mm-2069 === Subject: Re: A Simple Math Word Problem ? > Is this simple word problem easy enough for you to do in your head, Yes. or do > you need either pencil and paper, or some kind of a calculator ? No. > What is a Hundred times a Thousand times a Million times a Billion times a > Trillion ? 10^32 === Subject: Re: Yet another Math word problem >You are all do doubt well aware of the very old saying Another day another >Dollar and the witty reply to the same A million days a million dollars >Now my question is this; >January 1st 2000 (AD) was in a leap year, so counting from that start date, >what would be the calendar date One Million days later ? >This was and still is a question that was put to High School students, to >test their math skill, it never ceased to amaze me just how many students >forgot to take into account the leap year factor. There are a number of Gregorian <--> Julian Day Number calculators on the net which, for your question, give Sunday, November 28, 4737 Also, a February 1999 posting: === Subject: Re: Yet another Math word problem > You are all do doubt well aware of the very old saying Another day another > Dollar and the witty reply to the same A million days a million dollars > Now my question is this; > January 1st 2000 (AD) was in a leap year, so counting from that start date, > what would be the calendar date One Million days later ? 4743 November 29 gwh > This was and still is a question that was put to High School students, to > test their math skill, it never ceased to amaze me just how many students > forgot to take into account the leap year factor. === Subject: Re: Yet another Math word problem another > Dollar and the witty reply to the same A million days a million dollars > Now my question is this; > January 1st 2000 (AD) was in a leap year, so counting from that start > date, > what would be the calendar date One Million days later ? > 4743 November 29 How did you get it wrong by 6 years and a day? > gwh > This was and still is a question that was put to High School students, to > test their math skill, it never ceased to amaze me just how many students > forgot to take into account the leap year factor. === Subject: Re: Yet another Math word problem > You are all do doubt well aware of the very old saying Another day another > Dollar and the witty reply to the same A million days a million dollars > Now my question is this; > January 1st 2000 (AD) was in a leap year, so counting from that start date, > what would be the calendar date One Million days later ? > This was and still is a question that was put to High School students, to > test their math skill, it never ceased to amaze me just how many students > forgot to take into account the leap year factor. Is the year 4000 a leap year? === Subject: Re: Yet another Math word problem You are all do doubt well aware of the very old saying Another day another Dollar and the witty reply to the same A million days a million dollars Now my question is this; January 1st 2000 (AD) was in a leap year, so counting from that start date, what would be the calendar date One Million days later ? This was and still is a question that was put to High School students, to test their math skill, it never ceased to amaze me just how many students forgot to take into account the leap year factor. > Is the year 4000 a leap year? I seem to recall that there are leap centuries too. Maybe it's the case where they're not leapt when they should be or vice versa. === Subject: Re: Yet another Math word problem You are all do doubt well aware of the very old saying Another day another Dollar and the witty reply to the same A million days a million dollars Now my question is this; January 1st 2000 (AD) was in a leap year, so counting from that start date, what would be the calendar date One Million days later ? This was and still is a question that was put to High School students, to test their math skill, it never ceased to amaze me just how many students forgot to take into account the leap year factor. > Is the year 4000 a leap year? > I seem to recall that there are leap centuries too. Right, that's why 2000 was a leap year. Every century year is NOT a leap year unless divisible by 400. To make the math more accurate, this should be extended to millennia also, such that every 4th year is a leap year unless it's divisible by 100, in which case it's not unless it's divisible by 400, in which case it is unless it's divisible by 4000, in which case it's not But the millenia rule has not been officially adopted. Thus, Excel says that day 767070 is Feb 29, 4000. As of now, the answer to the question must consider the year 4000 to be a leap year, but it may turn out that by the time we reach that year, it won't be. > Maybe it's the case > where they're not leapt when they should be or vice versa. === Subject: predict 43 rd mersenne prime exponent.. exp(.4 * 43) = 29 502 925.9 exp(.4 * 43) = 29 502 925.9 More about calculator. predict 43 rd mersenne prime exponent.. exp(.4 * 43) = 29 502 925.9 extremely linear trend r = correlation coefft = .997. 43rd KNOWN mersenne prime exponent. LOG BASE10 mersenne exponents. the actual 43 rd known exponent is 30,402,457. M(p) = 2^p -1. don.mcdonald nz 29.12.2005 === Subject: Re: Santa Claus speed > Approximate the continental US by a 1500 x 3000-mile rectangle. > Santa Claus needs to make 20 million stops, uniformly > distributed within this rectangle. At each stop he > drops 8 feet through a chimney and climbs back out. > How fast must he travel to finish the journey in 12 hours, > not counting the return to the North Pole? There are 100 million households in the United States, not 20 million; and probably about 1500 million places to would need to be visited, worldwide. If there is an average of 10 meters, 100 meters or 1 km between stops that would come out to either 15 million, 150 million or 1500 million km over a 24 hour period. The minimum times (for those travelled at light speed) would be about 50, 500 or 5000 seconds. There are 86400 seconds in a day, so the average speed (depending on whether the mean distance is 10 meters, 100 meters or 1 km) would be somewhere around .05%, .5% or 5% of light speed). > If the US invests in Arctic oil drilling, will our vehicles > be as fuel efficient as Santa Claus's? If it invests in a competitive Manhattan project (like a combination of the X-prize and Manhattan project) to bring fusion on-line and to the market, that will (literally) blow everything out of the water and permanently resolve the energy issue. > The US must invest in interstate passenger rail. Amtrak needs high-speed > public railways, not private lines where it waits for freight trains to pass. There will be no substitute for exobionics and Aliens-style human-powered walkers and/or cyborg exoskeletons. === Subject: Re: Prime number theorem question into the bounds for pi(x) as well. I was looking for a asymptotic argument as to the number of primes in some arithmetic sequences that fit Dirichelet's theorem about arithmetic progressions. === Subject: matrix group question Let SU_n denote the Special Unitary group of finite degree n, and likewise SO_n refer to the Special Orthogonal group over the reals. A well-known result is that SU_2 modulo its center {1,-1}, is isomorphic to SO_3, the proper rotations of our Euclidean 3-space. SU_2 is also in one-one correspondence with the (three-dimensional) surface of the sphere in Euclidean 4-space, and that same factorization produces RP_3, the real projective three-space. The upshot would seem that there is a group structure defined on the points of RP_3, due to a pointwise correspondence with the elements of SO_3. Correct me if I am wrong about that. Now another known result is that the direct product SU_2 x SU_2, modulo (1,1), (-1,-1) } produces SO_4. And if I am correct, SU_2 x SU_2 modulo its four-element center { (1,1), (1,-1), (-1,1), (-1,-1) }, produces a copy of PSO_4, (SO_4 modulo its two element center). If this last one is correct, then my questions here would be whether the following group isomporphism is valid: 1) is PSO_4 = SO_3 x SO_3 ? and 2) does RP_6 = have this group structure ? === Subject: Re: matrix group question > Let SU_n denote the Special Unitary group of finite degree n, and likewise > SO_n refer to the Special Orthogonal group over the reals. > A well-known result is that SU_2 modulo its center {1,-1}, is isomorphic > to SO_3, the proper rotations of our Euclidean 3-space. SU_2 is also in > one-one correspondence with the (three-dimensional) surface of the sphere > in Euclidean 4-space, and that same factorization produces RP_3, the real > projective three-space. > The upshot would seem that there is a group structure defined on the > points of RP_3, due to a pointwise correspondence with the elements of > SO_3. Correct me if I am wrong about that. I think that is correct. The easiest way to view this, I believe, is to consider the action of the group H* of non-zero quaternions on the subspace of H by (q,x) -> q^{-1}xq . [I think the rest of what you said is also true.] -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: une fonction croissante > What does this mean in French? > x1 < x2 ==> f(x1) < f(x2) strictly increasing > or > x1 < x2 ==> f(x1) <= f(x2). increasing or non-decreasing -- I.N. Galidakis --- http://ioannis.virtualcomposer2000.com/ === Subject: Good Math Puzzle hi guys, i am guy interested in math,but cannot just crack some puzzle without help of guys like you.. here is how it goes There is an N*N square matrix where N is odd. The initial values of all the cells are '1'. We give value X as input to the centre of the matrix. Each time a value is given to a cell, it adds that value to the current cell value and passes on same stuff till the value X reaches ZERO i thought the value at the cente Pf = Pinitial + P(X),indepedent of N(could not find the reason...instinct..;)) P(X) = X + 8P(X-2)....but i was wrong. independent of N but I could not figure out the final value of the center cell given the initial value(1) and the X I found that for X,N = 3 it is 5 7 5 7 12 7 5 7 5 I thot some fibonacci (7+5 = 12) works for X=3 and greater N as well for X = 4,N = 3 24 32 24 32 45 32 24 32 24 Now how does 45 come here?? Please help me. Naren. === Subject: Re: Boolean logic, third value (true=false case) > As we cant give answer to liar's paradox using those boolean values [...] #1 That has no bearing on anything. The liar's paradox doesn't require anything more than the implication operator and the 2 rules (B) If B can be inferred from A, then A->B can be inferred The paradox: This statement implies X, where X can be anything you want. Theorem: X. Proof: For brevity, denote by S the state above. Then S is the statement S implies X. (b) By (B), since X can be inferred from S, then S->X is true (c) Thus, since S->X is true, which is restated as S, then by (A) X is true. QED #2 It's not called Boolean logic if it has 3 values. A boolean logic can only have 2^n values. These are the Boolean lattices. A appropriate generalization is Intuistionistic Logic (which is mainly characterized as the logic where Pierce's rule, double negative rule, and the equivalence of (A->B) = (not A or B) all fail). This also includes other generalization (like Fuzzy Logic) as special cases. The above rules are valid in Intuitionistic Logic. So everything follows through intact. Every finite lattice (i.e., a finite partially ordered set in which every 2 elements has a least upper bound and greatest lower bound) is a value system for an intuitionistic logic. In particular, the 3-element lattice (0->x->1) has 0&x = 0&1 = 0&0 = 1&0 = x&0 = 0; x&x = x&1 = 1&x = x; 1&1 = 1 0 or 0 = 0; 0 or x = x or 0 = x or x = x; 0 or 1 = x or 1 = 1 or 1 = 1 or x = 1 or 0 = 1 0->0 = 0->x = 0->1 = x->x = x->1 = 1->1 = 1; x->0 = 1->0 = 0; 1->x = x not(0) = 1; not(x) = not(1) = 0 The values for -> are implied by those for & and or by properties (A) and (B). The values for not come from the equivalence (not x = x->0). Double-negative yields a boolean sub-logic {0,1}, with not(not(0)) = 0; not(not(1)) = 1; not(not(x)) = 1. The infinite (and finite) Intuitionistic Logics are one and the same as the class of all topological spaces. But I forget how the operations are defined in terms of open sets, closed sets, etc. I think it goes like (a or b = a union b; not(a) = interior(complement(a)), etc.) A second generalization is Quantum Logic, (mainly characterized by the failure of the distributivity rule A&(B or C) = A&B or A&C). I don't know if the above self-referential paradox (or any like it) has been incorporated in Quantum Logic, but it probably can be done. The value systems for Quantum Logic consist of the projection operators of a Hilbert space. A third generalization is Categorical Logic, which emerged (almost by accident) back in the 1950's when Curry stumbled onto the famous Curry-Howard isomorphism, which breathed new life into the notions of typed logic. There are no self-referential paradoxes in categorical logic, since the terms are consistently typed. === Subject: Re: Non-Standard Analysis & Other Infinities <40rigtF1bli3iU1@individual.net You can [sic] define hypernaturals as omega-sequences of naturals > modulo an ultrafilter, but you don't have to do it with omega. > And I recall being told that sometimes (for model-theoretic > reasons) doing it with a bigger ordinal is required. I > don't, alas, recall why---it had something to do with saturation. The purpose of the ultrafilter construction is to prove that the theory *exists*, not to define it. The explication serves no more as a definition than saying, for instance, that a complex number is an ordered pair of reals, or a real an infinite sequence of integers. That's a confusion between is and can be represented by -- which is a cardinal sin in (modern) mathematics. Non-standard analysis is simply defined as the free extension of ordinary arithmetic, modulo the addition of a new distinguished constant (w) and a set of axioms (0 <40rigtF1bli3iU1@individual.net> |Non-standard analysis is simply defined as the free extension of |ordinary arithmetic, modulo the addition of a new distinguished |constant (w) and a set of axioms (0|Non-standard analysis is simply defined as the free extension of >|ordinary arithmetic, modulo the addition of a new distinguished >|constant (w) and a set of axioms (0I'm not fond of your terminology here. Ordinary arithmetic >seems like it *might* mean what is sometimes called true >arithmetic, i.e. the sentences holding in N. But if it does, >then free extension is a poor description for what you are >doing to it. I think that he is talking about the syntactic approach, which deals with a nonstandard *theory*, rather than a nonstandard model. The theory is, as Mark describes, just true arithmetic, plus a new constant w, plus the axioms n|What Robinson did was not define a new theory but merely prove that >the >|existence and consistency of the theory above (which is >|omega-inconsistent) is at least as high up on the >|existence/consistency hierarchy as ultrafilter existence, but >|probably below the level of axiom of choice. >What are you basing this on? It seems to me you're >trivializing his work. The consistency of the above theory >is, assuming your description of it is something like what >you meant to say, the simplest consequence of the compactness >theorem for first-order logic, which was known for a long time >before Robinson's work. Yes, I'm curious as to what exactly Robinson's model does, above and beyond the model that one gets from compactness and completeness. One difference is that Robinson's model is saturated, but I'm not sure how important that is. Also, Robinson's model satisfies the transfer principle, but I'm not sure that the model one gets from compactness and completeness does not. -- Daryl McCullough Ithaca, NY === Subject: Re: Non-Standard Analysis & Other Infinities You can [sic] define hypernaturals as omega-sequences of naturals modulo an ultrafilter, but you don't have to do it with omega. And I recall being told that sometimes (for model-theoretic reasons) doing it with a bigger ordinal is required. I don't, alas, recall why---it had something to do with saturation. I can see where my earlier remarks may be construed to mean that the ultrafilter construction is essentially the only one, but I did not intend to imply that. > The purpose of the ultrafilter construction is to prove that the theory > *exists*, not to define it. The explication serves no more as a > definition than saying, for instance, that a complex number is an > ordered pair of reals, or a real an infinite sequence of integers. > That's a confusion between is and can be represented by -- which is > a cardinal sin in (modern) mathematics. > Non-standard analysis is simply defined as the free extension of > ordinary arithmetic, modulo the addition of a new distinguished > constant (w) and a set of axioms (0 Robinson did not define a new theory, that's huge misconception which > has caused a lot of undue confusion of representations vs. definitions; > conflation; and exposure of all the dangling wires under the hood out > in the open where they don't below. > What Robinson did was not define a new theory but merely prove that the > existence and consistency of the theory above (which is > omega-inconsistent) is at least as high up on the > existence/consistency hierarchy as ultrafilter existence, but > probably below the level of axiom of choice. What other properties does this w have? Surely it is not the same as the ordinal number omega. I take it you agree that w is infinite. Is it the smallest infinite number? Note that in Robinson's theory there is no such thing as a smallest infinite number, nor is there a largest. Since (Ex)(x>w) is true in Robinson's theory, I don't see why you claim that the theory is omega-inconsistent. There are natural numbers larger than w, whatever w may mean. You must be interpreting omega-inconsistency in an external sense, not the internal sense. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Non-Standard Analysis & Other Infinities Dave Seaman says... Non-standard analysis is simply defined as the free extension of ordinary arithmetic, modulo the addition of a new distinguished constant (w) and a set of axioms (0What other properties does this w have? Surely it is not the same as >the ordinal number omega. I take it you agree that w is infinite. Is it >the smallest infinite number? Note that in Robinson's theory there is no >such thing as a smallest infinite number, nor is there a largest. No, w is an arbitrary hyperfinite natural. It's not the largest hyperfinite (because w+1 is larger) and it's not the smallest hyperfinite (because w-1 is smaller). >Since (Ex)(x>w) is true in Robinson's theory, I don't see why you claim >that the theory is omega-inconsistent. The definition of omega-consistency is that if Phi(0) Phi(1) Phi(2) ... are all theorems, then exists x, not Phi(x) is not a theorem. This is false for the specific Phi(x) Phi(x) == x=w Being omega-inconsistent isn't a *bad* thing, it just means that the theory is nonstandard. -- Daryl McCullough Ithaca, NY === Subject: Re: Non-Standard Analysis & Other Infinities > Dave Seaman says... > Non-standard analysis is simply defined as the free extension of > ordinary arithmetic, modulo the addition of a new distinguished > constant (w) and a set of axioms (0 No, w is an arbitrary hyperfinite natural. It's not the largest > hyperfinite (because w+1 is larger) and it's not the smallest > hyperfinite (because w-1 is smaller). I agree that this should be the case if the theory is Robinson's, but I don't see how to derive the existence of w-1 or w+1 from only the assumptions stated above, and no others. Are we assuming that ordinary arithmetic satisfies the axioms of an integral domain, or perhaps a field? Since (Ex)(x>w) is true in Robinson's theory, I don't see why you claim that the theory is omega-inconsistent. > The definition of omega-consistency is that if > Phi(0) > Phi(1) > Phi(2) > ... > are all theorems, then > exists x, not Phi(x) > is not a theorem. This is false for the specific Phi(x) > Phi(x) == x=w Phi(0) is not a theorem in that case. Perhaps you meant Phi(x) == x Being omega-inconsistent isn't a *bad* thing, it just > means that the theory is nonstandard. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Non-Standard Analysis & Other Infinities Dave Seaman says... Non-standard analysis is simply defined as the free extension of ordinary arithmetic, modulo the addition of a new distinguished constant (w) and a set of axioms (0What other properties does this w have? Surely it is not the same as >the ordinal number omega. I take it you agree that w is infinite. Is it >the smallest infinite number? Note that in Robinson's theory there is no >such thing as a smallest infinite number, nor is there a largest. No, w is an arbitrary hyperfinite natural. It's not the largest hyperfinite (because w+1 is larger) and it's not the smallest hyperfinite (because w-1 is smaller). >I agree that this should be the case if the theory is Robinson's, but I >don't see how to derive the existence of w-1 or w+1 from only the >assumptions stated above, and no others. Are we assuming that ordinary >arithmetic satisfies the axioms of an integral domain, or perhaps a >field? It is a first-order extension, so all first-order theorems of arithmetic are satisfied. If one has the completeness axiom of the reals, it is still true that it holds for all bounded sets defined by first-order statements. >Since (Ex)(x>w) is true in Robinson's theory, I don't see why you claim >that the theory is omega-inconsistent. The definition of omega-consistency is that if Phi(0) Phi(1) Phi(2) ... are all theorems, then exists x, not Phi(x) is not a theorem. This is false for the specific Phi(x) Phi(x) == x=w >Phi(0) is not a theorem in that case. Perhaps you meant > Phi(x) == xBut I still wonder whether your statement actually holds, even for this >Phi. I suppose it depends on what ... means. For first-order results, it is absolutely meaningless. Ordinary sequences do not exist in any model of the axioms given; this is what confuses people. The sum of 1/x(x+1) from x=1, ..., N = 1 - 1/(N+1), even if N is bigger than w, and from OUTSIDE the model, there are at least c, the cardinality of the continuum, integers in the model less than N. But INSIDE the model, N is finite. Also, all things like the smallest set which contains 0, and whenever it contains n, it contains n+1, is meaningless in first-order mathematics. One cannot form the intersection of all sets of a given type. Being omega-inconsistent isn't a *bad* thing, it just means that the theory is nonstandard. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Non-Standard Analysis & Other Infinities Herman Rubin says... > The definition of omega-consistency is that if > Phi(0) > Phi(1) > Phi(2) > ... > are all theorems, then > exists x, not Phi(x) > is not a theorem. But I still wonder whether your statement actually holds, even for this Phi. I suppose it depends on what ... means. >For first-order results, it is absolutely meaningless. What I meant was that in an omega-inconsistent theory, there is some formula Phi(x) such that for each *standard* natural n, Phi(N_n) is a theorem (where N_n is the numeral corresponding to n) and also the formula exists x, not Phi(x) is a theorem. If you take the complete theory of arithmetic, add a new constant w, and add the countably many new axioms w > N_n for each natural number n, then the corresponding theory is by definition omega-inconsistent, because there is a theorem exists x, not (w > x) -- Daryl McCullough Ithaca, NY === Subject: Re: Non-Standard Analysis & Other Infinities Dave Seaman says... No, w is an arbitrary hyperfinite natural. It's not the largest hyperfinite (because w+1 is larger) and it's not the smallest hyperfinite (because w-1 is smaller). >I agree that this should be the case if the theory is Robinson's, but I >don't see how to derive the existence of w-1 or w+1 from only the >assumptions stated above, and no others. Are we assuming that ordinary >arithmetic satisfies the axioms of an integral domain, or perhaps a >field? Start with the complete theory of natural numbers (the language 0,1, +, *) and add the new constant w, and new axioms w > 0, w > 1, etc. Then since it is a theorem of arithmetic that forall x, x > 0 -> exists y, y+1=x then it follows that exists y, y+1 = w Similarly, it follows that exists y, w-1 <= y*2 <= w exists y, y*y <= w <= (y+1)*(y+1) etc. So there are infinitely many hyperfinite numbers whose existence is implied by the existence of w. The definition of omega-consistency is that if Phi(0) Phi(1) Phi(2) ... are all theorems, then exists x, not Phi(x) is not a theorem. This is false for the specific Phi(x) Phi(x) == x=w >Phi(0) is not a theorem in that case. I'm sorry. I meant Phi(x) == x < w >Perhaps you meant > Phi(x) == xBut I still wonder whether your statement actually holds, even for this >Phi. I suppose it depends on what ... means. I just mean that for every standard natural number n, there is a corresponding theorem N_n < w where N_n is a numeral for n. -- Daryl McCullough Ithaca, NY === Subject: Re: Non-Standard Analysis & Other Infinities <40rigtF1bli3iU1@individual.net> Non-standard analysis is simply defined as the free extension of > ordinary arithmetic, modulo the addition of a new distinguished > constant (w) and a set of axioms (0 No, w is an arbitrary hyperfinite natural. It's not the largest > hyperfinite (because w+1 is larger) and it's not the smallest > hyperfinite (because w-1 is smaller). > I agree that this should be the case if the theory is Robinson's, but I > don't see how to derive the existence of w-1 or w+1 from only the > assumptions stated above, and no others. Are we assuming that ordinary > arithmetic satisfies the axioms of an integral domain, or perhaps a > field? Since (Ex)(x>w) is true in Robinson's theory, I don't see why you claim that the theory is omega-inconsistent. > The definition of omega-consistency is that if > Phi(0) > Phi(1) > Phi(2) > ... > are all theorems, then > exists x, not Phi(x) > is not a theorem. This is false for the specific Phi(x) > Phi(x) == x=w > Phi(0) is not a theorem in that case. Perhaps you meant > Phi(x) == x But I still wonder whether your statement actually holds, even for this > Phi. I suppose it depends on what ... means. > Being omega-inconsistent isn't a *bad* thing, it just > means that the theory is nonstandard. > -- > Dave Seaman > U.S. Court of Appeals to review three issues > concerning case of Mumia Abu-Jamal. > Well, why not then consider an infinitesimal like iota? It's the smallest infinitesimal, Russell, there are smaller infinitesimals. You accept some infinity that has lesser and greater values that are yet infinite, and calling it a constant, do you not find that parallel in consideration to an infinitesimal that while there are lesser and greater infinitesimals that it is suitable as a constant? Consider Banach-Tarski, many accept the equidecomposability of shapes with different volumes with finitely many pieces, yet you can't have the unit interval equidecomposable (equidecomposible?) with itself with infinitely many pieces? While you're at it, well-order the reals. Ross === Subject: Re: Non-Standard Analysis & Other Infinities <40rigtF1bli3iU1@individual.net> Well, why not then consider an infinitesimal like iota? Please define iota. It's the smallest infinitesimal, Russell, there are smaller infinitesimals. There is no smallest infinitessimal. For any infinitessimal x, x/2 exists and is smaller. === Subject: Re: Non-Standard Analysis & Other Infinities <40rigtF1bli3iU1@individual.net> Well, why not then consider an infinitesimal like iota? > Please define iota. It's the smallest infinitesimal, Russell, there are smaller infinitesimals. > There is no smallest infinitessimal. For any infinitessimal x, x/2 > exists and is smaller. > Jon, that's not crankish talk. Consider the dichotomy of points and line segments, and how that is a staple of mathematical discussions since antiquity. Today there are frameworks of modern discussion of these issues, and they lay at the root of very meaningful considerations of mathematical considerations pure and applied. Iota is a least positive real. It's basically consideration that if the line segment is comprised of points, and each point on the line segment can only border at most two other points, then those points are its neighbors in a contiguous sequence. The reals are a complete ordered field. For reals x, y, (x+y)/2 is a distinct number. I think it's necessary to consider in the dichotomy that those distinct reals are distinct in a manner that applies to their distinction in the complete ordered field, yet at the same time it is necessary to consider all possible aspects of the reals, so there is a justified notion that the reals are as well a contiguous sequence of points, in an at least dual structure that allows the recognition of what are implicit features of these mathematical objects. Consider the variation of Cantor's first as applied to well-orderings of the reals. Are the reals not a set, or is choice inconsistent with ZF, or are there uncountably many nested intervals, and so on and so forth? Those are not very acceptable. Prior to Robinson with his hyperreals, there were a variety of considerations of post-Weierstrass nonstandard rationalizations of the real numbers. There are as well also today others. I suggest that the useful mathematical structure of the complete ordered field doesn't need to be abandoned to address, eg, a well-ordering of the reals, indeed it need not be, yet that instead there is a vital complement, in definition and utility. Ross === Subject: Re: Non-Standard Analysis & Other Infinities <40rigtF1bli3iU1@individual.net> Jon, that's not crankish talk. Consider the dichotomy of points and line segments, and how that is a staple of mathematical discussions since antiquity. Today there are frameworks of modern discussion of these issues, and they lay at the root of very meaningful considerations of mathematical considerations pure and applied. Viewing points on a line in the same way Euclid did I have no problem with. The issue is your inserted assumption that the points are sequential. This is a violation of the very principle you are suggesting, since even from antiquity they knew that between any two distinct points there is a third in between them. Iota is a least positive real. There is no least possible real. As stated before: if iota is any real (or hyper-real), then iota / 2 exists and is less than iota. It's basically consideration that if the line segment is comprised of points, and each point on the line segment can only border at most two other points, then those points are its neighbors in a contiguous sequence. Your consideration is incorrect. It is true that the line segment is comprised of points. It is incorrect to say that each point on the line segment borders ar most two other points. As is true with the real line, it is also true with the geometric line segment that between any two distinct points, there lies another between them. The reals are a complete ordered field. Correct. For reals x, y, (x+y)/2 is a distinct number. Correct again. I think it's necessary to consider in the dichotomy that those distinct reals are distinct in a manner that applies to their distinction in the complete ordered field, yet at the same time it is necessary to consider all possible aspects of the reals, so there is a justified notion that the reals are as well a contiguous sequence of points... None of the rest of this makes sense to me. I don't know what dichotomy you are talking about. The points on a line are not sequential (whether it's Euclidean or the real line). Your constant presumption that they are is provably false. Consider the variation of Cantor's first as applied to well-orderings of the reals. Are the reals not a set... The reals are a set. ... or is choice inconsistent with ZF... Choice is provably independent to ZF. ...or are there uncountably many nested intervals, and so on and so forth? There are only countably many open disjoint sets on R at any one time. In your construction, your intervals are not open disjoint sets. Prior to Robinson with his hyperreals, there were a variety of considerations of post-Weierstrass nonstandard rationalizations of the real numbers. There are as well also today others. I don't know what you are referring to here. But whatever it is, it is not related to the discussion between Dave & Daryl. Jonathan Hoyle Eastman Kodak === Subject: Re: Non-Standard Analysis & Other Infinities <40rigtF1bli3iU1@individual.net> Jon, that's not crankish talk. Consider the dichotomy of points and line segments, and how that is a staple of mathematical discussions since antiquity. Today there are frameworks of modern discussion of these issues, and they lay at the root of very meaningful considerations of mathematical considerations pure and applied. > Viewing points on a line in the same way Euclid did I have no problem > with. The issue is your inserted assumption that the points are > sequential. This is a violation of the very principle you are > suggesting, since even from antiquity they knew that between any two > distinct points there is a third in between them. Iota is a least positive real. > There is no least possible real. As stated before: if iota is any real > (or hyper-real), then iota / 2 exists and is less than iota. It's basically consideration that if the line segment is comprised of points, and each point on the line segment can only border at most two other points, then those points are its neighbors in a contiguous sequence. > Your consideration is incorrect. It is true that the line segment is > comprised of points. It is incorrect to say that each point on the > line segment borders ar most two other points. As is true with the > real line, it is also true with the geometric line segment that between > any two distinct points, there lies another between them. The reals are a complete ordered field. > Correct. For reals x, y, (x+y)/2 is a distinct number. > Correct again. I think it's necessary to consider in the dichotomy that those distinct reals are distinct in a manner that applies to their distinction in the complete ordered field, yet at the same time it is necessary to consider all possible aspects of the reals, so there is a justified notion that the reals are as well a contiguous sequence of points... > dichotomy you are talking about. The points on a line are not > sequential (whether it's Euclidean or the real line). Your constant > presumption that they are is provably false. Consider the variation of Cantor's first as applied to well-orderings of the reals. Are the reals not a set... > The reals are a set. ... or is choice inconsistent with ZF... > Choice is provably independent to ZF. ...or are there uncountably many nested intervals, and so on and so forth? > There are only countably many open disjoint sets on R at any one time. > In your construction, your intervals are not open disjoint sets. Prior to Robinson with his hyperreals, there were a variety of considerations of post-Weierstrass nonstandard rationalizations of the real numbers. There are as well also today others. > not related to the discussion between Dave & Daryl. > Jonathan Hoyle > Eastman Kodak First, about the nested intervals, they are non-degenerate closed intervals in the complete ordered field. The construction may as well have the nested interval endpoints indicate open intervals, or half-open intervals, as they are contained by nested, non-degenerate closed intervals. So, I still don't see the notion about the intervals being closed as affecting the progression of the argument, that a well-ordering, which as described over in Well-Ordering the Reals is a bijection between R and some ordinal O', has the same consequences as R bijecting to an ordinal N. If otherwise, please explain what you mean more fully for those reading our current conversation who are not necessarily up to speed on that variation of Cantor's first or nested intervals. I think above Daryl was talking about diagonalization instead of nested intervals, also known as Cantor's first proof the uncountability of the reals, or as I hope to show Cantor's first proof of sequential points in the reals. Well-order the reals. R is a complete ordered field except generally division by zero is undefined. It has no zero divisors, which is a different issue that that for no x, y =/= 0, xy = 0. Various considerations of division by zero lead to for example meromorphic functions as I heard about a while back in f(f(f(y))) and projectively extended real numbers with a one or two point compactification. (You're dividing by zero!) In talking about the hyperreals and various other nonstandard constructions of the reals, you may as well consider for further comprehension notions a la Schmieden and Laugwitz' dually partially ordered ring and complete ordered field reals, which predate Robinson's construction a few years. There are everywhere and only reals between zero and one. Ross === Subject: Re: Non-Standard Analysis & Other Infinities <40rigtF1bli3iU1@individual.net> First, about the nested intervals, they are non-degenerate closed intervals in the complete ordered field. You don't know that. Without the precise well ordering, you can't rule out that your a_n's would like: 3, 3.1, 3.14, 3.141, 3.1415, etc. and your b_n's look like 4, 3.2, 3.15, 3.142, 3.1416, etc. leading after a countable number of steps the degenerate closed interval containing just the single point of pi. (The point about using open intervals guarantees that you do not ever have these single point intervals.) You also cannot rule out that there is an uncountable number less than Continuum Hypothesis. Remember that using Dedekind cuts, you can (with only a countable number of entries on each side) uniquely determine any real number, of which there are an uncountable number. An arbitrary real number of the form 0.abcde... has only a countable number (Aleph_0) of decimal places, but there are an uncountable number (10^Aleph_0 = c) of possible reals. Hope that helps, Jonathan Hoyle Eastman Kodak === Subject: Re: Non-Standard Analysis & Other Infinities First, about the nested intervals, they are non-degenerate closed intervals in the complete ordered field. > out that your a_n's would like: 3, 3.1, 3.14, 3.141, 3.1415, etc. and > your b_n's look like 4, 3.2, 3.15, 3.142, 3.1416, etc. leading after a > countable number of steps the degenerate closed interval containing > just the single point of pi. In your example, the length of the nth interval is 10/10^n > 0, for all n, so there is never a degenerate interval of length zero in the sequence of such intervals. The same result will hold for intervals based on the upper and lower decimal approximations to n places of any irrational number and many rationals numbers as well. === Subject: Re: Non-Standard Analysis & Other Infinities <40rigtF1bli3iU1@individual.net> You don't know that. Without the precise well ordering, you can't rule out that your a_n's would like: 3, 3.1, 3.14, 3.141, 3.1415, etc. and your b_n's look like 4, 3.2, 3.15, 3.142, 3.1416, etc. leading after a countable number of steps the degenerate closed interval containing just the single point of pi. >In your example, the length of the nth interval is 10/10^n > 0, for all >n, so there is never a degenerate interval of length zero in the >sequence of such intervals. Not in any of the finite n's, but it can degenerate at limit ordinals. Ross is using a well-ordering of the reals which maps the reals against the countable ordinals. Essentially, Ross was trying to prove Cantor's nested interval proof of uncountability was invalid by substituting the ordinals for the natural numbers and attempting to arrive at the same result. Ross's argument fails because the argument from the naturals does not extend to the ordinals, primarily because there are limit ordinals. === Subject: Re: Non-Standard Analysis & Other Infinities <40rigtF1bli3iU1@individual.net> First, about the nested intervals, they are non-degenerate closed intervals in the complete ordered field. > out that your a_n's would like: 3, 3.1, 3.14, 3.141, 3.1415, etc. and > your b_n's look like 4, 3.2, 3.15, 3.142, 3.1416, etc. leading after a > countable number of steps the degenerate closed interval containing > just the single point of pi. (The point about using open intervals > guarantees that you do not ever have these single point intervals.) > You also cannot rule out that there is an uncountable number less than > Continuum Hypothesis. > Remember that using Dedekind cuts, you can (with only a countable > number of entries on each side) uniquely determine any real number, of > which there are an uncountable number. An arbitrary real number of the > form 0.abcde... has only a countable number (Aleph_0) of decimal > places, but there are an uncountable number (10^Aleph_0 = c) of > possible reals. > Hope that helps, > Jonathan Hoyle > Eastman Kodak It's not two convergent sequences, it's a well-ordered set of closed nested intervals. For each, in the complete ordered field, there are values between them, and were an element to be in range, then it would be an endpoint of a later interval. If there's a degenerate closed interval after countably many steps, then the reals are countable. If there's not a degenerate closed interval after countably many steps, then the rationals are uncountable. Either is contradictory to the standard understanding, why would not one or the other hold? If that degenerate interval exists in the set of nested intervals, then some eventual predecessor contains only two points. That's about a problem with infinite limit ordinals, in that they have no immediate predecessor, instead, an existential quantifier for that applies to an eventual predecessor, for were that not the case, there would not be a degenerate interval, in the complete ordered field. That, takes some more consideration. It does illustrate that the degenerate interval could _only_ map to an initial ordinal, in the well ordering of the nested intervals. I just think the initial ordinal there is zero. As well, that seems to lead to a maximal element in the ordinals, if you get my drift, but there is no universe in ZF. Consider a Dedekind cut a la 0.1, 0.01, 0.001, 0.0001, .... That's kind of irrelevant, my opinion is that there Dedekind / Cauchy / continued fractions is insufficient.The reason I carry that opinion is because reasoning as above leads me to believe so using the very specific tools of mathematical logic as it is standardly practiced. It seems you describe that using open intervals would make it a stronger argument. Is that an accurate statement? That leads into various reasonings why a single point is open or closed, or perhaps as I think, one- or two-sided in or on R, polydimensional points in R^2N. A degenerate interval is a point moreso than an interval. Anyways, what happens if I say _open_ intervals? Ross === Subject: Re: Non-Standard Analysis & Other Infinities <40rigtF1bli3iU1@individual.net> It's not two convergent sequences, it's a well-ordered set of closed nested intervals. For each, in the complete ordered field, there are values between them, and were an element to be in range, then it would be an endpoint of a later interval. You have defined what to do only at successor ordinals. You have not defined what to do at limit ordinals (which is where your construction breaks down). If there's a degenerate closed interval after countably many steps, then the reals are countable. Not true at all. Any real can be closed in with countably many steps: [3,4], [3.1, 3.2], [3.14,3.15], etc. However there are uncountably many reals. There are Aleph_0 many decimal places, yet there are 10 choices for each decimal position, making 10^Aleph_0 = c possible reals. If that degenerate interval exists in the set of nested intervals, then some eventual predecessor contains only two points. This is *precisely* your error in logic. (As you note later) Not all ordinals have a predecessor! Omega, for example, does not. Neither does omega * 2, nor omega^2, etc etc. Limit ordinals are those ordinals without an immediate predecessor, and it is at these ordinals that your argument fails. At all the successor ordinals the a_n's and b_n's close in on each other, but you do not define what to do at these limit ordinals when you run out of room. In my example, if the a_n's start off { 3, 3.1, 3.14, ... } and your b_n's start off { 4, 3.2, 3.15, ... }, your only choice of a_omega and b_omega is pi. What do you do after that? Plus, you have an uncountable number of these countable limit ordinals your proof does not account for. That's about a problem with infinite limit ordinals, in that they have no immediate predecessor, instead, an existential quantifier for that applies to an eventual predecessor, for were that not the case, there would not be a degenerate interval, in the complete ordered field. That, takes some more consideration. It does illustrate that the degenerate interval could _only_ map to an initial ordinal, in the well ordering of the nested intervals. So it seems you are aware of the hole in your argument. I just think the initial ordinal there is zero. The term is limit ordinal not initial ordinal. And of course there are other limit ordinals. For example, omega = { 0, 1, 2, ... } (the set of finite natural numbers) is a limit ordinal. If not, which of the finite natural numbers is its immediate predecessor? It can't be any of them since every finite natural number has a successor which is another finite natural number. It can't be some mythical infinite integer omega - 1, since omega contains only finite natural numbers. It seems you describe that using open intervals would make it a stronger argument. Is that an accurate statement? It does because it prevents degenerate intervals. For (a,b) an open interval, it must be that a <40rigtF1bli3iU1@individual.net> It's not two convergent sequences, it's a well-ordered set of closed nested intervals. For each, in the complete ordered field, there are values between them, and were an element to be in range, then it would be an endpoint of a later interval. > You have defined what to do only at successor ordinals. You have not > defined what to do at limit ordinals (which is where your construction > breaks down). If there's a degenerate closed interval after countably many steps, then the reals are countable. > Not true at all. Any real can be closed in with countably many steps: > [3,4], [3.1, 3.2], [3.14,3.15], etc. However there are uncountably > many reals. There are Aleph_0 many decimal places, yet there are 10 > choices for each decimal position, making 10^Aleph_0 = c possible > reals. If that degenerate interval exists in the set of nested intervals, then some eventual predecessor contains only two points. > This is *precisely* your error in logic. (As you note later) Not all > ordinals have a predecessor! Omega, for example, does not. Neither > does omega * 2, nor omega^2, etc etc. Limit ordinals are those > ordinals without an immediate predecessor, and it is at these ordinals > that your argument fails. At all the successor ordinals the a_n's and > b_n's close in on each other, but you do not define what to do at these > limit ordinals when you run out of room. In my example, if the a_n's > start off { 3, 3.1, 3.14, ... } and your b_n's start off { 4, 3.2, > 3.15, ... }, your only choice of a_omega and b_omega is pi. What do > you do after that? Plus, you have an uncountable number of these > countable limit ordinals your proof does not account for. That's about a problem with infinite limit ordinals, in that they have no immediate predecessor, instead, an existential quantifier for that applies to an eventual predecessor, for were that not the case, there would not be a degenerate interval, in the complete ordered field. That, takes some more consideration. It does illustrate that the degenerate interval could _only_ map to an initial ordinal, in the well ordering of the nested intervals. > So it seems you are aware of the hole in your argument. I just think the initial ordinal there is zero. > The term is limit ordinal not initial ordinal. And of course there > are other limit ordinals. For example, omega = { 0, 1, 2, ... } (the > set of finite natural numbers) is a limit ordinal. If not, which of > the finite natural numbers is its immediate predecessor? It can't be > any of them since every finite natural number has a successor which is > another finite natural number. It can't be some mythical infinite > integer omega - 1, since omega contains only finite natural numbers. It seems you describe that using open intervals would make it a stronger argument. Is that an accurate statement? > It does because it prevents degenerate intervals. For (a,b) an open > interval, it must be that a is a closed set. You avoid degenerate cases outright. Anyways, what happens if I say _open_ intervals? > It exposes the flaw in your reasoning sooner. > Jonathan Hoyle > Eastman Kodak Well, yes, that is rather confusing. For there to be a degenerate interval, as described in the construction for nested intervals as a well-ordered set of ordered pairs, then that would not seem to take place in the complete ordered field. The reason the structure of the container of the nested intervals is modified is so that instead of having two sequences of interval endpoints which would be argued to be countable, a set is generated instead. Via transfinite induction over some ordinal O' equivalent to R, if that so holds, then nested intervals are generated. A limit ordinal in the well-ordering of the reals does not necessarily correspond to a limit ordinal in the nested intervals, in terms of a well-ordering of the nested intervals. It is not thus so that a limit ordinal of the domain of g, the function well-ordering the nested intervals via bijection to an ordinal, corresponds in its generation to a limit ordinal in the domain of f, the function well-ordering the reals. So, defining what to