mm-585 === Subject: Re: a simple question about reductio ad absurdum[crossp] >(...) >I discovered I was simply unable as a I'm bit rusty with my analysis course >in making a absurd hypothesis for reductin ad absurdum for simple thesis >like the strong goldbach >conjecture. >(I mean simple becouse the hypothesis is quite simple. I'm not stating here >Goldbach is simple!) >I guess that would be something like: there's a even n>2 that's not the sum >of two primes. >Simple like that? >Above that, I assume (and seem to remember) that there's just an absurd >hypothesis for a given hypothesis. Yes, it is always just the negation of the given hypothesis (which should better be called what one wants to prove - because if one actually uses it as a hypothesis, then one makes the error of assuming what one wants to prove !) Then you use formal logic to transform this negation - if needed and helping >(...) > I remember one must swap exist operand with for every but didn't >succeeded with Goldbach >(...) this is indeed because the there exists a ... such that and for every swap by taking the negation === Subject: Re: possible book titles for this course > What would be a possible or possible book titles for a course called > Strategies of Proof which has a prerequisite of semester number 2 of > undergraduate Calculus? The description of this Strategies of Proof course is, > ...set theory, and methods for constructing proofs of mathematical statements. > A bridge to the rigors of upper-division mathematics courses containing > significant abstract content. > I ask because I have never seen a textbook called anything like strategies of > proof. > G C I am an undergraduate so take this with a grain of salt. There is a book called Conjecture and Proof, by Laczkovich which has lots of interesting stuff in it (Banach-Tarski Paradox) and might serve as a good book for the course you speak of. Adam === Subject: Re: possible book titles for this course For the people who have been answering my question: The obvious should have been the guide. The word proof would usually be in the title; practically any short, conservative title with proof in it might be a corresponding book. What I did not recognize was the transitional status of the course called strategies of proof. G C === Subject: Re: How to do non-uniform (sampling) DFT? You could try Googling for the Lomb-Scargle Periodogram method. It works with unevenly sampled data. I believe it is used in the astrophysics community. I'm told that the Numerical Recipes in (FORTRAN,C,...) books have an implementation. I've never tried using it, but filed the reference away, just in case. Let me know how you get on. Jon > Hi all, > Can anybody point me to some helps/tutorials/references/stories on how > to do non-uniform (sampling) DFT? > -Walala === Subject: Re: how to decide a good sampling rate for sampling a function without obvious frequency? >Lucy is trying to fit a round peg into a square hole > gaussian(x, y)=1/(2*pi*sigmax*sigmay)*exp(-0.5*(x^2/sigmax^2+y^2/sigmay^2 )); > In my experiments using Matlab, I am using square grids to deiscretize the > above function. This is your problem right here. Why are you using square grids? What's so special about them, especially since what you're discretising is naturally a cylindrical polar shape? As well as that, you need greater resolution near the centre and less at the edges, so your cylindrical polar grid needs to have radial spacing that is a function of the radius. Perhaps you could tell us what your application is? What will you do with the Gaussian, once you've discretised it? === Subject: Re: how to decide a good sampling rate for sampling a function without obvious frequency? > Hi all, > I am having trouble with my sampling problem: > Basically I want to discretize a 2D Gaussian function f=gaussian(x, y) > gaussian(x, y)=1/(2*pi*sigmax*sigmay)*exp(-0.5*(x^2/sigmax^2+y^2/sigmay^2 )); > In my experiments using Matlab, I am using square grids to deiscretize the > above function. > Suppose I have a grid -- [-N..N, -N..N] where N is the number of samples in > one axis. So all together there are (2N+1)^2 samples. > Suppose the stepsize for discretization is dx. > The support of the discretized function will be [-N*dx..N*dx, -N*dy..N*dy]. > Since Gaussian function has an infinite support itself, I have to make N*dx > and N*dy sufficiently large in order to capture most of the energy of the > function f(x, y). > I remember for Gaussian distribution, it should be at least 3*sigma. > So I decide that the following conditions must be satisfied: > N*dx >= 4*sigma_x; > N*dy >= 4*sigma_y; > (sigma_x and sigma_y are known). > If the stepsize is large, the discretization will not be accurate. So dx and > dy should be small compared with sigma_x and sigma_y... > For example, I can choose dx=sigma_x / 100, dy= sigma_y /100, > but then N should be >= 400. > But then > [-N..N, -N..N] grids will have (2N+1)^2 = 801^2=641601 samples... > This is too large and will be very bad for subsequent operations... > After many experiments, I want to ask how to determine a good stepsize > taking into consideration of the storage size, while maintain good accuracy? > I know Nyquist sampling theory, but after many thinking, I found nowhere > here I can apply that theory... > Please help me and give me a hand! > -Lucy Use the Nyquist sampling theory on the frequency _components_ of the signal. Decide how much inaccuracy you can stand, and at what frequencies. Now look at the power spectral density of the signal you want to sample. Due to aliasing any signal energy above the Nyquist frequency will be folded back into your real signal, corrupting it. Because you know how much inaccuracy you can stand you can locate the Nyquist frequency to give you no more than your desired accuracy limit. Disclaimer: That made perfect sense coming out, but on re-reading it I think it would be a good growth enhancer in the garden. None the less I'm going to hit the send button... -- Tim Wescott Wescott Design Services http://www.wescottdesign.com === Subject: Re: how to decide a good sampling rate for sampling a function without obvious frequency? > Hi all, > I am having trouble with my sampling problem: > Basically I want to discretize a 2D Gaussian function f=gaussian(x, y) > gaussian(x, y)=1/(2*pi*sigmax*sigmay)*exp(-0.5*(x^2/sigmax^2+y^2/sigmay^2 )); > In my experiments using Matlab, I am using square grids to deiscretize the > above function. > Suppose I have a grid -- [-N..N, -N..N] where N is the number of samples in > one axis. So all together there are (2N+1)^2 samples. > Suppose the stepsize for discretization is dx. > The support of the discretized function will be [-N*dx..N*dx, -N*dy..N*dy]. > Since Gaussian function has an infinite support itself, I have to make N*dx > and N*dy sufficiently large in order to capture most of the energy of the > function f(x, y). > I remember for Gaussian distribution, it should be at least 3*sigma. > So I decide that the following conditions must be satisfied: > N*dx >= 4*sigma_x; > N*dy >= 4*sigma_y; > (sigma_x and sigma_y are known). > If the stepsize is large, the discretization will not be accurate. So dx and > dy should be small compared with sigma_x and sigma_y... > For example, I can choose dx=sigma_x / 100, dy= sigma_y /100, > but then N should be >= 400. > But then > [-N..N, -N..N] grids will have (2N+1)^2 = 801^2=641601 samples... > This is too large and will be very bad for subsequent operations... > After many experiments, I want to ask how to determine a good stepsize > taking into consideration of the storage size, while maintain good accuracy? > I know Nyquist sampling theory, but after many thinking, I found nowhere > here I can apply that theory... > Please help me and give me a hand! > -Lucy > Use the Nyquist sampling theory on the frequency _components_ of the > signal. Decide how much inaccuracy you can stand, and at what > frequencies. Now look at the power spectral density of the signal you > want to sample. Due to aliasing any signal energy above the Nyquist > frequency will be folded back into your real signal, corrupting it. > Because you know how much inaccuracy you can stand you can locate the > Nyquist frequency to give you no more than your desired accuracy limit. > Disclaimer: That made perfect sense coming out, but on re-reading it I > think it would be a good growth enhancer in the garden. None the less > I'm going to hit the send button... > -- > Tim Wescott > Wescott Design Services > http://www.wescottdesign.com Hi Tim, For the 2D gaussian function: gaussian(x, y)=1/(2*pi*sigmax*sigmay)*exp(-0.5*(x^2/sigmax^2+y^2/sigmay^2 )); I've computed its FT to be GAUSSIAN(u, v)=exp(-0.5*(u^2/sigmau^2+v^2/sigmav^2)); where sigma_u=1/(2*pi*sigma_x), sigma_v=1/(2*pi*sigma_y)... In my numerical calculation, I have chosen sigma_x = 1 and sigma_y... So sigma_u=1/(2*pi), sigma_v=1/(2*pi)... Since the frequency components are still in Gaussian shape, if I want to 99% of the freuqency components, I should keep the frequency ranging between [-4*sigma_u .. 4*sigma_u, -4*sigma_v .. 4*sigma_v] If I pick 4*sigma_u = 4*sigma_v to be my Nyquist frequency, and double it, my sampling rate should be 89493 === Subject: Abstract Algebra--Group Theory Suppose that H and K are subgroups of a group G s/t both subgroups have finite index in G. I am having major difficulty showing that (H intersect K) is also of finite order in G. Please help. This is not HW. Mkajuma === Subject: Re: Abstract Algebra--Group Theory Oops! Suppose that H and K are subgroups of a group G s/t both subgroups have finite index in G. I am having major difficulty showing that (H intersect K) is also of finite INDEX in G. Please help. This is not HW. Mkajuma > Suppose that H and K are subgroups of a group G s/t both subgroups have > finite index in G. I am having major difficulty showing that (H intersect K) > is also of finite order in G. Please help. This is not HW. > Mkajuma === Subject: Re: Abstract Algebra--Group Theory >Oops! >Suppose that H and K are subgroups of a group G s/t both subgroups have >finite index in G. I am having major difficulty showing that (H intersect K) >is also of finite INDEX in G. Please help. This is not HW. >Mkajuma > Suppose that H and K are subgroups of a group G s/t both subgroups have > finite index in G. I am having major difficulty showing that (H intersect > is also of finite order in G. Please help. This is not HW. > Mkajuma Well |G : H ^ K| = |G:H| |H: H ^ K|, so try proving that |H: H ^ K| <= |G:K|. And as a hint for doing that, show that, if T is a set of coset representatives of H ^ K in H, then the elements of T all lie in distinct cosets of K. Derek Holt. === Subject: Re: Abstract Algebra--Group Theory >Suppose that H and K are subgroups of a group G s/t both subgroups have >finite index in G. I am having major difficulty showing that (H intersect K) >is also of finite order in G. Please help. This is not HW. >Mkajuma I'm guessing that you meant to say that H intersect K is of finite index in G. If that is the case, here is a hint. If you prove the following inequality, the problem is done. | G : H intersect K | <= | G : H | | G : K | Hope this helps, Brian === Subject: Re: Abstract Algebra--Group Theory >Suppose that H and K are subgroups of a group G s/t both subgroups have >finite index in G. I am having major difficulty showing that (H intersect K) >is also of finite order in G. Please help. This is not HW. >Mkajuma > I'm guessing that you meant to say that H intersect K is of finite > index in G. If that is the case, here is a hint. > If you prove the following inequality, the problem is done. > | G : H intersect K | <= | G : H | | G : K | Ok, I agree if I could prove your inequality then I'm done. But isn't | G : H intersect K | > | G : H | and | G : H intersect K | > | G : K | if H=/ K???? > Hope this helps, > Brian === Subject: Re: DC Proof software > I'm sorry, Doug, but after thinking about it for a while, DC Proof is > probably NOT the best*sigma_u=8*sigma_v = 8/(2*pi)=4/pi=1.27, so I pick 2. So am I correct that I should only need to sample the gaussian function twice? -Lucy === Subject: Re: how to decide a good sampling rate for sampling a function without obvious frequency? >Hi all, >I am having trouble with my sampling problem: >Basically I want to discretize a 2D Gaussian function f=gaussian(x, y) >gaussian(x, > y)=1/(2*pi*sigmax*sigmay)*exp(-0.5*(x^2/sigmax^2+y^2/sigmay^2 )); >In my experiments using Matlab, I am using square grids to deiscretize > the >above function. -- snip -- >-Lucy >Use the Nyquist sampling theory on the frequency _components_ of the >signal. Decide how much inaccuracy you can stand, and at what >frequencies. Now look at the power spectral density of the signal you >want to sample. Due to aliasing any signal energy above the Nyquist >frequency will be folded back into your real signal, corrupting it. >Because you know how much inaccuracy you can stand you can locate the >Nyquist frequency to give you no more than your desired accuracy limit. >Disclaimer: That made perfect sense coming out, but on re-reading it I >think it would be a good growth enhancer in the garden. None the less >I'm going to hit the send button... >-- >Tim Wescott >Wescott Design Services >http://www.wescottdesign.com > Hi Tim, I know; I just didn't have time last night to go over it and figure out why, or how to make it more clear. Some time spent with graph paper and a pencil will be required to really understand the subject, no matter how well I explain it. > For the 2D gaussian function: > gaussian(x, y)=1/(2*pi*sigmax*sigmay)*exp(-0.5*(x^2/sigmax^2+y^2/sigmay^2 )); > I've computed its FT to be > GAUSSIAN(u, v)=exp(-0.5*(u^2/sigmau^2+v^2/sigmav^2)); > where sigma_u=1/(2*pi*sigma_x), sigma_v=1/(2*pi*sigma_y)... > In my numerical calculation, I have chosen sigma_x = 1 and sigma_y... > So sigma_u=1/(2*pi), sigma_v=1/(2*pi)... > Since the frequency components are still in Gaussian shape, if I want to 99% > of the freuqency components, I should keep the frequency ranging between > [-4*sigma_u .. 4*sigma_u, -4*sigma_v .. 4*sigma_v] > If I pick 4*sigma_u = 4*sigma_v to be my Nyquist frequency, > and double it, my sampling rate should be 8*sigma_u=8*sigma_v = > 8/(2*pi)=4/pi=1.27, so I pick 2. > So am I correct that I should only need to sample the gaussian function > twice? This gets back to how much accuracy you need. If you're truncating the function at 3 sigma then it certainly indicates that sampling it twice will be adequate. On the other hand that seems a terribly small sample to me. Are you sure that your 4 sigma/99% assumption is true in 2-D, and are you sure that truncating at 3 sigma is appropriate? If your intuition is telling you that two samples is too small then perhaps the error assumptions that you started with are too generous, and you need to tighten them up? Getting 99% of all the energy sounds good, but it only puts your Nyquist energy down 20 decibels -- a sampling system wouldn't be considered high quality unless it were getting the aliasing down to at least 40 to 60dB or more, which implies getting 99.9% or even 99.99% of your energy. If I did this calculation from the precept I stated above and came up with these numbers I would do one of two things: One, I would try it with the indicated number of samples, and 2x that, and perhaps 10x that. Two, I would go over my math to make sure that I really knew what I was talking about. > -Lucy -- Tim Wescott Wescott Design Services http://www.wescottdesign.com === Subject: Re: how to decide a good sampling rate for sampling a function without obvious frequency? >big snip< > So am I correct that I should only need to sample the gaussian function > twice? Well, yes you should be able to reconstruct an approximation of a gaussian with 2 sinusoids - 4 would be better - more even better. I think as you decrease the sample spacing geometrically the resolution will increase exponentially so you will not need a very large number of samples. Just make your sampling algorithm scalable. Start with a small managable size and if you need better resolution later you can do that. -jim http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- === Subject: Re: Invariant Galilean Transformations (FAQ) On All Laws > > > > > Another reason to update my kill file... > > > > David A. Smith > > I take that (correctly?) as meaning you cannot show Maxwell >non-invariant > under the galilean translatory transformation without the spurious > atheoretical time transform, t'=t, being asserted? > [snip crap] > Hey stooopid, Galilean transforms are empirically invalid in a > relativistic universe. You are playing solitaire with pinochle > deck - and losing every game. Idiot. >I do wish you wouldn't sign your material so un-self-ßatteringly, however >appropriately. >I do apreciate you submitting the above proof that you are so completely >dishonest intellectually. >How can it hurt in such a relativistic universe to admit that maxwell is/are >indeed invariant under the galilean translatory transformation when the >atheoretical t'=t transform is not (spuriously) imposed? It only hurts because it is wrong, which is good enough for me. t = t' is crap. Mostly because it is wrong... >BTW, tell us again all about the many spatial/object contractions you see >every day a million times. >Child: Want to see the world's fastest wild west draw? >Kid: Yes. >Child: Want to see it again? >eleaticus === Subject: Re: Invariant Galilean Transformations (FAQ) On All Laws > > in > > > > Another reason to update my kill file... > > > > David A. Smith > > > > I take that (correctly?) as meaning you cannot show Maxwell >non-invariant > > under the galilean translatory transformation without the spurious > > atheoretical time transform, t'=t, being asserted? > [snip crap] > > Hey stooopid, Galilean transforms are empirically invalid in a > relativistic universe. You are playing solitaire with pinochle > deck - and losing every game. Idiot. >I do wish you wouldn't sign your material so un-self-ßatteringly, however >appropriately. >I do apreciate you submitting the above proof that you are so completely >dishonest intellectually. >How can it hurt in such a relativistic universe to admit that maxwell is/are >indeed invariant under the galilean translatory transformation when the >atheoretical t'=t transform is not (spuriously) imposed? > It only hurts because it is wrong, which is good enough for me. > t = t' is crap. Mostly because it is wrong... galilean transform is wrong and (b) that even if you don't Maxwell is not invariant under those transforms. eleaticus === Subject: Re: Random number generation using radioactivity >[...] > If the random starting value is free from any bias and is reseeded > (an xor would be optimal) with a frequency greater than the quantity at > which bias/correlation is detectable from the CRC then this would remove > any remaining bias. > > The random starting value *is* the result from the previous CRC. That > *is* a high-quality random value: the previous processed end result. > > By simply not initializing the CRC each time, it is automatically > re-seeded with a high-quality random value. > > If we change the initial CRC register to every possible value, we also > change the CRC result to every possible value. No bias is introduced. > There is no need for XOR. > But your procedure is only needed in the case that the outputs of the CRC with > a constant start value are biased. If we model a hash as the remainder from a division, the usual modulo effects appear. The input data are wrapped around the smaller result. Biased values appear more often than expected, and each occurrence is only compensated if appropriate less-frequent values happen to map to the same point. Since that is unlikely, some bias seems likely to remain. The usual conclusion is that any reasonable hash function, CRC or cryptographic, will fail to hide some amount of the input bias. That probably over-simplifies reality. With CRC each bit has the same power to affect the result, but the noise samples are binary codings, not single bits. The bias is in the structure of the coding, rather than an unbalance between 1's and 0's. When the exact same 16-bit value occurs in different positions in the data, it is unlikely to map to the same CRC result. So the value bias does tend to be hidden. >If that is the case then your proposal will > indeed reduce the bias, but it will also cause successive outputs to be > statistically dependent on each other. This is a rather mild form of dependence, similar to the dependence found in the CBC encryption mode. Nobody cries about how each CBC value is dependent on the previous one, because that is the intention. And it does not seem to represent a cryptographic weakness. This dependence is also very similar to the process of subtracting the previous sample from the current one to get differential data and emphasize change in the numerical result. Extensive characterization on my pages shows that the (normal) distribution, and more especially the autocorrelation, are greatly improved by this process. (See, for example, http://www.ciphersbyritter.com/NOISE/ZENER1ZY.HTM versus http://www.ciphersbyritter.com/NOISE/ZENER1P7.HTM from http://www.ciphersbyritter.com/NOISE/NOISCHAR.HTM .) That makes sense, because the very definition of audio frequency implies correlation over time, and with fast sampling, lower frequency components will inject similar values into many, many samples. So if we subtract the previous sample from the current one, we can remove much of that interference. Without this processing, autocorrelation structure is a vastly larger issue. > So your proposal does not really help in getting really better outputs because > the output random numbers should be both unbiased and independent. Really? Who says? Who says that this particular type of dependence is a problem, and if so, exactly why? The whole point of de-skew computations is to change a distribution in real time. In the real world, we can only know the distribution as an approximation, and only as it was, not how it is, because it may vary. Since we cannot know the true distribution, we also cannot numerically take the inverse without introducing numerical distortions. --- Terry Ritter 1.2MB Crypto Glossary http://www.ciphersbyritter.com/GLOSSARY.HTM === Subject: Re: Random number generation using radioactivity > >[...] > If the random starting value is free from any bias and is reseeded > (an xor would be optimal) with a frequency greater than the quantity > at which bias/correlation is detectable from the CRC then this would > remove any remaining bias. > >The random starting value *is* the result from the >previous CRC. That *is* a high-quality random value: >the previous processed end result. > >By simply not initializing the CRC each time, it is >automatically re-seeded with a high-quality random value. > >If we change the initial CRC register to every possible >value, we also change the CRC result to every possible >value. No bias is introduced. There is no need for XOR. > You missunderstand what I am saying. Some initial random starting > value is necessary even with your scheme. If the range of inputs is > limited (as is usually the case with physical RNGs) the final output > may be baised. By periodically introducing random data that is > totally free from bias. The bais is eliminated. >I assume you misunderstand what *I* am saying: Each >CRC processing starts with the immediate preceding >processed random value, the result of processing a >whole sequence of biased somewhat-random data. The >resulting value is the finished, ßattened, directly >usable ultimate result. And if that result is good >enough for direct use thing to do geometry with -- not for the beginner, > anyway. See the thread, Any 1st order formalizations of geometry? I'm > afraid it would take hundreds of lines to establish even the most trivial > results in the Euclidean geometry starting from Tarski's axioms. That said, > DC Proof would probably be helpful in establishing for you the fundamentals > of logic and proof if you are weak in these areas. Working through the > Tutorial would probably be a good Ôwarm up' for any introductory course in > abstract mathetmatics. > OK, Dan you've save me quite a bit of time. I won't attempt to do > geometry in DC Proof. > How is DC Proof at number theory? I've been exploring the Euler > totient function lately. I think DC Proof would be quite good for elementary theorems in number theory. The intermediate user should find them to be a reasonable challenge. See, for example, the simple Induction proofs in the Samples directory. (Make sure you have the latest update.) This is a proof (two versions included) of the fact that no natural number can be its own successor. I haven't done much in number theory beyond this in DC Proof. So, I don't know how far you can go before formal, number theoretic proofs become unmanageable. Remember, too, that you would have to formally establish an number preliminary results before starting on any more advanced theorems. You can cite previously established theorems by including them as initial premises in a new proof. My guess is that abstract algebra (group theory, etc.) would be another rich source of ideas for the beginning or intermediate user. See, for example, Group Theory 1 - Unique Identities.proof in the Samples directory. Advanced users with much stamina can tackle projects such as the construction of the real numbers starting from Peano's Axioms. This is my own current project. DC Proof was, however, intended primarily to be a learning tool for the undergraduate non-specialist or advanced high school student to give them insights into the ultimate nature of logic and mathematics. Once the student has worked through several formal proofs in DC Proof, he/she should be able to better tackle the vagaries of everyday, informal proofs that they will typically find in textbooks, and it should help them structure their own informal proofs in just about any field of mathematics. Dan Download DC Proof at http://www.dcproof.com === Subject: Re: Problem demonstrating that the set of binary strings is uncountable. In nonstandard analysis, the integers are extended to the hyperintegers, which in some senses are just the integers like the hyperreals are just the reals, to allow something like Sum 1/2^n = 1 besides that lim Sum 1/2^n = 1 You might have heard of my argument that the Natural/Unit Equivalency Function is a bijective mapping between the natural integers and each element of the unit interval of reals, because infinite sets are equivalent and also because Cantor's results 1: that no mapping between the integers and reals could have a binary antidiagonal not on the list, and 2: that no mapping between the integers and reals could generate an infinite sequence of nested intervals, do not apply. If you accept that, then the reals are computable by a simple Turing machine that increments from zero indefinitely. Besides for zero, you can't say which definite real each would represent: just that they do represent, and that they are totally ordered. Consider functions that on average have a slope of 1/2 and in the limit equal f(x)=x. There are issues with this notion. If there may be any mapping between the integers and each of any given interval of reals, then there exists via composition a mapping between the integers and any set of reals. I've claimed something along the lines of an implicit composition of EF within any of those functions, in a similar way as to how any mapping between an open interval and a closed interval of different measure is an implicit composition. Then there is the consideration of Cantor's powerset result, with the selection of a powerset element via properties of the mapping precluding its place in the mapping, and Cantor-Bernstein about the mappings between the reals and the powerset of integers, leading to model changes and f(x)=x+1, and the dual representation of zero and infinity, in an axiom-free system. That leads into the notions that the set of all sets is the set of every value is represented as a bit string on the register, and then via understanding of a commonly accepted context that value may represent values, for example a real number in a symbolic computer algebra system. I argue the infinite sets are equivalent because I believe that. One reason I can rationally accept that EF is a bijectve mapping between N and R[0,1) or R[0,1] is because I've developed arguments to present to you that affirm that belief. If you've read this, you know some of them. Ross === Subject: Re: Why this is so difficult to do curve fitting? > I am mad at this dataset. I have spent a few days but it is so difficult to > fit... > I have the following data: > x: [0 36.4286 72.8571 109.2857 145.7143 182.1429 218.5714 255] > y: [0.0184 0.0361 0.0931 0.1918 0.3280 0.5173 0.7450 1] > How to fit this data smartly into the form of y=a*(x/255).^b+c and as a ßat random value, it is >also good enough to use as the unbiased start of the >next processing. There is thus no need to find some >other random source and somehow process that ßat >(the recursive turtles all the way down problem, >now avoided). Perhaps a demonstration using a small CRC would be helpful. Imagine an 5 bit crc. Now imagine a stream which is heavily biased. Say only two possible values are present. Ideally these two values are evenly distributed. Each sample contains one bit. Additionally assume that the CRC is taken as output after every 6 samples. There will be some set of values and crc which interact badly. In a different example one sample contains 5 bits of randomness completely unbiased and uncorrelated. The other 5 samples contain zero randomness. In this case even if we don't know which sample contains the randomness the result is completely unbiased and uncorrelated. The first is a somewhat random source. The second is a somewhere random source. These are discussed in literature. The CRC is a special case of a hash function which the literature also goes in to extensively. Even with ideal hash functions the turtle problem exists. Special constructions with these hash functions reduce the amount of additional unbiased random data needed to very small amounts. Perhaps a better analogy is chaos theory. The added unbiased random data act as the initial uncertainty. In any case CRCs work extraordinarily well but if you are worried about a bias of 1 in 2^32 then added precautions are needed. Or you could just use a bigger CRC. >As for the single original init, we do not care. >Bias is the collection, the distribution, not any >single value or any single processing. Whatever the >initial value, we then have hundreds or thousands >or however many processings, each of which start out >with an unbiased value, the result of the previous >processing. But the weenies among us can simply >discard the first value and use the rest, thus >improving confidence at almost no cost. >--- >Terry Ritter 1.2MB Crypto Glossary >http://www.ciphersbyritter.com/GLOSSARY.HTM Leslie ÔMack' McBride remove text between _ marks to respond via e-mail === Subject: Re: How to compute the addition of millions of functions efficiently and dynamically? If the signal is bandwidth-limited, you might not need full resolution, so taking every 10th point (for example) may be sufficient. That would give you a factor of 100 in processing time... Just at thought! BR, Andre >I don't keep up on MatLab, but 5 years ago their symbolic package used >the Maple engine (I think everyone uses the Maple engine...). >If you can't express it in closed form, or if you cannot wrestle the >sums into closed form then you are back to finding the fastest computer >you can and doing it discretely. >-- >Tim Wescott >Wescott Design Services >http://www.wescottdesign.com > will be non-collapsable millions of terms adding together. And I cannot find > a way to shorten it into one closed-form... > Then I need to compute its Inverse Fourier Transform. > Can any symbolical engine handle such big symbolical expression and such > IFT? > However, I want to know if for this special form of sum, there is any tricks > existing: > The FT is: > Z_SUM(u, v)=SUM( > r(x1, y1)*exp(-2*pi*i*(u*x1+v*y1)), > r(x2, y2)*exp(-2*pi*i*(u*x2+v*y2)), > ... > ... > r(x1000000, y1000000)*exp(-2*pi*i*(u*x1000000+v*y1000000)) > )*Z(u, v) > There are millions of such terms, u, v are frequencies, (x1, y1), > ...(x1000000, y1000000) are known constants... > Z(u, v) is some function. > Now I want to compute Z_SUM's Inverse FT z_sum... any efficient way of doing > this? > -Net -- Please change no_spam to a.lodwig when replying via email! === Subject: Mod I have following two questions. If a=b (mod m) for a,b in Z, m in N, Then ord_m(a) = ord_m(b) ? My lecturer say above is hold only if g,c,d (a,m)=1. But, I think this is true because, a=b (mod m) => g.c.d(a,m) = g.c.d(b,m) => = in Z_m => ord_m(a) = ord_m(b) So, I'm so confused what is true. Would you please help me ? 2> Can anybody give me example of a in Z, prime p satisfying g.c.d(a,p)=/=1 and a^(p-1) = 1 (mod p) ? (That is, counterexample of inverse of Fermat's little Thm) === Subject: Re: Mod > I have following two questions. > So, I'm so confused what is true. This depends on what you mean by ord_m(a). If you mean the order of the coset a+mZ in the additive group Z/mZ, then ord_m(a)=ord_m(b), because a+mZ=b+mZ, as you pointed out (me thinks). If, OTOH you mean the order of a+mZ in the multiplicative group (often written Z_m^*), then even though still a+mZ=b+mZ, the order of a+mZ isn't defined unless gcd(a,m)=1. So, which is it (Z/mZ,+) or (Z/mZ^*,.) ? > Would you please help me ? > 2> Can anybody give me example of a in Z, prime p > satisfying g.c.d(a,p)=/=1 and a^(p-1) = 1 (mod p) ? > (That is, counterexample of inverse of Fermat's little Thm) This is a nice (but not at all difficult) homework problem, so I'm not gonna spoil your fun! Jyrki === Subject: Re: More readable prime counting function > It turns out that I can write my prime counting function in such a way > that it's easier to read: > dS(x,y) = (p(x/y, y-1) - p(y-1, y-1))( p(y, y) - p(y-1, y-1)), > S(x,1) = 0, p(x, y) = ßoor(x) - S(x, y) - 1, and S(x,y) is the sum of > dS from dS(x,2) to dS(x,y). > p(x,x) = pi(x) so it's the count of prime numbers. > And yes, it still counts prime numbers and is in fact the exact same > function that I've been giving before as all that happens is that for > the higher ranges certain things just zero out. OOPS! It turns out that it won't work to push beyond y=sqrt(x) like I tried to do here, as there are a lot of problems with it. Part of what happened is that I had this idea in mind that if my prime counting function *looked* just a little bit simpler, then maybe that would help people, and in one sense the idea above should work, as S(x,x) does equal S(x,sqrt(x)) in the descriptive form. That is, given that S(x,sqrt(x)) is the sum of composites up to and including x, it makes since that S(x,x) would just be the same sum, but the dS(x,y) function as I defined it can start going negative when you push y past sqrt(x), or at least that's what I saw when checking yesterday! So it was an idea that doesn't pan out unless I missed something. Having my own research at the forefront where I can't reference a textbook, leaves me in the position of figuring things out for myself. It's neat in many ways, but at times I just screw-up, as yeah, I can't just go pick up a math reference to read about something that I discovered that hasn't yet made it into the references and textbooks. James Harris === Subject: Re: More readable prime counting function >[...] >So it was an idea that doesn't pan out unless I missed something. >Having my own research at the forefront where I can't reference a >textbook, leaves me in the position of figuring things out for myself. >It's neat in many ways, but at times I just screw-up, as yeah, I can't >just go pick up a math reference to read about something that I >discovered that hasn't yet made it into the references and textbooks. those of us who do research that we actually publish in journals are in the same position - we can't check our work by opening a textbook. but we end up retracting published errors only very rarely - in your case it seems to happen eventually with -most- of the things you post. i wonder what the reason for the difference is? >James Harris David C. Ullrich ************************** As far as I'm concerend you're trying to wait until I die, so I figure maybe you should die instead. How about that, eh? Wouldn't that be a better twist? You refuse to follow the math, so the great Powers that control reality and *speak* in mathematics decide to kill you instead of me. So what do you think about that, eh? Oh, can't hear Them talking? Well, I guess that's because you don't really understand Mathematics, the true language, which is THE language. They're talking about you now, and They agree with my assessment, and will not penalize me as They allowed the others like Galois and Abel to be penalized. They will kill you instead. James Harris speaking on Weird factorization, genius === Subject: Re: HOW CAN I SOLVE THIS EQUATION SYSTEM ? > It sounds like this should really be more complicated than just a system > of two differential equations. But anyway: > d(LI)/dt = L dI/dt + I dL/dt > so it would be good to be able to calculate dL/dt as well as just L(t). > Of course you could approximate that using a difference quotient (say > (L(t+h)-L(t-h))/(2 h)). And then you just have an ordinary system of > two differential equations in two dependent variables, to which you > could apply Runge-Kutta. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 This could be the solution for the first problem, but now another one arises : I always saw the R-K method applied to problem in the form : dy/dt=F(t,y) like say : y'=2t+ty+y in which, ones the time step (h) is given (and the initial condition y(0) too), it's very simple to apply the R-K method, because we have just to substitute the time step for t and the initial condition in the equation : F(t,y)=2t+ty+y so that y(t+h)=y(t)+g(k(i)) where g(k(i)) is a linear function depending on the order of the R-K method (e.g. 1/6 h [k(1)+4k(2)+k(3)] for the third order). But what happens when I don't have any analytical form for F(t,y) ? I mean : how can I substitute the values of F(t+h,y+g[k(i)]) for all the k(i) if I know these values only in the time step ? I know that my english is very bad, so this explanation is not so clear (even to me),anyway I hope you understood my questions. If it's not so, I will try to give a more readable explanation. === Subject: Re: Uncle Al's missing logic. > > Uncle Al never responds to actual factual material, so please (y'all) > let me > > know how it can be correct in one single, continuous, Ôlogical' > derivation > > process to assert: > > > > a=bc, a=dc, a=-dc, -d < b < d, d > 0, c >= 0. > > > > Meaning that b and d can independently take any of the values in the > > specified ranges. > > > If c = 0 then for a=bc, a=dc, a=-dc, a = 0 with values of b and d in the > specified ranges. > its specified range. > So, what I mean by those statements is that the process is to derive an > equation that applies to any and all values of b, c, d. > Sorry. > eleaticus > PH Actually c can't be anything but zero, since a = dc = -dc sets a = 0, but d > 0, so c = -c = 0. === Subject: Re: expressions equivalence via DNF >In other words, to compare 2 Boolean expressions, is it sufficient to >convert then both to DNF and compare them? > > but here it's clear that you're really asking about the converse. > yes, two wffs in the predicate calculus are equivalent if and only > if their [suitably normalized, ie Ôsorted'] dnf's are equal. >disagrees with your statement? >if i recall correctly the author of that message disagrees >with a lot of things [like he says the reals are countable.] >in that message he says that >(A AND B) OR (B AND C) >is dnf, which indicates he simply doesn't know what dnf >is. he also says ÔYou should be aware that there is usually >no single minimal DNF form,' which is nonsense [unless >he's simply talking about permuting the terms, which i >do not think is what he means.] >If A, B, and C are propositional variables then that formula is definitely >in DNF. >It is also the case that there is often no single minimal DNF. For instance, >the following formulas are expressed in minimal DNF and are equivalent to >one another: >(!b & !c) | (b & c) | (a & c) >(!b & !c) | (b & c) | (a & !b) >There is a notion of a canonical DNF that would be equivalent to listing >all the rows of a truth table for which the function in question evaluates >to true. That form is unique up to permutation of the clauses. For the >above formulas, the equivalent canonical DNF would be >(!a & !b & !c) | (a & !b & !c) | (a & !b & c) | (!a & b & c) | (a & b & c) . well, looking it up a few places it seems you're right, at least in many authors' terminology. i was taking dnf to mean what you call canonical dnf. [i don't think that what we're calling dnf above -should- be called dnf, since there's nothing Ônormal' about it... in any case i could swear that i was taught years ago that dnf means canonical dnf, although i don't have a reference at hand.] ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: expressions equivalence via DNF > If you describe your problem explicitly, the matters will become much > clearer. For example, are you considering predicate calculus formulas or > propositional formulas? >Jeffrey, >really makes sense. I should have seen it myself, it is so trivial! >As to my original question: I think I am dealing with predicate >calculus, though >I think I have some limitations which may allow me to reduce it to >propositional >logic for this particular task. >I have 2 formulas which consists of set of unary Boolean functions >(predicates?) combined with operations XOR, AND, XOR and NOT. Both > ^^^ I'm sure you mean or. > y=a*(x/255+b).^c+d? The problem is that this don't fit very well. > a, b, c, d are paramters that need to be determined... On demand, using heuristics, fitting to the first polynome it gives me a b c 0.652270279169733187 1.45279404390058744 1.65066115700720538e-05 error abs= 0.000116164344485998802 0 total=0.000116164344485998802 > With a ZERO in x, I did try polyfit with respect to log(x) and log(y)... > but it gave very bad result... > how do I fit this data smartly? > -Lucy > I am using Matlab... -- http://www.telecable.es/personales/gamo/ perl -e'print phone .hex(825475818),n;' perl -e'print 111_111_111*111_111_111,n;' === Subject: Re: Why this is so difficult to do curve fitting? > a, b, c, d are paramters that need to be determined... > On demand, using heuristics, fitting to the first polynome it gives me > a b c > 0.652270279169733187 1.45279404390058744 1.65066115700720538e-05 > error abs= 0.000116164344485998802 0 total=0.000116164344485998802 Correction: I get better results... a b c 0.660227103768863402 1.45187451027629845 3.87222984865106413e-07 error abs= 2.72031067725875068e-06 0 total=2.72031067725875068e-06 -- http://www.telecable.es/personales/gamo/ perl -e'print phone .hex(825475818),n;' perl -e'print 111_111_111*111_111_111,n;' === Subject: Re: what is the analytical function for this 2D shape? > This graph looks like a 3D surface since you are mapping RxR->R. In any > case, I don't know if you mean real-analytic when you say analytical, > Define g(x) = exp(-1/x) when x > 0 and 0 when x <= 0. g is C^oo, so let > g(4-x^2-y^2) > f(x,y) = ---------------------------------- > g(x^2-1) + g(y^2-1) + g(4-x^2-y^2) > This puts a square plateau with side 2 and altitude 1 on a hill with a > circular base of radius 2. For a rectangular plateau, elliptical base, > and variable height, use z = c f(x/a,y/b). > Rob Johnson > take out the trash before replying Hi Rob, have one more question: can your function be modified to obtain a infinite support version? I mean, the top is still square or rectangular plateau, but the bottom is a radially symmetrial( with the same aspect ratio of the top plateau), but with infinite support, something like a 2D Gaussian function... -Lucy === Subject: Re: Dividing Power Series >Can I divide power series by long division as follows: > yes, at least for z in a disk where both series converge > and the divisor has no zero. If the divisor has a zero at the origin, then you get a Laurent series about 0(for example, cos z / sin z > pf: the two series define analytic functions f, g in > this disk. now h = f/g is analytic in this disk, so > it has a power series. it's easy to show you can > get the series for f = g*h by multiplying the series > for g and h formally, and long division is just the > inverse of formal power-series multiplication [at > each step the criterion you use to choose a > coefficient for the quotient is exactly Ômake > the product g*h come out right'.] That was basically the proof I was alluding to, except I didnt want to work out the details in make the product g*h come out right thing. -Greg Magarshak Computers are useless. They can only give you answers. - Pablo Picasso. === Subject: Re: Dividing Power Series >Can I divide power series by long division as follows: > yes, at least for z in a disk where both series converge > and the divisor has no zero. If the divisor has a zero at the origin, then you get a Laurent series about 0(for example, cos z / sin z > pf: the two series define analytic functions f, g in > this disk. now h = f/g is analytic in this disk, so > it has a power series. it's easy to show you can > get the series for f = g*h by multiplying the series > for g and h formally, and long division is just the > inverse of formal power-series multiplication [at > each step the criterion you use to choose a > coefficient for the quotient is exactly Ômake > the product g*h come out right'.] That was basically the proof I was alluding to, except I didnt want to work out the details in make the product g*h come out right thing. -Greg Magarshak Computers are useless. They can only give you answers. - Pablo Picasso. === Subject: Re: Rotation in N dimensions > This is a long shot, but is there a nice way to express the > transformation matrix for a rotation in N dimensions about a given N-2 > dimensional line (all spaces are Euclidean)? Any particular > expression for the N-2 d space would most likely work, but the spanning > space of N-2 orthonormal vectors is probably easiest for everyone. > -Peter Here is a nice way. The rotation operator can be expressed as R=exp(i*angle*ab*I) where ab is the normalised invariant line of the rotation and I is the elliptic polarity (linear map from points to hyperplanes). The theory is a text A. N. Whitehead's Geometric Algebra that I'm writing that you can download from http://homepage.ntlworld.com/stebla/Whitehead.html and the actual equation you want is (4.26). Stephen Blake -- http://homepage.ntlworld.com/stebla === Subject: Re: Rotation in N dimensions > This is a long shot, but is there a nice way to express the > transformation matrix for a rotation in N dimensions about a given N-2 > dimensional line (all spaces are Euclidean)? Any particular > expression for the N-2 d space would most likely work, but the spanning > space of N-2 orthonormal vectors is probably easiest for everyone. > -Peter The rotation by angle theta in the xy-plane is described by the matrix | cos(theta) -sin(theta) | | sin(theta) cos(theta) | If you want to rotate N-space leaving the span of the first N-2 canonical basis vectors fixed, place the above matrix below and to the right of an (N-2)x(N-2) identity matrix, like this: | 1 | | 1 | R = | . | | . | | . | | 1 | | cos(theta) -sin(theta) | | sin(theta) cos(theta) | If you want to hold the N-2 plane P fixed, and rotate its orthogonal complement, let A be formed by taking an ortho- normal basis of P as its first N-2 columns, and an ortho- normal basis of the orthogonal complement of P as the final two columns. Then, your matrix for the rotation you want is formed by this product: X = A R A^T (where ^T: transpose). I believe that's what you want. Dale. === Subject: Re: Means: Arithmetic < (?) < Quadratic You can also add other well known interesting means (powers in parentheses), harmonic mean(-1), geometric mean(0) and maximum(infinity). These are also called generalized means and apply to sets of positive real numbers. See http://www.wikisearch.net/en/wikipedia/g/ge/generalized_ mean.html Ernie Adorio > As > {[(Q_1)^1 + (Q_2)^1 + (Q_3)^1 ... + (Q_ut)^1]/UT}^(1/1) > = Arithmetic Mean; > and > {[(Q_1)^2 + (Q_2)^2 + (Q_3)^2 ... + (Q_ut)^2]/UT}^(1/2) > = Quadratic Mean = Root-Mean-Square; > does > {[(Q_1)^1.5 + (Q_2)^1.5 + (Q_3)^1.5 ... + (Q_ut)^1.5]/UT}^(1/1.5) > = ? Mean = Root-Mean-Half-Cube?; > Since 1 = 2/2 and 2 = 4/2 and QUADRatic means 4 -atics--in the same > way quadrangle means 4 angles--as 1.5 = 3/2, could that be the > triatic mean? > Yeah, I know, using the same logic that means the arithmetic mean could > also be called the diatic mean--? > If not, is there a legitimate name for the Root-Mean-Half-Cube? > ~Kaimbridge~ > ----- > Wanted?Kaimbridge (w/mugshot!): > http://www.angelfire.com/ma2/digitology/Wanted_KMGC.html > ---------- > Digitology?The Grand Theory Of The Universe: > http://www.angelfire.com/ma2/digitology/index.html > ***** Void Where Permitted; Limit 0 Per Customer. ***** === Subject: Random Walk Query I've been trying to work this out for ages, but can't do it; it involves a very nasty convolution. It seems as if it should be a well-known result; the fact that it isn't suggests it may be too hard to get a closed-form result. Consider the standard unbiassed random walk on the plane lattice:- probability 1/4 each of going to the points 1 unit N, E, S, W of where you are. It is standard bookwork that (starting at the origin) one will eventually return to the x-axis sooner or later, and the probability distribution for when this happens is standard as well. What I want to know is, what is the probability distribution for the x-coordinate of where it occurs? Can anyone help me out please? ------------------------------------------------------------- --------------- -- Bill Taylor W.Taylor@math.canterbury.ac.nz ------------------------------------------------------------- --------------- -- Have you paid your random road tax recently? If not - then visit your local speed camera! ------------------------------------------------------------- --------------- -- === Subject: Re: Help me remember a book/mathematician > Some time ago I was reading a glossy biography of sorts of some US > mathematician, described as a) mostly publishing with a partner b) > usually providing the creative material and leaving the drudgery to > his partner c) he was delighted at the advent of computers, as he > could finally explore mathematics interactively, I think he called > experimental mathematics. Any idea what was the book or > mathematician? There are collective biographies with the titles Mathematical People and More Mathematical People. Although you were looking for a biography of a single person, perhaps one of these two books would help you to narrow the search. Try your library's interlibrary loan system. David Ames === Subject: Re: Brian Garry Symak - January 25th 1956 SKDH - Same Kook, Different Handle... > 11 12 13 14 15 <-Brian's positions > S Y M A K > 19 25 13 1 11 = 69 <-value > Brian says he is the second of five kids. > 69+ Dad 28 11 27 332/33 +10674 > 69+ Mom 1 2 31 32/333 +9513 > 69+ Kid > 182 Brian 25 1 56 25/341 +389 > Brian 44 Garry 69 Symak 69 > 69+ Kid > 69+ Kid > 69+ Kid > Brian's day, month and year of birth adds to 82, his name adds to > 182. His names add to 44, 69 and 69, corresponding to Genesis 44, > Exodus 19 and Exodus 19, together for 82. > Mom 1st <-Genesis with 1533 verses > Dad 28th <-Hosea with 197 verses > Brian 25th <-Lamentations with 154 verses > The parents were born on days of the year averaging 182 while > Brian's name adds to 182. Brian and his parents were born on days and > in months and years adding to 182. Dad and Brian were born on days 28 > and 25, together these Bible Books contain 351 verses, while mom was > born on the 1st, corresponding to formulas are using same set of functions all operating on same >variables. I need to determine their equivalence for any values of >these variables. >For example: >Variables: a,b,c >Functions: f1(a), f2(b), f3(c), f4(a) >Formulae 1: (f1(a) OR f2(b)) AND f4(a) >Formulae 2: (f1(a) AND f4(a)) OR (f2(b) AND f4(a)) Sometimes in logic terminology varies quite a bit. Since the variables are not quantified, think of them as *constants*: so a, b, c are constants. Your f1, f2, f2 and f4 are usually called 1-place *predicates*. Then f1(a), f2(b), etc., are called *atomic formulas*. >My reasoning is that I should be able to treat each of >function/variable pair here as propositional logic variable: >f1(a)=P1 >f2(b)=P2 >f3(c)=P3 >f4(a)=P4 >(other combinations, like f1(b) are omitted since they not used in >neither formulae) Yes, you can treat each atomic formula as