, IE4 or later Also available at the Tucows download site: http://www.tucows.com Dan Christensen Toronto, Canada === Subject: Can anyone assist in this geometry problem? Ill denote Ô=derivative and * usual \ multiplication on R. 1. let q(t) be a regular curve with | q| =a where a is a \ fixed positive constant. Show that if s is arc length measured from some point, then t=(s/a) + c for some constant c. Proof: So suppose we measure from a point t0. Then: By definition s(t)=int(| q`| dt, t0,t )=int( a dt , t0,t) = at - at0 =s so at=s+at0 hence t=(s/a)+t0 = (s/a) + C where C=t0 Is this correct? I`m not sure if the constant C must depend on the same point we are measuring from. 2. Let a(s) be a unit speed curve with curvature k!=0 and torsion t!=0, prove that a(s) lies on a sphere if and only if t/k= (t /(t*k^2)) The hint of the book is to combine the following two results (already proved in the book) : I. Let a(s) be a unit speed curve with curvature k!=0 and torsion t!=0. Assume that (1/k)^2 + ( (1/k)` * (1/t) )^2= constant=a^2 where a>0, then the image of a lies on a sphere of radius a. II. Let a(s) be a unit speed curve whose image lies on a sphere of radius m and center m. If torsion t!=0 then a - m = -(1/k)N - (1/k)` * (1/t) * B where B=binormal vector. I have been playing with this conditions to prove 2. without any progress, I dont see how to combine them. === Subject: re:Can anyone assist in this geometry problem? To solve your problem note that a(s) lies on a sphere if and only if (a-x,a-x) is constant for some x. Differentiate this and use the Frenet equations to show that this is equivalent to the required conditions. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Generating complete rows in Pascals Triangle independent of previous rows Im sure Im not the first one to \ notice this, but I didnt find programmer, not a math major, so please forgive my trampling on any formal naming/notation conventions. ) To generate a row m of Pascals triangle independent of any other row, do the following: set P[m,0] = 1 Now, for all P[m,n] for n=1 to m (really ~m/2 + 1 since the second half is just the first reversed, though it will generate properly) : P[m,n]= ((m -(n-1))/n) * P[m,n-1]) For large values of m youd need to make sure \ youve got enough precision decimal points in the division to make it accurate. Usually 1.5 * #_of_digits_in_P[m,n-1] is enough, but most math packages you can specify a default amount of precision and setting that to 500 wont hurt. This doesnt appear to trivially follow from anything \ Ive seen re: binomial coeffiecents, but again Im not a math \ major so there may be something Ive missed that would make it trivial. It also doesnt pick out a specific entry like \ P(m,n) = m!/( n! * (m-n)! ) will, but you can generate the complete row for ms > 20000 a lot faster than you can compute 3 factorials, multiply and then divide 20000 times. For example, Using a horribly inefficent math library I was able in 3 seconds to produce the 90th entry of the 351898290th row (its : 10129857767674636747591715322426976439568292354720284879396795 73326790098524 80075265177967090011964251748906315861712415089788916048209633 332384995422798 65917300632822299942582978665864465063715053835347823698068931 645644817749132 32139981420963935335207734993136460826018379533733273411586009 388781254888170 24634705976954989280552613876621873325170455478235898314757129 276595594340040 6405383138372010 16426057126596962645028630251951568269184930176854726542871052 80461646277571 18655495787218562152636183884631802187831999095036973236309337 450255527461872 29999578781739447814725368108971602158598821201398567827365924 791379238942737 1). This would be so expensive computationally using the factorial method as to render it useless. I hope this post was at least somewhat informative / useful / interesting and not redundant. Tom === Subject: Re: Generating complete rows in Pascals Triangle independent of previous rows > Im sure Im not the first one \ to notice this, but I didnt find > programmer, not a math major, so please forgive my trampling on any > formal naming/notation conventions. ) > To generate a row m of Pascals triangle independent of any other row, > do the following: in base 10: 11^0 = 1 11^1 = 11 11^2 = 121 11^3 = 1331 11^4 = 14641 101^5 = 010510100501 101^6 = 01061520150601 ... in base 2 101^11 = 01 11 11 01 (5^3=125) 1001^100 = 001 100 110 100 001 (9^4=6561) 10001^101 = 0001 0101 1010 1010 0101 0001 (17^5=1419857) works in any fixed radix base. Russell - 2 many 2 count === Subject: Re: Generating complete rows in Pascals Triangle independent of previous rows > To generate a row m of Pascals triangle independent of any other row, > do the following: > in base 10: > 11^0 = 1 > 11^1 = 11 > 11^2 = 121 > 11^3 = 1331 > 11^4 = 14641 > 101^5 = 010510100501 > 101^6 = 01061520150601 > ... > in base 2 > 101^11 = 01 11 11 01 (5^3=125) > 1001^100 = 001 100 110 100 001 (9^4=6561) > 10001^101 = 0001 0101 1010 1010 0101 0001 (17^5=1419857) > works in any fixed radix base. To add to this: If you assume multiplication is constant-time, you can calculate the entire nth row in O(log(n)) time as x^y requires O(log(y)) multiplications. This has been used in a proof that you can factorize any composite number in O(log(n)) time. Obviously, the premise is wrong: You cant do arbitray multiplication in constant time. Still, it is a fairly fast way of computing an entire row if you need it, but it is not a good way of computing a single entry in the triangle. Torben === Subject: Re: Generating complete rows in Pascals Triangle independent of previous rows > Im sure Im not the first one \ to notice this.... Well, Pascal noticed it, and Newton used it in his explanation. > .... > To generate a row m of Pascals triangle independent of any other row, > do the following: > set P[m,0] = 1 > Now, for all P[m,n] for n=1 to m (really ~m/2 + 1 since the second > half is just the first reversed, though it will generate properly) : > P[m,n]= ((m -(n-1))/n) * P[m,n-1]).... Arab and Chinese mathematicians generated the triangle recursively from the top. Pascal (1665) considered how to jump into the middle and find any of the binomial coefficients directly, \ without climbing all the way down from the top. The resulting formula is in text-books about combinations and the binomial theorem. A library might help you more than Google. :-) Ken Pledger. === Subject: Re: Generating complete rows in Pascals Triangle independent of previous rows > Im sure Im not the first one \ to notice this, but I didnt find > programmer, not a math major, so please forgive my trampling on any > formal naming/notation conventions. ) > To generate a row m of Pascals triangle independent of any other row, > do the following: > set P[m,0] = 1 > Now, for all P[m,n] for n=1 to m (really ~m/2 + 1 since the second > half is just the first reversed, though it will generate properly) : > P[m,n]= ((m -(n-1))/n) * P[m,n-1]) > For large values of m youd need to make sure \ youve got enough > precision decimal points in the division to make it accurate. Usually > 1.5 * #_of_digits_in_P[m,n-1] is enough, but most math packages you > can specify a default amount of precision and setting that to 500 > wont hurt. > This doesnt appear to trivially follow from anything \ Ive seen re: > binomial coeffiecents, but again Im not a \ math major so there may be > something Ive missed that would make it trivial. > It also doesnt pick out a specific entry like \ P(m,n) = m!/( n! * > (m-n)! ) will, but you can generate the complete row for ms > 20000 > a lot faster than you can compute 3 factorials, multiply and then > divide 20000 times. > For example, Using a horribly inefficent math library I was able in 3 > seconds to produce the 90th entry of the 351898290th row (its : > 10129857767674636747591715322426976439568292354720284879396795 733267900985248 0 > 07526517796709001196425174890631586171241508978891604820963333 238499542279865 9 > 17300632822299942582978665864465063715053835347823698068931645 644817749132321 3 > 99814209639353352077349931364608260183795337332734115860093887 812548881702463 4 > 70597695498928055261387662187332517045547823589831475712927659 559434004064053 8 > 3138372010 > 16426057126596962645028630251951568269184930176854726542871052 804616462775711 8 > 65549578721856215263618388463180218783199909503697323630933745 025552746187229 9 > 99578781739447814725368108971602158598821201398567827365924791 3792389427371). > This would be so expensive computationally using the factorial method > as to render it useless. No kidding. Thats why most people would take advantage of the formula, C(n, r) = product of the numbers from n down to n - r + 1, divided by product of the numbers from 1 up to r. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Generating complete rows in Pascals Triangle independent of previous rows > For example, Using a horribly inefficent math library I was able in 3 > seconds to produce the 90th entry of the 351898290th row (its : > 10129857767674636747591715322426976439568292354720284879396795 733267900985248 00752651779670900119642517489063158617124150897889160482096333 323849954227986 59173006328222999425829786658644650637150538353478236980689316 456448177491323 21399814209639353352077349931364608260183795337332734115860093 887812548881702 46347059769549892805526138766218733251704554782358983147571292 765955943400406 405383138372010 > 16426057126596962645028630251951568269184930176854726542871052 804616462775711 86554957872185621526361838846318021878319990950369732363093374 502555274618722 99995787817394478147253681089716021585988212013985678273659247 913792389427371 ). Are you sure about the result ? I get the follwoing different result 10129857767674636747591715322426976439568292354720284879396795 73326790098524 800 75265177967090011964251748906315861712415089788916048209633332 38499542279865 917 30063282229994258297866586446506371505383534782369806893164564 48177491323213 998 14209639353352077349931364608260183795337332734115860093887812 54888170246347 059 76954989280552613876621873325170455478235898314757129276595594 34004064053831 383 72010470500701236152580418596762906826650952815296177528484615 79047524805240 534 87757097042902362513701453040655427058407434590707201196437008 76653209133355 448 74094541573509257343053578961311171507812371178006744366663608 53317352100792 520 with the small Prolog program comb(N,M,Res) :- comb(N,M,1,1,Res). comb(N,M,D,I,O) :- (M =:= 0 -> O = I ; NN is N - 1, MM is M - 1, DD is D + 1, II is (I*N)//D, comb(NN,MM,DD,II,O) ). and query ?- comb(351898290,90,X). But maybe my implementation is wrong ? Bart Demoen === Subject: Re: a little question of non-math ... >By the 1950s and 1960s corporal punishment >in US public schools was still widely used but was not >routine for minor infractions. In the schools I was >in (in Alabama, Virginia, Florida, New Mexico, and >Arizona) between 1953 and 1964, there were perhaps >one or two cases per year of physical punishment of >students known to me. By the late 1960s and early >1970s corproal punishment in the schools had pretty >nearly vanished. It is unlikely to come back. But strangely enough, the practice of capital punishment is still widespread in the USA. -- Jeremy Boden === Subject: Re: a little question of non-math > On a related note, I fail to see how corporal punishment could have any > positive effect on academics, the putative object of school. If a small child touches the hot stove and burns his fingers, he remembers it VERY well, and doesnt do it again. Because of the way our brains are wired, pain is the quickest and surest route to learning. So since our society decided that using pain for teaching is cruel and unacceptible, we have to rely on less quick and less sure methods. [There is some anecdote I am trying to remember, maybe Erasmus and the Salamander, or something... Anyone know what I mean?] -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Autocorrelation function for Brown noise What is the autocorrelation function for brown noise? Is it only expressible in terms of the defining integral, or is said integral DG === Subject: Nice problem There are n red points and n blue points on the plane. Prove that there are n non-intersecting segments each having one red and one blue endpoint. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: Nice problem > There are n red points and n blue points on the plane. > Prove that there are n non-intersecting segments each having one red > and one blue endpoint. > ---------------------------------------------------------- > ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** > ---------------------------------------------------------- > http://www.usenet.com As stated, it is false. If n > 1 , all the points are on a line in the plane, and there is a point on that line separating the reds from the blues, then that one point must be on any line between any red and any blue. Some additional condition on the location of the points is necessary. Did you mean to require in addition that, say, no three of the points are colinear? === Subject: Re: Nice problem > There are n red points and n blue points on the plane. > Prove that there are n non-intersecting segments each having one red > and one blue endpoint. How about the case where all 2n points are collinear? For example the red points are at (0,1), (0,2), & (0,3) and the blues are at (0,4), (0,5), & (0,6). tom -- We have discovered a therapy ( NOT a cure ) for the common cold. Play tuba for an hour. === Subject: Re: Nice problem > There are n red points and n blue points on the plane. > Prove that there are n non-intersecting segments each having one red > and one blue endpoint. > How about the case where all 2n points are collinear? > For example the red points are at (0,1), (0,2), & (0,3) > and the blues are at (0,4), (0,5), & (0,6). 1979 Putnam problem A4. There the condition was given that no three points be collinear. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Complex Integration Im currently studying complex analysis and I have been \ given as a problem to find the value of a complex integral (over a contour). The complex function is defined as f(x+iy)=U+iV where U(x,y), V(x,y) are two-dimensional real functions. I noticed that dU/dx = dV/dy how can I use this to help me find the solution. TIA === Subject: Re: system of exponential equations > x^y=a, y^x=a+1 = x=a^(1/y), y^x=a+1,= x=a^(1/y), y^(a^(1/y))=a+1,= x=a^(1/y), y^a=(a+1)^y. Sorry, Doesnt work. The following step is wrong: y^(a^(1/y))=a+1 => y^a=(a+1)^y, It appears that equation: y^(a^(1/y))=a+1 cannot be solved by the LambertW. Pity. -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: Re: system of exponential equations > Sorry, Doesnt work. The following step is wrong: > y^(a^(1/y))=a+1 => y^a=(a+1)^y, > It appears that equation: y^(a^(1/y))=a+1 cannot be solved by the LambertW. > Pity. Amanda === Subject: Re: system of exponential equations \.91Ä Amanda \.91[CapitalEth]\.91ñ\.93»[EDouble Dot]±\.93Ì\.91[Micro] \.93¡\.93.b3\.91Ë \.91.b9\.91¬\.91ü\.93á[EDo ubleDot].b9\.91± > Sorry, Doesnt work. The following step is wrong: > y^(a^(1/y))=a+1 => y^a=(a+1)^y, > It appears that equation: y^(a^(1/y))=a+1 cannot be solved by the LambertW. > Pity. solution. > Amanda Whats damaging the analytical solution is the damn +1 factor. Had it been: x^y=a, y^x=a, it would be easier to handle. In particular it would reduce to x^(1/x)=y^(1/y), (or equivalently to x^y=y^x) for which there are extensive analyses for the pairs of (x,y) for which this holds. One such analysis shows up on my pages on infinite exponentials, by looking at the graph of the function: f(x)=x^(1/x): http://users.forthnet.gr/ath/jgal/math/exponents.html Other analyses (including pairs of integers for which the above holds) can be found in my references page: http://users.forthnet.gr/ath/jgal/math/IERefs.html The last resulting equation, y^(a^(1/y))=a+1, cannot be brought into this form, however, because of the +1 factor. And the fact that it cannot be solved by the LambertW, can be also seen easily as follows: set z=1/y, so the last equation becomes: (1/z)^(a^z)=a+1, => z^(a^z)=1/(a+1). taking logs: log(z)*a^z=log(1/(a+1)) The last one is of the form: log(x)*e^x=A, which cannot be solved by the LambertW, since the later solves analytically only the closely matching equations of the form: x*e^x=A. So your best bet would probably be to try to study the two functions numerically. -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: Re: system of exponential equations \.91Ä Amanda \.91[CapitalEth]\.91ñ\.93»[EDouble Dot]±\.93Ì\.91[Micro] \.93¡\.93.b3\.91Ë \.91.b9\.91¬\.91ü\.93á[EDo ubleDot].b9\.91± > Hello > Im trying to find the set of values of a>0 \ for which the system x^y = > a and y^x = a+1 has a solution. I\.8cÇm trying to use the Implicit > Function Theorem. If we define f:R^3 -> R^2 by the vector valued > function whose coordinates are f1(x,y,a) = x^y - a and f2(x) = y^x - > (a-1), then its Jacobian determinant with respect to x \ and y is > J(x,y) = x^y * (1 - ln(x) * ln(y)). If ln(x)*ln(y)<>1, then J(x,y)<>0. > We also see that the partial derivatives of f with respect to x and y > are continuous. By inspection we see f(2, 3, 8) = (0,0). Therefore, > the Implicit Function Theorem shows theres a neighborhood of 8 in the > reals where the system has a unique solution. > But this is not a great conclusion, it seems the I. F. Theorem doesnt > help much solve this proble, right? Right. You need to swim a little into deep waters here: x^y=a, y^x=a+1 => x=a^(1/y), y^x=a+1,=> x=a^(1/y), y^(a^(1/y))=a+1,=> x=a^(1/y), y^a=(a+1)^y. The last equation can be solved for a with the help of the LambertW function: y=-a/ln(a+1)*LambertW(-1/a*ln(a+1)). The LambertW function LW(x) is real valued for arguments x in [-1/e,+oo), so we need to make sure that this restriction is taken into account. So you need to make sure that: -ln(a+1)/a >= 1/e. Additionally you need also the restrictions: a =/=0 (for the denominator) and a + 1 > 0 (for the ln), => a > -1. (1) The last equation is again solved by the LambertW, this time by the non-prinicpal branch: LW(-1,x) which is real valued for x in [-1/e,0). Solving the last one with Maple: a >= -(LambertW(-1,-e^(-1-1/e))+1/e)*e The last expression evaluates with Maple to: ~ 4.759131734. Since this a satisfies conditions (1), => a >=4.759131734 > Amanda -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: Re: This Weeks Finds in Mathematical Physics (Week 203) >> Topologists know David Joyce as the inventor of the quandle - an >> algebraic structure that captures most of the information in a knot. > >Kewl... but does encompass handedness? No - but thats the *only* thing it cant \ detect. More precisely: two knots are unoriented equivalent if there is a diffeomorphism of R^3, possibly orientation-reversing, that maps one knot to the other. In 1979, David Joyce proved that knots are unoriented equivalent if and only if they have isomorphic quandles. === Subject: a digital game Two players define a number x = 0. d_1 d_2 d_3 ... (base 2) as follows: First, one player separates {1, 2, 3, ...} into two disjoint infinite sets and defines the d_i for all i in one of the sets; the other player then defines all the remaining d_i. By what strategy can one of the players ensure that x is transcendental, regardless of who plays first? --r.e.s. === Subject: Re: a digital game >Two players define a number x = 0. d_1 d_2 d_3 ... (base 2) as follows: >First, one player separates {1, 2, 3, ...} into two disjoint infinite >sets and defines the d_i for all i in one of the sets; the other player >then defines all the remaining d_i. By what strategy can one of the >players ensure that x is transcendental, regardless of who plays first? This is rather reminiscent of the Banach-Mazur game, where instead of just two turns you have an infinite sequence of turns: at each stage the player whose turn it is chooses a finite number of digits. In that case, either player can force x to be in any dense G_delta. However, thats not true for r.e.s.s game: the \ first player can force x to be in any dense G_delta, but there are also nowhere dense closed sets that the first player can force. 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: a digital game > Two players define a number x = 0. d_1 d_2 d_3 ... (base 2) as follows: > First, one player separates {1, 2, 3, ...} into two disjoint infinite > sets and defines the d_i for all i in one of the sets; the other player > then defines all the remaining d_i. By what strategy can one of the > players ensure that x is transcendental, regardless of who plays first? Thats quite an interesting puzzle! Kudos! But I wonder whether it can be trivially solved by the following conjectured solution (which looks not trivial at all, but I could easily be wrong and it could be completely trivially false): Let whichever player is trying to make Ôx \ transcendental simply list off the binary digits of pi, unless that would make the number algebraic, in which case he should list off the digits, inverted (0 to 1 and vice versa). The first player might as well choose every other digit. These both seem like really common-sense things to do, but neither of them seems obviously correct. I wonder whether a first-player strategy exists such that the above second-player strategy fails. -Arthur === Subject: Re: a digital game So, in summary, each player can force transcendental if he wants to; so neither player can force algebraic. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: a digital game > Two players define a number x = 0. d_1 d_2 d_3 ... (base 2) as follows: > First, one player separates {1, 2, 3, ...} into two disjoint infinite > sets and defines the d_i for all i in one of the sets; the other player > then defines all the remaining d_i. By what strategy can one of the > players ensure that x is transcendental, regardless of who plays first? It seems to me that the 1st player can guarantee a transcendental by going .010?00100?0000001000000?00...00100..00?.... where she is giving long stretches of 0 broken up by single 1s (the ? are the ones she leaves for the other player). No matter what the other player does, a number results with rational approximations to good to be algebraic. The second player can also guarantee a transcendental, by cardinality, although Im not so sure theres a practical \ strategy. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: a digital game > Two players define a number x = 0. d_1 d_2 d_3 ... (base 2) as follows: > First, one player separates {1, 2, 3, ...} into two disjoint infinite > sets and defines the d_i for all i in one of the sets; the other player > then defines all the remaining d_i. By what strategy can one of the > players ensure that x is transcendental, regardless of who plays first? > It seems to me that the 1st player can guarantee a transcendental > by going .010?00100?0000001000000?00...00100..00?.... > where she is giving long stretches of 0 broken up by single 1s > (the ? are the ones she leaves for the other player). No matter > what the other player does, a number results with rational > approximations to good to be algebraic. > The second player can also guarantee a transcendental, by cardinality, > although Im not so sure theres a practical \ strategy. If the second player is faced with, for example, 0.0101001?1010?1?001001... Just get a nice list of all the algebraics which match that pattern and make like Cantor. Not sure if this counts as a practical strategy. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: min area to ßip 2 hinged rods > ... > 2 rigid rods of unit length are hinged to each other. Initially they are > parallel to each other, much like a closed pair of divider. > > This divider is placed on a piece of paper. What is the minimum area of the > paper which allow the divider to be opened such that the angle between the > two legs extend from 0 to 360 degrees. The divider is to touch the paper > and no part of the divider is to extend beyond the paper throughout the > whole process. >>There is a related problem about passing a ladder of length L around a >>right angled corner from a corridor of width A to another of width B. >>Assuming the ladder is kept horizontal and that its thickness is >>negligible, it transpires that if L^(2/3) <= A^(2/3) + B^2/3), the >>desired passage of the ladder is just possible. >>Thus we may simplify the problem of the hinged rod by providing a much >>smaller definite upper bound on the area that has been proposed yet. >>Assuming each rod to be of length 1, the boundary curve need be no >>larger that (x^2)^(1/3) + (y^2)^(1/3) = 1 >>Since this bounds an area of 3*pi/32 in each quadrant, the total area >>required will be less than 3*pi/8 ~ 1.7881 square units. >>Since the rods are not constrained by the axes but only by the curve >>itself, and not, as was the ladder, constrained by the walls, a slightly >>smaller boundary is possible, possibly one of the similar form |x|^u + >>|y|^u = 1 with u < 2/3. >>Can anyone come up with an area less that 3*pi/8? >Yes! >I cant compute the area exactly (maybe someone else can help?) >but it is somewhere between sqrt(3)/4 = .43 and 3/4 = .75. Im >guessing its about .56, which is less than half of 3pi/8. >Here is the construction of the paper. Start with the origin O. >Then place three segments with endpoints O and A = (1,0), O and >B = (-1/2, sqrt(3)/2), and O and C = (-1/2, -sqrt(3)/2), so that >there is an angle of 2pi/3 between each pair of segments. Now >play the ladder game on each pair. For example, start with a >ladder of length 1 with endpoints O and A. \ Allow the endpoint >that is at O to move along the segment towards B, while keeping >the other endpoint on the x-axis. The area that this ladder (and >the three others) sweeps out is the entire paper. Ive \ found a >parameterization of the curve, but I cant find \ a direct formula. >This is why I cant give an exact area. The estimates are easy >to see, however. >Now, here is how we can actually get this divider to go through >360 degrees. Let the ends of the divider be R,S,T, with S in the >middle. These points will stay on the three lines (O-A, O-B, O-C) >the whole time. Start with > R,T at O, S at B. > R->A and T->C while S->O. > Let R,S stay on the x-axis as S->A and T->O. (so R->2A) > T->B as S->O (so R->A). > R->O and T->O while S->C. >I thought about the ladder problem when I saw this, too. I think >this construction may be improved (because of some asymmetry), >but I dont know the best way to proceed. Id \ be happy if anyone >could find a direct formula for the curve! If I am not mistaken, the family of lines you are considering, at least for the line sliding from OA to BO, is sin(t) x + cos(t) y = 1/sqrt(3) (cos(pi/3-2t) - 1/2) [1] for 0 <= t <= pi/3. For a given t, the line in [1] passes through the points 2/sqrt(3) sin(pi/3-t) (1,0) [2a] which is on OA and 2/sqrt(3) sin(t) (-1/2,sqrt(3)/2) [2b] which is on BO. Differentiating [1], we get cos(t) x - sin(t) y = 2/sqrt(3) sin(pi/3-2t) [3] Solving [1] and [3] simultaneously gives a point on the envelope of the family given by [1] x = sqrt(3)/6 (3 sin(pi/3-t) - sin(t) + sin(pi/3-3t)) [4a] y = sqrt(3)/6 (3 cos(pi/3-t) - cos(t) - cos(pi/3-3t)) [4b] I have put an image of this envelope at http://www.whim.org/nebula/math/images/deltoid.gif Computing the area of this deltoid gives (10 pi - 3 sqrt(3))/48 = 0.5462452940248187605 which is by far the best I have seen, though as has been noted by Brian Chandler, it is not optimal. Rob Johnson take out the trash before replying === Subject: Alternatives to Abraham/Marsden & Arnold Im looking for a modern mathematical treatment of classical mechanics that is brief without sacrificing mathematical rigor (even if this means referring the reader to other texts for proofs, etc.). I checked out Abraham and Marsdens tome, which is not \ brief, but I thought that at least it would be pretty rigurous. Well, it sure looks rigurous, but I find their exposition excessively formal (very French/Bourbaki-like). I get lost in the endless formalism, lose all hope of an intuitive understanding, and quickly miss the forest for the trees. I like Arnolds book on classical mechanics \ better, but sometimes I have a hard time with his notation. bill === Subject: Re: Alternatives to Abraham/Marsden & Arnold > Im looking for a modern mathematical treatment of \ classical > mechanics that is brief without sacrificing mathematical rigor > (even if this means referring the reader to other texts for proofs, > etc.). The latest book by Marsden & Ratiu (2nd edition) is quite nice (many illustrations, nice layout...). Youll find the \ reference plus a PDF of another (earlier) book here: A book I havent read is the one by Singer, which is \ reviewed by Montgomery here: . Finally there are also Dolgachevs lectures on physics (the first part is on classical mechanics): . And I would recommend going back to Arnold (second edition here too) after reading these. -- thomas. === Subject: Re: Alternatives to Abraham/Marsden & Arnold > Im looking for a modern mathematical treatment of \ classical > mechanics that is brief without sacrificing mathematical rigor.... D.E. Rutherford, Classical Mechanics. Ive been told that Rutherford was a pure mathematician who was asked to teach mechanics because of a staff shortage at Edinburgh University (long ago). Perhaps his crisp pure \ mathematicians style is what makes this little book appeal to me. Ken Pledger. === Subject: Re: Alternatives to Abraham/Marsden & Arnold > Im looking for a modern mathematical treatment of \ classical > mechanics that is brief without sacrificing mathematical rigor > (even if this means referring the reader to other texts for proofs, > etc.). > I checked out Abraham and Marsdens tome, which is not brief, but > I thought that at least it would be pretty rigurous. Well, it sure > looks rigurous, but I find their exposition excessively formal > (very French/Bourbaki-like). I get lost in the endless formalism, > lose all hope of an intuitive understanding, and quickly miss the > forest for the trees. > I like Arnolds book on classical mechanics \ better, but sometimes > I have a hard time with his notation. There are a number of alternatives. I know of the following: Classical Mathematical Physics: Dynamical Systems and Field Theories by Walter Thirring - again the notation may be difficult; - chapter 5 is a very nice introduction to the abstract version of mechanics; Classical Dynamics : A Contemporary Approach by by Jorge V. Jos.8e and Eugene J. Saletan - the abstract geometrical version of mechanics is woven into the text in a way that most of this material can be omitted, if desired (which you dont), and thus may provide a bridge between standard and geometrical approaches. George === Subject: Re: Multiplication of a constant mod 2^32 > classes, but nothing above linear algebra yet. Ive got a question I > am hoping someone can answer. I have a 32-bit constant K that I want > to multiply with any value between 0x00 and 0xFFFFFFFF inclusive and > take the result mod 2^32. How can I determine how many unique results > will be returned (besides trying every combination and filtering out > -=PJ=- There are only 2^9 numbers between 0 and 0xFFFFFFFF, so at most your answer is 0xFFFFFFFF, when all the numbers are different. The numbers 0, k, 2k, 3k, ... form the cyclic subgroup of Z_(2^32) generated by k. The order of this aubgroup is n = 2^32 / gcd(2^32, k). If n <= 0xFFFFFFFF + 1 = 2^9, then your solution is n; if n > 0xFFFFFFFF + 1, then your solution is 2^9. Because of the very special nature of 2^32 (i.e. it is a power of a prime number) you can calculate n = 2^32 / gcd(k, 2^32) in a very efficient way: long int n(long int k) { int p = 32; long int j; if((k % 2) == 1) { fprintf(stderr, Going to cause an integer overßow, giving up.n); return 0; } for(j = k; (j % 2) == 0; j /= 2) p--; return 1< classes, but nothing above linear algebra yet. Ive got a question I > am hoping someone can answer. I have a 32-bit constant K that I want > to multiply with any value between 0x00 and 0xFFFFFFFF inclusive and > take the result mod 2^32. How can I determine how many unique results > will be returned (besides trying every combination and filtering out A way to attack this problem: Given two numbers x and y, when will (x * K) mod 2^32 be equal to (y * K) mod 2^32? Once you know which numbers will produce the same result, it will be easy to find out how many different results are possible. === Subject: Re: Multiplication of a constant mod 2^32 > classes, but nothing above linear algebra yet. Ive got a question I > am hoping someone can answer. I have a 32-bit constant K that I want > to multiply with any value between 0x00 and 0xFFFFFFFF inclusive and > take the result mod 2^32. How can I determine how many unique results > will be returned (besides trying every combination and filtering out If K is odd, then its coprime to 2^32, so its multiples cover all of the values, so the answers 2^32. (Its a \ generator in the additive group, if I have my terminology not twisted up.) If K is even, then look at the number of zeros at the end of the binary representation. Obviously multiplying can never change them. But think of it shifted right so the zeros all fall off - now you have an odd number, and will generate all of the bit patterns in the remaining bits. So the answers 2^p, where p is the number \ of bits from the least significant 1 bit to bit 31. (Can no doubt be calculated in around three machine instructions given enough ingenuity.) I think this means that if k = m * 2^n, m odd, then 2^(32-n) values. Brian Chandler ---------------- Jigsaw puzzles from Japan http://imaginatorium.org/shop/ === Subject: Re: can someone check my maths here.... > (also xposted to uk.education.maths) > Can someone check my maths here.... > Im multiplying out and simplifying a denominator of a fraction > Is this correct - J denotes complex number. > (-W^2 + 25*JW)*(-W^2 + 25*JW) > (-W^2)*(-W^2 + 25*JW) + 25*JW*(-W^2 + 25*JW) > (W^4) - (W^2*25JW) - (W^2*25JW) + (625 W^2) ....[{ J x J = 1 }] so > dissapears in last term ? > (W^4) + (625 W^2) - (W^2*25JW) - (W^2*25JW) .....[{- (W^2*25JW) - > (W^2*25JW) = 0 }] correct?...so > this equals > W^4 + 625 W^2 Lets do a simpler sum: (7-2)*(7-2) = 7*7 - 7*2 - 2*7 + 2*2 = 49 - 14 - 14 + 4 = 49 - 28 + 4 = 21 + 4 = 25 Which is the right answer, as the original sum was (7-2)*(7-2), that is, 5*5. Notice that the -14-14 becomes -28, not 0. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: can someone check my maths here.... > (also xposted to uk.education.maths) > Can someone check my maths here.... > Im multiplying out and simplifying a denominator of a fraction > Is this correct - J denotes complex number. > (-W^2 + 25*JW)*(-W^2 + 25*JW) > (-W^2)*(-W^2 + 25*JW) + 25*JW*(-W^2 + 25*JW) > (W^4) - (W^2*25JW) - (W^2*25JW) + (625 W^2) ....[{ J x J = 1 }] so > dissapears in last term ? Are you sure you dont mean ...[{ J x J = -1 }] ??? -- 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: can someone check my maths here.... Ok J^2 is -1 so it should be: (W^4) - (W^2*25JW) - (W^2*25JW) - (625 W^2) and hence will be W^4 - 625 W^2 Is that now correct?? Im not convinced that Im right..... === Subject: Help! Need a book recommendation... Am doing a paper on Linear Algebra - fine, Probability - \ fine, and Differential Equations - dont know yet, but looks ok. But \ Im getting stuck on things like not being able to solve a third order polynomial, or being able to integrate e^f(x), where f(x) is a second order polynomial. What do I need to read up on? TIA. === Subject: Re: Help! Need a book recommendation... Originator: grubb@lola >Am doing a paper on Linear Algebra - fine, Probability - \ fine, and >Differential Equations - dont know yet, but looks ok. But Im >getting stuck on things like not being able to solve a third order >polynomial, or being able to integrate e^f(x), where f(x) is a second >order polynomial. What do I need to read up on? TIA. There is a general solution to a cubic equation, but it is seldom used because it is complicated. Look for ÔCardanos \ formula. The difficulty is that it doesnt give answers in \ usuable form for many problems. There is a version by Vieta using trig functions, also. As for thew integral of e^{f(x)} for f(x) a quadratic, you cant do it in closed form with elementary functions. For example, e^(-x^2) cannot be done in this way. The solution is to invent a new function for this purpose. --Dan Grubb === Subject: Re: Help! Need a book recommendation... >>Am doing a paper on Linear Algebra - fine, Probability - fine, and >>Differential Equations - dont know yet, but looks ok. But Im >>getting stuck on things like not being able to solve a third order >>polynomial, or being able to integrate e^f(x), where f(x) is a second >>order polynomial. What do I need to read up on? TIA. >There is a general solution to a cubic equation, but it is seldom used >because it is complicated. Look for ÔCardanos \ formula. The >difficulty is that it doesnt give answers in \ usuable form for >many problems. There is a version by Vieta using trig functions, also. Ive heard of Cardanos. Ill \ look into it. Would you have come across this in the course of doing a bachelors degree? This paper is part of an application for a masters. >As for thew integral of e^{f(x)} for f(x) a quadratic, you cant >do it in closed form with elementary functions. For example, >e^(-x^2) cannot be done in this way. The solution is to invent >a new function for this purpose. Ok. Ill look at it. === Subject: Re: Help! Need a book recommendation... > Am doing a paper on Linear Algebra - fine, Probability - fine, and > Differential Equations - dont know yet, but looks ok. But Im > getting stuck on things like not being able to solve a third order > polynomial, or being able to integrate e^f(x), where f(x) is a second > order polynomial. What do I need to read up on? TIA. Id recommend Shakespeare, or Dostoevsky. If what you want \ is a closed form expression in terms of elementary functions for the antiderivative of e^f where f is quadratic, those two authors will help you as much as anyone, and youll get some \ culture, to boot. Solving polynomial equations of degree three - any book on Theory of Equations. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Help! Need a book recommendation... >> Am doing a paper on Linear Algebra - fine, Probability - fine, and >> Differential Equations - dont know yet, but looks ok. \ But Im >> getting stuck on things like not being able to solve a third order >> polynomial, or being able to integrate e^f(x), where f(x) is a second >> order polynomial. What do I need to read up on? TIA. >Id recommend Shakespeare, or Dostoevsky. If what you want \ is >a closed form expression in terms of elementary functions for >the antiderivative of e^f where f is quadratic, those two authors >will help you as much as anyone, and youll get some \ culture, >to boot. You calling me thick? Do one for me, and let me see how you did it. Choose something that doesnt have simple roots. >Solving polynomial equations of degree three - any book on >Theory of Equations. I want the most concise book there is. I dont want to read loads of words when a short deduction will do. Any suggestions? === Subject: Re: Help! Need a book recommendation... >being able to integrate e^f(x), where f(x) is a second > order polynomial. > Do one for me, and let me see how you did it. > Choose something that doesnt have simple roots. No one can do one for you since it is impossible to integrate such a function (e.g. e^(-x^2)) in closed form. === Subject: Re: Help! Need a book recommendation... >>being able to integrate e^f(x), where f(x) is a second >> order polynomial. >> Do one for me, and let me see how you did it. >> Choose something that doesnt have simple roots. >No one can do one for you since it is impossible to integrate such a >function (e.g. e^(-x^2)) in closed form. Impossible? Weird. I dont know what closed form means. So how do I deal with it? === Subject: Re: Help! Need a book recommendation... >> >>being able to integrate e^f(x), where f(x) is a second >> order polynomial. >> Do one for me, and let me see how you did it. >> Choose something that doesnt have simple roots. >No one can do one for you since it is impossible to integrate such a >function (e.g. e^(-x^2)) in closed form. > Impossible? Weird. I dont know what closed form means. So how do > I deal with it? Depends on the context. If the question is What is the derivative with respect to t of the integral from 17 to t of e^(x^2) dx? then only the student who hasnt got a clue thinks she has to work out the antiderivative of e^(x^2). If the question is Find the integral from minus infinity to plus infinity of e^(-x^2) dx then again this can be done but the \ way to do it does not involve the impossible task of finding a \ nice formula for the antiderivative of e^(-x^2). So, what is it that you really want to do? Please try to be polite in your response. I know its hard for you, but I bet you can do it if you really try. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Help! Need a book recommendation... >> Impossible? Weird. I dont know what closed form means. \ So how do >> I deal with it? >Depends on the context. >If the question is What is the derivative with respect to t of >the integral from 17 to t of e^(x^2) dx? then only the student >who hasnt got a clue thinks she has to work out the antiderivative >of e^(x^2). >If the question is Find the integral from minus infinity to plus >infinity of e^(-x^2) dx then again this can be done but the way >to do it does not involve the impossible task of finding a nice >formula for the antiderivative of e^(-x^2). >So, what is it that you really want to do? >Please try to be polite in your response. I know its hard >for you, but I bet you can do it if you really try. which you are not one. === Subject: Re: Help! Need a book recommendation... >> Am doing a paper on Linear Algebra - fine, Probability - fine, and >> Differential Equations - dont know yet, but looks ok. \ But Im >> getting stuck on things like not being able to solve a third order >> polynomial, or being able to integrate e^f(x), where f(x) is a second >> order polynomial. What do I need to read up on? TIA. >Id recommend Shakespeare, or Dostoevsky. If what you want \ is >a closed form expression in terms of elementary functions for >the antiderivative of e^f where f is quadratic, those two authors >will help you as much as anyone, and youll get some \ culture, >to boot. > You calling me thick? Do one for me, and let me see how you did it. > Choose something that doesnt have simple roots. It dont matter whether it got simple roots or not. Think about what I said. I said that a book which has no mathematics in it at all will do as much good as a book filled with mathematics. Thats not a comment about you - its a \ comment about the problem you are posing. >Solving polynomial equations of degree three - any book on >Theory of Equations. > I want the most concise book there is. I dont want to \ read loads of > words when a short deduction will do. Any suggestions? There is no royal road to mathematics. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Help! Need a book recommendation... >> You calling me thick? Do one for me, and let me see how you did it. >> Choose something that doesnt have simple roots. >It dont matter whether it got simple roots or not. >Think about what I said. >I said that a book which has no mathematics in it at all >will do as much good >as a book filled with mathematics. >Thats not a comment about you - its a \ comment about >the problem you are posing. >>Solving polynomial equations of degree three - any book on >>Theory of Equations. >> I want the most concise book there is. I dont want to read loads of >> words when a short deduction will do. Any suggestions? >There is no royal road to mathematics. Jesus! You arent here to help at all are you? It looks like you are here to act pompous, and be condescending. Next time you see someone asking for help, ie. they put the word Help in the subject of their post, DONT ANSWER. Go sort your confidence \ problems out first. === Subject: Re: Help! Need a book recommendation... >> You calling me thick? Do one for me, and let me see how you did it. >> Choose something that doesnt have simple roots. >It dont matter whether it got simple roots or not. >Think about what I said. >I said that a book which has no mathematics in it at all >will do as much good >as a book filled with mathematics. >Thats not a comment about you - its a \ comment about >the problem you are posing. >>Solving polynomial equations of degree three - any book on >>Theory of Equations. >> >> I want the most concise book there is. I dont want to read loads of >> words when a short deduction will do. Any suggestions? >There is no royal road to mathematics. > Jesus! You arent here to help at all are you? It looks like you are > here to act pompous, and be condescending. Next time you see someone > asking for help, ie. they put the word Help in the subject of their > post, DONT ANSWER. Youre no fun at all. > Go sort your confidence problems out first. You asked for help. I gave you help. You didnt understand. \ I tried again. You reply with blasphemy, insults, and an attempt to silence me. Ill only respond to the last of these. Ive \ been posting here since about 1992. Youve been posting here since about Tuesday. Ill still be posting here long after youre just a bad memory. Deal with it. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Help! Need a book recommendation... >> Jesus! You arent here to help at all are you? It looks like you are >> here to act pompous, and be condescending. Next time you see someone >> asking for help, ie. they put the word Help in the subject of their >> post, DONT ANSWER. >Youre no fun at all. Yeah, well, Im not asking these questions for fun. >> Go sort your confidence problems out first. >You asked for help. I gave you help. You didnt understand. I tried >again. You reply with blasphemy, insults, and an attempt to silence >me. No you didnt. You made some silly quip about reading \ classic literature being as much help as anything else, in reply to a direct question about the integral of e^f(x), f(x) is a quadratic. Then, instead of recommending a book on theory of equations, you come up with another silly remark, about there being no royal road to maths. What on earth does that mean? I reckon you didnt understand the question. - If there is no concise book on this topic, due to its nature, state that. Dont give cryptic answers to straightforward questions. Its like talking to a crossword puzzle. >Ill only respond to the last of these. Ive \ been posting here since >about 1992. Youve been posting here since about Tuesday. Ill still >be posting here long after youre just a bad memory. Deal with it. Yeah great. What you mean is youll be here using your maths knowledge to make you sound clever. While Ill be doing something useful with the maths I learn, and possibly passing on my knowledge in a format easily understood by others who may not have the same level of mathematical sophistication as me. Who do you think is more useful to society? === Subject: Re: Help! Need a book recommendation... > I reckon you didnt understand the question. Following this thread from the outside, its pretty obvious to me that Gerry understood the question. The fact that you didnt understand his answer explains something, though. Doug === Subject: Re: Help! Need a book recommendation... >> I reckon you didnt understand the question. >Following this thread from the outside, its pretty obvious to me that >Gerry understood the question. >The fact that you didnt understand his answer explains something, though. >Doug Yeah Doug, I can tell youre a real important guy. === Subject: Re: Help! Need a book recommendation... >> I reckon you didnt understand the question. >Following this thread from the outside, its pretty obvious to me that >Gerry understood the question. >The fact that you didnt understand his answer explains something, though. > Yeah Doug, I can tell youre a real important guy. Kudos! For my part, I can tell that you like others to do your homework for you. Doug === Subject: Re: Help! Need a book recommendation... > Jesus! You arent here to help at all are you? It looks like you are > here to act pompous, and be condescending. Next time you see someone > asking for help, ie. they put the word Help in the subject of their > post, DONT ANSWER. Go sort your confidence \ problems out first. Yeah, youre going to help here *real fast* like that. Doug === Subject: Re: Help! Need a book recommendation... >>Solving polynomial equations of degree three - any book on >>Theory of Equations. >I want the most concise book there is. I dont want to read loads of >words when a short deduction will do. Any suggestions? And something complete if possible. I dont really want to \ be asking more questions about the topic than the text answers. TIA. === Subject: Re: Lipschitz condition on a measurable set >I have a HW question that I need help in answering. I would love a >good hint so that I can solve it myself, it just makes me feel good. >Here it is: Let f be a function from a set E to R and let E be a >measurable set with measure equal to zero. Show that the measure of >f(E) is equal to zero also. This does not seem that difficult but I >can not put my hand