mm-1083 === Subject: Re: sums of prime pairs Formalizing the definition of density is exactly what I tried to do in my last posting which does (I hope) provide a precise definition. However, there are different sorts of density which is why my last density. What do you think of the density definition in my previous posting? Is it clear, precise etc? Paul > I be precise in my statements. I mean density in the sense, for > example, that the probability that an integer n is prime is > approximately 1/ln(n). If I am not mistaken, this is equivalent, in my > case, to stating that if I have a set of elements Ksubk (or Kk), then > the density of the set in the vicinity of Kk is the number of members > K(i,j), of the set between i-1 and j+1 divided by |Kj-Ki|, i sufficiently large K and small |j-i|. This is still fuzzy, but I think > it is correct, and I hope it is sufficiently cogent for you get the > idea. I will think more about how to formalize this. === Subject: Re: Normal sequences of braids >>It almost surely means distinct from. Patrick, despite his >>first name, is French, and although his English (written and >>spoken) is very very good, it is not always perfectly idiomatic >>American English. > >No, you don't understand, I wanted to know what he meant by distinct >in this context! Non-equal and non-equivalent seem equally valid. Non-equal (I would bet a dime). Lee Rudolph === Subject: Re: A problem on probability No. The problem is you model this as if the concentration remains > constant. Imagine an urn containing 100 white balls and 1000 black > balls. 200 people keep drawing balls from this urn, each drawing until > s/he grabs a black ball. Whether you model them as grabbing all at once > or sequentially, it is not as if they keep having a 1 in 11 chance of > drawing a white ball. Since there are twice as many people as white > balls, this will have an effect regardless of the scale. I kind of see what you say here. I will rethink my analysis. > >> The fact that at least half of the finished molecules will have >> length 0 is also shown to be valid from the formula derived for >> P(B|C), which evaluates to 20/33 (>1/2). > > But P(0) = 20/1089 by your formula below. There was a typo error in the formula presented in my last mail. Actually, it is as follows (by mistake the power of 33 was written as (n-1) instead of (n+1)): P(n) = 2/33 * 1/2^n * 1/33^n * 10/33 = 5/(2^(n-2)*33^(n+1)) This yields P(0) = 20/33. Also, your P(n) is ambiguous. It looks like a geometric distribution > with parameter 65/66, but the constant is wrong; i.e., sum(n=0..infty, > P(n)) != 1. What is P(n) supposed to be? The length of randomly > selected completed molecule? Then, if this were geometric(65/66), the > expected total number of A compund in the completed molecules is > (2/65)x, leaving an expected (63/65)x unattached to B molecules. I > guess that in itself is not a priori impossible, since all the C > molecules may preclude most of the A molecules from attaching to B > molecules. > This is a good point. I will think more on this problem and get back to you. Partha. === Subject: Re: A problem on probability > No. The problem is you model this as if the concentration remains > constant. Imagine an urn containing 100 white balls and 1000 black > balls. 200 people keep drawing balls from this urn, each drawing until > s/he grabs a black ball. Whether you model them as grabbing all at once > or sequentially, it is not as if they keep having a 1 in 11 chance of > drawing a white ball. Since there are twice as many people as white > balls, this will have an effect regardless of the scale. I kind of see what you say here. I will rethink my analysis. > >> The fact that at least half of the finished molecules will have >> length 0 is also shown to be valid from the formula derived for >> P(B|C), which evaluates to 20/33 (>1/2). But P(0) = 20/1089 by your formula below. There was a typo error in the written formula in my last mail. Actually, it is as follows (by mistake the power of 33 was written as (n-1) instead of (n+1)): P(n) = 2/33 * 1/2^n * 1/33^n * 10/33 = 5/(2^(n-2)*33^(n+1)) This yields P(0) = 20/33. Also, your P(n) is ambiguous. It looks like a geometric distribution > with parameter 65/66, but the constant is wrong; i.e., sum(n=0..infty, > P(n)) != 1. What is P(n) supposed to be? The length of randomly > selected completed molecule? Then, if this were geometric(65/66), the > expected total number of A compund in the completed molecules is > (2/65)x, leaving an expected (63/65)x unattached to B molecules. I > guess that in itself is not a priori impossible, since all the C > molecules may preclude most of the A molecules from attaching to B > molecules. This is a good point. I will think more on this problem and get back to you. Or is P(n) supposed to be the distribution of the longest completed > molecule? I don't see how that follows, and the dsitrbution does still > not add up to 1. Partha. === Subject: Re: N-gon problem <5qj7f1t6q0nbg0a34b92dd1ppdiel1pmur@4ax.com> <9vk7f1h0a9naja49a13bni7j14lqcfmi2t@4ax.com> I'm not saying it can't be done... I'm saying I've never been taught how to solve these math problems correctly. Unfortunately, this math professor is an idiot who literally spends half the class time fixing his mistakes on the board. I am not exaggerating. But, I do know what you mean... I feel the same way with people and computers. I can solve any technology-related problem I'm given, because I can learn almost anything, while other people look dumbfounded. Unfortunately, mathematics is not my strong point. I am not a whiner or complainer... I only come here to ask your assistance with understanding this particular problem because this is YOUR speciality, not mine, and I thank Quasi for the help. Beyond this class, I don't give a rat's ass about math (Well that just sounds nice; I know I'll be using it throughout my life in various ways). In other words: lay off; you shouldn't turn everyone you don't know into a stereotype. I ask just one question, and you people have a whole attitude about it. Well, don't worry, I won't be coming back here. === Subject: Re: I have correct proof of 4-color theorem Proginoskes .8eæ.93.b9.81F > The only correct, simple and elegant proof of > the 4-color theorem [...] > > I can stop reading right here, because you're a crank. In fact, I could > have stopped at the first word (the) because there are many proofs of > a result. (A few dozen of the Pythagorian Theorem, for instance.) But for 4CT there is only one computer proof, not any one by hand. > If you claim you have a correct, elegant proof of the 4CT, feel free > to post it, though. Just as long as you're not saying that it's > impossible for five regions to share a common boundary edge. (The > 5-regions result was proved in 1840 by Augustus Moebius, before the > history of the 4CT, and is a RESULT of the 4CT, but it doesn't PROVE > it.) Do you nean every plane graph is 5-colorable? In fact my proof is for any plane graph is 4-colorable. If not so, would you please say more detail as my English is poor, can not catch you your meaning. You may use some refering book. > > , > who is in the acknowledgement section of > Robertson/Sanders/Seymour/Thomas's 4CT proof. === Subject: Re: I have correct proof of 4-color theorem > Proginoskes .8eæ.93.b9.81F > > The only correct, simple and elegant proof of > the 4-color theorem [...] > > I can stop reading right here, because you're a crank. In fact, I could > have stopped at the first word (the) because there are many proofs of > a result. (A few dozen of the Pythagorian Theorem, for instance.) > > But for 4CT there is only one computer proof, not any one by hand. That is incorrect; there are at least two computer proofs, one by Appel and Haken (which is presumably the one you're referring to), and another one by Robertson, Sanders, Seymour, and Thomas, which came out in the mid 1990s. (I did some grunt work on this one.) I also know of another alleged proof, which doesn't use computers, and yet another one which uses spiral chains to 4-color a planar graph. > If you claim you have a correct, elegant proof of the 4CT, feel free > to post it, though. Just as long as you're not saying that it's > impossible for five regions to share a common boundary edge. (The > 5-regions result was proved in 1840 by Augustus Moebius, before the > history of the 4CT, and is a RESULT of the 4CT, but it doesn't PROVE > it.) > > Do you nean every plane graph is 5-colorable? No. Among amateurs, there is some confusion between the two statements any plane graph can be 4-colored, and you cannot have 5 regions of the plane, every pair of which are adjacent. I'm calling the second statement the 5-regions result, to avoid repetitively repeating myself over and over again. > In fact my proof is for any plane graph is 4-colorable. That statement is false and is technically not what's referred to as the 4 Color Theorem; in fact, I have a counter example to it: a graph with one vertex, with a loop connecting that vertex to itself. That graph is planar and not colorable at all . I suspect you meant to say: Every loopless plane graph is 4-colorable, which is the 4 Color Theorem. > If not so, would you please say more detail as my English is poor, can > not catch you your meaning. You may use some refering book. It's just being careful and saying what you mean, and meaning what you say. === Subject: Re: I am the poster of I have correct proof of 4-color theorem <42f25616$4$fuzhry+tra$mr2ice@news.patriot.net> Contact me name: Cui Shitai address: Room 104, No. 26, Lane 842, Li Yuan Road, Shanghai 200023 China. Phon : (8621)63043909 Send E-mail to domain 263 dot net user cuishitai to contact me. === Subject: Re: I am the poster of I have correct proof of 4-color theorem You post here, everyone who is interested in the 4-color theorem will read it and search for the flaw in it for some period, and if no flaw is found you will have undying fame and glory and your choice of jobs at the world's most prestigious math departments, rewards worth far more than the money anyone reading sci.math could pay you. if no flaw is found I'll get nothing because my paper is still not accepted and published by journal. You post and forward the above post anywhere and any one, then I have more chance, you also have more chance to read it. === Subject: Re: I am the poster of I have correct proof of 4-color theorem >You post here, everyone who is interested in the 4-color theorem >will read it and search for the flaw in it for some period, and if >no flaw is found you will have undying fame and glory and your choice >of jobs at the world's most prestigious math departments, rewards >worth far more than the money anyone reading sci.math could pay you. > >if no flaw is found I'll get nothing because my paper is still not >accepted and published >by journal. >You post and forward the above post anywhere and any one, then I have >more chance, >you also have more chance to read it. Hi cuishitai, Out of curiosity, I have 2 questions: (1) What is your native language? (1) Is the proof in English? quasi === Subject: Re: I am the poster of I have correct proof of 4-color theorem native language is Chinese. the proof is in English. === Subject: Re: I am the poster of I have correct proof of 4-color theorem >native language is Chinese. >the proof is in English. It seems to me that even if the proof is correct, your difficulty with the English language would make the proof hard to understand. Why not do the proof first in Chinese and submit it to a Chinese journal? === Subject: limit of i^i^i^i .... What is the limit of i^i^i^i .... (i=imag unit) to 8 figures? === Subject: Re: limit of i^i^i^i .... > What is the limit of i^i^i^i .... (i=imag unit) to 8 figures? It's precisely W(pi/(2 i)) / (pi/(2 i)), where W denotes the Lambert W function, and approximately 0.43828293672703211163 + 0.36059247187138548595 i . David === Subject: Re: limit of i^i^i^i .... Ì David W. Cantrell ó.8d.98.87.8b.8c .97.99.95 .92.86.94.9d.92.87 > > What is the limit of i^i^i^i .... (i=imag unit) to 8 figures? > > It's precisely W(pi/(2 i)) / (pi/(2 i)), where W denotes the Lambert W > function, and approximately > > 0.43828293672703211163 + 0.36059247187138548595 i . Strictly technically speaking, none of the answers is correct. The answer is the above *IF* we know that i^i^i^... converges, first. And I don't see anyone showing that. It has been discussed here in the past, but suppose the OP asked, what is the limit of (3i)^(3i)^(3i)^...? Then solving (3i)^z=z, gives, via Lambert's W function, W(-log(3i))/(-log(3i))~=.3009549343+.5277490261 i However, (3i)^(3i)^(3i)^... does not converge, so the limit does not exist. > David -- I. N. Galidakis http://users.forthnet.gr/ath/jgal/ Eventually, _everything_ is understandable === Subject: Re: limit of i^i^i^i .... <20050805212927.352$ME@newsreader.com> as initial guess to FindRoot [I^z==z , ... ] and 64-digit w.p. I got 0.4382829367270321116269751635512648242678973516463946036092212404957914 8283653 + 0.3605924718713854859529405269060006538265770307860270047414512983804600 9638619 i Its for a sophomore HW, answer TBG. === Subject: Re: limit of i^i^i^i .... > as initial guess to FindRoot [I^z==z , ... ] and 64-digit w.p. I got > 0.4382829367270321116269751635512648242678973516463946036092212404957914 > 8283653 + > 0.3605924718713854859529405269060006538265770307860270047414512983804600 > 9638619 i > Its for a sophomore HW, answer TBG. FWIW, the last eight digits shown above, in both the real and imaginary parts, are incorrect, presumably. Since you're apparently using Mathematica, a more straightforward way to get a highly accurate answer should be something like In[4]:= N[ProductLog[Pi/(2 I)]/(Pi/(2 I)), 77] Out[4]= 0.43828293672703211162697516355126482426789735164639460360922124049579153222 270 + 0.36059247187138548595294052690600065382657703078602700474145129838046019521 151*I David === Subject: Re: limit of i^i^i^i .... <20050805212927.352$ME@newsreader.com> <20050806005535.589$Ms@newsreader.com> Correct. I will keep only 16 in the HW. It is about Newtons method for scalar equations, coded by hand in matlab (which is incapable of arbitrary precision) . === Subject: Re: limit of i^i^i^i .... > What is the limit of i^i^i^i .... (i=imag unit) to 8 figures? > x = i^i^i... i^x = x solve for x === Subject: Re: limit of i^i^i^i .... >What is the limit of i^i^i^i .... (i=imag unit) to 8 figures? I suspect it requires infinite imagination. === Subject: Re: limit of i^i^i^i .... > >What is the limit of i^i^i^i .... (i=imag unit) to 8 figures? > > I suspect it requires infinite imagination. Exactly. 