do at limit ordinals has multiple implications. I don't understand why open nested intervals would break down. It seems that a case-based reasoning with each of the conditions should be described, so that from a template of sorts the variations could be enumerated for specializations of the logical progression. Were the degenerate interval g(c) for a countable ordinal, then the reals would not be shown uncountable. Thus, g(c) where c is a countable ordinal is not a degenerate interval, in the standard reals where the complete ordered field is assumed throughout in this discussion unless otherwise specifically noted. When g(o) is a degenerate interval, then it's also the maximal element of AB. Where the well-ordering of the reals has no maximal element in ZF, the well-ordering of the nested intervals thus generated has a maximal element. Where g(o) is a degenerate interval, o has no successor in the domain of g. So, for g(o) a degenerate interval, o has no (immediate) predecessor, else g(0-1) would be an interval with two points, and no successor, else it would not be a degenerate interval. These are basically comments about various conditions on the mathematical structures there. I think there is more to consider about these things. Here we have left disussion about hyperreals for the moment, excuse me. If the original poster has more comments or questions about hyperreal numbers they would find a ready response among the interested and knowledgeable parties who read these things. Ross === Subject: Re: Non-Standard Analysis & Other Infinities Ross A. Finlayson says... >Jon, that's not crankish talk... >Iota is a least positive real. *That* is crankish talk. As Jon points out, if iota is a positive real, then iota/2 is a *smaller* positive real. >Consider the variation of Cantor's first as applied to well-orderings >of the reals. That's more crankish talk. Cantor's proof isn't about well-orderings, it's about *countability*. If the well-ordering is an uncountable well ordering, then Cantor's diagonalization procedure doesn't produce a real that is guaranteed not to be on the list. >Are the reals not a set Yes, they are a set. >or is choice inconsistent with No, it is not. >or are there uncountably many nested intervals You are confused. Cantor's proof has the following structure: Given any mapping from the naturals to the reals, Cantor shows how to produce a real that is not in the image of that mapping. It doesn't say Given any well ordering of the reals, Cantor shows how to produce a real that is not in the well-ordering. -- Daryl McCullough Ithaca, NY === Subject: Re: Non-Standard Analysis & Other Infinities Jon, that's not crankish talk... >Iota is a least positive real. > *That* is crankish talk. As Jon points out, if iota is a positive > real, then iota/2 is a *smaller* positive real. >Consider the variation of Cantor's first as applied to well-orderings >of the reals. > That's more crankish talk. Cantor's proof isn't about well-orderings, > it's about *countability*. If the well-ordering is an uncountable > well ordering, then Cantor's diagonalization procedure doesn't > produce a real that is guaranteed not to be on the list. >Are the reals not a set > Yes, they are a set. >or is choice inconsistent with >ZF > No, it is not. >or are there uncountably many nested intervals > You are confused. Cantor's proof has the following > structure: > Given any mapping from the naturals to the reals, > Cantor shows how to produce a real that is not in > the image of that mapping. > It doesn't say > Given any well ordering of the reals, Cantor > shows how to produce a real that is not in the > well-ordering. > -- > Daryl McCullough > Ithaca, NY Daryl, that's not crankish talk. Consider the dichotomy of points and line segments, and how that is a staple of mathematical discussions since antiquity. Today there are frameworks of modern discussion of these issues, and they lay at the root of very meaningful considerations of mathematical considerations pure and applied. Iota is a least positive real. It's basically consideration that if the line segment is comprised of points, and each point on the line segment can only border at most two other points, then those points are its neighbors in a contiguous sequence. The reals are a complete ordered field. For reals x, y, (x+y)/2 is a distinct number. I think it's necessary to consider in the dichotomy that those distinct reals are distinct in a manner that applies to their distinction in the complete ordered field, yet at the same time it is necessary to consider all possible aspects of the reals, so there is a justified notion that the reals are as well a contiguous sequence of points, in an at least dual structure that allows the recognition of what are implicit features of these mathematical objects. Consider the variation of Cantor's first as applied to well-orderings of the reals. Are the reals not a set, or is choice inconsistent with ZF, or are there uncountably many nested intervals, and so on and so forth? Those are not very acceptable. Prior to Robinson with his hyperreals, there were a variety of considerations of post-Weierstrass nonstandard rationalizations of the real numbers. There are as well also today others. I suggest that the useful mathematical structure of the complete ordered field doesn't need to be abandoned to address, eg, a well-ordering of the reals, indeed it need not be, yet that instead there is a vital complement, in definition and utility. Ross === Subject: Re: Non-Standard Analysis & Other Infinities Ross A. Finlayson says... >Consider the variation of Cantor's first as applied to well-orderings >of the reals. There is no such variation. -- Daryl McCullough Ithaca, NY === Subject: Re: Fubini's theorem Since the integrand is continuous, I didn't think measurability was an issue. === Subject: Re: Known lengths of edges of an abitrary polygon, seek for area answer both ways for a (1,1,1,1,5) pentagon and then tried something with a >I can't believe I did that again. >1,1,1,1,3 >1,1,1,1,3 >1,1,1,1,3 >--Keith Lewis klewis {at} mitre.org >The above may not (yet) represent the opinions of my employer. In the alternate universe, (1,1,1,1,5) exists and (1,1,1,1,3) does not. === Subject: Re: Question on absolute continuity, L^1 convergence > Suppose that f_n is a sequence of absolutely continuous (abbreviated AC) > functions on (a,b) and that both f_n and f_n ' are Cauchy sequences in > L^1(a,b). Show that f_n converges uniformly on (a,b) to an AC function and > (lim f_n) ' = lim (f_n '). > I have gotten everything except proving that f_n converges uniformly to its > limit function f. I don't even know why this is a pointwise convergence. hint: Fix a point c in (a,b). Then f_n(c) - f_m(c) = f_n(x) - f_m(x) - int_[c,x] (f_n'-f_m'). Now integrate over (a,b) to get a bound on |f_n(c) - f_m(c)| that is independent of c. === Subject: Looking for a good number theory book with lots of theorems. What is a good number theory book that begins at the foundations and proceeds to more complex theorems that can't be easily visualized? I am looking for the most important results of number theory, plus more. === Subject: Re: Looking for a good number theory book with lots of theorems. Hi guy's, There is a fifth edition of Number theory and its application by the great author kenneth rosen. I think this is very nice book especially for those who have a good computer science background ( who of course did study a course in analysis and design of algorithm ) but my advice this the book you should look for. === Subject: Re: Looking for a good number theory book with lots of theorems. There is not one book, but a few that can be most important which will give you a complete picture. The two most important book in number theory that are luckily in print are Gauss' Disquositione Arithmeticae and Dirichelet's Lectures on number theory. IF you want to actually learn number theory the way it was made, written by those that created the subject get these books. There exists NO substitute. Dirichlet's book is actually an explanation of Gauss's book, which is without a doubt the most important book in the subject and thank God it is in print for the moment. So get it now. http://www.amazon.com/gp/product/0821820176/104-0503592-1793556?%5Fencoding= UTF8&v=glance&n=283155 http://www.amazon.com/gp/product/0387962549/ref=ord_cart_shr/104-0503592-179 3556?%5Fencoding=UTF8&m=ATVPDKIKX0DER&v=glance&n=283155 If you want to learn theorems and number theory look for Hardy's An intorduction to number theory. Very good book, excelent start. Just as good, but more to the point is Edmund Landau's book Number theory. http://www.amazon.com/gp/product/0198531710/qid=1135755576/sr=1-3/ref=sr_1_3 /104-0503592-1793556?s=books&v=glance&n=283155 To give you a picture of how good and inclusive is Landau's book, he actually proves Dirichelet's theorem on primes in an arithmetic progression. Such a proof is usually left of for much more advanced texts, but that is Landau's philosophy to include advanced but important result early. It is hard for me to pick between Hardy and Landau, as each brings something different to the subject. At last i will add the most important book for reference in number theory, namely Dickson's History of the theory of numbers. If you plan to do anything in the subject GET these now before they go out of print. They are priceless. A special mention should be given to Shanks' Number theory- the amazon title is actually not wrong but not complete. Very good on the historical side, much like Hardy. Hope it helps. http://www.amazon.com/gp/product/082182824X/104-0503592-1793556?%5Fencoding= UTF8&v=glance&n=283155 === Subject: Re: Looking for a good number theory book with lots of theorems. > What is a good number theory book that begins at the foundations and > proceeds to more complex theorems that can't be easily visualized? I am > looking for the most important results of number theory, plus more. I'm not mathematican, but was quite impressed reading Elementary Number Theory and its Applications Kenneth H. Rosen, 2nd edition, Addison-Wesley 1988. -- Dave K http://www.southminster-branch-line.org.uk/ It is always of the form: month-year@domain. Hitting reply will work for a couple of months only. Later set it manually. The month is always written in 3 letters (e.g. Jan, not January etc) === Subject: Re: Looking for a good number theory book with lots of theorems. <43b1e559@212.67.96.135> Wasn't there a third edition of this book? Ray Steiner === Subject: Re: Looking for a good number theory book with lots of theorems. On 27 Dec 2005 15:05:54 -0800, scott7261@hotmail.com >What is a good number theory book that begins at the foundations and >proceeds to more complex theorems that can't be easily visualized? I am >looking for the most important results of number theory, plus more. A classic text, with lots of good exercises is: An Introduction to the Theory of Numbers by Niven & Zuckerman Another text that looks very good and is available as a free download is: A Computational Introduction to Number Theory and Algebra by Victor Shoup You can download it at the following link: quasi === Subject: Least Absolute Deviation Regression Hi all, I have been trying to find an algorithm for Least Absolute Deviation regression, and have been having the hardest time. I am writing the algorithm in Java. I am wondering if someone could describe the algorthm, have pseudocode, code, a Java lib/Jar, or anything useful! I tried a stat group, I figured I would try the almighty math group. TIA if possible, Dave === Subject: Re: Least Absolute Deviation Regression > Hi all, > I have been trying to find an algorithm for Least Absolute Deviation regression, and have been having the hardest time. I am writing the algorithm in Java. I am wondering if someone could describe the algorthm, have pseudocode, code, a Java lib/Jar, or anything useful! I tried a stat group, I figured I would try the almighty math group. If you have access to EXCEL, there is a built-in lp solver (in the Solver Tool). You can formulate and solve problems with up to about 100 data points. Be warned, however: do NOT attempt to solve the problem directly, using a summed absolute-value criterion, since Solver will choke on the problem (due to its non-differentiability). Use the standard modelling tricks to convert to a linear program. If you don't know how to do it, I would be happy to lend a hand. R.G. Vickson > TIA if possible, > Dave === Subject: Re: Least Absolute Deviation Regression You can formulate this as an lp (linear programming problem). I believe the lp solver lp_solve has a Java interface. Another possibility is to use iteratively a weighted ls procedure, as in: http://www.american.edu/academic.depts/cas/econ/gaussres/regress/LAD.SRC > Hi all, > I have been trying to find an algorithm for Least Absolute Deviation regression, and have been having the hardest time. I am writing the algorithm in Java. I am wondering if someone could describe the algorthm, have pseudocode, code, a Java lib/Jar, or anything useful! I tried a stat group, I figured I would try the almighty math group. > TIA if possible, > Dave ---------------------------------------------------------------- Erwin Kalvelagen erwin@gams.com, http://www.gams.com/~erwin ---------------------------------------------------------------- === Subject: Maximizing function This is a problem from image processing. An image has pixel values p(x,y), and it is desired to find the values of x0, y0 and r for which the sum of p(x,y) is a maximum on the circle (x-x0)^2 + (y-y0)^2 = r^2. The function to be maximized depends on a large number of constants p(x,y), and the 3 variable parameters x0, y0 and r. To apply most maximization techniques, you need the derivatives of the function wrt the parameters. I don't see how to find these analytically. Can someone point me in the right direction? === Subject: Re: Maximizing function >This is a problem from image processing. An image has pixel values >p(x,y), and it is desired to find the values of x0, y0 and r for which >the sum of p(x,y) is a maximum on the circle (x-x0)^2 + (y-y0)^2 = r^2. Sum over what? Perhaps you mean something like this: each pixel is a rectangle, say x_i-h/2 < x <= x_i+h/2, y_j-h/2 < y <= y_j+h/2, and your objective is the sum of the values of p on the pixels that the circle intersects. >The function to be maximized depends on a large number of constants >p(x,y), and the 3 variable parameters x0, y0 and r. To apply most >maximization techniques, you need the derivatives of the function wrt >the parameters. I don't see how to find these analytically. Can someone >point me in the right direction? If my interpretation of the question is correct, it's much worse than not being able to find derivatives analytically: the function is discontinuous. The classical optimization techniques are not very useful under these conditions. What you might do, however, is to use an approximate objective that will be differentiable: e.g. integrate sum_{i,j} f(x-x_i, y-y_j) p(x_i,y_j) over the circle where, e.g., f(x,y) = exp(-k (x^2+y^2)) for some k > 0. Use the classical techniques to maximize this. Then perhaps a local search near the solution to that problem to find an approximate maximum for your original objective. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Maximizing function not being able to find derivatives analytically: the function is > discontinuous. I think the sum of pixels function can be made continuous by multiplying each pixel value by a number related to the distance from the center of the pixel to the circle. === Subject: Re: Maximizing function p(x,y), and it is desired to find the values of x0, y0 and r for which >the sum of p(x,y) is a maximum on the circle (x-x0)^2 + (y-y0)^2 = r^2. > Sum over what? Perhaps you mean something like this: > each pixel is a rectangle, say x_i-h/2 < x <= x_i+h/2, > y_j-h/2 < y <= y_j+h/2, and your objective is the sum of the values of > p on the pixels that the circle intersects. That is correct === Subject: Re: can all F(x,y) = G(x,y) type relations be implicitly differentiated? >How would I check for the continuity of implicit equations ( H(x,y) = 0 >My somewhat educated guess would be to take these limits of the >original relation: >1) substitute x0 for x, and make sure as y->y0, then H(x0,y) approaches >2) substitute y0 for y, and make sure as x->x0, then H(x,y0) approaches >I'm trying to extend the definition of continuity from a simple y=f(x) >setting to a H(x,y) = 0 setting. Is the above two step test okay? It >makes sense to me because we are now testing continuity in two >variables now, but can't test both at the same time (at least I don't >think we can?!?!?!). No, you must do it in both variables at the same time. In terms of epsilons and deltas, for every epsilon > 0 there is delta > 0 such that whenever |x - x0| < delta and |y - y0| < delta, |H(x,y) - H(x0,y0)| < epsilon. >One other quick question. In dealing with true functions of the form >y=f(x), we are pretty much using the same IMPLICIT DIFFERENTIATION >(differentiate both sides with respect to x), yet we separate the two >processes. Is this because all the tests associated with implicit >functions are proven by someone as unecessary when we are sure that the >function behaves as a true function of the said form? >Here is my attempt at understanding the separation: >when I convert y = f(x) to the H(x,y) = 0 form, I get: >y - f(x) = 0 >differentiating implicitly, we get H_y = 1! ... and H_x = -f'(x). > Is that a valid proof >that y=f(x) forms are safe and fall outside the realm of the two >implicit function conditions you stated? Outside? Do you mean inside? Since constants are always continuous, for the implicit function theorem as I presented it to apply you want f' to be continuous near x0. > How does this >intrinsically/automatically imply continuity (the second condition you >mentioned when takling about H(x,y) = 0 equations)? I don't think I understand the question. What automatically implies continuity? >I can see how the first condition [H_y <> 0 is always satisfied by >y=f(x) ] but can't see how differentiating both sides of a y=f(x) using >our well proven chain and other rules automatically guarantees >continuity It doesn't. There are functions f(x) whose derivatives are not continuous. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Standard Deviation of PISA jinny whimpered in message A simple random sampling of 3,000 Americans is the MOST accurate way to do this. > Um, no. > There is no set number that would be the most accurate, short of 100% > of the > target population. I can get a very good result with 3000, depending on > the > parameters and size of the target. You would get absolute rubbish with > 295 > million due to your ineptitude. More to follow. > Please continue to astound us with your imbecility. I need something to > laugh at on long winter nights. John Knight ps--I'm going to post here the search engine results so that you can get a SENSE, not an understanding,but a SENSE, of how WRONG you are. > Are you sure you know how to use a search engine, jinny? > James Powell And the amusement continues! > More than 80% of the hits are people wh've never visited that page > before. One of the search engines that people use to locate that poll > is Google. The top 52 key words people searched on Google are listed > below, along with the number of times each word was searched, and its > percentage of the total number of search words. Do you have a point with this nonsense? Are you trying to state that everyone who took your poll arrived randomly through a search engine? Your unsubstantiated attempt to justify your made up poll results is hereby noted. I would pint out that your max hit < your claimed polled population. census of all 295 million Americans would be more accurate than this > internet-based poll. The first problem with this STUPID idea is that > it would take more than a year to complete it, and the public opinion > would have changed dramatically between the first answer and the 295 > millionth answer. In your opinion. How long it would take depends on the amount of energy put in completing the poll. The time is irrelevant, however, as it would not affect the accuracy. You are concerned with opinion changing in a year, yet you claim to have aan overwhelming consensus today? Take your meds, jinny, your schizo is showing. BTW - Seems to me that this is attempted every time we have one of those election thingies. We got a pretty good count of 120 million in less than 24 hours in the last one. > The second problem is the hundreds or thousands of government employees > who would be handling the data, any ONE of whom could make a simple > spreadsheet error (like the simple SINGLE spreadsheet error which cost > Citibank ONE BILLION BUCKS), and COMPLETELY invalidate it. As opposed to you making up the entire dataset? Besides, who would need to handle the data. We have new things these days called com-pu-ters. Your 'poll' could be turned into a simple ScanTron sheet and fed directly into a computer. > The third problem is collecting and collating millions of forms, even > WITH competent employees (which US government employees are NOT), > introduces numerous opportunities for fraud, key omissions, and like > with the Michigan affirmative action board which was ORDERED by the > court to certify the state ballot on affirmative action (and refused), > plain flat REFUSAL to follow the LAW. You just like creating problems that do not exist to justify your nonsense! See above, re: election. > The fourth and killer problem is that IF you ever completed such a > task, and the results showed that only 88% favored exiling the niggers, > compared to 90% on thisto SIMPLE poll--not a single competent > mathematician in the WORLD would dare claim that the government poll > was MORE accurate than this internet poll. Why? Because it's > STATISTICALLY IMPROBABLE for so many human [read: government] hands to > handle so much data WITHOUT making key errors. Well, I am sure the results would show nothing of the kind. In fact, I would be willing to bet on it. You go ahead and set all of this up, to my satisfaction as to reliability and accuracy. If the results are remotedly near yours, I will pay for the expense. As to your asinine conclusion ... well, it is classic Jinny Knight. > And what would it prove? Absolutely nothing. Whether or not 88% or > 90% is the actual figure--both are large enough to pass a > constitutional amendment anyway. I see you are unclear on the process of amending the Constitution as well. Not surprising. Feel free to try and float the proposal. It will pass right after you finishing with your repeal of the 19th Amendment. BTW - will that be before or after you complete your lawsuit against the State of California. How was Madagascar, nincompoop? Did they throw you out? James Powell === Subject: Re: Standard Deviation of PISA <1-6dnaaBov7SXjPeRVn-vA@wideopenwest.com> census of all 295 million Americans would be more accurate than this > internet-based poll. The first problem with this STUPID idea is that > it would take more than a year to complete it, and the public opinion > would have changed dramatically between the first answer and the 295 > millionth answer. hy'mie whined: In your opinion. How long it would take depends on the amount of energy put in completing the poll. The time is irrelevant, however, as it would not affect the accuracy. You are concerned with opinion changing in a year, yet you claim to have aan overwhelming consensus today? Take your meds, jinny, your schizo is showing. BTW - Seems to me that this is attempted every time we have one of those election thingies. We got a pretty good count of 120 million in less than 24 hours in the last one. less than 24 hours? To you, the entire election process is less than 24 hours? Bureaucrats, politicians, jewsmediots, pollsters, corporations, and government workers work on ONE election for decades before the election is finally held. You claimed that a poll of the entire population was more accurate than an internet poll, and you already know (or at least SHOULD know) that, with 3,000 respondents to the poll, the error is plus or minus 1%. Yes, public opinion WILL change by at least 1% over a year, and on some key issues has changed by 6-8% over the course of several decades. What you're claiming is that a census of ALL 295 million Americans would be more accurate than plus or minus 1%. Most of Gallup's polls are less than 1,000 respondents, which produces an error of plus or minus 2%. Yet they do predict election outcomes fairly accurately (accurately, that is, once you take into account their over-representation of Democrats by 4% and their under-representation of Republicans by 3%). As pointed out before, since niggers were only 7% of the respondents, but 12% of the population, it's POSSIBLE (but not CERTAIN) that if they'd not been under-represented, and IF they voted in the same way that the 7% who DID respond voted, then rather than 90% agreeing to exile, it MIGHT have been as low as 88% (a difference almost