around. I was thinking about using some G-delta >sets and an epsilon argument but I have not gotten anywhere. Any help >would be appreciated. Well of course this is false - since you mentioned Lipschitz in the subject line probably you also meant to assume that f \ satisfies a Lipschitz condition? If so then this is _very_ easy. Forget G-delta, for any epsilon > 0 E is contained in an open set of measure less than epsilon. That open set is a disjoint union of intervals... >Simon Cisneros ************************ David C. Ullrich === Subject: Re: what is the z-transform of sinc function? > Which infinity do you mean as the indeterminant form sin(x) / x^2 at > 0 does not have the same limit from both sides so it would not seem to > fit any plausible notion of removable singularity. Maybe, as Jerry indicates below, he was thinking of the unsigned infinity of R*, the one-point compactification of the reals. If R* is taken to be the codomain of sin(x) / x^2, then at x = 0 we do have the same limit from both sides, and so the singularity is removable. > Usually the notion > of computable value would be expected to have a well determined sign > for the value when the value is away from 0. But that infinity and 0 are both unusual, being unsigned. > you are trying to give meaning to 1 / x at 0. In the neighbourhood of > 0 the values 1 / x can be arbitarily large and of either sign. It is > a standard example of when the notion of a value gets into trouble. In R*, 1/0 is unsigned infinity. > Does infinity need to have a sign? I like to think of tan(x) and 1/x as > wrapping around and coming out the other side. Right. You might like the diagram of R* shown at . If the codomain of tan(x) is taken to be R*, then it is continuous on R, and its graph is simply a helix (on the surface of a cylinder, with the y-axis wrapped around as shown at the link above). For f(x) = 1/x, if we take both domain and range to be R*, then we can visualize its graph nicely on the surface of a torus. (Wrap the y-axis around to get a cylinder, then wrap the x-axis around, giving a torus.) The graph forms a simple closed curve. David Cantrell === Subject: Re: what is the z-transform of sinc function? ... > But that infinity and 0 are both unusual, being unsigned. >you are trying to give meaning to 1 / x at 0. In the neighbourhood of >0 the values 1 / x can be arbitarily large and of either sign. It is >a standard example of when the notion of a value gets into trouble. > In R*, 1/0 is unsigned infinity. >>Does infinity need to have a sign? I like to think of tan(x) and 1/x as >>wrapping around and coming out the other side. > Right. You might like the diagram of R* shown at > . > If the codomain of tan(x) is taken to be R*, then it is continuous on R, > and its graph is simply a helix (on the surface of a cylinder, with the > y-axis wrapped around as shown at the link above). > For f(x) = 1/x, if we take both domain and range to be R*, then we can > visualize its graph nicely on the surface of a torus. (Wrap the y-axis > around to get a cylinder, then wrap the x-axis around, giving a torus.) > The graph forms a simple closed curve. > David Cantrell Neat! Powerful! I like it. Theres something Smith-charty about it. Jerry -- Engineering is the art of making what you want from things you can get. ¿¿¿¿¿\ ¿¿¿[OS lash]¿¿¿¿\[DownQuest\ ion]¿¿¿ ¿¿¿¿¿\ ¿¿¿[OSl ash]¿¿¿¿\[DownQuesti\ on]¿¿¿ ¿¿¿¿¿\ ¿¿¿[OSl ash]¿¿¿¿\[DownQuesti\ on]¿¿¿ ¿¿¿¿¿\ ¿¿¿[OSl ash]¿¿¿¿\[DownQuesti\ on]¿¿¿ ¿¿¿ === Subject: Re: what is the z-transform of sinc function? > be expressed as an infinite product ~ > cos(x)*cos(x/2)*cos(x/4)*cos(x/8)......cos(x/2^n) as n goes to infinity. In this > form it is not particularly easy to evaluate except at multiples of pi/2. At x=0 > it will clearly be one and at any other multiple of pi/2 it will be zero. Close. Omit the first term. sin(x)/x = cos(x/2) * cos(x/4) * ... * cos(x/2^n) * ... Scott -- Scott Hemphill hemphill@alumni.caltech.edu This isnt ßying. This is falling, with style. -- Buzz Lightyear === Subject: Re: what is the z-transform of sinc function? > be expressed as an infinite product ~ > cos(x)*cos(x/2)*cos(x/4)*cos(x/8)......cos(x/2^n) as n goes to infinity. In this > form it is not particularly easy to evaluate except at multiples of pi/2. At x=0 > it will clearly be one and at any other multiple of pi/2 it will be zero. > Close. Omit the first term. > sin(x)/x = cos(x/2) * cos(x/4) * ... * cos(x/2^n) * ... Ah yes, of course, what I had was sin(2x)/2x. -jim http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- === Subject: Re: what is the z-transform of sinc function? >> ... >> from what they taught me, sin(x)/x clearly has a singularity at >> 0 but it >> is a removable singularity. sin(x)/x and *all* its derivatives >> exist >> everywhere and it is analytic everywhere. in fact, a decent >> McLaurin series can be determined for it. > If I consider sin(x) to just be a handy notation for > x-(x^3)/3!+(x^5)/5! ..., > the question of singularity of sin(x)/x is obviously the same as the > question if > f(x) = x/x > has a singularity at x=0. youre preaching to the choir (at least regarding me). both g(x) = sin(x)/x and f(x) = x/x have exactly the same degree of singularity at x=0. its called a removable singularity. r b-j === Subject: Pizza-Cutting With Circular Arcs Instead of the typical circle-cut-by-straight-lines problem, where we want the maximum number of sections a circle can be cut into by n straight lines, here I am asking, what is the maximum number of sections a circle can be divided into by n circular arcs, each arc with the same radius as the original circle? I imagine a pizza of radius r being cut by a giant compass with, instead of a pencil, a pizza-cutter on the end, the compass having a radius of r. And the point of this compass can be placed anywhere, either inside the pizzas circumference or outside it. I get, by hand the first 2 or 3 terms: 2, 5, 9 (or 10 or more?),... (Sequence is starting at one arc-cut.) I am not sure, but I believe this sequence is not in the EIS. And also, what would the sequence be if we required that the pizza/circle be cut so that the pieces all have the same area (but not necessarily the same shape)? Leroy Quet === Subject: Re: Pizza-Cutting With Circular Arcs > Instead of the typical circle-cut-by-straight-lines problem, where we > want the maximum number of sections a circle can be cut into by n > straight lines, > here I am asking, what is the maximum number of sections a circle can > be divided into by n circular arcs, each arc with the same radius as > the original circle? > I imagine a pizza of radius r being cut by a giant compass with, > instead of a pencil, a pizza-cutter on the end, the compass having a > radius of r. > And the point of this compass can be placed anywhere, either inside > the pizzas circumference or outside it. > I get, by hand the first 2 or 3 terms: > 2, 5, 9 (or 10 or more?),... > (Sequence is starting at one arc-cut.) > I am not sure, but I believe this sequence is not in the EIS. > And also, what would the sequence be if we required that the > pizza/circle be cut so that the pieces all have the same area (but not > necessarily the same shape)? > Leroy Quet Cant see an answer yet; heres another \ question -- what if we require the pieces to be of the same area and the cuts to be constructible? (Im not absolutely sure if we could construct an arc that cuts the circle into two equal pieces, but we could certainly construct a concentric circle with half the area, dividing it into a circle and an annulus.) How many slices can you get then? Back when I was in high school and the world was still in black-and-white with low-fidelity sound, I remember this damned old math competition problem which was phrased How many pieces can you cut a cake into with 3 straight cuts? The answer was not 7 but since cakes are three-dimensional and you can slice through each of the xy, yz, and xz planes. Darn those tricky clever people... come to think of it, I ended up winning a pretty sizeable (by high school standards) cash prize in that competition anyway though, so I guess they were all right after all. === Subject: Re: Interesting problem > I have the following problem: > A and B are two disjoint sets whose union is |R+. Both A and B are closed > under sum and multiplication. Is it possible that neither A nor B is the > VOID set? > {V} Hi Whip Youll have to fill in the details, but I think \ I got this one. The motivation for this came from thinking of A as an image of the positive reals, and B of the negative. Notation: e^a = e raised to the power a Let {Ua} be a Hamel Basis for R+ over Q+ (Then {-Ua} is a Hamel Basis for R- over Q+) Now define f: R -> R>=0 by f(x) = [SUM I](qi e^Uai) where x>0 and x = [SUM I](qi Uai) some finite I f(x) = [SUM I] (qi e^(-Uai)) where x<0 and x = -[SUM I](qi Uai) some finite I f(0) = 0 (not needed, just thrown in to be complete) Define A = f(R+) and B = f(R-) Then A and B obviously partition R+, since e^x:R->R+ is a bijection So, now closure: If x, y in A then xy = [SUM K](qk e^Uak) x [SUM J](qj e^Uaj) = [SUM J,K](qk qj e^(Uaj+Uak)) = f(z) where z = [SUM J,K] (qk qj (Uaj+Uak)) i.e. xy is in f(R+) = A x+y = [SUM K](qk e^Uak) + [SUM J] (qj e^Uaj) = [SUM L](ql e^Ual) + [SUM M] ((qm + qm) e^Uam) where L = (K-J) U (J-K) and M = K INTERSECT J So x+y = f(z) where z = [SUM L](ql Ual) + [SUM M] ((qm + \ qm) Uam) i.e. x+y is in f(R+) = A Similarly for closure of B This is kind of rough and my notation is atrocious (sigh \.9a no tools, no time, no talent) Rick Ostrander === Subject: Re: a logic problem | > Ive been trying to prove this last formula actually, starting from | > |- y=y_1 -> (p -> p*) | | Im thinking that semantically this means that formulas that are true with a | variable are also true with that variable replaced with a renamed variable? | As opposed to a constant. | > where p* is p in which some, but not necessarily all, occurrences of y are | > replaced by y_1. | | This cant be true in general (although possible in some specific systems). | Assuming Ive interpreted it correctly above. Or are the replacements | dependent upon the sentence? No, it expresses the fundamental property of =, namely that you can replace to equal elements wherever you want. So this is not substitution! You can really replace any occurrence, or none at all, or some... | As you mention in another post that this is from Keisler, its easy to | assume that Im misunderstanding (and maybe you are too). Have you | considered contacting him? a response. That could of course be because by accident I sent it four times... | | > This is an axiom in the system, in which I take p* = p, | > which can be translated to | > |- ~(y=y_1 & p & ~p) | > where a part of the desired formula already occurs... the problem is to | get | > rid of the ~p (~ meaning not), which can be trasformed to an | implication, | > which delivers me only the version with and, as mentioned below. | > | Message posted in German to z-netz.wissenschaft.mathematik and | > | maus.wissenschaft.mathematik also. | > | If anybody interested in a full answer, I did get one in de.sci.mathematik. My big mistake was trying to prove this whole thing without the Ex-quantifier, which is, as William Elliot pointed out. Hendrik === Subject: Integer-Alterations On a Grid (fun) This is a game I made up today. It is unoriginal, in that, not only do I steal ideas from my own games I have posted here in the past, but from other peoples games also. But, anyway, this game seems as if it would be SO fun to play that I HAD to post it! 2 players. Played on n-by-n grid (as many of my games are). One player makes up n integer-alteration rules (see below for examples) for the n rows, one rule per row. And the other player makes up n rules for the n columns, one rule per column. Rule: a (meta)mathematical rule which changes any input integer into an output integer. At the games beginning, after the rules are created, each player gets 3 random integers, from 1 to n. A players first 2 random integer give the \ coordinate of the square the player starts in, the last integer gives the players starting integer. And what player moves first is then chosen randomly. (But no more randomnes is in the game once play begins.) At each turn, a player can choose to apply the row-rule OR the column-rule to his/her integer. (Each player has their own integer