8i = OOi by rotating the number 8 90 degree, we have infinite imaginary number. === Subject: Re: Nonconvex areas > [...] > The boundary of a non-concave set (between two points on the boundary) > may be concave or convex. > > You mean non-convex? Concave = bulging in here, and convex = bulging out. === Subject: Q: Minimizing partial sums of the columns of a matrix by permuting the rows Content-Length: 1734 Originator: rusin@vesuvius The problem I would like to solve is this: I have a matrix A with n rows (n > 10,000) and m columns (10 < m < 50) such that each element A(i, j) is from -1 to 1 and the sum of each column is zero. Define c(i,j) to be the sum of the first j elements of column i (so that c(i, n) = 0), d(i) to be the max of the absolute value of c(i, j) for j = 1 to n, and dmax to be the max of all the d(i). I would like to find the permutation of the rows that minimizes dmax. This arises in the minimizing of the size of a sum relating to the matrix - by permuting the rows properly, the growth of the sum is minimized so it can be more easily and accurately computed on a fixed-precision machine. Another characteristic of the matrix is that only about 10% of the elements in each column are non-zero. Interestingly, the matrix as provided to me is in a pessimal form - each column has very large contiguous blocks of constant sign, so that dmax is very large (usually about 100). The number of times the partial column sums cross zero is from 1 through 5 (usually 2 or 3), while the expected number of returns to origin (as in Feller) should be about .8*sqrt(n) or about 80. I have tried generating random permutations of the rows, and the number of zero crossings agrees quite will with the result above. Also, dmax decreases from about 250 for the original matrix to about 20 for the randomly permutated matrix (the law of the iterated logarithm may come into play here). So, are any algorithms known that would minimize dmax in a reasonable amount of time? Perhaps a Remez-type, that would take the worst and best rows and swap them, repeating until no improvement can be found. TIA for any comments or suggestions, Martin Cohen === Subject: Re: ALTERNATE FUELS <09vke1df13rk325ulpp6qll53608q3i0u4@4ax.com> > On 29 Jul 2005 07:37:03 -0700, Hydro Fuel > >I HAVE PLANS AND COMPLETE SKETCHES FOR THE DESIGN OF AN ENGINE SYSTEM >THAT OPERATES ON WATER...........THE HYDROGEN TO BE MORE PRECISE. IF >YOUR INTERESTED, EMAIL ME, WE CAN DO BUSINESS, THESE PLANS WILL ONLY GO >TO A VERY LIMITED, AND QUALIFIED NUMBER OF PEOPLE, DONT WAIT, ACT NOW. > > Do you actually think that in a math newsgroup, you would find people > stupid enough to be lured into your scheme? He probably did. That's why he didn't cross-post it to sci.physics or other such groups. === Subject: Re: ALTERNATE FUELS <09vke1df13rk325ulpp6qll53608q3i0u4@4ax.com> > In message <09vke1df13rk325ulpp6qll53608q3i0u4@4ax.com>, quasi >On 29 Jul 2005 07:37:03 -0700, Hydro Fuel > >>I HAVE PLANS AND COMPLETE SKETCHES FOR THE DESIGN OF AN ENGINE SYSTEM >>THAT OPERATES ON WATER...........THE HYDROGEN TO BE MORE PRECISE. IF >>YOUR INTERESTED, EMAIL ME, WE CAN DO BUSINESS, THESE PLANS WILL ONLY GO >>TO A VERY LIMITED, AND QUALIFIED NUMBER OF PEOPLE, DONT WAIT, ACT NOW. > >Do you actually think that in a math newsgroup, you would find people >stupid enough to be lured into your scheme? > >Then again, maybe you noticed some of the more nonsensical recent >posts. > >I take it back. > > When I was at school (early 'teens) a kid (from the foundation stream) > proudly showed us all a beautifully drawn blueprint for a device which > electrolysed water, burnt the resulting oxygen and hydrogen in a steam > engine which in turn drove a generator to provide the power to > electrolyse water... > > There was a take-off spindle for the excess power generated. Some inventors of perpetual motion machines put in a braking system, so the device wouldn't run too quickly ... === Subject: Re: a fair die problem... > Of course! (Essentially, it boils down to calculating the number of > nonnegative solutions to the equation > > x1 + x2 + x3 + x4 + x5 + x6 = n, > > with the additional conditions that > > x2 = 10, x4 < 10, and x6 < 10.) > > > You've lost me here. I think you are trying to calculate the number of > ways the dice can fall in terms of the total number of 1's, 2's, 3's, > 4's, 5's and 6's, but these outcomes are not equiprobable so I don't > see how enumerating them helps. > > Secondly, even though you know that x2 = 10, you don't actually know > that the run finishes at roll n. The last 2 could have occurred > anywhere (anywhere after roll 9 anyway). (This is more easily fixable > though.) > > I see in your follow-up post that you say actually ... this doesn't > work, but I'm not sure if you are referring to the general concept or > just the follow-up calculations. I was referring to the follow-up calculations at the time, but because order is important, maybe a recurrence is better-suited as a general concept. So: maybe both. === Subject: Re: a fair die problem... >> >> Out[4]= 44.6894 >> > > times. My results for three runs were: Mean 44.99 44.82 44.02 StdDev 10.13 9.15 8.87 My interpretation of the rules is that you roll until you accumulate 10 2's OR 10 4's OR 10 6's. Paul === Subject: Re: Call Bush to Action on Science, Math, and Engineering Education > [...] >No way is Bush is capable of reading this. > > And even if he could read it, you're talking about a president who, in > my view, is as anti-academic as any president we've ever had. I would > bet that Bush was never any good in math and probably hates math. If > you remind him that he's funding it, he might just cut the the funding > rather than increase it, just to get back at all the math teachers who > made him feel stupid. Bush had the world at his hands: the best schools and the best opportunities. But he wasted it all, learned very little, and more importantly, he doesn't care. (Kind of an anti-Lincoln.) It's this attitude of his that really ticks me off. === Subject: Re: Call Bush to Action on Science, Math, and Engineering Education The clown hasn't got the necessary capabilities to understand the petition. At any rate, he seems to be much more interested to push a Christian fundamentalist agenda, in an effort to bring back institutionalized superstition and obscurantism. === Subject: Re: Call Bush to Action on Science, Math, and Engineering Education > >Why does it require another Manhattan Project to spur Math & Science >in the USA? > ... > In terms of foreign relations, we could start to improve our image by >working towards ends that benefit the world as a whole, not just us. > Sadly, US foreign policy is very much influenced by the interests of the US multinational corporations, more so than ever under Bush, and while there is the pretense that US government cares about the rest of the world, that's just image politics and most of the world sees right through that. Working towards ends that would benefit the world as a whole -- that's such a wonderfully pure and simple ideal, but it could only happen if that ideal evolves into an ideology which inspires a worldwide movement. quasi === Subject: Can't find the difference between 'Converge almost uniformly' and 'Uniform convergent almost everywhere'. I'm reading 'Lebesgue measure and integration' (P K JAIN V P GUPTA) book. 10.2 Theorem states, Let E be a measurable set with m(E)0 and delta>0, there is a set A < E with m(A)= N. 10.4 Theorem says (Egoroff's Theorem) Let E be a measurable set with m(E)0, there is a set A < E with m(A)0, there is a measurable set A in E with m(A) > > If the functions are orthogonal, or equivalently statistically independent, > there is no problem > > Integral(xy) = integral(x)*integral(y) > > or in probabilities > E(xy) = E(x)E(y) > > That is precisely it. I have an integral of the kind > > integral( l(t) . x^2(t) dt ) > > where l(t) is a positive function from 0 to 1, and x(t) is a gaussian. I > want to write it as > > integral( l(t) dt ) . integral( x^2(t) dt ) > > but I would like to see a formal proof of the fact it can be done. Papoulis? > Cover? Haykin? I know I have seen the proof at some point. > If x^2(t) is a Gaussian distribution function of t like integralfrom -infinity to t of e^-t^2 (times a constant) then no way is your integral of the product going to be the product of the === Subject: Re: Update: Objections to Cantor's Theory >Did Zermelo's proof use Zorn's Lemma? That seems like a natural way to >do it: Use the subset ordering on the set of all well-orderings of >subsets of R. By Zorn's lemma there is a maximal element, which has to >well-order all of R. Surely the subset ordering is insufficient for an application of Zorn's Lemma. It is quite simple to contruct a chain without an upper bound. Let A_n = {-n,-n+1,...,-1,0} for n = 0,1,2,3,...., under the usual ordering of the reals, then {A_n : n = 0,1,2,3,...} is a chain without an upper bound. On the other hand, a partial order on the set of all well-orderings of subsets of R can be introduced such that (A,<<_A) < (B,<<_B) (where A and B are subsets of R, <<_A is a well-ordering on A, and <<_B is a well- ordering on q) if A is a subset of B, <<_A is the restriction of <<_B to A (so that for elements x and y of A, x <<_A y iff x <<_B y), and A is an initial segment of B under the ordering <<_B (i.e. if x is an element of B, y is an element of A, and x <<_B y, then x is an element of A). Under this new order, every chain does have an upper bound, and so Zorn's Lemma can be applied, and the existence of a well-ordering of R follows immediately. ----- === Subject: Re: Update: Objections to Cantor's Theory >>Did Zermelo's proof use Zorn's Lemma? That seems like a natural way to >>do it: Use the subset ordering on the set of all well-orderings of >>subsets of R. By Zorn's lemma there is a maximal element, which has to >>well-order all of R. >Surely the subset ordering is insufficient for an application of Zorn's >Lemma. It is quite simple to contruct a chain without an upper bound. >Let A_n = {-n,-n+1,...,-1,0} for n = 0,1,2,3,...., under the usual >ordering of the reals, then {A_n : n = 0,1,2,3,...