equivalent to the error). It was quite a surprise to discover how many niggers agreed. Following is an email from one who does: whites keep blacks from going to Africa ? that's right homo. LMAO I LAUGH BECAUSE IT IS SO FUNNY, HEY FAGGOT, NOW U GOT A STAIN ON YOUR BRIEFS BUDDY!!!! NOBODY TOLD YOU TO BRING US OVER HERE YOU WHITE DEVIL!!!!!!! BECAUSE WE WOULD STILL BE SUPERIOR WITH GREAT CIVILIZATIONS, YOU WOULD BE IN A CAVE!!!!!!!! IT!!!!! THATS WHAT IT GET FOR BRINGING US HERE(WHICH SHOULD BE GREAT 4 YOU BECAUSE HAD WE NOT COME HERE YOU WOULD HAVE BEEN NOTHING!!!!!!! He claims that niggers WANT to go to Africa, but us honkies won't let 'em!! Well, open de do'. We're gonna finally let the niggers out so they'z kin go HOME!! John Knight http://christianparty.net/pollblacks.htm ps--I looked at how many people had searched for the word hate in the midst of thousands looking for rate, or statistics--EXACTLY ONE!!! GUFFAW!! === Subject: Re: Standard Deviation of PISA jinny sobbed in message Now let's compare this to your UTTERLY RIDICULOUS belief that a census of all 295 million Americans would be more accurate than this internet-based poll. The first problem with this STUPID idea is that it would take more than a year to complete it, and the public opinion would have changed dramatically between the first answer and the 295 millionth answer. > hy'mie whined: > In your opinion. How long it would take depends on the amount of energy > put > in completing the poll. The time is irrelevant, however, as it would > not > affect the accuracy. > You are concerned with opinion changing in a year, yet you claim to > have aan > overwhelming consensus today? Take your meds, jinny, your schizo is > showing. > BTW - Seems to me that this is attempted every time we have one of > those > election thingies. We got a pretty good count of 120 million in less > than 24 > hours in the last one. > less than 24 hours? To you, the entire election process is less > than 24 hours? Bureaucrats, politicians, jewsmediots, pollsters, > corporations, and government workers work on ONE election for decades > before the election is finally held. But the election itself last less than 24 hours. You know, the casting and counting of ballots. The 'polling' of the voters on their decision. In fact, localities are able to pull that together as often as twice a year! Let me know when you begin to approach reality. > You claimed that a poll of the entire population was more accurate than > an internet poll, and you already know (or at least SHOULD know) that, > with 3,000 respondents to the poll, the error is plus or minus 1%. Well, jinny pulls another number out of its ass! There is no way you can show a 1% margin for your idiotic poll! You do not have a scientifically selected sample, so how can you possibly determine an error margin? Remember to show all of your work. > Yes, public opinion WILL change by at least 1% over a year, and on some > key issues has changed by 6-8% over the course of several decades. > What you're claiming is that a census of ALL 295 million Americans > would be more accurate than plus or minus 1%. Most of Gallup's polls > are less than 1,000 respondents, which produces an error of plus or > minus 2%. Yet they do predict election outcomes fairly accurately > (accurately, that is, once you take into account their > over-representation of Democrats by 4% and their under-representation > of Republicans by 3%). Yep. A poll of all 295 million Americans would have a sample error rate of less than 1%. Other error rates still apply, based upon how the poll is conducted. Gallup takes great pains to scientifically select their sample and methodology, as well as how the questions are phrased. However, I doubt the margin of error on almost every poll. Human nature being what it is. Their accuracy in elections has been accused of being a self-fulfilling projection, but that discussion is way out of your league. > As pointed out before, since niggers were only 7% of the respondents, > but 12% of the population, it's POSSIBLE (but not CERTAIN) that if > they'd not been under-represented, and IF they voted in the same way > that the 7% who DID respond voted, then rather than 90% agreeing to > exile, it MIGHT have been as low as 88% (a difference almost equivalent > to the error). I won't even ask you to attempt to show your work for this assinine calculation. > It was quite a surprise to discover how many niggers agreed. Following > is an email from one who does: And how do you know the race of the person posting? Because they say so? (Any bets on how long before jinny gets busted by an online sting?) James Powell === Subject: Re: Standard Deviation of PISA <1-6dnaaBov7SXjPeRVn-vA@wideopenwest.com> <4cKdnVLIfr64ii7eRVn-rw@wideopenwest.com> hy'mie whined: Yep. A poll of all 295 million Americans would have a sample error rate of less than 1%. Other error rates still apply, based upon how the poll is conducted. Gallup takes great pains to scientifically select their sample and methodology, as well as how the questions are phrased. However, I doubt the margin of error on almost every poll. Human nature being what it is. hy'mie, you're CORRECT that the sampling error for a sample size which is 0.001% of the US population, or 2,950, would be less than 1%, but this is NOT the case with a poll of everyone. The errors introduced by a poll of all 295 million Americans would make sampling errors TRIVIAL! All the other errors which would be introduced in such a monstrous government undertaking can't even be measured. But we CAN use the recent US Census Bureau Survey in 2000 as an example, because it's RIFE with errors which simple social science polls reveal EVERY DAY.(not the LEAST of which is their error in not counting jews separately from Whites, a COLOSSAL HOAX if there ever was one). You're correct as far as you went when you said Gallup takes great pains to scientifically select their sample and methodology, as well as how the questions are phrased, but you SHOULD have finished the sentence. Gallup is THE polling organization which sticks up like a pole in the tent with faggot issues. At EXACTLY the same time they were shouting from the rooftops that only 43% of Americans support laws against faggots, voters in state after state, up to 86% of them, passed EXACTLY the laws Gallup CLAIMED the voters wouldn't vote for. Yes, they ARE selective with their sample and methodology. Yes, they DO hang up on people who don't answer the first question correctly. Yes, they CAN get any result to any poll you want, if you pay them enough, just by the way the jews there weasel word the questions. What Gallup DOES do cannot be done on an internet poll, any of them, even from wiberals like vote.com Furthermore, you have YET to cite ANY poll, statistic, study, survey, or ANY other source, to support your FALSE claim that this poll is inaccurate or biased in ANY way. And the reason you HAVE not is because you CAN not, and you NEVER will, because it's dead nuts accurate, and YOU KNOW IT. eh, hy'mie? > As pointed out before, since niggers were only 7% of the respondents, > but 12% of the population, it's POSSIBLE (but not CERTAIN) that if > they'd not been under-represented, and IF they voted in the same way > that the 7% who DID respond voted, then rather than 90% agreeing to > exile, it MIGHT have been as low as 88% (a difference almost equivalent > to the error). I won't even ask you to attempt to show your work for this assinine calculation. Good. I'm glad you didn't ask. So now I can show you anyway (and I'm sure that after you confirm it with your husband and he agrees, that you'll accept it at face value). Only 7% of the poll respondents were niggers. Only 25% of them, or 1.75% of all the respondents, Strongly Disagree or Disagree with all three exile options. IF the 5% who were under-represented had voted in the same ratio as the niggers who DID vote, then another 5% x 25% or 1.25% of ALL respondents MIGHT have disagreed with all three exile options. Since 89.96% currently AGREE with or DON'T KNOW about at least one exile option, the missing link of 5% MIGHT have decreased that to as low as 88.71% had they participated, a difference which is very close to the error anyway. I rounded the 88.71% off to 88% for the same reason that I round off my tax payments so that when I get audited I get money BACK. hy'mie whined: And how do you know the race of the person posting? Because they say so? (Any bets on how long before jinny gets busted by an online sting?) It's impossible for a honkey to write like a nigger. John Knight http://blackexile.com === Subject: Re: Standard Deviation of PISA jinny pissed aloud in message > hy'mie whined: > Yep. A poll of all 295 million Americans would have a sample error rate > of > less than 1%. Other error rates still apply, based upon how the poll is > conducted. Gallup takes great pains to scientifically select their > sample > and methodology, as well as how the questions are phrased. However, I > doubt > the margin of error on almost every poll. Human nature being what it > is. > hy'mie, you're CORRECT that the sampling error for a sample size which > is 0.001% of the US population, or 2,950, would be less than 1%, but > this is NOT the case with a poll of everyone. The errors introduced > by a poll of all 295 million Americans would make sampling errors > TRIVIAL! All the other errors which would be introduced in such a > monstrous government undertaking can't even be measured. Um, no. A sample size of 2950 would have a greater sampling error than 1% unless strict, scientifically designed, selection processes were followed. A poll of the entire population would have a 0% sampling error. All other errors of such a poll can be easily quantified and controlled to make them trivial. > But we CAN use the recent US Census Bureau Survey in 2000 as an > example, because it's RIFE with errors which simple social science > polls reveal EVERY DAY.(not the LEAST of which is their error in not > counting jews separately from Whites, a COLOSSAL HOAX if there ever was > one). Yeah, right. Name one that is backed by scientific data rather than your own bemused wanting. > You're correct as far as you went when you said Gallup takes great > pains to scientifically select their sample > and methodology, as well as how the questions are phrased, but you > SHOULD have finished the sentence. Gallup is THE polling > organization which sticks up like a pole in the tent with faggot > issues. At EXACTLY the same time they were shouting from the rooftops > that only 43% of Americans support laws against faggots, voters in > state after state, up to 86% of them, passed EXACTLY the laws Gallup > CLAIMED the voters wouldn't vote for. Yes, they ARE selective with > their sample and methodology. Yes, they DO hang up on people who > don't answer the first question correctly. Yes, they CAN get any > result to any poll you want, if you pay them enough, just by the way > the jews there weasel word the questions. And, as usual, you talk without any clue. There is a difference between the surveyed people and those that turned out and voted. Did the poll ask if they would vote in such an election? Furthermore, most of those laws have been challenged and other localities voted down the same type of initiative. > What Gallup DOES do cannot be done on an internet poll, any of them, > even from wiberals like vote.com Well, a truthful comment for once. You are correct. Internet polls cannot be conducted in a way that is scientifically valid. > Furthermore, you have YET to cite ANY poll, statistic, study, survey, > or ANY other source, to support your FALSE claim that this poll is > inaccurate or biased in ANY way. And the reason you HAVE not is > because you CAN not, and you NEVER will, because it's dead nuts > accurate, and YOU KNOW IT. This poll is inaccurate by definition. Get over it. You cannot even prove that you have bonafide data and are not making it up out of the air. Can you tell me how many times I took your poll? How about Bob? Carey's results? You cannot prove that anyone actually submitted a form to your computer. > eh, hy'mie? As pointed out before, since niggers were only 7% of the respondents, but 12% of the population, it's POSSIBLE (but not CERTAIN) that if they'd not been under-represented, and IF they voted in the same way that the 7% who DID respond voted, then rather than 90% agreeing to exile, it MIGHT have been as low as 88% (a difference almost equivalent to the error). > I won't even ask you to attempt to show your work for this assinine > calculation. > Good. I'm glad you didn't ask. So now I can show you anyway (and I'm > sure that after you confirm it with your husband and he agrees, that > you'll accept it at face value). > Only 7% of the poll respondents were niggers. Only 25% of them, or > 1.75% of all the respondents, Strongly Disagree or Disagree with > all three exile options. IF the 5% who were under-represented had > voted in the same ratio as the niggers who DID vote, then another 5% x > 25% or 1.25% of ALL respondents MIGHT have disagreed with all three > exile options. Since 89.96% currently AGREE with or DON'T KNOW about > at least one exile option, the missing link of 5% MIGHT have decreased > that to as low as 88.71% had they participated, a difference which is > very close to the error anyway. I rounded the 88.71% off to 88% for > the same reason that I round off my tax payments so that when I get > audited I get money BACK. Any statisticians care to explain to the loser the number of different errors in this explanation? He never believes my explanations. > hy'mie whined: > And how do you know the race of the person posting? Because they say > so? > (Any bets on how long before jinny gets busted by an online sting?) > It's impossible for a honkey to write like a nigger. Anything you say, boss. James Powell === Subject: Re: Standard Deviation of PISA <1-6dnaaBov7SXjPeRVn-vA@wideopenwest.com> How many nigger leaders have condemned Kambon for the following call for GENOCIDE???????? Now how do I know that the white people know that we are going to come up with a solution to the problem. I know it because they have retina scans, racial profiling, DNA banks, and they're monitoring our people to try to prevent the ONE person from coming up with the ONE idea. And the one idea is, how we are going to exterminate white people because that in my estimation is the only conclusion I have come to. We have to exterminate white people off the face of the planet to solve the problem. *tepid applause* Now I don't care whether you clap or not but I'm saying to you that we need to solve this problem because they are going to kill us. And I will leave on that. So we just have to set up our own system and stop playing and get very serious and not get diverte from coming up with a solution to the problem and the problem on the planet is white people. http://christianparty.net/kamaukambon.htm John Knight === Subject: Statement of Kato's Conjecture http://news.yahoo.com/s/ap/20051227/ap_on_re_us/math_problem Steven Hofmann, Pascal Auscher, Michael Lacey, John Lewis, Alan McIntosh and Philippe Tchamitchian have written a paper proving Kato's Conjecture. Does anybody have a clear statement of what this says? -- Daryl McCullough Ithaca, NY === Subject: Re: Statement of Kato's Conjecture > http://news.yahoo.com/s/ap/20051227/ap_on_re_us/math_problem > Steven Hofmann, Pascal Auscher, Michael Lacey, John Lewis, Alan McIntosh and > Philippe Tchamitchian have written a paper proving Kato's Conjecture. Does > anybody have a clear statement of what this says? I have even attended a series of seminars given by Steve Hofmann on the subject, and I still cannot remember exactly what the conjecture says. The short answer is that both the statement and proof (especially the proof) are rather technical. The slightly bigger answer. Consider an operator on functions on R^n of the form Jf(x) = sum_{i=1}^n sum_{j=1}^n d/dxi(a_ij df/dxj) where the derivatives are partial, and the coefficients a_ij, depending on x=(x1,...,xn), are bounded and satisfy a uniform ellipticity (positive definite) type condition. By the ellipticity condition the operator has a positive square root. Conjecture: the domain of sqrt(J) is the Sobolev space H^1. I think that they first had a proof in the case n=2, or when there was some kind of extra hypothesis (maybe Gaussian bounds on semigroup generated by J?). Then they finally managed to get rid of the Gaussian bounds hypothesis. I think it is related to problems of solving heat equations on domains whose boundaries only satisfy some kind of Lipschitz condition. Anyway, if you aren't totally immersed in this subject, it will take you many years to get to the point of understanding why the problem is important. Best Stephen === Subject: Re: Statement of Kato's Conjecture On 27 Dec 2005 15:26:31 -0800, stevendaryl3016@yahoo.com (Daryl >http://news.yahoo.com/s/ap/20051227/ap_on_re_us/math_problem >Steven Hofmann, Pascal Auscher, Michael Lacey, John Lewis, Alan McIntosh and >Philippe Tchamitchian have written a paper proving Kato's Conjecture. Does >anybody have a clear statement of what this says? With Google search on Kato.89s Conjecture, I found this: Title: The solution of Kato.89s conjecture (after Auscher, Hofmann, Lacey, McIntosh and Tchamitchian) Author: Philippe Tchamitchian Abstract: Kato.89s conjecture, stating that the domain of the square root of any accretive operator ... (details snipped) has recently been obtained by Auscher, Hofmann, Lacey, McIntosh and the author. These notes present the result and explain the strategy of proof. === Subject: logarithmic differentiation - flaw? On logarithmic differentiation, how is it that a lot of people just take the ln of both sides without giving any care to check that both sides are always > 0 ? Shouldn't logarithmic differentiation be avoided? Example: y = (x^2 - 8)^|x| is not guaranteed to be positive on it's implied domain (R). Is there something deeper going on allowing people to use logarithmic differentiation on everything and anything? === Subject: Re: logarithmic differentiation - flaw? - Certainly not >On logarithmic differentiation, how is it that a lot of people just >take the ln of both sides without giving any care to check that both >sides are always > 0 ? >Shouldn't logarithmic differentiation be avoided? >Example: >y = (x^2 - 8)^|x| >is not guaranteed to be positive on it's implied domain (R). >Is there something deeper going on allowing people to use logarithmic >differentiation on everything and anything? Do =not= avoid logarithmic differentiation, but just do not apply it indiscriminately to everything and anything. Wherever a real-valued function F on the reals is differentiable and F(x) > 0, the function ln(F) is differentiable, and d(ln (F(x)) / dx = F'(x)/F(x). No strings attached: just the chain rule for differentiable functions. For analytic complex functions the same formula holds wherever F(z) <> 0. When treating real functions, just avoid areas where f(x) <= 0, or give such areas special consideration. When treating complex functions, avoid poles and zeros, and take the multi-valuedness of the ln function into account. Ciao: Johan E. Mebius === Subject: Re: logarithmic differentiation - flaw? On 27 Dec 2005 16:36:26 -0800, Kenneth Bull take the ln of both sides without giving any care to check that both >sides are always > 0 ? >Shouldn't logarithmic differentiation be avoided? >Example: >y = (x^2 - 8)^|x| >is not guaranteed to be positive on it's implied domain (R). >Is there something deeper going on allowing people to use logarithmic >differentiation on everything and anything? So far there are many replies, none including what seems to me is the right answer. So maybe what I think of as logarithmic differentiation is something different, or I'm reading the question wrong, or something. Anyway, what seems to me to be the point is that doing logarithmic differentiation doesn't necessarily have anything to do with logarithms! Let's define (*) L(f) = f'/f. So if there _is_ such a thing as log(f) then Lf is the derivative of log(f). But officially we don't worry about that, we define (*) for any non-vanishing differentiable function f. Now a reason this is useful is that (**) L(fg) = L(f) + L(g). Remembering that L(f) is _sometimes_ the derivative of log(f) is one way to remember the formula (**). But in fact there's no need for logarithms in proving (**); if f and g are both non-vanishing differentiable functions then (**) is immediate, just from the product rule. Then by induction (**) implies a formula just like (**) except with n factors. This is very convenient for doing some calculations. For example if p(z) = c * product (z-z_j) (c a non-zero constant) is a complex polynomial then it's clear that p'/p = sum 1/(z-z_j) (without worry about complex logarithms.) ************************ David C. Ullrich === Subject: Re: logarithmic differentiation - flaw? |On 27 Dec 2005 16:36:26 -0800, Kenneth Bull | |>On logarithmic differentiation, how is it that a lot of people just |>take the ln of both sides without giving any care to check that |>both sides are always > 0 ? ... |So far there are many replies, none including what seems to me is the |right answer. So maybe what I think of as logarithmic |differentiation is something different, or I'm reading the question |wrong, or something. | |Anyway, what seems to me to be the point is that doing logarithmic |differentiation doesn't necessarily have anything to do with |logarithms! Let's define | |(*) L(f) = f'/f. | |So if there _is_ such a thing as log(f) then Lf is the derivative of |log(f). But officially we don't worry about that, we define (*) for |any non-vanishing differentiable function f. | |Now a reason this is useful is that | |(**) L(fg) = L(f) + L(g). furthermore, L has an interesting conceptual interpretation as follows: in real life, the idea of growth rate is ambiguous depending on whether the amount of growth is compared to an absolute standard (the stock's value grew at a rate of 3 cents per day) or a relative standard (the stock's value grew at a rate of 2% per day). ordinary differentiation is a formalization of the first kind of growth rate while logarithmic differentiation is a formalization of the second kind. this is pretty clear just from staring at the formula: f'/f says to take the absolute growth rate f', but then to compare it to f, the size of the thing whose growth is being measured. the advantage of using the relative growth rate, that it can be meaningful even when you have no idea of the value of the currency in which the stock's price is given, is often explained in stock market for dummies books under the heading of why logarithmic scale is good for representing stock prices. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: logarithmic differentiation - flaw? > On logarithmic differentiation, how is it that a lot of people just > take the ln of both sides without giving any care to check that both > sides are always > 0 ? > Shouldn't logarithmic differentiation be avoided? > Example: > y = (x^2 - 8)^|x| > is not guaranteed to be positive on it's implied domain (R). The implied domain is |x| >= sqrt(8), not all of R. On the interior of this domain, the function is positive. > Is there something deeper going on allowing people to use logarithmic > differentiation on everything and anything? I'm not sure what you're referring to. === Subject: Re: logarithmic differentiation - flaw? take the ln of both sides without giving any care to check that both > sides are always > 0 ? > Shouldn't logarithmic differentiation be avoided? > Example: > y = (x^2 - 8)^|x| > is not guaranteed to be positive on it's implied domain (R). > The implied domain is |x| >= sqrt(8), not all of R. On the > interior of this domain, the function is positive. > Is there something deeper going on allowing people to use logarithmic > differentiation on everything and anything? > I'm not sure what you're referring to. Referring to all the tutorials I've seen that explain logarithmic differentiation as take the ln of both sides, and differentiate implicity, solve for dy/dx How can we take the ln of both sides, when ln (although one-on-one) is only defined for positive arguments. both sides might be negative, yet I still see logarithmic differentiation explained without this warning I can see how it works for something like: y = sqrt(x) + x^8 Because both sides are always positive, no matter what x is. But for other equations, we can't assume this. === Subject: Re: logarithmic differentiation - flaw? y = sqrt(x) + x^8 > Because both sides are always positive, no matter what x is. Really? Even when x = -1? --- Christopher Heckman === Subject: Re: logarithmic differentiation - flaw? On logarithmic differentiation, how is it that a lot of people just take the ln of both sides without giving any care to check that both sides are always > 0 ? Shouldn't logarithmic differentiation be avoided? Example: y = (x^2 - 8)^|x| is not guaranteed to be positive on it's implied domain (R). The implied domain is |x| >= sqrt(8), not all of R. On the interior of this domain, the function is positive. Is there something deeper going on allowing people to use logarithmic differentiation on everything and anything? I'm not sure what you're referring to. >Referring to all the tutorials I've seen that explain logarithmic >differentiation as take the ln of both sides, and differentiate >implicity, solve for dy/dx >How can we take the ln of both sides, when ln (although one-on-one) is >only defined for positive arguments. both sides might be negative, >yet I still see logarithmic differentiation explained without this >warning If you used complex numbers, the problem would go away. If you insist on using the reals, then any quantity like f(x)^(g(x)) with g(x) not an integer is undefined when f(x) < 0. On the other hand, if you want to apply logarithmic differentiation to a product such as y(x) = f(x) g(x) h(x), you can handle any negative factors: take absolute values before taking the logarithm, so ln |y(x)| = ln |f(x)| + ln |g(x)| + ln |h(x)| and since d/dt ln |t| = 1/t, you get y'(x)/y(x) = f'(x)/f(x) + g'(x)/g(x) + h'(x)/h(x) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: logarithmic differentiation - flaw? > On logarithmic differentiation, how is it that a lot of people just > take the ln of both sides without giving any care to check that both > sides are always > 0 ? > Shouldn't logarithmic differentiation be avoided? > Example: > y = (x^2 - 8)^|x| > is not guaranteed to be positive on it's implied domain (R). > The implied domain is |x| >= sqrt(8), not all of R. On the > interior of this domain, the function is positive. > Is there something deeper going on allowing people to use logarithmic > differentiation on everything and anything? > I'm not sure what you're referring to. > Referring to all the tutorials I've seen that explain logarithmic > differentiation as take the ln of both sides, and differentiate > implicity, solve for dy/dx > How can we take the ln of both sides, when ln (although one-on-one) is > only defined for positive arguments. both sides might be negative, > yet I still see logarithmic differentiation explained without this > warning We're still waiting for an example. The example you gave above is faulty, as you incorrectly identified the implied domain as all of R. State an exercise from a tutorial or textbook that has the student taking the log of a function that is negative at points of the domain. === Subject: Re: logarithmic differentiation - flaw? :> How can we take the ln of both sides, when ln (although one-on-one) is :> only defined for positive arguments. both sides might be negative, :> yet I still see logarithmic differentiation explained without this :> warning : We're still waiting for an example. The example you gave above is : faulty, as you incorrectly identified the implied domain as all : of R. State an exercise from a tutorial or textbook that has the : student taking the log of a function that is negative at points : of the domain. Some (many?) textbooks encourage the use of logarithmic differentiation for functions like f(x) = (x-2)^5 * (x+1)^4 / ((x+1)^7 (x-3)^8), or with even more terms in the numerator and denominator. These functions can be negative. Ted === Subject: Re: logarithmic differentiation - flaw? : differentiation as take the ln of both sides, and differentiate : implicity, solve for dy/dx : How can we take the ln of both sides, when ln (although one-on-one) is : only defined for positive arguments. both sides might be negative, : yet I still see logarithmic differentiation explained without this : warning The reason that logarithmic differentation does not produce nonsense results, in spite of what you say above, is because the function ln x can be extended to a large portion of the complex plane, including the negative numbers. Its derivative remains 1/x. Therefore, when we take ln of both sides, we are still performing a legitimate operation even if both sides not necessarily positive. That wasn't a completely precise explanation, but a common case for logarithmic differentation can be checked directly. This is the case where f is a product and/or quotient of various terms, such as f(x) = (x-2)^7 * x^5 / ((x+2)^3 (x-1)^2). Let's just check the case y = f*g. If you do logarithmic differentiation on this, you get ln y = ln f + ln g y'/y = f'/f + g'/g y' = (f'/f + g'/g)*(fg) y' = f'g + fg' The result is correct (by the product rule). Ted === Subject: Re: logarithmic differentiation - flaw? : implicity, solve for dy/dx > : How can we take the ln of both sides, when ln (although one-on-one) is > : only defined for positive arguments. both sides might be negative, > : yet I still see logarithmic differentiation explained without this > : warning > The reason that logarithmic differentation does not produce nonsense > results, in spite of what you say above, is because the function ln x > can be extended to a large portion of the complex plane, including the > negative numbers. Its derivative remains 1/x. Therefore, when we take > ln of both sides, we are still performing a legitimate operation even > if both sides not necessarily positive. Okay, I needed to hear this to actually get confidence in log differentiation. Just a side question, does ln 0 also have meaning in advanced mathematics? === Subject: Re: logarithmic differentiation - flaw? :> The reason that logarithmic differentation does not produce nonsense :> results, in spite of what you say above, is because the function ln x :> can be extended to a large portion of the complex plane, including the :> negative numbers. Its derivative remains 1/x. Therefore, when we take :> ln of both sides, we are still performing a legitimate operation even :> if both sides not necessarily positive. : Okay, I needed to hear this to actually get confidence in log : differentiation. Just a side question, does ln 0 also have meaning in : advanced mathematics? No, it does not (e^z is never 0). However, even if a function has zeroes, logarithmic differentiation still works if the zeroes are isolated, because one can differentiate on intervals where the function has constant sign and then piece the results together. An important point: This argument using the complex logarithm only works under certain conditions, for example, if the function you are differentiating extends to an analytic function on an open subset of the complex plane. In other words, even though you're only interested in differentiating the function over the real numbers, it must actually be defined and differentiable for (most) complex numbers. For polynomials, rational functions, trig functions, exponentials, logarithms, etc. etc. (i.e. just about any 'common' function that you encounter in an ordinary calculus course), this is no problem. For other cases, you may be able to use the product rule argument I gave before. So logarithmic differentiation is a pretty robust technique. Ted === Subject: Re: Differentiation under the integral sign > Let (X,T,m) be a measured space, R the real numbers, E a normed vector > space, f: X x E -> R a measurable function such that: > 0) x->f(x,e) is integrable for every e in E > 1) u -> f(x,u) is differentiable on an open set O C E, for almost all x > in X > 2) For all u in O, |df(x,u)/du| <= g(x) for almost all x, where g is > integrable on X. By |df(x,u)/du| I take it you mean the norm of the continuous linear functional df(x,u)/du? > Does it follow that I(u)= int(f(x,u) dx) is differentiable and > dI(u).h=int(d_2f(x,u).h dx) > where d_2 is the differentiation with > respect to x ? You've changed notation and made a typo. But I think the answer is yes. I'll use the simpler notation D in place of d/du, and | | for norms on the underlying spaces. First note that for each u in O, h -> int Df(x,u)[h] dx is a bounded linear functional on E of norm at most int g. Fix u in O. It's enough to show int |f(x,u+h) - f(x,u) - Df(x,u)[h]|/|h| dx -> 0 as h -> 0. Certainly the integrand -> 0 pointwise. If we show that the integrand is bounded by 2g(x) a.e. for all small h, we'll be done by dominated convergence. By the ordinary MVT, for each small nonzero h in E there is a c_h in (0,1) such that f(x,u+h) - f(x,u) = Df(x,u+c_h*h)[h]. The integrand therefore equals | Df(x,u+c_h*h)[h/|h|] - Df(x,u)[h/|h|] | <= |Df(x,u+c_h*h)| + |Df(x,u)| <= 2g(x) a.e., by hyypothesis. === Subject: Re: Differentiation under the integral sign > Let (X,T,m) be a measured space, R the real numbers, E a normed vector > space, f: X x E -> R a measurable function such that: > 0) x->f(x,e) is integrable for every e in E > 1) u -> f(x,u) is differentiable on an open set O C E, for almost all x > in X > 2) For all u in O, |df(x,u)/du| <= g(x) for almost all x, where g is > integrable on X. > Does it follow that I(u)= int(f(x,u) dx) is differentiable and > dI(u).h=int(d_2f(x,u).h dx), where d_2 is the differentiation with > respect to x ? > Put in a nutshell, does the usual theorem of differentiation under the > Lebesgue integral sign still hold when the parameter belongs to a > general normed vector space ? > The proof I have in the case E = R uses a quotient and the mean value > theorem, which cannot be generalized to higher dimensions, that's why > I'm wondering. > -- > Julien Santini Fix u and for t a real number let g(x,t) =f(x,u+tv) ,then dg(x,t)/dt|_(t=0) =(df(x,u)/du).v reducing the problem to the real case .smn === Subject: Re: Differentiation under the integral sign > Let (X,T,m) be a measured space, R the real numbers, E a normed vector > space, f: X x E -> R a measurable function such that: > 0) x->f(x,e) is integrable for every e in E > 1) u -> f(x,u) is differentiable on an open set O C E, for almost all x > in X > 2) For all u in O, |df(x,u)/du| <= g(x) for almost all x, where g is > integrable on X. > Does it follow that I(u)= int(f(x,u) dx) is differentiable and > dI(u).h=int(d_2f(x,u).h dx), where d_2 is the differentiation with > respect to x ? > Put in a nutshell, does the usual theorem of differentiation under the > Lebesgue integral sign still hold when the parameter belongs to a > general normed vector space ? > The proof I have in the case E = R uses a quotient and the mean value > theorem, which cannot be generalized to higher dimensions, that's why > I'm wondering. > -- > Julien Santini > Fix u and for t a real number let g(x,t) =f(x,u+tv) ,then > dg(x,t)/dt|_(t=0) =(df(x,u)/du).v reducing the problem to the real > case .smn differentiable. This involves more than just directional derivatives. === Subject: Re: Differentiation under the integral sign Let (X,T,m) be a measured space, R the real numbers, E a normed vector > space, f: X x E -> R a measurable function such that: > 0) x->f(x,e) is integrable for every e in E > 1) u -> f(x,u) is differentiable on an open set O C E, for almost all x > in X > 2) For all u in O, |df(x,u)/du| <= g(x) for almost all x, where g is > integrable on X. > Does it follow that I(u)= int(f(x,u) dx) is differentiable and > dI(u).h=int(d_2f(x,u).h dx), where d_2 is the differentiation with > respect to x ? > Put in a nutshell, does the usual theorem of differentiation under the > Lebesgue integral sign still hold when the parameter belongs to a > general normed vector space ? > The proof I have in the case E = R uses a quotient and the mean value > theorem, which cannot be generalized to higher dimensions, that's why > I'm wondering. > > -- > Julien Santini > Fix u and for t a real number let g(x,t) =f(x,u+tv) ,then > dg(x,t)/dt|_(t=0) =(df(x,u)/du).v reducing the problem to the real > case .smn > differentiable. This involves more than just directional > derivatives. Yes of course you are right and have given the details very nicely in the next post.I should also have mentioned for G(x,t) =f(x,u+tv) that dG(x,t)/dt =df/du(x,u+tv).v (a very useful fact) so as you worked out f(x,u+v)-f(x,u)= G(x,1)-G(x,0)=df/du(x,u+t*u).v by the ordinary mean value theorem giving the required upper bound to apply dominated convergence.Looking at it now,I guess it does look pretty similar to the case u is a real variable except you divide by |v| instead of just v . Nice job .smn === Subject: fitting data hi everyone i have this data (38,37) (88,63) (138,134) (188,182) (238,233) (288,268) (338,325) (388,360) (438,424) I made a linear fit to this data to obtain the next point i've been told it is (488,471) but when i fit it i obtain (488,467) y = 0.969x - 5.5109 R(squared)= 0.9957 is there any way i can fit this data to make it result (488,471) === Subject: Re: fitting data >i have this data >(38,37) >(88,63) >(138,134) >(188,182) >(238,233) >(288,268) >(338,325) >(388,360) >(438,424) >I made a linear fit to this data to obtain the next point i've been told it is >(488,471) >but when i fit it i obtain (488,467) >y = 0.969x - 5.5109 >R(squared)= 0.9957 >is there any way i can fit this data to make it result (488,471) So you have calculated a least squares best fit of a straight line to your data. Your data does not fit exactly on a straight line, and the best fit line does not exactly hit any of your data points, nor would you expect it to. So why would you expect it to be exact on the next point. --Lynn === Subject: Re: fitting data Consider the data (x_k,y_k) , k in {1,2,...,n}. In your case n=9 , x_1=38, y_1=37 ,...,x_9=438 ,y_9=424. Select arbitrary positive numbers w_1,w_2,...,w_n such that w_1+w_2+...+w_n = 1 . Let a:= SUM_{k=1 to k=n} w_k*x_k b:= SUM_{k=1 to k=n} w_k*y_k P:= SUM_{k=1 to k=n} w_k*x_k*y_k D:= SUM_{k=1 to k=n} w_k*(x_k-a)^2 . In this case the linear fit is (#) y:=L(x)= b + P(x-a)/D . Because until now w_1,w_2,..., w_n are positive free-parameters subject only to w_1+w_2+...+w_n=1 , try to find a system of such weights W:=(w_1,w_2,...,w_n) having the property [see (#) ] L(488) = 467 . Note: In your case , when w_1=w_2=...=w_9=1/9 , you have a=238 , b=225.1111111 . Shortly,you will work with weighted- means [e.g., see a:= ] instead of arithmetic-means. === Subject: Re: Another question about derivatives >There is something wrong with your proof.The vital islet a->0, then >divide by b, then let b->0 to conclude f'(0)=0.It's wrong,I believe. > It's not clear to me which proof you're referring to, He's referring to Rouben's proof, which has a flaw, although not the one mentioned above. === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> Find a diagonal of a box with length =10, width=7, and height 2. > It is easy to generalize the distance formula, or metric in R^n. > For R^3 it is d^2= (x2-x1)^2+(y2-y1)^2+(z2-z1)^2 > For the numbers you gave you have (x1,y1,z1)=(0,0,0) and > (x2,y2,z2)=(10,7,2); order does not matter in this case. Take the > square root of 10^2+7^2+2^2. find it easily and would have to read a bunch of Math that I didn't need and so here I am ..giving some people a chance to utilize their judgemental ability:)- === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> I remembered seeing it somewhere yesterday but I can't remember where. >Google serach done just now, didn't help. Do you know by heart? > Well, see if you can generalize it from the 2 dimensional case. > Do you know the formula for the distance between 2 points in the > plane? > To add to this: this 2D distance is the diagonal of a rectangle. > The diagonal of a rectangle forms the hypotenuse of a right > triangle whose sides are the length and width of the > rectangle. >Do you remember the relationship between the sides of a right triangle? Sure, it's 7th grade Geometry back home. Will never froget. But, I was looking for short cut formula. === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> Well, see if you can generalize it from the 2 dimensional case. > Do you know the formula for the distance between 2 points in the > plane? > To add to this: this 2D distance is the diagonal of a rectangle. > The diagonal of a rectangle forms the hypotenuse of a right > triangle whose sides are the length and width of the > rectangle. >Do you remember the relationship between the sides of a right triangle? > Sure, it's 7th grade Geometry back home. Will never froget. But, I was > looking for short cut formula. I do not know the poster, and mean no personal diatribe against him/her. However, the last quoted sentence shows what is wrong with education in the U.S. today (and perhaps elsewhere). It also shows what is wrong with the instant gratification generation. Rather than thinking about the problem, the poster asks for a formula and a short cut. I suggest to this person that this attitude is wrong. Not all problems have a short cut answer. Part of educating oneself is learning how to solve problems that do NOT have a shortcut answer. === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> Well, see if you can generalize it from the 2 dimensional case. > Do you know the formula for the distance between 2 points in the > plane? > To add to this: this 2D distance is the diagonal of a rectangle. > The diagonal of a rectangle forms the hypotenuse of a right > triangle whose sides are the length and width of the > rectangle. >Do you remember the relationship between the sides of a right triangle? > Sure, it's 7th grade Geometry back home. Will never froget. But, I was > looking for short cut formula. > I do not know the poster, and mean no personal diatribe against > him/her. > However, the last quoted sentence shows what is wrong with education > in > the U.S. today (and perhaps elsewhere). It also shows what is wrong > with the > instant gratification generation. > Rather than thinking about the problem, the poster asks for a > formula and > a short cut. I suggest to this person that this attitude is wrong. Yeah..yeah..yeah..you didn't try to find out what this poster's goal (and immediate need) was before making this comment, did you? > Not all > problems have a short cut answer. Part of educating oneself is > learning how > to solve problems that do NOT have a shortcut answer. === Subject: Re: Finding a diagonal of a box Find a diagonal of a box with length =10, width=7, and height 2. do you know the distance formula in R^3? I remembered seeing it somewhere yesterday but I can't remember where. Google serach done just now, didn't help. Do you know by heart? Well, see if you can generalize it from the 2 dimensional case. Do you know the formula for the distance between 2 points in the plane? To add to this: this 2D distance is the diagonal of a rectangle. The diagonal of a rectangle forms the hypotenuse of a right triangle whose sides are the length and width of the rectangle. Do you remember the relationship between the sides of a right triangle? >Sure, it's 7th grade Geometry back home. Will never froget. But, I was >looking for short cut formula. If you try drawing a picture of the box, you'll see that the diagonal of the box is the hypotenuse of a right triangle whose sides are a height of the box and a diagonal of the base rectangle. However, the distance formula is the shortcut you seek -- it applies the pythagorean theorem for you, giving you just the distance. quasi === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> Do you remember the relationship between the sides of a right triangle? >Sure, it's 7th grade Geometry back home. Will never froget. But, I was >looking for short cut formula. > If you try drawing a picture of the box, you'll see that the diagonal > of the box is the hypotenuse of a right triangle whose sides are a > height of the box and a diagonal of the base rectangle. > However, the distance formula is the shortcut you seek -- it applies > the pythagorean theorem for you, giving you just the distance. Back home, while the Text/ teacher concentrated in triangles, it failed to give problems with the box and hence I didn't remember the formula easily. === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> of the box is the hypotenuse of a right triangle whose sides are a > height of the box and a diagonal of the base rectangle. > However, the distance formula is the shortcut you seek -- it applies > the pythagorean theorem for you, giving you just the distance. The original poster might find the results of the following search _very_ helpful, if I haven't angered him to the point where he doesn't read this post. Dave L. Renfro === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> of the box is the hypotenuse of a right triangle whose sides are a > height of the box and a diagonal of the base rectangle. > However, the distance formula is the shortcut you seek -- it applies > the pythagorean theorem for you, giving you just the distance. > The original poster might find the results of the following > search _very_ helpful, if I haven't angered him to the point > where he doesn't read this post. > Dave L. Renfro === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> Then drop the course. === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> Are you talking about y=mx+b? >While all this is very interesting, I really don't have time to become >a mathematician. Can I ask what is your future course of study? In what field? quasi === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> Find a diagonal of a box with length =10, width=7, and height 2. do you know the distance formula in R^3? I remembered seeing it somewhere yesterday but I can't remember where. Google serach done just now, didn't help. Do you know by heart? Well, see if you can generalize it from the 2 dimensional case. Do you know the formula for the distance between 2 points in the plane? >Are you talking about y=mx+b? >While all this is very interesting, I really don't have time to become >a mathematician. > Can I ask what is your future course of study? In what field? > quasi Sorry, I was away from PC the whole day. My plan of study is Bioinformatics. That's the combination of Programing + Biology mainly with some Statistics, specifically, Bayesian which I will take once I get to grad school. I had taken Intro to Probability a while back but it's kind of stale right now and will review that later. Currently, after GRE, needs to get knowledge in Cell and Molecular Biology (will take a course at a 2 year college in Spring) and Biochemistry (on my own basically - I am good in Chemsitry - since this is junior level course and I can't register for it as a non-student). Got enough programming knowledge but will concentrate on Perl (on my own) + take Unix programming next semester at a 2 year college. Then, hopefully will be moving out of this town in Fall to start grad program. === Subject: Re: Finding a diagonal of a box Find a diagonal of a box with length =10, width=7, and height 2. do you know the distance formula in R^3? I remembered seeing it somewhere yesterday but I can't remember where. >Google serach done just now, didn't help. Do you know by heart? Well, see if you can generalize it from the 2 dimensional case. Do you know the formula for the distance between 2 points in the > plane? Are you talking about y=mx+b? While all this is very interesting, I really don't have time to become a mathematician. Can I ask what is your future course of study? In what field? quasi >Sorry, I was away from PC the whole day. >My plan of study is Bioinformatics. That's the combination of >Programing + Biology mainly with some Statistics, specifically, >Bayesian which I will take once I get to grad school. I had taken >Intro to Probability a while back but it's kind of stale right now I see that. >and will review that later. Well, if you have time, rather than trying to learn piecemeal by looking at selected problems, I would find a text on introductory probability and review the basics. I think it's also advisable to review a text on intermediate algebra or precalculus. > Currently, after GRE, needs to get knowledge in Cell and Molecular >Biology (will take a course at a 2 year college in Spring) and >Biochemistry (on my own basically - I am good in Chemsitry - since this >is junior level course and I can't register for it as a non-student). >Got enough programming knowledge but will concentrate on Perl (on my >own) + take Unix programming next semester at a 2 year college. >Then, hopefully will be moving out of this town in Fall to start grad >program. I really think my suggestion is the right way, namely -- spend a few weeks with some texts. quasi. === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> <1m74r1potl4t9o31r5nlc2em0jt9ot34oo@4ax.com> [..] > I really think my suggestion is the right way, namely -- spend a few > weeks with some texts. > quasi. I decided to join grad program only at the beginning of December and was planning to sepnd time on GRE preparation right after the final exam period but I got sick - have a medical condition - and lost 2 1/2weeks. So when trying some problems w/o reviewing any material closely, I ended up posting questions here. === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> Are you talking about y=mx+b? While all this is very interesting, I really don't have time to become a mathematician. > Can I ask what is your future course of study? > In what field? I'm beginning to think Usenet trolling. His/her responses are so off the mark and so incongruous that it's hard for me to believe they're real. For example, where did