which is altered.) the square he/she is currently at. The player can then choose to move either up, down, left, or right, without regard to what integers (s)he must jump over, that number (m) of grid-squares. And the grid is considered topologically a torus, ie. going off the right edge, for example, brings you back to the left side (and the same (mod n)-topology for right-to-left or vertical moves). (So, for example, a move from column 3 to the right 7 squares, on an n = 6 grid, gets the player to column 2, which is one square to the *left* of where (s)he started.) But a player must only move to a square *without* an integer (written there by either player) already in it. And the games winner is the last player able to move. Notes: *Be Creative in coming up with rules! *The output integers signs are, in practice, \ insignificant, since you can move right OR left, up OR down. But an output of 0 is to be avoided! *Start out with easy rules as you first learn how to play this game, or if you simply are poor at math. It would be less fun to play this game if you must spend an hour calculating each of your integers! Example of rules: (For n=6) Rows: A) m = m+1; B) m = 2*m; C) m = m + sum of distinct primes dividing all already-written neighboring integers; D) m = d(|m|)*phi(|m|) -m; (d(m) is number-of-divisors function, phi(m) is Euler-phi function.) E) m = (numerator)-(denominator) of reduced rational, sum{k=1 to |m|} 1/k; F) m = m + number of already-written integers in row F. Columns: 1) m = ceiling(sqrt(|m|)) 2) m = ceiling(secant(m)), m in radians; 3) m = number of primes <= |m|; 4) m = 2^|m| - m; 5) m = number of divisors in product of all aready-written integers in column 1; 6) m = ceiling((|m|th Fibonacci number)/(|m|_th prime)). (Yes, computers might be needed for some kinds of rules, and are allowed.) And a grid to aid visualization: ------------------ A! ! ! ! ! ! ! ------------------ B! ! ! ! ! ! ! ------------------ C! ! ! ! ! ! ! ------------------ D! ! ! ! ! ! ! ------------------ E! ! ! ! ! ! ! ------------------ F! ! ! ! ! ! ! ------------------ 1 2 3 4 5 6 Leroy Quet === Subject: Re: Sum(1/|cos(n)|,n>=1) >The question was indeed : >Can you prove that, for a real a>1, the series >Sum(1/(|cos(n)|*n^a),n>=1) is divergent or convergent? >Sorry for having given before a trivial question (I know how to prove >the divergence, if a<=1). The question is basically about how well pi can be approximated by rational numbers with even numerators and odd denominators. There is a constant r such that there are only finitely many pairs (n,k) with |n - k pi/2| < 1/n^r. If |n - k pi/2| > 1/n^r for all k, we must have |cos(n)| > 1/(2 n^r). Thus the tail of your series is bounded by that of sum_n 2 n^(r-a), and converges if a > r+1. I believe the best estimate that has been proven (due to M. Hata) will tell you it will converge for a > 8.016045... But its probably true for all a > 2. In fact, for any a > \ 2, sum_n 1/(cos(nx) n^a) converges for almost every x in the sense of Lebesgue measure. 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: Sum(1/|cos(n)|,n>=1) > Sum a(n) can only ever be convergent if a(n) has the limit zero. Is the above true? Consider the series with a(n) = 1/n. I think what you want to tell here is that if Sum a(n) is convergent then lim a(n) is zero. Consequently, Sum a(n) is divergent if lim a(n) is not zero. === Subject: Re: Sum(1/|cos(n)|,n>=1) Matlab User escribi.97: >> Sum a(n) can only ever be convergent if a(n) has the limit zero. > Is the above true? Consider the series with a(n) = 1/n. > I think what you want to tell here is that if Sum a(n) is convergent > then lim a(n) is zero. Consequently, Sum a(n) is divergent if lim a(n) > is not zero. Im not a native English speaker, but I think that it is \ just that Christian said ... -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: all the roots of a nonlinear equation rajarshi@presidency.com says... > I realize this is probably a dumb question but Ive looked at numerical > analysis books and racked my brain but I must be missing something obvious. > I have equations which are sums of sines and cosines an example being: > 2 Cos[x] - Cos[2 x] - Sin[x] + 3 Sin[2 x] - Sin[3 x] = 0 > I need to find all the roots of this equation in the range -pi to pi. > The algorithms Ive looked at only give me one root and if based on a > bisection technique dont always give an answer in the range I mentioned > above. > I tried using NSolve in mathematica for this above equation and it doesnt > return any roots (there are 4 when looking at the plot). > I am interested in an algorithm that I could implement in C / Fortran for > later use that would give me all the roots of the above types of functions > automatically. > Is this possible? Or have I missed something very obvious? The best way to solve this type of equations is: 1. Expand the sin(n x) and cos(n x) terms so that you get an equation in sin(x) anc cos(x) 2. Substitute t=tan(x/2), sin(x)=2t/(1+t^2), cos(x)= (1-t^2)/(1+t^2) 3. Solve the resulting equation in t 4. Solve for x === Subject: Re: a little - big problem > May you give me an hint to prove that all partial derivatives of f of > order > < k at 0 vanish implies f is in J^k ? > Use the multivariable version of Taylors Theorem. > I think i should use Taylor Ôs formula with integral rest for function in > many real variables. > I know Taylor Ôs formula with integral \ Lagranges rest for function in many > real variables, it is the following. May you suggest me Taylor Ôs formula > with integral rest for function in many real variables, please ? tern, the Taylor theorems in several variables follow from the Taylor theorems in one variable. Just use the chain rule carefully. (I am not sure why you continue to post this excessively complicated notation to sci.math; sorry but I didnt read the below.) Now that I think of it, we dont need Taylors \ theorem for this result. Let V^k denote those functions whose partial derivatives of order < k at 0 vanish. Let O^k be the functions that are O(|x|^k) as x -> 0. Proposition: V^k is contained in O^k (obviously O^k is contained in J^k). Proof: Let f be in V^k. Then for each x there is a c = c(x) in (0,1) such that f(x) - f(0) = f(x) = sum xj*Djf(cx), by the one variable mean value theorem. It follows that V^1 is contained in O^1. Now use induction. > ----------------------------------------- > exists t in[0,1] such that: > f (x_0+x) = sum_{|alpha| <= k} frac{D^{alpha}f(x_0)}{alpha !} > x^alpha+ sum_{|alpha|=k+1} frac{D^alpha f(x_0+t x)}{alpha !}x^alpha > = sum_{|alpha| <= k} frac{D^alpha f(x_0)}{alpha !} > x^alpha+ sum_{alpha_1+...+alpha_n=k+1}frac{D^{(alpha_1,...,alpha_n)} > f(x_0+t x)}{alpha_1! ... alpha_n!} (x^1)^{alpha_1}...(x^n)^{alpha_1} > and, if you prefer to compile the above formula in a nice pdf file, lets > compile the following tex file: > exists $tin[0,1]$ such that: > begin{equation} > begin{split} > f(text{x}_0+text{x})& =sum_{|alpha|leq k}frac{displaystyle > D^{alpha}f(text{x}_0)}{displaystyle alpha !} > :text{x}^{alpha}+sum_{|alpha|=k+1}frac{displaystyle D^alpha > f(text{x}_{0}+t:text{x})}{displaystyle alpha !}:text{x}^alpha > & =sum_{|alpha|leq k}frac{displaystyle > D^{alpha}f(text{x}_0)}{displaystyle alpha !} > :text{x}^{alpha}+sum_{alpha_1+cdots+alpha_n=k+1}frac{displa ystyle > D^{(alpha_1,ldots,alpha_n)} f(text{x}_{0}+t:text{x})}{displaystyle > alpha_1!cdotsalpha_n!}:(x^1)^{alpha_1}cdots(x^n)^{alpha_1} > end{split} > end{equation} === Subject: Re: Elementary Arithmetical Method to prove that .... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i240aSJ01961; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i23NFYi24675 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i23NFYd27155; Could be there some connection to: p^2 = 1 mod 6 from p=5 : see: 5^2 = 25 = 1 mod 6 7^2 = 49 ... 11^2 = 121 ... ............... however it is also working for all composite numbers except of pi=2 and pi=3 ( P pi )^2 = 1 mod 6 (5*7)^2 = 1225 = 1 mod 6 Ro-man >> Im looking for the simplest elementary arithmetical proof, which >> doesnt use legendre symbol, that >> Let p, a prime number >> 2^((p-1)/2) = 1 (mod p) => p = \.8c±1 (mod 8) >> Could you give me a tip >See Gausss Disquisitiones Arithmeticae. >SPOILER SPACE >I claim first that if m = +- 3 (mod 8) then x^2 = 2 (mod m) is insoluble. >Suppose false and m is the smallest natual number with x^2 = 2 (mod m) >soluble. Then there is an odd a with 0 < a < m and a^2 = 2 (mod m). >That is, a^2 - 2 = mn. As a^2 - 2 equiv 7 (mod 8) then n = -+3 (mod 8) >too. We cant have a = 1, and so 0 < mn < m^2 so n < m. As a^2 = 2 >(mod n) then we have a contradiction. >As in standard theory, if 2^{(p-1)/2} = 1 (mod p) then 2 is a >square modulo p. (The equation x^{(p-1)/2} = 1 (mod p) has <= (p-1)/2 >solutions but they include 1^2, 2^2, ...., ((p-1)/2)^2). >-- >Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html >Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 >Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Two thieves rpoblem. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i241STe07967; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i241HSi06767 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i241HSH08989; >This is homework problem. >Two thieves have >robbed a warehouse and have to split a pile >of items without prices on them. How do >they do this so that each thief believes he >got at least one half of the value of the pile? >Answer: one of the two thieves splits the >pile in two parts so that he believes that both >parts are of equal value. The other one >chooses the part he believes is better or >equal than one half. >Problem I have to solve: Assume that 10 >thieves have robbed a warehouse. How do >they split the pile of items so that each thief >believes that he got at least one tenth of the >total value of the pile? >Could someone suggest where to start? Ithink I found the solution for 3 people. Let person 1 divide the loot into 3 equal parts. Then, persons 2 and 3 label two pieces out of 3 they want. case 1: If persons 2 and 3 chose the same two pieces, the put these 2 pieces back together and divide the whole thing fairly as 2 people did earlier and person 1 left with the last piece. ( in this case person 1 is interested in dividing the loot into 3 equal parts risking missing out if one piece was smaller and two other peoplke chose the bigger pieces). case 2: Now, If persons 2 and 3 chose differnt pieces, one piece will be in intersection eg they both want the same piece then they divide this wanted piece between them as 2 2 people would failry share it. Then, each person 2 and 3 would share their own piece failry with person 1 giving him a fiar 1/3 too. In all cases, person 1 is interested in dividing the loot in 3 equal pieces and so do persons 2 and 3. i wonder if this method could be applied to n number of people recursively === Subject: Re: Two thieves rpoblem. > Let person 1 divide the loot into 3 equal parts. Then, persons 2 and 3 > label two pieces out of 3 they want. What happens if 2 and 3 agree that a certain part is larger than one-third of the original and the other two parts are each smaller than one-third? -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Two thieves rpoblem. >This is homework problem. >Two thieves have >robbed a warehouse and have to split a pile >of items without prices on them. How do >they do this so that each thief believes he >got at least one half of the value of the pile? >Answer: one of the two thieves splits the >pile in two parts so that he believes that both >parts are of equal value. The other one >chooses the part he believes is better or >equal than one half. >Problem I have to solve: Assume that 10 >thieves have robbed a warehouse. How do >they split the pile of items so that each thief >believes that he got at least one tenth of the >total value of the pile? >Could someone suggest where to start? > Ithink I found the solution for 3 people. > Let person 1 divide the loot into 3 equal parts. Then, persons 2 and 3 > label two pieces out of 3 they want. > case 1: > If persons 2 and 3 chose the same two pieces, the put these 2 pieces back together and divide the whole thing fairly as 2 people did earlier and person 1 left with the last piece. ( in this case person 1 is interested in dividing the loot into 3 equal parts risking missing out if one piece was smaller and two other peoplke chose the bigger pieces). > case 2: > Now, If persons 2 and 3 chose differnt pieces, one piece will be in intersection eg they both want the same piece then they divide this wanted piece between them as 2 2 people would failry share it. Then, each person 2 and 3 would share their own piece failry with person 1 giving him a fiar 1/3 too. > In all cases, person 1 is interested in dividing the loot in 3 equal pieces and so do persons 2 and 3. > i wonder if this method could be applied to n number of people recursively The original statement for 2 parties can be done for any number recursively if it is just restated: One person divides up the loot, then he gets the last pile. === Subject: sum of tan^n(2x+1) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i241STd07975; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i2411xi05125 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i2411x806868; Some days ago I asked assitance in proving Sum(tan^2(2x+1))= 4005 where x goes from 0 to 44 and the angles are in degrees. Larossa gave me a beautiful solution using DeMoivre. Now my second question, How do you derive the general formula for sum(tan^n(2x+1)), where x goes from 0 to 44 and the angles are in degrees? === Subject: Re: sum of tan^n(2x+1) > Some days ago I asked assitance in proving > Sum(tan^2(2x+1))= 4005 where x goes from 0 to 44 and the angles are > in degrees. Larossa gave me a beautiful solution using DeMoivre. ÔLarrosa ... > Now my second question, > How do you derive the general formula for > sum(tan^n(2x+1)), where x goes from 0 to 44 and the angles are in > degrees? Let S(n) = Sum(tan^(2n)((2x+1).bc), x, 0, 44) The square of the tangents of the angles (2x+1).bc, x = 0, 1, .., 44 verify the equation 0 = Sum((-1)^kComb(90, 2k)u^k, k, 0, 45) This allow you to express u^45 as function of u^1, u^2, ..., u^44. Multiplying by u, u^2, ..., you get u^(45+k) as a function of u^(1+k), u^(2+k), ..., u^(44+k) Writting that for the 45 values and adding, you get S(45) = Sum((-1)^mComb(90, 2m)S(m), m, 0, 44) S(k+45) = Sum((-1)^mComb(90, 2m)S(k+m), m, 0, 44) a pretty recurrence relation that give S(n), known the previous 45 values S(n-1), ..., S(n-45) ... If you try to obtain an explicit formula using the standard lineal requrrence technics, you will get S(n) = tan^(2n)(1.bc) + tan^(2n)(2.bc) + ... + tan^(2n)(89.bc) no very interesting ... I dont know if something better can be obtained. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: (-2/3)^(-2/3) = (3/2)^(2/3)? ... > You effectively claimed that x^y could be defined for real negative x > and at least some values of y. So it seems reasonable to ask you exactly > how you plan to define it. > If y is expressible as p/q where p and q are integers and q is odd, then > x^y can be defined as the pth power of the real qth root of x, and this > is valid for all real x. > As the set of such y = p/q with q odd is dense in the reals, one could > extend that definition. Not in any sensible way. If x is negative, x^y > 0 would be dense over y and x^y < 0 would also be dense. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: (-2/3)^(-2/3) = (3/2)^(2/3)? > ... > > You effectively claimed that x^y could be defined for real negative x > > and at least some values of y. So it seems reasonable to ask you > > exactly > > how you plan to define it. > > > > If y is expressible as p/q where p and q are integers and q is odd, then > > x^y can be defined as the pth power of the real qth root of x, and this > > is valid for all real x. > > > > As the set of such y = p/q with q odd is dense in the reals, one could > > extend that definition. > Not in any sensible way. If x is negative, x^y > 0 would be dense over y > and x^y < 0 would also be dense. Then limit it to cases where p/q is a quotient of odd integers. I think one still gets density, from which one can fill in the gaps by continuity. === Subject: Re: (-2/3)^(-2/3) = (3/2)^(2/3)? ... > > As the set of such y = p/q with q odd is dense in the reals, one could > > extend that definition. > > Not in any sensible way. If x is negative, x^y > 0 would be dense over y > and x^y < 0 would also be dense. > Then limit it to cases where p/q is a quotient of odd integers. > I think one still gets density, from which one can fill in the gaps by > continuity. In that case your limiting process as you described will work the wrong way in a number of cases. Or do you really want (-1)^2 = -1? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: (-2/3)^(-2/3) = (3/2)^(2/3)? > Then limit it to cases where p/q is a quotient of odd integers. > I think one still gets density, from which one can fill in the gaps by > continuity. The gaps would include even integers. Do you want (-2)^2 to equal -4? -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: (-2/3)^(-2/3) = (3/2)^(2/3)? > Then limit it to cases where p/q is a quotient of odd integers. > I think one still gets density, from which one can fill in the gaps by > continuity. Thats true, of course. > The gaps would include even integers. Do you want (-2)^2 to equal -4? Exactly. I should think he doesnt want that. The operation which would be obtained by the procedure described by Virgil is, I think, merely sgn(x) |x|^y That is indeed a useful operation, but I dont think we want to call it just x^y. BTW, one example of its use: Suppose youre dealing with software which gives an error if you try to raise a negative number to a power, but you need to be able to get real cube roots. Then use sgn(x) |x|^(1/3). David Cantrell === Subject: Re: (-2/3)^(-2/3) = (3/2)^(2/3)? > Then limit it to cases where p/q is a quotient of odd integers. > I think one still gets density, from which one can fill in the gaps by > continuity. > Thats true, of course. > The gaps would include even integers. Do you want (-2)^2 to equal -4? > Exactly. I should think he doesnt want that. If I recollect, the question was not whether one wanted to do it but whether one could do it. > The operation which would be obtained by the procedure described by Virgil > is, I think, merely > sgn(x) |x|^y > That is indeed a useful operation, but I dont think we want to call it > just x^y. > BTW, one example of its use: > Suppose youre dealing with software which gives an error if you try to > raise a negative number to a power, but you need to be able to get real > cube roots. Then use sgn(x) |x|^(1/3). > David Cantrell === Subject: Re: (-2/3)^(-2/3) = (3/2)^(2/3)? >> Why (in general) operation exponentiation on reals is not defined >> for such cases? >> What is the reason 8^(1/3) to be defined and (-8)^(1/3) to not be? >> 1/3 is not just real, its rational. How, pray tell, \ would you >> define (-8)^x, where x is irrational? Can you clarify the connection between my question and your response? > You effectively claimed that x^y could be defined for real negative x > and at least some values of y. So it seems reasonable to ask you > exactly how you plan to define it. > Derek Holt. > If y is expressible as p/q where p and q are integers and q is odd, then > x^y can be defined as the pth power of the real qth root of x, and this > is valid for all real x. > As the set of such y = p/q with q odd is dense in the reals, one could > extend that definition. I cant think of anything reasonable which would allow that definition to be extended. Specifically, how were you thinking it could be extended? David Cantrell === Subject: Re: (-2/3)^(-2/3) = (3/2)^(2/3)? >> Why (in general) operation exponentiation on reals is not defined >> for such cases? >> What is the reason 8^(1/3) to be defined and (-8)^(1/3) to not be? >> 1/3 is not just real, its rational. How, pray tell, \ would you >> define (-8)^x, where x is irrational? Can you clarify the connection between my question and your response? You effectively claimed that x^y could be defined for real negative x > and at least some values of y. So it seems reasonable to ask you > exactly how you plan to define it. Derek Holt. If y is expressible as p/q where p and q are integers and q is odd, then > x^y can be defined as the pth power of the real qth root of x, and this > is valid for all real x. > As the set of such y = p/q with q odd is dense in the reals, one could > extend that definition. > I cant think of anything reasonable which would allow \ that definition to > be extended. Specifically, how were you thinking it could be extended? > David Cantrell Consider the following as a possible path to such an extension: Let S be the subset of Q, the rationals, of those rationals representable as a quotient, m/n, of integers with odd denomnator n, so that each real has a unique real nth root. Since S is easily shown to be closed under addition subtraction and multiplication, it is a subring of Q. S is also dense in Q and in R, the set of reals. For an arbitrary negative real, r, define f_r: S -> R, s -> r^s, with x^s being defined as explained above. For every real number, x, there are sequences in S converging to x, Let s_n be such a sequence and consider f_r(s_n). Do all such sequences f_r(s_n), with s_n converging to x in R , converge to the same value in R? I suspect so. If so, then let that value be the desired definition of r^x. And if that one wont work, Let S be the set of rationals \ with denominators powers of some fixed odd prime, and follow the above pattern. Now I have not really investigated these possibilities closely, but I see no immediate reason why they would not work. Do you? === Subject: Re: (-2/3)^(-2/3) = (3/2)^(2/3)? > Consider the following as a possible path to such an extension: > Let S be the subset of Q, the rationals, of those rationals > representable as a quotient, m/n, of integers with odd denomnator n, so > that each real has a unique real nth root. > Since S is easily shown to be closed under addition subtraction and > multiplication, it is a subring of Q. S is also dense in Q and in R, the > set of reals. > For an arbitrary negative real, r, define f_r: S -> R, s -> r^s, with > x^s being defined as explained above. > For every real number, x, there are sequences in S converging to x, > Let s_n be such a sequence and consider f_r(s_n). > Do all such sequences f_r(s_n), with s_n converging to x in R , converge > to the same value in R? I suspect so. If so, then let that value be the > desired definition of r^x. Consider the real number (sqrt(5)-1)/2 . There are many sequences of rational numbers that converge to it, but the most interesting ones are based on the ratios of consecutive Fibonacci numbers. Take these two in particular, stepping three at a time: 1/1, 3/5, 13/21, ... 2/3, 8/13, 34/55, ... In general, any term s is followed by (x+2)/(2x+3). If you use the first sequence as the exponents of a negative real, the results will all be negative. If you use the second sequence as the exponents of a negative real, the results will all be positive. In either case, theyll be far enough from zero to eliminate \ the easy way to resolve the sign. > And if that one wont work, Let S be the set of rationals with > denominators powers of some fixed odd prime, and follow the above > pattern. > Now I have not really investigated these possibilities closely, but I > see no immediate reason why they would not work. Do you? Because you cant reconcile the results of an exponent with an even numerator and with an odd numerator. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: (-2/3)^(-2/3) = (3/2)^(2/3)? > In general, any term s is followed by (x+2)/(2x+3). Oops. Any term s is followed by (s+2)/(2s+3). -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: question re product measures Let (X, R, mu) and (Y, S, nu) be sigma-finite measure spaces. A is a set in the hereditary sigma-ring generated by R, B is a set in the hereditary sigma-ring generated by S. Prove (mu x nu)*(A x B) = mu*(A)nu*(B). Note that * denotes the outer measure generated by. === Subject: Re: question re product measures I can show that (mx x nu)*(A x B) >= mu*(A)nu*(B), but am stuck on how === Subject: Re: question re product measures > I can show that (mx x nu)*(A x B) >= mu*(A)nu*(B), but am stuck on how Hmmm, this question sounds oddly familiar :) I can show that (mu x nu)*(A x B) <= mu*(A)nu*(B), so maybe we can help each other out :) Fix e > 0. Choose a countable covering {U_i} of A such that sum_i mu(U_i) < mu*(A) + e (the U_i are in R) and a countable covering {V_j} of B such that sum_j nu(V_i) < nu*(B) + e (the V_i are in S). Then the countable collection {U_i x V_j} is a covering of A x B and thus (mu x nu)*(A x B) <= sum_i,j (mu x nu)(U_i x V_j) = sum_i,j mu(U_i)nu(V_j) = sum_i mu(U_i) sum_j nu(V_j) < sum_i mu(U_i) (nu*(B) + e) < (mu*(A) + e) (nu*(B) + e) = mu*(A)nu*(B) + e(mu*(A) + nu*(B) + e). Now take e -> 0 to get the inequality we want. Hopefully you can help me with the other direction :) TM === Subject: Re: Science Without Math? (model-free common sense steering) > There are processors which have 1000 eyes and sensory inputs limited > to the < 10^-6 Hz range, which humans dont. One microHertz? ...approximately once every eleven and a half days? > Youre right, I dont know any humans with \ vision that poor! > Obviously you dont, since 1uHz is not even > in the visible spectrum. But, then again, > those of with working machines have > never really cared what philosophers > babble about. > Are you garunteeing that there are absolutely no such machine workers? Can > you certify that no machine workers care what philisophers say? What about a > warrenty in case I buy a sale of goods, money back if I find one? No, I dont say that. Hollywood film studios \ care what philosophers think. Since if you havent noticed Arnold Scwartznegger is now their lead spokesmen, rather than Adolf Hilter, so the world changes everyday. === Subject: Arclength Calculation Angle x subtends known arclength A and chord B (B/A)(x/2)=sin(x/2) trigonometric representation sin(x/2)=x/2-(x/2)^3/3!+(x/2)^5/5!-... Maclaurin sine t=x/2 x=2t Suppose it is desired to have 7th order precision for x a[0] =1-B/A a[1] = 0 a[2] = -1/3! a[3] = 0 a[4] = 1/5! a[5] = 0 a[6] = -1/7! a[7] = 0 a[8] = 1/9! a[9] = 0 a[10]= -1/11! a[11]= 0 a[12]= 1/13! a[13]= 0 a[14]= -1/15! +( a[ 1] +a[ 3]t^2)a[1] =0 +( a[ 2]-a[ 3]t+a[ 4]t^2)a[2] +( a[ 3]-a[ 4]t+a[ 5]t^2)a[3] =0 +( a[ 4]-a[ 5]t+a[ 6]t^2)a[4] +( a[ 5]-a[ 6]t+a[ 7]t^2)a[5] =0 +( a[ 6]-a[ 7]t+a[ 8]t^2)a[6] +( a[ 7]-a[ 8]t+a[ 9]t^2)a[7] =0 +( a[ 8]-a[ 9]t+a[10]t^2)a[8] +( a[ 9]-a[10]t+a[11]t^2)a[9] =0 +( a[10]-a[11]t+a[12]t^2)a[10] +( a[11]-a[12]t+a[13]t^2)a[11]=0 +( a[12]-a[13]t+a[14]t^2)a[12] +( a[13]-a[14]t )a[13]=0 +( a[14] )a[14] ------------------------------ +( a[ 2]-a[ 3]t+a[ 4]t^2)a[ 2] +( a[ 4]-a[ 5]t+a[ 6]t^2)a[ 4] +( a[ 6]-a[ 7]t+a[ 8]t^2)a[ 6] +( a[ 8]-a[ 9]t+a[10]t^2)a[ 8] +( a[10]-a[11]t+a[12]t^2)a[10] +( a[12]-a[13]t+a[14]t^2)a[12] +( a[14] )a[14] ------------------------------ a[3]=a[5]=a[7]=a[9]=a[11]=a[13]=0 +( a[ 2]-a[ 4]t^2)a[ 2] +( 2a[ 4]-a[ 6]t^2)a[ 4] +( 2a[ 6]-a[ 8]t^2)a[ 6] +( 2a[ 8]-a[10]t^2)a[ 8] +( 2a[10]-a[12]t^2)a[10] +( 2a[12]-a[14]t^2)a[12] +( 2a[14]-a[ 0]t^2)a[14] +( a[ 0] )a[16] ------------------------ = a[0]^2 multiplying across and adding the columns, ( -a[2]a[4]-a[4]a[6]-a[6]a[8]-a[8]a[10]-a[10]a[12] -a[12]a[14]-a[0]a[14] )t^2 + a[2]^2+2a[4]^2+2a[6]^2+2a[8]^2+2a[10]^2+2a[12]^2+ +2a[14]^2+a[0]a[16]-a[0]^2 t= { ( a[2]^2+2a[4]^2+2a[6]^2+2a[8]^2+2a[10]^2+2a[12]^2+ +2a[14]^2+a[0]a[16]-a[0]^2 ) / ( a[2]a[4]+a[4]a[6]+a[6]a[8]+a[8]a[10]+a[10]a[12]+ +a[12]a[14]+a[0]a[14] ) }^1/2 x=2{C/D}^1/2 C=(-1/3!)^2+2(1/5!)^2+2(-1/7!)^2+2(1/9!)^2+2(-1/11!)^2+.. ..+2(1/13!)^2+2(-1/15!)^2+(1/17!)(1-B/A)-(1-B/A)^2 D=(-1/3!)(1/5!)+(1/5!)(-1/7!)+(-1/7!)(1/9!)+(1/9!)(-1/11!)+.. ..+(-1/ll!)(1/13!)+(1/13!)(-1/15!)+(1/15!)(1-B/A) Selecting the first term of C and the first term \ of D, x~ 2{5!3!(1-B/A)^2 - 5!/3!}^1/2 B A x calculated ----------------------------------------------- 0 0 0 (2.8)pi *2 ~ 6pi = 0 2 pi pi (5.5)pi *2 ~ 11pi = pi 2^1/2 pi/2 pi/2 (2.3)pi *2 ~ 9/2pi = pi/2 . . === Subject: Arclength Calculation Angle x subtends known arclength A and chord B (B/A)(x/2)=sin(x/2) trigonometric representation sin(x/2)=x/2-(x/2)^3/3!+(x/2)^5/5!-... Maclaurin sine t=x/2 x=2t Suppose it is desired to have 7th order precision for x a[0] =1-B/A a[1] = 0 a[2] = -1/3! a[3] = 0 a[4] = 1/5! a[5] = 0 a[6] = -1/7! a[7] = 0 a[8] = 1/9! a[9] = 0 a[10]= -1/11! a[11]= 0 a[12]= 1/13! a[13]= 0 a[14]= -1/15! +( a[ 1] +a[ 3]t^2)a[1] =0 +( a[ 2]-a[ 3]t+a[ 4]t^2)a[2] +( a[ 3]-a[ 4]t+a[ 5]t^2)a[3] =0 +( a[ 4]-a[ 5]t+a[ 6]t^2)a[4] +( a[ 5]-a[ 6]t+a[ 7]t^2)a[5] =0 +( a[ 6]-a[ 7]t+a[ 8]t^2)a[6] +( a[ 7]-a[ 8]t+a[ 9]t^2)a[7] =0 +( a[ 8]-a[ 9]t+a[10]t^2)a[8] +( a[ 9]-a[10]t+a[11]t^2)a[9] =0 +( a[10]-a[11]t+a[12]t^2)a[10] +( a[11]-a[12]t+a[13]t^2)a[11]=0 +( a[12]-a[13]t+a[14]t^2)a[12] +( a[13]-a[14]t )a[13]=0 +( a[14] )a[14] ------------------------------ +( a[ 2]-a[ 3]t+a[ 4]t^2)a[ 2] +( a[ 4]-a[ 5]t+a[ 6]t^2)a[ 4] +( a[ 6]-a[ 7]t+a[ 8]t^2)a[ 6] +( a[ 8]-a[ 9]t+a[10]t^2)a[ 8] +( a[10]-a[11]t+a[12]t^2)a[10] +( a[12]-a[13]t+a[14]t^2)a[12] +( a[14] )a[14] ------------------------------ a[3]=a[5]=a[7]=a[9]=a[11]=a[13]=0 +( a[ 2]-a[ 4]t^2)a[ 2] +( 2a[ 4]-a[ 6]t^2)a[ 4] +( 2a[ 6]-a[ 8]t^2)a[ 6] +( 2a[ 8]-a[10]t^2)a[ 8] +( 2a[10]-a[12]t^2)a[10] +( 2a[12]-a[14]t^2)a[12] +( 2a[14]-a[ 0]t^2)a[14] +( a[ 0] )a[16] ------------------------ = a[0]^2 multiplying across and adding the columns, ( -a[2]a[4]-a[4]a[6]-a[6]a[8]-a[8]a[10]-a[10]a[12] -a[12]a[14]-a[0]a[14] )t^2 + a[2]^2+2a[4]^2+2a[6]^2+2a[8]^2+2a[10]^2+2a[12]^2+ +2a[14]^2+a[0]a[16]-a[0]^2 t= { ( a[2]^2+2a[4]^2+2a[6]^2+2a[8]^2+2a[10]^2+2a[12]^2+ +2a[14]^2+a[0]a[16]-a[0]^2 ) / ( a[2]a[4]+a[4]a[6]+a[6]a[8]+a[8]a[10]+a[10]a[12]+ +a[12]a[14]+a[0]a[14] ) }^1/2 x=2{C/D}^1/2 C=(-1/3!)^2+2(1/5!)^2+2(-1/7!)^2+2(1/9!)^2+2(-1/11!)^2+.. ..+2(1/13!)^2+2(-1/15!)^2+(1/17!)(1-B/A)-(1-B/A)^2 D=(-1/3!)(1/5!)+(1/5!)(-1/7!)+(-1/7!)(1/9!)+(1/9!)(-1/11!)+.. ..+(-1/ll!)(1/13!)+(1/13!)(-1/15!)+(1/15!)(1-B/A) Selecting the first term of C and the first term \ of D, x~ 2{5!3!(1-B/A)^2 - 5!/3!}^1/2 B A x calculated ----------------------------------------------- 0 0 0 (2.8)pi *2 ~ 6pi = 0 2 pi pi (5.5)pi *2 ~ 11pi = pi 2^1/2 pi/2 pi/2 (2.3)pi *2 ~ 9/2pi = pi/2 . . === Subject: Is this proof of Hadwigers Conjecture right? http://arxiv.org/abs/math.GM/0311475 Its idea is very strange and simple. It said that those conjectures are topological problems, the proof is out of graph theory. Is it right? === Subject: Re: Is this proof of Hadwigers Conjecture right? shealy > http://arxiv.org/abs/math.GM/0311475 > Its idea is very strange and simple. It said that those conjectures > are topological problems, the proof is out of graph theory. Is it > right? He doesnt prove this critical statement on page 4: Any planar map is in C4R2, so there does not exist a solid C5. But the scheme is worth considering. In R3 take any five points w,x,y,z, and the origin, such that no 3 of the five are collinear. The set of points u in R3 of the form aw+bx+cy, a,b,c all positive, form a solid blob, and there are three more such blobs made from {w,x,z}, {w,y,z}, {x,y,z} respectively. The four are pairwise adjacent in an obvious sense. Now given a plane map, or (more convenient) a map on the sphere, can we embed the sphere in R3 such that its intersections with these blobs are the given countries? Yes _if_ the graph is four-colorable! I leave it as an exercise :) LH === Subject: Re: Is this proof of Hadwigers Conjecture right? Larry Hammick > shealy > http://arxiv.org/abs/math.GM/0311475 > Its idea is very strange and simple. It said that those conjectures > are topological problems, the proof is out of graph theory. Is it > right? > He doesnt prove this critical statement on page 4: > Any planar map is in C4R2, so there does not exist a solid C5. > But the scheme is worth considering. In R3 take any five points w,x,y,z, and > the origin, such that no 3 of the five are collinear. Sorry, we need to assume a little more. It is enough that the inner product of any two of w,x,y,z is <0. > The set of points u in > R3 of the form aw+bx+cy, a,b,c all positive, form a solid blob, and there > are three more such blobs made from {w,x,z}, {w,y,z}, {x,y,z} respectively. > The four are pairwise adjacent in an obvious sense. Now given a plane map, > or (more convenient) a map on the sphere, can we embed the sphere in R3 such > that its intersections with these blobs are the given countries? Yes _if_ > the graph is four-colorable! I leave it as an exercise :) > LH === Subject: re:Is this proof of Hadwigers Conjecture right? Maybe theyre using some facts I dont know \ of, but to me it looks like a load of BS. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Question about Partial Derivatives Let u= f(x,y) let x= g(s,t) let y= h(s,t) du/ds = du/dx* dx/ds+ du/dy*dy/ds now when you take a second derivatives, d2u/ds2 = d/du[du/dx* dx/ds+ du/dy*dy/ds] And you have to use the chain rule again d2u/dx2*dx/dt+d2u/dx*dy (I am assuming that dx/dt and dy/dt are constant and can be pulled out of the equation.) Now, part of the trouble is on the above line when i have d2u/dx*dy, does the order of dx*dy matters? I know that from cholorus theorem you can swap them, but I am not sure that does it matter or not? === Subject: Re: Question about Partial Derivatives > Let u= f(x,y) > let x= g(s,t) > let y= h(s,t) > du/ds = du/dx* dx/ds+ du/dy*dy/ds > now when you take a second derivatives, > d2u/ds2 = d/du[du/dx* dx/ds+ du/dy*dy/ds] > And you have to use the chain rule again > d2u/dx2*dx/dt+d2u/dx*dy (I am assuming that dx/dt and dy/dt are > constant and can be pulled out of the equation.) > Now, part of the trouble is on the above line when i have d2u/dx*dy, > does the order of dx*dy matters? I know that from cholorus theorem > you can swap them, but I am not sure that does it matter or not? I believe the theorem states that the order of differentiation is ummaterial if u is continuous. -- === Subject: S_n How do I find all the possible orders of elements in S_n, the group of permutations of {1,...,n} ? My knowledge that the order of an element as the l.c.m. of the lengths of its disjoint cyles isnt helping me much. === Subject: Re: S_n >How do I find all the possible orders of elements in S_n, the group of >permutations of {1,...,n} ? >My knowledge that the order of an element as the l.c.m. of the lengths of >its disjoint cyles isnt helping me much. Well it does help - it tells you that S_n has elements of order p1^a1 p2^a2 ... pr^ar, for distinct primes p1,...,pr if and only if p1^a1 + p2^a2 + ... + pr^ar <= n. I dont believe that there is any better way to solve this problem than some kind of brute force search using that rule. A related question, which is occasionaly asked here is what is the largest order of an element of S_n. There is no simple formual known for this, although there are asymptotic answers. Derek Holt. === Subject: Re: S_n Adjunct Assistant Professor at the University of Montana. >A related question, which is occasionaly asked here is what is the largest >order of an element of S_n. There is no simple formual known for this, >although there are asymptotic answers. Dont know if my dads conjecture was ever \ disproven, but MR0480067 (58 #266) Magidin, Mario The best partition of an integer. Nordisk Tidskr. Informationsbehandling (BIT) 14 (1974), 203--208. 05A17 Authors summary: An algorithm, based on a conjecture, to compute a permutation whose repeated application to a given set will yield a maximum number of different orderings of that set is presented. The algorithm gives the lengths of the cycles required. This problem turns out to be equivalent to the problem of determining a partition $B(n)$ of $n$ for which the least common multiple (l.c.m.) of the numbers of $B(n)$ is maximal. -- Its not denial. Im just very selective \ about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: S_n > How do I find all the possible orders of elements in S_n, the group of > permutations of {1,...,n} ? > My knowledge that the order of an element as the l.c.m. of the lengths of > its disjoint cyles isnt helping me much. Consider S_3 = {(1)(2)(3), (1,2)(3), (1)(2,3), (1,2,3), (1,3)(2), (3,2,1)}. If P(n) is the partition function for n, then there are P(n) ways of writing permutations as disjoint cycles. Add the identity permutation and you have P(n) + 1 ways of writing permutations in S_n as different length cycles: 3 + 0 = 2 + 1 = 1 + 1 + 1 (2 + 1 ways) The orders of the permutations can then be: LCM(2,1) = 2 LCM(1,1,1) = 1 LCM(3,0) = 3 I dont know any short formula for calculating the possible orders for elements of, say, S_9831. === Subject: Some help is needed please Im sorry to ask but can someone help me with this problem. The problem is as follows: Prove that if H is a subgroup of G with the property that bH=Hb for every b E G, then H is a normal subgroup.