} is a chain without an >upper bound. >On the other hand, a partial order on the set of all well-orderings of >subsets of R can be introduced such that (A,<<_A) < (B,<<_B) (where A and >B are subsets of R, <<_A is a well-ordering on A, and <<_B is a well- >ordering on q) This bit should have read that <<_B is a well-ordering of B. >if A is a subset of B, <<_A is the restriction of <<_B to A >(so that for elements x and y of A, x <<_A y iff x <<_B y), and A is an >initial segment of B under the ordering <<_B (i.e. if x is an element of >B, y is an element of A, and x <<_B y, then x is an element of A). >Under this new order, every chain does have an upper bound, and so Zorn's >Lemma can be applied, and the existence of a well-ordering of R follows >immediately. ----- === Subject: Re: infinity > He says it all has to do with Cantor's set theory, cardinality etc..., but > browsing the internet didn't really help me much. > Any information or relevant links are very welcome, This is an old philosophical favorite from the fifties and sixties. Googling for supertask+philosophy will lead you to a number of papers and discussions. === Subject: Re: infinity > Googling for supertask+philosophy will lead you to a number of papers > and discussions. The Stanford Encyclopedia gives a good overview: http://plato.stanford.edu/entries/spacetime-supertasks/ === Subject: Re: infinity > The proof that you can put ten unlabelled balls into a vase take one > out, repeatedly for an infinite number of times and end up with an > empty vase is eluding me. > > Without labelling the balls at all: after adding the balls in step n, > A can remove one of the balls added in step (n/10 rounded up). If A > does so for all n, then every ball added will be removed before noon. > > Note that A doesn't have to do so deliberately. Accidently will > suffice, if you merely want to show that it is possible. > > > We can actually show more than that though: we can demonstrate that > even if A loses track of which balls were added when and just picks > one at random from all the balls in the vase, then the vase will > almost always be empty at noon. > > However, the proof in that case does rely on how the balls are added > and removed. If we add n balls at step n, then it is still possible > to have an empty vase at noon, but randomly picking out balls will > have probability 0 of doing so. > > > Here's a nasty twist to the problem: suppose the balls are labelled, > put in the vase in sequence, and A always removes the lowest numbered > ball from the vase. However, B is a miser who doesn't want to buy any > more balls than necessary. He has been removing the label from each > ball taken out and relabelling it as the first ball yet to be put in. > Ouch very nasty. Note there are two seemingly correct arguments. i) A cannot tell where B get the balls from, so the answer is not changed. The vase is empty at noon. ii) The first ball (indeed every ball) gets added to and removed from the vase an infinite number of times in any interval (noon-t,noon). Thus we cannot say whether or not the first ball is in the vase at noon. Thus the number of balls in the vase at noon is indeterminate. I would attempt to resolve the seeming paradox as follows. If we consider not balls but labels, it is clear that there is no label in the vase at noon (labels are not reused). Can we conclude from this that there are no balls in the vase at noon? i) if every ball has one and only one label, Yes. i') if every ball has only a finite number of labels, Yes. ii) if any ball has an inifinite number of labels, No. In the original problem we make the assumption that every label corresponds to exactly one ball, i.e. we assume that the balls are not reused. With this assumption the arguments used are correct. If we make the assumption that every ball is reused an infinite number of times, then every ball will correspond to an infinite number of labels and the arguments used are not correct. Still, if the original result is counterintuitive, this is counterintuitive squared. -William Hughes === Subject: Re: infinity > The original question: > Problem 1 > Suppose you have a giant vase and a bunch of ping pong balls with an > integer written on each one, e.g. just like the lottery, so the balls > are numbered 1, 2, 3, ... and so on. At one minute to noon you put > balls 1 to 10 in the vase and take out number 1. At half a minute to > noon you put balls 11 - 20 in the vase and take out number 2. At one > quarter minute to noon you put balls 21 - 30 in the vase and take out > number 3. Continue in this fashion. Obviously this is physically > impossible, but you get the idea. Now the question is this: At noon, > how many ping pong balls are in the vase? > > We can change the above question into an equivalent problem. > Problem 2. > Suppose you have a giant vase and a bunch of ping pong balls with an > integer written on each one, e.g. just like the lottery, so the balls > are numbered 1, 2, 3, ... and so on. > At minute 1 you put balls 1 to 10 in the vase and take out number 1. > At minute 2 you put balls 11 - 20 in the vase and take out number 2. > At minute 3 you put balls 21 - 30 in the vase and take out number 3. > Continue in this fashion. Obviously this is physically impossible, but > you get the idea. Now the question is this: At an infinite time from > now, how many ping pong balls are in the vase? > > According to this problem, noon is not reachable. > > Wrong. The experiment described is quite well defined. We know > which balls are added to the vase and when and we can consider > the different outcomes by choosing which balls are removed from > the vase and when. Noon is not reachable in any physical sense, > but we know that the experiment described has no physical realization > or approximation. > > I hope you can see that 'noon' in problem 1 is equivalent to 'an > infinite time from now' in problem 2. Yes. >They are un-reachable. Why? An infinite time from now is un-reachable in any physical sense, but there is no problem defining it in a mathematical sense. > Again I > have to emphasis that in order to prove the vase is empty, we have to > prove that the last ball is taken out from the vase. No, there is no such thing as the last ball. To prove that the vase is empty we have to prove that every ball that is put into the vase is taken out of the vase. > > Yes, you can claim from problem 2 that ball n is removed from the vase > at the nth minute. The vase can only be claimed to be empty if ball n > is the last ball in the vase. Obviously by the time the nth ball is > removed there are many more balls in the vase. So the vase cannot be > empty even though every ball currently in the vase will be removed > later in the future. > And thus at any finite t the vase will not be empty. This does not tell us what will hapen after an infinite amount of time. > By mathematical induction. n=1 is true. n=n+1 is also true. It implies > that it is true for all n. The vase is not empty for all n > 0. > Yes, after any finite amount of time the vase is not empty. However, this does not tell us what will happen after an infinite amount of time. > Mathematical induction still cannot tell you what happens exactly when > n is infinite which does not exist physically or mathematically. Small error here. Inifinity does not exist physically, but it does exist mathematically. > But > it can claim that the vase is not empty as n approaches to infinity. > And as this does not tell us what happens at infinity, it does not help. > In the vase example, if we only look at the withdrawal portion of the > argument then the claim is not strong. If both putting in and taken > out of balls are taken into account. In order to prove that the vase > is empty at noon, you must have to show at certain time the rate of > taking out is greater than the rate of putting in. > > No. The rate at which we add balls is infinite. So is the rate at > which we remove them. We cannot conclude anything from this fact. True, > at any time before noon the rate at which we are adding balls is greater > than the rate at which we are removing them. This tells us nothing about > what happens at noon. > -William Hughes > In the equivalent problem (problem 2), the rate of adding the balls to > the vase is 10 balls/minute. The withdrawal rate is 1 ball/minute. > So we can have an infinite rate and a finite time, or a finite rate and an infinite time. We still have an infinite number of balls. > Assume f(x) is an increasing function and f(0) > 0. Even though f(oo) > is undefined but I am quite sure that f(oo) cannot be 0. Why? -William Hughes === Subject: Re: infinity > The original question: > Problem 1 > Suppose you have a giant vase and a bunch of ping pong balls with an > integer written on each one, e.g. just like the lottery, so the balls > are numbered 1, 2, 3, ... and so on. At one minute to noon you put > balls 1 to 10 in the vase and take out number 1. At half a minute to > noon you put balls 11 - 20 in the vase and take out number 2. At one > quarter minute to noon you put balls 21 - 30 in the vase and take out > number 3. Continue in this fashion. Obviously this is physically > impossible, but you get the idea. Now the question is this: At noon, > how many ping pong balls are in the vase? > We can change the above question into an equivalent problem. > Problem 2. > Suppose you have a giant vase and a bunch of ping pong balls with an > integer written on each one, e.g. just like the lottery, so the balls > are numbered 1, 2, 3, ... and so on. > At minute 1 you put balls 1 to 10 in the vase and take out number 1. > At minute 2 you put balls 11 - 20 in the vase and take out number 2. > At minute 3 you put balls 21 - 30 in the vase and take out number 3. > Continue in this fashion. Obviously this is physically impossible, but > you get the idea. Now the question is this: At an infinite time from > now, how many ping pong balls are in the vase? >> According to this problem, noon is not reachable. >> Wrong. The experiment described is quite well defined. We know >> which balls are added to the vase and when and we can consider >> the different outcomes by choosing which balls are removed from >> the vase and when. Noon is not reachable in any physical sense, >> but we know that the experiment described has no physical realization >> or approximation. > I hope you can see that 'noon' in problem 1 is equivalent to 'an > infinite time from now' in problem 2. They are un-reachable. Again I > have to emphasis that in order to prove the vase is empty, we have to > prove that the last ball is taken out from the vase. Noon may be un-reachable in a physical sense, but this is not a physical problem. It's a mathematical problem. The question makes perfect sense from a mathematical standpoint. Your final sentence above makes no sense at all. There is no such thing as a last ball. All that is necessary to show that the vase is empty is to show that each ball is removed from the vase before noon and is not subsequently replaced. > Assume f(x) is an increasing function and f(0) > 0. Even though f(oo) > is undefined but I am quite sure that f(oo) cannot be 0. Why? -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: infinity > > Noon may be un-reachable in a physical sense, but this is not a physical > problem. It's a mathematical problem. The question makes perfect sense > from a mathematical standpoint. > > Your final sentence above makes no sense at all. There is no such thing > as a last ball. All that is necessary to show that the vase is empty > is to show that each ball is removed from the vase before noon and is not > subsequently replaced. > > Assume f(x) is an increasing function and f(0) > 0. Even though f(oo) > is undefined but I am quite sure that f(oo) cannot be 0. > > Why? > > Dave Seaman This is a paradox because it is not a physical nor a mathematical problem but rather a logical problem!!! Physically or mathematical it is not difficult to prove that the vase is not empty at noon. Mathematically, # of balls in the vase can be expressed by the equation f(t) = 9+9*log(1/t)/log(2) where t is the time in minute before noon. 't' can be 1, 1/2, 1/4, 1/8, ... 1/OO. f(1) = 9 f(1/2) = 18 f(1/4) = 27 f(1/8) = 36 Number of balls in the vase at noon is f(0) = OO. Try it :) The statement: All that is necessary to show that the vase is empty is to show that each ball is removed from the vase before noon and is not subsequently replaced. is a logical argument. Unfortunately, if infinity gets involved, this statement alone is not sufficient to claim the vase is empty before noon. The vase can only be claimed to be empty at a particular time t is when all the balls are removed at that time. At t = 1, ball 1 is removed but ball 10 (the last ball) is added to the vase. At t = 1/2, ball 2 is removed but ball 20 (the new last ball) is in the vase. There is no such time that all balls are removed. Therefore, the vase cannot be empty. If the problem is changed to remove the last ball each time then the argument would be much simplier - ball 1 is always in the base. === Subject: Re: infinity > > Noon may be un-reachable in a physical sense, but this is not a physical > problem. It's a mathematical problem. The question makes perfect sense > from a mathematical standpoint. > > Your final sentence above makes no sense at all. There is no such thing > as a last ball. All that is necessary to show that the vase is empty > is to show that each ball is removed from the vase before noon and is not > subsequently replaced. > > Assume f(x) is an increasing function and f(0) > 0. Even though f(oo) > is undefined but I am quite sure that f(oo) cannot be 0. > > Why? > > Dave Seaman > > > This is a paradox because it is not a physical nor a mathematical > problem but rather a logical problem!!! > > Physically or mathematical it is not difficult to prove that the vase > is not empty at noon. Mathematically, # of balls in the vase can be > expressed by the equation f(t) = 9+9*log(1/t)/log(2) where t is the > time in minute before noon. 