y=mx+b come from? Is the poster really serious about all this being interesting? (Let alone the fact that nothing has been discussed, so one is left scratching their head wondering what all this refers to.) And not having time to become a mathematician, what's that supposed to mean? Is he/she saying they could become a mathematician if they had more time? I think there's more to it than that, and trying to explain that you're too busy to spend an hour or two looking at something (which could very well be the case, by the way) by saying you don't have time to do something which takes about a decade of your life to accomplish just doesn't make sense. Dave L. Renfro === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> Are you talking about y=mx+b? While all this is very interesting, I really don't have time to become a mathematician. > Can I ask what is your future course of study? > In what field? > I'm beginning to think Usenet trolling. His/her > responses are so off the mark and so incongruous > that it's hard for me to believe they're real. > For example, where did y=mx+b come from? Is the > poster really serious about all this being interesting? > (Let alone the fact that nothing has been discussed, > so one is left scratching their head wondering what > all this refers to.) And not having time to become > a mathematician, what's that supposed to mean? > Is he/she saying they could become a mathematician if > they had more time? Do you think you are the only one who can be a mathematician? Anyway, I don't have time for judgemental people saying blah..blah.. >I think there's more to it than > that, and trying to explain that you're too busy > to spend an hour or two looking at something (which > could very well be the case, by the way) by saying > you don't have time to do something which takes > about a decade of your life to accomplish just > doesn't make sense. Sensitive, aren't you? > Dave L. Renfro === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> do you know the distance formula in R^3? > I remembered seeing it somewhere yesterday > but I can't remember where. Google serach done > just now, didn't help. Do you know by heart? Are you kidding?? Googling just about anything at all related to what you asked about should work. For example, the search gives these two web pages among the top three hits: http://www.horacemann.pvt.k12.ny.us/academics/math/pcbch/vect/3d.html http://www.mathwords.com/d/distance_formula.htm Unless you just learned about the internet last week or something, I find it hard to believe that you find the distance formula for three dimensions. Dave L. Renfro === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> do you know the distance formula in R^3? > I remembered seeing it somewhere yesterday > but I can't remember where. Google serach done > just now, didn't help. Do you know by heart? > Are you kidding?? Googling just about anything at > all related to what you asked about should work. > For example, the search Well, we didn't use English back home and so excuse me for not coming up with the term distance formual and 3 domension when doing goole search. I simply put formula for diagonal for a box and got links with heavy duty explanation. === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> Sure I got many hits too but what's your point? === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> do you know the distance formula in R^3? > I remembered seeing it somewhere yesterday > but I can't remember where. Google serach done > just now, didn't help. Do you know by heart? Are you kidding?? Googling just about anything at all related to what you asked about should work. For example, the search gives these two web pages among the top three hits: http://www.horacemann.pvt.k12.ny.us/academics/math/pcbch/vect/3d.html http://www.mathwords.com/d/distance_formula.htm Unless you just learned about the internet last week or something, I find it hard to believe that you find the distance formula for three dimensions. > Sure I got many hits too but what's your point? Quasi asked if you knew the distance formula for three dimensional space. You didn't answer the question (which was, did you know the formula), but instead went on to say you saw it didn't give you the formula. I pointed out that googling the most obvious thing gives many hits with the distance formula. Not only did I say this, I proved it by giving you the URL for the search I used and the #1 and #3 hits for that search (both of which give the distance formula for R^3). Did you even bother to look at the web pages I gave? I note that while you claimed to have conducted a search and come up empty, you didn't provide any evidence to back this up, so for all we know you might just be saying you did a search but actually didn't do one. It also occurs to me that maybe you saw the distance formula in some web pages but didn't know how to apply it to the specific problem you were asking about. However, if this was the case, you need to tell us. We can't read your mind. This part falls under the things I have in mind in the last sentence of the next paragraph, by the way. I know this sounds a little testy, but I was hoping you'd get the hint that if you want someone to put any effort towards helping you, then you don't want to come across as if you spent all of 30 seconds thinking about the problem and 10 more seconds writing up a request for help. Many people in this group are, or have been, college professors, and so they're more than willing to help out if someone appears as if they really want to learn. However, you'll find that the amount and quality of help you get goes down a lot if you come across as someone who doesn't care. Dave L. Renfro === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com[..] > It also occurs to me that maybe you saw the distance formula > in some web pages but didn't know how to apply it to the specific > problem you were asking about. However, if this was the case, > you need to tell us. We can't read your mind. This part falls > under the things I have in mind in the last sentence of the > next paragraph, by the way. > I know this sounds a little testy, but I was hoping you'd get the > hint that if you want someone to put any effort towards helping > you, then you don't want to come across as if you spent all of > 30 seconds thinking about the problem and 10 more seconds writing > up a request for help. Quasi knew that that was what I was doing because from the beginning he was aware that I was practising for GRE. > Many people in this group are, or have been, > college professors, and so they're more than willing to help Good. I was a part-adjunct at a 2 year college temporarily too but don't like to bring the work home. > out if someone appears as if they really want to learn. Frankly, I'd rther sepnd time learning Perl or System programming but got to take GRE, what can one do? >However, you'll > find that the amount and quality of help you get goes down a lot > if you come across as someone who doesn't care. > Dave L. Renfro Okay, I didn't mean to offend anyone in case. Got a little amused last night with someone telling me to drop t he copurse, etc. ANyway, === Subject: Re: Finding a diagonal of a box <4mh1r1l1ok1kdpscttq5v3vqtpeditq78f@4ax.com> do you know the distance formula in R^3? > I remembered seeing it somewhere yesterday > but I can't remember where. Google serach done > just now, didn't help. Do you know by heart? Are you kidding?? Googling just about anything at all related to what you asked about should work. For example, the search gives these two web pages among the top three hits: http://www.horacemann.pvt.k12.ny.us/academics/math/pcbch/vect/3d.html http://www.mathwords.com/d/distance_formula.htm Unless you just learned about the internet last week or something, I find it hard to believe that you find the distance formula for three dimensions. > Sure I got many hits too but what's your point? > Quasi asked if you knew the distance formula for three > dimensional space. > You didn't answer the question (which was, did you know > the formula), Excuse me for being in a rush and not spelling it out which implied I didn't remember at that moment. >but instead went on to say you saw it > didn't give you the formula. > I pointed out that googling the most obvious thing gives many hits > with the distance formula. Not only did I say this, I proved it by > giving you the URL for the search I used and the #1 and #3 hits for > that search (both of which give the distance formula for R^3). Yes you did and I appreciated that but I was too tired to write too much after I replied to quasi. > Did you even bother to look at the web pages I gave? Of course I did but like I said, I was too tired to write too much after I replied to quasi. >I note that while > you claimed to have conducted a search and come up empty, you > didn't provide any evidence to back this up, I didn't know that I was at trial needing to ptrovide proof of my claim. >so for all we know > you might just be saying you did a search but actually didn't > do one. Assuming hat I am a 20-22 year old college kid, right? > It also occurs to me that maybe you saw the distance formula > in some web pages but didn't know how to apply it to the specific > problem you were asking about. No, I already applied it and got the answer correctly but I forgot to write the formula down in my note. However, if this was the case, > you need to tell us. We can't read your mind. This part falls > under the things I have in mind in the last sentence of the > next paragraph, by the way. > I know this sounds a little testy, but I was hoping you'd get the > hint that if you want someone to put any effort towards helping > you, then you don't want to come across as if you spent all of > 30 seconds thinking about the problem and 10 more seconds writing > up a request for help. Many people in this group are, or have been, > college professors, and so they're more than willing to help out > if someone appears as if they really want to learn. However, you'll > find that the amount and quality of help you get goes down a lot > if you come across as someone who doesn't care. > Dave L. Renfro === Subject: Announcement: Jakarta Commons Math 1.1 Released The Jakarta Commons Math team is pleased to announce the release of Commons Math 1.1. Commons Math is a Java library of lightweight, self-contained mathematics and statistics components. For more information, see the Commons Math web site: http://jakarta.apache.org/commons/math/. The new release contains bug fixes and enhancements. A full list of changes since the 1.0 release can be found in the change log at http://jakarta.apache.org/commons/math/changes-report.html. Commons Math is available in either binary or source form from the Commons Math downloads page on the Apache mirrors: http://jakarta.apache.org/site/downloads/downloads_commons-math.cgi. The commons-math 1.1 release jar has also been deployed to the Apache Maven repository: http://www.apache.org/dist/java-repository/ and the Maven main repository: http://www.ibiblio.org/maven/. Please remember to verify the signatures of the files you download using the keys available on the download page. Jakarta Commons welcomes community participation and contributions from all interested parties. User feedback or questions related to Commons Math should be directed to the commons-user mailing list. Development-related topics are discussed on the commons-dev list. See the Commons mailing list page for instructions on how to subscribe to or view the archives of these lists. Please start the subject line of math-related posts to either of these lists with [math]. To submit patches or bug reports, follow the directions on this page: http://jakarta.apache.org/commons/math/issue-tracking.html -The Commons Math Team === Subject: Re: a game -- piles of stones Given an initial pile of n stones. 2 players alternate turns. At each turn, the player must choose one of the remaining piles of stones and split it into 2 or 3 piles (not necessarily of equal size). Since stones are not breakable, any pile of size 1 can't be split, and hence survives intact for the rest of the game. If there are no piles that can be split, the player whose turn it is loses. For which values of n is the game a win for player 1? >All n >= 2. For those values of n, specify an optimal strategy. >Split the pile into two equal piles or two equal piles + one single >stone. Then copy the opponent's moves. Ok, Robert -- you nailed that one almost instantly. Here's a variation, much harder, I think. Same basic rules, except for the options for replacing piles (In this game, piles get replaced, not split). Here's the game: Given an initial pile of n stones, n>=2. 2 players alternate turns. At each turn, the player must choose exactly one of the remaining piles of stones, let's say a pile of m stones, and replace the pile of m stones by 2 piles with a stones and b stones respectively, where a*b=m and aGiven an initial pile of n stones. >2 players alternate turns. >At each turn, the player must choose one of the remaining piles of >stones and split it into 2 or 3 piles (not necessarily of equal size). >Since stones are not breakable, any pile of size 1 can't be split, and >hence survives intact for the rest of the game. If there are no piles >that can be split, the player whose turn it is loses. >For which values of n is the game a win for player 1? All n >= 2. >For those values of n, specify an optimal strategy. Split the pile into two equal piles or two equal piles + one single stone. Then copy the opponent's moves. >Ok, Robert -- you nailed that one almost instantly. >Here's a variation, much harder, I think. >Same basic rules, except for the options for replacing piles (In this >game, piles get replaced, not split). >Here's the game: >Given an initial pile of n stones, n>=2. >2 players alternate turns. >At each turn, the player must choose exactly one of the remaining >piles of stones, let's say a pile of m stones, and replace the pile of >m stones by 2 piles with a stones and b stones respectively, where >a*b=m and aIf a player can't move, the player loses. >Who wins, player 1 or playe 2? >I'll start it off with the most obvious case: >If n is prime, player 2 wins. >The general case is open for everyone. >As a test case for a particular n, who wins if n = 1000? >quasi Ignore this game -- too easy -- I posed it without much thought. There's no strategy. The game is an automatic win or loss depending on whether the total number of prime factors (not necessarily distinct) in the prime factorization of n is odd or even. Back to the drawing board. I'll return with a harder variation when I think of one. quasi === Subject: Re: a game -- piles of stones : Ignore this game -- too easy -- I posed it without much thought. : There's no strategy. The game is an automatic win or loss depending on : whether the total number of prime factors (not necessarily distinct) : in the prime factorization of n is odd or even. : Back to the drawing board. : I'll return with a harder variation when I think of one. Google for octal games -- these include lots of games of the type you are describing (splitting and removing piles of stones), and many of them are quite difficult to solve. The general technique is Sprague-Grundy theory, but it's not always easy to see a pattern in the values. Ted === Subject: Re: a game -- piles of stones >: Ignore this game -- too easy -- I posed it without much thought. >: There's no strategy. The game is an automatic win or loss depending on >: whether the total number of prime factors (not necessarily distinct) >: in the prime factorization of n is odd or even. >: Back to the drawing board. >: I'll return with a harder variation when I think of one. >Google for octal games -- these include lots of games of the type you >are describing (splitting and removing piles of stones), and many of them >are quite difficult to solve. The general technique is Sprague-Grundy >theory, but it's not always easy to see a pattern in the values. >Ted Actually, I have a method in mind which seems somewhat deep, but it's a method looking for a game. It's likely though that all I've done is rediscovered some known method, perhaps related to what you describe above. quasi === Subject: all inclusive proof of power rule I'd appreciate it if anyone could point me to a proof of the power rule for all real exponents. I have tried looking for some on the internet, but haven't found a general power rule proof. By power rule I mean: d x^n = n x^(n-1) for any n /in R dx === Subject: Re: all inclusive proof of power rule >I'd appreciate it if anyone could point me to a proof of the power rule >for all real exponents. I have tried looking for some on the internet, >but haven't found a general power rule proof. >By power rule I mean: >d x^n = n x^(n-1) for any n /in R If x is non-zero, one can prove it for positive and negative integers n by the power rule and the rule for reciprocals. For x = 0, only for positive n. For other n, there is a problem for x < 0; there is no agreement on what x^n is. Also, for x = 0, x^n is not defined for n < 0 in any formulation. For x > 0, one can use Newton's expansion for (1+u)^n, |u| < 1; this can be proved by using the basic properties of real numbers. Or one can use the monotonicity of powers, increasing for x > 1 and decreasing for x < 1, to get the result firts for rational n, and then by passing to the limit for all real n. It is just using the definition and limiting arguments. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: all inclusive proof of power rule On 27 Dec 2005 20:12:29 -0800, R. Colacitti for all real exponents. I have tried looking for some on the internet, >but haven't found a general power rule proof. >By power rule I mean: >d x^n = n x^(n-1) for any n /in R Before you can prove this you need a _definition_ of x^n for real n. If x > 0 the usual definition is x^n [ exp(n*log(x)), from which the formula for the derivative follows easily. ************************ David C. Ullrich === Subject: Re: all inclusive proof of power rule > Before you can prove this you need a _definition_ > of x^n for real n. If x > 0 the usual definition is > x^n [ exp(n*log(x)) Shhh ... they aren't supposed to know about [ until they join the Pythagoras Society. === Subject: Re: all inclusive proof of power rule May be derivating a product d/dx x * x * x * x * ..x = 1*x*x*x* ..+ x*1*x*x* ...+ x*x*1*x*x... n times = (1+1+.....)*(x*x... ) = n*x^n At school we ,of yore, started from: ( x^n - a^n) / (x -a ) = x^(n-1) + a*x^(n-2) + ......a^(n-1) as a -->x = n*x^(n-1) ... Happy New year !! Alain. === Subject: Re: all inclusive proof of power rule > I'd appreciate it if anyone could point me to a proof of the power rule > for all real exponents. I have tried looking for some on the internet, > but haven't found a general power rule proof. > By power rule I mean: > d x^n = n x^(n-1) for any n /in R > dx Let y =x^n Then ln(y) =n ln(x) Then (1/y)dy/dx=n/x Then dy/dx=y*(n/x) But y=x^n So dy/dx=(x^n)*(n/x)=n x^(n-1) === Subject: Re: all inclusive proof of power rule > I'd appreciate it if anyone could point me to a proof of the power rule > for all real exponents. I have tried looking for some on the internet, > but haven't found a general power rule proof. It's as easy as rolling off a log. ;-) > By power rule I mean: > d x^n = n x^(n-1) for any n /in R > dx and x restriced to positive reals. d/dx x^n = d/dx e^(n.log x) = e^(n.log x) d/dx (n.log x) = x^n n/x = n.x^(n-1) === Subject: Tables of basic Derivation and Integration formulae If you could help me provide the complete Tables of basic Derivation and Integration formulae in Maths. === Subject: Re: Tables of basic Derivation and Integration formulae > If you could help me provide the complete Tables of basic Derivation > and Integration formulae in Maths. have a look at this link: http://www.sosmath.com/tables/tables.html Maybe that helps. Ciao Karl === Subject: Re: velocity problem in two dimensions >Ok, let's say I have a ship in space, but let's only deal with a >two-dimensional plane. A spaceship, I mean, not the Love Boat. >Here's the given information about the ship: >current position: x1, y1 >destination: x2, y2 >current heading: h >current velocity: v0 >maximum possible velocity: v(max) [because of the ship's structural >integrity, say] >maximum possible acceleration: a(max) >maximum possible deceleration: d(max) [a different set of engines, >less powerful, to decelerate] >maximum rate of turn: rt(max) >maximum rate of turn acceleration (and deceleration): rta(max) >If the destination is too close to the current position, then the ship >will obviously not make it to its maximum possible velocity (v(max)). >Sorry, I should say speed because I'm not dealing with vectors. >So, what's the formula to determine the maximum velocity the ship may >reach, limited by v(max), in order to reach its destination?! I tried >for days to figure this out once but my math skills are too rusty and I >can't find an answer online. Start with a simpler problem and work your way up. Assume h is the correct heading to get from your current position to v(max) and d(max), you can calculate the time and distance covered in decelerating to a stop. If the sum of the two distances does not exceed the distance to the destination, the answer is v(max). If the sum does, you want the intermediate velocity, v(int), such that the sum of the distance covered accelerating to v(int) and the distance covered decelerating to a stop equals the distance to the destination. The problem where h is not the correct heading is more complicated. For example, is it better to accelerate while turning or turn first and then accelerate (or possibly even decelerate while turning). Or maybe accelerate once the current heading is within some tolerance of h. < Ok, let's say I have a ship in space, but let's only deal with a > two-dimensional plane. A spaceship, I mean, not the Love Boat. > Here's the given information about the ship: > current position: x1, y1 > destination: x2, y2 > current heading: h > current velocity: v0 > maximum possible velocity: v(max) [because of the ship's structural > integrity, say] > maximum possible acceleration: a(max) > maximum possible deceleration: d(max) [a different set of engines, > less powerful, to decelerate] > maximum rate of turn: rt(max) > maximum rate of turn acceleration (and deceleration): rta(max) > If the destination is too close to the current position, then the ship > will obviously not make it to its maximum possible velocity (v(max)). > Sorry, I should say speed because I'm not dealing with vectors. > So, what's the formula to determine the maximum velocity the ship may > reach, limited by v(max), in order to reach its destination?! I tried > for days to figure this out once but my math skills are too rusty and I > can't find an answer online. This is a problem in optimal control. The math associated with such problems is quite hefty, and goes well beyond introductory calculus. Problems like this have been studied in great detail, and many of the solutions to simple cases have likely been published; I would bet that some of the solutions to more detailed and realistic versions remain unpublished, due to secrecy concerns. Try the aerospace or optimal control literature. Question: does your ship have an infinite fuel supply? If not, then modelling fuel loss, maximum fuel capacity, and change of mass through fuel burn will be imortant. A heavier ship, full of fuel, will be more sluggish; a lighter one, with an almost-empty fuel tank will be more agile! R.G. Vickson > Scott McMasters === Subject: What is the norm of a gaussian variable? Given a random vector X in an n-dimensional space. Assume that X has a normal distribution in that space with known mean vector M and covariance matrix C. What is the expected norm-square of X, i.e. E(X'.X) where X' is the transpose of X and E is expectation operator? Thnx --Hossein === Subject: Re: What is the norm of a gaussian variable? >Given a random vector X in an n-dimensional space. Assume that X has a >normal distribution in that space with known mean vector M and >covariance matrix C. What is the expected norm-square of X, i.e. >E(X'.X) where X' is the transpose of X and E is expectation operator? By definition the covariance C = E[(X - M)(X - M)'] = E[X X'] - M M' and X' X = trace(X' X) = trace(X X'), so E[X' X] = trace(C) + M' M. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: What is the norm of a gaussian variable? Given a random vector X in an n-dimensional space. Assume that X has a normal distribution in that space with known mean vector M and covariance matrix C. What is the expected norm-square of X, i.e. E(X'.X) where X' is the transpose of X and E is expectation operator? >By definition the covariance C = E[(X - M)(X - M)'] = E[X X'] - M M' >and X' X = trace(X' X) = trace(X X'), so >E[X' X] = trace(C) + M' M. Also seen from X'.X = sum(i, (X_i)^2): The expectation is sum(i, Var(X_i) + (EX_i)^2). Note that normality is needed nowhere. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: an open ball <251220052112188403%bruck@math.usc.edu> X is a subset of R^n such that if x and y are two arbitrary points in X, then there exists an open ball included in X that contains x and y. Prove that X is an open ball. R is the set of real numbers R^n itself satisfies the hypothesis and is _not_ an open ball ... > so you'll need to impose further conditions on your X. If X is bounded and nonempty, the claim is true. This is an interesting argument. My gut tells me it should be true in >any Banach space (perhaps even any NLS), but of course you can't say >the centers Ci can be assumed to converge in that case. Does anyone have a proof or counterexample in a B-space? This proof works in any Banach space: Let d = diam(X). Thus for any n, there exist points x_n, y_n > in X with ||x_n - y_n|| > d-1/n, and thus z_n and r_n such that > the ball B_{r_n}(z_n) = {w: ||w - z_n|| < r_n} contains x_n and y_n > and is contained in X. In particular, since > diam(B_{r_n}(z_n}) = 2 r_n, d-1/n <= 2 r_n <= d. > Now for any m and n, > d > diam(B_{r_n}(z_n) union B_{r_m}(z_m)) >= r_m + r_n + ||z_n - z_m|| > so ||z_n - z_m|| < (1/n + 1/m)/2. In particular, {z_n} is a Cauchy > sequence, and converges to some z, and X contains B_{d/2}(z). Moreover, > if x is not in B_{d/2}(z) we have ||x - y|| > d for some y in B_{d/2}(z), > so x can't be in X. Thus X = B_{d/2}(z). The result isn't true in incomplete normed linear spaces. Namely, if > V is such a space let Z be the completion of V, z a member of Z V, > and consider X = B_1(z) intersect V. Since V is dense in Z it's > easy to see that X has eugene's property. Note also that the result for unbounded X (being the whole space or a half-space) does not extend to arbitrary Banach spaces (even finite-dimensional ones). Consider e.g. R^2 with the sup norm ||(x,y)|| = max(|x|,|y|). Then X = {(x,y): x > 0, y > 0} has the property. Question: what additonal property of the Banach space would make the result on unbounded X true? The inversion trick should work in a Hilbert space. But, for example, what about a uniformly convex Banach space? >Noting that this is just a wild-assed guess: That doesn't seem >like the right condition to me. In the counterexample the >straight lines in the boundary of the ball are not the problem, >the problem is the _corner_. So I'd conjecture the condition >is something like the existence of a unique supporting >hyperplane at each point of the boundary of the unit ball >(ie a unique norming bounded linear functional.) Yes I think you're right. > That necessary and sufficient, at least in the finite-dimensional > case (good of you to put off posting a proof for a few hours, > heh. You can do the infinite-dimensional case...) I can, eh? Well, I tried. But I think I need to assume uniform convexity too for the hard part. > First the easy half. Suppose that ||x_0||=1, L_1 and L_2 > are distince linear functionals of norm 1 and > L_1(x_0) = L_2(x_0) = 1. Let > E = union_{r > 0} r B(x_0, 1). > The fact that L_1 > 0 on E shows that E <> X, the > fact that L_2 is also > 0 on E and L_1 and L_2 are > independent shows that E is not a half-space. And > it's easy to see that 0 < r_1 < r_2 implies that > r_1 B(x_0,1) is contained in r_2 B(x_0,1), which > implies that E is eugene. That part is OK in any normed linear space. > Now suppose that X is finite dimensional, every > point of the boundary of the unit ball has a > unique norming functional, E is eugene, E is > unbounded and E <> X. Same without the finite-dimensional, but assuming X is a uniformly convex Banach space... > Lemma. If 0 < r_1 < r_2 and p is a boundary > point of B(x, r_2) then there exists y such > that p is a boundary point of B(y, r_1) > and B(y, r_1) is contained in B(x, r_2). > (Hint: Let y = p + (r_1/r_2)(x-p).) Also OK in any Banach space. > Now suppose wlog that 0 is a boundary point of E. > E is certainly convex, and open. > so by the Hahn-Banach theorem > there is a linear functional L of norm 1 such that > L > 0 on E. > Now, there are arbitrarily large balls contained > in E that contain points approaching 0. Using > the lemma we see that there are balls of any > size we like containing points approaching 0, > and now by compactness (applied for each > fixed r > 0) we see that for every r > 0 there > exists x_r in E such that B(x_r, r) is contained > in E and 0 is a boundary point of B(x_r, r) > (ie ||x_r|| = 1.) You mean ||x_r|| = r. OK, so I have to do something different here. I'll use uniform convexity. For every r > 0 and any positive integer n there is x = x_{r,n} such that B(x, r) is contained in E and ||x|| < r + 1/n. Moreover, L(x) >= r (otherwise there would be points y of B(x, r) with L(y) < 0). By uniform convexity, for any epsilon > 0 there is delta > 0 such that if ||u|| <= 1, ||v|| <= 1 and ||u - v|| > epsilon, then ||(u + v)/2|| < 1-delta. Consider u = x_{r,n}/(r + 1/n) and v = x_{r,m}/(r + 1/n) with m > n, and note that if n is sufficiently large ||(u+v)/2|| >= L((u + v)/2) >= r/(r + 1/n) > 1 - delta. Therefore for n sufficiently large, ||u - v|| <= epsilon, i.e. ||x_{r,m} - x_{r,n}|| < (r + 1/n) epsilon, and {x_{r,n}} is a Cauchy sequence. If x_r is its limit, we have B(x_r, r) contained in E, and 0 is a boundary point of B(x_r, r) (i.e. ||x_r|| = r). Moreover, L(x_r) = r. So L is the unique linear functional of norm 1 such that L(x_r) = r, while (again by the uniform convexity) x_r is the only x with ||x|| = r and L(x) = r. But this implies that x_r = r x_1. > Suppose that E is not equal to {x : Lx > 0}. > Then again by the Hahn-Banach theorem there > exists L' <> L such that L' has norm 1 and > L' > c on E. > Since ||L'|| = 1 it follows that > L'(x_r) - r >= c. > Since ||x_r|| = r it follows that |L'(x_r)| <= r. So > 1 >= L'(x_r/r) >= 1 + c/r. > Let r -> infinity, and let z be an accumulation > point of (x_r/r) as r -> infinity. Then |z| = 1 > and L'(z) = 1. But the same argument shows that > L(z) = 1, contradicting the fact that there is > a unique norming functional at z. Since x_r/r = x_1, this still works. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth > ... > No, it does not prevent that. You are free to label the numbers created > by the Peano axioms as you wish. A common misconception is that the > Peano axioms define arithmetic. They do not. > ... Okay, then when in the process of knowing whether 0,5,10,15,..... is the Natural Numbers or whether 0,1,2,3, .... are the Natural Numbers, when does that moment of understanding suppose to take place? Not after reading the Peano Axioms but days later when addition, sequence, function are defined? Dik, I think the very biggest problem in all of this is the fact that every human knows the Counting Numbers much better than any formalization could ever try to axiomatize. If children were first taught the Peano axioms before ever doing any numbers, they would never get it. > It is best to think of the Peano axioms as defining a sequence (and *not* > a sequence of numbers). When there is an element, there is a successor. > There is a base element that is not the successor of any other element. > And if two elements have the same successor they are the same (that is, > no element may have more than one predecessor). > And see this. No mention of 0 at all, I call it the base element. Okay, I accept that for it is in keeping with your insistence that this is Set theory. Okay. But where does Set theory have Successor? Is not Sequence and successor already that of arithmetic. Does not set theory have very primitive concepts such as membership, union, intersection, but no concept of successor nor a concept of sequence. Is not 0 arithmetic and not set theory for if it was set theory should that not be the empty set rather than a number of 0. With #5, if that is set theory then the empty set seldom if ever holds for any property, and 0 is involved there are many cases in which Math Induction is true if you start with 1 and not 0. And thus already you have arithmetic involved here. > The Fibonacci numbers F_i (with i >= 0, and i finite) are conformant to > #1 to #5. > And see this. Okay, I looked up this sequence to be 1,1,2,3,5,8,13,..... Okay, if you put into the #1 that 0 and 1 exists and change #2 to read that Successor is the endless adding of the metric of 0 to 1, then the Fibonacci no longer contends for the Natural Numbers and in fact is a unique for #1 to #5. So where does the Fibonacci get thrown out as not the Natural Numbers? > being gone has made Chris Heckman go heywire. > Well, if you welcome me, you better also read what I write. You know reading and understanding what the other is saying is two different things. > Anyway, I doubt it is a mere labeling. For if it were mere labeling > then you would not even need 0 existence postulate. > The 0 postulate does *not* postulate a number '0' with the properties > as you know them. It only postulates a starting point for a sequence. But I believe that is not enough. That the axioms for the Natural Numbers needs a distance metric or parameter that guages the spacing between the Natural Numbers in their sequence. Surely a sequence has a concept of spacing. And this is what is missing in the Peano axioms. That the spacing from one Natural Number to the next is all the same and derived from an axiom. Peano's axioms do not eliminate the sequence 0,5,10,15,20.... as the Natural Numbers and should be able to eliminate every contender except 0,1,2,3..... Axiom systems can be abstract, but not so abstract that they cease to clarify. > I contend that > unless you have the 1 existence alongside the 0, the Naturals could be > 0,5,10,15, ..... just as well as 0,1,2,3,..... > call the Fibonacci numbers the naturals according to the axioms. This is a flaw in the Peano Axioms and there must be a inclusion of 1 in #1 and a clarification in #2 that the Successor is this metric of the existence of 0 and 1 in #1. Physicists reading this thread would instantly sympathize with me, because they have had a long history of knowing that existence usually comes in pairings and that mathematicians who have no physics background would easily fall into such a trap of thinking that a postulate of 0 exists is sufficient when in fact it is deficient. all that when mathematicians doing pure mathematics of axiomatizing the Natural Numbers would get it wrong and flawed. > And in the Successor postulate which we know is tantamount to Endless > adding of 1, > Nope. Adding is *not* involved. Addition has not even been defined yet, > so how can you say that it is tantamount to adding 1? Dik, tell me where Successor is defined in Set theory or even a part of set theory. Tell me where sequence is innate and inherent in Set theory? Can you have a sequence or successor without addition? I think you are quibbling over nothing. > So if it were mere labeling then you would dispense with even that of 0 > existing and just start with Successor. > You also have to start with a starting point for the sequence, and that > is what the postulating of 0 does do. Yes I accept that and agree with that. But you cannot have a Successor with only one point existing. You need at least 2 points to have a Successor. Take for example tape-rulers that carpenters and builders use. There is a meter rule, a centimeter rule, a mm ruler. You have to have 2 marks to mark a meter or centimeter or mm. You have to have a mark of 0 and a mark of 1 meter or cm or mm. You cannot have a ruler with the existence of only one mark, the 0 mark. Each of those rulers is a ruler of the Natural Numbers, but they do not exist unless you have a postulate claiming the existence of two marks. > So none of the other axioms eliminates the series 0,5,10,15,.... as the > Natural Numbers except for if you postulate 0 and 1 both exist and that > this parameter of a distance of 0 to 1 is what the Successor function > applies. > But that series is *not* eleminated. I know even of formulations of the > Peano axioms where the naturals are 1, 2, 3, 4, ..., see, no 0 involved. > The 0 postulate in what you did learn is the 1 postulate in that system. Dik, tell me whether Successor is a sequence or a function. Tell me if you can have a sequence or function or successor without the concept of addition. Seems to me that the Fibonacci sequence is built from the addition of previous two numbers. As far as I know, set theory does not even have Successor but that Arithmetic has successor in the form of sequence and function and so the Peano Axioms already have addition. - Hide quoted text - - Show quoted text - > Without such of 1 existing postulated then 0,5,10,15, ...... are the > Natural Numbers according to Peano axioms. > Indeed, with different labels. Once you start defining arithmetic with > the Peano axioms in mind you will find that the arithmetic rules become > quite different. Let me define addition and multiplication (both > recursive, and with 0 and 1 based naturals in mind), and assume predecessor > and successor functions: > 0 based: > op + = (number a, b) if a = 0 then b else pred(a) + succ(b). > op * = (number a, b) if a = 0 then 0 else pred(a) * b + b. > 1 based > op + = (number a, b) if a = 1 then succ(b) else pred(a) + succ(b). > op * = (number a, b) if a = 1 then b else pred(a) * b + b. > Using this definitions and your sequecne 0, 5, 10, 15, ..., I find the > following rules: > if the sequence is 0 based: > 5 * 10 = 10 > if the sequence is 1 based: > 5 * 10 = 30. > Note that the *arithmetic* is not different, it is the *labelling* that > is different. > And I think that you will find in N. Bourbaki (and in a lot of French > mathematics) the naturals are defined as starting with 1. Which (when > you know about the history of mathematics) makes much sense. Of course > in anglo-saxon mathematics you can, when you have the naturals, > immediately define all the integers. In Bourbaki arithmetic you have > first to introduce '0'. > Dik, could you please comment on this question. Given the Adics, > whether a specific p-adic or all the adics unioned, if you apply > Peano's axiom #3 and #4 that 0 has no predeccessor and successors are > unique to the Adics. What would you have left once you applied #3 and > #4? > You first have to define the successor function. Once you have done > that you can define arithmetic. When you have defined arithmetic you > can define a metric, and once you have defined a metric you can find > the difference between the different p-adics etc. There are many > systems that conform to all of #1 to #4. The p-adics do not, as does > the set of all integers (with the successor function being adding 1). > Because all of them have a predecessor of 0. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Dik, can you prove this statement: If a given set of numbers is built from the Successor function of endless adding of 1 such as the p-adic integers, then that set is Countable. In other words, Counting is equivalent to a Successor function of endless adding of 1. If that is provable, which I think it is easily proveable then the Reals are Countable. --- end quoting my post of yesterday --- The old thread was getting too long, over 130 posts, and it is cumbersome. So start a new thread. In a sense, what Dik Winter is saying is that Peano axioms are flawed and it is known that they are flawed for a long time, since we end up with a labelling issue to sort out whether the Natural Numbers are 1,1,2,3,5,8,.... or 0,5,10,15,20,.... or 0,1,2,3,...... For what I gather of Dik's reply is that we cannot know which of those above three sets is the Natural Numbers until we define addition and other things and then the labeling condenses the contenders and leaves us with the set 0,1,2,3.... If that is a fair interpretation of what Dik is saying? But I would contend that Dik and the rest of the math community should have done something about that because it is a flaw and the flaw is easily remedied by saying two numbers exist in postulate #1 where both 0 and 1 exists. And in postulate #2 use the fact of 0 and 1 existing as the successor is the endless adding of the metric of 1-0 = a spacing distance of 1. Thus, label away. Because the revised Peano axioms thus answer all questions immediately. Now Dik contends that the Peano Axioms have no arithmetic and that they are set theory. But I would argue with him that Successor is not a set theory concept. I would argue that Successor is chock full of addition for it is a concept of sequence and sequence is way deep inside arithmetic. So this stance that Peano axioms has no arithmetic, no addition seems rather untenable because Successor is arithmetic and it is unclear what Successor is unless you clarify it by saying it is the endless adding of 1, begot from the previous postulate where you have the existence of 0 and 1. Now I checked the history of Peano to see when he did his Natural Numbers Axioms and I get a date of 1889 and I was interested in knowing when Set Theory came into history and I get a date of 1874 by Cantor. So could Peano have modeled his axioms upon the newly arrived Cantor set theory? That is a space of 15 years? And would Peano have been confident that the concepts of set theory, paltry and paucity as they are-- membership, union, intersection, could they have been enough for Peano to place his new axioms? Was the concept of Successor in 1889 part of set theory? Or was the concept of Successor not to be found anywhere in mathematics and that Peano had simply taken a common language term and filled it in his #2 and #5 postulates. It would be interesting for math historians to check to see where Peano derived the concept of Successor. I am confident that successor was not in set theory in 1889 or ever. But simply a term drawn out of the air. Maybe Dik is correct in that we should simply ignore the flaws and errors of Peano Axioms and never alter, change or improve them. Thinking about this issue, I believe I have struck a near-perfect analogy of what a Mathematical Axiom System is to the rest of the world. The Linnaeus biology classification of species is what a math axiom system is. I remember a cat in Linnaeus classification in that it has all the other definitions of the classification and what separates the cat species is retractable claws. The unique feature. I forgot the unique feature of humans compared to apes in the Linnaeus classification-- perhaps opposable thumb. Anyway, the Peano Axioms is to the Numbers of 0,1,2,3,4..... what Linnaeus classification in biology is to Homo sapiens. A list of what is unique to our species compared to all the other animals-- backbone, warm-blooded, etc etc. So instead of postulate #1 we have backbone, and instead of postulate #2 we have warm-blooded etc etc. Now where this analogy really gets interesting is that we all know Linnaeus Classification is not as good as a species DNA. So we can on one hand enumerate the Linnaeus Classification differences and on the other hand simple hold out a beaker of human DNA and say this is our species. Likewise in mathematics, we can enumerate Peano Axioms as describing Natural Numbers, or on the other hand we simply write out 0,1,2,3,4,.... and say These are the Natural Numbers. See how easy that was. But the interesting thing is that the Axioms are often wrong and flawed and gap ridden with error, whereas saying 0,1,2,3,..... are the Natural Numbers is indubitable. Likewise, Linnaeus Classifications are prone to error, whereas decoding the DNA of a species and holding up the DNA and saying this is Homo sapiens is error free. And I mean, well, the Natural Numbers were known as 1,2,3,.... before human history, whereas the axiomatics of Natural Numbers is 1889. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth ... > No, it does not prevent that. You are free to label the numbers > created by the Peano axioms as you wish. A common misconception > is that the Peano axioms define arithmetic. They do not. > ... > Okay, then when in the process of knowing whether 0,5,10,15,..... is > the Natural Numbers or whether 0,1,2,3, .... are the Natural Numbers, > when does that moment of understanding suppose to take place? As Chris alluded to, which of the following are the natural numbers: sifiri, d'aya, biyu, uku, hud'u, biyar,... nul, een, twee, drie, vier, vyf, ses, sewe, ... sefr, pandzh, dah, ponzdah, bist, ... and how do you decide which of the following two are the natural numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ... 0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, ... ? And if both the last can be the natural numbers, why can: 0, 5, 10, 15, 20, 25, 30, 35, ... *not* be the natural numbers? > Not after > reading the Peano Axioms but days later when addition, sequence, > function are defined? Not even at that stage. You are still confusing labels (the letters by which we describe something) with numbers (more abstract ideas). With 0, 5, 10, 15, ... as natural numbers we get the following multiplication (depending on whether we view the axioms as 0-based or 1 based): 0-based: 5 * 5 = 5 1-based: 5 * 5 = 30. Even when you use the sequence 0, 1, 2, 3, ... there is a difference between whether you see the axoims as 0-based, or 1-based. In the latter case you get: 0 + 1 = 2 and 1 * 1 = 3. Note, that these things are only labels. You should *not* confuse them with actual numbers. > Dik, I think the very biggest problem in all of this is the fact that > every human knows the Counting Numbers much better than any > formalization could ever try to axiomatize. If children were first > taught the Peano axioms before ever doing any numbers, they would never > get it. You do not seem to get it. The axioms are *not* meant to be taught to children to teach them arithmetic. They are the result of the research of mathematicians to the smallest number of axioms that can be used to define arithmetic such that it conforms to standard usage. Everything else can be done by using definitions, and when you apply the proper labels to your objects you get the standard arithmetic. I wonder that you completed in any way your mathematical education. > And see this. No mention of 0 at all, I call it the base element. > Okay, I accept that for it is in keeping with your insistence that this > is Set theory. Okay. But where does Set theory have Successor? Is not > Sequence and successor already that of arithmetic. No, why should it? > Does not set theory > have very primitive concepts such as membership, union, intersection, > but no concept of successor nor a concept of sequence. Indeed, so when there *is* a successor function on the set, the Peano axioms describe how you create a sequence from them. And they define how to create a sequence with a starting point but no terminating point. > Is not 0 > arithmetic and not set theory for if it was set theory should that not > be the empty set rather than a number of 0. No. In this case 0 is purely notational and denotes some member of the set (what I called the base element). There is *nothing* arithmetic about it. > With #5, if that is set theory then the empty set seldom if ever holds > for any property, and 0 is involved there are many cases in which Math > Induction is true if you start with 1 and not 0. And thus already you > have arithmetic involved here. You simply do not understand. The empty set does not conform to the Peano axioms, as there is no base element. And indeed, every finite set can not conform to the Peano axioms, as there is either a base element and an element with no successor, or there is no base element. In a set with successor function there is *no* last element if it is to conform to the Peano axioms. But no arithmetic involved at all. > The Fibonacci numbers F_i (with i >= 0, and i finite) are conformant > to #1 to #5. (Should have been, as Chris noted, starting with the second 1.) > Okay, if you put into the #1 that 0 and 1 exists and change #2 to read > that Successor is the endless adding of the metric of 0 to 1, then the > Fibonacci no longer contends for the Natural Numbers and in fact is a > unique for #1 to #5. So where does the Fibonacci get thrown out as not > the Natural Numbers? But I do not understand any of this. > Well, if you welcome me, you better also read what I write. > You know reading and understanding what the other is saying is two > different things. But I think you do not even read large parts of what I write. > Anyway, I doubt it is a mere labeling. For if it were mere labeling > then you would not even need 0 existence postulate. > The 0 postulate does *not* postulate a number '0' with the properties > as you know them. It only postulates a starting point for a sequence. > But I believe that is not enough. That the axioms for the Natural > Numbers needs a distance metric or parameter that guages the spacing > between the Natural Numbers in their sequence. Surely a sequence has a > concept of spacing. Why do you think so? The only concept a sequence has is that one follows the other. > Axiom systems can be abstract, but not so abstract that they cease to > clarify. Axioms are not there to clarify, they are there to mention those things that can not be proven within a system. > Nope. Adding is *not* involved. Addition has not even been defined yet, > so how can you say that it is tantamount to adding 1? > Dik, tell me where Successor is defined in Set theory or even a part of > set theory. Tell me where sequence is innate and inherent in Set > theory? Can you have a sequence or successor without addition? I think > you are quibbling over nothing. Oh, well. Let me have the ordered set (aap, noot, mies, wim, zus, jet, teun, vuur, gijs, lam, kees, bok, weide, does, hok, duif, schapen), where the successor is defined as shown. What arithmetic is involved when a child is asked to tell the successor of noot? There are generations of people in the Netherlands that learned reading with that sequence. Where is the arithmetic? Set theory does not define a successor function, the Peano axioms require you to define one. And if you define one it has to have some particular properties in order to be able to conform to the axioms. > You also have to start with a starting point for the sequence, and that > is what the postulating of 0 does do. > Yes I accept that and agree with that. But you cannot have a Successor > with only one point existing. You need at least 2 points to have a > Successor. Eh? If you have a set of only one element: {foobar} you can create the successor function: successor(foobar) = foobar a perfectly valid successor function. But it does not conform to the Peano axioms. > But that series is *not* eleminated. I know even of formulations of the > Peano axioms where the naturals are 1, 2, 3, 4, ..., see, no 0 involved. > The 0 postulate in what you did learn is the 1 postulate in that system. > Dik, tell me whether Successor is a sequence or a function. As I call it the successor function on occasion, the answer should be clear. > As far as I know, set theory does not > even have Successor but that Arithmetic has successor in the form of > sequence and function and so the Peano Axioms already have addition. You are wrong. > Dik, can you prove this statement: > If a given set of numbers is built from the Successor function of > endless adding of 1 such as the p-adic integers, then that set is > Countable. In other words, Counting is equivalent to a Successor > function of endless adding of 1. If that is provable, which I think it > is easily proveable then the Reals are Countable. You have to define what endless adding of 1 means. With the definitions I know, the p-adic integers are *not* obtained by endless adding of 1. And with (what I surmise is) your understanding, being countable is *not* equivalent to endless adding 1. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth children to teach them arithmetic. They are the result of the research > of mathematicians to the smallest number of axioms that can be used to > define arithmetic such that it conforms to standard usage. Everything > else can be done by using definitions, and when you apply the proper > labels to your objects you get the standard arithmetic. And the smallest number of axioms to distinguish the Naturals from any other contender is to include the existence of 1 in #1 and the inclusion that the Successor is the same as adding 1 in #2. I get the feeling that your labeling is base representation. Tell me what the Peano axioms deliver if you include 1 in #1 and replace Successor with endless adding of 1? It is a more powerful axiom set. > > If a given set of numbers is built from the Successor function of > endless adding of 1 such as the p-adic integers, then that set is > Countable. In other words, Counting is equivalent to a Successor > function of endless adding of 1. If that is provable, which I think it > is easily proveable then the Reals are Countable. > You have to define what endless adding of 1 means. With the definitions > I know, the p-adic integers are *not* obtained by endless adding of 1. > And with (what I surmise is) your understanding, being countable is *not* > equivalent to endless adding 1. A webpages lists the definition of p-adic integer as a power series. Later on it shows that a p-adic integer is a sequence of sums and explains these two are equivalent. So are we just quibbling, in that P-adic-Integers are derived as a series of endless adding of 1. And that the Peano axiom of Successor is equivalent to series or sequence. And the Successor is the only creator axiom of the Natural Numbers so that Successor of either series or sequence is equivalent to Counting. The Natural Numbers are Countable because of the Successor. Thus the P-adics-Integers are built in the same way and thus they are Countable. Now since the P-adic-Integers are equinumerous to the Reals then the Reals are a countable set. Seems fairly clear and straightforward to me. And of course beautiful. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: the axioms of Eucl geom has addition and 0,1, and infinity Re: axiom systems in math is like Linnaeus classification in biology children to teach them arithmetic. They are the result of the research > of mathematicians to the smallest number of axioms that can be used to > define arithmetic such that it conforms to standard usage. Everything > else can be done by using definitions, and when you apply the proper > labels to your objects you get the standard arithmetic. I disagree with Dik and Chris on the issue of whether Peano axioms are deficient and flawed. I did make a mistake. The mistake is that Peano not only missed including the existence of 1 alongside the existence of 0 but forgot to include the existence of infinity. As the old Peano axioms do not tell us whether the Successor is finite or infinite. I believe most if not all of Dik's and Chris's argument is centered around the fact that Peano axioms are base independent, which Dik calls label. Base independence is not concerning the flaws of Peano. The flaws are that Peano needs to include the existence of 1 and infinity. Also, there is another storm issue involving Field theory in mathematics in that the old notion is that arithmetic is separate from axiom systems and that you define things of arithmetic and a metric separate from the axioms such as Peano or Euclid axioms. This is flawed and erroneous because Field theory overstepped its domain. Field theory is so flawed that it is incapable of even recognizing that Riemann geom and Lobachevsky geom do not obey commutative, associative and transitive properties, yet every modern day mathematicians except myself do not understand that. They do not understand that Field theory has been falsely applied. But that is another issue or another thread. Dik seems to believe that arithmetic is defined on a axiom set after the axioms are established. But that is a falsity because we only need examine a different axiom system in mathematics and compare it to the Peano axioms. Let us take Euclid axioms for Eucl geom. Now if Dik is correct there should not be any arithmetic in those axioms and arithmetic defined later. But we immediately see there is the number 0 and 1 and also 2 and infinity. The 2 points determine a line. The number of lines parallel to a given line is 0 in Riem geom and infinity in Loba geom and 1 in Eucl geom. Please pardon my mistake Dik, in that Peano is missing not only 1 to go alongside 0 but is missing infinity for the Peano Axioms fail to address the issue that Natural Numbers are infinite set. So you have to postulate the existence of infinity in the Peano axioms in the first postulate by stating the existence of three numbers not just 0. And in the second postulate you have to incorporate those three numbers into the Successor function. Dik, a question, why is base representation seem to come into action only once we postulate the existence of the number 3 and beyond. In other words if the world had only the numbers 0,1,2 and infinity, the 2-adics, that everything would be base independent. But once you enter 3 and beyond, we have to consider the base. Why is that Dik? Is it because once 3 enters that we have to have an algebra? Just curious? But back to the main issue of the flaws of Peano axioms. There is not another axiom system and it is impossible to have a mathematical axiom system without having a postulate concerning the existence of three numbers 0,1, infinity. And since you have those three numbers existing in any axiom system of mathematics that you automatically have a addition and multiplication already embedded within those axioms because 0,1,infinity are the additive and multiplicative inverses and thus addition already exists. Despite the objection of Dik or Chris. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth Note, that these things are only labels. You should *not* confuse them > with actual numbers. [...] This is one of AP's recurring errors, mistaking the map for the territory, the menu for the food, etc. > You do not seem to get it. The axioms are *not* meant to be taught to > children to teach them arithmetic. They are the result of the research > of mathematicians to the smallest number of axioms that can be used to > define arithmetic such that it conforms to standard usage. Everything > else can be done by using definitions, and when you apply the proper > labels to your objects you get the standard arithmetic. I was going to add a thread about this tonight, about how the Peano Axioms are about the minimum amount of structure on a set needed to turn it into the natural numbers. (Also, the set of axioms itself is minimal: If you remove any of the conditions, you let in something which cannot be turned into the natural numbers.) Now looking at AP's axioms --- > 1. There is a natural number 0. > 2. There is a natural number 1. > 3. There is a successor series of endless adding of 1 > 4. The number 0 has no predecessor. --- There is a lot of baggage here that isn't immediately obvious. First AP talks about adding without saying which properties it's supposed to have. Clearly, he wants (N,+) to be a monoid (closed, associtiave, neutral element), commutative, and he evidently is also assuming that 1 + 1 + 1 + ... + 1 is never 0, for any number of 1's. Having a binary operation with all of these properties is more baggage than a successor function with two or three basic properties. Something more surprising is that condition (1) in AP's axioms can be removed (!), with the alterning of the properties of + which are needed. (Note that if + is closed on {1,2,3,...}, then 0 cannot be n + 1, for any natural number n, so (4) is removed as well.) The reason for this has to do with how negative numbers are formalized in Peano Arithmetic: The set of ordered pairs (a,b) is considered, where a and b are allowed to be any natural numbers, and (a,b) is equivalent to (c,d) if a + d = b + c. In Peano Arithmetic, this defines -1, as (0, S(0)) = (S(0), S(S(0))) = etc., 0 as (0, 0), (S(0),S(0)) = etc,1 as (S(0), 0) = etc., and you get negative numbers. For AP's system, this same construction gives you the nonpositive integers ({0,-1,-2,-3,...}). > I wonder that you completed in any way your mathematical education. I suspect he only minored in math, which would mean he only needed math up to Calculus, Linear Algebra, Differential Equations, and maybe a logic course. The fact that he didn't remember or didn't learn what a vector space was has only lowered the bar in my mind. > And see this. No mention of 0 at all, I call it the base element. > > Okay, I accept that for it is in keeping with your insistence that this > is Set theory. Okay. But where does Set theory have Successor? Is not > Sequence and successor already that of arithmetic. > No, why should it? Set theory doesn't have addition, either. AP doesn't seem bothered by THAT, though. > With #5, if that is set theory then the empty set seldom if ever holds > for any property, and 0 is involved there are many cases in which Math > Induction is true if you start with 1 and not 0. And thus already you > have arithmetic involved here. > You simply do not understand. The empty set does not conform to the > Peano axioms, as there is no base element. And indeed, every finite set > can not conform to the Peano axioms, as there is either a base element > and an element with no successor, or there is no base element. Or the successor function is not one-to-one, viz. S(x) = S(y) for some x and y which are distinct. > Okay, if you put into the #1 that 0 and 1 exists and change #2 to read > that Successor is the endless adding of the metric of 0 to 1, then the > Fibonacci no longer contends for the Natural Numbers and in fact is a > unique for #1 to #5. So where does the Fibonacci get thrown out as not > the Natural Numbers? > But I do not understand any of this. He's saying 1 + 1 = 2, 2 + 1 = 3, but 3 +1 is not 5; that is, if you add 1 to a Fibonacci number you don't necessarily get another Fibonacci number. (Remember, for AP, the successor function is defined in terms of addition: S(x) = x + 1.) > Well, if you welcome me, you better also read what I write. > > You know reading and understanding what the other is saying is two > different things. > But I think you do not even read large parts of what I write. I don't think he reads a lot of what I write, either. He has complained in the past about long posts, and I'll sometimes summarize the important parts at the top of my reply. But a lot of his posts need a line-by-line commentary, and he doesn't have the patience to read it all and do the calculations that I'm aiming him towards. > As far as I know, set theory does not > even have Successor but that Arithmetic has successor in the form of > sequence and function and so the Peano Axioms already have addition. > You are wrong. Indeed. Addition is mentioned nowhere in the axioms. The successor function should be thought of as a pointwise definition; that S(0) is something, S(1) is something, etc., without any attempt to write a formula for the general case. > Dik, can you prove this statement: > > If a given set of numbers is built from the Successor function of > endless adding of 1 such as the p-adic integers, then that set is > Countable. In other words, Counting is equivalent to a Successor > function of endless adding of 1. If that is provable, which I think it > is easily proveable then the Reals are Countable. > You have to define what endless adding of 1 means. With the definitions > I know, the p-adic integers are *not* obtained by endless adding of 1. > And with (what I surmise is) your understanding, being countable is *not* > equivalent to endless adding 1. AP has trouble with infinity, as well. Lots of people do. Even if the 10-adics were obtained by endless adding of 1, AP still has the problem that 0 has a predecessor, namely ...999_10. So in either case, the 10-adics aren't the natural numbers. (Same argument works for Infinite Integers, which AP considers a different concept, but he's never explained how.) --- Christopher Heckman === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth ... No, it does not prevent that. You are free to label the numbers created by the Peano axioms as you wish. A common misconception is that the Peano axioms define arithmetic. They do not. ... >Okay, then when in the process of knowing whether 0,5,10,15,..... is >the Natural Numbers or whether 0,1,2,3, .... are the Natural Numbers, >when does that moment of understanding suppose to take place? Not after >reading the Peano Axioms but days later when addition, sequence, >function are defined? >Dik, I think the very biggest problem in all of this is the fact that >every human knows the Counting Numbers much better than any >formalization could ever try to axiomatize. If children were first >taught the Peano axioms before ever doing any numbers, they would never >get it. This is a very good point. However the issue is, what simple set of properties of the natural numbers can we specify as axioms so we can prove things formally, as opposed to just asserting they're obviously true by inspection. No one claims the Peano axioms are sufficient to pin down the natural numbers completely. In fact, Godel proved that no finite set of axioms in the language of first order logic can uniquely characterize the natural numbers. What the Peano axioms represent is a simple set of axioms sufficient to prove all the standard laws of arithmetic. They don't depend on the concept of a real number line. Your desire to model the naturals physically as points on the real line, spaced 1 apart is a natural idea, and is close in spirit to the way most people probably picture the natural numbers. In fact, I would guess that the ancient Greeks thought about the natural numbers in this way, since they tended to view numbers geometrically. Probably they would have laughed at the Peano axioms and rejected them, at least initially. But there's nothing really wrong with the Peano axioms. They're just a set of carefully chosen, simple facts about the naturals, declared as axioms. The fact the axioms are not complete enough to force the picture you want is not really a problem. As long as they're obviously true and not contradictory, they can be used as a base. As far as trying to replace them with a more geometrically based set of axioms, there are some problems. The main problem is the apparent need to define the real number line first. That's kind of hard to do without having first the naturals, then the integers, then the rationals. I guess it could be done, but we're talking foundations here, so the goal is to start with axioms that are self evident and close to minimal. No need to axiomatize something that could be defined and/or proved later. The Peano axioms have the desired minimal feel. Thus, for example, no need to axiomatize 1 since we can simply define it. Similarly, we don't need the real number line to prove the laws of arithmetic. The reals and the real number line can come much later, all in good time. The need to axiomatize is dramatized clearly by some of the pseudo-math controversies that often appear in sci.math. Without a standard set of axioms and a clear concept of what constitutes a proof, there would be endless arguments with no resolution. Well axiom systems are useful for they are a condensation of the traits and characteristics of a set of something. But it is foolish to think they are something special. They are needed of course because they are a shorthand of what we agree something is. But they are prone to error and need mending and fixing every now and then. It is far better and truthful, just to hold up a set and say, here, these are the Natural Numbers as 0,1,2,3,..... But now when you are proving things about Natural Numbers you want a consensus of what is characteristic about them that everyone agrees with and can refer to that characteristic as postulate #2 or #5 etc. Do not be fooled that Natural Numbers never existed until they were axiomatized in 1889. They always existed and their very best and truthful display is not the axioms but simply the display of 0,1,2,3,..... To create an axiom system about something, that something already exists and thus the axiom system is merely a coded shorthand of what the overall general characteristics of that something. Too much of modern mathematics has reverence to axioms, were a better viewpoint of axioms is that they are useful but always a mere approximation of the underlying reality that we are trying to learn about. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies. === Subject: think in sets Hi Maybe someone can give me a hand with this, Let E be a set, with subsets A in E and B in E, and * be the operation A * B = (E - A) intersect with (E - B), now using A, B and *, how can I express A U B the best I thought of is this; A * B = (Not A) intersect with (Not B) = not ( A U B), any more than that will need some explanations. === Subject: Re: think in sets > Let E be a set, with subsets A in E and B in E, and * be > the operation A * B = (E - A) intersect with (E - B), > now using A, B and *, how can I express A U B > the best I thought of is this; > A * B = (Not A) intersect with (Not B) = not ( A U B), > any more than that will need some explanations. You have it, if complement (your not) is allowed. A*B = (A complement) intersect (B complement) = (A U B) complement So, A U B = A*B complement. === Subject: Re: think in sets > Hi > Maybe someone can give me a hand with this, > Let E be a set, with subsets A in E and B in E, and * be the operation A * B > = (E - A) intersect with > (E - B), now using A, B and *, how can I express A U B > the best I thought of is this; > A * B = (Not A) intersect with (Not B) = not ( A U B), any more than that > will need some explanations. Consider what C * C equals... === Subject: Re: think in sets > Maybe someone can give me a hand with this, Only if you stop using in to mean subset. > Let E be a set, with subsets A in E and B in E, and * be the operation A A is a subset of E, ie A subset E. A is not in E, ie A is not an element of E. > * B = (E - A) intersect with (E - B), now using A, B and *, how can I > express A U B A * nulset = E - A A / B = (E - A)*(E - B) A / B = E - (E - A)/(E - B) = E - A*B = (A * B) * nulset A * B = B * A (A * B) * C /= A * (B * C) nulset * (nulset * A) = A (nulset * nulset) * A = E * A = nulset === Subject: Re: think in sets > Maybe someone can give me a hand with this, > Only if you stop using in to mean subset. > Let E be a set, with subsets A in E and B in E, and * be the operation A > A is a subset of E, ie A subset E. > A is not in E, ie A is not an element of E. > * B = (E - A) intersect with (E - B), now using A, B and *, how can I > express A U B > A * nulset = E - A > A / B = (E - A)*(E - B) > A / B = E - (E - A)/(E - B) > = E - A*B = (A * B) * nulset A * A = E - A A / B = (A*A) * (B*B) A / B = (A*B) * (A*B) A - B = A / (E - B) = (A*A) * B nulset = A * (A*A) E = (A * (A*A)) * (A * (A*A)) > A * B = B * A > (A * B) * C /= A * (B * C) > nulset * (nulset * A) = A > (nulset * nulset) * A = E * A = nulset === Subject: You are nominated for the MBA by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id jBS8g5313500; by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, $Revision: 1.6 secondary) with SMTP id jBS8emFA016488; --------------------------------------------------------------------- UNIVERSITY DIPLOMAS NON-ACCREDITED UNIVERSITIES BASED ON YOUR PRESENT KNOWLEDGE AND LIFE EXPERIENCE. If you qualify, no tests, study, books or exams. We have Bachelor's, MBA's, Doctorate & PhD degrees available in your field. CONFIDENTIALITY ASSURED CALL NOW TO GET YOUR DIPLOMA WITHIN 2 WEEKS 1-206-984-0106 CALL 24 HOURS, 7 DAYS A WEEK === Subject: Re: math in WinEdit > By the way ,why you write Latex and spell it Latek? Presumably you meant 'pronounce it Latek'. It is written LaTeX: the La is the first two letters of Lamport, the TeX is the TeX of Knuth, and the X in TeX is not an English eks, but a Greek chi. http://en.wikipedia.org/wiki/LaTeX has a section on the pronunciation. === Subject: Re: math in WinEdit >Question: Suppose that I get some papers written in WinEdit with the >extension pdf,dvi or ps.Is there a method or a software which shows the >source code of the papers? No, not really. For the same reason that you cannot reconstruct the egg from an omelette. Well, that's a bit of an exaggeration, but it's almost true. Now I want to make helpful comment not related to your question but regarding your statement: >I am meticulous in editing so I pay attention to the details. I have pasted your message in its entirety at the end of this note. Your spelling and grammar are very good. However since you are so meticulous, you should pay more attention to the placement of punctuation marks. DON'T leave a space before a comma or a period. DO leave a space after a comma or a period. Aside: For those of you non-Americans, a period is what our British cousins call a full stop. It's the dot that indicates the end of a sentence. extension pdf,dvi or ps.Is there a method or a software There should be a space after the comma and another space after the period. which ,consequently, have similar notations. There should be no space before the first comma and there should be a space after it. As to: >By the way ,why you write Latex and spell it Latek? LaTeX is spelled LaTeX and is commonly pronounced la-tech. The tech in TeX is derived from the Greek tau-epsilon-chi which is the root of words such as technology and technique. By the way, note the misplaced comma in the sentence I have quoted above. Happy TeXing! Rouben Rostamian --- original message ------------------------------------------------- >Question: Suppose that I get some papers written in WinEdit with the >extension pdf,dvi or ps.Is there a method or a software which shows the >source code of the papers? >The question's reason: I am a beginner in using winedit. I am meticulous >in editing so I pay attention to the details.The problem is that the >help either unless I spend huge amounts of time searching for answers. >That's why I think the best way is to have models(templates) in source >code. Moreover I am interested in papers related to my subject of study >which ,consequently, have similar notations. >By the way ,why you write Latex and spell it Latek? >Happy holidays, Bogdan