a propositional atom. >Now, all I need to prove, is equivalence of: >(P1 OR P2) AND P4) >and >(P1 AND P4) OR (P2 AND P4) Right. Your formulas are Boolean combinations of atomic formulas. >Does it make sense? I apologise if my reasoning is naive, but I am >willing to >learn more and will appreciate pointers in right direction, so I can >read more >about the subject and arrive to the right solution. What you're doing makes perfect sense. There are at least two simple ways to prove the equivalence of propositional formulas. (i) by constructing truth tables (this gives you DNF) (ii) by a semantic tableau (you may not have learnt about this yet). --- Jeff === Subject: Re: Open Letter to Serre Regarding Grothendieck I have the (full, I believe) translation of the autobiography to Russian. If anyone is interested, please contact me at i76r@yahoo.com. Joseph Shtok === Subject: Re: Upper bounding a transcendental function with Polynomials. > Jan C. Hoffmann schrieb im Newsbeitrag > schrieb im Newsbeitrag > > schrieb im Newsbeitrag > > > Hello everyone, > > > > > > I am trying to solve the following problem : > > > > > > I have a function f[x] given as an unsolvable integral > > > which is : > > > > > > f[x] = - Integrate[Exp[-y^2/2]/Sqrt[2*Pi] * > > > Ln[ (1-B) + B/D * Exp[(y+A*x)^2/2/(1+C)] ], > > > {y,-Infinity,Infinity}] > > > > > > where A>0 , 0 < B < 1 , C = 1/A^2 , D = Sqrt[1+A^2] > > > > > > now define g[x] = f[x] - f[0]. > > > > > > It is easy to see that f[x] and g[x] are even functions of x. > > > Also g[x=0] = 0 , and it can be seen that g[x] < 0 for all > > > other x ( g'[x] < 0 for all x>0) and finally > > > lim_{x->Infinity} (g[x]) -> -Infinity > > > > > > g[x] is monotonically decreasing for x>=0 , from 0 to > > > -Infinity. > > > > > > > > > Now I'm looking for the LARGEST number E>0 such that > > > (I need to do this for SPECIFIC values of A,B) > > > > > > g[x] + E*x^2 <= 0 for all x. > > > > > > > > > Does anyone have any idea how to attack this problem please ? > > > > > > If f[x] was a simple function then by common calculus tools > > > this could be solved, BUT the unsolvable integral complicates > > > things a great deal. > > > > > > EVEN if I can find a number that is close enough to this optimal > > > E from BELOW it will be great (just so the above equation is NEVER > > > larger than 0). > > > > > > > > > If anyone knows how to find a polynomial that will be an upper > > > bound on g[x] at least in the interval [0, z) , for z close to 1 or > > > smaller, this can be very helpful as well (I couldn't do this using > > > taylor series because knowing the sign of the the error is very > > > problematic). > > > > > > > > > Your function > > > > z = -Exp(-y ^ 2 / 2) / Sqrt(2 * pi) * Log((1 - B_) + B_ / D_ * Exp((y > + > A_ * > > x) ^ 2 / 2 / (1 + C_))) > > > > Example values > > > > A_ = 0.5 > > B_ = 0.5 > > C_ = 1 / A_ ^ 2 > > D_ = Sqrt(1 + A_ ^ 2) > > > > Have a least square polynomial function approximation > > > > z = f (x, y) = 4.06440168454817*10^-03*x^0*y^0 > > + -3.09292015753578*10^-07*x^1*y^0 + -3.33405592787114*10^-03*x^2*y^0 > > + -1.29496691491341*10^-06*x^3*y^0 + -1.85929697392827*10^-05*x^4*y^0 > + > > 6.21558610087785*10^-09*x^0*y^1 + -1.89758446536286*10^-03*x^1*y^1 + > > 4.48123783330665*10^-07*x^2*y^1 + -2.15564446786301*10^-05*x^3*y^1 + > > 3.25800980134267*10^-07*x^4*y^1 + -7.77554611994807*10^-04*x^0*y^2 + > > 2.77828932482955*10^-08*x^1*y^2 + 3.55854425430838*10^-04*x^2*y^2 + > > 1.15351834872955*10^-07*x^3*y^2 + 2.10917380269816*10^-06*x^4*y^2 > > + -2.19336285936576*10^-10*x^0*y^3 + 6.73395796611381*10^-05*x^1*y^3 > > + -1.5798490603598*10^-08*x^2*y^3 + 7.68155639890942*10^-07*x^3*y^3 > > + -1.14608524858889*10^-08*x^4*y^3 + 2.08631471802857*10^-05*x^0*y^4 > > + -5.56456152799723*10^-10*x^1*y^4 + -7.87273816815265*10^-06*x^2*y^4 > > + -2.29770862224741*10^-09*x^3*y^4 + -4.79893346366327*10^-08*x^4*y^4 > > > > Integral ~ -3.0267E-02 > > > > Graph and boundaries you find in > > http://homepages.compuserve.de/Jan390906/news/z-ng-04-08-13- 07.htm > > > > > > Hi Jan, > > > > I'm sorry but I fail to see how this approximation helps me in > > solving my initial problem ? > > I don't need just an approximation of g[x], I also MUST know if this > > approximation constitutes an upper bound on my real function. > > > > The problem I'm trying to solve is to find the largest E, such that > > g[x] + E*x^2 <= 0 for all x. > So far I integrated f[x] for a specified domain. > For finding integration f[x=0] I had to set x=10^-6 for avoiding trouble > with integration. > f[10^-6]= 3.296E-09 ~ 0. > That is g[x] ~ f[x] > Now you could compute table values like > x g[x] > 0 ~ 0 > 1 -1.26685267106262E-03 > 2 -3.02674846613907E-02 > 3 -0.116469436214612 > 4 -0.292684349167664 > 5 -0.596404538356658 > 10 -5.724148314286670 > Least square approximated data for [0, 10] > g[x] = 3.718173667589*10^-03 + -1.919570149511*10^-02 * x^1 + > 1.457880028900*10^-02 * x^2 + -6.993542610978*10^-03 * x^3 > Find E > g[x] + E*x^2 = 0 > E = -g[x]/x^2 > For this example you get a local max E(0.417) = 0.0129 > See also graph and max > http://homepages.compuserve.de/Jan390906/news/z-ng-04-08-14- 11.htm > Remark > You stated {y,-Infinity,Infinity}. You can reduce infinity to a plausible > domain that can be evaluated by an increase of the result e.g. +/-10^-6. > CORRECTION: > My program didn't find the solution via polynomial (degree 4). The program > was accidenly swiched to direct numerical integration to your function. > Sorry, I wasn't aware of that. But the result is ok. > A polynomial solution is possible. But you need a much higher degree. Sorry but I'm not looking for an approximation polynomial, I'm looking for a polynomial that is a good UPPER BOUND on the function g[x] which I defined above. I'm still waiting for some assistance on this, if anyone has any idea what I can do please advise. === Subject: Re: LinAlg: Vector Spaces - need help understanding an example > Yes, it's the same as the original system, but with the constant of > each equation set to 0, i.e. one sets the right side to zeros. So, I > know what a homogenous system is but it seems as if I still haven't > fully grasped all its implications. > Consider the single homogeneous equation x + y = 0 > versus the single non-homogeneous equation x + y = 1. > For the first (homogeneous) equation, x + y = 0, > x = 1, y = -1 and x = 3, y = 3 are solutions, > and x = 1 + 3 = 4, y = -1 + -3 = -4 is also a solution, so adding > those solutions gives another solution. This always happens. This is interesting and I see that this is the case by doing the algebra behind the sum of solutions, i.e. for the homogenous: x + y = 0 ; y = -x (x_1, -x_1) + (x_2, -x_2) = (x_1 + x_2, -x_1 + -x_2) = (x_1 + x_2, -(x_1 + x_2)) = (x, -x) and for the non-homogenous: x + y = 1 ; y = a - x (x_1, a - x_1) + (x_2, a - x_2) = (x_1 + x_2, (a - x_1) + (a - x_2)) = (x_1 + x_2, 2a - x_1 - x_2) = (x_1 + x_2, 2a - (x_1 + x_2)) = (x, 2a - x) but what is the intuition behind this? Is it that all vectors in the solution set of the homogenous equation lie in the line making up the set and thus the sum of two vectors will also lie in the line? And that adding vectors from the solution set of the non-homogenous equation will give a vector which is offset from a solutions by the constant in the equation? Is this as intuitive as this gets or is there some even more intutive explanation? > For the second (non-homogeneous) equation, x + y = 1, > x = 2, y = -1 and x = 3, y = -2 are solutions. > but x = 2 + 3 = 5, y = -1 + -2 = -3 is not a solution, so that adding > solutions does not always give a solution. Hmm, you say does not always. Does this mean that there are non-homoegenous equations where two solutions actually add to another solution? > Does this simple example help sort things out for you? Yes, it was a very nice example that got me thinking even harder about these things! /Henrik === Subject: Re: You're still a ing nut-case, Daryl... > Looks like the kook changed his handle and e-mail address to avoid > killfiling... [snip huge repost of daryl randomness] I'll be killfiling you if you don't stop reposting Daryl's madness in its entirety with your 1-liner replies! === Subject: Re: how to evaluate the addition of millions of functions dynamically and efficiently? > I am also interested in the RBF monopole you've mentioned in your > previous post. > Do you think it will give me a drastic performance improvement? If so, I am > going to dive into it and learn how to make my program a lot faster... I have never implemented it, but the idea is rather similar to the FFT - break the problem into smaller simpler subproblems, and recurse. I understand that the implemetation is rather involved, so I'd suggest that you consider using it only if - your centres are genuinely scattered, - you basis function does not have compact support. Otherwise I would guess that a local search would be quicker and a lot easier to implement. As I say, I'm no expert on this, others may correct me. There are quite a few papers in the (mathematics) literature, I'd recommend that you look at those before proceeding. -j === Subject: Re: how to evaluate the addition of millions of functions dynamically and efficiently? > Is your algorithm supposed only to work only for regularly laid-out > functions? I mean for the support of functions, I can make them truncated > into rectangular-shaped finite support, even for infinite support functison. Yes, the trick works best for - translates of the same function with small support - arranged on a regular grid (might be square, ractangular, triangular) I'd recommend against truncating a function with infinite support -- this will lead to a discontinuity and so to a slow decay in the fourier transform. Instead (if possible) use a basis function which is compactly supported and smooth. Search for Wendland or Lewitt along with radial basis functions. > These functions are nicely behaved. So that's ok. But for the layout of the > center of these functions, for this function array, your dynamic local > summation only work for horizontally and vertically alligned function-center > array layouts? Not necessarily. You just need to be able to work out what the indicies of the nearby points are *without* doing a test of every possible grid node. This can be done for a triangular grid, for example. If your grid nodes are genuinely scattered then the easy way wont work. > How to decide if a point is covered by which overlapped neighboring > functions, for those irregular layout? I guess there should be some fast > algorithm and fast data structure for handling this in the literature? Possibly you could use a 2-stage method 1) find the neighbours for each grid node & store in a file 2) read neighbours into memory and for each node do the local search with these neighbours Then 1) (which is non-trivial, but there is some code out there for doing it) has (if I remember correctly) complexity O(N^2 log(N)), but 2) has O(N), and 1) only need to be done once. Note that this may require *lots* of memory. Work out how much you will need before you start to code! -j === Subject: Re: books for self study, number theory, set theory, analysis... > Hi all, > I have a free couple of months coming up here in a week or so. I'd like to > do some self-study and have access to a decent university math library. > I am interested in number theory, set theory, and perhaps real and complex > analysis. > Can anyone reccomend some *fairly basic* texts on above topics? > I have a fairly good handle on calculus, geometry and trig, linear algebra, > kW For number theory, a previous posting listed 23 books on number Re: 341 Problems in Elementary Number Theory in order to get that list. There is also a university in Australia that has/did have on-line lecture notes for Number Theory and Algebraic Number Theory. David Ames === Subject: Re: books for self study, number theory, set theory, analysis... >Hi all, >I have a free couple of months coming up here in a week or so. I'd like to >do some self-study and have access to a decent university math library. >I am interested in number theory, set theory, and perhaps real and complex >analysis. >Can anyone reccomend some *fairly basic* texts on above topics? >I have a fairly good handle on calculus, geometry and trig, linear algebra, > kW For set theory, P. Halmos: Naive set theory is the best introduction. And if you like original sources, get G. Cantor: Contributions to the Founding of the Theory of Transfinite Numbers For number theory, it is difficult to make any sense of it without understanding how it evolved. So make sure you get a book that treats number theory in its historical context. For example, O. Ore: Number theory and its history or A. Weil: Number theory: an approach through history or H.M. Edwards: Fermat's Last Theorem (A genetic introduction to algebraic number theory) === Subject: multiples in sequence I need help to understand a solution to the following problem Problem:How many multiples of s are there in the sequence r,2r,3r,...,sr? Solution:Apply the following theorem to the problem: Th:ax ==b (mod m) has exactly (a,m) solutions if (a,m)|b. Answer:(r,s) Now I don't understand this. Would someone kindly explain this for me? === Subject: Re: multiples in sequence > I need help to understand a solution to the following problem > Problem:How many multiples of s are there in the sequence r,2r,3r,...,sr? > Solution:Apply the following theorem to the problem: > Th:ax ==b (mod m) has exacGenesis with 1533 verses > (1533-351=1182). Chapter 182 is Deuteronomy 29 with 29 verses, it > exceeds it's 140th non-prime position by 42 (the 29th non-prime), > Bible Book 29 is Joel (42). The parents were born on days of the month > adding to 29 and in years averaging 29. The parents were born in years > adding to 58, it's the 42nd non-prime while 42 in turn is the 29th > non-prime, while Bible Book 29 is Joel (42). The parents were born on > days and in months adding to 42 (29th non-prime). Dad was born 299 > days closer to the end of the year than to the beginning of the year. > Brian was born 58 (29+29) days after dad's birthday. The parents were > born in 27 and 31, the former is 87% (29+29+29%) of the latter. Mom > and Brian were born in years adding to 87 (29+29+29). Dad's 332nd day > of birth is the 265th non-prime while chapter 265 is First Samuel 29. > The parents were born on days of the century adding to a multiple of > 29 (21547=29x743). Brian's consonants exceed his vowels by 58 (29+29 > and is the 29th non-prime in non-prime position). His odd valued > letters exceed his even valued letters by 42 (29th non-prime). Brian's > prime and square valued letters together add with their positions for > 182 (Deuteronomy 29 with 29 verses). His letters with upsidedown > and/or ßippable attributes (A, A, A, I, K, M, R, R, R and S) add to > 109 (29th prime). His unrepeated letters add with their positions for > 129 (the primes up to 29 add to 129). The parents are separated by > 1161 days (First John 2 with 29 verses... it's the last hour). Brian > and I were born on days of the year adding to the 73 verses of Bible > Book 29 (and is the Lucas numbers up to 29). > Lucas > 1 > 3 > 4 > 7 > 11 > 18 > 29 > -- > 73 <-the Lucas numbers up to 29 add to > the 73 verses of Bible Book 29 > J O E L <-Bible Book 29 > 10 15 5 12 = 42 <-29th non-prime > C O P P E R <-29th element > 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas > numbers up to 29, there > is a copper riding a > horse on the 1973 > Canadian 25 cent piece > C E N T <-made out of 29th element > 3 5 14 20 = 42 <-29th non-prime > Non-Primes > 1 57 110 158 > 4 58 111 159 > 6 60 112 160 > 8 62 114 161 > 9 63 115 162 > 10 64 116 164 > 12 65 117 165 > 14 66 118 166 > 15 68 119 168 > 16 69 120 169 > 18 70 121 170 > 20 72 122 171 > 21 74 123 172 > 22 75 124 174 > 24 76 125 175 > 25 77 126 176 > 26 78 128 177 > 27 80 129 178 > 28 81 130 180 > 30 82 132 182 <-140th > 32 84 133 183 > 33 85 134 184 > 34 86 135 185 > 35 87 136 186 > 36 88 138 187 > 38 90 140 188 > 39 91 141 189 > 40 92 142 190 > 42 93 143 192 > 44 94 144 194 > 45 95 145 195 > 46 96 146 196 > 48 98 147 198 > 49 99 148 200 > 50 100 150 201 > 51 102 152 202 > 52 104 153 203 > 54 105 154 204 > 55 106 155 205 > 56 108 156 206 > The parents were born in months adding to 13 and on days of the > month adding to the 29 chapters of Bible Book 13. Brian and his > parents were born on days of the month adding to 54 (13 plus the 13th > prime), Bible Book 13 opens with 54 verses while Bible Book 54 > contains 113 verses. The parents were together born 339 (113+113+113) > days further into their years than Brian, prettier as Bible Book 13 > opens at chapter 339 (113+113+113). Dad and Brian were born on days of > the year adding to 351 (Bible Book 13 chapter 13), it's the numbers 1 > through to 26 (13+13). Mom was 9124 days old when she gave birth to > Brian, it's 4 times the 339th (113+113+113th) prime (2281). The > parents were together 53.13 years old when Brian was born. His given > names add together for 113, his common name adds to 113. His > unrepresented letters exceed his represented letters by 113. Mom was > born on the 11354th day of the century (Book 54 contains 113 verses), > I wonder if she first gave birth in 54 (13 plus the 13th prime). Brian > and his parents were born on days of the century adding to 42025, it's > a multiple of the 13th prime squared (5x5x41x41). Dad was born on the > 10193rd day of the century, it's the 1252nd (313+313+313+313th) prime. > Brian was born on the 25th, corresponding to Lamentations with 154 > verses (the first 13 non-primes). > Primes Non-Primes Fibonacci Lucas Numbers > 2 1 0 1 1 > 3 4 1 3 2 > 5 6 1 4 3 > 7 8 2 7 4 > 11 9 3 11 5 > 13 10 5 18 6 > 17 12 8 29 7 > 19 14 13 47 8 > 23 15 21 76 9 > 29 16 34 123 10 > 31 18 55 199 11 > 37 20 89 322 12 > 41 <-13th-> 21 <-13th-> 144 <-13th-> 521 <-13th-> 13 > --- --- -- > 238 154 <-Lamentations 91 > Brian was born on the 25th, his given names differ in value by 25, > his first and last names differ in value by 25. His even valued > letters are in positions adding to 25 (16th non-prime) while his odd > valued letters add to a multiple of 16 (112=7x16). His 16 missing > letters add to 116+116. The parents were together 53 (16th prime) > years old when Brian was born on the 25th (16th non-prime). Mom was > born on the 16+16th day of the year and gave birth to Brian on the > 25th (16th non-prime). Dad and Brian were born on days of the month > adding to 53 (16th prime). Dad and Brian were born in years adding to > the 83 verses of the 16th Book of the New Testament. Symak adds to 69 > (16 plus the 16th prime). Brian was born 316 days closer to the > beginning of the year than to the end of the year, corresponding to > the 25th (16th non-prime) chapter of The Kings. > Primes Non-Primes Fibonacci Lucas Numbers > 2 1 0 1 1 > 3 4 1 3 2 > 5 6 1 4 3 > 7 8 2 7 4 > 11 9 3 11 5 > 13 10 5 18 6 > 17 12 8 29 7 > 19 14 13 47 8 > 23 15 21 76 9 > 29 16 34 123 10 > 31 18 55 199 11 > 37 20 89 322 12 > 41 21 144 521 13 > 43 22 233 843 14 > 47 24 377 1364 15 > 53 <-16th-> 25 <-16th-> 610 <-16th-> 2207 <-16th-> 16 > --- --- ---- ---- --- > 381 225 1596 5775 136 > Mom was born on the 1st (Genesis with 50 chapters) and married > Symak, pretty as the name adds to 69 (the 50th non-prime). > Brian's middle name adds to 69 and also his last name adds to 69. > His 44 valued first name adds with his 25th day of birth for 69. > The parents were born on days 332 and 32. Brian's day, month and > year of birth adds to 82 (Exodus 32). His repeating letters exceed his > unrepeated letters by 32. > Primes Non-Primes > 2 1 > 3 4 > 5 6 > 7 8 > 11 <-5th-> 9 > -- -- > 28 28 > Dad was born on the 28th (the first 5 primes or the first5 > non-primes) day of the 11th (5th prime) month. Mom was born in 31 (the > 5th prime in prime position). There are 5 kids in the family. Brian > has 5, 5 and 5 lettered names. He was born on the 25th, his names > differ in value by 25, it's 5x5 and is 5 plus the 5th prime (11) plus > the 5th non-prime (9). He was born in year 56, it's the first 5 primes > plus the first 5 non-primes. His first name adds to 44 (Acts, the 5th > Book of the New Testament). Brian's repeating letters exceed his > unrepeated letters by 32 (2 to the 5th). The different letters in his > given names add to 76 (55th non-prime). Dad (born on the 28th) was > 10285 days old when Brian was born (28.15 years). Brian is 55.57 weeks > older than me. > Primes > 2 61 149 239 347 > 3 67 151 241 349 > 5 71 157 251 353 > 7 73 163 257 359 > 11 79 167 263 367 > 13 83 173 269 373 > 17 89 179 271 379 > 19 97 181 277 383 > 23 101 191 281 389 <-77th > 29 103 193 283 397 > 31 107 197 293 401 > 37 109 199 307 409 > 41 113 211 311 419 > 43 127 223 313 421 > 47 131 227 317 431 > 53 137 229 331 433 > 59 139 233 337 439 > Brian is coming out of a family of 7. Brian and his parents were > born on days of the year adding to 389 (77th prime), and Brian is 389 > days older than me (77th prime). Brian and his parents were together > born with 707 days remaining in their years. They were born in months > adding to 14 (7+7). Brian was born 7 days before mom's birthday. Brian > has 7 different letters in his given names. The vowels in his given > names add to 36 (7+7p+7np), the consonants in his given names add to > 77. He was born in 56 (77 is the 56th non-prime). > The first 4 primes plus the first > 4 non-primes add together for the > 36 chapters of Bible Book 4 Numbers: > Primes Non-Primes > 2 1 > 3 4 > 5 6 > 7 <-4th-> 8 > -- -- > 17 19 > The first 4 primes in prime > positions add together for the > 36 chapters of Bible Book 4: > Primes > In Prime > Primes Positions > 1 2 > 2 3 <- 3 > 3 5 <- 5 > 4 7 > 5 11 <- 11 > 6 13 > 7 17 <- 17 > -- > 36 > Seven plus the 7th prime plus the > 7th non-prime adds together for > the 36 chapters of Bible Book 4, > pretty as 7 is the 4th prime: > Primes Non-Primes > 2 1 > 3 4 > 5 6 > 7 8 > 11 9 > 13 10 > 17 <-7th-> 12 > -- -- > 58 50 > The parents were born in years adding to 58 (the 7 primes up to > 17). The parents were born in 27 and 31, these Bible Books contain an > average of 189 verses (the first 17 primes minus the first 17 > non-primes). Brian was botly (a,m) solutions if (a,m)|b. > Answer:(r,s) > Now I don't understand this. Would someone kindly explain this for me? In the theorem take m=s, a=r, b=0. === Subject: Re: Tangle: The Game > Here is a game which *might* be fun to play, and might be fun to > analyze mathematically and for strategy. > > > For 2 or more players, each with a different colored pencil. > > Start with a large square (or any convex polygon) drawn on paper. > > Player 1 starts by drawing a straight line-segment, using a > straight-edge, from any corner of the square/polygon through the > interior of the polygon until hitting an edge of the polygon. > > Players together draw a path by alternatingly drawing line segments, > using a straight-edge (and making each segment the color associated > with the player drawing the segment), each segment which is from the > end of the last segment drawn by the previous player, in any > direction(with restrictions*), and ending at the first previously > drawn line-segment intersecting the segment being drawn, or ending at > a boundry of the square/polygon. > > (*) If the last line-segment drawn ended at the edge of the > square/polygon, the path must bounce, ie. the next segment must be > towards the interior of the polygon. > If the last segment drawn ended at a previously drawn segment (of any > color), the path must cross this older segment, ie. the new segment > must be on the opposite side of the older crossed segment from the > immediately previous segment. > > Segments cannot be drawn to vertexes (where the path crosses itself or > bounces off the edge of the polygon.) > > The players draw a predetermined number of segments (say, a total of > 50). > Each player gets a point every time the path crosses a pre-drawn > segment of his/her color. > The winner has the most number of points. > > A clarification: > The path refered to above is the total collection of line-segments. > The line-segments are drawn end-to-end and alternate in color. > I will attempt some ascii-art. > (View with fixed-width font.) > If the game is like so: > < older line-segment > > -- < last line drawn by previous player > > > you continue on the opposite side of the older line-segment in any > direction: > /< your new segment > / > -- > > > And your new segment continues only until the first line-segment or > edge of the square/polygon encountered, where your line-segment then > terminates. > If the game is like so: > > < last line drawn by previous player > > You may bounce in any direction (either to left or right), as long as > the path stays within the square/polygon: > / > / < your new segment > / > One final note: > No segment is allowed to be drawn along a previous segment. > (No co-linear segments.) > Leroy Quet 4 more things: 1) It is evident after I played this that it would probably be better, rather than making a path of a predetermined number of line-segments, that the game simply continue until one player first gets a predetermined score, say 10 or 20. In this way, players do not have to worry about keeping track of the number of moves made. And there is also, with this new rule, no chance of a tie. 2) One purpose of playing this game, aside from winning, is to hopefully get a cool abstract design. :) 3) I was a little unclear about when a point is scored pre-drawn segment of his/her color.). Whenever a line-segment is drawn by anyone to another segment of color c, and player c is using pencil/pen color c, then player c gets a point. By path crosses, I mean 2 segments come together at the previously drawn line-segment. Like so: < previously drawn segment of color c, path crosses this segment. _______ / / / 4) As is noted elsewhere in this thread, there is an advantage to being player 1 (and so players should play multiple rounds, trading off who is player 1). There is a likely set of opening moves. As an easy puzzle, try to give the best next move for player 2: Playing on a square board. Moves: 1) Player 1 draws segment from lower-left (Southwest) corner East-Northeast to right edge of square. 2) Player 2: NW to upper edge of square. 3) Player 1: South to player 1's first segment. 4) Player 2: SE to right edge of square. 5) Player 1: West to player 1's first segment. Now, where should player 2 move? (Easy, but this will help you familiarize yourself with the game, if you choose to try to solve this puzzle.) Leroy Quet === Subject: Re: Tangle: The Game > 4 more things: > 1) It is evident after I played this that it would probably be better, > rather than making a path of a predetermined number of line-segments, > that the game simply continue until one player first gets a > predetermined score, say 10 or 20. > In this way, players do not have to worry about keeping track of the > number of moves made. And there is also, with this new rule, no chance > of a tie. We've played (2 players) until a player reaches 50, starting with a square approx. 6x6 inches; although I find that if you increase one diagonal to 8 inches (and start on a short diagonal corner), you've got more space. I've wondered whether the game would be more balanced if the two diagonals were pre-drawn in a neutral color. Michael -- Feel the stare of my burning hamster and stop smoking! === Subject: Re: A Simpler More General Recursion Puzzle > First I will give a more specific case, then the more general recursion. > Let n = any positive integer. > If a(1) = 1; > a(m) = (1/m)*(ßoor(m^(1/n)) + sum{k=1 to m-1} a(k)) > for all m >= 2, > then give a non-recursive form for {a(m)}. > More generally, > if {b(k)} is any sequence defined for all positive integer indexes k, > and a(1) = b(1), > and > a(m) = (1/m)*(b(m) + sum{k=1 to m-1} a(k)), > for all m >= 2, > then give a non-recursive form, in terms of {b(k)}, for {a(m)}. > I will give the answer in a few days if no one else gets it sooner. > Leroy Quet Michael Mendelsohn already solved this on rec.puzzles. Here is a link to a post there: Leroy Quet === Subject: Re: Ingeger solutions of algebraic function >I am trying to solve the ecuation >x^2+y^2+z^2 = 2xyz >in the integers. >I reduce it Mod 3 and i reduced it to >x^2+y^2+z^2=6xyz factoring the 3. >I think that this equiation only has the trivial solutions is it >right? do you have a hint to proof it? > Why do you think so? > You might try your equation mod 4, and then by induction > mod 2^k for k >= 2. > 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: Convolution Sum Over Divisors Let a(1) = 1. Let, for m >= 2, a(m) = sum{k|m, k escribi.97: > I'm currently trying to find a solution as to how to find all the > tangents that touch two ellipses. The ellipses are independent of each > other and are not rotated. And yes, the ellipses could be the same, > could be intersecting, and so forth. > Would anyone know of a solution or at least know of a good web site > that contains code or an actual mathematical procedure to solve the > tangent-ellipse problem? Any help would be greatly appreciated. Let the two ellipses, conic sections in general, to be: c1: Ax^2 + By^2 + Cxy + Dx +Ey + F = 0 c2: A'x^2 + B'y^2 + C'xy + D'x +E'y + F' = 0 and the tangent t: y = mx + n, with m and n unknown. In order to t be tangent to c1, the system of two equations, of t an c1, must have only one solution. Then replace y = mx + n in the equation of c1 to get a second degree equation in x, and set its discriminat equal to zero. You get a second degree equation in m and n. Do the same with c2; you get a second eauation of degree 2 in m and n. That system of two eauations of degree 2 in m and n can have among 0 and four (real) solutions, the wanted tangents. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: 3D Plane Supersedes: >Yes, I was after the vertical shadow (if you consider Z to be the up >axis). I thought the normal is the a, b and c values of the equation: >ax + by + cz + d = 0 >So you're saying I re-arrange that to be: >(-ax - bx - d) / c = z >by plugging in (a,b,c) which is the normal of the plane and plugging in >x,y and leaving z on the other side of the equation? >I can't speak for Lynn but this looks right to me. >z_projected = (-ax - bx - d) / c >distance = |z - z_projected| Correction: z_projected = (-ax - by - d) / c --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: measurable sets and functions > Hi everybody > > I've just started studying measure theory and chose Rudin's Real and > Complex Analysis. I'm confused about some points and any help is > welcome. > > If X is a set and M is a sigma-algebra in X, then, according to Rudin, > a subset of X (which Rudin assumes to be a topological space) is > measurable if it belongs to M. If the pair (X, M) is a measurable > space (a pair composed of a topological space and a sigma-algebra > defined on it, according to Rudin) and Y is a topological space, then > Rudin defines f:X->Y as a mesurable function if f^(-1)(V) is in M > whenever V is an open set of Y. To Rudin, a measure is any real or > complex valued set function defined on M that is countably additive. > > But in his book, Charambolos D. Alipranti uses also the term > sub-additivity of the measure. He considers the concept of an outer > measure, which doesn't show in Rudin's book. He says a set E of X is > measurable according to an outer measure u if u(E) = u(E inter A) + > u(E inter A') - A' being the complement of A-for every set A of the > sigma-algebra defined in X. > > Rudin's definition implies the measurability of a set depends only on > the sigma-algebra, and Alipranti's one implies it also depends on an > outer measure. > > Are these 2 approaches eventually the same? > The same? No. Rudin does restrict his measure spaces to those arising > from topologies because he's an analyst, and I would be hard-pressed to > give a USEFUL example which arises any other way. Also, one of his > goals is to show how to use the Riesz Representation Theorem as the > DEFINITION of integral. > This is nonstandard (in the sense of being unusual). Aliprantis's > approach (you misspelled his name) is the standard one, in the sense of > being the most commonly used. Go ahead and read both books, and keep > in mind that there's more than one way to skin a cat. I'll try to read both books and bartle's too. Yes, I mispelled his name, it's Charalambos (seems Greek). And I also gave a wrong definition of measurable set..The correct is, a set E of X is measurable if for every subset A of X we have u(A) = u(A inter E) + u(A inter E'). Artur === Subject: Re: measurable sets and functions > > I've just started studying measure theory and chose Rudin's Real and > > Complex Analysis. I'm confused about some points and any help is > > welcome. Great book but unless your IQ is high in the sky I wouldn't recommend it as an introduction (focusing on too many technical details, which makes the essential hard to abstract, especially when you reach the Riesz representation theorem). > > Rudin's definition implies the measurability of a set depends only on > > the sigma-algebra, and Alipranti's one implies it also depends on an > > outer measure. > > > > Are these 2 approaches eventually the same? I must admit I really didn't understand other answers, and I also think that's because the question is a little confusing: measurability depends only on the sigma-algebra of the definition set (and on the topology or sigma-algebra structure of the image set, even though, as pointed by Rudin, the essential point is the presence of a topology). But the construction of a sigma-algebra may be done using outer measures -and we know easy ways to construct outer measures given a nonnegative mapping defined over a covering class- or metric outer measures, which can be constructed rather easily too -which ensure that the Borel sets are measurable-. I don't see how these two points were linked in the above question. -- Julien Santini === Subject: Re: JSH: Unskilled and unaware? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7GEP8l11760; (snip) >We don't matter to the rest of the world. >We're social misfits. You speak only for yourself (and your buddy Quinn Tyler Jackson, your closet list pal). >I may not be mathematically skilled in many ways, but I'm aware of >that and am aware of my misfit status. Ah, the truth doth shine out of all your drivel. >The clearest demonstration of our kinship is that you cannot just >ignore me, even when I request it. According to your reasoning (if I dare dignify your crap as reasoning), you can't ignore me either, you Quinn-loving nutcase. >Still, I'll try one more time: Take down the webpages. Stop the >cyberstalking. Stop the obsessive replying to my posts. I bet you can't follow your own instructions. I dare you. I DOUBLE DARE YOU! >Let me just be one more misfit on sci.math along with the other >misfits. Stay in your misfit oblivion, crank! >I'd like to retire the JSH at this point, though not saying I am >definitely, and I fear some one of you will keep dragging it out, time >after time, obsessively. We can only dream, but your constant coming out of retirement is a continual nightmare. But then, coming out (of your closet list) is what you and Quinn do best. >But again, at least I can put the idea out there. >I don't want to stop posting. I LIKE posting. I get attention ... because you are a troll (by definition), as well as a crank and crackpot. >If I had any power at all I'd have sued some of you by now welfare checks can't pay for even a greedy ambulance chaser. What happened, your parents can't afford a lawyer for you this month? They can barely afford your anti-psychotic medication (which you don't take, anyhow). You just continue to leech and steal from your parents. You are some disappointment to them. And after all the times they bailed you out of jail, and paid your dues for your high-IQ membership (ha, what a joke that group is - it includes both you and Quinn Tyler Jackson!) >To those of you who have worked for years to try and drive me off the >newsgroup, SCREW YOU Quinn's AIDS. >James Harris ... is a crank, crackpot, troll, and nuts! Yet Quinn still loves his closet list friend James. === === Subject: Re: Explaining the prime distribution > I've contacted professional mathematicians off of Usenet, and mostly > been ignored. judgment of professional mathematicians. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Explaining the prime distribution > Your first points where you define functions are all derived from > dS(x,y), so here I'm only going to talk about your function for that, > since that must be correct in order for the rest to work. >p(x,sqrt(x)) = ßoor(x) - S(x,sqrt(x)) - 1 > Just a side note, this doesn't really matter, but since sqrt(x) is a > function of x and not a seperate independent varaible, you only need > one input. Also, ßoor(x) is unnecessary because ßoor(x) = x when x > is natural, which it will always be. Actually it won't as can be seen further down, where I end up with p(x/y,y-1) and the ßoor can conveniently be put in that one place and handle everything. Also, the function is p(x,y) where p(x,sqrt(x)) just happens to count primes! So it's a multi-variable function that can be evaluated over a certain path to count prime numbers, but it makes sense, as evaluating it requires both variables to write it as a I have. > p(x) = x - S(x,sqrt(x)) - 1 > Anyway, on to the meat. > So I've finally outlined all the necessary functions, and it's now > time to get the full dS(x,y) function, which surprisingly enough is > easy to the point of trivial. > > Remember, dS(x,y) is the count of composites up to and including x > that have y as a factor which do not have any other primes less than y > as a factor. > > As a first step to calculating it, I use the fact that the total > number of naturals with y as a factor up to and including x is given > by > > ßoor(x/y) > > like with 10 again, the naturals with 3 as a factor are > > 3, 6, 9 > > but wait, 3 gets included, but it is prime and not a composite, so > with primes I need to use > > ßoor(x/y) - 1 > > to get a count of all composites up to and including x with y as a > factor, and here y is prime. > > But, what about primes less than y? Well I can count those out with > > p(y-1,sqrt(y-1)) > Right here, in your definition for dS(x,y), you use the function > p(x,y) = ßoor(x) - S(x,y) - 1. S(x,y) is the sum of dS(x,i) where i > = every prime from 2 to y. In your definition of dS(x,y) you have a > function that uses dS(x,y) in its definition. Unless you can take > p(x,y) out of your definition of dS(x,y), it will be a circular > definition and therefore be impossible to evaluate. It's a recurrence relationship. The recursion does not prevent evaluation. Evaluating dS(x,y) from y =2 to the sqrt(x), will naturally force S(x,1) = 0 and p(x,1) = x-1 so you will not infinitely recurse. Notice that S(x,1) can be talked out as saying that there are no composites with primes less than 1 as a factor as in fact there are no primes less than 1. > which you'll notice works up to 10, as > > p(3-1,sqrt(3-1)) = p(2,sqrt(2)) = 1 > as, of course, 2 is the only prime less than 3, so I now have > > ßoor(x/y) - 1 - p(y-1,sqrt(y-1)), with y prime, > > and notice that gives me the correct count up to 10 for 3, as that > just leaves > > 9 > as 3 got subtracted off by 1, and 6 got subtracted off by the count of > primes less than 3. > > But I'm not done in general, as there may be composites now that > multiply times y that are less than x, which have primes less than y > as factors. > > For instance, up to 12 I have 2(2)(3), and what I've done so far would > still leave that in the count for dS(12,3). > Ok, just trying for clarification. So far, if we took dS(12,3) we'd > get ßoor(12/3) = 4 (Count(3, 6, 9, 12)). Subtract one for the prime, > 3 (6, 9, 12). Subtract another for p(2,sqrt(2)), and we get 2 (9 and > 12 or 6 and 9). Yup, and 2(2)(3) is still left after the first corrections, though it has 2 as a factor but 2 is, of course, a prime less than 3, so more work is needed. > But I have a way to handle that possibility! > > I simply use S(x/y, y-1) as the y-1 tells it to only count primes less > than y, and it's counting the composites which can multiply times y > and still be less than x, and now I'm nearly done as I get > Here you use another function of dS(x,y) to define dS(x,y). Also, I > know it doesn't matter for the function, but it's just more clear if > you say S(ßoor(x/y),y-1). That could be an alternate way to go, I'll admit. I just like p(x,y) = ßoor(x) - S(x,y) - 1 as the one place where ßoor() shows up. As for the recursion, as I explained above, it's not a problem. In case the recursion makes you really nervous you can look at some of the explicit forms of the dS(x,y) function. For instance, if x=N, where N is an even natural, then dS(N,2) = N/2 - 1 dS(N,3) = ßoor((N-4)/6) (to see calculation go to my blog http://mathforprofit.blogspot.com/ and look for tidbits which will be near bottom) and explicit forms exist for every prime out there in idea space you could say. But the recursion can be handled anyway by noting that S(x,1) = 0. > So I finally have a working dS function: > > dS(x,y) = (p(x/y,y-1) - p(y-1,sqrt(y-1))(p(y,sqrt(y)) - > p(y-1,sqrt(y-1))) > > and to help readability I usually use brackets: > > dS(x,y) = [p(x/y,y-1) - p(y-1,sqrt(y-1)][p(y,sqrt(y)) - > p(y-1,sqrt(y-1))] > > and just like that I have my dS function rigorously defined using some > very basic algebra and a few good ideas. > James Harris > I checked the function with your examples and they all worked. > However, it's hard to know whether they work for all natural numbers > or not. With a function like this, programming the whole thing then > doing it on the computer would be faster than running through it 3 or > 4 times by hand, but I can't picture a way to program it because of > its circular nature. You can find posted programs implemented it, but I'd prefer you understand in detail from the derivation. The derivation is supposed to show that it works: why and how. Quite simply I count composites for a particular prime p, where I keep from double counting by subtracting off the count of composites that have primes less than p as a factor in two ways: 1. Subtract off the number of primes less than p 2. Subtract off the number of composites with primes less than p as a factor That route leads to recursion, which can definitely be handled, and it stops at S(x,1) = 0, as there are no composites with primes less than 1 as a factor as there are no primes less than 1. If something is not clear then I'd be happy to expand out on the derivation. > If worst comes to worst, this will still be very strong, though. The > ultimate goal is an independent function that will give you a list of > primes without having to know any previous primes and without huge > factoring. Although your circular definitions mean that you don't > quite have that, if you reworked your equations you'd very likely have > a way to find primes through iterations using previous primes, which > would be still be an amazing accomplishment. What you call a circular definition is technically called a partial difference equation. And you don't need previous primes. My prime counting function finds them on its own, which is one of those obvious facts that sci.math posters dismiss and I wonder about the mind-control. Like, that feature is really neat, never been seen before, and clearly proves that my work is important, but sci.math posters would just claim it's useless and stupid! They did that for YEARS. So you have a function that recurses to automatically select out only prime numbers, something never before seen in the math literature, but not only do mathematicians dismiss it as unimportant, posters from the sci.math newsgroup specifically attack that unique feature as useless hundred years old!!! I think my research is neat, fascinating, and clearly something worth recording for posterity, but over the last two years when I've talked about it on newsgroups I've been insulted and my work dismissed, and when I talk to mathematicians, I've been told it's of no interest or just ignored. I think they're weird. And mathematicians aren't acting at all like the Hollywood version. They're not quite acting like curious human beings. They're being strangely incurious. It's bizarre. It's like something is missing inside their heads as I'd think that they'd be excited about more math to play with, but instead, like on sci.math, they get angry and abusive. James Harris === Subject: Re: Explaining the prime distribution >The prime distribution is important for many reasons and supposedly >mathematiicans have some interest in it, but I'm going to show you in >this post a rather fascinating and basic derivation of one of the most >fascinating functions in mathematical history, which mathematicians >have for some reason fought against recognizing. >no, it was recognized by mathematicians 200 years ago. >Since there are physicists doing research on the prime distribution >and what they see as a connection to the physics world, >i could be wrong, but my impression is that physicists don't >actually think this distribution has anything to do with >physics - rather some feel that certain methods that people >wish they could make woek might also be useful in physics, >-if- they could be made to work. You raise an excellent point! What DOES any of this have to do with physics? === Subject: Re: Explaining the prime distribution > > ... > > > > I found this route to counting primes over two years ago > > Legendre found it two hundred years ago... > I'll answer this post to show you how sci.math posters so diligently > lie about my work, and in fact Christian Bau was on sci.math over two > years ago when I first started talking about my prime counting > function, he lied about it then, and he's still lying about it now. You are a complete idiot. That formula that you are so proud of _is_ a trivial variation of Legendre's formula. See: L. E. Dickson, History of the Theory of Numbers, Chelsea Publ. Co. (Reprint), Volume 1, Chapter XVIII. You may also have a look at the paper Computing pi (x): The Meissel-Lehmer Method by Lagarias, Miller and Odlyzko. There your formula can be found on the very first page and it is called not of immediate utility in calculating pi (x) because it involves 6/pi^2 * (1 - log 2) x nonzero terms. Your formula can be explicitely seen as formula 2.4 on page 8 of that paper, which is still the historical part in which they describe outdated methods. The ratio of your ego to your capabilities seems to be around 1000 to 1. By a strange coincidence, that is also how much faster my implementation that can be found at www.cbau.freeserve.co.uk is compared to your fastest implementation so far. === Subject: Re: Explaining the prime distribution > you must be new here: > nobody disputes that the Ôfunction' works [that is, > the code james has written correctly counts primes], > not that one would necessarily be able to see why > this is so from the explanation above. > the Ôdispute' is over the mathemaical significance. > it's claimed that it's a brand new method, when in Well, I'm disputing whether the function works - it doesn't, yet. But I can't believe you would dispute the mathematical significance if it would work! A function that outputs prime numbers, with no factoring, and no need for previous primes!? That would be the biggest breakthrough in number theory since Pascal's Triangle. Look, his function doesn't work. And my follow-up comment wasn't an attempt to show that it did. It was my opinion on his function. But if his function did work like he wants it to, it would be so significant that saying it wouldn't makes me question your understanding of what he's saying. === Subject: Re: Explaining the prime distribution > you must be new here: > nobody disputes that the Ôfunction' works [that is, > the code james has written correctly counts primes], > not that one would necessarily be able to see why > this is so from the explanation above. > the Ôdispute' is over the mathemaical significance. > it's claimed that it's a brand new method, when in >Well, I'm disputing whether the function works - it doesn't, yet. But >I can't believe you would dispute the mathematical significance if it >would work! A function that outputs prime numbers, with no factoring, >and no need for previous primes!? what gave you the idea that it does that? it -counts- primes. >That would be the biggest >breakthrough in number theory since Pascal's Triangle. >Look, his function doesn't work. the description in this thread may well not work - the code he's given -does- work [for counting primes, not generating them.] > And my follow-up comment wasn't an >attempt to show that it did. It was my opinion on his function. But >if his function did work like he wants it to, it would be so >significant that saying it wouldn't makes me question your >understanding of what he's saying. uh, right. read his post again, then get back to us on who's misunderstanding what. [come to think of it, maybe the problem is you think pi[n] is the n-th prime? it's not, pi[n] is -standard- notation for the number of primes <= n.] ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Explaining the prime distribution > > you must be new here: > > nobody disputes that the Ôfunction' works [that is, > the code james has written correctly counts primes], > not that one would necessarily be able to see why > this is so from the explanation above. > > the Ôdispute' is over the mathemaical significance. > it's claimed that it's a brand new method, when in > Well, I'm disputing whether the function works - it doesn't, yet. But > I can't believe you would dispute the mathematical significance if it > would work! A function that outputs prime numbers, with no factoring, > and no need for previous primes!? That would be the biggest > breakthrough in number theory since Pascal's Triangle. No, it doesn't output prime numbers. It calculates the number of primes <= N without determining the actual primes, which is a nice little trick that Legendre figured out 200 years ago. > Look, his function doesn't work. And my follow-up comment wasn't an > attempt to show that it did. It was my opinion on his function. But > if his function did work like he wants it to, it would be so > significant that saying it wouldn't makes me question your > understanding of what he's saying. === Subject: single word for enquiring sequence order A small matter, may be it has some mathematical relevance: There appears no single word as an interrogative pronoun when enquiring about sequential order. E.g., in: Answer Question She came first in the race. In what place did she come? She is the second issue in the family. What is her birth rank? He was seated in the third row. Which position did he occupy? He cleared eight feet jump in a In which attempt did he clear? second attempt. Generally we ask what, which? or which one of? etc. when asking to select or point out a particular element in a set, but these words are not used specifically to describe consecutive order. Such an enquiring word may shorten a sentence and could be useful in faster speech/chat. Comments for a new word? === Subject: HELP - most important in Algebraic Geometry I posted this request before, but did not get any reply. I am a graduate student starting on my research in Algebraic Geometry. I should really be thankful to the kind soul who can give me a list of important research papers that are a must-read. A list of dozen or so classical anny times === Subject: Re: a simple question about reductio ad absurdum[crossp] >(...) >I discovered I was simply unable as a I'm bit rusty with my analysis course >in making a absurd hypothesis for reductin ad absurdum for simple thesis >like the strong goldbach >conjecture. >(I mean simple becouse the hypothesis is quite simple. I'm not stating here >Goldbach is simple!) >I guess that would be something like: there's a even n>2 that's not the sum >of two primes. >Simple like that? >Above that, I assume (and seem to remember) that there's just an absurd >hypothesis for a given hypothesis. Yes, it is always just the negation of the given hypothesis (which should better be called what one wants to prove - because if one actually uses it as a hypothesis, then one makes the error of assuming what one wants to prove !) Then you use formal logic to transform this negation - if needed and helping >(...) > I remember one must swap exist operand with for every but didn't >succeeded with Goldbach >(...) this is indeed because the there exists a ... such that and for every swap by taking the negation === Subject: Re: possible book titles for this course > What would be a possible or possible book titles for a course called > Strategies of Proof which has a prerequisite of semester number 2 of > undergraduate Calculus? The description of this Strategies of Proof course is, > ...set theory, and methods for constructing proofs of mathematical statements. > A bridge to the rigors of upper-division mathematics courses containing > significant abstract content. > I ask because I have never seen a textbook called anything like strategies of > proof. > G C I am an undergraduate so take this with a grain of salt. There is a book called Conjecture and Proof, by Laczkovich which has lots of interesting stuff in it (Banach-Tarski Paradox) and might serve as a good book for the course you speak of. Adam === Subject: Re: possible book titles for this course For the people who have been answering my question: The obvious should have been the guide. The word proof would usually be in the title; practically any short, conservative title with proof in it might be a corresponding book. What I did not recognize was the transitional status of the course called strategies of proof. G C === Subject: Re: How to do non-uniform (sampling) DFT? You could try Googling for the Lomb-Scargle Periodogram method. It works with unevenly sampled data. I believe it is used in the astrophysics community. I'm told that the Numerical Recipes in (FORTRAN,C,...) books have an implementation. I've never tried using it, but filed the reference away, just in case. Let me know how you get on. Jon > Hi all, > Can anybody point me to some helps/tutorials/references/stories on how > to do non-uniform (sampling) DFT? > -Walala === Subject: Re: how to decide a good sampling rate for sampling a function without obvious frequency? >Lucy is trying to fit a round peg into a square hole > gaussian(x, y)=1/(2*pi*sigmax*sigmay)*exp(-0.5*(x^2/sigmax^2+y^2/sigmay^2 )); > In my experiments using Matlab, I am using square grids to deiscretize the > above function. This is your problem right here. Why are you using square grids? What's so special about them, especially since what you're discretising is naturally a cylindrical polar shape? As well as that, you need greater resolution near the centre and less at the edges, so your cylindrical polar grid needs to have radial spacing that is a function of the radius. Perhaps you could tell us what your application is? What will you do with the Gaussian, once you've discretised it? === Subject: Re: how to decide a good sampling rate for sampling a function without obvious frequency? > Hi all, > I am having trouble with my sampling problem: > Basically I want to discretize a 2D Gaussian function f=gaussian(x, y) > gaussian(x, y)=1/(2*pi*sigmax*sigmay)*exp(-0.5*(x^2/sigmax^2+y^2/sigmay^2 )); > In my experiments using Matlab, I am using square grids to deiscretize the > above function. > Suppose I have a grid -- [-N..N, -N..N] where N is the number of samples in > one axis. So all together there are (2N+1)^2 samples. > Suppose the stepsize for discretization is dx. > The support of the discretized function will be [-N*dx..N*dx, -N*dy..N*dy]. > Since Gaussian function has an infinite support itself, I have to make N*dx > and N*dy sufficiently large in order to capture most of the energy of the > function f(x, y). > I remember for Gaussian distribution, it should be at least 3*sigma. > So I decide that the following conditions must be satisfied: > N*dx >= 4*sigma_x; > N*dy >= 4*sigma_y; > (sigma_x and sigma_y are known). > If the stepsize is large, the discretization will not be accurate. So dx and > dy should be small compared with sigma_x and sigma_y... > For example, I can choose dx=sigma_x / 100, dy= sigma_y /100, > but then N should be >= 400. > But then > [-N..N, -N..N] grids will have (2N+1)^2 = 801^2=641601 samples... > This is too large and will be very bad for subsequent operations... > After many experiments, I want to ask how to determine a good stepsize > taking into consideration of the storage size, while maintain good accuracy? > I know Nyquist sampling theory, but after many thinking, I found nowhere > here I can apply that theory... > Please help me and give me a hand! > -Lucy Use the Nyquist sampling theory on the frequency _components_ of the signal. Decide how much inaccuracy you can stand, and at what frequencies. Now look at the power spectral density of the signal you want to sample. Due to aliasing any signal energy above the Nyquist frequency will be folded back into your real signal, corrupting it. Because you know how much inaccuracy you can stand you can locate the Nyquist frequency to give you no more than your desired accuracy limit. Disclaimer: That made perfect sense coming out, but on re-reading it I think it would be a good growth enhancer in the garden. None the less I'm going to hit the send button... -- Tim Wescott Wescott Design Services http://www.wescottdesign.com === Subject: Re: how to decide a good sampling rate for sampling a function without obvious frequency? > Hi all, > I am having trouble with my sampling problem: > Basically I want to discretize a 2D Gaussian function f=gaussian(x, y) > gaussian(x, y)=1/(2*pi*sigmax*sigmay)*exp(-0.5*(x^2/sigmax^2+y^2/sigmay^2 )); > In my experiments using Matlab, I am using square grids to deiscretize the > above function. > Suppose I have a grid -- [-N..N, -N..N] where N is the number of samples in > one axis. So all together there are (2N+1)^2 samples. > Suppose the stepsize for discretization is dx. > The support of the discretized function will be [-N*dx..N*dx, -N*dy..N*dy]. > Since Gaussian function has an infinite support itself, I have to make N*dx > and N*dy sufficiently large in order to capture most of the energy of the > function f(x, y). > I remember for Gaussian distribution, it should be at least 3*sigma. > So I decide that the following conditions must be satisfied: > N*dx >= 4*sigma_x; > N*dy >= 4*sigma_y; > (sigma_x and sigma_y are known). > If the stepsize is large, the discretization will not be accurate. So dx and > dy should be small compared with sigma_x and sigma_y... > For example, I can choose dx=sigma_x / 100, dy= sigma_y /100, > but then N should be >= 400. > But then > [-N..N, -N..N] grids will have (2N+1)^2 = 801^2=641601 samples... > This is too large and will be very bad for subsequent operations... > After many experiments, I want to ask how to determine a good stepsize > taking into consideration of the storage size, while maintain good accuracy? > I know Nyquist sampling theory, but after many thinking, I found nowhere > here I can apply that theory... > Please help me and give me a hand! > -Lucy > Use the Nyquist sampling theory on the frequency _components_ of the > signal. Decide how much inaccuracy you can stand, and at what > frequencies. Now look at the power spectral density of the signal you > want to sample. Due to aliasing any signal energy above the Nyquist > frequency will be folded back into your real signal, corrupting it. > Because you know how much inaccuracy you can stand you can locate the > Nyquist frequency to give you no more than your desired accuracy limit. > Disclaimer: That made perfect sense coming out, but on re-reading it I > think it would be a good growth enhancer in the garden. None the less > I'm going to hit the send button... > -- > Tim Wescott > Wescott Design Services > http://www.wescottdesign.com Hi Tim, For the 2D gaussian function: gaussian(x, y)=1/(2*pi*sigmax*sigmay)*exp(-0.5*(x^2/sigmax^2+y^2/sigmay^2 )); I've computed its FT to be GAUSSIAN(u, v)=exp(-0.5*(u^2/sigmau^2+v^2/sigmav^2)); where sigma_u=1/(2*pi*sigma_x), sigma_v=1/(2*pi*sigma_y)... In my numerical calculation, I have chosen sigma_x = 1 and sigma_y... So sigma_u=1/(2*pi), sigma_v=1/(2*pi)... Since the frequency components are still in Gaussian shape, if I want to 99% of the freuqency components, I should keep the frequency ranging between [-4*sigma_u .. 4*sigma_u, -4*sigma_v .. 4*sigma_v] If I pick 4*sigma_u = 4*sigma_v to be my Nyquist frequency, and double it, my sampling rate should be 8*sigma_u=8*sigma_v = 8/(2*pi)=4/pi=1.27, so I pick 2. So am I correct that I should only need to sample the gaussian function twice? -Lucy === Subject: Re: how to decide a good sampling rate for sampling a function without obvious frequency? >Hi all, >I am having trouble with my sampling problem: >Basically I want to discretize a 2D Gaussian function f=gaussian(x, y) >gaussian(x, > y)=1/(2*pi*sigmax*sigmay)*exp(-0.5*(x^2/sigmax^2+y^2/sigmay^2 )); >In my experiments using Matlab, I am using square grids to deiscretize > the >above function. -- snip -- >-Lucy >Use the Nyquist sampling theory on the frequency _components_ of the >signal. Decide how much inaccuracy you can stand, and at what >frequencies. Now look at the power spectral density of the signal you >want to sample. Due to aliasing any signal energy above the Nyquist >frequency will be folded back into your real signal, corrupting it. >Because you know how much inaccuracy you can stand you can locate the >Nyquist frequency to give you no more than your desired accuracy limit. >Disclaimer: That made perfect sense coming out, but on re-reading it I >think it would be a good growth enhancer in the garden. None the less >I'm going to hit the send button... >-- >Tim Wescott >Wescott Design Services >http://www.wescottdesign.com > Hi Tim, I know; I just didn't have time last night to go over it and figure out why, or how to make it more clear. Some time spent with graph paper and a pencil will be required to really understand the subject, no matter how well I explain it. > For the 2D gaussian function: > gaussian(x, y)=1/(2*pi*sigmax*sigmay)*exp(-0.5*(x^2/sigmax^2+y^2/sigmay^2 )); > I've computed its FT to be > GAUSSIAN(u, v)=exp(-0.5*(u^2/sigmau^2+v^2/sigmav^2)); > where sigma_u=1/(2*pi*sigma_x), sigma_v=1/(2*pi*sigma_y)... > In my numerical calculation, I have chosen sigma_x = 1 and sigma_y... > So sigma_u=1/(2*pi), sigma_v=1/(2*pi)... > Since the frequency components are still in Gaussian shape, if I want to 99% > of the freuqency components, I should keep the frequency ranging between > [-4*sigma_u .. 4*sigma_u, -4*sigma_v .. 4*sigma_v] > If I pick 4*sigma_u = 4*sigma_v to be my Nyquist frequency, > and double it, my sampling rate should be 8*sigma_u=8*sigma_v = > 8/(2*pi)=4/pi=1.27, so I pick 2. > So am I correct that I should only need to sample the gaussian function > twice? This gets back to how much accuracy you need. If you're truncating the function at 3 sigma then it certainly indicates that sampling it twice will be adequate. On the other hand that seems a terribly small sample to me. Are you sure that your 4 sigma/99% assumption is true in 2-D, and are you sure that truncating at 3 sigma is appropriate? If your intuition is telling you that two samples is too small then perhaps the error assumptions that you started with are too generous, and you need to tighten them up? Getting 99% of all the energy sounds good, but it only puts your Nyquist energy down 20 decibels -- a sampling system wouldn't be considered high quality unless it were getting the aliasing down to at least 40 to 60dB or more, which implies getting 99.9% or even 99.99% of your energy. If I did this calculation from the precept I stated above and came up with these numbers I would do one of two things: One, I would try it with the indicated number of samples, and 2x that, and perhaps 10x that. Two, I would go over my math to make sure that I really knew what I was talking about. > -Lucy -- Tim Wescott Wescott Design Services http://www.wescottdesign.com === Subject: Re: how to decide a good sampling rate for sampling a function without obvious frequency? >big snip< > So am I correct that I should only need to sample the gaussian function > twice? Well, yes you should be able to reconstruct an approximation of a gaussian with 2 sinusoids - 4 would be better - more even better. I think as you decrease the sample spacing geometrically the resolution will increase exponentially so you will not need a very large number of samples. Jd modern papers (preferrably in English) would be wonderful. Leo === Subject: Re: 3D Plane by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7GFi2018958; >I have a 3d plane, and have worked out its normal. I want to know the >distance from (x,y,z) to where z intersects the plane. i.e. x and y will > be the same value, but z falls on the plane. A minor nitpick; You say 3d plane, but in fact you really are working with a 2d plane embedded in E^3. This mistake seems to be carried along by others in their responses. Indeed, one post referred to a Ô3d' line. AFAIK, a line is always 1-dimensional. It might be embedded in E^3, but it is still 1 dimensional..... === Subject: Re: books for self study, number theory, set theory, analysis... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7GFi2V18921; >For number theory, it is difficult to make any sense of it without >understanding how it evolved. So make sure you get a book that treats >number theory in its historical context. >For example, >O. Ore: Number theory and its history >A. Weil: Number theory: an approach through history >H.M. Edwards: Fermat's Last Theorem (A genetic introduction to >algebraic number theory) A superb (IMO) book that treats things from a historical persepective and is the often overlooked: D. Shanks Solved & Unsolved Problems in Number Theory. BTW, Dan used to call me Agman. Think about it..... === Subject: How to eliminate scale factors on a pair of matrices.... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7GFi2B18941; I'm stuck with the following problem that I can't solve. I consider two 3-by-3 (real) matrices defined as B1 = s1*H*A1*H^T B2 = s2*H*A2*H^T where s1 and s2 are non zero real numbers,H is a full-rank 3-by-3 (real) matrix (H^T denotes its transpose) and A1 and A2 are two 3-by-3 (real) matrices such that rank(A1)=3 and rank(A2)<=3. I'm seeking some assumptions on B1, B2 that will ensure that s2/s1 equals a known quantity ie, only depending on A1 and A2. When rank(A2) = 3, it suffices to assume that det(B1) = det(B2) = 1 so det(inv(B2))*det(B1) = (s1/s2)^3*det(A1)/det(A2) = 1 yields (s1/s2)^3 = det(A2)/det(A1) When rank(A2)<=3, I could not think of any assumption on B1, B2 that will help me to solve the problem in a similar way. === Subject: Re: Atheist MorituriMax > Our great Einstein wanted to know God; how He created the universe; > the rest were just details. > Mitch Raemsch http://www.skeptic.com/archives50.html I believe in Spinoza's God who reveals himself in the orderly harmony of what exists, not in a God who concerns himself with fates and actions of human beings. -- Einsetin alex === Subject: Re: Atheist MorituriMax > Our great Einstein wanted to know God; how He created the universe; > the rest were just details. > Mitch Raemsch > http://www.skeptic.com/archives50.html > I believe in Spinoza's God who reveals himself in the orderly harmony of what > exists, not in a God who concerns himself with fates and actions of human beings. > -- Einsetin > alex More folks than you know are a theist, Alex... what's that have to do with physics? See: http://www.csicop.org/si/9907/ === Subject: Re: Projective geometry >Projective geometry is something I know next to nothing about. > Neither do I. >I have seen the construction of RP^2 by identifying antipodes >on S^2. The point that always confuses me is of course the equator. > S^2 is the surface of a unit sphere. S^1 a circle. > Is RP^2 given a verbal name or intuitive description? RP^n: Real Projective n-space (for n=1: Real projective line, for n=2: real projective plane): RP^n is the quotient of R^(n+1){0} under the action of the multiplicative group R* = R {0}. RP^n is the space of lines in R^(n+1) passing through the origin, or equivalently the space of 1-dimensional linear subspaces. >I also wonder about RP^1. It seems that it would be formed by >identifying >opposite points on S^1, the circle--which is the part of the >RP^2 construction that I find confusing. > Because of the spherical symmetry of the construction, a circle would be > the same in any orientation. >It seems that S^1 is 1-1 with the half-open interval [0,1), >and then 1 must be identified with 0 to get S^1. Or should >one say it is 1-1 with [0,1] with 0 and 1 identified. > That's absurd, as 1 isn't in [0,1). Yes, if 0 and 1 of [0,1] are > identified, then you get S^1, the unit circle. As for cardinality, S^1 is > 1-1 with R, R^n, (0,1), [0,1]x[3,5)x(-2,1]. The points of [0,1] with 0 > and 1 identified can be constrewed as (0,1) and the identified { 0,1 }, > namely { {r}, { 0,1 } | r in (0,1) }, the partitions of [0,1] >It seems that if one identifies the points a and a + pi on S^1 >(a = angle), one just gets S^1 back again: you could collapse >the top half of the circle onto an open interval (0,1), then >identify 0 and 1 to get [0,1), the same as for S^1. > I'm inclined to agree, you get the circle. Then would not RP^2 be S^2 ? RP^1 is the circle; the canonical map from S^1 to RP^1 is a double cover (identifying antipodal points yields a 2:1 mapping). RP^2 is *not* S^2. S^2 is simply-connected, but the fundamental group of RP^2 is cyclic of order 2. >PS: I also want to look at CP^1 and compare to RP^2, and then >HP^1. > What's H ? H (for Hamilton) refers to quaternions. CP^1 is homeomorphic to S^2 (the 2-sphere). If I recall correctly, HP^1 is S^4 (the 4-sphere). Dale. === Subject: Re: How unimportant is sci.math? > I've been pointing out for a while that Usenet doesn't matter, when I > think what's clear from my research is that sci.math doesn't matter to > the world at large... > Perhaps sci.math does not matter to the world at large, and it is > certainly not where legitimate research mathematicians submit results for > peer review, but for amateurs like myself, I find it a tremendously valuable > resource. I have often submitted questions, and, within a few hours, > received many thoughtful and thorough replies. It helps if I try to pose the > question accurately, have looked elsewhere first, and show that I've given > the matter a least a little thought beforehand. Obeying a small set of > self-imposed ground rules seems to help a great deal: > 1) Display no arrogance > 2) Assume you're is misstating the question > 3) Apologize in advance for poor terminology, weakly grasped concepts, and > profuse errors > 4) Remember that if you're an amateur, and you've figured out something, the > odds are extremely high that it's been discovered before, probably a long > time ago, and the mathematician who did so had a lot more interesting things > to say than you did, proved it with a great deal more rigor, understood it > in context to the rest of mathematics, and usually went on to other > fascinating and useful results. > Big problems in math are usually resolved incrementally - standing on the > shoulders of giants - as per the Wiles FLT proof. Even a small but > legitimate contribution to say, the Riemann Hypothesis is vitally important. > If an amateur posts a solution to a hard problem, especially one that > clearly shows no attention to the history of attempts to solve said problem, > one's crank alert is likely to be triggered - and legitimately; the > solution is almost certainly in error, and the poster should express due > modesty if he/she wants genuine assistance. A good set of rules. I suggest a parallel set of rules for the expert responders: 1) Display no arrogance. 2) Restate the question, if doing so seems helpful. 3) Cut the asker a lot of slack about terminology, imperfectly grasped concepts, and profuse errors. 4) An amateur who makes an independent discovery, no matter how modest, has already gone far beyond the majority of students, who limit themselves to passively imbibing what Teacher says. Give him or her credit for this. -- Chris Henrich God just doesn't fit inside a single religion. === Subject: Re: How unimportant is sci.math? > I've been pointing out for a while that Usenet doesn't matter, when I > think what's clear from my research is that sci.math doesn't matter to > the world at large... > Perhaps sci.math does not matter to the world at large, and it is > certainly not where legitimate research mathematicians submit results for > peer review, but for amateurs like myself, I find it a tremendously valuable > resource. I have often submitted questions, and, within a few hours, > received many thoughtful and thorough replies. It helps if I try to pose the > question accurately, have looked elsewhere first, and show that I've given > the matter a least a little thought beforehand. Obeying a small set of > self-imposed ground rules seems to help a great deal: > 1) Display no arrogance > 2) Assume you're is misstating the question > 3) Apologize in advance for poor terminology, weakly grasped concepts, and > profuse errors > 4) Remember that if you're an amateur, and you've figured out something, the > odds are extremely high that it's been discovered before, probably a long > time ago, and the mathematician who did so had a lot more interesting things > to say than you did, proved it with aust make your sampling algorithm scalable. Start with a small managable size and if you need better resolution later you can do that. -jim http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- === Subject: Re: Invariant Galilean Transformations (FAQ) On All Laws > > > > > Another reason to update my kill file... > > > > David A. Smith > > I take that (correctly?) as meaning you cannot show Maxwell >non-invariant > under the galilean translatory transformation without the spurious > atheoretical time transform, t'=t, being asserted? > [snip crap] > Hey stooopid, Galilean transforms are empirically invalid in a > relativistic universe. You are playing solitaire with pinochle > deck - and losing every game. Idiot. >I do wish you wouldn't sign your material so un-self-ßatteringly, however >appropriately. >I do apreciate you submitting the above proof that you are so completely >dishonest intellectually. >How can it hurt in such a relativistic universe to admit that maxwell is/are >indeed invariant under the galilean translatory transformation when the >atheoretical t'=t transform is not (spuriously) imposed? It only hurts because it is wrong, which is good enough for me. t = t' is crap. Mostly because it is wrong... >BTW, tell us again all about the many spatial/object contractions you see >every day a million times. >Child: Want to see the world's fastest wild west draw? >Kid: Yes. >Child: Want to see it again? >eleaticus === Subject: Re: Invariant Galilean Transformations (FAQ) On All Laws > > in > > > > Another reason to update my kill file... > > > > David A. Smith > > > > I take that (correctly?) as meaning you cannot show Maxwell >non-invariant > > under the galilean translatory transformation without the spurious > > atheoretical time transform, t'=t, being asserted? > [snip crap] > > Hey stooopid, Galilean transforms are empirically invalid in a > relativistic universe. You are playing solitaire with pinochle > deck - and losing every game. Idiot. >I do wish you wouldn't sign your material so un-self-ßatteringly, however >appropriately. >I do apreciate you submitting the above proof that you are so completely >dishonest intellectually. >How can it hurt in such a relativistic universe to admit that maxwell is/are >indeed invariant under the galilean translatory transformation when the >atheoretical t'=t transform is not (spuriously) imposed? > It only hurts because it is wrong, which is good enough for me. > t = t' is crap. Mostly because it is wrong... galilean transform is wrong and (b) that even if you don't Maxwell is not invariant under those transforms. eleaticus === Subject: Re: Random number generation using radioactivity >[...] > If the random starting value is free from any bias and is reseeded > (an xor would be optimal) with a frequency greater than the quantity at > which bias/correlation is detectable from the CRC then this would remove > any remaining bias. > > The random starting value *is* the result from the previous CRC. That > *is* a high-quality random value: the previous processed end result. > > By simply not initializing the CRC each time, it is automatically > re-seeded with a high-quality random value. > > If we change the initial CRC register to every possible value, we also > change the CRC result to every possible value. No bias is introduced. > There is no need for XOR. > But your procedure is only needed in the case that the outputs of the CRC with > a constant start value are biased. If we model a hash as the remainder from a division, the usual modulo effects appear. The input data are wrapped around the smaller result. Biased values appear more often than expected, and each occurrence is only compensated if appropriate less-frequent values happen to map to the same point. Since that is unlikely, some bias seems likely to remain. The usual conclusion is that any reasonable hash function, CRC or cryptographic, will fail to hide some amount of the input bias. That probably over-simplifies reality. With CRC each bit has the same power to affect the result, but the noise samples are binary codings, not single bits. The bias is in the structure of the coding, rather than an unbalance between 1's and 0's. When the exact same 16-bit value occurs in different positions in the data, it is unlikely to map to the same CRC result. So the value bias does tend to be hidden. >If that is the case then your proposal will > indeed reduce the bias, but it will also cause successive outputs to be > statistically dependent on each other. This is a rather mild form of dependence, similar to the dependence found in the CBC encryption mode. Nobody cries about how each CBC value is dependent on the previous one, because that is the intention. And it does not seem to represent a cryptographic weakness. This dependence is also very similar to the process of subtracting the previous sample from the current one to get differential data and emphasize change in the numerical result. Extensive characterization on my pages shows that the (normal) distribution, and more especially the autocorrelation, are greatly improved by this process. (See, for example, http://www.ciphersbyritter.com/NOISE/ZENER1ZY.HTM versus http://www.ciphersbyritter.com/NOISE/ZENER1P7.HTM from http://www.ciphersbyritter.com/NOISE/NOISCHAR.HTM .) That makes sense, because the very definition of audio frequency implies correlation over time, and with fast sampling, lower frequency components will inject similar values into many, many samples. So if we subtract the previous sample from the current one, we can remove much of that interference. Without this processing, autocorrelation structure is a vastly larger issue. > So your proposal does not really help in getting really better outputs because > the output random numbers should be both unbiased and independent. Really? Who says? Who says that this particular type of dependence is a problem, and if so, exactly why? The whole point of de-skew computations is to change a distribution in real time. In the real world, we can only know the distribution as an approximation, and only as it was, not how it is, because it may vary. Since we cannot know the true distribution, we also cannot numerically take the inverse without introducing numerical distortions. --- Terry Ritter 1.2MB Crypto Glossary http://www.ciphersbyritter.com/GLOSSARY.HTM === Subject: Re: Random number generation using radioactivity > >[...] > If the random starting value is free from any bias and is reseeded > (an xor would be optimal) with a frequency greater than the quantity > at which bias/correlation is detectable from the CRC then this would > remove any remaining bias. > >The random starting value *is* the result from the >previous CRC. That *is* a high-quality random value: >the previous processed end result. > >By simply not initializing the CRC each time, it is >automatically re-seeded with a high-quality random value. > >If we change the initial CRC register to every possible >value, we also change the CRC result to every possible >value. No bias is introduced. There is no need for XOR. > You missunderstand what I am saying. Some initial random starting > value is necessary even with your scheme. If the range of inputs is > limited (as is usually the case with physical RNGs) the final output > may be baised. By periodically introducing random data that is > totally free from bias. The bais is eliminated. >I assume you misunderstand what *I* am saying: Each >CRC processing starts with the immediate preceding >processed random value, the result of processing a >whole sequence of biased somewhat-random data. The >resulting value is the finished, ßattened, directly >usable ultimate result. And if that result is good >enough for direct use as a ßat random value, it is >also good enough to use as the unbiased start of the >next processing. There is thus no need to find some >other random source and somehow process that ßat >(the recursive turtles all the way down problem, >now avoided). Perhaps a demonstration using a small CRC would be helpful. Imagine an 5 bit crc. Now imagine a stream which is heavily biased. Say only two possible values are present. Ideally these two values are evenly distributed. Each sample contains one bit. Additionally assume that the CRC is taken as output after every 6 samples. There will be some set of values and crc which interact badly. In a different example one sample contains 5 bits of randomness completely unbiased and uncorrelated. The other 5 samples contain zero randomness. In this case even if we don't know which sample contains the randomness the result is completely unbiased and uncorrelated. The first is a somewhat random source. The second is a somewhere random source. These are discussed in literature. The CRC is a special case of a hash function which the literature also goes in to extensively. Even with ideal hash functions the turtle problem exists. Special constructions with these hash functions reduce the amount of additional unbiased random data needed to very small amounts. Perhaps a better analogy is chaos theory. The added unbiased random data act as the initial uncertainty. In any case CRCs work extraordinarily well but if you are worried about a bias of 1 in 2^32 then added precautions are needed. Or you could just use a bigger CRC. >As for the single original init, we do not care. >Bias is the collection, the distribution, not any >single value or any single processing. Whatever the >initial value, we then have hundreds or thousands >or however many processings, each of which start out >with an unbiased value, the result of the previous >processing. But the weenies among us can simply >discard the first value and use the rest, thus >improving confidence at almost no cost. >--- >Terry Ritter 1.2MB Crypto Glossary >http://www.ciphersbyritter.com/GLOSSARY.HTM Leslie ÔMack' McBride remove text between _ marks to respond via e-mail === Subject: Re: How to compute the addition of millions of functions efficiently and dynamically? If the signal is bandwidth-limited, you might not need full resolution, so taking every 10th point (for example) may be sufficient. That would give you a factor of 100 in processing time... Just at thought! BR, Andre >I don't keep up on MatLab, but 5 years ago their symbolic package used >the Maple engine (I think everyone uses the Maple engine...). >If you can't express it in closed form, or if you cannot wrestle the >sums into closed form then you are back to finding the fastest computer >you can and doing it discretely. >-- >Tim Wescott >Wescott Design Services >http://www.wescottdesign.com > will be non-collapsable millions of terms adding together. And I cannot find > a way to shorten it into one closed-form... > Then I need to compute its Inverse Fourier Transform. > Can any symbolical engine handle such big symbolical expression and such > IFT? > However, I want to know if for this special form of sum, there is any tricks > existing: > The FT is: > Z_SUM(u, v)=SUM( > r(x1, y1)*exp(-2*pi*i*(u*x1+v*y1)), > r(x2, y2)*exp(-2*pi*i*(u*x2+v*y2)), > ... > ... > r(x1000000, y1000000)*exp(-2*pi*i*(u*x1000000+v*y1000000)) > )*Z(u, v) > There are millions of such terms, u, v are frequencies, (x1, y1), > ...(x1000000, y1000000) are known constants... > Z(u, v) is some function. > Now I want to compute Z_SUM's Inverse FT z_sum... any efficient way of doing > this? > -Net -- Please change no_spam to a.lodwig when replying via email! === Subject: Mod I have following two questions. If a=b (mod m) for a,b in Z, m in N, Then ord_m(a) = ord_m(b) ? My lecturer say above is hold only if g,c,d (a,m)=1. But, I think this is true because, a=b (mod m) => g.c.d(a,m) = g.c.d(b,m) => = in Z_m => ord_m(a) = ord_m(b) So, I'm so confused what is true. Would you please help me ? 2> Can anybody give me example of a in Z, prime p satisfying g.c.d(a,p)=/=1 and a^(p-1) = 1 (mod p) ? (That is, counterexample of inverse of Fermat's little Thm) === Subject: Re: Mod > I have following two questions. > So, I'm so confused what is true. This depends on what you mean by ord_m(a). If you mean the order of the coset a+mZ in the additive group Z/mZ, then ord_m(a)=ord_m(b), because a+mZ=b+mZ, as you pointed out (me thinks). If, OTOH you mean the order of a+mZ in the multiplicative group (often written Z_m^*), then even though still a+mZ=b+mZ, the order of a+mZ isn't defined unless gcd(a,m)=1. So, which is it (Z/mZ,+) or (Z/mZ^*,.) ? > Would you please help me ? > 2> Can anybody give me example of a in Z, prime p > satisfying g.c.d(a,p)=/=1 and a^(p-1) = 1 (mod p) ? > (That is, counterexample of inverse of Fermat's little Thm) This is a nice (but not at all difficult) homework problem, so I'm not gonna spoil your fun! Jyrki === Subject: Re: More readable prime counting function > It turns out that I can write my prime counting function in such a way > that it's easier to read: > dS(x,y) = (p(x/y, y-1) - p(y-1, y-1))( p(y, y) - p(y-1, y-1)), > S(x,1) = 0, p(x, y) = ßoor(x) - S(x, y) - 1, and S(x,y) is the sum of > dS from dS(x,2) to dS(x,y). > p(x,x) = pi(x) so it's the count of prime numbers. > And yes, it still counts prime numbers and is in fact the exact same > function that I've been giving before as all that happens is that for > the higher ranges certain things just zero out. OOPS! It turns out that it won't work to push beyond y=sqrt(x) like I tried to do here, as there are a lot of problems with it. Part of what happened is that I had this idea in mind that if my prime counting function *looked* just a little bit simpler, then maybe that would help people, and in one sense the idea above should work, as S(x,x) does equal S(x,sqrt(x)) in the descriptive form. That is, given that S(x,sqrt(x)) is the sum of composites up to and including x, it makes since that S(x,x) would just be the same sum, but the dS(x,y) function as I defined it can start going negative when you push y past sqrt(x), or at least that's what I saw when checking yesterday! So it was an idea that doesn't pan out unless I missed something. Having my own research at the forefront where I can't reference a textbook, leaves me in the position of figuring things out for myself. It's neat in many ways, but at times I just screw-up, as yeah, I can't just go pick up a math reference to read about something that I discovered that hasn't yet made it into the references and textbooks. James Harris = great deal more rigor, understood it > in context to the rest of mathematics, and usually went on to other > fascinating and useful results. > Big problems in math are usually resolved incrementally - standing on the > shoulders of giants - as per the Wiles FLT proof. Even a small but > legitimate contribution to say, the Riemann Hypothesis is vitally important. > If an amateur posts a solution to a hard problem, especially one that > clearly shows no attention to the history of attempts to solve said problem, > one's crank alert is likely to be triggered - and legitimately; the > solution is almost certainly in error, and the poster should express due > modesty if he/she wants genuine assistance. > Barnaby Very well said... and completely lost on JSH, of course === Subject: Re: Rado's Sigma and the Halting Problem for Programs >function. Functions don't /have/ solutions. It's like Shaquille > Function SolveForSqrRoot(N: Extended) : Extended; > Begin > Result:=Sqrt(N); > end; >Arthur is right about mathematical functions not having solutions. Very >informally, any function f is just a set of ordered pairs > { (a[0],b[0]), (a[1],b[1]), (a[2],b[2]), ... } >where we _define_ the notation f(a[i]) to mean b[i]. So given an a[i], it >doesn't quite make sense to say b[i] is the Ôsolution' to f(a[i]), because >f(a[i]) is really just another way of writing b[i]. >Although you've mentioned you don't want to get into too much math, to >really get the distinction you'll want to take a look at >http://en.wikipedia.org/wiki/Function_(mathematics) I suggest you read these 2,000+ examples of functions having solutions: Larry === Subject: Re: Rado's Sigma and the Halting Problem for Programs Distribution: inet > function. Functions don't /have/ solutions. It's like Shaquille [...O'Neal, saying something stupid.] > Arthur is right about mathematical functions not having solutions. [...Explanation snipped.] > Although you've mentioned you don't want to get into too much math, to > really get the distinction you'll want to take a look at > http://en.wikipedia.org/wiki/Function_(mathematics) > I suggest you read these 2,000+ examples of functions having solutions: You know, trolling doesn't make you look smart. Michael and I have been politely explaining that you made a little terminological slip-up, and what the correct terminology is. The proper response would be been more like Naw, math sucks! I like to contradict people! Not too hot, you know? If you'd like to start contributing /productively/ to the thread again, I'll be waiting. -Arthur, signing off === Subject: Hyperbolic geometry: loci of points that see a segment with a given angle Hi! In Euclidean Geometry, given a segment AB and an angle alpha, the set of points P such that the angle APB = alpha is an arc of circle. What is this loci in hyperbolic geometry? === Subject: Re: The most elegant method to prove an identity with binomial coefficients === Subject: Re: What is a differential field? > You lost me here. I understand connected in the topological case, but > from where's the topology when working with discrete finite fields? And > what do you mean by extension? Some sort of group product? I went a-googling and found this URL: in which group extension is defined. Let G be a group, and N a normal subgroup; let K be (isomorphic to) G/N; then G is an extension of N by K. -- Chris Henrich The total lack of evidence is the surest sign that the conspiracy is working. === Subject: Re: What is a differential field? | | | |> : What are algebraic groups exactly? Is that some fancy way of saying |> : matrix groups? What intrinsic property makes them algebraic? |> An algebraic group is an affine algebraic variety (set of solutions to a |> finite number of polynomials over a finite number of variables, over some |> field) together with a group structure, where multiplication and inversion |> are rational maps. |> There are two very different kinds of algebraic groups: |> (1) matrix groups, which are subgroups of GL(n,F) defined by polynomials |> in |> the matrix entries. These are also called linear algebraic groups. The |> differential Galois group of a Picard-Vessiot extension is always a |> linear algebraic group (forgot to mention that in the previous post). | |I've only recently learned what an affine algebraic variety is. So let me |see if I can prove that GL(n,F) satisfies the requirements to be one. |First, n x n matrices are elements of a n^2 dimensional vector space over |F, which is an affine space. So GL(n,F) would certainly be an affine |variety, since it is embedded in an affine space. The condition for a |matrix M to be in the group is GL(n,F) is that det(M) != 0. Interesting, |this is an inequality and not an equation. Does that qualify GL(n,F) to |be a variety at all? SL(n,F) would definitely qualify, since the only |condition on its elements is det(M)=1. | |About rationality of group operations. If M=NP, then the matrix elements |of M are certainly rational (polynomial even) functions of matrix elements |of N and P. To get the inverse we can use the formula M^(-1) = M*/det(M), |where M* is the matrix adjoint, hence inversion is also rational. |Everything checks out here. However I'm a little bothered by the |requirement of rationality here. I've been indoctrinated by modern |differential geometry that intrinsic geometric properties/notions must be |coordinate independent. Yet rationality is always defined with respect to |some coordinates (variables). Granted that rationality here is invariant |under linear transformations of the affine space in which the algebraic |group is embedded, but I would still like to see some sort of intrinsic |condition on the group that shows it to be an algebraic group. the vague impression that i have about this is that a lie group's being algebraic (i think affine algebraic in the strict sense actually, meaning that the group operation is not just rational but polynomial) is secretly very closely related to the representation theory of the group. if the representation theory of the group is sufficiently nice, meaning basically that the group has enough representations on finite-dimensional vector spaces, then the so-called representative functions on the group will form an affine coordinate ring making the group into an algebraic group. a representative function on a group is a polynomial function of the matrix by which a group element is represented under a finite-dimensional representation of the group. the nice trick that shows that the representative functions on a group form a (commutative) ring is that the sum of representative functions is representative because you can take the direct sum of group representations, while the product of representative functions is representative because you can take the tensor product of group representations. -- [e-mail address jdolan@math.ucr.edu] === Subject: Algebraic groups (was Re: What is a differential field?) > |I've only recently learned what an affine algebraic variety is. So let me > |see if I can prove that GL(n,F) satisfies the requirements to be one. > |First, n x n matrices are elements of a n^2 dimensional vector space over > |F, which is an affine space. So GL(n,F) would certainly be an affine > |variety, since it is embedded in an affine space. The condition for a > |matrix M to be in the group is GL(n,F) is that det(M) != 0. Interesting, > |this is an inequality and not an equation. Does that qualify GL(n,F) to > |be a variety at all? SL(n,F) would definitely qualify, since the only > |condition on its elements is det(M)=1. > |About rationality of group operations. If M=NP, then the matrix elements > |of M are certainly rational (polynomial even) functions of matrix > |elements of N and P. To get the inverse we can use the formula M^(-1) = > |M*/det(M), where M* is the matrix adjoint, hence inversion is also > |rational. Everything checks out here. However I'm a little bothered by > |the requirement of rationality here. I've been indoctrinated by modern > |differential geometry that intrinsic geometric properties/notions must be > |coordinate independent. Yet rationality is always defined with respect to > |some coordinates (variables). Granted that rationality here is invariant > |under linear transformations of the affine space in which the algebraic > |group is embedded, but I would still like to see some sort of intrinsic > |condition on the group that shows it to be an algebraic group. > the vague impression that i have about this is that a lie group's being > algebraic (i think affine algebraic in the strict sense actually, meaning > that the group operation is not just rational but polynomial) is secretly > very closely related to the representation theory of the group. if the > representation theory of the group is sufficiently nice, meaning basically > that the group has enough representations on finite-dimensional vector > spaces, then the so-called representative functions on the group will > form an affine coordinate ring making the group into an algebraic group. I'm not completely sure what an affine coordinate ring is, but I would guess it would be the analog of say polynomials in x,y,z for R^3. For example, polynomials in exp(i t) over the complex numbers would form a coordinate ring for U(1), parametrized as usual with t in [0,2pi[. Right? > a representative function on a group is a polynomial function of the > matrix by which a group element is represented under a finite-dimensional > representation of the group. the nice trick that shows that the > representative functions on a group form a (commutative) ring is that the > sum of representative functions is representative because you can take the > direct sum of group representations, while the product of representative > functions is representative because you can take the tensor product of > group representations. One condition on an abstract group would certainly determine whether it is algebraic or not. Namely if that group can be embedded inside another group that is known to be algebraic (say GL(n,F)) in such a way that the embedding gives you an algebraic variety. This condition certainly would have links to representation theory. That's about as far as my train of thought is taking me. :-) If we can say that all the coordinate dependent properties of algebraic groups can be recast into the language of some sort of intrinsic coordinates on the group itself (such as the affine coordinate ring formed by matrix elements of representations) then I can see that the notion of being algebraic is intrinsic to the group. The real question as to a group's algebraic character would be whether such intrinsic coordinates exist. Igor === Subject: Re: What is a differential field? : I've only recently learned what an affine algebraic variety is. So let me : see if I can prove that GL(n,F) satisfies the requirements to be on== Subject: Re: More readable prime counting function >[...] >So it was an idea that doesn't pan out unless I missed something. >Having my own research at the forefront where I can't reference a >textbook, leaves me in the position of figuring things out for myself. >It's neat in many ways, but at times I just screw-up, as yeah, I can't >just go pick up a math reference to read about something that I >discovered that hasn't yet made it into the references and textbooks. those of us who do research that we actually publish in journals are in the same position - we can't check our work by opening a textbook. but we end up retracting published errors only very rarely - in your case it seems to happen eventually with -most- of the things you post. i wonder what the reason for the difference is? >James Harris David C. Ullrich ************************** As far as I'm concerend you're trying to wait until I die, so I figure maybe you should die instead. How about that, eh? Wouldn't that be a better twist? You refuse to follow the math, so the great Powers that control reality and *speak* in mathematics decide to kill you instead of me. So what do you think about that, eh? Oh, can't hear Them talking? Well, I guess that's because you don't really understand Mathematics, the true language, which is THE language. They're talking about you now, and They agree with my assessment, and will not penalize me as They allowed the others like Galois and Abel to be penalized. They will kill you instead. James Harris speaking on Weird factorization, genius === Subject: Re: HOW CAN I SOLVE THIS EQUATION SYSTEM ? > It sounds like this should really be more complicated than just a system > of two differential equations. But anyway: > d(LI)/dt = L dI/dt + I dL/dt > so it would be good to be able to calculate dL/dt as well as just L(t). > Of course you could approximate that using a difference quotient (say > (L(t+h)-L(t-h))/(2 h)). And then you just have an ordinary system of > two differential equations in two dependent variables, to which you > could apply Runge-Kutta. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 This could be the solution for the first problem, but now another one arises : I always saw the R-K method applied to problem in the form : dy/dt=F(t,y) like say : y'=2t+ty+y in which, ones the time step (h) is given (and the initial condition y(0) too), it's very simple to apply the R-K method, because we have just to substitute the time step for t and the initial condition in the equation : F(t,y)=2t+ty+y so that y(t+h)=y(t)+g(k(i)) where g(k(i)) is a linear function depending on the order of the R-K method (e.g. 1/6 h [k(1)+4k(2)+k(3)] for the third order). But what happens when I don't have any analytical form for F(t,y) ? I mean : how can I substitute the values of F(t+h,y+g[k(i)]) for all the k(i) if I know these values only in the time step ? I know that my english is very bad, so this explanation is not so clear (even to me),anyway I hope you understood my questions. If it's not so, I will try to give a more readable explanation. === Subject: Re: Uncle Al's missing logic. > > Uncle Al never responds to actual factual material, so please (y'all) > let me > > know how it can be correct in one single, continuous, Ôlogical' > derivation > > process to assert: > > > > a=bc, a=dc, a=-dc, -d < b < d, d > 0, c >= 0. > > > > Meaning that b and d can independently take any of the values in the > > specified ranges. > > > If c = 0 then for a=bc, a=dc, a=-dc, a = 0 with values of b and d in the > specified ranges. > its specified range. > So, what I mean by those statements is that the process is to derive an > equation that applies to any and all values of b, c, d. > Sorry. > eleaticus > PH Actually c can't be anything but zero, since a = dc = -dc sets a = 0, but d > 0, so c = -c = 0. === Subject: Re: expressions equivalence via DNF >In other words, to compare 2 Boolean expressions, is it sufficient to >convert then both to DNF and compare them? > > but here it's clear that you're really asking about the converse. > yes, two wffs in the predicate calculus are equivalent if and only > if their [suitably normalized, ie Ôsorted'] dnf's are equal. >disagrees with your statement? >if i recall correctly the author of that message disagrees >with a lot of things [like he says the reals are countable.] >in that message he says that >(A AND B) OR (B AND C) >is dnf, which indicates he simply doesn't know what dnf >is. he also says ÔYou should be aware that there is usually >no single minimal DNF form,' which is nonsense [unless >he's simply talking about permuting the terms, which i >do not think is what he means.] >If A, B, and C are propositional variables then that formula is definitely >in DNF. >It is also the case that there is often no single minimal DNF. For instance, >the following formulas are expressed in minimal DNF and are equivalent to >one another: >(!b & !c) | (b & c) | (a & c) >(!b & !c) | (b & c) | (a & !b) >There is a notion of a canonical DNF that would be equivalent to listing >all the rows of a truth table for which the function in question evaluates >to true. That form is unique up to permutation of the clauses. For the >above formulas, the equivalent canonical DNF would be >(!a & !b & !c) | (a & !b & !c) | (a & !b & c) | (!a & b & c) | (a & b & c) . well, looking it up a few places it seems you're right, at least in many authors' terminology. i was taking dnf to mean what you call canonical dnf. [i don't think that what we're calling dnf above -should- be called dnf, since there's nothing Ônormal' about it... in any case i could swear that i was taught years ago that dnf means canonical dnf, although i don't have a reference at hand.] ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: expressions equivalence via DNF > If you describe your problem explicitly, the matters will become much > clearer. For example, are you considering predicate calculus formulas or > propositional formulas? >Jeffrey, >really makes sense. I should have seen it myself, it is so trivial! >As to my original question: I think I am dealing with predicate >calculus, though >I think I have some limitations which may allow me to reduce it to >propositional >logic for this particular task. >I have 2 formulas which consists of set of unary Boolean functions >(predicates?) combined with operations XOR, AND, XOR and NOT. Both >formulas are using same set of functions all operating on same >variables. I need to determine their equivalence for any values of >these variables. >For example: >Variables: a,b,c >Functions: f1(a), f2(b), f3(c), f4(a) >Formulae 1: (f1(a) OR f2(b)) AND f4(a) >Formulae 2: (f1(a) AND f4(a)) OR (f2(b) AND f4(a)) Sometimes in logic terminology varies quite a bit. Since the variables are not quantified, think of them as *constants*: so a, b, c are constants. Your f1, f2, f2 and f4 are usually called 1-place *predicates*. Then f1(a), f2(b), etc., are called *atomic formulas*. >My reasoning is that I should be able to treat each of >function/variable pair here as propositional logic variable: >f1(a)=P1 >f2(b)=P2 >f3(c)=P3 >f4(a)=P4 >(other combinations, like f1(b) are omitted since they not used in >neither formulae) Yes, you can treat each atomic formula as a propositional atom. >Now, all I need to prove, is equivalence of: >(P1 OR P2) AND P4) >and >(P1 AND P4) OR (P2 AND P4) Right. Your formulas are Boolean combinations of atomic formulas. >Does it make sense? I apologise if my reasoning is naive, but I am >willing to >learn more and will appreciate pointers in right direction, so I can >read more >about the subject and arrive to the right solution. What you're doing makes perfect sense. There are at least two simple ways to prove the equivalence of propositional formulas. (i) by constructing truth tables (this gives you DNF) (ii) by a semantic tableau (you may not have learnt about this yet). --- Jeff === Subject: Re: Open Letter to Serre Regarding Grothendieck I have the (full, I believe) translation of the autobiography to Russian. If anyone is interested, please contact me at i76r@yahoo.com. Joseph Shtok === Subject: Re: Upper bounding a transcendental function with Polynomials. > Jan C. Hoffmann schrieb im Newsbeitrag > schrieb im Newsbeitrag > > schrieb im Newsbeitrag > > > Hello everyone, > > > > > > I am trying to solve the following problem : > > > > > > I have a function f[x] given as an unsolvable integral > > > which is : > > > > > > f[x] = - Integrate[Exp[-y^2/2]/Sqrt[2*Pi] * > > > Ln[ (1-B) + B/D * Exp[(y+A*x)^2/2/(1+C)] ], > > > {y,-Infinity,Infinity}] > > > > > > where A>0 , 0 < B < 1 , C = 1/A^2 , D = Sqrt[1+A^2] > > > > > > now define g[x] = f[x] - f[0]. > > > > > > It is easy to see that f[x] and g[x] are even functions of x. > > > Also g[x=0] = 0 , and it can be seen that g[x] < 0 for all > > > other x ( g'[x] < 0 for all x>0) and finally > > > lim_{x->Infinity} (g[x]) -> -Infinity > > > > > > g[x] is monotonically decreasing for x>=0 , from 0 to > > > -Infinity. > > > > > > > > > Now I'm looking for the LARGEST number E>0 such that > > > (I need to do this for SPECIFIC values of A,B) > > > > > > g[x] + E*x^2 <= 0 for all x. > > > > > > > > > Does anyone have any idea how to attack this problem please ? > > > > > > If f[x] was a simple function then by common calculus tools > > > this could be solved, BUT the unsolvable integral complicates > > > things a great deal. > > > > > > EVEN if I can find a number that is close enough to this optimal > > > E from BELOW it will be great (just so the above equation is NEVER > > > larger than 0). > > > > > > > > > If anyone knows how to find a polynomial that will be an upper > > > bound on g[x] at least in the interval [0, z) , for z close to 1 or > > > smaller, this can be very helpful as well (I couldn't do this using > > > taylor series because knowing the sign of the the error is very > > > problematic). > > > > > > > > > Your function > > > > z = -Exp(-y ^ 2 / 2) / Sqrt(2 * pi) * Log((1 - B_) + B_ / D_ * Exp((y > + > A_ * > > x) ^ 2 / 2 / (1 + C_))) > > > > Example values > > > > A_ = 0.5 > > B_ = 0.5 > > C_ = 1 / A_ ^ 2 > > D_ = Sqrt(1 + A_ ^ 2) > > > > Have a least square polynomial function approximation > > > > z = f (x, y) = 4.06440168454817*10^-03*x^0*y^0 > > + -3.09292015753578*10^-07*x^1*y^0 + -3.33405592787114*10^-03*x^2*y^0 > > + -1.29496691491341*10^-06*x^3*y^0 + -1.85929697392827*10^-05*x^4*y^0 > + > > 6.21558610087785*10^-09*x^0*y^1 + -1.89758446536286*10^-03*x^1*y^1 + > > 4.48123783330665*10^-07*x^2*y^1 + -2.15564446786301*10^-05*x^3*y^1 + > > 3.25800980134267*10^-07*x^4*y^1 + -7.77554611994807*10^-04*x^0*y^2 + > > 2.77828932482955*10^-08*x^1*y^2 + 3.55854425430838*10^-04*x^2*y^2 + > > 1.15351834872955*10^-07*x^3*y^2 + 2.10917380269816*10^-06*x^4*y^2 > > + -2.19336285936576*10^-10*x^0*y^3 + 6.73395796611381*10^-05*x^1*y^3 > > + -1.5798490603598*10^-08*x^2*y^3 + 7.68155639890942*10^-07*x^3*y^3 > > + -1.14608524858889*10^-08*x^4*y^3 + 2.08631471802857*10^-05*x^0*y^4 > > + -5.56456152799723*10^-10*x^1*y^4 + -7.87273816815265*10^-06*x^2*y^4 > > + -2.29770862224741*10^-09*x^3*y^4 + -4.79893346366327*10^-08*x^4*y^4 > > > > Integral ~ -3.0267E-02 > > > > Graph and boundaries you find in > > http://homepages.compuserve.de/Jan390906/news/z-ng-04-08-13- 07.htm > > > > > > Hi Jan, > > > > I'm sorry but I fail to see how this approximation helps me in > > solving my initial problem ? > > I don't need just an approximation of g[x], I also MUST know if this > > approximation constitutes an upper bound on my real function. > > > > The problem I'm trying to solve is to find the largest E, such that > > g[x] + E*x^2 <= 0 for all x. > So far I integrated f[x] for a specified domain. > For finding integration f[x=0] I had to set x=10^-6 for avoiding trouble > with integration. > f[10^-6]= 3.296E-09 ~ 0. > That is g[x] ~ f[x] > Now you could compute table values like > x g[x] > 0 ~ 0 > 1 -1.26685267106262E-03 > 2 -3.02674846613907E-02 > 3 -0.116469436214612 > 4 -0.292684349167664 > 5 -0.596404538356658 > 10 -5.724148314286670 > Least square approximated data for [0, 10] > g[x] = 3.718173667589*10^-03 + -1.919570149511*10^-02 * x^1 + > 1.457880028900*10^-02 * x^2 + -6.993542610978*10^-03 * x^3 > Find E > g[x] + E*x^2 = 0 > E = -g[x]/x^2 > For this example you get a local max E(0.417) = 0.0129 > See also graph and max > http://homepages.compuserve.de/Jan390906/news/z-ng-04-08-14- 11.htm > Remark > You stated {y,-Infinity,Infinity}. You can reduce infinity to a plausible > domain that can be evaluated by an increase of the result e.g. +/-10^-6. > CORRECTION: > My program didn't find the solution via polynomial (degree 4). The program > was accidenly swiched to direct numerical integration to your function. > Sorry, I wasn't aware of that. But the result is ok. > A polynomial solution is possible. But you need a much higher degree. Sorry but I'm not looking for an approximation polynomial, I'm looking for a polynomial that is a good UPPER BOUND on the function g[x] which I defined above. I'm still waiting for some assistance on this, if anyone has any idea what I can do please advise. === Subject: Re: LinAlg: Vector Spaces - need help understanding an example > Yes, it's the same as the original system, but with the constant of > each equation set to 0, i.e. one sets the right side to zeros. So, I > know what a homogenous system is but it seems as if I still haven't > fully grasped all its implications. > Consider the single homogeneous equation x + y = 0 > versus the single non-homogeneous equation x + y = 1. > For the first (homogeneous) equation, x + y = 0, > x = 1, y = -1 and x = 3, y = 3 are solutions, > and x = 1 + 3 = 4, y = -1 + -3 = -4 is also a solution, so adding > those solutions gives another solution. This always happens. This is interesting and I see that this is the case by doing the algebra behind the sum of solutions, i.e. for the homogenous: x + y = 0 ; y = -x (x_1, -x_1) + (x_2, -x_2) = (x_1 + x_2, -x_1 + -x_2) = (x_1 + x_2, -(x_1 + x_2)) = (x, -x) and for the non-homogenous: x + y = 1 ; y = a - x (x_1, a - x_1) + (x_2, a - x_2) = (x_1 + x_2, (a - x_1) + (a - x_2)) = (x_1 + x_2, 2a - x_1 - x_2) = (x_1 + x_2, 2a - (x_1 + x_2)) = (x, 2a - x) but what is the intuition behind this? Is it that all vectors in the solution set of the homogenous equation lie in the line making up the set and thus the sum of two vectors will also lie in the line? And that adding vectors from the solution set of the non-homogenous equation will give a vector which is offset from a solutions by the constant in the equation? Is this as intuitive as this gets or is there some even more intutive explanation? > For the second (non-homogeneous) equation, x + y = 1, > x = 2, y = -1 and x = 3, y = -2 are solutions. > but x = 2 + 3 = 5, y = -1 + -2 = -3 is not a solution, so that adding > solutions does not always give a solution. Hmm, you say does not always. Does this mean that there are non-homoegenous equations where two solutions actually add to another solution? > Does this simple example help sort things out for you? Yes, it was a very nice example that got me thinking even harder about these things! /Henrik === Subject: Re: You're still a ing nut-case, Daryl... > Looks like the kook changed his handle and e-mail address to avoid > killfiling... [snip huge repost of daryl randomness] I'll be killfiling ye. : First, n x n matrices are elements of a n^2 dimensional vector space over : F, which is an affine space. So GL(n,F) would certainly be an affine : variety, since it is embedded in an affine space. The condition for a : matrix M to be in the group is GL(n,F) is that det(M) != 0. Interesting, : this is an inequality and not an equation. The trick is to take an additional variable z, different from the matrix entries. The condition for GL(n,F) is then det(M)*z=1. : I've been indoctrinated by modern : differential geometry that intrinsic geometric properties/notions must be : coordinate independent. Yet rationality is always defined with respect to : some coordinates (variables). Granted that rationality here is invariant : under linear transformations of the affine space in which the algebraic : group is embedded, but I would still like to see some sort of intrinsic : condition on the group that shows it to be an algebraic group. I understand what you're asking for, but I don't know of a good answer to this. Maybe someone else can help... ? :> (2) abelian varieties, which are algebraic groups where the group :> structure :> extends to a complete algebraic variety. These are always commutative :> (as the name suggests). For example, 1 dimensional abelian varieties are :> elliptic curves. : I know very little about elliptic curves. I do know that there is a group : law associated to the points on an elliptic curve. But this fact always : seemed like a coincidence to me. When is it possible to associate a group : law to the points of an algebraic variety? I presume there would be some : conditions on the polynomials whose zero-sets define the variety. This isn't an easy question to answer, as far as I know. To take a Ôsimple' example, suppose we are considering a _complete_ algebraic _curve_ X. If X can be given the structure of an algebraic group, then for every nonzero x in X, the translation by x map T_x, is fixed-point free. By the Lefschetz fixed point theorem, the alternating sum of the traces of the maps in cohomology induced by T_x is 0. By a sort of Ôcontinuity' argument, the same is true for T_0, the identity map. But the alternating sum of traces for the identity map is the Euler characteristic. So X has Euler characteristic 0, which means X has genus 1. So only genus 1 curves can have a group structure, which are elliptic curves. :> In general, a connected algebraic group is an extension of an abelian :> variety by a linear algebraic group. : You lost me here. I understand connected in the topological case, but : from where's the topology when working with discrete finite fields? And : what do you mean by extension? Some sort of group product? G is an extension of A by B if there is an exact sequence of groups 1 -> A -> G -> B -> 1 The topology we are using is the Zariski topology (closed sets are the algebraic sets). In the case where the field is the complex numbers, it is also okay to use the usual topology to determine connectedness. The reason for insisting on connectedness is to avoid groups like G x F, where G is a connected algebraic group and F is a finite group. Ted === Subject: Algebraic groups (was Re: What is a differential field?) > : I've only recently learned what an affine algebraic variety is. So let > : me see if I can prove that GL(n,F) satisfies the requirements to be one. > : First, n x n matrices are elements of a n^2 dimensional vector space > : over F, which is an affine space. So GL(n,F) would certainly be an > : affine variety, since it is embedded in an affine space. The condition > : for a matrix M to be in the group is GL(n,F) is that det(M) != 0. > : Interesting, this is an inequality and not an equation. > The trick is to take an additional variable z, different from the matrix > entries. The condition for GL(n,F) is then det(M)*z=1. Hmm, wouldn't that technically make GL(n,F) a _projective_ algebraic variety? I may be confused by what a projective variety is supposed to be. Igor === Subject: Shalosh B. Ekhad As far as I know, Zeilberger calls his computer Shalosh B. Ekhad. In his book A=B, he cites: One must not always invert Carl G. J. Jacobi [Shalosh B. Ekhad] Was that citation found out on the computer called Ekhad or was Shalosh B. Ekhad the jewish name of Jacobi? === Subject: Paper published by Algebraic and Geometric Topology A paper has been published and also a provisional erratum for a previously published paper (see end). --- The following paper has been published. Algebraic and Geometric Topology URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-27.abs.html Title: Commutators and squares in free groups Author(s): Sucharit Sarkar Abstract: prove that the commutator of x^m and y^n is a product of two squares if and only if mn is even. We also show using topological methods that there are infinitely many obstructions for an element in F_2 to be a product of two squares. Secondary: 57M07 Keywords: Commutators, free groups, products of commutators Author(s) address(es): Stat-Math Unit, Indian Statistical Institute Bangalore, India Email: bmat0212@isibang.ac.in --- An error has been found in the following paper. A full erratum will be published in due course. In the meanwhile a provisional erratum is available (linked from the URL below). K-theory of virtually poly-surface groups by S.K. Roushon URL: http://www.maths.warwick.ac.uk/agt/AGTVol3/agt-3-4.abs.html === Subject: Re: A question about finite sets of integrally independent vectors >How can one verify if a set of vectors are integral dependent? that >is: if they are linearly independent with integral coefficients. >(Actually I'm looking for a method to reduce a finite set of vectors >to a set of integrally independent vectors) First of all, integrally dependent is the same as rationally dependent, i.e. if a linear combination with rational coefficients is 0, you multiply by the denominators and get one with integer coefficients. I'll suppose your vectors are in F^n where F is some field of characteristic 0. We need to assume you have a way of checking whether finite subsets of F are rationally related, and finding such relations if they exist. Then you can look at all the coordinates of all your vectors and find a maximal rationally independent set, say S = {s_1,...,s_m}. This will be a basis for the vector space (over the rationals Q) spanned by those coordinates. Using that basis, there's a Q-linear one-to-one correspondence from Q^{nm} to a linear subspace of F^n that contains your vectors. Thus the problem is reduced to the case of Q^{nm}, where the standard methods of Gaussian elimination can be used to check for linear (which in this case would be Q-linear) independence and find bases. ' === Subject: Re: some questions about the algebraic closure of Q_p Epigone-thread: twyswinply >These questions will seem rather naive to ots of people, I bet. So I >apologize in advance. >Let K be the field of algebraic p-adic numbers. (So it is a subfield >of Q_p, and it is the Henselization of Q with respect to the p-adic >valuation). >Let K_1 be the valued field extension of K obtained by addition of the >roots of all X^(p^n)-X. So the residue field of K_1 is the algebraic >closure of F_p. K_1 is the inertia field of K. The valuation group of >K_1 is still Z with v(p)=1. >Let K_2 be the extension of K_1 obtained by addition of the roots of >all X^n-p. Let say we keep v(p)=1, and the valuation group of K_2 is Q. >The question is : is K_2 algebraically closed ? If not, why ? >Can someone give me an exemple, for a given p, of a polynomial which >does not split in K_2 ? >Other question : what happens if K=Q_p ? It follows from property 5 listed in my previous posting that the extension from V_1 to Q_p^~ is non-abelian, thus K_2 cannot be the algebraic closure of Q_p^~. So the answer to your question is NO. H === Subject: Re: Infinitely divisible probability distributions Epigone-thread: phoisterdsloo >I find in the second volume of Feller's famous book the >attribution of the concept to Bruno de Finetti, in 1929. >But I couldn't find that in any list of references in the >book. -- Mike Hardy > I don't have a copy available to check (I've only ever read it > on loan from a library), but I'm *reasonably* sure de Finetti > discusses the history of the idea (and possibly even reprints > @book {MR92b:60009a, > AUTHOR = {de Finetti, Bruno}, > TITLE = {Theory of probability. {V}ol. 1}, > SERIES = {Wiley Classics Library}, > PUBLISHER = {John Wiley & Sons Ltd.}, > YEAR = {1990}, > MRNUMBER = {92b:60009a}, > } > @book {MR92b:60009b, > AUTHOR = {de Finetti, Bruno}, > TITLE = {Theory of probability. {V}ol. 2}, > PUBLISHER = {John Wiley & Sons Ltd.}, > YEAR = {1990}, > MRNUMBER = {92b:60009b}, > } . > Looking for that information in MathSciNet, I also found > @book {MR86i:01060, > AUTHOR = {de Finetti, Bruno}, > TITLE = {Scritti (1926--1930)}, > YEAR = {1981}, > MRNUMBER = {86i:01060}, > } > which from its title would appear to include the paper you want > (but may not help you if you don't read Italian, and/or cannot > get your hands on what might be a reasonably obscure book). >Scritti does indeed contain several papers on what we would nowadays >call processes with stationary independent increments, including one >dating from 1929. (My Italian is too non-existant to say more.) You >might also have a look at a survey paper by M. Fisz: > AUTHOR = {Fisz, Marek}, > TITLE = {Infinitely divisible distributions: recent results and > applications}, > JOURNAL = {Ann. Math. Statist.}, > VOLUME = {33}, > YEAR = {1962}, > PAGES = {68--84}, > MRCLASS = {60.20 (60.30)}, > MRNUMBER = {25 #2624}, >MRREVIEWER = {M. Lo{`e}ve}, >The bibliography of this survey includes one paper by De Finetti. The >De F paper is the earliest of the 104 papers cited, lending support to >Feller's assertion. >-- Would you mind giving us an idea ? How does Ôinfinitely divisible probability distributions'look like! Do you know a simple case and its mathematical representation . === Subject: Re: Infinitely divisible probability distributions ................... >Would you mind giving us an idea ? >How does Ôinfinitely divisible probability distributions'look like! >Do you know a simple case and its mathematical representation . Every infinitely divisible distribution is the convolution of a normal distribution and the convolution of a countable number of Poisson convolutions of identical distributions. It is easier to express in random variables. X = Z + sum sum_1^{N_i} Z_{ij}, all the random variables on the right side are independent, Z is normal, the N_i are Poisson, and the Z_{ij} have the same distribution for each i. Some of the terms can be absent, and the normal term can be a constant. Examples are normal, Poisson, Gamma, t, and a huge variety of others. Conversely, if the sum converges almost surely, the random variable X has an infinitely divisible distribution. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Riemann-Roch for reducible P^2-plane curves Epigone-thread: glangswongheld >Let C be a plane complex projective curve that is a union of two >smooth >curves C_1 and C_2 of degrees d_1, resp. d_2, and D a divisor on C, >consisting of smooth points only. What is in this setting the analogy >of the classical >Riemann-Roch theorem for the dimension of |D|, i.e. the space of >divisors linearly equivalent to D ? >Strangely enough, I cannot find it in obvious literature. >Some theory in such setting can be found in the paper >Theta-characteristics on algebraic curves by J.Harris >Dmitrii >(dima[1-3]pasechnik[1-4] at msri dot org,ou if you don't stop reposting Daryl's madness in its entirety with your 1-liner replies! === Subject: Re: how to evaluate the addition of millions of functions dynamically and efficiently? > I am also interested in the RBF monopole you've mentioned in your > previous post. > Do you think it will give me a drastic performance improvement? If so, I am > going to dive into it and learn how to make my program a lot faster... I have never implemented it, but the idea is rather similar to the FFT - break the problem into smaller simpler subproblems, and recurse. I understand that the implemetation is rather involved, so I'd suggest that you consider using it only if - your centres are genuinely scattered, - you basis function does not have compact support. Otherwise I would guess that a local search would be quicker and a lot easier to implement. As I say, I'm no expert on this, others may correct me. There are quite a few papers in the (mathematics) literature, I'd recommend that you look at those before proceeding. -j === Subject: Re: how to evaluate the addition of millions of functions dynamically and efficiently? > Is your algorithm supposed only to work only for regularly laid-out > functions? I mean for the support of functions, I can make them truncated > into rectangular-shaped finite support, even for infinite support functison. Yes, the trick works best for - translates of the same function with small support - arranged on a regular grid (might be square, ractangular, triangular) I'd recommend against truncating a function with infinite support -- this will lead to a discontinuity and so to a slow decay in the fourier transform. Instead (if possible) use a basis function which is compactly supported and smooth. Search for Wendland or Lewitt along with radial basis functions. > These functions are nicely behaved. So that's ok. But for the layout of the > center of these functions, for this function array, your dynamic local > summation only work for horizontally and vertically alligned function-center > array layouts? Not necessarily. You just need to be able to work out what the indicies of the nearby points are *without* doing a test of every possible grid node. This can be done for a triangular grid, for example. If your grid nodes are genuinely scattered then the easy way wont work. > How to decide if a point is covered by which overlapped neighboring > functions, for those irregular layout? I guess there should be some fast > algorithm and fast data structure for handling this in the literature? Possibly you could use a 2-stage method 1) find the neighbours for each grid node & store in a file 2) read neighbours into memory and for each node do the local search with these neighbours Then 1) (which is non-trivial, but there is some code out there for doing it) has (if I remember correctly) complexity O(N^2 log(N)), but 2) has O(N), and 1) only need to be done once. Note that this may require *lots* of memory. Work out how much you will need before you start to code! -j === Subject: Re: books for self study, number theory, set theory, analysis... > Hi all, > I have a free couple of months coming up here in a week or so. I'd like to > do some self-study and have access to a decent university math library. > I am interested in number theory, set theory, and perhaps real and complex > analysis. > Can anyone reccomend some *fairly basic* texts on above topics? > I have a fairly good handle on calculus, geometry and trig, linear algebra, > kW For number theory, a previous posting listed 23 books on number Re: 341 Problems in Elementary Number Theory in order to get that list. There is also a university in Australia that has/did have on-line lecture notes for Number Theory and Algebraic Number Theory. David Ames === Subject: Re: books for self study, number theory, set theory, analysis... >Hi all, >I have a free couple of months coming up here in a week or so. I'd like to >do some self-study and have access to a decent university math library. >I am interested in number theory, set theory, and perhaps real and complex >analysis. >Can anyone reccomend some *fairly basic* texts on above topics? >I have a fairly good handle on calculus, geometry and trig, linear algebra, > kW For set theory, P. Halmos: Naive set theory is the best introduction. And if you like original sources, get G. Cantor: Contributions to the Founding of the Theory of Transfinite Numbers For number theory, it is difficult to make any sense of it without understanding how it evolved. So make sure you get a book that treats number theory in its historical context. For example, O. Ore: Number theory and its history or A. Weil: Number theory: an approach through history or H.M. Edwards: Fermat's Last Theorem (A genetic introduction to algebraic number theory) === Subject: multiples in sequence I need help to understand a solution to the following problem Problem:How many multiples of s are there in the sequence r,2r,3r,...,sr? Solution:Apply the following theorem to the problem: Th:ax ==b (mod m) has exactly (a,m) solutions if (a,m)|b. Answer:(r,s) Now I don't understand this. Would someone kindly explain this for me? === Subject: Re: multiples in sequence > I need help to understand a solution to the following problem > Problem:How many multiples of s are there in the sequence r,2r,3r,...,sr? > Solution:Apply the following theorem to the problem: > Th:ax ==b (mod m) has exactly (a,m) solutions if (a,m)|b. > Answer:(r,s) > Now I don't understand this. Would someone kindly explain this for me? In the theorem take m=s, a=r, b=0. === Subject: Re: Tangle: The Game > Here is a game which *might* be fun to play, and might be fun to > analyze mathematically and for strategy. > > > For 2 or more players, each with a different colored pencil. > > Start with a large square (or any convex polygon) drawn on paper. > > Player 1 starts by drawing a straight line-segment, using a > straight-edge, from any corner of the square/polygon through the > interior of the polygon until hitting an edge of the polygon. > > Players together draw a path by alternatingly drawing line segments, > using a straight-edge (and making each segment the color associated > with the player drawing the segment), each segment which is from the > end of the last segment drawn by the previous player, in any > direction(with restrictions*), and ending at the first previously > drawn line-segment intersecting the segment being drawn, or ending at > a boundry of the square/polygon. > > (*) If the last line-segment drawn ended at the edge of the > square/polygon, the path must bounce, ie. the next segment must be > towards the interior of the polygon. > If the last segment drawn ended at a previously drawn segment (of any > color), the path must cross this older segment, ie. the new segment > must be on the opposite side of the older crossed segment from the > immediately previous segment. > > Segments cannot be drawn to vertexes (where the path crosses itself or > bounces off the edge of the polygon.) > > The players draw a predetermined number of segments (say, a total of > 50). > Each player gets a point every time the path crosses a pre-drawn > segment of his/her color. > The winner has the most number of points. > > A clarification: > The path refered to above is the total collection of line-segments. > The line-segments are drawn end-to-end and alternate in color. > I will attempt some ascii-art. > (View with fixed-width font.) > If the game is like so: > < older line-segment > > -- < last line drawn by previous player > > > you continue on the opposite side of the older line-segment in any > direction: > /< your new segment > / > -- > > > And your new segment continues only until the first line-segment or > edge of the square/polygon encountered, where your line-segment then > terminates. > If the game is like so: > > < last line drawn by previous player > > You may bounce in any direction (either to left or right), as long as > the path stays within the square/polygon: > / > / < your new segment > / > One final note: > No segment is allowed to be drawn along a previous segment. > (No co-linear segments.) > Leroy Quet 4 more things: 1) It is evident after I played this that it would probably be better, rather than making a path of a predetermined number of line-segments, that the game simply continue until one player first gets a predetermined score, say 10 or 20. In this way, players do not have to worry about keeping track of the number of moves made. And there is also, with this new rule, no chance of a tie. 2) One purpose of playing this game, aside from winning, is to hopefully get a cool abstract design. :) 3) I was a little unclear about when a point is scored pre-drawn segment of his/her color.). Whenever a line-segment is drawn by anyone to another segment of color c, and player c is using pencil/pen color c, then player c gets a point. By path crosses, I mean 2 segments come together at the previously drawn line-segment. Like so: < previously drawn segment of color c, path crosses this segment. _______ / / / 4) As is noted elsewhere in this thread, there is an advantage to being player 1 (and so players should play multiple rounds, trading off who is player 1). There is a likely set of opening moves. As an easy puzzle, try to give the best next move for player where >aword[1-3] means awo) >Reducible and multiple curves<, in which the Riemann-Roch-theorem is treated in the general situation. However I fear that it is out of print and unfortunately cannot remember the publisher ... H === Subject: Re: Riemann-Roch for reducible P^2-plane curves >Reducible and multiple curves<, in which the Riemann-Roch-theorem > is treated in the general situation. > However I fear that it is out of print and unfortunately cannot > remember the publisher ... it's his PhD thesis. Easy to find in Dutch libraries. Dima === Subject: Re: Question re: the two 2_21's in 3_21 and series parallel posets >If you want to pick arbitrary extremum vertices and edge directions >of 2_21 that make a poset and an algebraic lattice, you can, >but I can't quite put my finger on why. I don't think that you can make a poset from the 2_21. The Schlaeßi graph is not transitively orientable nor is its complement. Not even, when you add edges from the top-vertex to the 10 bottom-vertices. --Guenter === Subject: Re: Question re: the two 2_21's in 3_21 and series parallel posets ------------------------------------------------------------- --------- > Each vertex of a polytope can be paired with a diametrically opposed > vertex if the polytope has central symmetry. This isn't possible > with 2_21 since it has 27 vertices, an odd number ... snip > Orienting 2_21 with a simplex on top results in a simplex being > on the bottom too. Here's the distribution matrix for this case: > 6 5 10 1 > 15 4 8 4 > 6 1 10 5 > If you want to pick arbitrary extremum vertices and edge directions > of 2_21 that make a poset and an algebraic lattice, you can, > but I can't quite put my finger on why. ------------------------------------------------------------- ------- I am grateful to SF for even bothering to respond after my inexcusably careless and loose terminology concerning the central symmetry of 2_21. I can only plead typing without thinking in an attempt to simplify the phrasing of the question. With respect to the why one might one to make a lattice out of 2_21, I would not trouble S. Fortescue with outlining a possible why unless I knew how he would answer the question raised in my geometry. === Subject: super-Hopf algebras Epigone-thread: werskendtweld I want enough materials about this topic super-Hopf algebras === Subject: Keep it simple: Inf and -Inf and extending exponents by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7RJL6E16758; I sent the note below to alt.math.undergrad discussion forum under the to make it simple, clarify matters, and make common sense arguments about a bit of sometimes troublesome math notation. Since it provoked no public response at all, I thought I would try it again under a new subject line. Surely someone will want to object to my suggestions! ------------------------------------------------------------- I must say that this thread, and other similar ones about uses or meanings of the symbols Inf and -Inf and also ad hoc definitions of extensions of functions like division, powers, exponential and log always strike me as being strange and unhelpful, and indeed likely to cause confusion among novice or casual students or users of math. 0. I think it would be much better in elementary math (even up to advanced calculus or beginning analysis at least) to stick with not giving any meaning to Inf or -Inf other than as part of useful abbreviations (easily remembered and written nicknames) for long somewhat complicated phrases involving certain specified behaviors of functions defined on the Reals or a tail thereof. Discourage and disavow any uses of these symbols that even suggest they are somehow just new extended versions of numbers. As far as I am aware, there is no utility and lots of likelihood for causing confusion to do otherwise. What could be the motivation of someone to want to consider Inf and -Inf as extended numbers? That is not helpful - notational conventions are put in place for reasons of efficiency of thinking and writing and an aid to instant recognition/apprehension of complicated expressions, to simplify and clean up our writing of math. About exponents: 1. We all know that for any element B in any ring, B^n makes perfectly good elementary common sense for any positive integer n and is just a shorthand notation for writing out a longer expression involving n appearances of B and (n-1) multiplications parenthesized in any way you like. 2. In the 1600's people started to systematically use this exponent notation and noted the usual Ôrules of exponents' - all quite useful for computation and compact notation. Later someone noticed that in fact it was convenient for the writing of polynomials and power series in B using the sum notation to write the constant term C0 instead as C0*B0 with the value of B0 being 1 by convention. Fortunately the rules of exponents still worked ok, and there was then just one standard form (in general sum notation) for computing the coefficient of B^n in products of polynomials/series, even when n is 0. This is all unmitigated good news; we've gained in efficiency of writing math and calculating. 3. But then some nagging student or stickler for details noted that if we are doing this for a polynomial/series expression in B, say P(B) which begins C0*B0 +C1*B1 + ... , then if we evaluate the expression when B is 0, one expects to get the constant term C0 of P, but the expression suggests that what we get is the value of C0*00. The obvious thing to do is to insist that the convention B0 = 1 continue to hold even if B=0. And again everything still works fine. See the and examples showing that the convention 00 = 1 is indeed very useful - especially for combinatorics and whenever formal power series (generating functions) are used. 4. We also know that while we can write an expression (compact description of a step-by-step process to calculate something) involving B^(-n)for n a positive integer, when we get around to actually evaluating the expression we should get an ÔError message' whenever B is not invertible in the ring. But it worked out nicely that if B is invertible then B^(-n) could conveniently be defined to be the inverse of B^n which turned out to be the same as (B^[-1])^n - and, very important, the usual rules of exponents still worked ok. However, since 0 is not invertible, 0^[-n] is undefined - and I know of no good reason to try to assign it some meaning. 5. Notice that the above 1-4 all work quite generally in a ring (well at least if B is in the center of the ring if we are multiplying polynomials in B). But if we are working in a special ring like the field of 2: Playing on a square board. Moves: 1) Player 1 draws segment from lower-left (Southwest) corner East-Northeast to right edge of square. 2) Player 2: NW to upper edge of square. 3) Player 1: South to player 1's first segment. 4) Player 2: SE to right edge of square. 5) Player 1: West to player 1's first segment. Now, where should player 2 move? (Easy, but this will help you familiarize yourself with the game, if you choose to try to solve this puzzle.) Leroy Quet === Subject: Re: Tangle: The Game > 4 more things: > 1) It is evident after I played this that it would probably be better, > rather than making a path of a predetermined number of line-segments, > that the game simply continue until one player first gets a > predetermined score, say 10 or 20. > In this way, players do not have to worry about keeping track of the > number of moves made. And there is also, with this new rule, no chance > of a tie. We've played (2 players) until a player reaches 50, starting with a square approx. 6x6 inches; although I find that if you increase one diagonal to 8 inches (and start on a short diagonal corner), you've got more space. I've wondered whether the game would be more balanced if the two diagonals were pre-drawn in a neutral color. Michael -- Feel the stare of my burning hamster and stop smoking! === Subject: Re: A Simpler More General Recursion Puzzle > First I will give a more specific case, then the more general recursion. > Let n = any positive integer. > If a(1) = 1; > a(m) = (1/m)*(ßoor(m^(1/n)) + sum{k=1 to m-1} a(k)) > for all m >= 2, > then give a non-recursive form for {a(m)}. > More generally, > if {b(k)} is any sequence defined for all positive integer indexes k, > and a(1) = b(1), > and > a(m) = (1/m)*(b(m) + sum{k=1 to m-1} a(k)), > for all m >= 2, > then give a non-recursive form, in terms of {b(k)}, for {a(m)}. > I will give the answer in a few days if no one else gets it sooner. > Leroy Quet Michael Mendelsohn already solved this on rec.puzzles. Here is a link to a post there: Leroy Quet === Subject: Re: Ingeger solutions of algebraic function >I am trying to solve the ecuation >x^2+y^2+z^2 = 2xyz >in the integers. >I reduce it Mod 3 and i reduced it to >x^2+y^2+z^2=6xyz factoring the 3. >I think that this equiation only has the trivial solutions is it >right? do you have a hint to proof it? > Why do you think so? > You might try your equation mod 4, and then by induction > mod 2^k for k >= 2. > 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: Convolution Sum Over Divisors Let a(1) = 1. Let, for m >= 2, a(m) = sum{k|m, k escribi.97: > I'm currently trying to find a solution as to how to find all the > tangents that touch two ellipses. The ellipses are independent of each > other and are not rotated. And yes, the ellipses could be the same, > could be intersecting, and so forth. > Would anyone know of a solution or at least know of a good web site > that contains code or an actual mathematical procedure to solve the > tangent-ellipse problem? Any help would be greatly appreciated. Let the two ellipses, conic sections in general, to be: c1: Ax^2 + By^2 + Cxy + Dx +Ey + F = 0 c2: A'x^2 + B'y^2 + C'xy + D'x +E'y + F' = 0 and the tangent t: y = mx + n, with m and n unknown. In order to t be tangent to c1, the system of two equations, of t an c1, must have only one solution. Then replace y = mx + n in the equation of c1 to get a second degree equation in x, and set its discriminat equal to zero. You get a second degree equation in m and n. Do the same with c2; you get a second eauation of degree 2 in m and n. That system of two eauations of degree 2 in m and n can have among 0 and four (real) solutions, the wanted tangents. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: 3D Plane Supersedes: >Yes, I was after the vertical shadow (if you consider Z to be the up >axis). I thought the normal is the a, b and c values of the equation: >ax + by + cz + d = 0 >So you're saying I re-arrange that to be: >(-ax - bx - d) / c = z >by plugging in (a,b,c) which is the normal of the plane and plugging in >x,y and leaving z on the other side of the equation? >I can't speak for Lynn but this looks right to me. >z_projected = (-ax - bx - d) / c >distance = |z - z_projected| Correction: z_projected = (-ax - by - d) / c --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: measurable sets and functions > Hi everybody > > I've just started studying measure theory and chose Rudin's Real and > Complex Analysis. I'm confused about some points and any help is > welcome. > > If X is a set and M is a sigma-algebra in X, then, according to Rudin, > a subset of X (which Rudin assumes to be a topological space) is > measurable if it belongs to M. If the pair (X, M) is a measurable > space (a pair composed of a topological space and a sigma-algebra > defined on it, according to Rudin) and Y is a topological space, then > Rudin defines f:X->Y as a mesurable function if f^(-1)(V) is in M > whenever V is an open set of Y. To Rudin, a measure is any real or > complex valued set function defined on M that is countably additive. > > But in his book, Charambolos D. Alipranti uses also the term > sub-additivity of the measure. He considers the concept of an outer > measure, which doesn't show in Rudin's book. He says a set E of X is > measurable according to an outer measure u if u(E) = u(E inter A) + > u(E inter A') - A' being the complement of A-for every set A of the > sigma-algebra defined in X. > > Rudin's definition implies the measurability of a set depends only on > the sigma-algebra, and Alipranti's one implies it also depends on an > outer measure. > > Are these 2 approaches eventually the same? > The same? No. Rudin does restrict his measure spaces to those arising > from topologies because he's an analyst, and I would be hard-pressed to > give a USEFUL example which arises any other way. Also, one of his > goals is to show how to use the Riesz Representation Theorem as the > DEFINITION of integral. > This is nonstandard (in the sense of being unusual). Aliprantis's > approach (you misspelled his name) is the standard one, in the sense of > being the most commonly used. Go ahead and read both books, and keep > in mind that there's more than one way to skin a cat. I'll try to read both books and bartle's too. Yes, I mispelled his name, it's Charalambos (seems Greek). And I also gave a wrong definition of measurable set..The correct is, a set E of X is measurable if for every subset A of X we have u(A) = u(A inter E) + u(A inter E'). Artur === Subject: Re: measurable sets and functions > > I've just started studying measure theory and chose Rudin's Real and > > Complex Analysis. I'm confused about some points and any help is > > welcome. Great book but unless your IQ is high in the sky I wouldn't recommend it as an introduction (focusing on too many technical details, which makes the essential hard to abstract, especially when you reach the Riesz representation theorem). > > Rudin's definition implies the measurability of a set depends only on > > the sigma-algebra, and Alipranti's one implies it also depends on an > > outer measure. > > > > Are these 2 approaches eventually the same? I must admit I really didn't understand other answers, and I also think that's because the question is a little confusing: measurability depends only on the sigma-algebra of the definition set (and on the topology or sigma-algebra structure of the image set, even though, as pointed by Rudin, the essential point is the presence of a topology). But the construction of a sigma-algebra may be done using outer measures -and we know easy ways to construct outer measures given a nonnegative mapping defined over a covering class- or metric outer measures, which can be constructed rather easily too -which ensure that the Borel sets are measurable-. I don't see how these two points were linked in the above question. -- Julien Santini === Subject: Re: JSH: Unskilled and unaware? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i real numbers R, then as we know, Newton and others of his generation popularized some useful extensions of the exponent notation, first to rational number exponents and then to real number exponents - when the base is a positive number, all of which are now incorporated into calculators and computer arithmetic. Some also have the special extension (-B)^r = {-(B^[1/q])}^p for a rational number r in reduced fraction form p/q with q odd and B a positive number. I don't remember now ever hearing any good arguments for any other extensions of exponent notation, except for the corresponding situation for the complex numbers. (I'm ignoring here the occasional use of notation B^A in set theoretical discussions in discrete math texts with the meaning Ôset of all functions from A into B' - that doesn't seem to have wide usage.) Ladnor Geissinger, Math Prof, Univ NC at Chapel Hill, NC 27599 === Subject: Re: Keep it simple: Inf and -Inf and extending exponents > I sent the note below to alt.math.undergrad discussion forum under the > to make it simple, clarify matters, and make common sense arguments > about a bit of sometimes troublesome math notation. Since it provoked > no public response at all, I thought I would try it again under a new > subject line. Surely someone will want to object to my suggestions! It's too too long on talk, talk, talk and not simple upon what you're suggesting. Instead of making us guess by wading thru reams of verbiage, just present your point in 100 words, an abstract of your thesis, and then proceed with the details. Again, rather than rambling, present your points succently. Elaborate, if you must, later after we know in the first place what the heck you're attempting. > I must say that this thread, and other similar ones about uses or > meanings of the symbols Inf and -Inf and also ad hoc definitions > of extensions of functions like division, powers, exponential and log > always strike me as being strange and unhelpful, and indeed likely to > cause confusion among novice or casual students or users of math. > 0. I think it would be much better in elementary math (even up to > advanced calculus or beginning analysis at least) to stick with not > giving any meaning to Inf or -Inf other than as part of useful > abbreviations (easily remembered and written nicknames) for long > somewhat complicated phrases involving certain specified behaviors of > functions defined on the Reals or a tail thereof. Discourage and > disavow any uses of these symbols that even suggest they are somehow > just new extended versions of numbers. As far as I am aware, there is > no utility and lots of likelihood for causing confusion to do > otherwise. What could be the motivation of someone to want to > consider Inf and -Inf as extended numbers? That is not helpful - > notational conventions are put in place for reasons of efficiency of > thinking and writing and an aid to instant recognition/apprehension of > complicated expressions, to simplify and clean up our writing of math. > About exponents: > 1. We all know that for any element B in any ring, B^n makes > perfectly good elementary common sense for any positive integer n and > is just a shorthand notation for writing out a longer expression > involving n appearances of B and (n-1) multiplications parenthesized > in any way you like. > 2. In the 1600's people started to systematically use this exponent > notation and noted the usual Ôrules of exponents' - all quite useful > for computation and compact notation. Later someone noticed that in > fact it was convenient for the writing of polynomials and power series > in B using the sum notation to write the constant term C0 instead as > C0*B0 with the value of B0 being 1 by convention. Fortunately the > rules of exponents still worked ok, and there was then just one > standard form (in general sum notation) for computing the coefficient > of B^n in products of polynomials/series, even when n is 0. This is > all unmitigated good news; we've gained in efficiency of writing math > and calculating. > 3. But then some nagging student or stickler for details noted that > if we are doing this for a polynomial/series expression in B, say P(B) > which begins C0*B0 +C1*B1 + ... , then if we evaluate the expression > when B is 0, one expects to get the constant term C0 of P, but the > expression suggests that what we get is the value of C0*00. The > obvious thing to do is to insist that the convention B0 = 1 continue > to hold even if B=0. And again everything still works fine. See the > and examples showing that the convention 00 = 1 is indeed very useful > - especially for combinatorics and whenever formal power series > (generating functions) are used. > 4. We also know that while we can write an expression (compact > description of a step-by-step process to calculate something) > involving B^(-n)for n a positive integer, when we get around to > actually evaluating the expression we should get an ÔError message' > whenever B is not invertible in the ring. But it worked out nicely > that if B is invertible then B^(-n) could conveniently be defined to > be the inverse of B^n which turned out to be the same as (B^[-1])^n - > and, very important, the usual rules of exponents still worked ok. > However, since 0 is not invertible, 0^[-n] is undefined - and I know > of no good reason to try to assign it some meaning. > 5. Notice that the above 1-4 all work quite generally in a ring (well > at least if B is in the center of the ring if we are multiplying > polynomials in B). But if we are working in a special ring like the > field of real numbers R, then as we know, Newton and others of his > generation popularized some useful extensions of the exponent > notation, first to rational number exponents and then to real number > exponents - when the base is a positive number, all of which are now > incorporated into calculators and computer arithmetic. Some also have > the special extension (-B)^r = {-(B^[1/q])}^p for a rational number r > in reduced fraction form p/q with q odd and B a positive number. > I don't remember now ever hearing any good arguments for any other > extensions of exponent notation, except for the corresponding > situation for the complex numbers. (I'm ignoring here the occasional > use of notation B^A in set theoretical discussions in discrete math > texts with the meaning Ôset of all functions from A into B' - that > doesn't seem to have wide usage.) > Ladnor Geissinger, Math Prof, Univ NC at Chapel Hill, NC 27599 === Subject: STAT Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7S51Fk29392; Heres the question... A student is planning on taking three different classes. Two of the classes, calc I and Computer Science, are offered at different time periods. The third class, Physics, is at a time that will conßict with Calc I. Probability of getting into calc I = 1/2 , Probability of getting into Computer Science = 1/3, which is the same probability as getting into Physics (1/3). Probability that he will get into both Calc I and CS = 1/6. Probability that he will get into both CS and Physics is 1/6. Find the prob that he will be able to get into at least one of the three classes. I think the answer goes like this. P(AuBuC)=P(a) + P(b) + P(c) - P(a Intersect B) - P(a intersect c) - P(b intersect c) + P(a intersect b intersect c). Therfore, P(a) = 1/2 , Prob of getting into calc I P(b) = 1/3 , Prob of getting into CS P(c) = 1/3, Prob of getting into Physics P(a intersect b) = 1/6 P(b intersect c) = 1/6 So, 1/2 + 1/3 + 1/3 - 2/6 = 5/6 If you could tell me if i'm anywhere close, i would appreciate it. === Subject: Re: STAT Problem in alt.math.undergrad: > Heres the question... > A student is planning on taking three different classes. Two of the > classes, calc I and Computer Science, are offered at different time > periods. The third class, Physics, is at a time that will conßict > with Calc I. Probability of getting into calc I = 1/2 , Probability > of getting into Computer Science = 1/3, which is the sa7GEP8l11760; (snip) >We don't matter to the rest of the world. >We're social misfits. You speak only for yourself (and your buddy Quinn Tyler Jackson, your closet list pal). >I may not be mathematically skilled in many ways, but I'm aware of >that and am aware of my misfit status. Ah, the truth doth shine out of all your drivel. >The clearest demonstration of our kinship is that you cannot just >ignore me, even when I request it. According to your reasoning (if I dare dignify your crap as reasoning), you can't ignore me either, you Quinn-loving nutcase. >Still, I'll try one more time: Take down the webpages. Stop the >cyberstalking. Stop the obsessive replying to my posts. I bet you can't follow your own instructions. I dare you. I DOUBLE DARE YOU! >Let me just be one more misfit on sci.math along with the other >misfits. Stay in your misfit oblivion, crank! >I'd like to retire the JSH at this point, though not saying I am >definitely, and I fear some one of you will keep dragging it out, time >after time, obsessively. We can only dream, but your constant coming out of retirement is a continual nightmare. But then, coming out (of your closet list) is what you and Quinn do best. >But again, at least I can put the idea out there. >I don't want to stop posting. I LIKE posting. I get attention ... because you are a troll (by definition), as well as a crank and crackpot. >If I had any power at all I'd have sued some of you by now welfare checks can't pay for even a greedy ambulance chaser. What happened, your parents can't afford a lawyer for you this month? They can barely afford your anti-psychotic medication (which you don't take, anyhow). You just continue to leech and steal from your parents. You are some disappointment to them. And after all the times they bailed you out of jail, and paid your dues for your high-IQ membership (ha, what a joke that group is - it includes both you and Quinn Tyler Jackson!) >To those of you who have worked for years to try and drive me off the >newsgroup, SCREW YOU Quinn's AIDS. >James Harris ... is a crank, crackpot, troll, and nuts! Yet Quinn still loves his closet list friend James. === Subject: Re: JSH: Unskilled and unaware? Discussion, linux) >The clearest demonstration of our kinship is that you cannot just >ignore me, even when I request it. > According to your reasoning (if I dare dignify your crap > as reasoning), you can't ignore me either, you Quinn-loving > nutcase. Why's that? When's the last time he responded to *you*? He seems to be doing a fine job of ignoring you. (I feel a dark moment of introspection coming on.) -- The papers are currently at journals. [When published,] make no mistake, there will be no place on this planet where you can hide. Remember, I'm not talking about something vague here. I'm talking about publication in journals. James S. Harris. Wow. Journals. === Subject: Re: 1+i > i > hello.....doctor~ Sorry, but I'm getting more curious as time goes on: who is this doctor you refer to? :) alex === Subject: Re: Explaining the prime distribution > I've contacted professional mathematicians off of Usenet, and mostly > been ignored. judgment of professional mathematicians. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Explaining the prime distribution > Your first points where you define functions are all derived from > dS(x,y), so here I'm only going to talk about your function for that, > since that must be correct in order for the rest to work. >p(x,sqrt(x)) = ßoor(x) - S(x,sqrt(x)) - 1 > Just a side note, this doesn't really matter, but since sqrt(x) is a > function of x and not a seperate independent varaible, you only need > one input. Also, ßoor(x) is unnecessary because ßoor(x) = x when x > is natural, which it will always be. Actually it won't as can be seen further down, where I end up with p(x/y,y-1) and the ßoor can conveniently be put in that one place and handle everything. Also, the function is p(x,y) where p(x,sqrt(x)) just happens to count primes! So it's a multi-variable function that can be evaluated over a certain path to count prime numbers, but it makes sense, as evaluating it requires both variables to write it as a I have. > p(x) = x - S(x,sqrt(x)) - 1 > Anyway, on to the meat. > So I've finally outlined all the necessary functions, and it's now > time to get the full dS(x,y) function, which surprisingly enough is > easy to the point of trivial. > > Remember, dS(x,y) is the count of composites up to and including x > that have y as a factor which do not have any other primes less than y > as a factor. > > As a first step to calculating it, I use the fact that the total > number of naturals with y as a factor up to and including x is given > by > > ßoor(x/y) > > like with 10 again, the naturals with 3 as a factor are > > 3, 6, 9 > > but wait, 3 gets included, but it is prime and not a composite, so > with primes I need to use > > ßoor(x/y) - 1 > > to get a count of all composites up to and including x with y as a > factor, and here y is prime. > > But, what about primes less than y? Well I can count those out with > > p(y-1,sqrt(y-1)) > Right here, in your definition for dS(x,y), you use the function > p(x,y) = ßoor(x) - S(x,y) - 1. S(x,y) is the sum of dS(x,i) where i > = every prime from 2 to y. In your definition of dS(x,y) you have a > function that uses dS(x,y) in its definition. Unless you can take > p(x,y) out of your definition of dS(x,y), it will be a circular > definition and therefore be impossible to evaluate. It's a recurrence relationship. The recursion does not prevent evaluation. Evaluating dS(x,y) from y =2 to the sqrt(x), will naturally force S(x,1) = 0 and p(x,1) = x-1 so you will not infinitely recurse. Notice that S(x,1) can be talked out as saying that there are no composites with primes less than 1 as a factor as in fact there are no primes less than 1. > which you'll notice works up to 10, as > > p(3-1,sqrt(3-1)) = p(2,sqrt(2)) = 1 > as, of course, 2 is the only prime less than 3, so I now have > > ßoor(x/y) - 1 - p(y-1,sqrt(y-1)), with y prime, > > and notice that gives me the correct count up to 10 for 3, as that > just leaves > > 9 > as 3 got subtracted off by 1, and 6 got subtracted off by the count of > primes less than 3. > > But I'm not done in general, as there may be composites now that > multiply times y that are less than x, which have primes less than y > as factors. > > For instance, up to 12 I have 2(2)(3), and what I've done so far would > still leave that in the count for dS(12,3). > Ok, just trying for clarification. So far, if we took dS(12,3) we'd > get ßoor(12/3) = 4 (Count(3, 6, 9, 12)). Subtract one for the prime, > 3 (6, 9, 12). Subtract another for p(2,sqrt(2)), and we get 2 (9 and > 12 or 6 and 9). Yup, and 2(2)(3) is still left after the first corrections, though it has 2 as a factor but 2 is, of course, a prime less than 3, so more work is needed. > But I have a way to handle that possibility! > > I simply use S(x/y, y-1) as the y-1 tells it to only count primes less > than y, and it's counting the composites which can multiply times y > and still be less than x, and now I'm nearly done as I get > Here you use another function of dS(x,y) to define dS(x,y). Also, I > know it doesn't matter for the function, but it's just more clear if > you say S(ßoor(x/y),y-1). That could be an alternate way to go, I'll admit. I just like p(x,y) = ßoor(x) - S(x,y) - 1 as the one place where ßoor() shows up. As for the recursion, as I explained above, it's not a problem. In case the recursion makes you really nervous you can look at some of the explicit forms of the dS(x,y) function. For instance, if x=N, where N is an even natural, then dS(N,2) = N/2 - 1 dS(N,3) = ßoor((N-4)/6) (to see calculation go to my blog http://mathforprofit.blogspot.com/ and look for tidbits which will be near bottom) and explicit forms exist for every prime out there in idea space you could say. But the recursion can be handled anyway by noting that S(x,1) = 0. > So I finally have a working dS function: > > dS(x,y) = (p(x/y,y-1) - p(y-1,sqrt(y-1))(p(y,sqrt(y)) - > p(y-1,sqrt(y-1))) > > and to help readability I usually use brackets: > > dS(x,y) = [p(x/y,y-1) - p(y-1,sqrt(y-1)][p(y,sqrt(y)) - > p(y-1,sqrt(y-1))] > > and just like that I have my dS function rigorously defined using some > very basic algebra and a few good ideas. > James Harris > I checked the function with your examples and they all worked. > However, it's hard to know whether they work for all natural numbers > or not. With a function like this, programming the whole thing then > doing it on the computer would be faster than running through it 3 or > 4 times by hand, but I can't picture a way to program it because of > its circular nature. You can find posted programs implemented it, but I'd prefer you understand in detail from the derivation. The derivation is supposed to show that it works: why and how. Quite simply I count composites for a particular prime p, where I keep from double counting by subtracting off the count of composites that have primes less than p as a factor in two ways: 1. Subtract off the number of primes less than p 2. Subtract off the number of composites with primes less than p as a factor That route leads to recursion, which can definitely be handled, and it stops at S(x,1) = 0, as there are no composites with primes less than 1 as a factor as there are no primes less than 1. If something is not clear then I'd be happy to expand out on the derivation. > If worst comes to worst, this will still be very strong, though. The > ultimate goal is an independent function that will give you a list of > primes without having to know any previous primes and without huge > factoring. Although your circular definitions mean that you don't > quite have that, if you reworked your equations you'd very likely have > a way to find primes through iterations using previous primes, which > would be still be an amazing accomplishment. What you call a circular definition is technically called a partial difference equation. And you don't need previous primes. My prime counting function finds them on its own, which is one of those obvious facts that sci.math posters dismiss and I wonder about the mind-control. Like, that feature is really neat, never been seen before, and clearly proves that my work is important, but sci.math posters would just claim it's useless and stupid! They did that for YEARS. So you have a function that recurses to automatically select out only prime numbers, something never before seen in the math literature, bme probability > as getting into Physics (1/3). Probability that he will get into both > Calc I and CS = 1/6. Probability that he will get into both CS and > Physics is 1/6. Find the prob that he will be able to get into at > least one of the three classes. > I think the answer goes like this. > P(AuBuC)=P(a) + P(b) + P(c) - P(a Intersect B) - P(a intersect c) - > P(b intersect c) + P(a intersect b intersect c). > Therfore, P(a) = 1/2 , Prob of getting into calc I > P(b) = 1/3 , Prob of getting into CS > P(c) = 1/3, Prob of getting into Physics > P(a intersect b) = 1/6 > P(b intersect c) = 1/6 To use the formula, you also need to know P(a intersect c) and P(a int. b int. c), which of course are both 0 because of the scheduling conßict. I think that you probably realized this, but your solution would be clearer if you actually said it ... > So, 1/2 + 1/3 + 1/3 - 2/6 = 5/6 P(a U b U c) = 1/2 + 1/3 + 1/3 - 1/6 - 1/6 - 0 + 0 = 5/6. Otherwise, everything's fine. Brian === Subject: Please Help with this problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7U5XBp22906; I am sure this is not a hard problem for the smart people here. However, I am not one of those smart people so I would really appreciate some help if you don't mind. Here is the question... A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed? Trent === Subject: Re: Please Help with this problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UG2GK09438; === Subject: Re: Please Help with this problem >I am sure this is not a hard problem for the smart people here. >However, I am not one of those smart people so I would really >appreciate some help if you don't mind. >Here is the question... >A farmer with 2000 meters of fencing wants to enclose a rectangular >plot that borders on a straight highway. If the farmer does not fence >the side along the highway, what is the largest area that can be >enclosed? >Trent I won't tell you the answer. In fact I don't think people should just tell you the answer. But if you get a good hint then you can often figure out the answer for yourself, and that you can use the next time a problem appears. so I'll begin with simple things and you can see where it helps. It doesn't say but you are almost certainly expected to have two sides the same length and have both those at right angles to the edge of the highway. Then the third side will perhaps be of a different length but it will be parallel to the highway. Now you already have two different lengths and an area that is going to depend on those two lengths. That means at least two variables. And working with one variable is usually easier than working with two, or more. So, first task, can you come up with a formula for the area that only involves one variable? You need to make use of the other information given in the problem. But coming up with that formula is a key step at this point. It will involve the formula for the area of a rectangle and to use that will involve the lengths and needs, probably, to only have one variable in it, other than area. Then, once you have tried and figured out a formula, you might want to check it. Try to think of a way to check it. Once you have checked it then stop thinking about fence for a moment and just look at the equation, to see if it looks like any kind of equation that you might have seen, something you might be able to get a maximum for. Or can you perhaps rearrange it to turn it into something where you could do that. That should be enough of a hint for you to discover how to do this. I hope it works out === Subject: Re: Please Help with this problem > I am sure this is not a hard problem for the smart people here. > However, I am not one of those smart people so I would really > appreciate some help if you don't mind. > Here is the question... > A farmer with 2000 meters of fencing wants to enclose a rectangular > plot that borders on a straight highway. If the farmer does not fence > the side along the highway, what is the largest area that can be > enclosed? Draw a sketch. Let the two equal sides (perpendicular to the road) each be Ôx.' The length of the third side is therefore: (total fencing available) - (fencing already used) 2000 - 2x A = L*W = x(2000 - 2x) = 2000x - 2x^2 Proceed to find x that maximizes A, then evaluate A for that x. When you get bored doing these, generalize. What can be said about any rectangle that maximizes area given a certain perimeter? P = 2L + 2W Solve for one of L or W, in terms of the other: P-2L = 2W (P-2L)/2 = W Plug that into the area formula: A = LW = L(P-2L)/2 Find the L that maximizes A (remember P is constant.) You should get L=P/4 which implies a square. But most farmers already know the rectangle enclosing the maximum area will be a square (they are smarter than most think.) Already have one natural Ôside' such as a highway, river, etc? No prob. just make one side twice as long as the other two (ie take the fencing you would have used for the fourth side and give it to opposite side, such that L=P/4 and W=P/2). -- Darrell === Subject: Re: Please Help with this problem > A farmer with 2000 meters of fencing wants to enclose a rectangular > plot that borders on a straight highway. If the farmer does not fence > the side along the highway, what is the largest area that can be > enclosed? If the sides of the rectangle are x and y, and y is the one along the highway, then the fence length is 2x + y and the area is x*y. You need to find the maximum of A = x*y given the constraint 2x + y = 2000. If you have learned about Lagrange multipliers, then this is a very simple problem. If you haven't, then you can express y in terms of x and substitute to get A = x(2000 - 2x) = 2000x - 2x^2, for which you can then find a maximum using derivatives. Finally, if you haven't learned derivatives, simply use the property of a quadratic that its extremum is halfway between its zeros; from A = x(2000 - 2x), the zeros of this one are clearly 0 and 1000. hth meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Re: Please Help with this problem > I am sure this is not a hard problem for the smart people here. > However, I am not one of those smart people so I would really > appreciate some help if you don't mind. > Here is the question... > A farmer with 2000 meters of fencing wants to enclose a rectangular > plot that borders on a straight highway. If the farmer does not fence > the side along the highway, what is the largest area that can be > enclosed? > Trent The area of a rectangle of sides x and y is x*y. Right? If y is the side the borders on the highway what is the formula for the amount of fencing? Are you taking calculus? Do you know how to find maximums? Bill === Subject: Question of finite element of a group If x(=/= identity) is in a group G and ord(x) is finite, Then, this fact imply G is a finite group ? If it is true then, Would you please give me some hint for proving that? === Subject: Re: Question of finite element of a group > If x(=/= identity) is in a group G and ord(x) is finite, > Then, this fact imply G is a finite group ? > If it is true then, No, it is not true. Counterexample: Z x Z/2Z. J. === Subject: Re: Question of finite element of a group > If x(=/= identity) is in a group G and ord(x) is finite, > Then, this fact imply G is a finite group ? > If it is true then, > No, it is not true. Counterexample: Z x Z/2Z. > J. You mean Z x Z/2Z ~= Z x Z_2 ? If then, What element is finite in Z x Z_2 - {(0,0)}? I don't know such element (a,b) in Z x Z_2 - {(0,0)}. Sorry, if this question is little stupid. === Subject: Re: Question of finite element of a group >If x(=/= identity) is in a group G and ord(x) is finite, >Then, this fact imply G is a finite group ? >If it is true then, >No, it is not true. Counterexample: Z x Z/2Z. >J. > You mean Z x Z/2Z ~= Z x Z_2 ? > If then, > What element is finite in Z x Z_2 - {(0,0)}? > I don't know such element (a,b) in Z x Z_2 - {(0,0)}. Z/2Z is the finite group with two elements 0 and 1 with 1+1=0. It coincides with the quotient of Z with respect to the subgroup 2Z. In Z x Z/2Z, (0,1) is an element of order 2, whereas (1,0) has infinite order. === Subject: Re: Question of finite element of a group > > If x(=/= identity) is in a group G and ord(x) is finite, > > Then, this fact imply G is a finite group ? > No, it is not true. Counterexample: Z x Z/2Z. > You mean Z x Z/2Z ~= Z x Z_2 ? Z_2 is defined as Z/2Z. > What element is finite in Z x Z_2 - {(0,0)}? Surly you mean What element has finite order in ... (0,1) > I don't know such element (a,b) in Z x Z_2 - {(0,0)}. > Sorry, if this question is little stupid. Yah. Here's a harder one. If every element in a group has finite order is the group finite? === Subject: Re: Question of finite element of a group days. My association with the Department is that of an alumnus. >Yah. Here's a harder one. >If every element in a group has finite order > is the group finite? Not even true if you also require the group to be finitely generated, though it is true if every element is of exponents 2, 3, 4, or 6. This is related to ut not only do mathematicians dismiss it as unimportant, posters from the sci.math newsgroup specifically attack that unique feature as useless hundred years old!!! I think my research is neat, fascinating, and clearly something worth recording for posterity, but over the last two years when I've talked about it on newsgroups I've been insulted and my work dismissed, and when I talk to mathematicians, I've been told it's of no interest or just ignored. I think they're weird. And mathematicians aren't acting at all like the Hollywood version. They're not quite acting like curious human beings. They're being strangely incurious. It's bizarre. It's like something is missing inside their heads as I'd think that they'd be excited about more math to play with, but instead, like on sci.math, they get angry and abusive. James Harris === Subject: Re: Explaining the prime distribution >The prime distribution is important for many reasons and supposedly >mathematiicans have some interest in it, but I'm going to show you in >this post a rather fascinating and basic derivation of one of the most >fascinating functions in mathematical history, which mathematicians >have for some reason fought against recognizing. >no, it was recognized by mathematicians 200 years ago. >Since there are physicists doing research on the prime distribution >and what they see as a connection to the physics world, >i could be wrong, but my impression is that physicists don't >actually think this distribution has anything to do with >physics - rather some feel that certain methods that people >wish they could make woek might also be useful in physics, >-if- they could be made to work. You raise an excellent point! What DOES any of this have to do with physics? === Subject: Re: Explaining the prime distribution > > ... > > > > I found this route to counting primes over two years ago > > Legendre found it two hundred years ago... > I'll answer this post to show you how sci.math posters so diligently > lie about my work, and in fact Christian Bau was on sci.math over two > years ago when I first started talking about my prime counting > function, he lied about it then, and he's still lying about it now. You are a complete idiot. That formula that you are so proud of _is_ a trivial variation of Legendre's formula. See: L. E. Dickson, History of the Theory of Numbers, Chelsea Publ. Co. (Reprint), Volume 1, Chapter XVIII. You may also have a look at the paper Computing pi (x): The Meissel-Lehmer Method by Lagarias, Miller and Odlyzko. There your formula can be found on the very first page and it is called not of immediate utility in calculating pi (x) because it involves 6/pi^2 * (1 - log 2) x nonzero terms. Your formula can be explicitely seen as formula 2.4 on page 8 of that paper, which is still the historical part in which they describe outdated methods. The ratio of your ego to your capabilities seems to be around 1000 to 1. By a strange coincidence, that is also how much faster my implementation that can be found at www.cbau.freeserve.co.uk is compared to your fastest implementation so far. === Subject: Re: Explaining the prime distribution > you must be new here: > nobody disputes that the Ôfunction' works [that is, > the code james has written correctly counts primes], > not that one would necessarily be able to see why > this is so from the explanation above. > the Ôdispute' is over the mathemaical significance. > it's claimed that it's a brand new method, when in Well, I'm disputing whether the function works - it doesn't, yet. But I can't believe you would dispute the mathematical significance if it would work! A function that outputs prime numbers, with no factoring, and no need for previous primes!? That would be the biggest breakthrough in number theory since Pascal's Triangle. Look, his function doesn't work. And my follow-up comment wasn't an attempt to show that it did. It was my opinion on his function. But if his function did work like he wants it to, it would be so significant that saying it wouldn't makes me question your understanding of what he's saying. === Subject: Re: Explaining the prime distribution > you must be new here: > nobody disputes that the Ôfunction' works [that is, > the code james has written correctly counts primes], > not that one would necessarily be able to see why > this is so from the explanation above. > the Ôdispute' is over the mathemaical significance. > it's claimed that it's a brand new method, when in >Well, I'm disputing whether the function works - it doesn't, yet. But >I can't believe you would dispute the mathematical significance if it >would work! A function that outputs prime numbers, with no factoring, >and no need for previous primes!? what gave you the idea that it does that? it -counts- primes. >That would be the biggest >breakthrough in number theory since Pascal's Triangle. >Look, his function doesn't work. the description in this thread may well not work - the code he's given -does- work [for counting primes, not generating them.] > And my follow-up comment wasn't an >attempt to show that it did. It was my opinion on his function. But >if his function did work like he wants it to, it would be so >significant that saying it wouldn't makes me question your >understanding of what he's saying. uh, right. read his post again, then get back to us on who's misunderstanding what. [come to think of it, maybe the problem is you think pi[n] is the n-th prime? it's not, pi[n] is -standard- notation for the number of primes <= n.] ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Explaining the prime distribution > > you must be new here: > > nobody disputes that the Ôfunction' works [that is, > the code james has written correctly counts primes], > not that one would necessarily be able to see why > this is so from the explanation above. > > the Ôdispute' is over the mathemaical significance. > it's claimed that it's a brand new method, when in > Well, I'm disputing whether the function works - it doesn't, yet. But > I can't believe you would dispute the mathematical significance if it > would work! A function that outputs prime numbers, with no factoring, > and no need for previous primes!? That would be the biggest > breakthrough in number theory since Pascal's Triangle. No, it doesn't output prime numbers. It calculates the number of primes <= N without determining the actual primes, which is a nice little trick that Legendre figured out 200 years ago. > Look, his function doesn't work. And my follow-up comment wasn't an > attempt to show that it did. It was my opinion on his function. But > if his function did work like he wants it to, it would be so > significant that saying it wouldn't makes me question your > understanding of what he's saying. === Subject: single word for enquiring sequence order A small matter, may be it has some mathematical relevance: There appears no single word as an interrogative pronoun when enquiring about sequential order. E.g., in: Answer Question She came first in the race. In what place did she come? She is the second issue in the family. What is her birth rank? He was seated in the third row. Which position did he occupy? He cleared eight feet jump in a In which attempt did he clear? second attempt. Generally we ask what, which? or which one of? etc. when asking to select or point out a particular element in a set, but these words are not used specifically to describe consecutive order. Such an enquiring word may shorten a sentence and could be useful in faster speech/chat. Comments for a new word? === Subject: HELP - most important in Algebraic Geometry I posted this request before, but did not get any reply. I am a graduate student starting on my research in Algebraic Geometry. I should really be thankful to the kind soul who can give me a list of important research papers that are a must-read. A list of dozen or so classical and modern papers (preferrably in English) would be wonderful. Leo === Subject: Re: 3D Plane by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7GFi2018958; >I have a 3d plane, and have worked out its normal. I want to know the >distance from (x,y,z) to where z intersects the plane. i.e. x and y will > be the same value, but z falls on the plane. A minor nitpick; You say 3d plane, but in fact you really are working with a 2d plane embedded in E^3. This mistake seems to be carried along by others in their responses. Indeed, one post referred to a Ô3d' line. AFAIK, a line is always 1-dimensional. It might be embedded in E^3, but it is still 1 dimensional..... === Subject: Re: books for self study, number theory, set theory, analysis... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7GFi2V18921; >For number theory, it is difficult to make any sense of it without >understanding how it evolved. So make sure you get a book that treats >number theory in its historical context. >For example, >O. Ore: Number theory and its history >A. Weil: Number theory: an approach through history >H.M. Edwards: Fermat's Last Theorem (A genetic introduction to >algebraic number theory) A superb (IMO) book that treats things from a historical persepective and is the often overlooked: D. Shanks Solved & Unsolved Problems in Number Theory. BTW, Dan used to call me Agman. Think about it..... === Subject: How to eliminate scale factors on a pair of matrices.... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7GFi2B18941; I'm stuck with the following problem that I can't solve. I consider two 3-by-3 (real) matrices defined as B1 = s1*H*A1*H^T B2 = s2*H*A2*H^T where s1 and s2 are non zero real numbers,H is a full-rank 3-by-3 (real) matrix (H^T denotes its transpose) and A1 and A2 are two 3-by-3 (real) matrices such that rank(A1)=3 and rank(A2)<=3. I'm seeking some assumptions on B1, B2 that will ensure that s2/s1 equals a known quantity ie, only depending on A1 and A2. When rank(A2) = 3, it suffices to assume that det(B1) = det(B2) = 1 so det(inv(B2))*det(B1) = (s1/s2)^3*det(A1)/det(A2) = 1 yields (s1/s2)^3 = det(A2)/det(A1) When rank(A2)<=3, I could not think of any assumption on B1, B2 that will help me to solve the problem in a similar way. === Subject: Re: Atheist MorituriMax > Our great Einstein wanted to know God; how He created the universe; > the rest were just details. > Mitch Raemsch http://www.skeptic.com/archives50.html I believe in Spinoza's God who reveals himself in the orderly harmony of what exists, not in a God who concerns himself with fates and actions of human beings. -- Einsetin alex === Subject: Re: Atheist MorituriMax > Our great Einstein wanted to know God; how He created the universe; > the rest were just details. > Mitch Raemsch > http://www.skeptic.com/archives50.html > I believe in Spinoza's God who reveals himself in the orderly harmony of what > exists, not in a God who concerns himself with fates and actions of human beings. > -- Einsetin > alex More folks than you know are a theist, Alex... what's that have to do with physics? See: http://www.csicop.org/si/9907/ === Subject: Re: Projective geometry >Projective geometry is something I know next to nothing about. > Neither do I. >I have seen the construction of RP^2 by identifying antipodes >on S^2. The point that always confuses me is of course the equator. > S^2 is the surface of a unit sphere. S^1 a circle. > Is RP^2 given a verbal name or intuitive description? RP^n: Real Projective n-space (for n=1: Real projective line, for n=2: real projective plane): RP^n is the quotient of R^(n+1){0} under the action of the multiplicative group R* = R {0}. RP^n is the space of lines in R^(n+1) passing through the origin, or equivalently the space of 1-dimensional linear subspaces. >I also wothe Burnside Problem. For sufficiently high prime p, there exist finitely generated infinite groups such that ever proper subgroup is either cyclic of order p, or infinite. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Question of finite element of a group > Yah. Here's a harder one. > If every element in a group has finite order > is the group finite? No, let F_n the field with p^n elements (p prime) - clearly of characteristic p. Then F_n is contained in F_{n+1}. Let F be the union of all F_n's. Then every non-zero element in F has finite order (w.r.t. to multiplication), but Fsetminus {0} is not finite. J. === Subject: Re: Question of finite element of a group > Yah. Here's a harder one. > If every element in a group has finite order > is the group finite? > No, let F_n the field with p^n elements (p prime) - clearly of > characteristic p. Then F_n is contained in F_{n+1}. Let F be the union > of all F_n's. Then every non-zero element in F has finite order (w.r.t. > to multiplication), but Fsetminus {0} is not finite. > J. Do you mean F_n is a subfield of F{n+1}? Hanford === Subject: Re: Question of finite element of a group > Yah. Here's a harder one. > If every element in a group has finite order > is the group finite? > No, let F_n the field with p^n elements (p prime) - clearly of > characteristic p. Then F_n is contained in F_{n+1}. Let F be the union > of all F_n's. Then every non-zero element in F has finite order (w.r.t. > to multiplication), but Fsetminus {0} is not finite. A simpler construction is the subgroup of G = Z_2 x Z_3 x Z_5 x ... of elements of finite support, ie elements (a2, a3, a5, ... ) for which all but finitely many a's are nonzero. === Subject: Re: Question of finite element of a group by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UFHPt05054; > Yah. Here's a harder one. > If every element in a group has finite order > is the group finite? > No, let F_n the field with p^n elements (p prime) - clearly of > characteristic p. Then F_n is contained in F_{n+1}. Let F be the union > of all F_n's. Then every non-zero element in F has finite order (w.r.t. > to multiplication), but Fsetminus {0} is not finite. >A simpler construction is the subgroup of > G = Z_2 x Z_3 x Z_5 x ... >of elements of finite support, ie elements > (a2, a3, a5, ... ) >for which all but finitely many a's are nonzero. In other words, the direct sum of Z_2, Z_3, etc. The prime orders of the cyclic groups are of course irrelevant. Snappier example: Q/Z. There are *lots* of torsion groups after all. Todd Trimble === Subject: Re: Question of finite element of a group > > > If every element in a group has finite order > > is the group finite? > >A simpler construction is the subgroup of > G = Z_2 x Z_3 x Z_5 x ... >of elements of finite support, ie elements > (a2, a3, a5, ... ) >for which all but finitely many a's are nonzero. This was corrected to read, for which only finitely many a's are nonzero. > In other words, the direct sum of Z_2, Z_3, etc. The prime orders > of the cyclic groups are of course irrelevant. That will not do, as the identity (1,1,1,... ) has infinite order. It has to be a subset of the direct sum to avoid such elements, ie, elements (a2,a3,... ) where all but finitely many a's are zero. > Snappier example: Q/Z. There are *lots* of torsion groups after all. Indeed, but not R/Z. === Subject: Re: Question of finite element of a group Correction below. > > If every element in a group has finite order > > is the group finite? > No, let F_n the field with p^n elements (p prime) - clearly of > characteristic p. Then F_n is contained in F_{n+1}. Let F be the union > of all F_n's. Then every non-zero element in F has finite order (w.r.t. > to multiplication), but Fsetminus {0} is not finite. > A simpler construction is the subgroup of > G = Z_2 x Z_3 x Z_5 x ... > of elements of finite support, ie elements > (a2, a3, a5, ... ) > for which all but finitely many a's are nonzero. for which only finitely many a's are nonzero. === Subject: Re: i need a riddle solved!!! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UBmoI18259; >3 candidates were told that they would be blindfolded and that a hat >would be placed on the head of each one. Each hat could be either >black or white. At a signal, the blindfolds would be removed and each >canidate could see the hats of the other 2. Anyone who saw a black hat >was to raise his or her hand. Once the blindfolds had been put on, all >three were given black hats. When the blindfolds were removed all >three raised their hands. If you were one of the 3 candidates, would >you be able to determine the color of your hat. How would you know? >You may assume that the other 2 candidates are reasonably bright >people. This is a classic. Think of it like this: suppose you hadn't said all three were given black hats, and suppose you had a white hat, and all three candidates raised their hands. Now visualize the situation from the perspective of a black-hat wearer. Since s/he is bright, what would s/he deduce almost immediately? (To make the problem more compelling, you should have said these are candidates for a job and the first to say which color hat s/he is wearing gets the job.) If s/he didn't say immediately, what would you deduce? Todd Trimble === Subject: Re: water pelican === Subject: subbasis for the topology of a set I'm trying to prove the following: If X is a set and G an arbitrary set of parts of X, there is exactly one topology T(G) on X such that G is a subbasis for T(G) (the topology``generated'' by G). Proof: Let X be a topological space. Let G = {} so that G is in X. G is open by the definition of topological space. Claim G is a subbasis for T(G). Now, we must show that every open set of G is the union of finite intersections of sets in G. Since G = {}, {} is in G. So {} / {} is in G. G is a subbasis for T(G) because since G = {}, there's no other set that we could try to intersect in G. Now we must show that T(G) is the only topology on X. By way of contradiction, assume that there is a T'(G) on X such that T'(G) is different than T(G). Let a, b be two open subsets of G. a U b is in G, a / b is in G. a U b and a / b must both be open. Specifically, since G = {}, then a = b = {}. Now, I don't really know how to show that T(G) = T'(G), although they must be because G = {}. I would try showing that all axioms hold for both, but that only says that that they are topologies; how would show that they are the same topology? Perhaps I couldn't have let G = {}. Advices are welcome. === Subject: Re: integrals of y = 1 / (1 + x^n) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UGfEI12906; >I wanted to calculate the area 0 --> +infinite below the functions >like 1 / (1 + x^2), changing the exponent of x to integers larger than >if n = 2, the area is = pi/2 >if n = 3, the area I calculated is (pi / (2 * sqrt(3))) (correct?) >I tried several times to calculate the integral for n = 4, but always >got stumped. >Anyone knows the integral for n = 4? Or just the area value? Or, even >better, knows of a webpage with integrals and areas for n = 2 to 10 ?? >(the language I studied maths in, is not English...) >Roberto N. I'll assume you are using methods of elementary calculus to do these problems. I got 2pi/(3*sqrt(3)) for the case n = 3 (the arctan term in the indefinite integral is (1/sqrt(3))arctan((x - 1/2)/(sqrt(3)/2)); the limit at infinity produces your pi/(2sqrt(3)), and the value at 0 is -pi/(6sqrt(3)). Subtract to get my answer. Did you factorize 1 + x^4 into quadratic factors? It may not be obvious without using complex numbers (basically, find the complex roots of x^4 = -1, using (cos t + i sin t)^n = cos nt + i sin nt; here n = 4, and nt = pi + multiples of 2pi gives -1 on the right). One gets (x^2 - sqrt(2)x + 1)(x^2 + sqrt(2)x + 1) = 1 + x^4, where the first quadratic factor comes from the complex roots in the first and fourth quadrants, and the other from the roots in the second and third quadrants. It is then straightforward but tedious to work out the indefinite integral of 1/(1 + x^4); my back-of-the-envelope calculation gives (1/2sqrt(2))((1/2)ln(x^2 + sqrt(2)x + 1) - (1/2)ln(x^2 - sqrt(2)x + 1) + arctan(x + sqrt(2)/2) + arctan(x - sqrt(2)/2 ) + C which if correct gives pi/(2sqrt(2)) as the definite integral. In general, here's how x^n + 1 factorizes, based on the idea of using complex numbers as above. n odd: (x + 1)*Product_{i=1}^{(n-1)/2} (x^2 - 2(cos t_i)x + 1) n even: Product_{i=1}^{n/2} (x^2 - 2(cos t_i)x + 1) where in both cases t_i = (2i - 1)pi/n. In principle one can work out the indefinite integrals from this, but it's a slog. Better to use Mathematica or Maple or something. To get the definite integrals, it is probably more efficient to use the method of complex residues. For n even, it suffices to calculate Integral_{-inf}^{inf} 1/(1 + x^n) dx by considering a contour going from -R to R on the real axis, followed by a semicircle of radius R in the upper half-plane taken counter-clockwise. For large R, the contribution to the integral from the semicircle part is vanishingly small, so 1/(2pi)i times this integral is the sum of the residues at the poles which occur in the upper half-plane (i.e. at the roots of 1 + z^n with positive imaginary part). To save you the trouble, the answer turns out to be (2pi/n)/(sin pi/n)) i.e. for even n, your definite integral is (pi/n)/(sin (pi/n)). Cool! Todd Trimble === Subject: Re: integrals of y = 1 / (1 + x^n) > To get the definite integrals, it is probably more efficient to > use the method of complex residues. For n even, it suffices to > calculate > Integral_{-inf}^{inf} 1/(1 + x^n) dx > by considering a contour going from -R to R on the real axis, > followed by a semicircle of radius R in the upper half-plane > taken counter-clockwise. For large R, the contribution to the > integral from the semicircle part is vanishingly small, so 1/(2pi)i > times this integral is the sum of the residues at the poles which > occur in the upper half-plane (i.e. at the roots of 1 + z^n with > positive imaginary part). To save you the trouble, the answer > turns out to be > (2pi/n)/(sin pi/n)) > i.e. for even n, your definitender about RP^1. It seems that it would be formed by >identifying >opposite points on S^1, the circle--which is the part of the >RP^2 construction that I find confusing. > Because of the spherical symmetry of the construction, a circle would be > the same in any orientation. >It seems that S^1 is 1-1 with the half-open interval [0,1), >and then 1 must be identified with 0 to get S^1. Or should >one say it is 1-1 with [0,1] with 0 and 1 identified. > That's absurd, as 1 isn't in [0,1). Yes, if 0 and 1 of [0,1] are > identified, then you get S^1, the unit circle. As for cardinality, S^1 is > 1-1 with R, R^n, (0,1), [0,1]x[3,5)x(-2,1]. The points of [0,1] with 0 > and 1 identified can be constrewed as (0,1) and the identified { 0,1 }, > namely { {r}, { 0,1 } | r in (0,1) }, the partitions of [0,1] >It seems that if one identifies the points a and a + pi on S^1 >(a = angle), one just gets S^1 back again: you could collapse >the top half of the circle onto an open interval (0,1), then >identify 0 and 1 to get [0,1), the same as for S^1. > I'm inclined to agree, you get the circle. Then would not RP^2 be S^2 ? RP^1 is the circle; the canonical map from S^1 to RP^1 is a double cover (identifying antipodal points yields a 2:1 mapping). RP^2 is *not* S^2. S^2 is simply-connected, but the fundamental group of RP^2 is cyclic of order 2. >PS: I also want to look at CP^1 and compare to RP^2, and then >HP^1. > What's H ? H (for Hamilton) refers to quaternions. CP^1 is homeomorphic to S^2 (the 2-sphere). If I recall correctly, HP^1 is S^4 (the 4-sphere). Dale. === Subject: Re: How unimportant is sci.math? > I've been pointing out for a while that Usenet doesn't matter, when I > think what's clear from my research is that sci.math doesn't matter to > the world at large... > Perhaps sci.math does not matter to the world at large, and it is > certainly not where legitimate research mathematicians submit results for > peer review, but for amateurs like myself, I find it a tremendously valuable > resource. I have often submitted questions, and, within a few hours, > received many thoughtful and thorough replies. It helps if I try to pose the > question accurately, have looked elsewhere first, and show that I've given > the matter a least a little thought beforehand. Obeying a small set of > self-imposed ground rules seems to help a great deal: > 1) Display no arrogance > 2) Assume you're is misstating the question > 3) Apologize in advance for poor terminology, weakly grasped concepts, and > profuse errors > 4) Remember that if you're an amateur, and you've figured out something, the > odds are extremely high that it's been discovered before, probably a long > time ago, and the mathematician who did so had a lot more interesting things > to say than you did, proved it with a great deal more rigor, understood it > in context to the rest of mathematics, and usually went on to other > fascinating and useful results. > Big problems in math are usually resolved incrementally - standing on the > shoulders of giants - as per the Wiles FLT proof. Even a small but > legitimate contribution to say, the Riemann Hypothesis is vitally important. > If an amateur posts a solution to a hard problem, especially one that > clearly shows no attention to the history of attempts to solve said problem, > one's crank alert is likely to be triggered - and legitimately; the > solution is almost certainly in error, and the poster should express due > modesty if he/she wants genuine assistance. A good set of rules. I suggest a parallel set of rules for the expert responders: 1) Display no arrogance. 2) Restate the question, if doing so seems helpful. 3) Cut the asker a lot of slack about terminology, imperfectly grasped concepts, and profuse errors. 4) An amateur who makes an independent discovery, no matter how modest, has already gone far beyond the majority of students, who limit themselves to passively imbibing what Teacher says. Give him or her credit for this. -- Chris Henrich God just doesn't fit inside a single religion. === Subject: Re: How unimportant is sci.math? > I've been pointing out for a while that Usenet doesn't matter, when I > think what's clear from my research is that sci.math doesn't matter to > the world at large... > Perhaps sci.math does not matter to the world at large, and it is > certainly not where legitimate research mathematicians submit results for > peer review, but for amateurs like myself, I find it a tremendously valuable > resource. I have often submitted questions, and, within a few hours, > received many thoughtful and thorough replies. It helps if I try to pose the > question accurately, have looked elsewhere first, and show that I've given > the matter a least a little thought beforehand. Obeying a small set of > self-imposed ground rules seems to help a great deal: > 1) Display no arrogance > 2) Assume you're is misstating the question > 3) Apologize in advance for poor terminology, weakly grasped concepts, and > profuse errors > 4) Remember that if you're an amateur, and you've figured out something, the > odds are extremely high that it's been discovered before, probably a long > time ago, and the mathematician who did so had a lot more interesting things > to say than you did, proved it with a great deal more rigor, understood it > in context to the rest of mathematics, and usually went on to other > fascinating and useful results. > Big problems in math are usually resolved incrementally - standing on the > shoulders of giants - as per the Wiles FLT proof. Even a small but > legitimate contribution to say, the Riemann Hypothesis is vitally important. > If an amateur posts a solution to a hard problem, especially one that > clearly shows no attention to the history of attempts to solve said problem, > one's crank alert is likely to be triggered - and legitimately; the > solution is almost certainly in error, and the poster should express due > modesty if he/she wants genuine assistance. > Barnaby Very well said... and completely lost on JSH, of course === Subject: Re: Rado's Sigma and the Halting Problem for Programs >function. Functions don't /have/ solutions. It's like Shaquille > Function SolveForSqrRoot(N: Extended) : Extended; > Begin > Result:=Sqrt(N); > end; >Arthur is right about mathematical functions not having solutions. Very >informally, any function f is just a set of ordered pairs > { (a[0],b[0]), (a[1],b[1]), (a[2],b[2]), ... } >where we _define_ the notation f(a[i]) to mean b[i]. So given an a[i], it >doesn't quite make sense to say b[i] is the Ôsolution' to f(a[i]), because >f(a[i]) is really just another way of writing b[i]. >Although you've mentioned you don't want to get into too much math, to >really get the distinction you'll want to take a look at >http://en.wikipedia.org/wiki/Function_(mathematics) I suggest you read these 2,000+ examples of functions having solutions: Larry === Subject: Re: Rado's Sigma and the Halting Problem for Programs Distribution: inet > function. Functions don't /have/ solutions. It's like Shaquille [...O'Neal, saying something stupid.] > Arthur is right about mathematical functions not having solutions. [...Explanation snipped.] > Although you've mentioned you don't want to get into too much math, to > really get the distinction you'll want to take a look at > http://en.wikipedia.org/wiki/Function_(mathematics) > I suggest you read these 2,000+ examples of functions having solutions: You know, trolling doesn't make you look smart. Michael and I have been politely explaining that you made a little terminological slip-up, and what the correct terminology is. The proper response would be been more like Naw, math sucks! I like to contradict people! Not too hot, you know? If you'd like to start contributing /productively/ to the thread again, I'll be waiting. -Arthur, signing off === Subject: Hyperbolic geometry: loci of points that see a segment with a given angle Hi! In Euclidean Geometry, given a segment AB and an angle alpha, the set of points P such that the angle APB = alpha is an arc of circle. What is this loci in hyperbolic geometry? === Subject: Re: The most elegant method to prove an identity with binomial coefficients === Subject: Re: What is a differential field? > You lost me here. I understand connected in the topological case, but > from where's the topology when working with discrete finite fields? And > what do you mean by extension? Some sort of group product? I went a-googling and found this URL: in which group extension is defined. Let G be a group, and N a normal subgroup; let K be (isomorphic to) G/N; then G is an extension of N by K. -- Chris Henrich The total lack of evidence is the surest sign that the conspiracy is working. === Subject: Re: What is a differential field? | | | |> : What are algebraic groups exactly? Is that some fancy way of saying |> : matrix groups? What intrinsic property makes them algebraic? |> An algebraic group is an affine algebraic variety (set of solutions to a |> finite number of polynomials over a finite number of variables, over some |> field) together with a group structure, where multiplication and inversion |> are rational maps. |> There are two very different kinds of algebraic groups: |> (1) matrix groups, which are subgroups of GL(n,F) defined by polynomials |> in |> the matrix entries. These are also called linear algebraic groups. The |> differential Galois group of a Picard-Vessiot extension is always a |> linear algebraic group (forgot to mention that in the previous post). | |I've only recently learned what an affine algebraic variety is. So let me |see if I can prove that GL(n,F) satisfies the requirements to be one. |First, n x n matrices are elements of a n^2 dimensional vector space over |F, which is an affine space. So GL(n,F) would certainly be an affine |variety, since it is embedded in an affine space. The condition for a |matrix M to be in the group is GL(n,F) is that det(M) != 0. Interesting, |this is an inequality and not an equation. Does that qualify GL(n,F) to |be a variety at all? SL(n,F) would definitely qualify, since the only |condition on its elements is det(M)=1. | |About rationality of group operations. If M=NP, then the matrix elements |of M are certainly rational (polynomial even) functions of matrix elements |of N and P. To get the inverse we can use the formula M^(-1) = M*/det(M), |where M* is the matrix adjoint, hence inversion is also rational. |Everything checks out here. However I'm a little bothered by the |requirement of rationality here. I've been indoctrinated by modern |differential geometry that intrinsic geometric properties/notions must be |coordinate independent. Yet rationality is always defined with respect to |some coordinates (variables). Granted that rationality here is invariant |under linear transformations of the affine space in which the algebraic |group is embedded, but I would still like to see some sort of intrinsic |condition on the group that shows it to be an algebraic group. the vague impression that i have about this is that a lie group's being algebraic (i think affine algebraic in the strict sense actually, meaning that the group operation is not just rational but polynomial) is secretly very closely related to the representation theory of the group. if the representation theory of the group is sufficiently nice, meaning basically that the group has enough representations on finite-dimensional vector spaces, then the so-called representative functions on the group will form an affine coordinate ring making the group into an algebraic group. a representative function on a group is a polynomial function of the matrix by which a group element is represented under a finite-dimensional representation of the group. the nice trick that shows that the representative functions on a group form a (commutative) ring is that the sum of representative functions is representative because you can take the direct sum of group representations, while the product of representative functions is representative because you can take the tensor product of group representations. -- [e-mail address jdolan@math.ucr.edu] === Subject: Algebraic groups (was Re: What is a differential field?) > |I've only recently learned what an affine algebraic variety is. So let me > |see if I can prove that GL(n,F) satisfies the requirements to be one. > |First, n x n matrices are elements of a n^2 dimensional vector space over > |F, which is an affine space. So GL(n,F) would certainly be an affine > |variety, since it is embedded in an affine space. The condition for a > |matrix M to be in the group is GL(n,F) is that det(M) != 0. Interesting, > |this is an inequality and not an equation. Does that qualify GL(n,F) to > |be a variety at all? SL(n,F) would definitely qualify, since the only > |condition on its elements is det(M)=1. > |About rationality of group operations. If M=NP, then the matrix elements > |of M are certainly rational (polynomial even) functions of matrix > |elements of N and P. To get the inverse we can use the formula M^(-1) = > |M*/det(M), where M* is the matrix adjoint, hence inversion is also > |rational. Everything checks out here. However I'm a little bothered by > |the requirement of rationality here. I've been indoctrinated by modern > |differential geomet integral is (pi/n)/(sin (pi/n)). > Cool! It's easier to use the contour 0 -> R -> R*exp(2pi/n) -> 0. Then you only enclose one pole, and the last leg of the integral is a multiple of the one we want. === Subject: Re: integrals of y = 1 / (1 + x^n) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UM3UZ09638; Trimble) > To get the definite integrals, it is probably more efficient to > use the method of complex residues. For n even, it suffices to > calculate > Integral_{-inf}^{inf} 1/(1 + x^n) dx > by considering a contour going from -R to R on the real axis, > followed by a semicircle of radius R in the upper half-plane > taken counter-clockwise. For large R, the contribution to the > integral from the semicircle part is vanishingly small, so 1/(2pi)i > times this integral is the sum of the residues at the poles which > occur in the upper half-plane (i.e. at the roots of 1 + z^n with > positive imaginary part). To save you the trouble, the answer > turns out to be > (2pi/n)/(sin pi/n)) > i.e. for even n, your definite integral is (pi/n)/(sin (pi/n)). > Cool! >It's easier to use the contour 0 -> R -> R*exp(2pi/n) -> 0. Then you only >enclose one pole, and the last leg of the integral is a multiple of the one >we want. Todd === Subject: Re: integrals of y = 1 / (1 + x^n) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UFgCg07311; >I wanted to calculate the area 0 --> +infinite below the functions >like 1 / (1 + x^2), changing the exponent of x to integers larger than >if n = 2, the area is = pi/2 >if n = 3, the area I calculated is (pi / (2 * sqrt(3))) (correct?) >I tried several times to calculate the integral for n = 4, but always >got stumped. >Anyone knows the integral for n = 4? Or just the area value? Or, even >better, knows of a webpage with integrals and areas for n = 2 to 10 ?? >(the language I studied maths in, is not English...) >Roberto N. inf dx/(1 + x^n) = .a6/(nsin(.a6/n)) 0 n = 2: = .a6/2 = 0.5.a6 n = 3: = .a6/(3sin(60)) = = 2.a6/3^(3/2) = 0.38490....a6 n = 4: = .a6/(4sin(45)) = = .a6/(2sqrt(2)) = 0.35355....a6 n = 5: = .a6/(5sin(36)) = = 2.a6/5^(5/4)sqrt[(sqrt(5)+1)/2] = 0.34026....a6 n = 6: = .a6/(6sin(30)) = .a6/3 = 0.33333....a6 etc. n -> inf: -> 1 === Subject: Re: integrals of y = 1 / (1 + x^n) >I wanted to calculate the area 0 --> +infinite below the functions >like 1 / (1 + x^2), changing the exponent of x to integers larger > than >2. >if n = 2, the area is = pi/2 >if n = 3, the area I calculated is (pi / (2 * sqrt(3))) (correct?) >I tried several times to calculate the integral for n = 4, but always >got stumped. >Anyone knows the integral for n = 4? Or just the area value? Or, even >better, knows of a webpage with integrals and areas for n = 2 to 10 > ?? >(the language I studied maths in, is not English...) >Roberto N. > inf > dx/(1 + x^n) = .a6/(nsin(.a6/n)) > 0 ??? It's best to use simply ASCII. Otherwise, you risk illegibility. Just in case the result above is illegible to someone else: The character .a6 above should be thought of as representing pi. Doing so, the result is correct. David > n = 2: = .a6/2 = 0.5.a6 > n = 3: = .a6/(3sin(60)) = > = 2.a6/3^(3/2) = 0.38490....a6 > n = 4: = .a6/(4sin(45)) = > = .a6/(2sqrt(2)) = 0.35355....a6 > n = 5: = .a6/(5sin(36)) = > = 2.a6/5^(5/4)sqrt[(sqrt(5)+1)/2] = 0.34026....a6 > n = 6: = .a6/(6sin(30)) > = .a6/3 = 0.33333....a6 > etc. > n -> inf: -> 1 === Subject: Re: Maths Newsgroups by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UHEiL16085; pls send me the past qustion for math === Subject: taylor series expansion by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UIY8k23548; Can anyone help me with this. taylor series expansion of sqrt((x+y)^2 +z^2)) === Subject: Re: taylor series expansion > Can anyone help me with this. > taylor series expansion of sqrt((x+y)^2 +z^2)) With respect to which variable? With what constraints on the others? Or do you mean a multi-variable expansion? === Subject: confusing riddle by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UMQO211195; This is a riddle that I need an answer to asap.....if anyone out there can help me please email me with what you have here it goes... One disturbs the peace and another keeps it. The third likes to take matters into its own hands. All three come from the same chrome horse corral. Name the corral and all three. anyone? === Subject: Re: confusing riddle > This is a riddle that I need an answer to asap..... Could you please send such things to rec.puzzles rather than a mathematical news group? Ken Pledger. === Subject: Re: Clashing of local variable names and symbolic inputs to procedures |> Now, here is my question. Suppose that I wanted QD to work for arbitrary |>arguments, be they symbolic, numeric or some mixture of the two, without any |>restriction whatsoever on the symbolic names chosen. Then, it seems to me, I |>may not use _any_ (valid!) local variables whatsoever, for fear of a |>possible clash with something symbolic in the input list. Such a clash |>cannot, of course, happen if the input has no names in it - just as in any |>(numeric) programming language I can think of, the names of local variables |>are unimportant, since they _cannot_ clash with anything outside their |>scope. > Here, however, the situation is different. How can this clash be >avoided? There shouldn't be any clash. Local variables and global variables of the same name are treated as completely different, even though they look the same. For example: > f:= proc(x) local y; x - y end proc; f(y); y - y The result doesn't simplify to 0 because there are two different y's here: the first is the global y (corresponding to the formal parameter x) and the second is the local y. I haven't checked to see what happens to your code with symbolic inputs, but whatever the problem is, it shouldn't be a clash between local and global variables. ' === Subject: Easy way to solve solvable QUINTICS by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7U2dFb03513; Hello all, For those interested in the quintic, this paper might be of some help: An Easy Way To Solve Solvable Quintics Using Two Sextics ABSTRACT: Using a method initially developed by George Young (1819- 1889), Arthur Cayley (1821-1895), and later by George Watson (1886- 1965), an explicit quartic is constructed to enable the solution in radicals of a quintic when it is a solvable equation. Not one, but two sextic resolvents are derived which are important to forming the coefficients of this quartic. Certain difficulties and their solutions as well as a novel consequence to the method are also addressed in this paper. Mathematics Subject Classification. Primary: 12E12; Secondary: 12F10. http://www.geocities.com/titus_piezas/ Just click on the link in the website to the pdf file. Titus === Subject: Re: Where do math symbols originate? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7U5XBG22912; >THE reference, albeit a little old, on this subject is Cajori, Florian, A >History of Mathematical Notations which is still available from Dover. would >immediately give a straight yes or no answer to your question, his book >might permit your two opposing teams to bring many examples and >counter-examples in your discussions. >Bernard Mass.8e hope to get down to Purdue and see if it or something like it might be there. The squabble down in the Lab has quieted down in the last 2 years. The abstract math fellows are in some type of deep catatonic state, stunned into silence by something that, to them, seems completely unpossible. It turns out that as each and every cell in the body lives and breathes it is running the general respiration reaction: organics + oxygen => water + carbon dioxide + ~energy (eq. 1) If you remember your biochemistry, that happens in processes like the Kreb's tri-carboxylic acid cycle and similar well-oiled channels. One point is, the water molecules are ~created in a regular sequence, forged in concert with our energizing experience in the world, extruded from each process in incremental units. Now, as you might also remember, the experimentalists locked in the sub-basement at the Lab have routinely been aligning rod magnets along the radii of a tetrahedron and sticking two north poles to two south poles. After grueling analysis or the resulting artifact, they have discovered that since that thing has six edges, there are at least six way such a thing can be oriented in some imposed field. This leads to a 6-to-the-nth type of math. This means that for a sequence of, say, 16 units, there are 6^16 or 2.82x10^12 ways to arrange those sixteen units in a stack or chain. Many of the fellows in the lab have not had that many different experiences, so they are somewhat excited. The logic goes, as best I can follow it, that this is some sort of smoking gun model of consciousness. The analog math don't lie. The water molecules are formed in distributed, parallel processing channels and as the units are extruded, they are jiggled into one of the six positions BY the subtle inßuences of the inbound (sensory) quantum gravitational vibrations. Once the chain is truncated at some length (perhap for some, a consistent length), it's like a little resonant reßection of the ~current environmental experience. If Mom was yelling, HOT!, then Ôhot!' it is. And so forth. Interestingly, the chain one person maps to Ôhot!' doesn't have to be the say as someone else does. Now, the fact is, they really don't know whether to invoke or apply a fancy call to quantum gravitation, or whether to just refer to inßuences of the surrounding, more unified field. The point is, they have a non-linearly expanding point. Water makes up something like 80% of just about everything important in and near widgets of consciousness. Water is FORMED and extruded in rational units, sequentially, during respiration, in concert with the vibration of our experience. Water does form in ordered water arrangements. So-called bound water, water that gets knit together in organic materials, is energetically very stable and persistance. Anharmonic oscillators, matrix math, torsion-induced sine and cosine tables, even wave-equations appear to be quite naturally at home in such a resonance-based computational/associative media. Plus, running such a water-based internal analog math system still allows for wild variations in the initializations and in the ways linguistics, perceptions, beliefs, values, etc., can be formed and held and changed. Com links to deeper levels of organizations and comminication are not automatically excluded. As well, dream imagery, prayer, imagination and, I suppose, even the creation of abstract math symbols and models can all be created using the same sort of media/relationhip -- analog math symbols. This has come as a bit of a shock to the abstract math folks. Once their denial shattered, they have just been sitting there, day after peaceful day, watching their breath, trying to figure out if there is anything mathematical that they can say. So far, they are still completely silent on the matter. to it. Ralph Frost Imagine consciousiness as a single internal analog language made of ordered water, and its variants. http://ßep.refrost.com ...and love your neighbor as yourself Matthew 19:19 >uqpg8nsacsr0c4@corp.supernews.com... > [Rated RFTOR in s.p.r.] > The normally well-mannered, docile fellows in the advanced symbolics and > unification division here at the lab started a bad sry that intrinsic geometric properties/notions must be > |coordinate independent. Yet rationality is always defined with respect to > |some coordinates (variables). Granted that rationality here is invariant > |under linear transformations of the affine space in which the algebraic > |group is embedded, but I would still like to see some sort of intrinsic > |condition on the group that shows it to be an algebraic group. > the vague impression that i have about this is that a lie group's being > algebraic (i think affine algebraic in the strict sense actually, meaning > that the group operation is not just rational but polynomial) is secretly > very closely related to the representation theory of the group. if the > representation theory of the group is sufficiently nice, meaning basically > that the group has enough representations on finite-dimensional vector > spaces, then the so-called representative functions on the group will > form an affine coordinate ring making the group into an algebraic group. I'm not completely sure what an affine coordinate ring is, but I would guess it would be the analog of say polynomials in x,y,z for R^3. For example, polynomials in exp(i t) over the complex numbers would form a coordinate ring for U(1), parametrized as usual with t in [0,2pi[. Right? > a representative function on a group is a polynomial function of the > matrix by which a group element is represented under a finite-dimensional > representation of the group. the nice trick that shows that the > representative functions on a group form a (commutative) ring is that the > sum of representative functions is representative because you can take the > direct sum of group representations, while the product of representative > functions is representative because you can take the tensor product of > group representations. One condition on an abstract group would certainly determine whether it is algebraic or not. Namely if that group can be embedded inside another group that is known to be algebraic (say GL(n,F)) in such a way that the embedding gives you an algebraic variety. This condition certainly would have links to representation theory. That's about as far as my train of thought is taking me. :-) If we can say that all the coordinate dependent properties of algebraic groups can be recast into the language of some sort of intrinsic coordinates on the group itself (such as the affine coordinate ring formed by matrix elements of representations) then I can see that the notion of being algebraic is intrinsic to the group. The real question as to a group's algebraic character would be whether such intrinsic coordinates exist. Igor === Subject: Re: What is a differential field? : I've only recently learned what an affine algebraic variety is. So let me : see if I can prove that GL(n,F) satisfies the requirements to be one. : First, n x n matrices are elements of a n^2 dimensional vector space over : F, which is an affine space. So GL(n,F) would certainly be an affine : variety, since it is embedded in an affine space. The condition for a : matrix M to be in the group is GL(n,F) is that det(M) != 0. Interesting, : this is an inequality and not an equation. The trick is to take an additional variable z, different from the matrix entries. The condition for GL(n,F) is then det(M)*z=1. : I've been indoctrinated by modern : differential geometry that intrinsic geometric properties/notions must be : coordinate independent. Yet rationality is always defined with respect to : some coordinates (variables). Granted that rationality here is invariant : under linear transformations of the affine space in which the algebraic : group is embedded, but I would still like to see some sort of intrinsic : condition on the group that shows it to be an algebraic group. I understand what you're asking for, but I don't know of a good answer to this. Maybe someone else can help... ? :> (2) abelian varieties, which are algebraic groups where the group :> structure :> extends to a complete algebraic variety. These are always commutative :> (as the name suggests). For example, 1 dimensional abelian varieties are :> elliptic curves. : I know very little about elliptic curves. I do know that there is a group : law associated to the points on an elliptic curve. But this fact always : seemed like a coincidence to me. When is it possible to associate a group : law to the points of an algebraic variety? I presume there would be some : conditions on the polynomials whose zero-sets define the variety. This isn't an easy question to answer, as far as I know. To take a Ôsimple' example, suppose we are considering a _complete_ algebraic _curve_ X. If X can be given the structure of an algebraic group, then for every nonzero x in X, the translation by x map T_x, is fixed-point free. By the Lefschetz fixed point theorem, the alternating sum of the traces of the maps in cohomology induced by T_x is 0. By a sort of Ôcontinuity' argument, the same is true for T_0, the identity map. But the alternating sum of traces for the identity map is the Euler characteristic. So X has Euler characteristic 0, which means X has genus 1. So only genus 1 curves can have a group structure, which are elliptic curves. :> In general, a connected algebraic group is an extension of an abelian :> variety by a linear algebraic group. : You lost me here. I understand connected in the topological case, but : from where's the topology when working with discrete finite fields? And : what do you mean by extension? Some sort of group product? G is an extension of A by B if there is an exact sequence of groups 1 -> A -> G -> B -> 1 The topology we are using is the Zariski topology (closed sets are the algebraic sets). In the case where the field is the complex numbers, it is also okay to use the usual topology to determine connectedness. The reason for insisting on connectedness is to avoid groups like G x F, where G is a connected algebraic group and F is a finite group. Ted === Subject: Algebraic groups (was Re: What is a differential field?) > : I've only recently learned what an affine algebraic variety is. So let > : me see if I can prove that GL(n,F) satisfies the requirements to be one. > : First, n x n matrices are elements of a n^2 dimensional vector space > : over F, which is an affine space. So GL(n,F) would certainly be an > : affine variety, since it is embedded in an affine space. The condition > : for a matrix M to be in the group is GL(n,F) is that det(M) != 0. > : Interesting, this is an inequality and not an equation. > The trick is to take an additional variable z, different from the matrix > entries. The condition for GL(n,F) is then det(M)*z=1. Hmm, wouldn't that technically make GL(n,F) a _projective_ algebraic variety? I may be confused by what a projective variety is supposed to be. Igor === Subject: Shalosh B. Ekhad As far as I know, Zeilberger calls his computer Shalosh B. Ekhad. In his book A=B, he cites: One must not always invert Carl G. J. Jacobi [Shalosh B. Ekhad] Was that citation found out on the computer called Ekhad or was Shalosh B. Ekhad the jewish name of Jacobi? === Subject: Paper published by Algebraic and Geometric Topology A paper has been published and also a provisional erratum for a previously published paper (see end). --- The following paper has been published. Algebraic and Geometric Topology URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-27.abs.html Title: Commutators and squares in free groups Author(s): Sucharit Sarkar Abstract: prove that the commutator of x^m and y^n is a product of two squares if and only if mn is even. We also show using topological methods that there are infinitely many obstructions for an element in F_2 to be a product of two squares. Secondary: 57M07 Keywords: Commutators, free groups, products of commutators Author(s) address(es): Stat-Math Unit, Indian Statistical Institute Bangalore, India Email: bmat0212@isibang.ac.in --- An error has been found in the following paper. A full erratum will be published in due course. In the meanwhile a provisional erratum is available (linked from the URL below). Kquabble the other day > that perhaps someone in SPR can speak into and put an end to ...so those > guys can shut up and get back to calculating -- to doing real work. > One group, the experimentalists, came out with the idea that since > abstract math symbols are linguistic artifacts, all abstract math symbols > and thus all abstract mathematics arise from and thus are secondary to a > very small number of very ßexible physical structures and relationships > down in the guts of human consciousness. > This made sense to the experimentalists, because they couldn't escape >the > notion that their plans for improved experimental rigs, as best they can > tell, ßow up out of various rearrangements of the same ßexible physical > structures. > However, as readers might well imagine, the abstract math contingent will > have none of this troublesome talk and taunting, even though they > can only grunt and groan ineffable noises when confronted with the > observation that the unified thing itself does all it's math ßawlessly > using the so-called analog math symbols. To complicate matters, some of > abstract folks are becoming a bit intrigued by the idea that the step up > the emerging, more unified models in fact very might require the > introduction and shift to using more robust, more synchronous math symbols > than folks use in the less unified models. At this moment, though, none of > them can venture a guess on which analog math symbol might be the absolute > best to be deployed or how such an awkward notion might be refined and > developed. > Anyway, I am asking for your help to resolve this squabble one way or >the > other. > -- > Ralph Frost > Looking for a desktop model to help you ponder this topic? > http://ßep.refrost.com -- now with secure online ordering > Use more robust symbols > Seek a thought worthy of speech. === Subject: Re: Does Magma Still Exist? Keywords: Magma > Recently, I have been doing web searches for information about BCH > codes. > A computer algebra package, Magma, developed at the University of > Sydney, and based on an earlier program called Cayley, was mentioned > several times. So, today, I looked for more information about it. > However, most of the references I saw referred to an old site, which > contained only a link to what was supposed to be the current site, but > which was broken. A newsgroup post in 2002 gave a link, which did not > work. > because their Internet resources are limited, and so perhaps some > of my difficulties are merely due to network congestion issues. Or > has something happened to Magma? Nothing so interesting, I'm afraid. The web server was unexpectedly out for a couple of days. It is working again now, and the correct link is: http://magma.maths.usyd.edu.au/ In particular, this segment of the online help http://magma.maths.usyd.edu.au/magma/htmlhelp/text1451.htm contains the part of the Magma Handbook specifically mentioning BCH codes, although there are uses in other parts of the Handbook. (You may wish to wander up a level or two in the documentation to look at more of the coding facilities offered by Magma.) Geoff. -- Geoff Bailey Email: geoff@maths.usyd.edu.au Computational Algebra Group Phone: +61 2 9351 3114 University of Sydney Fax: +61 2 9351 4534 === Subject: Re: Does Magma Still Exist? Did you not find this web site? http://magma.maths.usyd.edu.au/magma/MagmaInfo.html === Subject: Re: Computing Huffman codes for Fibonacci numbers on a Turing Machine > Computing Huffman codes for Fibonacci numbers on a Turing Machine Also at Here is a fragment from the log file: ##### < Run 1.2 > Input word(s) & head start position(s) on tape(s) ##### Tape#0 : Word = a * a * a a * a a a * a a a a a * a a a a a a a a ; Head pos = 0 Tape#1 : Word = b ; Head pos = 0 Tape#2 : Word = b ; Head pos = 0 Note. This input word is Fibonacci sequence (1, 1, 2, 3, 5, 8) ##### < Run 1.2 > Processing ##### ----- < Run 1.2 > Initial Configuration ----- State : q0 Tape#0 : [a] * a * a a * a a a * a a a a a * a a a a a a a a Tape#1 : [b] Tape#2 : [b] < Run 1.2 > Applied Rule# 0 : q0 [ a b b ] ---> q0 [ (a,L) (b, L) (b, L) ] ///////////// // Omitted // ///////////// ----- < Run 1.2 > Configuration#2712 ----- State : qf Tape#0 : [x] b Tape#1 : x 1 1 1 1 a a 1 1 1 0 1 a 1 1 1 0 0 a 1 1 0 a a a 1 0 a a a a a 0 a a a a a a a [a] Tape#2 : [x] b < Run 1.2 > Success : Current state (qf) is halting one Weights and their codes on Tape#1 Weight Code ------ ---- 1 11101 1 11100 2 1111 3 110 5 10 8 0 -- Alex Vinokur http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn === Subject: This integral blows my PC (Maple 9.5) Help ! While integration p1 from 0 to 1 (int(p1(lambda),lambda=0..1)) the PC works and works.. And memory usage increasing and increasing until the whole stops.. Tried on 2 different PCs (one w/ w2k one with XP). Same result... Can someone help? What do I wrong? May be there are other ways to integrate? the function to integrate: p1:=lambda-> c1/(lambda^5*(exp(beta/(lambda*T))-1)); where beta:=Constant(h)*Constant(c)/Constant(k); c1:=2*Pi*Constant(h)*(Constant(c))^2; T:=3200 This function p1 is from physics and known as Rayleight-Jeans formula. I can plot p1, it is continuous as it should be and the integral will be (is) finite.. Aran === Subject: Re: This integral blows my PC (Maple 9.5) Help ! |>While integration p1 from 0 to 1 (int(p1(lambda),lambda=0..1)) the PC |works |>and works.. ... |>p1:=lambda-> c1/(lambda^5*(exp(beta/(lambda*T))-1)); |>where |>beta:=Constant(h)*Constant(c)/Constant(k); |>c1:=2*Pi*Constant(h)*(Constant(c))^2; You don't need to write Constant(h) etc. Just h will do fine. |>T:=3200 I think you forgot to say that you were assuming h, c and k were positive. Without any assumption, int works fine for me in Maple 9.5, in 0.391 seconds on my computer, except that since Maple doesn't know those signs it gives an answer in terms of a limit as lambda -> 0+. Maple seems to have trouble evaluating that limit when h, c and k are assumed positive (or when values such as 1 are used for those constants). ' === Subject: Re: This integral blows my PC (Maple 9.5) Help ! > While integration p1 from 0 to 1 (int(p1(lambda),lamb-theory of virtually poly-surface groups by S.K. Roushon URL: http://www.maths.warwick.ac.uk/agt/AGTVol3/agt-3-4.abs.html === Subject: Re: A question about finite sets of integrally independent vectors >How can one verify if a set of vectors are integral dependent? that >is: if they are linearly independent with integral coefficients. >(Actually I'm looking for a method to reduce a finite set of vectors >to a set of integrally independent vectors) First of all, integrally dependent is the same as rationally dependent, i.e. if a linear combination with rational coefficients is 0, you multiply by the denominators and get one with integer coefficients. I'll suppose your vectors are in F^n where F is some field of characteristic 0. We need to assume you have a way of checking whether finite subsets of F are rationally related, and finding such relations if they exist. Then you can look at all the coordinates of all your vectors and find a maximal rationally independent set, say S = {s_1,...,s_m}. This will be a basis for the vector space (over the rationals Q) spanned by those coordinates. Using that basis, there's a Q-linear one-to-one correspondence from Q^{nm} to a linear subspace of F^n that contains your vectors. Thus the problem is reduced to the case of Q^{nm}, where the standard methods of Gaussian elimination can be used to check for linear (which in this case would be Q-linear) independence and find bases. ' === Subject: Re: some questions about the algebraic closure of Q_p Epigone-thread: twyswinply >These questions will seem rather naive to ots of people, I bet. So I >apologize in advance. >Let K be the field of algebraic p-adic numbers. (So it is a subfield >of Q_p, and it is the Henselization of Q with respect to the p-adic >valuation). >Let K_1 be the valued field extension of K obtained by addition of the >roots of all X^(p^n)-X. So the residue field of K_1 is the algebraic >closure of F_p. K_1 is the inertia field of K. The valuation group of >K_1 is still Z with v(p)=1. >Let K_2 be the extension of K_1 obtained by addition of the roots of >all X^n-p. Let say we keep v(p)=1, and the valuation group of K_2 is Q. >The question is : is K_2 algebraically closed ? If not, why ? >Can someone give me an exemple, for a given p, of a polynomial which >does not split in K_2 ? >Other question : what happens if K=Q_p ? It follows from property 5 listed in my previous posting that the extension from V_1 to Q_p^~ is non-abelian, thus K_2 cannot be the algebraic closure of Q_p^~. So the answer to your question is NO. H === Subject: Re: Infinitely divisible probability distributions Epigone-thread: phoisterdsloo >I find in the second volume of Feller's famous book the >attribution of the concept to Bruno de Finetti, in 1929. >But I couldn't find that in any list of references in the >book. -- Mike Hardy > I don't have a copy available to check (I've only ever read it > on loan from a library), but I'm *reasonably* sure de Finetti > discusses the history of the idea (and possibly even reprints > @book {MR92b:60009a, > AUTHOR = {de Finetti, Bruno}, > TITLE = {Theory of probability. {V}ol. 1}, > SERIES = {Wiley Classics Library}, > PUBLISHER = {John Wiley & Sons Ltd.}, > YEAR = {1990}, > MRNUMBER = {92b:60009a}, > } > @book {MR92b:60009b, > AUTHOR = {de Finetti, Bruno}, > TITLE = {Theory of probability. {V}ol. 2}, > PUBLISHER = {John Wiley & Sons Ltd.}, > YEAR = {1990}, > MRNUMBER = {92b:60009b}, > } . > Looking for that information in MathSciNet, I also found > @book {MR86i:01060, > AUTHOR = {de Finetti, Bruno}, > TITLE = {Scritti (1926--1930)}, > YEAR = {1981}, > MRNUMBER = {86i:01060}, > } > which from its title would appear to include the paper you want > (but may not help you if you don't read Italian, and/or cannot > get your hands on what might be a reasonably obscure book). >Scritti does indeed contain several papers on what we would nowadays >call processes with stationary independent increments, including one >dating from 1929. (My Italian is too non-existant to say more.) You >might also have a look at a survey paper by M. Fisz: > AUTHOR = {Fisz, Marek}, > TITLE = {Infinitely divisible distributions: recent results and > applications}, > JOURNAL = {Ann. Math. Statist.}, > VOLUME = {33}, > YEAR = {1962}, > PAGES = {68--84}, > MRCLASS = {60.20 (60.30)}, > MRNUMBER = {25 #2624}, >MRREVIEWER = {M. Lo{`e}ve}, >The bibliography of this survey includes one paper by De Finetti. The >De F paper is the earliest of the 104 papers cited, lending support to >Feller's assertion. >-- Would you mind giving us an idea ? How does Ôinfinitely divisible probability distributions'look like! Do you know a simple case and its mathematical representation . === Subject: Re: Infinitely divisible probability distributions ................... >Would you mind giving us an idea ? >How does Ôinfinitely divisible probability distributions'look like! >Do you know a simple case and its mathematical representation . Every infinitely divisible distribution is the convolution of a normal distribution and the convolution of a countable number of Poisson convolutions of identical distributions. It is easier to express in random variables. X = Z + sum sum_1^{N_i} Z_{ij}, all the random variables on the right side are independent, Z is normal, the N_i are Poisson, and the Z_{ij} have the same distribution for each i. Some of the terms can be absent, and the normal term can be a constant. Examples are normal, Poisson, Gamma, t, and a huge variety of others. Conversely, if the sum converges almost surely, the random variable X has an infinitely divisible distribution. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Riemann-Roch for reducible P^2-plane curves Epigone-thread: glangswongheld >Let C be a plane complex projective curve that is a union of two >smooth >curves C_1 and C_2 of degrees d_1, resp. d_2, and D a divisor on C, >consisting of smooth points only. What is in this setting the analogy >of the classical >Riemann-Roch theorem for the dimension of |D|, i.e. the space of >divisors linearly equivalent to D ? >Strangely enough, I cannot find it in obvious literature. >Some theory in such setting can be found in the paper >Theta-characteristics on algebraic curves by J.Harris >Dmitrii >(dima[1-3]pasechnik[1-4] at msri dot org, where >aword[1-3] means awo) >Reducible and multiple curves<, in which the Riemann-Roch-theorem is treated in the general situation. However I fear that it is out of print and unfortunately cannot remember the publisher ... H === Subject: Re: Riemann-Roch for reducible P^2-plane curves >Reducible and multiple curves<, in which the Riemann-Roch-theorem > is treated in the general situation. > However I fear that it is out of print and unfortunately cannot > remember the publisher ... it's his PhD thesis. Easy to find in Dutch libraries. Dima === Subject: Re: Question re: the two 2_21's in 3_21 and series parallel posets >If you want to pick arbitrary extremum vertices and edge directions >of 2_21 that make a poset and an algebraic lattice, you can, >but I can't quite put my finger on why. I don't think that you can make a poset from the 2_21. The Schlaeßi graph is not transitively orientable nor is its complement. Not even, when you add edges from the top-vertex to the 10 bottom-vertices. --Guenter === Subject: Re: Question re: the two 2_21's in 3_21 and series parallel posets ------------------------------------------------------------- --------- > Each vertex of a polytope can be paired with a diametrically opposeda=0..1)) the PC works > and works.. And memory usage increasing and increasing until the whole > stops.. Tried on 2 different PCs (one w/ w2k one with XP). Same > result... When I try, Maple can get the antiderivative, which is a sum of polylogs, but Maple cannot compute the limit at 0. Ignoring the constants, you can observe this in int(1/x^5/(exp(1/x)-1), x= 0..1) (same integral as yours except the constants are different). > Can someone help? What do I wrong? You did nothing wrong. > May be there are other ways to integrate? In this case you can make a substitution in the integral with the student[changevar] command. > the function to integrate: > p1:=lambda-> c1/(lambda^5*(exp(beta/(lambda*T))-1)); > where > beta:=Constant(h)*Constant(c)/Constant(k); > c1:=2*Pi*Constant(h)*(Constant(c))^2; > T:=3200 Rather than use the ScientifcConstants package, I'll just use symbolic h, c, and k, and assume they are positive. You can substitute in the actual constants at the end. > restart; > assume(h>0, k>0, c>0); > beta:= h*c/k; T:= 3200; c1:= 2*Pi*h*c^2; > p1:= lambda-> c1/lambda^5/(exp(beta/lambda/T)-1); > J:= Int(p1(lambda), lambda= 0..1); # Note capital I > student[changevar](u= exp(beta/lambda/T), J, u); > value(%); -6400*Pi*c*k*ln(A-1) -6400*I*Pi^2*c*k -Pi*h*c^2/2 +83886080000000*Pi^5*k^4/3/c^2/h^3 -61440000*Pi*k^2/h*polylog(2,A) +393216000000*Pi*k^3/c/h^2*polylog(3,A) -1258291200000000*Pi*k^4/c^2/h^3*polylog(4,A) where A = exp(c*h/3200/k); The imaginary I is cancelled by the imaginary components of the polylogs. === Subject: Re: Mathematica Vs. Matlab > Interesting reading is this comparison betwen Mathematica an Matlab. > Here are my observations: > I had to solve a system of four nonlinear equations (something like > that:-K1*V2*sin(th2-K2)-K3*V2*Vmid*sin(th2-thmid)-K4) the other three > are similar. > What I found was that the results I got with function Findroot in > Mathematica 5.0 was an order more accurate than fsolve of Matlab Were you using the default tolerances for FSOLVE? Very basically, as soon as the function values get Ôclose enough' (as defined by TolFun) to 0 and the variable values aren't changing by very much (TolX), FSOLVE says Good enough. Try changing the TolX and TolFun parameters in your call to FSOLVE to change the definition of Ôclose enough' and see if that improves the solution. If it doesn't, could you send the equations for the problem, the solution you received from Mathematica, and the solution you received with the default tolerance and the tighter TolX and TolFun using FSOLVE to our support staff so that we can investigate what's going on? -- Steve Lord slord@mathworks.com === Subject: Re: QUINTIC BY RADICALS > As several others have informed you, as long as there are _some_ quintics > that cannot be solved by radicals involving rational functions of the > coefficients, your example does not contradict the work of Abel and Galois. > I found a useful web site for this stuff, including solutions of quintics in > terms of hypergeometric functions: > http://icl.pku.edu.cn/yujs/MathWorld/math/q/q111.htm The current MathWorld link is http://mathworld.wolfram.com/QuinticEquation.html See also http://library.wolfram.com/examples/quintic/ Paul -- Paul Abbott Phone: +61 8 9380 2734 School of Physics, M013 Fax: +61 8 9380 1014 The University of Western Australia (CRICOS Provider No 00126G) 35 Stirling Highway Crawley WA 6009 mailto:paul@physics.uwa.edu.au AUSTRALIA http://physics.uwa.edu.au/~paul === Subject: Re: QUINTIC BY RADICALS by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7U2d8j03244; >For example, x^5+3*x+1 has Galois >group S_5, and therefore is not solvable in radicals. >Robert Israel israel@math.ubc.ca >Department of Mathematics href=http://www.math.ubc.ca/~israel>http://www.math.ubc.ca/~ israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2 >Let f(x)=x^5+(3+k1)*x+(1+k2) >What the value of k1 and k2 that make f(x) solvable in radicals >Such that > k1 and k2 are rational complex numbers > magnitude(k1)+ magnitude(k2) is minimum >M. A. Fajjal Actually there is no lower limit of k1 and k2. There are unlimited solutions of k1 and k2 to get x^5 + (3+k1)*x + 1+k2 = 0 solvable in radicals I have got for example k1= 44434258558174122962160348326776536558145581228690199781440/ 1442174741293 25277147705355136900000000000000000000000000000000000000000 k2= 14708205761097273214405059026344745912854711319982092916802495 50230760545 091499893691392/ 1442174741293252771477053551369000000000000000000000000000000 0000000000000000000000000000000000000000 u=-126291550529446/100000000000000 v=-664962210424963/1000000000000000 Then The real root = u*((4*v+3)/(v^2+1)/5)^(1/5)*((1-2*v+2*(v^2+1)^(1/2)+(5-4*v+8* v^2+(500*(v^2+1 )*(1-2*v)+(4*v+3)^3)/125/(1+v^2)^(1/2))^(1/2))^(1/5)-(-(1-2*v -2*(v^2+1)^(1/2) -(5-4*v+8*v^2-(500*(v^2+1)*(1-2*v)+(4*v+3)^3)/125/(1+v^2)^(1/ 2))^(1/2)))^(1/5 )-(-(1-2*v-2*(v^2+1)^(1/2)+(5-4*v+8*v^2-(500*(v^2+1)*(1-2*v)+ (4*v+3)^3)/125/( 1+v^2)^(1/2))^(1/2)))^(1/5)-(-(1-2*v+2*(v^2+1)^(1/2)-(5-4*v+8 *v^2+(500*(v^2+1 )*(1-2*v)+(4*v+3)^3)/125/(1+v^2)^(1/2))^(1/2)))^(1/5)) where u=-126291550529446/100000000000000 v=-664962210424963/1000000000000000 r1=-0.331989029584597 M. A. Fajjal === Subject: Another problem, LITTLE DIFFICULT I need to do some computation for a cache architecture. I have arrived at a simplified model and I need help to get it solved. Here is the problem. Balls are being put into and removed from a bucket, one arrival and one departure in each unit time. Arivals: Balls that are arriving have equal probability that they can have any color from b different colors. Departure: Balls from the bucket are departing in following way, a) All colors are selected, whose balls haven't been departed in last c unit times. b) Then from selected colors, a ball is departed of the color which has maximum number of balls. Thus, in every window of c unit time, balls of c different colors need to be sent. (c < b) The bucket has N balls initially, with equal number of balls of every b color. Now, when this process is carried out for very long time say T (T is trillions), how many times will the bucket be having balls of only c colors (c < b). In other words, after a long time and in steady state, whats is the probability that the bucket will have balls of only c colors (c < b). Sailesh === Subject: Help??? I am not an algebra person....my daughter needs to know..,,if you have 3 dice totaling 6....how many times can you get the total 12? What is the algorithm? === Subject: Re: help..... > Sorry.....I worded the previous question wrong.... > If you have 3 dice each has the numbers 1-6....how many times can you get > the total 12? What is the algorithm? A preliminary point: if you post to more than one news group, please cross-post (as I've now done) to save different people from duplicating one another's efforts. The simplest solution to your problem may be just to list all the possibilities and count them. However, since you ask What is the algorithm?, here we go! Here's a simpler example to start with. Suppose you have three coins, each marked 1 on one side and 2 on the other. If you toss all three, in how many ways does each total turn up? You can get a total of 3 in one way (111). You can get a total of 4 in three ways (112, 121, 211). You can get a total of 5 in three ways (122, 212, 221). You can get a total of 6 in one way (222). Now here's a more high-brow way to see that. Represent the two sides of each coin by the exponents in the function x^1 + x^2. Tossing all three coins then corresponds to (x^1 + x^2)^3; because multiplying that out gives for example the term in x^5 as (x^1)(x^2)(x^2) + (x^2)(x^1)(x^2) + (x^2)(x^2)(x^1) = 3(x^5). That coefficient 3 shows how many ways a total of 5 can occur. Now set up a similar generating function for your problem. Represent the six faces d > vertex if the polytope has central symmetry. This isn't possible > with 2_21 since it has 27 vertices, an odd number ... snip > Orienting 2_21 with a simplex on top results in a simplex being > on the bottom too. Here's the distribution matrix for this case: > 6 5 10 1 > 15 4 8 4 > 6 1 10 5 > If you want to pick arbitrary extremum vertices and edge directions > of 2_21 that make a poset and an algebraic lattice, you can, > but I can't quite put my finger on why. ------------------------------------------------------------- ------- I am grateful to SF for even bothering to respond after my inexcusably careless and loose terminology concerning the central symmetry of 2_21. I can only plead typing without thinking in an attempt to simplify the phrasing of the question. With respect to the why one might one to make a lattice out of 2_21, I would not trouble S. Fortescue with outlining a possible why unless I knew how he would answer the question raised in my geometry. === Subject: super-Hopf algebras Epigone-thread: werskendtweld I want enough materials about this topic super-Hopf algebras === Subject: Keep it simple: Inf and -Inf and extending exponents by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7RJL6E16758; I sent the note below to alt.math.undergrad discussion forum under the to make it simple, clarify matters, and make common sense arguments about a bit of sometimes troublesome math notation. Since it provoked no public response at all, I thought I would try it again under a new subject line. Surely someone will want to object to my suggestions! ------------------------------------------------------------- I must say that this thread, and other similar ones about uses or meanings of the symbols Inf and -Inf and also ad hoc definitions of extensions of functions like division, powers, exponential and log always strike me as being strange and unhelpful, and indeed likely to cause confusion among novice or casual students or users of math. 0. I think it would be much better in elementary math (even up to advanced calculus or beginning analysis at least) to stick with not giving any meaning to Inf or -Inf other than as part of useful abbreviations (easily remembered and written nicknames) for long somewhat complicated phrases involving certain specified behaviors of functions defined on the Reals or a tail thereof. Discourage and disavow any uses of these symbols that even suggest they are somehow just new extended versions of numbers. As far as I am aware, there is no utility and lots of likelihood for causing confusion to do otherwise. What could be the motivation of someone to want to consider Inf and -Inf as extended numbers? That is not helpful - notational conventions are put in place for reasons of efficiency of thinking and writing and an aid to instant recognition/apprehension of complicated expressions, to simplify and clean up our writing of math. About exponents: 1. We all know that for any element B in any ring, B^n makes perfectly good elementary common sense for any positive integer n and is just a shorthand notation for writing out a longer expression involving n appearances of B and (n-1) multiplications parenthesized in any way you like. 2. In the 1600's people started to systematically use this exponent notation and noted the usual Ôrules of exponents' - all quite useful for computation and compact notation. Later someone noticed that in fact it was convenient for the writing of polynomials and power series in B using the sum notation to write the constant term C0 instead as C0*B0 with the value of B0 being 1 by convention. Fortunately the rules of exponents still worked ok, and there was then just one standard form (in general sum notation) for computing the coefficient of B^n in products of polynomials/series, even when n is 0. This is all unmitigated good news; we've gained in efficiency of writing math and calculating. 3. But then some nagging student or stickler for details noted that if we are doing this for a polynomial/series expression in B, say P(B) which begins C0*B0 +C1*B1 + ... , then if we evaluate the expression when B is 0, one expects to get the constant term C0 of P, but the expression suggests that what we get is the value of C0*00. The obvious thing to do is to insist that the convention B0 = 1 continue to hold even if B=0. And again everything still works fine. See the and examples showing that the convention 00 = 1 is indeed very useful - especially for combinatorics and whenever formal power series (generating functions) are used. 4. We also know that while we can write an expression (compact description of a step-by-step process to calculate something) involving B^(-n)for n a positive integer, when we get around to actually evaluating the expression we should get an ÔError message' whenever B is not invertible in the ring. But it worked out nicely that if B is invertible then B^(-n) could conveniently be defined to be the inverse of B^n which turned out to be the same as (B^[-1])^n - and, very important, the usual rules of exponents still worked ok. However, since 0 is not invertible, 0^[-n] is undefined - and I know of no good reason to try to assign it some meaning. 5. Notice that the above 1-4 all work quite generally in a ring (well at least if B is in the center of the ring if we are multiplying polynomials in B). But if we are working in a special ring like the field of real numbers R, then as we know, Newton and others of his generation popularized some useful extensions of the exponent notation, first to rational number exponents and then to real number exponents - when the base is a positive number, all of which are now incorporated into calculators and computer arithmetic. Some also have the special extension (-B)^r = {-(B^[1/q])}^p for a rational number r in reduced fraction form p/q with q odd and B a positive number. I don't remember now ever hearing any good arguments for any other extensions of exponent notation, except for the corresponding situation for the complex numbers. (I'm ignoring here the occasional use of notation B^A in set theoretical discussions in discrete math texts with the meaning Ôset of all functions from A into B' - that doesn't seem to have wide usage.) Ladnor Geissinger, Math Prof, Univ NC at Chapel Hill, NC 27599 === Subject: Re: Keep it simple: Inf and -Inf and extending exponents > I sent the note below to alt.math.undergrad discussion forum under the > to make it simple, clarify matters, and make common sense arguments > about a bit of sometimes troublesome math notation. Since it provoked > no public response at all, I thought I would try it again under a new > subject line. Surely someone will want to object to my suggestions! It's too too long on talk, talk, talk and not simple upon what you're suggesting. Instead of making us guess by wading thru reams of verbiage, just present your point in 100 words, an abstract of your thesis, and then proceed with the details. Again, rather than rambling, present your points succently. Elaborate, if you must, later after we know in the first place what the heck you're attempting. > I must say that this thread, and other similar ones about uses or > meanings of the symbols Inf and -Inf and also ad hoc definitions > of extensions of functions like division, powers, exponential and log > always strike me as being strange and unhelpful, and indeed likely to > cause confusion among novice or casual students or users of math. > 0. I think it would be much better in elementary math (even up to > advanced calculus or beginning analysis at least) to stick with not > giving any meaning to Inf or -Inf other than as part of useful > abbreviations (easily remembered and written nicknames) for long > somewhat complicated phrases involving certain specified behaviors of > functions defined on the Reals or a tail thereof. Discourage and > disavow any uses of these symbols that even suggest they are somehow > just new extended versions of numbers. As far as I am aware, there is > no utility and lots of likelihood for causing confusion to do > otherwise. What could be the motivation of someone to want to > consider Inf and -Inf as extended numbers? That is not helpful - > notational conventions are put in place for reasons of efficiency of > thinking and writing and an aid to instant recognition/apprehension of > complicated expressions, to simplify and clean up our writing of math. > About exponents: > 1. We all know that for any element B in any ring, B^n makes > perfectly good elementary common sense for any positive integer n and > is just a shorthand notation for writing out a longer expression > involving n appearances of B and (n-1) multiplications parenthesized > in any way you like. > 2. In the 1600's people started to systematically use this exponent > notation and noted the usual Ôrules of exponents' - all quite useful > for computation and compact notation. Later someone noticed that in > fact it was convenient for the writing of polynomials and power series > in B using the sum notation to write the constant term C0 instead as > C0*B0 with the value of B0 being 1 by convention. Fortunately the > rules of exponents still worked ok, and there was then just one > standard form (in general sum notation) for computing the coefficient > of B^n in products of polynomials/series, even when n is 0. This is > all unmitigated good news; we've gained in efficiency of writing math > and calculating. > 3. Butof each die by the exponents in the function x^1 + x^2 + x^3 + x^4 + x^5 + x^6. Tossing all three dice then corresponds to (x^1 + x^2 + x^3 + x^4 + x^5 + x^6)^3 as above. The number of ways to get a total of 12 will be the coefficient of x^12 in the expansion of that. So much for the combinatorial theory. The rest is just algebra to calculate the coefficient of x^12. x^1 + x^2 + x^3 + x^4 + x^5 + x^6 = x(1 - x^6)/(1 - x) by the usual formula to sum a finite geometric series. Therefore (x^1 + x^2 + x^3 + x^4 + x^5 + x^6)^3 = (x(1 - x^6)/(1 - x))^3 = (x^3)((1 - x^6)^3)((1 - x)^(-3)) = (x^3)*(1 - 3(x^6) + 3(x^12) - x^18)* *(sum from n = 1 to oo of binomial(-3,n)*(-x)^n). The coefficient of x^12 is binomial(-3,9)*(-1)^9) - 3*binomial(-3,3)*(-1)^3) = 55 - 30 = 25. Well, you did ask. ;-( Ken Pledger. === Subject: Help..... Sorry.....I worded the previous question wrong.... If you have 3 dice each has the numbers 1-6....how many times can you get the total 12? What is the algorithm? === Subject: Re: Pls help with this expression by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7V4fxw08646; >((2*pi)^-1) * |[S]|^(-0.5) exp(0.5*transpose([V])*inv([S])*([V])) >Where >^ refers the power of, >[S], [V] represents a matrix or vector, >transpose and inv are matrix operations >| | represents determinant >I have a problem with this expression as the results of my >determinant >is negative and cause a sqrt(-ve number). >rdgs, >jax > That looks a heck of a lot like the Gaussian (normal) probability >density in n dimensions. I did a search on Gaussian probability Density. The above equation seems to be it. However, there is some things i do not understand. In the equation, there is |[s]|^ -0.5. That means the determinant is squared root. What will happen if the results of the determinant is negative? A normal C program will give an error for square rooting a negative value. I don't think complex numbers should be involved here either. Or should i always take the abs value of the determinant result? Within the exponential function, an inverse is being performed. This usually emit a large enough value to give the result of zero on the exponential computation (eg: exp(-ve large value)=0). What will this mean? I am sorry if my questions are too basic, but i am really weak in statistical computations. rdgs, jax === Subject: Re: Abstract/Modern Algebra by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7V4fx808626; >Does anyone know of a online/ distance learning course in >Abstract/Modern Algebra? I'm interested too. Did you receive any useful replies? === Subject: inequalities by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7VBf9s09369; Find the interval satisfying the following inequalities!!!! 1. -7<(2x-7)/5<1 or x-1>0 2. 2+|3-2x|>5 I have trouble with these!! === Subject: Re: inequalities > Find the interval satisfying the following inequalities!!!! > 1. -7<(2x-7)/5<1 or x-1>0 Break it into its component pieces: { -7 < (2x - 7)/5 AND (2x - 7)/5 < 1 } OR x - 1 > 0 Then see what interval each component implies and do the ANDs and ORs to see what the result is. Then check your result back in the original. > 2. 2+|3-2x|>5 Use the fact that |z| = z iff z>=0 |z| = -z iff z<0 to remove the absolute value bars, then do your interval analysis. > I have trouble with these!! Do what you can, then if you still can't do it, post here again, but *show the work you've been able to do*. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Question of finite element of a group by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7VBfAH09409; > > > If every element in a group has finite order > > is the group finite? > >A simpler construction is the subgroup of > G = Z_2 x Z_3 x Z_5 x ... >of elements of finite support, ie elements > (a2, a3, a5, ... ) >for which all but finitely many a's are nonzero. >This was corrected to read, for which only finitely many a's >are nonzero. > In other words, the direct sum of Z_2, Z_3, etc. The prime orders > of the cyclic groups are of course irrelevant. >That will not do, as the identity > (1,1,1,... ) >has infinite order. It has to be a subset of the direct sum to avoid >such elements, ie, elements > (a2,a3,... ) >where all but finitely many a's are zero. > Snappier example: Q/Z. There are *lots* of torsion groups after all. >Indeed, but not R/Z. In my previous reply to this, I referred to Lang's Algebra, pp. 36-37. That's the 3rd edition I was referring to for those page numbers. Todd === Subject: Re: Question of finite element of a group by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7VBfA009377; >A simpler construction is the subgroup of > G = Z_2 x Z_3 x Z_5 x ... >of elements of finite support, ie elements > (a2, a3, a5, ... ) >for which all but finitely many a's are nonzero. >This was corrected to read, for which only finitely many a's >are nonzero. > In other words, the direct sum of Z_2, Z_3, etc. The prime orders > of the cyclic groups are of course irrelevant. >That will not do, as the identity > (1,1,1,... ) >has infinite order. It has to be a subset of the direct sum to avoid >such elements, ie, elements > (a2,a3,... ) >where all but finitely many a's are zero. No, what you are referring to is called the direct *product*, which differs from the direct sum if the set of summands is infinite. Just to obviate any misunderstanding, a direct sum is also called a coproduct (in the category of abelian groups), and a direct product is just a product, with the usual categorical descriptions in terms of universal properties. Please see any graduate text in algebra for the correct definition of direct sum, e.g., Lang's Algebra, pp. 36-37. It's the usual name for what you were describing. By the way, the identity of the direct product or the direct sum above is (0, 0, ...), not (1, 1, ...) (surely you meant to use the additive groups Z_2, Z_3, etc., not the underlying multiplicative monoids of their ring structures). Todd === Subject: Re: integral calculas by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7VDJeq17856; i wnt to know definition of calculas so that i can understand easily thank u r.r 31/8/04 india === Subject: Re: integral calculas > i wnt to know definition of calculas so that i can understand > easily thank u Calculus is a branch of mathematics divided into two general fields: differential calculus and integral calculus. Differential calculus can be used to find rates of change, like orbits of planets, satellites, and spacecraft. Integral calculus is a method of calculating quantities by splitting them up into a large number of small parts. It can be used to find the surface area of irregular objects. You can find out the total surface area of your car (even the round parts) by using integral calculus. Source: Children's Encyclopedia Britannica vol. 3, p. 308-309, 1989. === Subject: Re: subbasis for the topology of a set alt.math.undergrad: > I'm trying to prove the following: > If X is a set and G an arbitrary set of parts of X, there is > exactly one topology T(G) on X such that G is a subbasis for T(G) > (the topology``generated'' by G). > Proof: Let X be a topological space. Let G = {} so that G is in > X. Two problems here. First, you don't get to decide what G is: the statement that G is an *arbitrary* set of subsets of X means that it can be absolutely any such collection, and you have to prove the result no matter which one it actually is. Secondly, G is not a subset of X (as you seem to be thinking when you write Ôso that G is in X'); it's a *collection* of subsets of X. Example: If X = {1, 2, 3}, G might be {{1}, {2}, {2, 3}}. Let's take a closer look at what you're trying to prove. The claim is that there is one and only one topology having G as a subbase. This means that you have two things to prove: (1) There is some topology on X that has G as a subbase. (2) If T_1 and T_2 are topologies on X that both have G as a subbase, then T_1 = T_2. Here (1) says that G is a subbase for at least one topology on X, and (2) says that G is a subbase for at most one topology on X; putting the two together gives the desired result. Now in fact (1) isn't quite true as you've stated the theorem: in order to prove (1), you have to know that G covers X, i.e., that every point of X is in at least one member of G. (Another way to say this: U{A : A in G} = X.) I don't know whether you've misstated the problem, or whether it was poorly stated in your source. However, (2) is still true, and I suspect that it's the main point of the exercise. A proof of (2) just about has to begin something like this: Let G be a collection of subsets of X. Suppose that T_1 and T_2 are two topologies on X, and that G is a subbase for both T_1 and T_2. Somehow you want to get from this starting point to the conclusion that T_1 and T_2 are really the same topology after all, i.e., that T_1 = T_2. T_1 and T_2 are just collections of subsets of X. In particular, T_1 and T_2 are themselves sets of things. One way to prove that two sets are equal is to show that each of then some nagging student or stickler for details noted that > if we are doing this for a polynomial/series expression in B, say P(B) > which begins C0*B0 +C1*B1 + ... , then if we evaluate the expression > when B is 0, one expects to get the constant term C0 of P, but the > expression suggests that what we get is the value of C0*00. The > obvious thing to do is to insist that the convention B0 = 1 continue > to hold even if B=0. And again everything still works fine. See the > and examples showing that the convention 00 = 1 is indeed very useful > - especially for combinatorics and whenever formal power series > (generating functions) are used. > 4. We also know that while we can write an expression (compact > description of a step-by-step process to calculate something) > involving B^(-n)for n a positive integer, when we get around to > actually evaluating the expression we should get an ÔError message' > whenever B is not invertible in the ring. But it worked out nicely > that if B is invertible then B^(-n) could conveniently be defined to > be the inverse of B^n which turned out to be the same as (B^[-1])^n - > and, very important, the usual rules of exponents still worked ok. > However, since 0 is not invertible, 0^[-n] is undefined - and I know > of no good reason to try to assign it some meaning. > 5. Notice that the above 1-4 all work quite generally in a ring (well > at least if B is in the center of the ring if we are multiplying > polynomials in B). But if we are working in a special ring like the > field of real numbers R, then as we know, Newton and others of his > generation popularized some useful extensions of the exponent > notation, first to rational number exponents and then to real number > exponents - when the base is a positive number, all of which are now > incorporated into calculators and computer arithmetic. Some also have > the special extension (-B)^r = {-(B^[1/q])}^p for a rational number r > in reduced fraction form p/q with q odd and B a positive number. > I don't remember now ever hearing any good arguments for any other > extensions of exponent notation, except for the corresponding > situation for the complex numbers. (I'm ignoring here the occasional > use of notation B^A in set theoretical discussions in discrete math > texts with the meaning Ôset of all functions from A into B' - that > doesn't seem to have wide usage.) > Ladnor Geissinger, Math Prof, Univ NC at Chapel Hill, NC 27599 === Subject: STAT Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7S51Fk29392; Heres the question... A student is planning on taking three different classes. Two of the classes, calc I and Computer Science, are offered at different time periods. The third class, Physics, is at a time that will conßict with Calc I. Probability of getting into calc I = 1/2 , Probability of getting into Computer Science = 1/3, which is the same probability as getting into Physics (1/3). Probability that he will get into both Calc I and CS = 1/6. Probability that he will get into both CS and Physics is 1/6. Find the prob that he will be able to get into at least one of the three classes. I think the answer goes like this. P(AuBuC)=P(a) + P(b) + P(c) - P(a Intersect B) - P(a intersect c) - P(b intersect c) + P(a intersect b intersect c). Therfore, P(a) = 1/2 , Prob of getting into calc I P(b) = 1/3 , Prob of getting into CS P(c) = 1/3, Prob of getting into Physics P(a intersect b) = 1/6 P(b intersect c) = 1/6 So, 1/2 + 1/3 + 1/3 - 2/6 = 5/6 If you could tell me if i'm anywhere close, i would appreciate it. === Subject: Re: STAT Problem in alt.math.undergrad: > Heres the question... > A student is planning on taking three different classes. Two of the > classes, calc I and Computer Science, are offered at different time > periods. The third class, Physics, is at a time that will conßict > with Calc I. Probability of getting into calc I = 1/2 , Probability > of getting into Computer Science = 1/3, which is the same probability > as getting into Physics (1/3). Probability that he will get into both > Calc I and CS = 1/6. Probability that he will get into both CS and > Physics is 1/6. Find the prob that he will be able to get into at > least one of the three classes. > I think the answer goes like this. > P(AuBuC)=P(a) + P(b) + P(c) - P(a Intersect B) - P(a intersect c) - > P(b intersect c) + P(a intersect b intersect c). > Therfore, P(a) = 1/2 , Prob of getting into calc I > P(b) = 1/3 , Prob of getting into CS > P(c) = 1/3, Prob of getting into Physics > P(a intersect b) = 1/6 > P(b intersect c) = 1/6 To use the formula, you also need to know P(a intersect c) and P(a int. b int. c), which of course are both 0 because of the scheduling conßict. I think that you probably realized this, but your solution would be clearer if you actually said it ... > So, 1/2 + 1/3 + 1/3 - 2/6 = 5/6 P(a U b U c) = 1/2 + 1/3 + 1/3 - 1/6 - 1/6 - 0 + 0 = 5/6. Otherwise, everything's fine. Brian === Subject: Please Help with this problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7U5XBp22906; I am sure this is not a hard problem for the smart people here. However, I am not one of those smart people so I would really appreciate some help if you don't mind. Here is the question... A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed? Trent === Subject: Re: Please Help with this problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UG2GK09438; === Subject: Re: Please Help with this problem >I am sure this is not a hard problem for the smart people here. >However, I am not one of those smart people so I would really >appreciate some help if you don't mind. >Here is the question... >A farmer with 2000 meters of fencing wants to enclose a rectangular >plot that borders on a straight highway. If the farmer does not fence >the side along the highway, what is the largest area that can be >enclosed? >Trent I won't tell you the answer. In fact I don't think people should just tell you the answer. But if you get a good hint then you can often figure out the answer for yourself, and that you can use the next time a problem appears. so I'll begin with simple things and you can see where it helps. It doesn't say but you are almost certainly expected to have two sides the same length and have both those at right angles to the edge of the highway. Then the third side will perhaps be of a different length but it will be parallel to the highway. Now you already have two different lengths and an area that is going to depend on those two lengths. That means at least two variables. And working with one variable is usually easier than working with two, or more. So, first task, can you come up with a formula for the area that only involves one variable? You need to make use of the other information given in the problem. But coming up with that formula is a key step at this point. It will involve the formula for the area of a rectangle and to use that will involve the lengths and needs, probably, to only have one variable in it, other than area. Then, once you have tried and figured out a formula, you might want to check it. Try to think of a way to check it. Once you have checked it then stop thinking about fence for a moment and just look at the equation, to see if it looks like any kind of equation that you might have seen, something you might be able to get a maximum for. Or can you perhaps rearrange it to turn it into something where you could do that. That should be enough of a hint for you to discover how to do this. I hope it works out === Subject: Re: Please Help with this problem > I am sure this is not a hard problem for the smart people here. > However, I am not one of those smart people so I would really > appreciate some help if you don't mind. > Here is the question... > A farmer with 2000 meters of fencing wants to enclose a rectangular > plot that borders on a straight highway. If the farmer does not fence > the side along the highway, what is the largest area that can be > enclosed? Draw a sketch. Let the two equal sides (perpendicular to the road) each be Ôx.' The length of the third side is therefore: (total fencing available) - (fencing already used) 2000 - 2x A = L*W = x(2000 - 2x) = 2000x - 2x^2 Proceed to find x that maximizes A, then evaluate A for that x. When you get bored doing these, generalize. What can be said about any rectangle that maximizes area given a certain perimeter? P = 2L + 2W Solve for one of L or W, in terms of the other: P-2L = 2W (P-2L)/2 = W Plug that into the area formula: A = LW = L(P-2L)/2 Find the L that maximizes A (remember P is constant.) You should get L=P/4 which implies a square. But most farmers already know the rectangle enclosing the maximum area will be a square (they are smarter than most think.) Already have one natural Ôside' such as a highway, river, etc? No prob. just make one side twice as long as the other two (ie take the fencing you would have used for the fourth side and give it to opposite side, such that L=P/4 and W=P/2). -- Darrell === Subject: Re: Please Help with this problem > A farmer with 2000 meters of fencing wants to enclose a rectangular > plot that borders on a straight highway. If the farmer does not fence > the side along the highway, what is the largest area that can be > enclosed? If the sides of the rectangle are x and y, and y is the one along the highway, then the fence length is 2x + y and the area is x*y. You need to find the maximum of A = x*y given the constraint 2x + y = 2000. If you have learned about Lagrange multipliers, then this is a very simple problem. If you haven't, then you can express y in terms of x and substitute to get A = x(2000 - 2x) = 2000x - 2x^2, for which you can then find a maximum using derivatives. Finally, if you haven't learned derivatives, simply use the property of a quadratic that its extremum is halfway between its zeros; from A = x(2000 - 2x), the zeros of this one are clearly 0 and 1000. hth meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Re: Please Help with this problem > I am sure this is not a hard problem for the smart people here. > However, I am not one of those smart people so I would really > appreciate some help if you don't mind. > Here is the question... > A farmer with 2000 meters of fencing wants to enclose a rectangular > plot that borde them is a subset of the other: T_1 is a subset of T_2, and T_2 is a subset of T_1. To show that T_1 is a subset of T_2, let V be an arbitrary element of T_1. Since G is a subbase for T_1, this means that V is built out of elements of G in a certain way (how?). But G is also a subbase for T_2, so anything built out of elements of G in that way is also in T_2; in particular, V is in T_2. Proving that T_2 is a subset of T_1 is essentially the same argument. What you have to do to make this a real proof is work out the details of the Ôhow?'. [...] Brian === Subject: Re: subbasis for the topology of a set === Subject: subbasis for the topology of a set >I'm trying to prove the following: >If X is a set and G an arbitrary set of parts of X, there is >exactly one topology T(G) on X such that G is a subbasis for T(G) >(the topology``generated'' by G). G subset P(X) = { A | A subset X } Yes that's correct. However for G to be a subbase, it is necessary and sufficient that /G = Union G = X. >Proof: Let X be a topological space. Let G = {} so that G is in >X. G is open by the definition of topological space. Claim G is a >subbasis for T(G). Now, we must show that every open set of G is >the union of finite intersections of sets in G. If G = nulset, then only by insisting that an intersection of no sets is X (an rather abstruse technical point of some debate and little merit) could G produce a base, which would include the nulset in the topology as the union of no sets. >Since G = {}, {} is in G. So {} / {} is in G. G is a subbasis >for T(G) because since G = {}, there's no other set that we could >try to intersect in G. This is not so. The nulset can contain nothing, not even the nulset. >Now we must show that T(G) is the only topology on X. By way of >contradiction, assume that there is a T'(G) on X such that T'(G) >is different than T(G). >Let a, b be two open subsets of G. a U b is in G, a / b is in G. >a U b and a / b must both be open. Specifically, since G = {}, >then a = b = {}. You haven't covered the case for an infinite union of open sets. >Now, I don't really know how to show that T(G) = T'(G), although >they must be because G = {}. I would try showing that all axioms >hold for both, but that only says that that they are topologies; >how would show that they are the same topology? The topology generated by G, is by definition the smallest topology containing all of G. As two smallest topologies are equal, you are done. >Perhaps I couldn't have let G = {}. Advices are welcome. That's bogus as you need to show that, not for just one G, but for any arbitrary G. ---- === Subject: Re: subbasis for the topology of a set > I'm trying to prove the following: > If X is a set and G an arbitrary set of parts of X, there is > exactly one topology T(G) on X such that G is a subbasis for T(G) > (the topology``generated'' by G). G is a set of subsets of X. A topology on X is also a set of subsets of X. > Proof: Let X be a topological space. Let G = {} so that G is in > X. Yikes! Who said G was empty? In fact, the empty set is the only set which is _not_ a subbase for a topology (unless X is also empty). Almost certainly G is not an element of X (in X). > G is open by the definition of topological space. Double yikes! G is not a subset of X (except in rare cases) open does not apply to G. > Claim G is a subbasis for T(G). Triple yikes! Any non-empty collection of subsets of X is a subbase for a topology on X. T(G) is all unions of finite intersections of elements of G which leaves you no choice about T(G) so T(G) is unique. You might need to show that T(G) is in fact a topology. > Now, we must show that every open set of G is > the union of finite intersections of sets in G. Quadruple yikes! We are not given a topology for G; open does not apply to subsets of G. > Since G = {}, {} is in G. So {} / {} is in G. G is a subbasis > for T(G) because since G = {}, there's no other set that we could > try to intersect in G. > Now we must show that T(G) is the only topology on X. Quuintuple yikes! There are probably a whole bunch of topologies on X. [...] We very much need to know the definitions you are using here. One common one : Define T(G) as the smallest topology containing G. In that case you need: 1) There _is_ a topology containing G. [The discrete topology works.] 2) There is a _smallest_ topology containing G. 3) There is a _unique_ smallest topology containing G. I'd suggest you mull over the definition of a topological space and then get back to us with the definitions of subbase and T(G) that you are using and the verbatim statement of the problem. -- Paul Sperry Columbia, SC (USA) === Subject: Caculator Help! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7VN25l05533; I am in PreCalculus, but I don't have a graphing calculator. Is there a place online that is basically a graphing calculator that you can use for free? === Subject: any book on Nonhomogeneous Poisson Process systematically? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I am studying NHPP lately, but it's very hard for me to find a book having detailed description on NHPP. I have tried to find online files, and what I found are just NHPP fragments. I think there should be one book on point processes or similar, which has summarize NHPP related properties together. Do you have any information about it? Neon === Subject: Re: Fractional Integral Equation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) A(d/dx)^(-1/3)f(x)+x^(-1/3)f(x)=1, where A is some constant (may be taken 1). The problem is to obtain a solution for the equation. There is some theory for constant-coefficient fractional integral equations, but in this particular case the problems arise from the non-constant coefficient x^(-1/3). Probably the solution is some special function, obviously the power function x^k is not the solution for any value of k. Does anyone know any systematic way to find out the solution? Numerically the solution can naturally be found, if the boundary conditions are given. A.L. >Does anyone the solution for the following fractional integral >equation: >A*D^{-b}f(x)+x^{-b}*f(x)=1 > I've read Ôyour fractional integral equation'. > you want to solve:A*d/dx ^(-1/3) f(x)+ x^(-1/3)*f(x)=1. > A way to consider fractional derivative /integral ,r real is > putting d/dx^[r] (x^k)=Gamma(k+1)/Gamma(k-r+1)*x^(k-r) a > generalization of d/dx^[n] (x^k)=k!/(k-n)!*x^(k-n) n,k integer. > For f(x)=c*x^k we've got: > A*c*Gamma(k+1)/Gamma(k+4/3)*x^(k+1/3)+c*x^(k-1/3)=1 > What means A?. > May be it helps... === Subject: Re: Fractional Integral Equation Originator: ilya@powdermilk Originator: bergv@math.uiuc.edu (Maarten Bergvelt) [A complimentary Cc of this posting was sent to AL > (d/dx)^(-1/3)f(x)+x^(-1/3)f(x)=1, > Probably the solution is some special function, obviously the power > function x^k is not the solution for any value of k. One (stupid?) idea is to look for a solution in the form F(x) = SUM_k Ak x^(k/3) [It is easy to find Ak, but the series may diverge. But even if it diverges, it may help: you could be able to find an ODE which is satisfied by the formal solution F(x), then solve the ODE for an ordinary function G(x); the result may give you hints about how f(x) behaves.] > Numerically the solution can naturally be found, if the boundary > conditions are given. Eh??? Boundary conditions for non-local equations do exist, but they are very different from ones for ODE (e.g., you assume that f(x)=0 for x<0). Moreover, it is most probably not easy to solve numerically even with such a condition... Hope this helps, Ilya === Subject: Re: Fractional Integral Equation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > (d/dx)^(-1/3)f(x)+x^(-1/3)f(x)=1, > Probably the solution is some special function, obviously the power > function x^k is not the solution for any value of k. > One (stupid?) idea is to look for a solution in the form > F(x) = SUM_k Ak x^(k/3) > Numerically the solution can naturally be found, if the boundary > conditions are given. > Eh??? Boundary conditions for non-local equations do exist, but > they are very different from ones for ODE If you are really only interested in a series representation under natural boundary conditions this is probably easy: You might require that the boundary conditions are such that (under these conditions) the equation is equivalent to f + J^{1/3}Mf =J^{1/3}e where e(x):=1, Mg(x):=x^{-1/3}g(x), and J^{1/3} is the fractional integral operator J^{1/3}g(x):=int_0^x(x-y)^{-2/3}g(y)dy/Gamma(1/3). The right-hand side is easily calculated (I guess something like 3x^{1/3}/Gamma(1/3), but better recalculate). Moreover, since -J^{1/3}M is a compact Volterra operator (at least in L_p[0,T] for sufficiently small p), it has spectral radius 0 and so the inverse of I-(-J^{1/3}M) exists (in these spaces) and can be calculated using the Neumann series. Since -J^{1/3}M is a regular integral operator, this Neumann series will be of the form I+K where K is an integral operator whose kernel is the series of the iterated kernels - I guess it is not hard to calculate an explicit formula for these kernels and hence a (necessarily a.e. convergent!) series for the solution. === Subject: Re: Maximal subgroups of S_n Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > > Are (Is) there (a) published paper(s) or book(s) about the following > question: > > Is it possible to determine all maximal subgroups of S_n, the symmetric > group of degree n. I mean by the existence if a complete description > of maximal subgroups in terms of only n !! > > What is the number of all maximal subgroups of S_n. > > Any information is appreciated. > > > A. Abdollahi > Let me just say a little more about the maximal subgroups of S_n. They > are of > three types: > 1. Intransitive maximals, > 2. Transitive but imprimitive maximals, > 3. Primitive maximals. > The first two of these types are easily described and well understood. > Type 1 consists of subgroups S_m x S_{n-m} for 1 <= m < n/2 > Type 2 consists of wreath products S_m wr S_{n/m} for factors m of n > with > 1 < m < n. > With very few exceptions, these types of subgroups are maximal in S_n. > Finite primitive permutation groups are described by the O'Nan-Scott > Theorem. The coset representations of the almost simple groups form > one category of primitive groups, and probably present the major > obstacle to describing all > maximals of S_n. There is another category, however, known as the > affine > primitive groups. These have degree a prime power p^m, and have a > normal > elementary abelian subgroup of order p^m with a complement isomorphic > to an > irreducible subgroup of GL(m,p). So, inrs on a straight highway. If the farmer does not fence > the side along the highway, what is the largest area that can be > enclosed? > Trent The area of a rectangle of sides x and y is x*y. Right? If y is the side the borders on the highway what is the formula for the amount of fencing? Are you taking calculus? Do you know how to find maximums? Bill === Subject: Question of finite element of a group If x(=/= identity) is in a group G and ord(x) is finite, Then, this fact imply G is a finite group ? If it is true then, Would you please give me some hint for proving that? === Subject: Re: Question of finite element of a group > If x(=/= identity) is in a group G and ord(x) is finite, > Then, this fact imply G is a finite group ? > If it is true then, No, it is not true. Counterexample: Z x Z/2Z. J. === Subject: Re: Question of finite element of a group > If x(=/= identity) is in a group G and ord(x) is finite, > Then, this fact imply G is a finite group ? > If it is true then, > No, it is not true. Counterexample: Z x Z/2Z. > J. You mean Z x Z/2Z ~= Z x Z_2 ? If then, What element is finite in Z x Z_2 - {(0,0)}? I don't know such element (a,b) in Z x Z_2 - {(0,0)}. Sorry, if this question is little stupid. === Subject: Re: Question of finite element of a group >If x(=/= identity) is in a group G and ord(x) is finite, >Then, this fact imply G is a finite group ? >If it is true then, >No, it is not true. Counterexample: Z x Z/2Z. >J. > You mean Z x Z/2Z ~= Z x Z_2 ? > If then, > What element is finite in Z x Z_2 - {(0,0)}? > I don't know such element (a,b) in Z x Z_2 - {(0,0)}. Z/2Z is the finite group with two elements 0 and 1 with 1+1=0. It coincides with the quotient of Z with respect to the subgroup 2Z. In Z x Z/2Z, (0,1) is an element of order 2, whereas (1,0) has infinite order. === Subject: Re: Question of finite element of a group > > If x(=/= identity) is in a group G and ord(x) is finite, > > Then, this fact imply G is a finite group ? > No, it is not true. Counterexample: Z x Z/2Z. > You mean Z x Z/2Z ~= Z x Z_2 ? Z_2 is defined as Z/2Z. > What element is finite in Z x Z_2 - {(0,0)}? Surly you mean What element has finite order in ... (0,1) > I don't know such element (a,b) in Z x Z_2 - {(0,0)}. > Sorry, if this question is little stupid. Yah. Here's a harder one. If every element in a group has finite order is the group finite? === Subject: Re: Question of finite element of a group days. My association with the Department is that of an alumnus. >Yah. Here's a harder one. >If every element in a group has finite order > is the group finite? Not even true if you also require the group to be finitely generated, though it is true if every element is of exponents 2, 3, 4, or 6. This is related to the Burnside Problem. For sufficiently high prime p, there exist finitely generated infinite groups such that ever proper subgroup is either cyclic of order p, or infinite. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Question of finite element of a group > Yah. Here's a harder one. > If every element in a group has finite order > is the group finite? No, let F_n the field with p^n elements (p prime) - clearly of characteristic p. Then F_n is contained in F_{n+1}. Let F be the union of all F_n's. Then every non-zero element in F has finite order (w.r.t. to multiplication), but Fsetminus {0} is not finite. J. === Subject: Re: Question of finite element of a group > Yah. Here's a harder one. > If every element in a group has finite order > is the group finite? > No, let F_n the field with p^n elements (p prime) - clearly of > characteristic p. Then F_n is contained in F_{n+1}. Let F be the union > of all F_n's. Then every non-zero element in F has finite order (w.r.t. > to multiplication), but Fsetminus {0} is not finite. > J. Do you mean F_n is a subfield of F{n+1}? Hanford === Subject: order to classify the maximal > subgroups > of S_n with n = p^m, you would also need to classify all maximal > irreducible > subgroups of GL(m,p), which is another problem on which much work has > been done > (there is a big theorem of Aschbacher describing these subgroups), but > with no > complete solution as yet. To solve it completely, you would probably > need to > find all irreducible representations of all almost simple groups over > the prime > field. I have realized that that last statement I made is nonsense. The maximal subgroup of A_n/S_n of affine type is of course the full semidirect product of an elementary group of order p^m with GL(m,p). The paper of Liebeck, Praeger and Saxl mentioned in Magidin's post essentially reduces the problem of finding the maximals of A_n or S_n to finding the maximal subgroups of the almost simple groups. It also lists all exceptional situations, where subgroups of a type that would normally be maximal in A_n or S_n are not maximal. Derek Holt === Subject: sheaf theory applications X-mailer: epigone Epigone-thread: plootincrur Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I'm intrested in applications of sheaf theory to computer science. somebody may kindly guide me in this direction. === Subject: Re: Under what circumstances is a complete metric space Hilbertizable? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Given a complete metric space X (with no given linear space structure), > under what conditions does there exist a Hilbert space structure on X which > induces the given metric? To put it another way, under what circumstances is > a complete metric space Hilbertizable? Further, how does one determine the > Hilbert space structure on X which induces the given metric? Do you want to know whether the metric space X *is* a Hilbert space or whether it can be embedded isometrically into one? In the latter case: there are simple conditions on the metric d which have been established in several papers by Schoenberg around 1938 or 1939 (eg, 522-536, 1938). Essentially, the negative squared metric -d^2 has to be conditionally positive semi-definite, that is sum_i c_i c_j d^2(x_i,x_j) <=0 for all x_i, x_j in X and for all c with sum c_i = 0. === Subject: Discrete Spectrum, compact perturbation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Hi all, I have a question on the discrete spectrum of a sum of a multiplication with an integral operator. Let X a compact metric space, with some (non-uniform) probability measure on it, k:X times X to R a nice kernelfunction (positive, continuous, smooth, whatever you want) and d a continuous positive function. Let M the multiplication operator with multiplier d, i.e. Mf(x) = d(x)f(x) and S the compact integral operator with kernel k, i.e. Sf(x) = int k(x,y)f(y) dP(y) (all operators defined on the space C(X), or also on L_2(P) if it makes things easier). I am interested in the discrete spectrum of the operator U = M - S. I know that the essential spectrum of U coincides with the (essential) spectrum of M (as S is a compact perturbation). But I want to find out under which conditions there are isolated eigenvalues in the spectrum of U. Does anyone know any methods to deal with such a question? Maybe there are general results on compact perturbation of multiplication operators? I checked some standard perturbation theory books, but I dind't find anything about the discrete spectrum. In case this might help, I am actually interested in a special case of this problem: X some compact subset of R^n, k(x,y) := exp(-||x-y||^2/t), d(x) = int k(x,y) dP(y), P for example a mixture of two Gaussians. Grateful for any hint, Ule === Subject: Re: Discrete Spectrum, compact perturbation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > I have a question on the discrete spectrum of a sum of a > multiplication with an integral operator. > Let X a compact metric space, with some (non-uniform) probability > measure on it, > k:X times X to R a nice kernelfunction (positive, continuous, > smooth, whatever you want) and d a continuous positive function. Some estimates for the spectrum (and a simple suggestion for obtaining estimates using approximations) for the case X=[a,b] (but more general X is not much different) can be found in the last section of E. A. Biberdorf, M. V.8ath, On the spectrum of orthomorphisms and Barbashin operators, Z. Anal. Anwendungen 18, No. 4 (1999), 859-873. === Subject: a multilinear Riesz-Thorin theorem? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Let otimes denote the projective tensor product of Banach spaces, let S be any set, and let p,q in (1,infty) be such that 1/p + 1/q =1. Suppose that f is a multiplier of both ell^p(S) otimes ell^q(S) and ell^q(S) otimes ell^p(S), i.e., f is a bounded function on S times S such that pointwise multiplication with f maps ell^p(S) otimes ell^q(S) into itself and the same for ell^q(S) otimes ell^p(S). Question: Is f then also a multiplier of ell^2(S) otimes ell^2(S)? This question is somewhat reminiscent of the Riesz-Thorin interpolation theorem, but the projective tensor product makes it impossible to apply complex interpolation directly. Any pertinent hints are appreciated! Volker Runde. === Subject: A new proof of the four color thorem Content-Length: 215 Originator: rusin@vesuvius Here is a link for an non-computer proof of the four color theorem which uses graph theory tools. It is a kind of Columbus egg! But certainly it is not trivial. http://arxiv.org/abs/math.CO/0408247 Cahit === Subject: Higher dimensional Klein/Pluecker terminology Content-Length: 345 Originator: rusin@vesuvius Does anyone know of standard names for the higher dimensional analogues of Pluecker coordinates, the Klein correspondence, or the Klein quadric? In particular, if you map a pair of n-space vectors, defining a plane through the origin and corresponding projective line, to the wedge product, it lies on a projective variety. Does it have a name? === Subject: Aldor loci Hi all, Where do Aldor enthusiasts and developers meet? There is a (now past) mention of a future mailing-list or group in aldor.org but I have not been able to find anything. Is this the right group? Also, is the source for Aldor available somewhere? Not that I need/want to examine it now, but I care for the continuity of the language. Somehow I feel uneasy, in that right now Aldor is just a testbed for ideas that will eventually go into commercial Maple. C. === Subject: Re: FFT Program > Hello brothers > I am planning to make FFT program. I am using Numerical reciepe in C > code and had also tried the RealFFT code. > I am facing problem with RealFFT code in sense that i am getting > values in the magnitude much higher, while another reference program > developed in DAOS platform- mathematical C like langauge (THE REAL > ENVIRONMENT) is having values in smaller ranges. > E.g. Range f Re: Question of finite element of a group > Yah. Here's a harder one. > If every element in a group has finite order > is the group finite? > No, let F_n the field with p^n elements (p prime) - clearly of > characteristic p. Then F_n is contained in F_{n+1}. Let F be the union > of all F_n's. Then every non-zero element in F has finite order (w.r.t. > to multiplication), but Fsetminus {0} is not finite. A simpler construction is the subgroup of G = Z_2 x Z_3 x Z_5 x ... of elements of finite support, ie elements (a2, a3, a5, ... ) for which all but finitely many a's are nonzero. === Subject: Re: Question of finite element of a group by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UFHPt05054; > Yah. Here's a harder one. > If every element in a group has finite order > is the group finite? > No, let F_n the field with p^n elements (p prime) - clearly of > characteristic p. Then F_n is contained in F_{n+1}. Let F be the union > of all F_n's. Then every non-zero element in F has finite order (w.r.t. > to multiplication), but Fsetminus {0} is not finite. >A simpler construction is the subgroup of > G = Z_2 x Z_3 x Z_5 x ... >of elements of finite support, ie elements > (a2, a3, a5, ... ) >for which all but finitely many a's are nonzero. In other words, the direct sum of Z_2, Z_3, etc. The prime orders of the cyclic groups are of course irrelevant. Snappier example: Q/Z. There are *lots* of torsion groups after all. Todd Trimble === Subject: Re: Question of finite element of a group > > > If every element in a group has finite order > > is the group finite? > >A simpler construction is the subgroup of > G = Z_2 x Z_3 x Z_5 x ... >of elements of finite support, ie elements > (a2, a3, a5, ... ) >for which all but finitely many a's are nonzero. This was corrected to read, for which only finitely many a's are nonzero. > In other words, the direct sum of Z_2, Z_3, etc. The prime orders > of the cyclic groups are of course irrelevant. That will not do, as the identity (1,1,1,... ) has infinite order. It has to be a subset of the direct sum to avoid such elements, ie, elements (a2,a3,... ) where all but finitely many a's are zero. > Snappier example: Q/Z. There are *lots* of torsion groups after all. Indeed, but not R/Z. === Subject: Re: Question of finite element of a group Correction below. > > If every element in a group has finite order > > is the group finite? > No, let F_n the field with p^n elements (p prime) - clearly of > characteristic p. Then F_n is contained in F_{n+1}. Let F be the union > of all F_n's. Then every non-zero element in F has finite order (w.r.t. > to multiplication), but Fsetminus {0} is not finite. > A simpler construction is the subgroup of > G = Z_2 x Z_3 x Z_5 x ... > of elements of finite support, ie elements > (a2, a3, a5, ... ) > for which all but finitely many a's are nonzero. for which only finitely many a's are nonzero. === Subject: Re: i need a riddle solved!!! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UBmoI18259; >3 candidates were told that they would be blindfolded and that a hat >would be placed on the head of each one. Each hat could be either >black or white. At a signal, the blindfolds would be removed and each >canidate could see the hats of the other 2. Anyone who saw a black hat >was to raise his or her hand. Once the blindfolds had been put on, all >three were given black hats. When the blindfolds were removed all >three raised their hands. If you were one of the 3 candidates, would >you be able to determine the color of your hat. How would you know? >You may assume that the other 2 candidates are reasonably bright >people. This is a classic. Think of it like this: suppose you hadn't said all three were given black hats, and suppose you had a white hat, and all three candidates raised their hands. Now visualize the situation from the perspective of a black-hat wearer. Since s/he is bright, what would s/he deduce almost immediately? (To make the problem more compelling, you should have said these are candidates for a job and the first to say which color hat s/he is wearing gets the job.) If s/he didn't say immediately, what would you deduce? Todd Trimble === Subject: Re: water pelican === Subject: subbasis for the topology of a set I'm trying to prove the following: If X is a set and G an arbitrary set of parts of X, there is exactly one topology T(G) on X such that G is a subbasis for T(G) (the topology``generated'' by G). Proof: Let X be a topological space. Let G = {} so that G is in X. G is open by the definition of topological space. Claim G is a subbasis for T(G). Now, we must show that every open set of G is the union of finite intersections of sets in G. Since G = {}, {} is in G. So {} / {} is in G. G is a subbasis for T(G) because since G = {}, there's no other set that we could try to intersect in G. Now we must show that T(G) is the only topology on X. By way of contradiction, assume that there is a T'(G) on X such that T'(G) is different than T(G). Let a, b be two open subsets of G. a U b is in G, a / b is in G. a U b and a / b must both be open. Specifically, since G = {}, then a = b = {}. Now, I don't really know how to show that T(G) = T'(G), although they must be because G = {}. I would try showing that all axioms hold for both, but that only says that that they are topologies; how would show that they are the same topology? Perhaps I couldn't have let G = {}. Advices are welcome. === Subject: Re: integrals of y = 1 / (1 + x^n) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UGfEI12906; >I wanted to calculate the area 0 --> +infinite below the functions >like 1 / (1 + x^2), changing the exponent of x to integers larger than >if n = 2, the area is = pi/2 >if n = 3, the area I calculated is (pi / (2 * sqrt(3))) (correct?) >I tried several times to calculate the integral for n = 4, but always >got stumped. >Anyone knows the integral for n = 4? Or just the area value? Or, even >better, knows of a webpage with integrals and areas for n = 2 to 10 ?? >(the language I studied maths in, is not English...) >Roberto N. I'll assume you are using methods of elementary calculus to do these problems. I got 2pi/(3*sqrt(3)) for the case n = 3 (the arctan term in the indefinite integral is (1/sqrt(3))arctan((x - 1/2)/(sqrt(3)/2)); the limit at infinity produces your pi/(2sqrt(3)), and the value at 0 is -pi/(6sqrt(3)). Subtract to get my answer. Did you factorize 1 + x^4 into quadratic factors? It may not be obvious without using complex numbers (basically, find the complex roots of x^4 = -1, using (cos t + rom code 10 pow +3, while DAOS one shows value 10 pow -3 > NOTE: DATA Pattern observed is similar. Kinks in data shapes is same. > i.e. plot of data is same > I have to plot Power Spectra of time domain data to Frequency domain: > Is the > Sandeep > 1. Writing an FFT program is not what this newsgroup (sci.math.symbolic) is about. > 2. Numerical recipes book generally has programs that are not robust. and in this case not particularly fast either. > 3. There are hundreds of FFT programs around. Find a good one. > 4. Your note is incoherent. > 5. Good luck. RonL === Subject: Re: new version of Fermat available by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8AGJ7r28070; > Two independent researchers have discovered that Fermat is the > fastest system in the world for multivariate polynomial GCD. Their > letter and data are also on my website. >Could you give the link to the precise data? >On your page, I see that the pb was to take the gcd of a*b and a*c >with e.g. total degree of a*b and a*c being 60 or 30 etc., but >it would be interesting to know what the total degree of a is >(is it half the total degree of a*b?) >Bernard Parisse The data is now at htti sin t)^n = cos nt + i sin nt; here n = 4, and nt = pi + multiples of 2pi gives -1 on the right). One gets (x^2 - sqrt(2)x + 1)(x^2 + sqrt(2)x + 1) = 1 + x^4, where the first quadratic factor comes from the complex roots in the first and fourth quadrants, and the other from the roots in the second and third quadrants. It is then straightforward but tedious to work out the indefinite integral of 1/(1 + x^4); my back-of-the-envelope calculation gives (1/2sqrt(2))((1/2)ln(x^2 + sqrt(2)x + 1) - (1/2)ln(x^2 - sqrt(2)x + 1) + arctan(x + sqrt(2)/2) + arctan(x - sqrt(2)/2 ) + C which if correct gives pi/(2sqrt(2)) as the definite integral. In general, here's how x^n + 1 factorizes, based on the idea of using complex numbers as above. n odd: (x + 1)*Product_{i=1}^{(n-1)/2} (x^2 - 2(cos t_i)x + 1) n even: Product_{i=1}^{n/2} (x^2 - 2(cos t_i)x + 1) where in both cases t_i = (2i - 1)pi/n. In principle one can work out the indefinite integrals from this, but it's a slog. Better to use Mathematica or Maple or something. To get the definite integrals, it is probably more efficient to use the method of complex residues. For n even, it suffices to calculate Integral_{-inf}^{inf} 1/(1 + x^n) dx by considering a contour going from -R to R on the real axis, followed by a semicircle of radius R in the upper half-plane taken counter-clockwise. For large R, the contribution to the integral from the semicircle part is vanishingly small, so 1/(2pi)i times this integral is the sum of the residues at the poles which occur in the upper half-plane (i.e. at the roots of 1 + z^n with positive imaginary part). To save you the trouble, the answer turns out to be (2pi/n)/(sin pi/n)) i.e. for even n, your definite integral is (pi/n)/(sin (pi/n)). Cool! Todd Trimble === Subject: Re: integrals of y = 1 / (1 + x^n) > To get the definite integrals, it is probably more efficient to > use the method of complex residues. For n even, it suffices to > calculate > Integral_{-inf}^{inf} 1/(1 + x^n) dx > by considering a contour going from -R to R on the real axis, > followed by a semicircle of radius R in the upper half-plane > taken counter-clockwise. For large R, the contribution to the > integral from the semicircle part is vanishingly small, so 1/(2pi)i > times this integral is the sum of the residues at the poles which > occur in the upper half-plane (i.e. at the roots of 1 + z^n with > positive imaginary part). To save you the trouble, the answer > turns out to be > (2pi/n)/(sin pi/n)) > i.e. for even n, your definite integral is (pi/n)/(sin (pi/n)). > Cool! It's easier to use the contour 0 -> R -> R*exp(2pi/n) -> 0. Then you only enclose one pole, and the last leg of the integral is a multiple of the one we want. === Subject: Re: integrals of y = 1 / (1 + x^n) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UM3UZ09638; Trimble) > To get the definite integrals, it is probably more efficient to > use the method of complex residues. For n even, it suffices to > calculate > Integral_{-inf}^{inf} 1/(1 + x^n) dx > by considering a contour going from -R to R on the real axis, > followed by a semicircle of radius R in the upper half-plane > taken counter-clockwise. For large R, the contribution to the > integral from the semicircle part is vanishingly small, so 1/(2pi)i > times this integral is the sum of the residues at the poles which > occur in the upper half-plane (i.e. at the roots of 1 + z^n with > positive imaginary part). To save you the trouble, the answer > turns out to be > (2pi/n)/(sin pi/n)) > i.e. for even n, your definite integral is (pi/n)/(sin (pi/n)). > Cool! >It's easier to use the contour 0 -> R -> R*exp(2pi/n) -> 0. Then you only >enclose one pole, and the last leg of the integral is a multiple of the one >we want. Todd === Subject: Re: integrals of y = 1 / (1 + x^n) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UFgCg07311; >I wanted to calculate the area 0 --> +infinite below the functions >like 1 / (1 + x^2), changing the exponent of x to integers larger than >if n = 2, the area is = pi/2 >if n = 3, the area I calculated is (pi / (2 * sqrt(3))) (correct?) >I tried several times to calculate the integral for n = 4, but always >got stumped. >Anyone knows the integral for n = 4? Or just the area value? Or, even >better, knows of a webpage with integrals and areas for n = 2 to 10 ?? >(the language I studied maths in, is not English...) >Roberto N. inf dx/(1 + x^n) = .a6/(nsin(.a6/n)) 0 n = 2: = .a6/2 = 0.5.a6 n = 3: = .a6/(3sin(60)) = = 2.a6/3^(3/2) = 0.38490....a6 n = 4: = .a6/(4sin(45)) = = .a6/(2sqrt(2)) = 0.35355....a6 n = 5: = .a6/(5sin(36)) = = 2.a6/5^(5/4)sqrt[(sqrt(5)+1)/2] = 0.34026....a6 n = 6: = .a6/(6sin(30)) = .a6/3 = 0.33333....a6 etc. n -> inf: -> 1 === Subject: Re: integrals of y = 1 / (1 + x^n) >I wanted to calculate the area 0 --> +infinite below the functions >like 1 / (1 + x^2), changing the exponent of x to integers larger > than >2. >if n = 2, the area is = pi/2 >if n = 3, the area I calculated is (pi / (2 * sqrt(3))) (correct?) >I tried several times to calculate the integral for n = 4, but always >got stumped. >Anyone knows the integral for n = 4? Or just the area value? Or, even >better, knows of a webpage with integrals and areas for n = 2 to 10 > ?? >(the language I studied maths in, is not English...) >Roberto N. > inf > dx/(1 + x^n) = .a6/(nsin(.a6/n)) > 0 ??? It's best to use simply ASCII. Otherwise, you risk illegibility. Just in case the result above is illegible to someone else: The character .a6 above should be thought of as representing pi. Doing so, the result is correct. David > n = 2: = .a6/2 = 0.5.a6 > n = 3: = .a6/(3sin(60)) = > = 2.a6/3^(3/2) = 0.38490....a6 > n = 4: = .a6/(4sin(45)) = > = .a6/(2sqrt(2)) = 0.35355....a6 > n = 5: = .a6/(5sin(36)) = > = 2.a6/5^(5/4)sqrt[(sqrt(5)+1)/2] = 0.34026....a6 > n = 6: = .a6/(6sin(30)) > = .a6/3 = 0.33333....a6 > etc. > n -> inf: -> 1 === Subject: Re: Maths Newsgroups by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UHEiL16085; pls send me the past qustion for math === Subject: taylor series expansion by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UIY8k23548; Can anyone help me with this. taylor series expansion of sqrt((x+y)^2 +z^2)) === Subject: Re: taylor series expansion > Can anyone help me with this. > taylor series expansion of sqrt((x+y)^2 +z^2)) With respect to which variable? With what constraints on the others? Or do you mean a multi-variable expansion? === Subject: confusing riddle by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7UMQO211195; This is a riddle that I need an answer to asap.....if anyone out there can help me please email me with what you have here it goes... One disturbs the peace and another keeps it. The third likes to take matters into its own hands. All three come from the same chrome horse corral. Name the corral and all three. anyone? === Subject: Re: confusing riddle > This is a riddle that I need an answer to asap..... Could you please send such things to rec.puzzles rather than a mathematical news group? Ken Pledger. === Subject: Re: Clashing of local variable names and symbolic inputs to procedures |> Now, here is my question. Suppose that I wanted QD to work for arbitrary |>arguments, be they symbolic, numeric or some mixture of the two, without any |>restriction whatsoever on the symbolic names chosen. Then, it seems to me, I |>may not use _any_ (valid!) local variables whatsoever, for fear of a |>possible clash with something symbolic in the input list. Such a clash |>cannot, of course, happen if the ip://www.bway.net/~lewis/fermat/gcdcode.tar.gz Robert H. Lewis === Subject: sybmbolic computation programs Hello. I need to find several sybmbolic computation programs, particulary a free Linux/Unix versions. I have now information about Maxima, MuPad and Mathomatic, but it's not enough for my work. Could you say me about any other programs like that? I mean only full-functional programs (actually, Mathomatic is not quite that what I need). P.S. Excuse me for my terrible English... === Subject: multiple summation stack limit problems, Maple 9.0 Why does Maple so quickly run out of stack space when I use multiple summations? For instance, when I type: > sum(sum(binomial(n,m),n=0..N),m=0..M); Maple takes over 15 seconds and returns something horrendously complex. Actually, I have an only slightly more complicated function than this involving a double summation and a few combinatoric terms, and when I try to define this function, Maple runs for over 10,000 seconds before giving me Kernel failure! Execution stopped: stack limit reached. have! === Subject: Re: multiple summation stack limit problems, Maple 9.0 > For instance, when I type: > sum(sum(binomial(n,m),n=0..N),m=0..M); > Maple takes over 15 seconds and returns something horrendously > complex. The time used and the complexity of the answer in this case can be greatly reduced if you make the assumptions: > assume(N::nonnegint, M::nonnegint, n::nonnegint, m::nonnegint); === Subject: A new book on Maple for users and developers On Fultus web site http://writers.fultus.com/aladjev/book01.html you can find information about a new book on Maple that can be useful enough both for beginners, and the advanced Maple users, as well as for Maple developers and developers of Maple appendices. === Subject: Re: Does MuPad 3.0 support Elliptic Integrals? > I particularly want a CAS which can accurately calculate elliptic integrals. > Maple fails to do this... > http://www.math.rwth-aachen.de/mapleAnswers/html/1346.html This bug has been fixed in Maple 9.5. -- Thomas Richard Maple Support Scientific Computers GmbH http://www.scientific.de === Subject: Re: 2D / 3D Differential Geometry Wizard > Hi! > I would like to know which Maple version this wizard fits to. I > work with Maple 6. I haven't actually tried it, but the download page mentions Maple 8: http://www.graphtree.com/Maple/atlasWizard/DownLoad/ download.htm Note that Maple 6 does not support the Maplets technology. -- Thomas Richard Maple Support Scientific Computers GmbH http://www.scientific.de nput has no names in it - just as in any |>(numeric) programming language I can think of, the names of local variables |>are unimportant, since they _cannot_ clash with anything outside their |>scope. > Here, however, the situation is different. How can this clash be >avoided? There shouldn't be any clash. Local variables and global variables of the same name are treated as completely different, even though they look the same. For example: > f:= proc(x) local y; x - y end proc; f(y); y - y The result doesn't simplify to 0 because there are two different y's here: the first is the global y (corresponding to the formal parameter x) and the second is the local y. I haven't checked to see what happens to your code with symbolic inputs, but whatever the problem is, it shouldn't be a clash between local and global variables. ' === Subject: Easy way to solve solvable QUINTICS by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7U2dFb03513; Hello all, For those interested in the quintic, this paper might be of some help: An Easy Way To Solve Solvable Quintics Using Two Sextics ABSTRACT: Using a method initially developed by George Young (1819- 1889), Arthur Cayley (1821-1895), and later by George Watson (1886- 1965), an explicit quartic is constructed to enable the solution in radicals of a quintic when it is a solvable equation. Not one, but two sextic resolvents are derived which are important to forming the coefficients of this quartic. Certain difficulties and their solutions as well as a novel consequence to the method are also addressed in this paper. Mathematics Subject Classification. Primary: 12E12; Secondary: 12F10. http://www.geocities.com/titus_piezas/ Just click on the link in the website to the pdf file. Titus === Subject: Re: Where do math symbols originate? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7U5XBG22912; >THE reference, albeit a little old, on this subject is Cajori, Florian, A >History of Mathematical Notations which is still available from Dover. would >immediately give a straight yes or no answer to your question, his book >might permit your two opposing teams to bring many examples and >counter-examples in your discussions. >Bernard Mass.8e hope to get down to Purdue and see if it or something like it might be there. The squabble down in the Lab has quieted down in the last 2 years. The abstract math fellows are in some type of deep catatonic state, stunned into silence by something that, to them, seems completely unpossible. It turns out that as each and every cell in the body lives and breathes it is running the general respiration reaction: organics + oxygen => water + carbon dioxide + ~energy (eq. 1) If you remember your biochemistry, that happens in processes like the Kreb's tri-carboxylic acid cycle and similar well-oiled channels. One point is, the water molecules are ~created in a regular sequence, forged in concert with our energizing experience in the world, extruded from each process in incremental units. Now, as you might also remember, the experimentalists locked in the sub-basement at the Lab have routinely been aligning rod magnets along the radii of a tetrahedron and sticking two north poles to two south poles. After grueling analysis or the resulting artifact, they have discovered that since that thing has six edges, there are at least six way such a thing can be oriented in some imposed field. This leads to a 6-to-the-nth type of math. This means that for a sequence of, say, 16 units, there are 6^16 or 2.82x10^12 ways to arrange those sixteen units in a stack or chain. Many of the fellows in the lab have not had that many different experiences, so they are somewhat excited. The logic goes, as best I can follow it, that this is some sort of smoking gun model of consciousness. The analog math don't lie. The water molecules are formed in distributed, parallel processing channels and as the units are extruded, they are jiggled into one of the six positions BY the subtle inßuences of the inbound (sensory) quantum gravitational vibrations. Once the chain is truncated at some length (perhap for some, a consistent length), it's like a little resonant reßection of the ~current environmental experience. If Mom was yelling, HOT!, then Ôhot!' it is. And so forth. Interestingly, the chain one person maps to Ôhot!' doesn't have to be the say as someone else does. Now, the fact is, they really don't know whether to invoke or apply a fancy call to quantum gravitation, or whether to just refer to inßuences of the surrounding, more unified field. The point is, they have a non-linearly expanding point. Water makes up something like 80% of just about everything important in and near widgets of consciousness. Water is FORMED and extruded in rational units, sequentially, during respiration, in concert with the vibration of our experience. Water does form in ordered water arrangements. So-called bound water, water that gets knit together in organic materials, is energetically very stable and persistance. Anharmonic oscillators, matrix math, torsion-induced sine and cosine tables, even wave-equations appear to be quite naturally at home in such a resonance-based computational/associative media. Plus, running such a water-based internal analog math system still allows for wild variations in the initializations and in the ways linguistics, perceptions, beliefs, values, etc., can be formed and held and changed. Com links to deeper levels of organizations and comminication are not automatically excluded. As well, dream imagery, prayer, imagination and, I suppose, even the creation of abstract math symbols and models can all be created using the same sort of media/relationhip -- analog math symbols. This has come as a bit of a shock to the abstract math folks. Once their denial shattered, they have just been sitting there, day after peaceful day, watching their breath, trying to figure out if there is anything mathematical that they can say. So far, they are still completely silent on the matter. to it. Ralph Frost Imagine consciousiness as a single internal analog language made of ordered water, and its variants. http://ßep.refrost.com ...and love your neighbor as yourself Matthew 19:19 >uqpg8nsacsr0c4@corp.supernews.com... > [Rated RFTOR in s.p.r.] > The normally well-mannered, docile fellows in the advanced symbolics and > unification division here at the lab started a bad squabble the other day > that perhaps someone in SPR can speak into and put an end to ...so those > guys can shut up and get back to calculating -- to doing real work. > One group, the experimentalists, came out with the idea that since > abstract math symbols are linguistic artifacts, all abstract math symbols > and thus all abstract mathematics arise from and thus are secondary to a > very small number of very ßexible physical structures and relationships > down in the guts of human consciousness. > This made sense to the experimentalists, because they couldn't escape >the > notion that their plans for improved experimental rigs, as best they can > tell, ßow up out of various rearrangements of the same ßexible physical > structures. > However, as readers might well imagine, the abstract math contingent will > have none of this troublesome talk and taunting, even though they > can only grunt and groan ineffable noises when confronted with the > observation that the unified thing itself does all it's math ßawlessly > using the so-called analog math symbols. To complicate matters, some of > abstract folks are becoming a bit intrigued by the idea that the step up > the emerging, more unified models in fact very might require the > introduction and shift to using more robust, more synchronous math symbols > than folks use in the less unified models. At this moment, though, none of > them can venture a guess on which analog math symbol might be the absolute > best to be deployed or how such an awkward notion might be refined and > developed. > Anyway, I am asking for your help to resolve this squabble one way or >the > other. > -- > Ralph Frost > Looking for a desktop model to help you ponder this topic? > http://ßep.refrost.com -- now with secure online ordering > Use more robust symbols > Seek a thought worthy of speech. === Subject: Re: Does Magma Still Exist? Keywords: Magma > Recently, I have been doing web searches for information about BCH > codes. > A computer algebra package, Magma, developed at the University of > Sydney, and based on an earlier program called Cayley, was mentioned > several times. So, today, I looked for more information about it. > However, most of the references I saw referred to an old site, which > contained only a link to what was supposed to be the current site, but > which was broken. A newsgroup post in 2002 gave a link, which did not > work. > because their Internet resources are limited, and so perhaps some > of my difficulties are merely due to network congestion issues. Or > has something happened to Magma? Nothing so interesting, I'm afraid. The web server was unexpectedly out for a couple of days. It is working again now, and the correct link is: http://magma.maths.usyd.edu.au/ In particular, this segment of the online help http://magma.maths.usyd.edu.au/magma/htmlhelp/text1451.htm contains the part of the Magma Handbook specifically mentioning BCH codes, although there are uses in other parts of the Handbook. (You may wish to wander up a level or two in the documentation to look at more of the coding facilities offered by Magma.) Geoff. -- Geoff Bailey Email: geoff@maths.usyd.edu.au Computational Algebra Group Phone: +61 2 9351 3114 University of Sydney Fax: +61 2 9351 4534 === Subject: Re: Does Magma Still Exist? Did you not find this web site? http://magma.maths.usyd.edu.au/magma/MagmaInfo.html === Subject: Re: Computing Huffman codes for Fibonacci numbers on a Turing Machine > Computing Huffman codes for Fibonacci numbers on a Turing Machine Also at Here is a fragment from the log file: ##### < Run 1.2 > Input word(s) & head start position(s) on tape(s) ##### Tape#0 : Word = a * a * a a * a a a * a a a a a * a a a a a a a a ; Head pos = 0 Tape#1 : Word = b ; Head pos = 0 Tape#2 : Word = b ; Head pos = 0 Note. This input word is Fibonacci sequence (1, 1, 2, 3, 5, 8) ##### < Run 1.2 > Processing ##### ----- < Run 1.2 > Initial Configuration ----- State : q0 Tape#0 : [a] * a * a a * a a a * a a a a a * a a a a a a a a Tape#1 : [b] Tape#2 : [b] < Run 1.2 > Applied Rule# 0 : q0 [ a b b ] ---> q0 [ (a,L) (b, L) (b, L) ] ///////////// // Omitted // ///////////// ----- < Run 1.2 > Configuration#2712 ----- State : qf Tape#0 : [x] b Tape#1 : x 1 1 1 1 a a 1 1 1 0 1 a 1 1 1 0 0 a 1 1 0 a a a 1 0 a a a a a 0 a a a a a a a [a] Tape#2 : [x] b < Run 1.2 > Success : Current state (qf) is halting one Weights and their codes on Tape#1 Weight Code ------ ---- 1 11101 1 11100 2 1111 3 110 5 10 8 0 -- Alex Vinokur http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn === Subject: This integral blows my PC (Maple 9.5) Help ! While integration p1 from 0 to 1 (int(p1(lambda),lambda=0..1)) the PC works and works.. And memory usage increasing and increasing until the whole stops.. Tried on 2 different PCs (one w/ w2k one with XP). Same result... Can someone help? What do I wrong? May be there are other ways to integrate? the function to integrate: p1:=lambda-> c1/(lambda^5*(exp(beta/(lambda*T))-1)); where beta:=Constant(h)*Constant(c)/Constant(k); c1:=2*Pi*Constant(h)*(Constant(c))^2; T:=3200 This function p1 is from physics and known as Rayleight-Jeans formula. I can plot p1, it is continuous as it should be and the integral will be (is) finite.. Aran === Subject: Re: This integral blows my PC (Maple 9.5) Help ! |>While integration p1 from 0 to 1 (int(p1(lambda),lambda=0..1)) the PC |works |>and works.. ... |>p1:=lambda-> c1/(lambda^5*(exp(beta/(lambda*T))-1)); |>where |>beta:=Constant(h)*Constant(c)/Constant(k); |>c1:=2*Pi*Constant(h)*(Constant(c))^2; You don't need to write Constant(h) etc. Just h will do fine. |>T:=3200 I think you forgot to say that you were assuming h, c and k were positive. Without any assumption, int works fine for me in Maple 9.5, in 0.391 seconds on my computer, except that since Maple doesn't know those signs it gives an answer in terms of a limit as lambda -> 0+. Maple seems to have trouble evaluating that limit when h, c and k are assumed positive (or when values such as 1 are used for those constants). ' === Subject: Re: This integral blows my PC (Maple 9.5) Help ! > While integration p1 from 0 to 1 (int(p1(lambda),lambda=0..1)) the PC works > and works.. And memory usage increasing and increasing until the whole > stops.. Tried on 2 different PCs (one w/ w2k one with XP). Same > result... When I try, Maple can get the antiderivative, which is a sum of polylogs, but Maple cannot compute the limit at 0. Ignoring the constants, you can observe this in int(1/x^5/(exp(1/x)-1), x= 0..1) (same integral as yours except the constants are different). > Can someone help? What do I wrong? You did nothing wrong. > May be there are other ways to integrate? In this case you can make a substitution in the integral with the student[changevar] command. > the function to integrate: > p1:=lambda-> c1/(lambda^5*(exp(beta/(lambda*T))-1)); > where > beta:=Constant(h)*Constant(c)/Constant(k); > c1:=2*Pi*Constant(h)*(Constant(c))^2; > T:=3200 Rather than use the ScientifcConstants package, I'll just use symbolic h, c, and k, and assume they are positive. You can substitute in the actual constants at the end. > restart; > assume(h>0, k>0, c>0); > beta:= h*c/k; T:= 3200; c1:= 2*Pi*h*c^2; > p1:= lambda-> c1/lambda^5/(exp(beta/lambda/T)-1); > J:= Int(p1(lambda), lambda= 0..1); # Note capital I > student[changevar](u= exp(beta/lambda/T), J, u); > value(%); -6400*Pi*c*k*ln(A-1) -6400*I*Pi^2*c*k -Pi*h*c^2/2 +83886080000000*Pi^5*k^4/3/c^2/h^3 -61440000*Pi*k^2/h*polylog(2,A) +393216000000*Pi*k^3/c/h^2*polylog(3,A) -1258291200000000*Pi*k^4/c^2/h^3*polylog(4,A) where A = exp(c*h/3200/k); The imaginary I is cancelled by the imaginary components of the polylogs. === Subject: Re: Mathematica Vs. Matlab > Interesting reading is this comparison betwen Mathematica an Matlab. > Here are my observations: > I had to solve a system of four nonlinear equations (something like > that:-K1*V2*sin(th2-K2)-K3*V2*Vmid*sin(th2-thmid)-K4) the other three > are similar. > What I found was that the results I got with function Findroot in > Mathematica 5.0 was an order more accurate than fsolve of Matlab Were you using the default tolerances for FSOLVE? Very basically, as soon as the function values get Ôclose enough' (as defined by TolFun) to 0 and the variable values aren't changing by very much (TolX), FSOLVE says Good enough. Try changing the TolX and TolFun parameters in your call to FSOLVE to change the definition of Ôclose enough' and see if that improves the solution. If it doesn't, could you send the equations for the problem, the solution you received from Mathematica, and the solution you received with the default tolerance and the tighter TolX and TolFun using FSOLVE to our support staff so that we can investigate what's going on? -- Steve Lord slord@mathworks.com === Subject: Re: QUINTIC BY RADICALS > As several others have informed you, as long as there are _some_ quintics > that cannot be solved by radicals involving rational functions of the > coefficients, your example does not contradict the work of Abel and Galois. > I found a useful web site for this stuff, including solutions of quintics in > terms of hypergeometric functions: > http://icl.pku.edu.cn/yujs/MathWorld/math/q/q111.htm The current MathWorld link is http://mathworld.wolfram.com/QuinticEquation.html See also http://library.wolfram.com/examples/quintic/ Paul -- Paul Abbott Phone: +61 8 9380 2734 School of Physics, M013 Fax: +61 8 9380 1014 The University of Western Australia (CRICOS Provider No 00126G) 35 Stirling Highway Crawley WA 6009 mailto:paul@physics.uwa.edu.au AUSTRALIA http://physics.uwa.edu.au/~paul === Subject: Re: QUINTIC BY RADICALS by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7U2d8j03244; >For example, x^5+3*x+1 has Galois >group S_5, and therefore is not solvable in radicals. >Robert Israel israel@math.ubc.ca >Department of Mathematics href=http://www.math.ubc.ca/~israel>http://www.math.ubc.ca/~ israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2 >Let f(x)=x^5+(3+k1)*x+(1+k2) >What the value of k1 and k2 that make f(x) solvable in radicals >Such that > k1 and k2 are rational complex numbers > magnitude(k1)+ magnitude(k2) is minimum >M. A. Fajjal Actually there is no lower limit of k1 and k2. There are unlimited solutions of k1 and k2 to get x^5 + (3+k1)*x + 1+k2 = 0 solvable in radicals I have got for example k1= 44434258558174122962160348326776536558145581228690199781440/ 1442174741293 25277147705355136900000000000000000000000000000000000000000 k2= 14708205761097273214405059026344745912854711319982092916802495 50230760545 091499893691392/ 1442174741293252771477053551369000000000000000000000000000000 0000000000000000000000000000000000000000 u=-126291550529446/100000000000000 v=-664962210424963/1000000000000000 Then The real root = u*((4*v+3)/(v^2+1)/5)^(1/5)*((1-2*v+2*(v^2+1)^(1/2)+(5-4*v+8* v^2+(500*(v^2+1 )*(1-2*v)+(4*v+3)^3)/125/(1+v^2)^(1/2))^(1/2))^(1/5)-(-(1-2*v -2*(v^2+1)^(1/2) -(5-4*v+8*v^2-(500*(v^2+1)*(1-2*v)+(4*v+3)^3)/125/(1+v^2)^(1/ 2))^(1/2)))^(1/5 )-(-(1-2*v-2*(v^2+1)^(1/2)+(5-4*v+8*v^2-(500*(v^2+1)*(1-2*v)+ (4*v+3)^3)/125/( 1+v^2)^(1/2))^(1/2)))^(1/5)-(-(1-2*v+2*(v^2+1)^(1/2)-(5-4*v+8 *v^2+(500*(v^2+1 )*(1-2*v)+(4*v+3)^3)/125/(1+v^2)^(1/2))^(1/2)))^(1/5)) where u=-126291550529446/100000000000000 v=-664962210424963/1000000000000000 r1=-0.331989029584597 M. A. Fajjal === Subject: Another problem, LITTLE DIFFICULT I need to do some computation for a cache architecture. I have arrived at a simplified model and I need help to get it solved. Here is the problem. Balls are being put into and removed from a bucket, one arrival and one departure in each unit time. Arivals: Balls that are arriving have equal probability that they can have any color from b different colors. Departure: Balls from the bucket are departing in following way, a) All colors are selected, whose balls haven't been departed in last c unit times. b) Then from selected colors, a ball is departed of the color which has maximum number of balls. Thus, in every window of c unit time, balls of c different colors need to be sent. (c < b) The bucket has N balls initially, with equal number of balls of every b color. Now, when this process is carried out for very long time say T (T is trillions), how many times will the bucket be having balls of only c colors (c < b). In other words, after a long time and in steady state, whats is the probability that the bucket will have balls of only c colors (c < b). Sailesh === Subject: Help??? I am not an algebra person....my daughter needs to know..,,if you have 3 dice totaling 6....how many times can you get the total 12? What is the algorithm? === Subject: Re: help..... > Sorry.....I worded the previous question wrong.... > If you have 3 dice each has the numbers 1-6....how many times can you get > the total 12? What is the algorithm? A preliminary point: if you post to more than one news group, please cross-post (as I've now done) to save different people from duplicating one another's efforts. The simplest solution to your problem may be just to list all the possibilities and count them. However, since you ask What is the algorithm?, here we go! Here's a simpler example to start with. Suppose you have three coins, each marked 1 on one side and 2 on the other. If you toss all three, in how many ways does each total turn up? You can get a total of 3 in one way (111). You can get a total of 4 in three ways (112, 121, 211). You can get a total of 5 in three ways (122, 212, 221). You can get a total of 6 in one way (222). Now here's a more high-brow way to see that. Represent the two sides of each coin by the exponents in the function x^1 + x^2. Tossing all three coins then corresponds to (x^1 + x^2)^3; because multiplying that out gives for example the term in x^5 as (x^1)(x^2)(x^2) + (x^2)(x^1)(x^2) + (x^2)(x^2)(x^1) = 3(x^5). That coefficient 3 shows how many ways a total of 5 can occur. Now set up a similar generating function for your problem. Represent the six faces of each die by the exponents in the function x^1 + x^2 + x^3 + x^4 + x^5 + x^6. Tossing all three dice then corresponds to (x^1 + x^2 + x^3 + x^4 + x^5 + x^6)^3 as above. The number of ways to get a total of 12 will be the coefficient of x^12 in the expansion of that. So much for the combinatorial theory. The rest is just algebra to calculate the coefficient of x^12. x^1 + x^2 + x^3 + x^4 + x^5 + x^6 = x(1 - x^6)/(1 - x) by the usual formula to sum a finite geometric series. Therefore (x^1 + x^2 + x^3 + x^4 + x^5 + x^6)^3 = (x(1 - x^6)/(1 - x))^3 = (x^3)((1 - x^6)^3)((1 - x)^(-3)) = (x^3)*(1 - 3(x^6) + 3(x^12) - x^18)* *(sum from n = 1 to oo of binomial(-3,n)*(-x)^n). The coefficient of x^12 is binomial(-3,9)*(-1)^9) - 3*binomial(-3,3)*(-1)^3) = 55 - 30 = 25. Well, you did ask. ;-( Ken Pledger. === Subject: Help..... Sorry.....I worded the previous question wrong.... If you have 3 dice each has the numbers 1-6....how many times can you get the total 12? What is the algorithm? === Subject: Re: Pls help with this expression by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7V4fxw08646; >((2*pi)^-1) * |[S]|^(-0.5) exp(0.5*transpose([V])*inv([S])*([V])) >Where >^ refers the power of, >[S], [V] represents a matrix or vector, >transpose and inv are matrix operations >| | represents determinant >I have a problem with this expression as the results of my >determinant >is negative and cause a sqrt(-ve number). >rdgs, >jax > That looks a heck of a lot like the Gaussian (normal) probability >density in n dimensions. I did a search on Gaussian probability Density. The above equation seems to be it. However, there is some things i do not understand. In the equation, there is |[s]|^ -0.5. That means the determinant is squared root. What will happen if the results of the determinant is negative? A normal C program will give an error for square rooting a negative value. I don't think complex numbers should be involved here either. Or should i always take the abs value of the determinant result? Within the exponential function, an inverse is being performed. This usually emit a large enough value to give the result of zero on the exponential computation (eg: exp(-ve large value)=0). What will this mean? I am sorry if my questions are too basic, but i am really weak in statistical computations. rdgs, jax === Subject: Re: Abstract/Modern Algebra by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7V4fx808626; >Does anyone know of a online/ distance learning course in >Abstract/Modern Algebra? I'm interested too. Did you receive any useful replies? === Subject: inequalities by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7VBf9s09369; Find the interval satisfying the following inequalities!!!! 1. -7<(2x-7)/5<1 or x-1>0 2. 2+|3-2x|>5 I have trouble with these!! === Subject: Re: inequalities > Find the interval satisfying the following inequalities!!!! > 1. -7<(2x-7)/5<1 or x-1>0 Break it into its component pieces: { -7 < (2x - 7)/5 AND (2x - 7)/5 < 1 } OR x - 1 > 0 Then see what interval each component implies and do the ANDs and ORs to see what the result is. Then check your result back in the original. > 2. 2+|3-2x|>5 Use the fact that |z| = z iff z>=0 |z| = -z iff z<0 to remove the absolute value bars, then do your interval analysis. > I have trouble with these!! Do what you can, then if you still can't do it, post here again, but *show the work you've been able to do*. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Question of finite element of a group by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7VBfAH09409; > > > If every element in a group has finite order > > is the group finite? > >A simpler construction is the subgroup of > G = Z_2 x Z_3 x Z_5 x ... >of elements of finite support, ie elements > (a2, a3, a5, ... ) >for which all but finitely many a's are nonzero. >This was corrected to read, for which only finitely many a's >are nonzero. > In other words, the direct sum of Z_2, Z_3, etc. The prime orders > of the cyclic groups are of course irrelevant. >That will not do, as the identity > (1,1,1,... ) >has infinite order. It has to be a subset of the direct sum to avoid >such elements, ie, elements > (a2,a3,... ) >where all but finitely many a's are zero. > Snappier example: Q/Z. There are *lots* of torsion groups after all. >Indeed, but not R/Z. In my previous reply to this, I referred to Lang's Algebra, pp. 36-37. That's the 3rd edition I was referring to for those page numbers. Todd === Subject: Re: Question of finite element of a group by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7VBfA009377; >A simpler construction is the subgroup of > G = Z_2 x Z_3 x Z_5 x ... >of elements of finite support, ie elements > (a2, a3, a5, ... ) >for which all but finitely many a's are nonzero. >This was corrected to read, for which only finitely many a's >are nonzero. > In other words, the direct sum of Z_2, Z_3, etc. The prime orders > of the cyclic groups are of course irrelevant. >That will not do, as the identity > (1,1,1,... ) >has infinite order. It has to be a subset of the direct sum to avoid >such elements, ie, elements > (a2,a3,... ) >where all but finitely many a's are zero. No, what you are referring to is called the direct *product*, which differs from the direct sum if the set of summands is infinite. Just to obviate any misunderstanding, a direct sum is also called a coproduct (in the category of abelian groups), and a direct product is just a product, with the usual categorical descriptions in terms of universal properties. Please see any graduate text in algebra for the correct definition of direct sum, e.g., Lang's Algebra, pp. 36-37. It's the usual name for what you were describing. By the way, the identity of the direct product or the direct sum above is (0, 0, ...), not (1, 1, ...) (surely you meant to use the additive groups Z_2, Z_3, etc., not the underlying multiplicative monoids of their ring structures). Todd === Subject: Re: integral calculas by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7VDJeq17856; i wnt to know definition of calculas so that i can understand easily thank u r.r 31/8/04 india === Subject: Re: integral calculas > i wnt to know definition of calculas so that i can understand > easily thank u Calculus is a branch of mathematics divided into two general fields: differential calculus and integral calculus. Differential calculus can be used to find rates of change, like orbits of planets, satellites, and spacecraft. Integral calculus is a method of calculating quantities by splitting them up into a large number of small parts. It can be used to find the surface area of irregular objects. You can find out the total surface area of your car (even the round parts) by using integral calculus. Source: Children's Encyclopedia Britannica vol. 3, p. 308-309, 1989. === Subject: Re: subbasis for the topology of a set alt.math.undergrad: > I'm trying to prove the following: > If X is a set and G an arbitrary set of parts of X, there is > exactly one topology T(G) on X such that G is a subbasis for T(G) > (the topology``generated'' by G). > Proof: Let X be a topological space. Let G = {} so that G is in > X. Two problems here. First, you don't get to decide what G is: the statement that G is an *arbitrary* set of subsets of X means that it can be absolutely any such collection, and you have to prove the result no matter which one it actually is. Secondly, G is not a subset of X (as you seem to be thinking when you write Ôso that G is in X'); it's a *collection* of subsets of X. Example: If X = {1, 2, 3}, G might be {{1}, {2}, {2, 3}}. Let's take a closer look at what you're trying to prove. The claim is that there is one and only one topology having G as a subbase. This means that you have two things to prove: (1) There is some topology on X that has G as a subbase. (2) If T_1 and T_2 are topologies on X that both have G as a subbase, then T_1 = T_2. Here (1) says that G is a subbase for at least one topology on X, and (2) says that G is a subbase for at most one topology on X; putting the two together gives the desired result. Now in fact (1) isn't quite true as you've stated the theorem: in order to prove (1), you have to know that G covers X, i.e., that every point of X is in at least one member of G. (Another way to say this: U{A : A in G} = X.) I don't know whether you've misstated the problem, or whether it was poorly stated in your source. However, (2) is still true, and I suspect that it's the main point of the exercise. A proof of (2) just about has to begin something like this: Let G be a collection of subsets of X. Suppose that T_1 and T_2 are two topologies on X, and that G is a subbase for both T_1 and T_2. Somehow you want to get from this starting point to the conclusion that T_1 and T_2 are really the same topology after all, i.e., that T_1 = T_2. T_1 and T_2 are just collections of subsets of X. In particular, T_1 and T_2 are themselves sets of things. One way to prove that two sets are equal is to show that each of them is a subset of the other: T_1 is a subset of T_2, and T_2 is a subset of T_1. To show that T_1 is a subset of T_2, let V be an arbitrary element of T_1. Since G is a subbase for T_1, this means that V is built out of elements of G in a certain way (how?). But G is also a subbase for T_2, so anything built out of elements of G in that way is also in T_2; in particular, V is in T_2. Proving that T_2 is a subset of T_1 is essentially the same argument. What you have to do to make this a real proof is work out the details of the Ôhow?'. [...] Brian === Subject: Re: subbasis for the topology of a set === Subject: subbasis for the topology of a set >I'm trying to prove the following: >If X is a set and G an arbitrary set of parts of X, there is >exactly one topology T(G) on X such that G is a subbasis for T(G) >(the topology``generated'' by G). G subset P(X) = { A | A subset X } Yes that's correct. However for G to be a subbase, it is necessary and sufficient that /G = Union G = X. >Proof: Let X be a topological space. Let G = {} so that G is in >X. G is open by the definition of topological space. Claim G is a >subbasis for T(G). Now, we must show that every open set of G is >the union of finite intersections of sets in G. If G = nulset, then only by insisting that an intersection of no sets is X (an rather abstruse technical point of some debate and little merit) could G produce a base, which would include the nulset in the topology as the union of no sets. >Since G = {}, {} is in G. So {} / {} is in G. G is a subbasis >for T(G) because since G = {}, there's no other set that we could >try to intersect in G. This is not so. The nulset can contain nothing, not even the nulset. >Now we must show that T(G) is the only topology on X. By way of >contradiction, assume that there is a T'(G) on X such that T'(G) >is different than T(G). >Let a, b be two open subsets of G. a U b is in G, a / b is in G. >a U b and a / b must both be open. Specifically, since G = {}, >then a = b = {}. You haven't covered the case for an infinite union of open sets. >Now, I don't really know how to show that T(G) = T'(G), although >they must be because G = {}. I would try showing that all axioms >hold for both, but that only says that that they are topologies; >how would show that they are the same topology? The topology generated by G, is by definition the smallest topology containing all of G. As two smallest topologies are equal, you are done. >Perhaps I couldn't have let G = {}. Advices are welcome. That's bogus as you need to show that, not for just one G, but for any arbitrary G. ---- === Subject: Re: subbasis for the topology of a set > I'm trying to prove the following: > If X is a set and G an arbitrary set of parts of X, there is > exactly one topology T(G) on X such that G is a subbasis for T(G) > (the topology``generated'' by G). G is a set of subsets of X. A topology on X is also a set of subsets of X. > Proof: Let X be a topological space. Let G = {} so that G is in > X. Yikes! Who said G was empty? In fact, the empty set is the only set which is _not_ a subbase for a topology (unless X is also empty). Almost certainly G is not an element of X (in X). > G is open by the definition of topological space. Double yikes! G is not a subset of X (except in rare cases) open does not apply to G. > Claim G is a subbasis for T(G). Triple yikes! Any non-empty collection of subsets of X is a subbase for a topology on X. T(G) is all unions of finite intersections of elements of G which leaves you no choice about T(G) so T(G) is unique. You might need to show that T(G) is in fact a topology. > Now, we must show that every open set of G is > the union of finite intersections of sets in G. Quadruple yikes! We are not given a topology for G; open does not apply to subsets of G. > Since G = {}, {} is in G. So {} / {} is in G. G is a subbasis > for T(G) because since G = {}, there's no other set that we could > try to intersect in G. > Now we must show that T(G) is the only topology on X. Quuintuple yikes! There are probably a whole bunch of topologies on X. [...] We very much need to know the definitions you are using here. One common one : Define T(G) as the smallest topology containing G. In that case you need: 1) There _is_ a topology containing G. [The discrete topology works.] 2) There is a _smallest_ topology containing G. 3) There is a _unique_ smallest topology containing G. I'd suggest you mull over the definition of a topological space and then get back to us with the definitions of subbase and T(G) that you are using and the verbatim statement of the problem. -- Paul Sperry Columbia, SC (USA) === Subject: Caculator Help! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7VN25l05533; I am in PreCalculus, but I don't have a graphing calculator. Is there a place online that is basically a graphing calculator that you can use for free? === Subject: any book on Nonhomogeneous Poisson Process systematically? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I am studying NHPP lately, but it's very hard for me to find a book having detailed description on NHPP. I have tried to find online files, and what I found are just NHPP fragments. I think there should be one book on point processes or similar, which has summarize NHPP related properties together. Do you have any information about it? Neon === Subject: Re: Fractional Integral Equation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) A(d/dx)^(-1/3)f(x)+x^(-1/3)f(x)=1, where A is some constant (may be taken 1). The problem is to obtain a solution for the equation. There is some theory for constant-coefficient fractional integral equations, but in this particular case the problems arise from the non-constant coefficient x^(-1/3). Probably the solution is some special function, obviously the power function x^k is not the solution for any value of k. Does anyone know any systematic way to find out the solution? Numerically the solution can naturally be found, if the boundary conditions are given. A.L. >Does anyone the solution for the following fractional integral >equation: >A*D^{-b}f(x)+x^{-b}*f(x)=1 > I've read Ôyour fractional integral equation'. > you want to solve:A*d/dx ^(-1/3) f(x)+ x^(-1/3)*f(x)=1. > A way to consider fractional derivative /integral ,r real is > putting d/dx^[r] (x^k)=Gamma(k+1)/Gamma(k-r+1)*x^(k-r) a > generalization of d/dx^[n] (x^k)=k!/(k-n)!*x^(k-n) n,k integer. > For f(x)=c*x^k we've got: > A*c*Gamma(k+1)/Gamma(k+4/3)*x^(k+1/3)+c*x^(k-1/3)=1 > What means A?. > May be it helps... === Subject: Re: Fractional Integral Equation Originator: ilya@powdermilk Originator: bergv@math.uiuc.edu (Maarten Bergvelt) [A complimentary Cc of this posting was sent to AL > (d/dx)^(-1/3)f(x)+x^(-1/3)f(x)=1, > Probably the solution is some special function, obviously the power > function x^k is not the solution for any value of k. One (stupid?) idea is to look for a solution in the form F(x) = SUM_k Ak x^(k/3) [It is easy to find Ak, but the series may diverge. But even if it diverges, it may help: you could be able to find an ODE which is satisfied by the formal solution F(x), then solve the ODE for an ordinary function G(x); the result may give you hints about how f(x) behaves.] > Numerically the solution can naturally be found, if the boundary > conditions are given. Eh??? Boundary conditions for non-local equations do exist, but they are very different from ones for ODE (e.g., you assume that f(x)=0 for x<0). Moreover, it is most probably not easy to solve numerically even with such a condition... Hope this helps, Ilya === Subject: Re: Fractional Integral Equation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > (d/dx)^(-1/3)f(x)+x^(-1/3)f(x)=1, > Probably the solution is some special function, obviously the power > function x^k is not the solution for any value of k. > One (stupid?) idea is to look for a solution in the form > F(x) = SUM_k Ak x^(k/3) > Numerically the solution can naturally be found, if the boundary > conditions are given. > Eh??? Boundary conditions for non-local equations do exist, but > they are very different from ones for ODE If you are really only interested in a series representation under natural boundary conditions this is probably easy: You might require that the boundary conditions are such that (under these conditions) the equation is equivalent to f + J^{1/3}Mf =J^{1/3}e where e(x):=1, Mg(x):=x^{-1/3}g(x), and J^{1/3} is the fractional integral operator J^{1/3}g(x):=int_0^x(x-y)^{-2/3}g(y)dy/Gamma(1/3). The right-hand side is easily calculated (I guess something like 3x^{1/3}/Gamma(1/3), but better recalculate). Moreover, since -J^{1/3}M is a compact Volterra operator (at least in L_p[0,T] for sufficiently small p), it has spectral radius 0 and so the inverse of I-(-J^{1/3}M) exists (in these spaces) and can be calculated using the Neumann series. Since -J^{1/3}M is a regular integral operator, this Neumann series will be of the form I+K where K is an integral operator whose kernel is the series of the iterated kernels - I guess it is not hard to calculate an explicit formula for these kernels and hence a (necessarily a.e. convergent!) series for the solution. === Subject: Re: Maximal subgroups of S_n Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > > Are (Is) there (a) published paper(s) or book(s) about the following > question: > > Is it possible to determine all maximal subgroups of S_n, the symmetric > group of degree n. I mean by the existence if a complete description > of maximal subgroups in terms of only n !! > > What is the number of all maximal subgroups of S_n. > > Any information is appreciated. > > > A. Abdollahi > Let me just say a little more about the maximal subgroups of S_n. They > are of > three types: > 1. Intransitive maximals, > 2. Transitive but imprimitive maximals, > 3. Primitive maximals. > The first two of these types are easily described and well understood. > Type 1 consists of subgroups S_m x S_{n-m} for 1 <= m < n/2 > Type 2 consists of wreath products S_m wr S_{n/m} for factors m of n > with > 1 < m < n. > With very few exceptions, these types of subgroups are maximal in S_n. > Finite primitive permutation groups are described by the O'Nan-Scott > Theorem. The coset representations of the almost simple groups form > one category of primitive groups, and probably present the major > obstacle to describing all > maximals of S_n. There is another category, however, known as the > affine > primitive groups. These have degree a prime power p^m, and have a > normal > elementary abelian subgroup of order p^m with a complement isomorphic > to an > irreducible subgroup of GL(m,p). So, in order to classify the maximal > subgroups > of S_n with n = p^m, you would also need to classify all maximal > irreducible > subgroups of GL(m,p), which is another problem on which much work has > been done > (there is a big theorem of Aschbacher describing these subgroups), but > with no > complete solution as yet. To solve it completely, you would probably > need to > find all irreducible representations of all almost simple groups over > the prime > field. I have realized that that last statement I made is nonsense. The maximal subgroup of A_n/S_n of affine type is of course the full semidirect product of an elementary group of order p^m with GL(m,p). The paper of Liebeck, Praeger and Saxl mentioned in Magidin's post essentially reduces the problem of finding the maximals of A_n or S_n to finding the maximal subgroups of the almost simple groups. It also lists all exceptional situations, where subgroups of a type that would normally be maximal in A_n or S_n are not maximal. Derek Holt === Subject: sheaf theory applications X-mailer: epigone Epigone-thread: plootincrur Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I'm intrested in applications of sheaf theory to computer science. somebody may kindly guide me in this direction. === Subject: Re: Under what circumstances is a complete metric space Hilbertizable? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Given a complete metric space X (with no given linear space structure), > under what conditions does there exist a Hilbert space structure on X which > induces the given metric? To put it another way, under what circumstances is > a complete metric space Hilbertizable? Further, how does one determine the > Hilbert space structure on X which induces the given metric? Do you want to know whether the metric space X *is* a Hilbert space or whether it can be embedded isometrically into one? In the latter case: there are simple conditions on the metric d which have been established in several papers by Schoenberg around 1938 or 1939 (eg, 522-536, 1938). Essentially, the negative squared metric -d^2 has to be conditionally positive semi-definite, that is sum_i c_i c_j d^2(x_i,x_j) <=0 for all x_i, x_j in X and for all c with sum c_i = 0. === Subject: Discrete Spectrum, compact perturbation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Hi all, I have a question on the discrete spectrum of a sum of a multiplication with an integral operator. Let X a compact metric space, with some (non-uniform) probability measure on it, k:X times X to R a nice kernelfunction (positive, continuous, smooth, whatever you want) and d a continuous positive function. Let M the multiplication operator with multiplier d, i.e. Mf(x) = d(x)f(x) and S the compact integral operator with kernel k, i.e. Sf(x) = int k(x,y)f(y) dP(y) (all operators defined on the space C(X), or also on L_2(P) if it makes things easier). I am interested in the discrete spectrum of the operator U = M - S. I know that the essential spectrum of U coincides with the (essential) spectrum of M (as S is a compact perturbation). But I want to find out under which conditions there are isolated eigenvalues in the spectrum of U. Does anyone know any methods to deal with such a question? Maybe there are general results on compact perturbation of multiplication operators? I checked some standard perturbation theory books, but I dind't find anything about the discrete spectrum. In case this might help, I am actually interested in a special case of this problem: X some compact subset of R^n, k(x,y) := exp(-||x-y||^2/t), d(x) = int k(x,y) dP(y), P for example a mixture of two Gaussians. Grateful for any hint, Ule === Subject: Re: Discrete Spectrum, compact perturbation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > I have a question on the discrete spectrum of a sum of a > multiplication with an integral operator. > Let X a compact metric space, with some (non-uniform) probability > measure on it, > k:X times X to R a nice kernelfunction (positive, continuous, > smooth, whatever you want) and d a continuous positive function. Some estimates for the spectrum (and a simple suggestion for obtaining estimates using approximations) for the case X=[a,b] (but more general X is not much different) can be found in the last section of E. A. Biberdorf, M. V.8ath, On the spectrum of orthomorphisms and Barbashin operators, Z. Anal. Anwendungen 18, No. 4 (1999), 859-873. === Subject: a multilinear Riesz-Thorin theorem? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Let otimes denote the projective tensor product of Banach spaces, let S be any set, and let p,q in (1,infty) be such that 1/p + 1/q =1. Suppose that f is a multiplier of both ell^p(S) otimes ell^q(S) and ell^q(S) otimes ell^p(S), i.e., f is a bounded function on S times S such that pointwise multiplication with f maps ell^p(S) otimes ell^q(S) into itself and the same for ell^q(S) otimes ell^p(S). Question: Is f then also a multiplier of ell^2(S) otimes ell^2(S)? This question is somewhat reminiscent of the Riesz-Thorin interpolation theorem, but the projective tensor product makes it impossible to apply complex interpolation directly. Any pertinent hints are appreciated! Volker Runde. === Subject: A new proof of the four color thorem Content-Length: 215 Originator: rusin@vesuvius Here is a link for an non-computer proof of the four color theorem which uses graph theory tools. It is a kind of Columbus egg! But certainly it is not trivial. http://arxiv.org/abs/math.CO/0408247 Cahit === Subject: Higher dimensional Klein/Pluecker terminology Content-Length: 345 Originator: rusin@vesuvius Does anyone know of standard names for the higher dimensional analogues of Pluecker coordinates, the Klein correspondence, or the Klein quadric? In particular, if you map a pair of n-space vectors, defining a plane through the origin and corresponding projective line, to the wedge product, it lies on a projective variety. Does it have a name? === Subject: Aldor loci Hi all, Where do Aldor enthusiasts and developers meet? There is a (now past) mention of a future mailing-list or group in aldor.org but I have not been able to find anything. Is this the right group? Also, is the source for Aldor available somewhere? Not that I need/want to examine it now, but I care for the continuity of the language. Somehow I feel uneasy, in that right now Aldor is just a testbed for ideas that will eventually go into commercial Maple. C. === Subject: Re: FFT Program > Hello brothers > I am planning to make FFT program. I am using Numerical reciepe in C > code and had also tried the RealFFT code. > I am facing problem with RealFFT code in sense that i am getting > values in the magnitude much higher, while another reference program > developed in DAOS platform- mathematical C like langauge (THE REAL > ENVIRONMENT) is having values in smaller ranges. > E.g. Range from code 10 pow +3, while DAOS one shows value 10 pow -3 > NOTE: DATA Pattern observed is similar. Kinks in data shapes is same. > i.e. plot of data is same > I have to plot Power Spectra of time domain data to Frequency domain: > Is the > Sandeep > 1. Writing an FFT program is not what this newsgroup (sci.math.symbolic) is about. > 2. Numerical recipes book generally has programs that are not robust. and in this case not particularly fast either. > 3. There are hundreds of FFT programs around. Find a good one. > 4. Your note is incoherent. > 5. Good luck. RonL === Subject: Re: new version of Fermat available by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8AGJ7r28070; > Two independent researchers have discovered that Fermat is the > fastest system in the world for multivariate polynomial GCD. Their > letter and data are also on my website. >Could you give the link to the precise data? >On your page, I see that the pb was to take the gcd of a*b and a*c >with e.g. total degree of a*b and a*c being 60 or 30 etc., but >it would be interesting to know what the total degree of a is >(is it half the total degree of a*b?) >Bernard Parisse The data is now at http://www.bway.net/~lewis/fermat/gcdcode.tar.gz Robert H. Lewis === Subject: sybmbolic computation programs Hello. I need to find several sybmbolic computation programs, particulary a free Linux/Unix versions. I have now information about Maxima, MuPad and Mathomatic, but it's not enough for my work. Could you say me about any other programs like that? I mean only full-functional programs (actually, Mathomatic is not quite that what I need). P.S. Excuse me for my terrible English... === Subject: multiple summation stack limit problems, Maple 9.0 Why does Maple so quickly run out of stack space when I use multiple summations? For instance, when I type: > sum(sum(binomial(n,m),n=0..N),m=0..M); Maple takes over 15 seconds and returns something horrendously complex. Actually, I have an only slightly more complicated function than this involving a double summation and a few combinatoric terms, and when I try to define this function, Maple runs for over 10,000 seconds before giving me Kernel failure! Execution stopped: stack limit reached. have! === Subject: Re: multiple summation stack limit problems, Maple 9.0 > For instance, when I type: > sum(sum(binomial(n,m),n=0..N),m=0..M); > Maple takes over 15 seconds and returns something horrendously > complex. The time used and the complexity of the answer in this case can be greatly reduced if you make the assumptions: > assume(N::nonnegint, M::nonnegint, n::nonnegint, m::nonnegint); === Subject: A new book on Maple for users and developers On Fultus web site http://writers.fultus.com/aladjev/book01.html you can find information about a new book on Maple that can be useful enough both for beginners, and the advanced Maple users, as well as for Maple developers and developers of Maple appendices. === Subject: Re: Does MuPad 3.0 support Elliptic Integrals? > I particularly want a CAS which can accurately calculate elliptic integrals. > Maple fails to do this... > http://www.math.rwth-aachen.de/mapleAnswers/html/1346.html This bug has been fixed in Maple 9.5. -- Thomas Richard Maple Support Scientific Computers GmbH http://www.scientific.de === Subject: Re: 2D / 3D Differential Geometry Wizard > Hi! > I would like to know which Maple version this wizard fits to. I > work with Maple 6. I haven't actually tried it, but the download page mentions Maple 8: http://www.graphtree.com/Maple/atlasWizard/DownLoad/ download.htm Note that Maple 6 does not support the Maplets technology. -- Thomas Richard Maple Support Scientific Computers GmbH http://www.scientific.de