't' can be 1, 1/2, 1/4, 1/8, ... 1/OO. > f(1) = 9 > f(1/2) = 18 > f(1/4) = 27 > f(1/8) = 36 > Number of balls in the vase at noon is f(0) = OO. Try it :) O.K, f(0) = 9 + 9*log(1/0)/log(2) which is undefined. I guess we will have to use a different way to find out how many balls are in the vase at noon. If this were a physical problem we could use the limit of f(t) as t approaches 0. However, this is not a physical problem. > > The statement: All that is necessary to show that the vase is empty is > to show that each ball is removed from the vase before noon and is not > subsequently replaced. is a logical argument. And a sound one. > Unfortunately, if > infinity gets involved, this statement alone is not sufficient to claim > the vase is empty before noon. True, but only because the vase is not empty at any time before noon. The statement is sufficient to claim the vase is empty at noon. > The vase can only be claimed to be > empty at a particular time t is when all the balls are removed at that > time. No, we need The vase can only be claimed to be empty at a particular time t is when all the balls are removed at or before that time. --------- In there are only a finite number of steps, adding or before is not needed, we can always find the last time at which balls are removed. However, if there are a infinite number of steps there may not be a last time at which balls are removed. > At t = 1, ball 1 is removed but ball 10 (the last ball) is added > to the vase. At t = 1/2, ball 2 is removed but ball 20 (the new last > ball) is in the vase. There is no such time that all balls are > removed. Therefore, the vase cannot be empty. > Yes, there is no time before noon at which all balls are removed. However, there are an infinite number of steps and this statement does not tell us anything about what happens after an infinite number of steps. > If the problem is changed to remove the last ball each time then the > argument would be much simplier - ball 1 is always in the base. As has been noted many times, if you change the order in which you remove the balls you can get any result from 0 to infinity. - William Hughes === Subject: Re: infinity > O.K, f(0) = 9 + 9*log(1/0)/log(2) which is undefined. Mathematically: f(0) = OO (infinity). In usual case we would say it is undefined but because we are dealing with infinities here so f(0) = OO is well defined. Physically: Infinity is undefined physically. > I guess we will have to use a different way to > find out how many balls are in the vase > at noon. If this were a physical problem we could use the limit > of f(t) as t approaches 0. However, this is not a physical problem. > > The statement: All that is necessary to show that the vase is empty is > to show that each ball is removed from the vase before noon and is not > subsequently replaced. is a logical argument. > > And a sound one. Agree, a sound one but still not sufficient to make the claim. > > Unfortunately, if > infinity gets involved, this statement alone is not sufficient to claim > the vase is empty before noon. > > True, but only because the vase is not empty at any time before > noon. The statement is sufficient to claim the vase is empty at noon. The operation of adding or removal of balls is undefined at noon. In order for the vase to be empty at noon, the vase has to be empty before noon. It is proved that the vase is not empty at any time before noon. How can the vase be empty at noon! > > The vase can only be claimed to be > empty at a particular time t is when all the balls are removed at that > time. > > No, we need > The vase can only be claimed to be > empty at a particular time t is when all the balls > are removed at or before that time. > --------- > > In there are only a finite number of steps, adding or before > is not needed, we can always find the last time at which balls > are removed. However, if there are a infinite number of steps there > may not be a last time at which balls are removed. > > At t = 1, ball 1 is removed but ball 10 (the last ball) is added > to the vase. At t = 1/2, ball 2 is removed but ball 20 (the new last > ball) is in the vase. There is no such time that all balls are > removed. Therefore, the vase cannot be empty. > > > Yes, there is no time before noon at which all balls are removed. > However, there are an infinite number of steps and this statement > does not tell us anything about what happens after an infinite number > of steps. Again, the operation of putting in or taking out balls is undefined at noon. You are not allow to add or remove any ball right at the noon time. In order for the vase to be empty at noon, the vase has to be empty some time before noon. You already agree that no time before noon at which all balls are removed. How can the vase be empty at noon? > If the problem is changed to remove the last ball each time then the > argument would be much simplier - ball 1 is always in the base. > > As has been noted many times, if you change the order in which > you remove the balls you can get any result from 0 to infinity. No, the result can never be 0. === Subject: Re: infinity >> Noon may be un-reachable in a physical sense, but this is not a physical >> problem. It's a mathematical problem. The question makes perfect sense >> from a mathematical standpoint. >> Your final sentence above makes no sense at all. There is no such thing >> as a last ball. All that is necessary to show that the vase is empty >> is to show that each ball is removed from the vase before noon and is not >> subsequently replaced. >> Assume f(x) is an increasing function and f(0) > 0. Even though f(oo) >> is undefined but I am quite sure that f(oo) cannot be 0. >> Why? >> Dave Seaman > This is a paradox because it is not a physical nor a mathematical > problem but rather a logical problem!!! Saying that it's a paradox merely means that the result is counterintuitive. It doesn't mean there is actually a contradiction. > Physically or mathematical it is not difficult to prove that the vase > is not empty at noon. Mathematically, # of balls in the vase can be > expressed by the equation f(t) = 9+9*log(1/t)/log(2) where t is the > time in minute before noon. 't' can be 1, 1/2, 1/4, 1/8, ... 1/OO. > f(1) = 9 > f(1/2) = 18 > f(1/4) = 27 > f(1/8) = 36 > Number of balls in the vase at noon is f(0) = OO. Try it :) Incorrect. You are confusing f(0) with lim_{t->0-} f(t). The former is 0, but the latter is +oo. The former is what the problem asks for. Once you get your notation straight, the paradox disappears. > The statement: All that is necessary to show that the vase is empty is > to show that each ball is removed from the vase before noon and is not > subsequently replaced. is a logical argument. Unfortunately, if > infinity gets involved, this statement alone is not sufficient to claim > the vase is empty before noon. The vase can only be claimed to be > empty at a particular time t is when all the balls are removed at that > time. Incorrect. Change at that time to at or before that time. Then the time t=noon fills the bill. > At t = 1, ball 1 is removed but ball 10 (the last ball) is added > to the vase. At t = 1/2, ball 2 is removed but ball 20 (the new last > ball) is in the vase. There is no such time that all balls are > removed. Therefore, the vase cannot be empty. All balls are removed before noon. > If the problem is changed to remove the last ball each time then the > argument would be much simplier - ball 1 is always in the base. If the problem is changed so that all the balls are in the vase at 11:00 and from then on balls are only removed, never added, then do you agree that it is possible for the vase to be empty at noon? -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: infinity >> Unfortunately, if >> infinity gets involved, badly working minds stop working entirely. We see it all the time. Lee Rudolph === Subject: Re: infinity > Physically or mathematical it is not difficult to prove that the vase > is not empty at noon. Mathematically, # of balls in the vase can be > expressed by the equation f(t) = 9+9*log(1/t)/log(2) where t is the > time in minute before noon. 't' can be 1, 1/2, 1/4, 1/8, ... 1/OO. > f(1) = 9 > f(1/2) = 18 > f(1/4) = 27 > f(1/8) = 36 > Number of balls in the vase at noon is f(0) = OO. Try it :) > > Incorrect. You are confusing f(0) with lim_{t->0-} f(t). The former is > 0, but the latter is +oo. The former is what the problem asks for. Once > you get your notation straight, the paradox disappears. > According to the fomula f(t) = 9+9*log(1/t)/log(2), f(0) = 9 + 9 log(1/0)/log(2) = 9 + 9 log(OO)/log(2) = 9 + 9 * OO / log(2) = OO not 0. In mathematical term, usually we refer infinity as undefined e.g 1/0 = OO (infinity) but very often we say 1/0 is undefined. I think you do not mean t->0- above. It should be t->0+, right? f(t) is undefined as t->0-. f(0+) = 9 + 9 log(1/0+)/log(2) = 9 + 9 log(OO)/log(2) = 9 + 9 * OO / log(2) = OO f(0-) = 9 + 9 log(1/0-)/log(2) = 9 + 9 log(-OO)/log(2) Cannot proceed because log(-n) is not valid > The statement: All that is necessary to show that the vase is empty is > to show that each ball is removed from the vase before noon and is not > subsequently replaced. is a logical argument. Unfortunately, if > infinity gets involved, this statement alone is not sufficient to claim > the vase is empty before noon. The vase can only be claimed to be > empty at a particular time t is when all the balls are removed at that > time. > > Incorrect. Change at that time to at or before that time. Then the > time t=noon fills the bill. > > At t = 1, ball 1 is removed but ball 10 (the last ball) is added > to the vase. At t = 1/2, ball 2 is removed but ball 20 (the new last > ball) is in the vase. There is no such time that all balls are > removed. Therefore, the vase cannot be empty. > > All balls are removed before noon. Again this argument alone is not sufficient to claim that the vase is empty at noon. Do you have any other argument? > If the problem is changed to remove the last ball each time then the > argument would be much simplier - ball 1 is always in the base. > > If the problem is changed so that all the balls are in the vase at 11:00 > and from then on balls are only removed, never added, then do you agree > that it is possible for the vase to be empty at noon? If 'all the balls' refers to only a finite number of balls, say a be empty before noon. Let n be the total number of ball at 11:59 t minute be the time before noon f(t) = n - log(1/t)/log(2) f(1) = n f(1/2) = n - 1 f(1/4) = n - 2 ... f(1/2^n) = n - n = 0 You can see that one of the requirements for the vase to be empty at noon is the ball withdrawal rate is greater than the adding rate. In this case the adding rate is 0 but the removal rate is greater than 0. The last ball (ball n) is to be removed at a time 1/2^n minute before noon. By the time the last ball is being removed, no other ball is left in the vase. This is exactly the criteria for the vase to be empty. Unfortunately, 'all the balls' in this problem refers to an infinite number of balls. f(t) = OO - log(1/t)/log(2) f(1) = OO f(1/2) = OO - 1 = OO f(1/4) = OO - 2 = OO ... There is no such a time that the vase would be empty. --------------------------------------------------------- Case 1. If the removal rate is 10 balls/each time and the adding rate is 1 ball/each time then the vase will be empty at noon provide that there are not infinitely number of balls in the vase initially. Case 2. If the removal rate is 1 ball/each time and the adding rate is 10 balls/each time then the vase will not be empty at noon. Case 3. If the removal rate is 1 ball/each time and the adding rate is 1 ball/each time then the vase will be empty at noon only if there is no ball in the vase initially. I hope you can see that the argument All balls are removed before noon. alone is not sufficient to claim that the vase is empty at noon. === Subject: Need help in Logarithm formula Can somebody tell me, why the following is valid: 1/(k+1) <= Integrate[1/t, {t, k, k + 1}] ( k is natural number grater that 1, {t, k, k + 1} means we integrate from k to k+1, mathematica syntax) Of cource i can see it on the graph of the functions 1/t and Log(t), but, a proof? PS: This formula (1/(k+1) <= Integrate[1/t, {t, k, k + 1}]) is used to prove that the euler's gamma exists. === Subject: Re: Greedy backpack filling > Now, my problem is, given a set of coin values, how can one determine > if the greedy algorithm is also optimal? Does this help? -- Don Reble djr@nk.ca === Subject: Re: Easy question > Hint: What is the geometric relation between a function and its > inverse? > You mean geometric relation between a function and its mirror image, > not inverse. Did you actually think through the hint, or just assume it was useless? - Tim === Subject: Re: Easy question <2rHqQnzmKPh=hXbSLCHWvsRgU0kB@4ax.com> The inverse of f=x^n is x=f^(1/n). However, if I plot f=x^(1/n) it is not a mirror image. I found hint useful but not helpful. I still need an answer. === Subject: Re: Easy question <2rHqQnzmKPh=hXbSLCHWvsRgU0kB@4ax.com> Never mind === Subject: Re: Bizarritudes >> <18313041.1122673658810.JavaMail.jakarta@nitrogen.math >> forum.org>, >>Is there a function as sum and product to calculate >> an expression elevated at an exponent >>(which could be itself and vary according to n) a >> number n of times ? >>Otherwise, how do you prove that in the case of such >> a function f(x,n) as n tends to the >>infinite we have f(2^0.5) = 2 ? I mean that >>(2^0.5)^(2^0.5)^(2^0.5)^(2^0.5)^(2^0.5)^(2^0.5)^(2^0. >> 5)...... = 2 ? >> You have to be careful with exponents, because they >> are not associative. I think you mean >> x^(x^(x^...))). Let f(x,n) be this function with n >> iterations (i.e. f(x,0) = x and >> f(x,n+1)= x^f(x,n)). If a limit exists, i.e. f(x,n) >> -> L as n -> infty, then x^L = L >> so (if x > 0 and L > 0) x = L^(1/L). However, the >> limit might not exist. >> The function g(t) = x^t has g'(t) = ln(x) x^t, so the >> fixed point L is stable if >> |ln(x)| L < 1, i.e. |ln(L)| < 1 if x = L^(1/L), and >> unstable if |ln(L)| > 1. >> The limit won't exist if it's unstable (except in the >> unlikely case that f(x,n) = L for some n). >> In your particular case x = sqrt(2), L = 2, |ln(2)| < >> 1 so it is stable. Note that L = 4 >> would also correspond to x = 4^(1/4) = sqrt(2), but 4 >> is an unstable fixed point. >f'(x,n)Are there rules when the exponent varies according to x or n ? I mean >what if the exponent is 1/n or x+1 or n+1 and so on ? Care to try a specific example? If it's in the form f(x,n+1) = g(x, f(x,n)) you can probably use the fixed-point stability theory. Otherwise, you may be able to do something else. >Do you think there are examples in the nature which use such functions ? I don't know what you mean. What would you count as nature? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Bizarritudes 1.01*1.02*1.03*1.04*1.05*1.06*1.07*1.08*1.09*1.1*1.11*1.12 *1.13*1.14*1.15*1.16*1.17*1.18*1.19*1.2*1.21*1.22*1.23*1.2 4*1.25*1.26*1.27*1.28*1.29*1.3*3.1*1.32*1.33*1.34*1.35*1.3 6*1.37*1.38*1.39*1.4*1.41*1.42*1.43*1.44*1.45*1.46*1.47*1. 48*1.49*1.5*1.51*1.52*1.53*1.54*1.55*1.56*1.57*1.58*1.59*1 6*1.61*1.62*1.63*1.64*1.65*1.66*1.67*1.68*1.69*1.7*1.71*1 72*1.73*1.74*1.75*1.76*1.77*1.78*1.79*1.8*1.81*1.82*1.83* 1.84*1.85*1.86*1.87*1.88*1.89*1.9*1.91*1.92*1.93*1.94*1.95 *1.96*1.97*1.98*1.99*2 : the result is e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^ e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^ e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^e^ e^e^e^e^e^e^e^e^ (9.2968677714896404158318070335902150926162938497960026420 9079451665461269589986755951033497183435605908212250585528 4745283589281538340366798370287098049868410113982558224047 5215299350463012070468200527.8710^7410) for my calculator. How would you make it understandable for the poor human I am ? Nature: I don't know, the number of molecules of air displaced by a superbioionic shuttle in relation with the light-years of the travel, or the number of potential humans, molecules, atoms, ions, contained in the universe. Other said, as the '^2' describe the surface, '^3' the volume, what would this '^x_n^x_n+1^x_n+2...' mean (if it is supposed to mean anything)? === Subject: Re: Mathematical concepts ... >SMSG was the group In fact, the School Mathematics Study Group, if I recall correctly. >producing high school materials, as I >understand it. It was sponsored by NSF, and unfortunately >by the educationists in NSF. Those were the ones responsible >for not adopting the fully modernized Euclid and keeping >students from learning that proofs are needed. Whence the other expansion of SMSG (at the time): Some Math, Some Garbage. Lee Rudolph === Subject: euclidean distance question if you have 2 vectors a and b of length n is it true that the euclidean distance between the two is greater than the euclidean distance between them if each of their elements contained their means. it seems this way.. if d( (a1,a2),(b1,b2) ) >= d( ((a1+a2)/2, (a1+a2)/2), ((b1+b2)/2, (b1+b2)/2) ) === Subject: Re: euclidean distance question > if you have 2 vectors a and b of length n is it true that the euclidean > distance between the two is greater than the euclidean distance between > them if each of their elements contained their means. it seems this > way.. > > if d( (a1,a2),(b1,b2) ) >= d( ((a1+a2)/2, (a1+a2)/2), ((b1+b2)/2, > (b1+b2)/2) ) Do you mean for R^2? The question depends on what dimensional space you are talking about... === Subject: Re: euclidean distance question > if you have 2 vectors a and b of length n is it true that the euclidean > distance between the two is greater than the euclidean distance between > them if each of their elements contained their means. it seems this > way.. > > if d( (a1,a2),(b1,b2) ) >= d( ((a1+a2)/2, (a1+a2)/2), ((b1+b2)/2, > (b1+b2)/2) ) > In what dimensional space? <3kvb4hF102qodU1@individual.net> >>>2. No continuum has been discovered in physics -- everything seems to >>>change in finite units called quanta. That's the real world. > Perhaps someone has pointed this out, but the energy > only binding forces that give rise to discrete states. > > Sufficiently high energy states of an electron > in a Hydrogen atom also have a continuous energy spectrum. > > See > > http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_1.html > > for example. > > In general, I am not sure. I don't even know if we > have a closed-form solution for any atom more complex than > hydrogen. > > Heck, we don't even have a closed-form solution for > the classical three-body problem. And in fact, we > know that there isn't one. Since a solution to > the quantum problem would presumably reduce to > one for the classical problem in the appropriate > limit, I'm pretty sure we don't have one for anything > more complicated than Hydrogen. I'll disagree with that. A solution to the three body problem exists, because otherwise there would not be more than two objects in the universe, and anyways when there are two objects in the universe, then functions among them and those among them are objects and the universe is infinite and thus also infinite sets are equivalent. I'm surprised to hear that there are not well-known closed form solutions to the three body problem. There might be considered the symplectic integration, I think there are some tools available to solve that problem besides the numerical approximations and their refinements, which I think also have well-known closed form solutions, of which I am ignorant. Will this thread about Cantor have a thousand posts and bring progress? There is reason, not just that it is intuitive but also provable, that infinite sets are equivalent. Where there is Cantor's powerset result, about powersets, and the antidiagonal argument, about the reals, and nested intervals of the complete reals, also about the reals, the continuum of the real numbers, then reeconciling the notions that infinite sets are equivalent with those of Cantor lead to a variety of considerations of the natural numbers, via natural deduction. About the reals, the reals are at once the complete ordered field and also the ring, they're a contiguous sequence of elements. For exact correspondence with the results of integral calculus and geometry, there are as many reals between zero and one as there are natural numbers. Then, where the reals are dimensionless points, it's also possible and as necessary necessary they are polydimensional points. Examine Vitali with respect to unidimensional points, with that as a tenet of a nonstandard measure theory, an infinital analytic measure theory, there are no non-measurable sets. Ross > >>>> >>>>>2. No continuum has been discovered in physics -- everything seems to >>>>>change in finite units called quanta. That's the real world. >>> >>>Perhaps someone has pointed this out, but the energy >>>only binding forces that give rise to discrete states. >> >>Sufficiently high energy states of an electron >>in a Hydrogen atom also have a continuous energy spectrum. >> >>See >> >>http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_1.html >> >>for example. >> >> >>>In general, I am not sure. I don't even know if we >>>have a closed-form solution for any atom more complex than >>>hydrogen. >> >>Heck, we don't even have a closed-form solution for >>the classical three-body problem. And in fact, we >>know that there isn't one. I'll disagree with that. That's probably because you don't understand the statement. There is no analytic function into which you can feed the initial positions and velocities of three bodies interacting via Newtonian gravitation and from which you can subsequently compute their position at arbitrary times. Of course, there are numerical and approximation schemes that allow you to work these quantities out to any precision you care about. <3kvb4hF102qodU1@individual.net> > Heck, we don't even have a closed-form solution for > the classical three-body problem. And in fact, we > know that there isn't one. Since a solution to > the quantum problem would presumably reduce to > one for the classical problem in the appropriate > limit, I'm pretty sure we don't have one for anything > more complicated than Hydrogen. > > I'll disagree with that. And you'll be wrong. > A solution to the three body problem exists, > because otherwise there would not be more than two objects in the > universe, The three-body problem is the problem of describing this motion mathematically, not the problem of this motion existing. There is no solution to the three-body problem. As Robert points out, it has been shown that the problem can not be solved in closed form. > I'm surprised to hear that there are not well-known closed form > solutions to the three body problem. There are solutions to some specific special cases, but none in general. As soon as you get more than two bodies, the problem is chaotic in the mathematical sense. > There might be considered the > symplectic integration, I think there are some tools available to solve > that problem besides the numerical approximations and their > refinements, You think wrong. > which I think also have well-known closed form solutions, Numerical approximation is the other choice when you don't have a closed-form solution. The notion of a closed form solution to a numerical approximation doesn't make any sense. - Randy <3kvb4hF102qodU1@individual.net> > > Heck, we don't even have a closed-form solution for > the classical three-body problem. And in fact, we > know that there isn't one. Since a solution to > the quantum problem would presumably reduce to > one for the classical problem in the appropriate > limit, I'm pretty sure we don't have one for anything > more complicated than Hydrogen. > > I'll disagree with that. > > And you'll be wrong. > > A solution to the three body problem exists, > because otherwise there would not be more than two objects in the > universe, > > The three-body problem is the problem of describing this > motion mathematically, not the problem of this motion > existing. > > There is no solution to the three-body problem. As Robert > points out, it has been shown that the problem can not be > solved in closed form. > > I'm surprised to hear that there are not well-known closed form > solutions to the three body problem. > > There are solutions to some specific special cases, but none > in general. As soon as you get more than two bodies, the > problem is chaotic in the mathematical sense. > > There might be considered the > symplectic integration, I think there are some tools available to solve > that problem besides the numerical approximations and their > refinements, > > You think wrong. > > which I think also have well-known closed form solutions, > > Numerical approximation is the other choice when you don't > have a closed-form solution. The notion of a closed form > solution to a numerical approximation doesn't make any > sense. > > - Randy How is it shown that there can be no mathematical closed form solution to an N>2 