mm-3515 === Subject: Long division I was reading a math book which had a proof of differentiation of a quotient usually written as u/v that I hadn't seen before. In essence the proof relied on doing a long division of the form a + b into c + d As the division proceeded the terms became smaller and smaller and so insignificant. The catch is I've never seen a long division of the form a+b into c+d so can't make sense of it. Please can someone explain how one does such a long division. Best wishes, -- Quentin Grady ^ ^ / / / ... and the blind dog was leading. http://homepages.paradise.net.nz/quentin === Subject: Re: Long division ----------------------------------------------------------------- One way that you can do this, a way that is often in older texts (such as those published before 1940), is by rationalizing the denominator. Beginning with (delta y/x) = [y + (delta y)] / [x + (delta x)], which is the change in y/x, multiply both the numerator and the denominator by x - delta(x). This gives you [y + (delta y)]*[x - (delta x)] / [ x^2 - (delta x)^2 ] [ xy + x*(delta y) - y*(delta x) - (delta x)(delta y)] divided by [ x^2 - (delta x)^2 ]. Keeping only the first order changes, which is what one does for linear approximations, we get [ xy + x*(delta y) - y*(delta x) ] / x^2 (xy)/(x^2) + [ x*(delta y) - y*(delta x) ] / x^2. Hence, up to first-order terms, (delta y/x) - y/x is equal to [ x*(delta y) - y*(delta x) ] / x^2, which is the quotient rule (in first-order finite form). ----------------------------------------------------------------- What you're suggesting is that we can get the same thing by long division. So let's just do the long division and see what happens. We want to divide x + (delta x) into y + (delta y). Ideally, you'll want to set this up the usual way when dividing polynomials: the column format, where leading term is divided into leading term, multiply and bring down, subtract, divide leading term into leading term again, etc. I'll describe what I'm doing, but I'm not going to attempt any ASCII art to display things the usual way one would write this in a high school algebra class. Dividing x into y gives y/x. Multiplying y/x by x + (delta x) gives y + (y/x)*(delta x). Subtracting y + (y/x)*(delta x) _from_ y + (delta y) gives (delta y) - (y/x)*(delta x). Dividing x into (delta y) gives (delta y)/x. Multiplying (delta y)/x by x + (delta x) gives (delta y) + (delta x)(delta y)/x. Subtracting (delta y) + (delta x)(delta y)/x _from_ (delta y) - (y/x)*(delta x) gives (-y/x)*(delta x) - (delta x)(delta y)/x. Dividing x into (-y/x)*(delta x) gives (-y/x^2)*(delta x). Multiplying (-y/x^2)*(delta x) by x + (delta x) gives (-y/x)*(delta x) - (y/x^2)*(delta x)^2. Subtracting (-y/x)*(delta x) - (y/x^2)*(delta x)^2 _from_ (-y/x)*(delta x) - (delta x)(delta y)/x leaves only terms of second order, and so to a first order approximation we're done. The quotient we've generated is y/x + (delta y)/x - (y/x^2)*(delta x). Hence, up to first-order terms, y/x - (delta y/x) is equal to (delta y)/x - (y/x^2)*(delta x), which is equal to [ x*(delta y) - y*(delta x) ] / x^2, and again we have the quotient rule. ----------------------------------------------------------------- Dave L. Renfro === Subject: Re: Long division Perhaps if you could reproduce a few lines of that long division, we might have a better idea of what is going on. === Subject: Re: Long division Consider (for instance) that all the math we do is essentialy with polynomials operating against polynomials. e.g.: 1234 * 5678 That could also be considered as (1000 * 1 + 100 * 2 + 10 * 3 + 4) * (1000 * 5 + 100 * 6 + 10 * 7 + 8) Of course, that grouping is arbitrary. We can collect the terms however we like and still get the same result. If you give a pointer to the proof in question, I guess that someone can help you to see it clearly. === Subject: Connections, Curvature, Torsion, Nonmetricity & Star Gate Time Travel bcc OK Jack, fire away. Give me your best shot. Here is a direct quote from Carroll: ...any set of connections can be expressed as some fiducial connection plus a tensorial correction. (Notes on General Relativity, p 59). Now explain to us why this doesn't apply to the LC connection? I told you a jillion times. Read the rest of what he says! Clearly, a zero correction is no correction at all. For Carroll's statement to be non-trivial, the tensorial correction must be non-vanishing. Yet you are saying, in effect, that in the case of the LC connection the tensorial correction is always exactly zero. I think you have to prove this mathematically, in the abstract, without any appeal to any particular physical theory. Irrelevant. Sean Carroll like many others points out that most generally Affine = LC + Torsion + Nonmetricity Torsion and non-metricity are GCT 3rd rank tensors. LC is not a tensor. GR is precisely that theory in the set of theories with different connections such that Affine = LC End of trivial story. Physically LC parallel transports vectors along worldlines. Going around a closed worldline the difference of orientation of the vector with itself is a measure of the average curvature over the area of the loop - shrunk to zero - i.e. crystal disclination defect. Limit of the ratio of angle deficit to area of shrinking loop is the sectional curvature dimensions 1/Area. If there is non-metricity the length of the vector will also change in the parallel transport and in a loop you will get a discrepancy in lengths of the vector unless you impose a kind of Bohr-Sommerfeld quantization on the closed loop. If there is torsion starting with two sides of an infinitesimal parallelogram and parallel transporting one against the other both ways the loop does not close to second order - there is a torsion gap, i.e. crystal dislocation defect. In my theory you have a set of 0-forms the Goldstone phases of the vacuum coherence ODLRO Higgs field. There are 10 1-forms. 4 tetrad 1-forms & 6 spin connection 1-forms. Then you have GR + torsion trivially. i.e. a theory with A = LC + Torsion Tensor beyond 1915 GR of course. dO-form = 1-form d^2)-form = 0 However that 1-form need not be exact if the 0-form has a singularity, i.e. Integral of the 1-form over a non-bounding 1-cycle without boundary = winding number integer Bohr-Sommerfeld condition from non-trivial 1-homotopy. Then you can pretend Stoke's theorem works i.e. flux without flux - these are 1D string vortex defects. The integral of the 1-form around the loop is the surface integral of a ghostly 2-form flux through the area of the loop. This is related to the Bohm-Aharonov effect. Similarly given 2 0-forms A & B, I define the non-closed 1-form as C = dA/B - A/dB 1-form dC = 2dA/dB =/= 0 2-form I can now play the same game one dimension higher! These are 0D point monopole stable defects. Now we have star gate wormholes! The mouth of the wormhole is the non-bounding 2-cycle without boundary - the portal like they see on the Skinwalker Ranch! The deRham integral of 2-form C over the closed 2-surface is now an integer wrapping number ~ Bekenstein BITS for the entropy of black hole event horizons, de Sitter dark energy future observer horizons, Unruh effect, Hawking radiation, you name it. I got it all topologically pre-metrically. There is again the world hologram ghostly volume 3-form without volume from the actual hologram area 2-form dC. OK everything I need comes from C as the template. C^a^b = dA^a/B^b - A^a/dB^b The 4 Einstein curvature tetrad 1-forms are the diagonal matrix elements C^a^a The 6 torsion field spin connection 1-forms are the anti-symmetrized off-diagonal matrix elements C^[a,b] = - C^[b,a] Einstein's 1915 GR is the limit C^[a,b] = 0 D = d + W^ac/ e^a = I^a + C^a^a The torsion 1-form is T^a = De^a = 0 This determines the non-dynamical zero torsion spin connections W^ac from de^a + W^ac/e^c = 0 The curvature 2-form is R^a^b = DW^a^b The Einstein-Hilbert action density is {a,b,c,d}R^a^b/e^c/e^d and the rest is history. See Rovelli Ch II for the details. Plus it's obvious how to generalize for torsion. So I have 8 0-form Goldstone phases from 9 real Higgs fields with vacuum manifold S^8. This fits most naturally in terms of stable defects into a 9+1 D spacetime. Hey that's interesting. Where have we seen that before? I want to compactify to 3 large space dimensions so I can use A^2 = A^aAa B^2 = B^aBa take square roots. And I am back to the Bekenstein world holography bits with point defects in the world crystal's spacelike slices. They are the lattice points! Jack Sarfatti sarfatti@pacbell.net If we knew what it was we were doing, it would not be called research, would it? - Albert Einstein http://www.authorhouse.com/BookStore/ItemDetail.aspx?bookid=23999 http://lifeboat.com/ex/bios.jack.sarfatti http://qedcorp.com/APS/Dec122006.ppt http://www.flickr.com/photos/lub/sets/72157594439814784 === Subject: Re: Cantor Confusion That is what I say, but according to set theory it is wrong. N is the infinite set of all *finite* numbers. Accordinfg to set theory there is no infinite number in it although the limit aleph 0 of numbers is defined by set theory. Yes, it is a mess. But it is exactly what everybody asserts when being talking about the infinite set of finite numbers. Give me Cantors diagonal constructed up to any digit d nn and I will show you the place where d nn can be fuond in the list. In case of Cantor's argument you apply the principle: Compare with all numers simultaneously. In case of N differing from its final segments you allow only comparison one after the other. That is bad logic. You say it above. Why do you use such ridiculous examples to demonstrate your non- quantifier dyslexia? The initial segments of N are not like your A1, A2, ... as you know well. Only wrong analogies like that above can lead to the opinion that forall A thereis x such that: x notin A and x in S could be possible for linear segments. === Subject: Re: Cantor Confusion Because you have a two step approach to showing that S cannot differ from every A_i I: there is no single x such that S differes from every A_i by x II: It is sufficient to show that there is not single x such that S differs from every A_i by x I is trivial, but you spend all your energy on I. You only mention II when someone points out that I is not in general sufficient. And asked for a proof of II you gave a proof of I. Here is a counterexample using sets you have described. Let E be the set of all even natural numbers. Let the A_i be the initial segments of E, (i.e. A_1 ={2}, A_2 = {2,4}, A_3 = {2,4,6}, A_i={2,4,6,...,2i}) Then I: There is no single x such that E differs from every A_i by x. II: E differs from every A_i - William Hughes === Subject: Re: Cantor Confusion According to WM's view of set theory that is wrong, but WM knows nothing of set theory. Not at all, there are all those finite rationals and negatives around that are not in N. N is the infinite set of all *natural* numbers, all of which are finite. According to set theory there is no infinite NATURAL number, but as aleph_0 is not a natural, it is quite acceptable that it be infinite. WM is getting desperate if he has to conflate naturals, which are all finite, with the more general cardinals or ordinals which are not so limited. No, it is only the mess WM tries, and fails, to create. You must first give us the list of reals on which the Cantor rule is to be applied. The Cantor proof is in the form of a claim: Cantor says, give me a list of reals and I will give you an real not listed in that list. So that WM, as challenger of that claim, must first produce a list. Only then need Cantor do anything. It is better logic that WM has yet exhibited. Cantor merely says that when you give him a list he can apply a rule which will produce a number not listed. Since his rule produces a number different from any in the list, his claim succeeds. The time it might take to put his rule into effect is immaterial. No one but WM and his fellow kooks says anything so stupid. It is WM's actual quantifier dyslexia at issue here. Given N as the set of all naturals and S_n, for n in N, as the set of all naturals less than n, It is true that: For each S_n, there is an x in N such that x is not a member of S_n. It is false that: There is an x in N such that for each S_n, x is not a member of S_n. WM habitually conflates these two statements, which is the logical fault called quantifier dyslexia. Except that it is necessarily the case in ZF and NBG and the wrongness exists only in WM's head, due to his quantifier dyslexia. For each and every finite initial segment of naturals, there is a natural in the set of all naturals not in that segment. This in no way requires that it be a same one for all segments. The construction: for all x there is a y should always read for all x there is a y whose value in general depends on x. WM wants it to mean for all x there is a y whose value is independent of x, but it rarely, if ever, does. At least in mathematics. Consider, for example , the delta-epsilon definition of a limit: For every positive epsilon there is a positive delta (depending on epsilon),... If one were to allow WM's meaning, calculus would still be mired in fluxions and fluents. For someone purporting to have produced a text on mathematics, WM's ignorance of his subject is appalling. === Subject: Re: Cantor Confusion As all the finite subsets like {p(0), p(1), p(2), ...} which belong to an infinite path like p(oo) form a countable set, the set of all unions is counable. This set of all unions is P(oo) = {p(oo), q(oo), ...}. It is easy to undertand. We are investigating this assertion. We find that every finite tree contaibns only finite paths. The union of some finite paths is an ifinite path. Thee are only countable many unions of finite paths which yieldinfinite paths. As the unions of finite subsets of paths like {0., 0.1, 0.10, 0.100, ...} form a countable set, the path 0. is used countably often. === Subject: Re: Cantor Confusion This only argues that /one/ path is countable, which no one disputes, not that the set of all paths is countable. This set of all unions is P(oo) = {p(oo), q(oo), Wrong! WM indicates it as a list, but it cannot be listed. Does WM claim to be able to list all endless binary sequences? It is trivial that there are as many paths as such sequences, so until WM proves the sequences listable, he cannot claim the paths to be listable. But then, WM is quite incapable of proving anything to mathematical standards, at least on the evidence of his futile attempts to do so. As there is one path for each of the uncountably many endless binary seqeunces in such an infinite tree, WM is wrong. Once in each of uncountably many paths. The only reason that WM objects to infiniteness is that his mental powers are just not up to dealing with it successfully. WM cannot reason it out, so he has to guess, and he guesses wrong. === Subject: Re: Cantor Confusion I cannot hinder you, but it is misleading. Rational numbers. === Subject: Re: Cantor Confusion Not to us. And WM manages to mislead himself without our help, so it won't affect him any more then he affects himself. 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Rappaport $8.00 for e-books(fix price) if buy 2 ebooks free 1 ebooks Ebooks Modern Control Engeneering by ogata Discrete Time Signal Processing by alan v.oppen Numerical Methods 3rd Edition by faires Calculus Early Transcendentals 5e by James Stewart Introduction to Thermal 5E : Thermodynamics, Fluid Mechanics, and Heat Transferby Michael J. Moran,Howard N. Shapiro,Bruce R. Munson and David P. DeWit Serway Jewett Phys for Scientist and Eng 6Ed Fluid Mechanics 4th ed - F. White Thermodynamics An Engineering Approach 5th Edition by cengel Applied Statistics And Probability For Engineers by D. Montgomery & Runger Numerical Methods in Engineering with MATLAB (2005) Numerical methods matlab by Won Young Yang, Wenwu Cao, Tae-Sang Chung and John Morris === Subject: Re: 'Generalized spiral points': further improvement Below is an informal comparison of the spiral longitude calculations of Bauer and Rakmanov et al. The similarity is really just below the surface. Let us denote the longitude as phi. 1. Rakmanov et al.: dphi(h)/dh = 3.6/sqrt(n)/r (n/2) = 1.8 sqrt(n)/r where the factor n/2 corrects for the non-infinitisimal step size. 2. Bauer: dphi(h)/dh = sqrt(pi n) darccos(h)/dh = sqrt(pi n)/sqrt(1-h^2) = sqrt(pi n)/r = 1.772(..) sqrt(n)/r Knud Thomsen === Subject: Re: 'Generalized spiral points': further improvement If we, in accordance with my previous posting, substitute the constant pi in Bauer's longitude calculation with the corresponding (semi- empirical) value (3.6/2)^2 = 3.24 from the spiral of Rakhmanov, Saff and Zhou, we get slightly better potential energies. In particular I me a bit, looks better. n dE(-1,n) dE( 0,n) dE(+1,n) --------------------------------- 4 1.029355 1.008939 0.984461 5 0.995598 0.983682 0.965456 6 1.013700 1.003542 0.983430 7 0.963860 0.961153 0.960249 8 0.978428 0.975673 0.973855 9 0.966625 0.964493 0.963916 10 0.934881 0.936988 0.943605 11 0.946011 0.945805 0.946071 12 0.958207 0.956867 0.954656 13 0.933802 0.935948 0.943261 14 0.917211 0.921943 0.937058 15 0.932295 0.936422 0.950096 16 0.940226 0.943116 0.953898 17 0.919972 0.925354 0.940490 20 0.915948 0.921589 0.933867 25 0.894147 0.905444 0.932603 30 0.863723 0.878911 0.910160 50 0.801400 0.830938 0.891700 100 0.716472 0.767679 0.865553 200 0.626588 0.702990 0.851920 500 0.391312 0.500397 0.767565 1000 0.247882 0.355191 0.699457 2000 0.148632 0.237709 0.629668 5000 0.070795 0.128808 0.534353 10000 0.038232 0.075659 0.455589 20000 0.019176 0.040852 0.365825 Knud Thomsen === Subject: Re: 'Generalized spiral points': further improvement Which brings out a good point to consider further. What is the best measure for the distribution of points? Many have used potential energy as a measure of uniformity of points on a sphere, lower energy implying better uniformity. But there are other measures. For example, consider the variance in the area of the Voronoi cells surrounding each point. Lower variance implies better uniformity. Or the difference between the max and min Voronoi area. Using a measure based on the Voronoi diagram represents a more localized measure. Only the closest points to a given point play a role in the measure. Of course, which measure is best depends on what you plan to do with the points. So how do you plan to use the set of points that you generate? One final point, if I recall correctly RSZ put a point at the south and north poles, whereas Bauer puts points half a step away from the poles. Not sure that this is any better. Just pointing out the difference. - MO === Subject: Re: 'Generalized spiral points': further improvement As pointed out earlier in this thread, I do not doubt that the Bauer spiral converges; its convergence just seemed a bit slow considering its impressive performance for even relatively big values of n. The main reason for my using the energy measure here is that my discussion started in the context of the RSZ spiral, which was specifically designed by the authors to have a low potential energy. One reason I'm interested in a good point spiral is that I have found it to be a very valuable component in an algorithm that generates approximately uniformly distributed points on the surface of a 4D unit hypersphere (these points can then be utilized as appr. uniformly distributed orientation/rotations). Using the spiral repeatedly this way is an interesting alternative to (for example) the recursive zonal equal area method devised by Paul Leopardi (1). It's correct that the RSZ spiral puts a point at each pole and then spaces the remaining points evenly in between. I actually started this thread because the RSZ spiral could be improved considerably by increasing the distance between the polar points and their immediate neighbors. 1. Paul Leopardi, A partition of the unit sphere into regions of equal area and small diameter, Electronic Transactions on Numerical Analysis, Volume 25, 2006, pp. 309-327. Knud Thomsen === Subject: Re: 'Generalized spiral points': further improvement Knud, I am still not understanding what you mean by convergence. Could you try to explain more fully. You are comparing the energy of RSK to the energy of Bauer, seeing that Bauer's energy converges more slowly to RSK's than other algorithms and then are implying that Bauer's slow convergence to RSK is an undesirable trait. Is convergence, fast or slow, to RSK a desirable feature? Also, if what you want is uniformly distributed points, then I would indeed argue that that are better measures of uniformity than potential energy. - MO === Subject: Re: 'Generalized spiral points': further improvement 1. In my tables, I list: the difference in the energy of a particular spiral from the optimum energy, as a fraction of the difference in the energy of the original RSZ spiral from that same optimum energy. More precisely, if E(s,n) is the energy of original RSZ spiral E'(s,n) is the energy of the particular spiral under discussion (e.g. Bauer's spiral) f(s,n) is the optimum energy, i.e. the energy of the optimal point configuration, as approximated in the RSZ paper. then my table lists the value: dE(s,n) = (E(s,n)-E'(s,n)) / (E(s,n)-f(s,n)) Consequently, we have the following special cases: dE<0: The spiral under discussion is wose energy-wise than the RSZ spiral dE=0: The spiral under discussion has the same energy as the RSZ spiral dE=1: The spiral under discussion has the optimum energy. 2. I'm fully aware that there may be better criteria than potential energy, depending on the exact application. Knud Thomsen === Subject: Re: 'Generalized spiral points': further improvement Shortly after my mentioning in this thread the usefulness of approximately uniformly distributed points on the surface of a 4D hypersphere, Bob Bauer kindly informed me about another interesting paper of his, where he describes a spiral on S4 (!!!): Robert Bauer: Uniform Sampling of SO3. Presented at the Flight Mechanics Symposium, 2001 June, Goddard Space Flight Center, Greenbelt, Maryland. Knud Thomsen === Subject: Re: Bit mapped Set Theory The package can be used to find e.g. simple rational approximations for the square root of 2 (within the Stern-Brocot tree): 2834355043780459462455531250463335487615531135454623613592667924060 < sqrt(2) < / 1417177521890229731227765625231667743807765567727311806796333962030 348 iterations Nice huh? Han de Bruijn === Subject: Re: Bit mapped Set Theory Oh yeah, the accuracy of this approximation is better than 130 decimals, and can be improved quite easily. Han de Bruijn === Subject: Re: continuum hypothesis and 0=1 Pierre-Yves.Gaillard@iecn.u-nancy.fr a .8ecrit : Not exactly, but... If there is a proof, let's construct a model (of ZFC) by forcing, where CH is false (Cohen's work). In this model, we have a contradiction, so the construction is impossible. Now, it is well known we must (for instance) suppose Consis(ZFC) to construct such a model, so by contraposition we have a proof of non Consis(ZFC) (and of weaker things, like there doesn't exists inaccessible cardinals) . This is not exactly a contradiction, but looks strongly like one. At least, we know then (Gentzen) that ZFC is w-inconsistent, i.e. we have things like P(0), P(1), P(2),... all true, and exists n integer such that (non P(n)) also true... === Subject: Re: continuum hypothesis and 0=1 I can write you a computer program, and, given a proof in ZFC of the continuum hypothesis, it will give you a proof in ZFC that 0=1. === Subject: Re: continuum hypothesis and 0=1 This is what I come up with: ZFC proves that if ZFC consistent then ZFC does not prove CH. So ZFC proves that if ZFC proves CH then ZFC inconsistent. So ZFC proves that if ZFC proves CH then ZFC proves 0=1. Suppose ZFC proves CH. Then ZFC proves that ZFC proves 0=1. But perhaps ZFC is consistent while proving false statements. So it may be that ZFC does not prove 0=1 while ZFC proves that ZFC proves 0=1. Or, if I'm not correct, where have I gone wrong? MoeBlee === Subject: Re: continuum hypothesis and 0=1 MoeBlee a .8ecrit : Exactly === Subject: Re: continuum hypothesis and 0=1 Rupert a .8ecrit : Are you sure (see my own answer). I believe this is not so clear, as we only know that if ZFC is supposed w-consistent. Or I am making some confusion? === Subject: Conformal Mapping and Curve Fitting Hello I would like to find the best conformal mapping from one image to another given a series of corresponding points in each image. In this context, I have some questions about conformal maps, analytic functions and curve fitting that I'd welcome some input on. If w is a complex function of a complex variable z [w=w(z)] then the function w can be thought of as a mapping from the complex plane onto the complex plane. Now, if w is an analytic function then (if I understand correctly) the mapping described by w is conformal. Furthermore, any function w which can be expressed as a power series of z will be analytic. Given this, is there any reason why the standard polynomial least-squares based curve fitting for scalar functions of one variable should not be extended to complex functions of complex variables? Is this a standard technique and if so does anyone have any references or keywords to search on? Jeff === Subject: relativity what is the relativity theory? === Subject: Re: relativity Google is your friend. Get acquainted with it. Bob Kolker === Subject: Re: relativity It is something to ask for at sci.physics === Subject: Re: Why Islam? why spam the channel ? === Subject: Re: 2 interesting quotes by famous mathematicians According to http://www-groups.dcs.st-and.ac.uk/~history/Search/historysearch.cgi?SUGGEST ION=Hardy&CONTEXT=1 Hardy was a pure mathematician who hoped his mathematics could never be applied. Also He was interested in the game of chess, but was frankly puzzled by something in its nature which seemed to come into conflict with his mathematical principles. http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Hardy.html Nick === Subject: bi-Unitarily invariant matrix I have often read that a standard random gaussian matrix is bi unitarily invariant, ie if we denote by U a unitary random matrix, X a standard gaussian matrix independent of U than UX has the same statistics as X. === Subject: Re: bi-Unitarily invariant matrix Effectively, by using the characteristic function, the result is easily derived. However, I have searched a proof for this result in order to try to use the same techniques for studying the general case where the gaussian vector has non iid entries. I have found by simulations that if X is a gaussian vector having independent zero mean entries with different variances and U is a unitary random matrix than UX is a gaussian vector with iid entries with variance equal to the mean of the variances of the initial vector X. I have tried to prove what I have observed through simulations by using the characteristic function but in vain. If you have some elements of response, I would be grateful if you mind to help me. === Subject: Re: bi-Unitarily invariant matrix You can easily show this by considering the characteristc function, in other words the Fourier transform of the distribution. ************************ David C. Ullrich === Subject: Re: Moduli of annuli, containing critical point of z^2 +c Does anyone have any ideas? === Subject: Re: Long Nose Plier Man, are you spamming us? Is he conspiring with Edwin A. Abbott? -- NO CARRIER === Subject: Re: Modular arithmetic in PARI/GP It's not true: ? bezout(123,2234) %1 = [781, -43, 1] ? bezout(2234,123) %2 = [-43, 781, 1] Makes sense. Phil -- Home taping is killing big business profits. We left this side blank so you can help. -- Dead Kennedys, written upon the B-side of tapes of /In God We Trust, Inc./. === Subject: Re: Average Distance to Circumference Huh? What is rho? If you are trying to say that the average distance from the origin to the random point is 2r/3, where r is the radius of the circle, then that's correct, but that wasn't the question. === Subject: Re: help on computing a numerically unstable summation of alternating series... Mike a .8ecrit : Using lcc-win32 I obtain: qfloat Calculate(void) { qfloat result = expq(qratiol(930640970593765LL,35184372088832LL))- expq(qratiol(8263351415649457LL,281474976710656LL))+ expq(qratiol(2217817423426967LL,70368744177664LL))- expq(qratiol(584311734835661LL,17592186044416LL))+ expq(qratiol(4864341440217035LL,140737488355328LL))- expq(qratiol(1253427415565995LL,35184372088832LL))+ expq(qratiol(5127716809645411LL,140737488355328LL))- expq(qratiol(5209607668443935LL,140737488355328LL))+ expq(qratiol(5261448879727287LL,140737488355328LL))- expq(qratiol(1321119012235495LL,35184372088832LL))+ expq(qratiol(659908689006993LL,17592186044416LL))- expq(qratiol(163932163860885LL,4398046511104LL))+ expq(qratiol(161986714262777LL,4398046511104LL))- expq(qratiol(5091270694146951LL,140737488355328LL))+ expq(qratiol(4966850064647737LL,140737488355328LL))- expq(qratiol(1201767032694329LL,35184372088832LL))+ expq(qratiol(4606806576659703LL,140737488355328LL))- expq(qratiol(8715066440833619LL,281474976710656LL))+ expq(qratiol(8086318748158631LL,281474976710656LL))- expq(qratiol(7247264814689091LL,281474976710656LL)); return result; } int main(void) { int i; for (i=0; i<1000;i++) Calculate(); printf(result=%80.70qfn,Calculate()); } Output: result= 32.462529730533044389336715771207815595845717638671784470781938405078708 The time needed to do this is 2.32 seconds, i.e. 2.32ms for each calculation. === Subject: Re: help on computing a numerically unstable summation of alternating series... says... How about trying summing all the +ve elements (odd ones) and all the -ve ones (even ones) and then combining. And how about combining elements in pairs, + and -, to get a smaller value, and then accumulate those... Those are some of the possible ways to get better numerical results than a straight sum. -- Loren http://blogs.mathworks.com/loren/ === Subject: Re: help on computing a numerically unstable summation of alternating series... snip Though I don't know what 'ee' is, have a look a 'convergence acceleration' techniques, like http://numbers.computation.free.fr/Constants/Miscellaneous/seriesacceleratio n.html It's especially helpful for alternating series. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany === Subject: Re: help on computing a numerically unstable summation of alternating series... Please notice that although the summands are really huge, but the final result should always be between 0 and 1. Thus numerically it is very strange... === Subject: Re: help on computing a numerically unstable summation of alternating series... What is meant by ee(i) ? ... === Subject: Re: help on computing a numerically unstable summation of alternating series... whatever trick you might use in damping the errors in summation (convergence acceleration as Helmut Jarausch proposed, Kahan_Babuska summation (not yet mentioned here) which will damp the errors in computing the sum): first of all you must have the summands with sufficient precision, since, with the summands once rounded, no trick wii bring back the precision lost there. therefore I would propose you try to rewrite the sum as sum_{j=0 to 9 } exp(ee(2*j))*(1-exp(ee(2*j+1)-ee(2*j))) !!!! provided ee(2*j-1)