body problem? When I say closed form solution to a numerical approximation, I mean a closed form solution for the building blocks of the numerical approximation, leading to an exact, one step closed form analytic solution. So I just think that there are methods to go from the discrete time serium approximations directly to continuous time approximations. Randy, infinite sets are equivalent. Divide by zero. Anyways, I hope you can enlighten me why there can not be closed form solutions to the three body problem so I can tell the people looking for a theory of everythnig about it. Consider objects A, B, and C in a straight line at relative rest, solve. Why are people always trying to stop the Hilbert program? Ross <3kvb4hF102qodU1@individual.net> > How is it shown that there can be no mathematical closed form solution > to an N>2 body problem? I'm a little fuzzy on the details. Here are a couple of hints: Poincar.8e and Bruns proved that the three-body problem is unsolvable in the precise sense that it does not admit 'enough' analytic integrals to integrate it. (http://www.mathaware.org/mam/05/) Bruns Theorem: The 10 classical integrals of the three-body problem (three for the position of the center of mass, three for the velocity of the center of mass, three for the angular momentum, and one for the energy) are the only algebraically independent integrals of this 18-degree-of-freedom system. (http://scienceworld.wolfram.com/physics/BrunsTheorem.html) > When I say closed form solution to a numerical approximation, I mean > a closed form solution for the building blocks of the numerical > approximation, leading to an exact, one step closed form analytic > solution. If I have n discrete points on a curve, found numerically, it is true that I can find a smooth curve that interpolates them, and write down a closed-form expression for that curve. However, there is no sense in which that can be called the solution to the numerically-derived results. There are infinitely many smooth curves which interpolate the same points. > So I just think that there are methods to go from the > discrete time serium approximations directly to continuous time > approximations. Precisely. You've just got another approximation. It's silly to describe that as the solution. It's just one more approximation. > Randy, infinite sets are equivalent. Divide by zero. Non sequitor, and infinite sets are provably not equivalent. > Anyways, I hope you can enlighten me why there can not be closed form > solutions to the three body problem so I can tell the people looking > for a theory of everythnig about it. What does the TOE have to do with the three-body problem? - Randy > approximations. Randy, infinite sets are equivalent. Divide by zero. Division by zero has nothing to do with the equivalence or non-equivlanece of sets. A set is infinite if and only if there exist a bijection from the set onto a proper subset of itself. This definite of infinite cardinality has nothing whatever to do with diving by zero. It is a proven theorem that a set cannot be put into one to one correspondence with its power set. This is true for -any- set, be it infinite or finite. If S is an infinite set (see above definition) its power set is also infinite yet S and its power set have not the same cardinal number. Do you know what you are talking about? Bob Kolker <3kvb4hF102qodU1@individual.net> > > approximations. > > Randy, infinite sets are equivalent. Divide by zero. > > Division by zero has nothing to do with the equivalence or > non-equivlanece of sets. A set is infinite if and only if there exist a > bijection from the set onto a proper subset of itself. This definite of > infinite cardinality has nothing whatever to do with diving by zero. > > It is a proven theorem that a set cannot be put into one to one > correspondence with its power set. This is true for -any- set, be it > infinite or finite. If S is an infinite set (see above definition) its > power set is also infinite yet S and its power set have not the same > cardinal number. > > Do you know what you are talking about? > > Bob Kolker Hi Bob, Yes, I do know to some extent that which I discuss. Consider a theory with a set of all sets, where all its elements are sets, as everything is a set in the theory. That set is its own powerset, and identity is a bijective function between it and itself. There are a variety of reasons why to consider that a theory that is to maintain a reference to any and every object of discourse has a universal element. As you're aware, I suggest reading about Hegel's descriptions of Being and Nothing and their conflation, and suggest that these are primitive, and true, notions openly discussed since the ancients, but that there are certain ways to formalize those notions that impact mathematical and logical, and physical, theories. You're partially right, division by zero doesn't really have much to do with infinite sets being equivalent, except to the extent that certain analytical descriptions, of basically the unit scalar infinity, are constructions. Consider the Schroedinger equation, using relatively modern methods of the last twenty years or so it's solveable, before that it was not, but that wasn't a proof that solutions did not exist, only that they had not been discovered. It's like the ancient Greeks where when one of their fellows suggested, and maintained, that there existed irrational numbers, they forthwith ejected him from the boat. Today it's generally maintained that the number that is the length of the diagonal of a unit square is an irrational number. Put A, B, and C in a line, A at -1, B at 0, and C at 1. The system is symmetric about the origin. Bob, the set of all sets is its own powerset. To some extent, I know what I'm talking about. Heh heh heh. What do you think that is? Ross <3kvb4hF102qodU1@individual.net> > Yes, I do know to some extent that which I discuss. > To some extent, I know what I'm talking about. Heh heh heh. What do > you think that is? That extent is zero. - Randy === Subject: Re: weak form of LLN and CLT? See the excellent text, Asymptotic Theory for Econometricians by Halbert White. > Hi all, > > I am very much interested in the weak forms of the Law of Large Numbers and > Central Limit Theorem. > > I want to know when can we weaken the condition of IID? > > I have seen that under some conditions, the IID in CLT can be weaken into > independent but not identically distributed random samples. Can we > further weaken the condition to only weakly correlated... > > I have never seen a complete statement and argument about these conditions. > > I am not satisfied with the current books they only talk about the simplest > form and leave the harder part to you alone. > > Could anybody elaborate on these conditions? And also point me to some > resources so I can have a very complete view of the conditions? > === Subject: Re: I couldn't make it > What the @#$@# are you talking about? What paper? Is @#$@# heck? > === Subject: Re: I couldn't make it > What the @#$@# are you talking about? What paper? Math/0507611, which was rejected from the NT class! This time, I thought I made it; I think I did some embarrassing breakthrough on the Riemann hypothesis. If the L-functions satisfy |L(s, chi)| = |L(1 - s, overline{chi})|, then the proof is easily extended to the generalized Riemann hypothesis. Was that because of the title? The application of the theorem is so straightforward that you can pinpoint an error. Was that because of some elementary facts that are so obvious that stating them could unqualify the paper for NT class? But I think the paper is correct. At least everyone can understand what I am trying to do; not some nonunderstandable work in GM class. Eldes Now you are trying to put this thread next to a crankish one, aren't you. > > === Subject: Re: I couldn't make it > What the @#$@# are you talking about? What paper? Math/0507611, which was rejected from the NT class! > This time, I thought I made it; I think I did some > embarrassing breakthrough on the Riemann hypothesis. > If the L-functions satisfy |L(s, chi)| = |L(1 - s, overline{chi})|, then the proof is easily extended to the generalized > Riemann hypothesis. Was that because of the title? The application of the > theorem is so straightforward that you can pinpoint > an error. No kidding. you can NOT pinpoint an error is what I really was going to write. Was that because of some elementary facts that are so > obvious that stating them could unqualify the paper > for NT class? But I think the paper is correct. At least everyone > can understand what I am trying to do; not some > nonunderstandable work in GM class. Eldes Now you are trying to put this thread next to a > crankish one, aren't you. > > > === Subject: Re: Can anyone download/view videos from the Clay Institute? > rouben@pc18.math.umbc.edu >The Clay Mathematics Institute has several videos in > its website at: > > http://www.claymath.org/video/ > >I can neither view nor download those videos. I > have tried it >on two different computers and operating systems. > Is it only >I who is having a problem or the web site is screwed > up? I just tried unsucessfully to view one. Maybe their > quote at the top > of the page: The CMI video catalogue is undergoing > revision and > format conversion. The full catalogue will be > available once again by > the end of the year. is relevant. --Lynn Are they taking new ones? Maybe, Perelman might appear. Eldes === Subject: Finite Groups a.) If G is a finite group with fewer than 100 elements and G has subgroups of orders 10 and 25, what is the order of G? b.) Let x and y belong to a group G. Assume x not equal to e (the identity), |y|=2, and yxy^(-1)=x^2. find the order of G. I have no idea where to begin-beginner mathematician. === Subject: Re: Finite Groups >a.) If G is a finite group with fewer than 100 elements and G has >subgroups of orders 10 and 25, what is the order of G? >b.) Let x and y belong to a group G. Assume x not equal to e (the >identity), |y|=2, and yxy^(-1)=x^2. find the order of G. >I have no idea where to begin-beginner mathematician. I saw that someone answered the first question, so I will talk about part (b). I am not certain what you are referring to, so I will guess that G is generated by x and y, where y has order 2 and you have the relation y x y^-1 = x^2. Then, using that equation we have ( y x y^-1 ) ( y x y^-1 ) = x^2 x^2 So, y x ( y^-1 y ) x y^-1 = x^4 y x^2 y^-1 = x^4 But x^2 = y x y^-1 so y x^2 y^-1 = x^4 is the same as y ( y x y^-1 ) y^-1 = x^4 y^2 x y^-2 = x^4 Since the order of y is 2, this last equation reduces to x = x^4 Thus, x^3 = e, and since x is not the identity element, the order of x is 3. So, if I am correct and G is generated by x and y, where x has order 3 and y has order 2. What can you say about G? Brian === Subject: Re: Finite Groups > a.) If G is a finite group with fewer than 100 elements and G has > subgroups of orders 10 and 25, what is the order of G? Lagrange's theorem states that the order of a group must be divisible by the orders of its subgroups. In your case, the order of the G is <=99. Given this upper bound, you know that the order of G must also be divisible by 10 and 25. The answer would occur to me as being the LCM of 10 and 25. (2.5.5 = 50) > b.) Let x and y belong to a group G. Assume x not equal to e (the > identity), |y|=2, and yxy^(-1)=x^2. find the order of G. > I have no idea where to begin-beginner mathematician. > === Subject: Re: History of Math > >Which societies developed algebra and trig - classical, arabic or european? > > Trigonometry, as we know it, was developed by the Hellenistic > Greeks. The modern understanding of the trigonometric > functions was mainly 18th century, and the applications to > other than surveying and astronomy mainly later. > > As for algebra, what does that mean? The earliest use of > a symbol for numbers seems due to Diophantus. The Muslims > spread this, but the systematic use of many symbols starts > with Viete near the end of the 16th century, and so is > Western European. > > -- > 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 If we were using the Hellenistic trigonometry we would be using chords for sines and we would not know what animal is cotangent and we would not have the trigonometric ratios as we know them. I respectfully suggest that as far as trigonometry goes you search the name Al-Battani over the Internet and find out about Al-Battani's book Kitab Al-Zij (the book of tables). The Latin translation of this book under the title De Motu Stellarum, by Plato of Tivoli, is accepted to be very influential in the Middle Ages and the Renaissance. It contains an elaborate set of astronomical tables and discusses a wide range of practical problems in spherical astronomy, some of which were devised for the purpose of solving related astrological problems. (The guy also corrects Ptolemy on many counts. These Muslims/Arabs were BAD!) Wikipedia has a nice account of the various innovations he made and from those it appears that he was indeed the grand DADDY of the trigonometry as we know it. For your convenience here is the link http://en.wikipedia.org/wiki/Al_Battani [I suggest Wikipedia, because it does not accept that Al-Battani was a Muslim, though his name says a lot about his religion.] In the case of algebra too sir, one can find out that the West did not get from the Muslims the same algebra that was done before them. It was an algebra on which the West could base its progress. It was an algebra upon which the Italians could base their earlier work on the cubic and the quartic equations. I agree that Diophantus, if we are talking about the same Diophantus, must have used symbols for numbers. With the kind of wretched number system they had, it would be insane not to use some kind of symbolism. But sir, that was very primitive. Now sir, looking at your earlier post and the one above I can safely say that you do not know a thing about this stuff, you jumped in, with all your Purdue weight, just to perpetuate the ignorant pride that is prevalent among American youngsters. I would not point it out but the falling standards at American universities concern me a lot. I am your humble servant sir, Muhammad Zafrullah === Subject: Four Properties of the Euler Phi Function The following properties of the Euler 'phi' function are troubling me: I am using the word phi instead of the standard notation, so forgive me in advance. (i) Prove that the Euler phi function is multiplicative: This isn't too bad, I think it is best done with the principle of inclusion exclusion. Although someone might have a method I have not considered? (ii) For any positive integers m and n prove phi(mn)= phi(m).phi(n).(d/phi(d)) where d is the gcd of m and n. I think the multiplicative nature makes the m and n parts doable, what happens with d/phi(d)? (iii) For any positive integers m and n prove phi(m).phi(n)= phi(gcd(m,n)).phi(lcm(m,n)) It might be possible to use the fact that gcd(m,n)lcm(m,n) = mn? Or does that move away from the multiplicativity? (iv) Take n to be a positive integer greater than 1. Prove that the sum of the integers m with 1<=m<=n that are coprime with n is (n.phi(n))/2 Not sure where this is going? === Subject: Re: Four Properties of the Euler Phi Function > (i) Prove that the Euler phi function is multiplicative: It's multiplicative only for coprime integers. phi(nm) = phi(n) phi(m) only when n,m are coprime. > This isn't too bad, I think it is best done with the principle of > inclusion exclusion. Although someone might have a method I have not > considered? That is a direct result of the isomorphisms in the theorem (m,n) = 1 ==> Z_mn = Z_m x Z_n, Z_mn^* = Z_m^* x Z_n^* Let f be the first isomorphism and show that the same f restricted to the multiplicative groups is an isomorphism. Yes, you can use the Chinese Remainder Theorem to show f is surjection. Alternatively you can use an injection between equinumerous sets is a surjection. > (ii) For any positive integers m and n > prove phi(mn)= phi(m).phi(n).(d/phi(d)) > where d is the gcd of m and n. This may be useful distinct primes p1,.. pk, n = p1^a1 p2^a2 ... pk^ak ==> phi(n) = n(1 - 1/p1)(1 - 1/p2)...(1 - 1/pk) > (iii) For any positive integers m and n > prove phi(m).phi(n)= phi(gcd(m,n)).phi(lcm(m,n)) (iii) follows from (ii) since nm = (n,m)[n,m] phi nm = phi m * phi n * (n,m)/phi (n,m) phi (n,m)[n,m] = phi (n,m) * phi [n,m] * ((n,m),[n,m])/phi ((n,m),[n,m]) = phi (n,m) * phi [n,m] * (n,m)/phi (n,m) > (iv) Take n to be a positive integer greater than 1. Prove that the > sum of the integers m with 1<=m<=n that are coprime with n is > (n.phi(n))/2 Do like Gauss by adding 1 2 ... n-2 n-1 n-1 n-2 .... 2 1 ------------------- n n ..... n n To get 1 +..+ n-1 = n(n - 1)/2 Show when a + b = n and a coprime n, so is b. Then count the number of additions, viz phi n. ---- === Subject: Re: Four Properties of the Euler Phi Function > The following properties of the Euler 'phi' function are troubling me: I am > using the word phi instead of the standard notation, so forgive me in > advance. (i) Prove that the Euler phi function is multiplicative: This isn't too bad, I think it is best done with the principle of inclusion > exclusion. Although someone might have a method I have not considered? I would use the Chinese remainder theorem. > (ii) For any positive integers m and n > prove phi(mn)= phi(m).phi(n).(d/phi(d)) > where d is the gcd of m and n. I think the multiplicative nature makes the m and n parts doable, what > happens with d/phi(d)? Note that the case d=1 is equivalent to the first property. > (iii) For any positive integers m and n > prove phi(m).phi(n)= phi(gcd(m,n)).phi(lcm(m,n)) It might be possible to use the fact that gcd(m,n)lcm(m,n) = mn? Or does > that move away from the multiplicativity? As a last resort, you could use the prime factorizations of m and n. > (iv) Take n to be a positive integer greater than 1. Prove that the sum of > the integers m with 1<=m<=n that are coprime with n is (n.phi(n))/2 > Not sure where this is going? If m is coprime with n, what can you say about n-m and n? -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Four Properties of the Euler Phi Function > > The following properties of the Euler 'phi' function are troubling me: I am > using the word phi instead of the standard notation, so forgive me in > advance. > > (i) Prove that the Euler phi function is multiplicative: > > This isn't too bad, I think it is best done with the principle of inclusion > exclusion. Although someone might have a method I have not considered? > > I would use the Chinese remainder theorem. Here I go: Suppose t = mn where m,n are coprime. Then the chinese remainder theorem states that (a,t) = 1 if and only if (a,m)=1 and (a,n)=1. Now the number of positive integers not greater than and coprime with is precisely phi(t), but it is also the number of pairs (u,v) where u is not greater than m and coprime with m and v not greater than n and coprime with n, thus phi(mn)=phi(m)phi(n) > (ii) For any positive integers m and n > prove phi(mn)= phi(m).phi(n).(d/phi(d)) > where d is the gcd of m and n. > > I think the multiplicative nature makes the m and n parts doable, what > happens with d/phi(d)? > > Note that the case d=1 is equivalent to the first property. > > (iii) For any positive integers m and n > prove phi(m).phi(n)= phi(gcd(m,n)).phi(lcm(m,n)) > > It might be possible to use the fact that gcd(m,n)lcm(m,n) = mn? Or does > that move away from the multiplicativity? > > As a last resort, you could use the prime factorizations of m and n. > > (iv) Take n to be a positive integer greater than 1. Prove that the sum of > the integers m with 1<=m<=n that are coprime with n is (n.phi(n))/2 > Not sure where this is going? > > If m is coprime with n, what can you say about n-m and n? > -- > Daniel W. Johnson > panoptes@iquest.net > http://members.iquest.net/~panoptes/ > 039 53 36 N / 086 11 55 W begin 666 img9.png === Subject: estimate Ax I am estimating Ax with a sum of the sort sigma{ u_i u_i^{T} x_i } (for i = 1..n). I can show that as n becomes large, the sum converges to Ax. I would also like to estimate A^{-1}x. Any ideas on how to do this using the iteration process. --j === Subject: estimate Ax >=20 > I am estimating Ax with a sum of the sort sigma{ u_i u_i^{T} x_i } > (for i =3D 1..n). I can show that as n becomes large, the sum = converges > to Ax. I would also like to estimate A^{-1}x. Any ideas on how to do > this > using the iteration process. Do you really mean x_i in the sum above ? if the u_i are column vectors the sumands are not defined.=20 Or should that be: sum_i s_i u_i u_i^T ??? === Subject: INCALCULABLE Definition: [adj] not capable of being calculated INCALCULABLE Definition: [adj] not capable of being calculated INFINITE 1 An infinity; an incalculable or very great number. RANDOM lacking any definite plan or order or purpose; governed by or depending on chance 1. Unpredictable (with no mathematical definition); weird. See; INCALCULABLE INFINITE RANDOM INFORMATION The system's been behaving pretty randomly. INFORMATION Definition: [n] (communication theory and or by definition or law ;that can be fact or fiction based on PERCEPTION THEORY Definition: [n] a tentative theory about that anything is posssible in theory; a concept that is not yet verified but that if true would explain certain facts. Broken by the laws of INCALCULABLE INFINITE RANDOM INFORMATION IMAGINATION Definition: [n] the formation of a mental image of something that is not os is perceived as real popular imagination created man and the world everything we see and study imagination reveals what the world could be or what information looks like in a visual matrix. FICTION Definition: [n] a literary work based on the imagination and not necessarily on fact [n] a deliberately false or improbable account PERCEPTION Definition: The process of organizing information received through the senses and interpreting it. process of knowing or being aware of information This is done by the a way of perceiving; something;knowledge gained by perceiving; and perceiving; The PERCEPTION of INFORMATION and application changed when i change my PERCEPTION and how i felt about the world more positive outlook can produce more INFORMATION by ways of IMAGINATION MATRIX [n] an enclosure within which something originates or develops 3D INFINITE PERCEPTION IMAGINATION MATRIX FOR RANDOM INFORMATION THEORY OF RANDOM QUANTITIVITY THAT EVERYTHING IS RANDOM.RANDOM giving new meaning to infinite outcome that is endless. Take a document then or a 3D matrix document change it two random or binary code or just a program for 0's and 1's and fold it over and over like a piece of paper then having the 1 and 0 add each other or you gave the folds addresses like on a spread sheet there would be no math. First A 1-24 would fold to k 1-24 down.(See Example A ) Then at F1-24 down two k 1-24 ( See example B ) If you written a very long letter and then change it two binary code it would look like this. 123456789.............24 a.01010101010101010101010 b.10010101010101010101010 c.01010101001010101010010 e.00010101000101010101010 f.10010101010100101010101 First A 24 would fold to k 24 down g.01010101010100001100101 See Example A h.01001010101010101010111 I.11110111001101010101010 j.01010101010101010101010 k.10101010101010101010101 See Example A 123456789.............24 f.10010101010100101010101 g.01010101010100001100101 h.01001010101010101010111 Then at F1-24 down two k 1-24 I.11110111001101010101010 j.01010101010101010101010 k.10101010101010101010101 See example B 123456789.............24 I.11110111001101010101010 j.01010101010101010101010 Then from I 1-24 to K 1-24 k.10101010101010101010101 123456789.............24 j.01010101010101010101010 Then from j-24 to j-1 123456789... j.010101010101 Then from j-12 to j-1 123456 j.010101 Then from j-6 to j1 123 j.010 Then from j-3 to j1 12 j.01 Then from j-2 to j1 j.0 Then you would have 1 bit to transfer over the Internet The bit sent would be 0 and the key code would be F1-24,k 1-24, I 1-24,K 1-24,j24,j1,j12,j1,j6,j1,j3,j1,j2,j1 and would unzip or be new encryption you could encrypt or compress 100 terabits down to 1 bit of information. Now if you take this idea from my web site you could make this allot more complex and unbreakable. Data encryption 360 degrees rotation document 90 degrees and encrypt on every angel then 45 degrees change it two binary code do it again and again and fold it over like a piece of paper then having the one's and zero program to someone's computer and have it unzip and run and install or make a large ram drive and buffer use one half of the 64 bit processor decode the message and the main 64 bit run the numbers. Another way of doing this is to have a parallel computers with using one of the processes run the compressed 1 bit of information give the uncompressed a address on the ram drive to change and not even go threw the processor and then with different information on each machine compare and run statistics on information on a 45 tflops supercomputer and turn that 45 tflops computer into a 1 bit = 100,000 terabits to infinite as long as you have the ram for storage! with my calculations 45 tflops wouldn't matter any more it would be how much data you have on a 32bit operating system changing that to a 1 bit system it would be 32 * 45tflops would = 1440 tflops RUN THEASE TWO IN NEW RANDOM QUANTITIVITY http://groups.yahoo.com/group/Quantum_AI/ www.beyond-science.com === Subject: Iterative Crank-N. with 1D advection Hello all, I am trying to solve an adection eq in a hydrologic model: d(Ac)/dt+d(Qc)/dx=S with c=sediment concentration A=flow cross-section Q=discharge S=source term Note that I have to propagate only c,because Q is known for for each grid-point and time-step. I used an iterative C-N, especially according to: http://www.sissa.it/~rezzolla/lnotes/hyperblc_pdes_lnotes/index.html the model is still highly unstable, even cutting down time-steps... A friend argued that I am dealing with a sort of Burger's eq; if that's the case, I give up, since that is too difficult for a simple engineer like me. Suggestions? Alessandro === Subject: Re: Iterative Crank-N. with 1D advection >Hello all, > I am trying to solve an adection eq in a hydrologic model: >d(Ac)/dt+d(Qc)/dx=S > >with c=sediment concentration > A=flow cross-section > Q=discharge > S=source term >Note that I have to propagate only c,because Q is known for for each grid-point and time-step. > >I used an iterative C-N, especially according to: >http://www.sissa.it/~rezzolla/lnotes/hyperblc_pdes_lnotes/index.html > >the model is still highly unstable, even cutting down time-steps... > >A friend argued that I am dealing with a sort of Burger's eq; if that's the case, I give up, since that is too difficult for a simple engineer like me. > > Suggestions? > > Alessandro no, you are simply using a nonsense method. there is proposed an implicit scheme (in order to obtain a second order accurate and unconditionally stable scheme...however : CFL remains behind the scene ... ) and then uses what is called direct iteration or picard iteration in order to solve the nonlinear equations instead of doing what one should do here: use Newton's method for solving the nonlinear equations. I propose you refer to a better source, namely randy le veque: numerical methods for conservation laws. (birkhaeuser 1992). you could for example try lax-wendroff combined with a flux limiter, for example the superbee or van Leer. in any case, for such a flow problem the numerics should model the flow, taking CFL into account. hence I wouldn't use an implicit scheme. hth peter === Subject: Re: incremental least squares > First of all, how do I recompute the covariance matrix of {P union p} > without touching P? (as a function of cov(P) and p?) > I right now have to recompute this (I can update the mean, but that > does not help since P is also involved). > You might consider using the singular value decomposition (SVD). There are algorithms for updating the SVD and have a look at (e.g.) http://web.mit.edu/be.400/www/SVD/Singular_Value_Decomposition.htm -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany === Subject: Re: incremental least squares <42E8ADA9.3060506@igpm.rwth-aachen.de> So if I try to use SVD (and if i understand what you are suggesting), this is the problem. Given M (nxd) where n is the number of points and d is the dimension of the points. Let m be a matrix of size nxd such that each row is the mean of the points in M (so each row is same). Now we do a SVD of (M-m) to find the best fitting plane. Now comes the update part. A new point p is added to M. Lets call the new matrix M' whose dimension is (n+1)xd. The new mean matrix m' is also (n+1)xd. What is the relation between the SVD of (M-m) and (M'-m')? Is there an easy way to compute SVD(M'-m') given M, m, m' and SVD(M-m)? I actually thought about the other approach (using covariance matrices). A Covariance matrix can indeed be updated using a new point p. But the catch is that the update rule seems weird enough that I dont see again how the SVD could use the previous covariance matrix result to get the new one... --j > First of all, how do I recompute the covariance matrix of {P union p} > without touching P? (as a function of cov(P) and p?) > I right now have to recompute this (I can update the mean, but that > does not help since P is also involved). > > > You might consider using the singular value decomposition (SVD). > There are algorithms for updating the SVD and have a look at (e.g.) > http://web.mit.edu/be.400/www/SVD/Singular_Value_Decomposition.htm > > > -- > Helmut Jarausch > > Lehrstuhl fuer Numerische Mathematik > RWTH - Aachen University > D 52056 Aachen, Germany === Subject: Re: incremental least squares >So if I try to use SVD (and if i understand what >you are suggesting), this is the problem. > >Given M (nxd) where n is the number of points and >d is the dimension of the points. Let m be a matrix >of size nxd such that each row is the mean of the >points in M (so each row is same). Now we do >a SVD of (M-m) to find the best fitting plane. > >Now comes the update part. A new point p is >added to M. Lets call the new matrix M' whose >dimension is (n+1)xd. The new mean matrix m' is >also (n+1)xd. What is the relation between >the SVD of (M-m) and (M'-m')? Is there an >easy way to compute SVD(M'-m') given M, m, m' and >SVD(M-m)? > >I actually thought about the other approach (using >covariance matrices). A Covariance matrix can indeed >be updated using a new point p. But the catch is that >the update rule seems weird enough that I dont see >again how the SVD could use the previous covariance >matrix result to get the new one... > >--j > > > >> First of all, how do I recompute the covariance matrix of {P union p} > without touching P? (as a function of cov(P) and p?) >> I right now have to recompute this (I can update the mean, but that > does not help since P is also involved). >> >> >> You might consider using the singular value decomposition (SVD). >> There are algorithms for updating the SVD and have a look at (e.g.) >> http://web.mit.edu/be.400/www/SVD/Singular_Value_Decomposition.htm > > >> -- >> Helmut Jarausch > >> Lehrstuhl fuer Numerische Mathematik >> RWTH - Aachen University > D 52056 Aachen, Germany > if your true problem is orthogonal least distance fitting, you could use a completely different approach, namely direct (nonlinear) smooth minimization I assume d (the dimension) will be small, hence , since the problem is also nicely smooth, any reasonable minimizer will run fast and reliable, given a reasonable initial guess. hence, just starting from d points with an unique solution you add points, modifying the least distance function (indeed, only the length of the loop in computing it changes) and start the minimization anew, with the previous result as initial guess. this should be much more efficient than svd updating, the more as in your situation the modification is not a small rank modification (m' not only gets a new row, all rows are modified!) hth peter === Subject: Minimum search algorithm I need to optimise the dimensions of a viscoelastic region designed to absorb outgoing elastic waves in a railway problem. As I don't know the function, it will be a search in one, two, or three dimensions for a point which would minimise the energy of the entire structure. I believe that there are theories both in classical numerical methods and artificial intelligence for this kind of search problem. Do you know of a reference or an academic department which would be able to help me? === Subject: Re: Minimum search algorithm >I need to optimise the dimensions of a viscoelastic region designed to > absorb outgoing elastic waves in a railway problem. As I don't know the >function, it will be a search in one, two, or three dimensions for a > point which would minimise the energy of the entire structure. I believe >that there are theories both in classical numerical methods and >artificial intelligence for this kind of search problem. > >Do you know of a reference or an academic department which would be able >to help me? > > if I understand right, you have a function total energy depending on one, two or three real variables which can be computed numerically (given values for the design parameters) and which you want to minimize. hopefully thsi function behaves continuously or better smoothly on the parameters. this is a standard problem in derivative free optimization and you might choose one of the algorithms in http://plato.la.asu.edu/topics/problems/nlounres.html under no gradient available for example newuoa, or praxis should fit your needs hth peter === Subject: Re: Minimum search algorithm Peter, The energy function is completely unknown beforehand. The mappings between the parameters and the outcome only unfold heuristically as the search progresses. Is this how you understood my question? Hakan Lane > if I understand right, you have a function total energy depending > on one, two or three real variables which can be computed numerically > (given values for the design parameters) and which you want to minimize. > hopefully thsi function behaves continuously or better smoothly on the > parameters. this is a standard problem in derivative free optimization > and you might choose one of the algorithms in > http://plato.la.asu.edu/topics/problems/nlounres.html > under no gradient available > for example > newuoa, or praxis > should fit your needs > hth > peter > === Subject: MatLab Version of Subr Period Needed I am wondering if anyone has a MatLab version of the Numerical Recipes program/subroutine PERIOD (that performs Lomb-Scargle periodogram method) that they would be willing to share? (I would be more than willing to acknowledge this person in a paper I am involved in writing using this program.) Matthew T. Brenneman Dept of Electrical and Computer Engineering 121 Kreger Hall / Miami University Oxford, OH 45056 513-529-7340 === Subject: Nonuniform time step Hi everybody, I am wondering if there is a well-established method for time integration of a time-dependent problem using a higher-order Adams-Bashford method with VARIABLE TIME STEP. I am aware that there is one with constant time step. Irfan === Subject: Multiplying by repeated addition I am looking for an optimized algorithm for multiplying by repeated addition. For example instead of doing result = 11 * 9 = 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 do temp = 11 + 11 temp = temp + temp //temp has 4 * 11 temp = temp + temp //temp has 8 * 11 result = temp + 11 which is 4 less additions. For large numbers the savings can be substantial by the use of temporary variables (i.e. registers in a machine) to hold intermediate results. Surely something like that exists since it's useful for machines without multiplication circuitry. === Subject: Re: Multiplying by repeated addition > I am looking for an optimized algorithm for multiplying by repeated > addition. For example instead of doing > > result = 11 * 9 = 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 > > do > > temp = 11 + 11 > temp = temp + temp //temp has 4 * 11 > temp = temp + temp //temp has 8 * 11 > result = temp + 11 > > which is 4 less additions. For large numbers the savings can be > substantial by the use of temporary variables (i.e. registers in a > machine) to hold intermediate results. > > Surely something like that exists since it's useful for machines > without multiplication circuitry. I think multiplication circuitry is just a hardware implementation of an addition-based algorithm. This is easy in base 2, because multiplying by powers of 2 is done by bitshifting (represented in C by the << operator). So 11 << 1 is 11*2, and 11 << 2 is 11*4, ... x << n is x * 2^n. Now suppose you wanted to multiply 11 by 9. Write 9 in base 2: 1001 = 2^3 + 2^0 So 11*9 = 11*2^3 + 11*2^0 Shift 11 by 3, shift 11 by 0 and add the two numbers. - Randy === Subject: Re: Multiplying by repeated addition Use the same algorithm that people use for modular exponentiation: the binary method, only instead of modular multiplication at each step, one does addition. See Knuth Vol 2. To multiple A*B, choose the smaller of the two and represent it in binary. (say A) Set a variable x = B. Delete the MSB in A. Then, for each 0 in the representation of A, add x to itself. For each 1, add x to itself, and also add B. This requires, on average 3/2 log_2(A) additions. === Subject: Re: Multiplying by repeated addition !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > I am looking for an optimized algorithm for multiplying by repeated > addition. For example instead of doing > > result = 11 * 9 = 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 > > do > > temp = 11 + 11 > temp = temp + temp //temp has 4 * 11 > temp = temp + temp //temp has 8 * 11 > result = temp + 11 > > which is 4 less additions. For large numbers the savings can be > substantial by the use of temporary variables (i.e. registers in a > machine) to hold intermediate results. Go to your library and get Donald Knuth, The Art of Computer programming Vol. 2, Seminumerical Algorithms. You won't easily find a more extensive analysis of this problem anywhere else. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum