>>Also, can we have things such as e^(quat) or ln(quat) where quat is a >>quaternion? Has anybody ever figured out how to do this? Exponentials, and logarithms of invertible elements, exist in any Banach >algebra. Look up holomorphic functional calculus. I should qualify that. The holomorphic functional calculus exists in complex Banach algebras, while the usual quaternions are an algebra over the reals. Of course there's no trouble complexifying to get an algebra of quaternions over the complex numbers (with the slight notational annoyance that the quaternion i and the complex number i are not the same). The exp of a (real) quaternion is a (real) quaternion; ln (using the principal branch) of a (real) quaternion whose spectrum does not intersect the nonpositive reals is a (real) quaternion. This is because exp and this branch of ln can be uniformly approximated on a neighbourhood of the spectrum by polynomials with real coefficients. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >Are there any future plans to 'phase out' complex numbers and replace >them with 2x2 matrixes which work exactly the same? No. > How do you know? He should NOT have answered. The secret committee for the supression of matrices will NOT be amused. ==== > Are there any future plans to 'phase out' complex numbers and replace > them with 2x2 matrixes which work exactly the same? > No. How do you know? The real question is, are there any plans to phase out the real numbers and replace them with 1x1 matrices? ==== > How long has the relation between complex numbers and particular 2x2 > matrixes been known? > Are there any future plans to 'phase out' complex numbers and replace > them with 2x2 matrixes which work exactly the same? > Complex numbers are isomorph 2x2-rotation-matrices are isomorph ordinary 2D-vectors (with a certain multiplication added) -You calculate them exactly the same - and you can calculate them in mixed mode too: see the mixed-mode calculator: http://i-z.eu.tt or http://i-is-no-longer-imaginary.gmxhome.de There is the geometric vector-space (without coordinate-system) too, which demonstrates, that you don't need this imaginarystuff any longer, nor the Argand diagramm nor the imaginary axis, just like Caspar Wessel started 2oo years ago without these. Have fun Hero ==== > You'll need to write a macro using VBA. > Most, if not all of the info you need, should be in one of the Word VBA > books I have listed in the list of WordVBA books at my URL below. I suggest getting Steve Roman's Writing Word Macros. > Thaqnks, MathType 5.2 is a best solution in my case. ==== Suppose we had the set of all integers which resembled an integer 2^n where n is any integer. This would start from 1 in the case where n = 0 to indefinately high. Let's call this set Q. There are an infinite amount of such numbers since the log of infinity is infinity. So the cardinality is aleph-x, where x is an as yet undetermined variable. Now, by definition, power sets are larger than the sets associated with them. So the power set of Q would be equal in size to the set of all positive integers. Or all non negative integers, if you included the null set. So while Q is not finite, its power set, which must be larger than it, is supposedly the smallest infinite set, accordding to a certain theory which guarantees that aleph notation is exact and there are no 'middle numbers' in between two consecutive alephs. Isn't this a contradiction? (...Starblade Riven Darksquall...) ==== > Suppose we had the set of all integers which resembled an integer 2^n > where n is any integer. This would start from 1 in the case where n = > 0 to indefinately high. Let's call this set Q. > No, let's call it K. Q is taken already by the set of rational numbers. However, I don't get your supposition. It appears that you're taking all integers that resemble an integer 2^n. What does it mean for a number to resemble an integer? I'll suppose for the moment that you really meant to say all integers of the form 2^n. That set certainly exists. It's just the set of integers 2^n for n a non-negative integer: {1,2,4,8,16,32,...,2^n, ...} > There are an infinite amount of such numbers since the log of infinity > is infinity. So the cardinality is aleph-x, where x is an as yet > undetermined variable. > Infinity is not a number, and log(infinity) is undefined. Surely, the limit lim_{x -->infinity} log(x) fails to exist, because the function grows without bound, and that fact is typically written lim_{x -->infinity} log(x) = infinity. That does not mean that one can substitute infinity for x in the expression log(x), and out pops the value infinity. The above expression for the limit is a *shorthand* expression for what I bound as x gets arbitrarily large. > Now, by definition, power sets are larger than the sets associated > with them. So the power set of Q would be equal in size to the set of > all positive integers. Or all non negative integers, if you included > the null set. > No, it isn't *by definition* that power sets are larger than the sets associated with them, if you're meaning to say that, for any set X, one has card(X) < card(2^X). This is a theorem, proven (if I recall correctly) by Cantor. Theorems are hardly ever definitions, and this theorem is surely not a definition. > So while Q is not finite, its power set, which must be larger than it, > is supposedly the smallest infinite set, accordding to a certain > theory which guarantees that aleph notation is exact and there are no > 'middle numbers' in between two consecutive alephs. So, you have the set that I've renamed K. True, it is not finite: by its construction, K is clearly of the same cardinality as N_o, the set of natural numbers with zero. Next, you claim that its power set 2^K (which must be of greater cardinality, according to the theorem I mentioned relating X and 2^X), is supposedly the smallest infinite set, accordding to a certain theory which guarantees that aleph notation is exact and there are no 'middle numbers' in between two consecutive alephs. Since the alephs are defined as the successive cardinals, it is *this* that is a definition (if I understand the definitions correctly). However, 2^K is surely not the smallest infinite set. The existence of the alephs has nothing to do with 2^K, and it is certain that they do *not* show that 2^K is the smallest infinite set. Isn't this a contradiction? Here's what I think you're doing: 1. Take N_o, the natural numbers with zero, and form the set K = {2^0, 2^1, 2^2, ..., 2^k ,... } of successive powers of 2. 2. K is clearly of the same cardinality as N_o. 3. K is clearly 2^N_o. 4. card(N_o) < card(2^N_o), contradiction. Your error is in step 3. The sequence of powers of 2 is not the same as the power set, appearances notwithstanding. The notation 2^X for the power set of X is (as with the limit expression I discussed earlier) simply a shorthand for a particular construction. In this case, it refers to the set of functions from the set X to the two-element set {0, 1}. That is, one can uniquely identify any subset S of X with a map M_S from X to {0,1} by this method: if x is in S, let M_S(x) = 1, otherwise let M_S(x) = 0: M_S(x) = 1 if x is in S, otherwise M_S(x) = 0. It is simple to verify that distinct subsets are associated to distinct functions, every subset yields a function, and every function comes from a subset. Simply taking 2^n for every element n of some set of naturals does not yield the power set. Note that the method fails already for finite sets. I'll take a small set X, and form your construction (I'll call it K again): X = {1,2,3} K = {2,4,8}. The power set 2^X is this: 2^X = { {}, {1},{2},{3},{1,2},{1,3},{2,3},{1,2,3} } Note that the power set 2^X doesn't have any elements that are integers (except {} for folks who use the von Neumann construction of the integers, in which case {} is 0). (...Starblade Riven Darksquall...) You might understand some of this if you took an introductory course in mathematics, or read an elementary set theory text (I've heard good things about Halmos's Naive Set Theory, currently in the Springer UTM series). Dale. CAVEAT: I'm no set theorist, nor do I play one on TV. There are plenty of regular contributors to sci.math who are more adept at this particular game than I am, and they may well have more to say on this topic. ==== > Suppose we had the set of all integers which resembled an integer 2^n > where n is any integer. This would start from 1 in the case where n = > 0 to indefinately high. Let's call this set Q. > > No, let's call it K. Q is taken already by the set of rational numbers. However, I don't get your supposition. It appears that you're taking all > integers that resemble an integer 2^n. What does it mean for a number > to resemble an integer? I'll suppose for the moment that you really > meant to say all integers of the form 2^n. That set certainly exists. > It's just the set of integers 2^n for n a non-negative integer: {1,2,4,8,16,32,...,2^n, ...} > There are an infinite amount of such numbers since the log of infinity > is infinity. So the cardinality is aleph-x, where x is an as yet > undetermined variable. > > Infinity is not a number, and log(infinity) is undefined. Surely, the > limit lim_{x -->infinity} log(x) fails to exist, because the function > grows without bound, and that fact is typically written lim_{x -->infinity} log(x) = infinity. That does not mean that one can substitute infinity for x in the > expression log(x), and out pops the value infinity. The above > expression for the limit is a *shorthand* expression for what I > bound as x gets arbitrarily large. > Now, by definition, power sets are larger than the sets associated > with them. So the power set of Q would be equal in size to the set of > all positive integers. Or all non negative integers, if you included > the null set. > > No, it isn't *by definition* that power sets are larger than the sets > associated with them, if you're meaning to say that, for any set X, > one has card(X) < card(2^X). This is a theorem, proven (if I recall correctly) by Cantor. Theorems > are hardly ever definitions, and this theorem is surely not a > definition. > So while Q is not finite, its power set, which must be larger than it, > is supposedly the smallest infinite set, accordding to a certain > theory which guarantees that aleph notation is exact and there are no > 'middle numbers' in between two consecutive alephs. So, you have the set that I've renamed K. True, it is not finite: by its > construction, K is clearly of the same cardinality as N_o, the set of > natural numbers with zero. Next, you claim that its power set 2^K (which must be of greater > cardinality, according to the theorem I mentioned relating X and 2^X), is supposedly the smallest infinite set, accordding to a > certain theory which guarantees that aleph notation is exact > and there are no 'middle numbers' in between two consecutive > alephs. Since the alephs are defined as the successive cardinals, it is *this* > that is a definition (if I understand the definitions correctly). > However, 2^K is surely not the smallest infinite set. The existence > of the alephs has nothing to do with 2^K, and it is certain that they > do *not* show that 2^K is the smallest infinite set. > However, 2^K is the set of all natural numbers. And the set of all natural numbers is of cardinality N_0. However, K is also of cardinality N_0. If the power set of any set is strictly larger (IE has a greater cardinality) then how is it possible? This would require that N_0 > N_0, which is the contradiction I'm talking about. > Isn't this a contradiction? Here's what I think you're doing: 1. Take N_o, the natural numbers with zero, and form > the set K = {2^0, 2^1, 2^2, ..., 2^k ,... } of > successive powers of 2. 2. K is clearly of the same cardinality as N_o. 3. K is clearly 2^N_o. 4. card(N_o) < card(2^N_o), contradiction. Your error is in step 3. The sequence of powers of 2 is not the same as > the power set, appearances notwithstanding. (*Snip the rest*) What I'm saying is that K is the set of all numbers which are a power of 2. If it terminates at, say, 2^X, then it has X+1 members. However, its power set has 2^(X+1) members, and is bijective with the set of nonnegative integers from 0 to 2^(X+1) if you count the empty set as part of its power set, and is bijective with the set of positive integers from 1 to 2^(X+1) if you don't count the empty set as part of its power set. Then take X as X -> inf. So K is of cardinality N_0, from what you said, but so is its power set, since its power set is bijective with the set of natural numbers. This is the contradiction I'm talking about. The power set is always of a larger cardinality than the set that generates it. But this would mean N_0 > N_0, and thus the contradiction. (...Starblade Riven Darksquall...) ==== > However, 2^K is the set of all natural numbers. What makes you think this is true? > Here's what I think you're doing: > 1. Take N_o, the natural numbers with zero, and form > the set K = {2^0, 2^1, 2^2, ..., 2^k ,... } of > successive powers of 2. > 2. K is clearly of the same cardinality as N_o. > 3. K is clearly 2^N_o. > 4. card(N_o) < card(2^N_o), contradiction. > Your error is in step 3. The sequence of powers of 2 is not the same as > the power set, appearances notwithstanding. > (*Snip the rest*) What I'm saying is that K is the set of all numbers which are a power > of 2. If it terminates at, say, 2^X, then it has X+1 members. Be careful with trying to extend finite reasoning to infinite sets. Infinite sets have different properties, for instance being in one-to-one-correspondence with proper subsets. Let's see where reasoning about the finite set K_X = {0, 2^1, ..., 2^X} goes, however. > However, > its power set has 2^(X+1) members, and is bijective with the set of > nonnegative integers from 0 to 2^(X+1) No. Having 2^(X+1) members (X finite) means that you have a bijection with 1, 2, ..., 2^(X+1), not with 0, 1, 2, ..., 2^(X+1). Having 3 members means you have a bijection with {1,2,3}. Having M members means you have a bijection with {1,2,..., M}. > if you count the empty set as > part of its power set, and is bijective with the set of positive > integers from 1 to 2^(X+1) if you don't count the empty set as part of > its power set. You've miscounted. Consider the following set: {1,2,3}. The power set has the following 2^3 = 8 elements: {} {1} {2} {3} {1,2} {1,3} {2,3} {1,2,3} Note that the empty set is one of the members of the power set, and is included in the count of 8, i.e., a one-to-one correspondence with the integers from 1 to 8. > Then take X as X -> inf. So K is of cardinality N_0, from what you said, but so is its power > set, since its power set is bijective with the set of natural numbers. You can't draw any conclusions from such a limit. This is one of the subtleties of dealing with infinity. Limit processes don't go to infinity. That's not what the symbol -> inf means. You don't ever reach infinity. What you're doing is noting that as X grows without bound, taking values in the natural numbers, then 2^X also grows without bound, also taking values in the natural numbers. What makes you believe in this bijection? - Randy ==== >Suppose we had the set of all integers which resembled an integer 2^n >where n is any integer. This would start from 1 in the case where n = >0 to indefinately high. Let's call this set Q. > No, let's call it K. Q is taken already by the set of rational numbers. >> ... stuff deleted ... >>Since the alephs are defined as the successive cardinals, it is *this* >>that is a definition (if I understand the definitions correctly). >>However, 2^K is surely not the smallest infinite set. The existence >>of the alephs has nothing to do with 2^K, and it is certain that they >>do *not* show that 2^K is the smallest infinite set. >> However, 2^K is the set of all natural numbers. And the set of all > natural numbers is of cardinality N_0. However, K is also of > cardinality N_0. If the power set of any set is strictly larger (IE > has a greater cardinality) then how is it possible? This would require > that N_0 > N_0, which is the contradiction I'm talking about. > This is simply not so. If K is the set of powers of 2, it is just not the case that 2^K is the set of all natural numbers. Since you're the one making this assertion, I'll ask you to tell me how one associates a natural number with an arbitrary subset of K, so that every natural number is matched with a subset, and so that no two different subsets are matched to the same natural number. According to Cantor's theorem it cannot be done. Given any set X, its collection of all subsets has more elements than X. > >Isn't this a contradiction? Here's what I think you're doing: 1. Take N_o, the natural numbers with zero, and form >> the set K = {2^0, 2^1, 2^2, ..., 2^k ,... } of >> successive powers of 2. 2. K is clearly of the same cardinality as N_o. 3. K is clearly 2^N_o. 4. card(N_o) < card(2^N_o), contradiction. Your error is in step 3. The sequence of powers of 2 is not the same as >>the power set, appearances notwithstanding. > What you say below suggests that the artifice of the powers of 2 is irrelevant to your argument. You could just as well take the set of natural numbers less than or equal to N: S_N = {1, 2, ..., N} and its power set P_N = 2^{1,2,...,N}. The one has N elements, and the other has 2^N elements. You're trying to state that since the individual steps are all finite, somehow one can force a 1:1 mapping in the limit. That logic does not apply. > (*Snip the rest*) What I'm saying is that K is the set of all numbers which are a power > of 2. If it terminates at, say, 2^X, then it has X+1 members. However, > its power set has 2^(X+1) members, and is bijective with the set of > nonnegative integers from 0 to 2^(X+1) if you count the empty set as > part of its power set, and is bijective with the set of positive > integers from 1 to 2^(X+1) if you don't count the empty set as part of > its power set. Then take X as X -> inf. > The limiting argument is invalid. Here's why: The standard mapping from {1, ..., N} with the set of numbers {0, 1, ..., 2^(N-1)} is achieved by taking the function X |----> sum(x(k)*2^(k-1); k =1, ..., N-1 ), where X is any subset of {1, ..., N}, and x(k) is the function from {1, ..., N} that is 1 when x is in X and 0 otherwise. If you attempt to force this procedure to the limit as N --> infinity, then you need to look at the function: X |---> sum( x(k)*2^(k-1); k = 1, 2, ... ) where the sum is over all positive integers. For the function to map to the positive integers, one needs the sum to converge, but note that it doesn't converge if X is infinite: it has infinitely many terms, each of which is a positive integer (in fact, the if the sum has infinitely many terms, then it has terms that grow without bound). The series cannot converge in the limit. This particular form of limiting argument therefore cannot succeed. If you have another limiting argument (by the way, the phrase as N --> infinity, the limit works is just *not* a valid limiting argument), then make it explicit. > So K is of cardinality N_0, from what you said, but so is its power > set, since its power set is bijective with the set of natural numbers. No, it isn't. There is no limiting argument here. > This is the contradiction I'm talking about. The power set is always > of a larger cardinality than the set that generates it. But this would > mean N_0 > N_0, and thus the contradiction. (...Starblade Riven Darksquall...) No, you're mistaken. The limiting argument that you suggest is no argument at all, since it is strewn with implicit steps that have no clear implementation. Dale. ==== >Suppose we had the set of all integers which resembled an integer 2^n >where n is any integer. This would start from 1 in the case where n = >0 to indefinately high. Let's call this set Q. > No, let's call it K. Q is taken already by the set of rational numbers. >> ... stuff deleted ... >>Since the alephs are defined as the successive cardinals, it is *this* >>that is a definition (if I understand the definitions correctly). >>However, 2^K is surely not the smallest infinite set. The existence >>of the alephs has nothing to do with 2^K, and it is certain that they >>do *not* show that 2^K is the smallest infinite set. > However, 2^K is the set of all natural numbers. And the set of all > natural numbers is of cardinality N_0. However, K is also of > cardinality N_0. If the power set of any set is strictly larger (IE > has a greater cardinality) then how is it possible? This would require > that N_0 > N_0, which is the contradiction I'm talking about. > > This is simply not so. If K is the set of powers of 2, it is just not > the case that 2^K is the set of all natural numbers. Since you're the > one making this assertion, I'll ask you to tell me how one associates > a natural number with an arbitrary subset of K, so that every natural > number is matched with a subset, and so that no two different subsets > are matched to the same natural number. According to Cantor's theorem > it cannot be done. Given any set X, its collection of all subsets has > more elements than X. > You're misusing cantor's theorum. I never said that K wasn't greater than K. I said N was equal in size to the power set of K. Cantor's theorum does not involve this kidn of thing. The power set of K is related to the natural numbers if you do this: For each element in the power set of K, add all the numbers together. You will always get a unique natural number. For instance, take K = (1, 2, 4). Take the power set of K = (Null, 1, 2, 1&2, 4, 1&4, 2&4, 1&2&4). Now this relates to the natural numbers Num = (0, 1, 2, 3, 4, 5, 6, 7) easily. Now K is smaller than the power set of K. However, the power set of K, as the size of the power set approaches infinity, also approaches the set of the natural numbers (Even if 0 is not a natural number, you can just add 1 to each member of the set so that it goes from 1 to 8 rather than 0 to 7.) and is therefore bijective. (...Starblade Riven Darksquall...) ==== > You're misusing cantor's theorum. I never said that K wasn't greater > than K. I said N was equal in size to the power set of K. Cantor's > theorum does not involve this kidn of thing. You're talking nonsense. How is K greater than K? > The power set of K is related to the natural numbers if you do this: > For each element in the power set of K, add all the numbers together. > You will always get a unique natural number. Try the example I gave before. Let K_p = { 2^p : p is prime }. Then K_p is a subset of K. What is the sum of all the members of K_p? Which natural number is that? -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== ... stuff deleted ... However, 2^K is the set of all natural numbers. And the set of all >natural numbers is of cardinality N_0. However, K is also of >cardinality N_0. If the power set of any set is strictly larger (IE >has a greater cardinality) then how is it possible? This would require >that N_0 > N_0, which is the contradiction I'm talking about. > This is simply not so. If K is the set of powers of 2, it is just not >>the case that 2^K is the set of all natural numbers. Since you're the >>one making this assertion, I'll ask you to tell me how one associates >>a natural number with an arbitrary subset of K, so that every natural >>number is matched with a subset, and so that no two different subsets >>are matched to the same natural number. According to Cantor's theorem >>it cannot be done. Given any set X, its collection of all subsets has >>more elements than X. >> > You're misusing cantor's theorum. I never said that K wasn't greater > than K. I said N was equal in size to the power set of K. Cantor's > theorum does not involve this kidn of thing. > I am certainly *not* misusing Cantor's theorEm. K is, by construction, of the same cardinality as N. After all, it is produced by taking the image of N via an invertible function. Just because I stepped from N to a set of the same cardinality, that does not invalidate the application of Cantor's theorem. Your statement that N is equal in size to the power set of K is not true. Recall the mapping I provided (I've corrected a typo: originally mapping): The standard mapping from 2^{1, ..., N} with the set of numbers {0, 1, ..., 2^(N-1)} is achieved by taking the function X |----> sum(x(k)*2^(k-1); k =1, ..., N-1 ), where X is any subset of {1, ..., N}, and x(k) is the function from {1, ..., N} that is 1 when x is in X and 0 otherwise. This is the mapping you're using here: > The power set of K is related to the natural numbers if you do this: > For each element in the power set of K, add all the numbers together. > You will always get a unique natural number. Do those sums exist for *every* subset of K? What about the set of all elements of K? Every other element of K? You should realize that the only subsets you can map to N via this mapping are the FINITE ones. It is true that the collection of FINITE subsets of N has the same cardinality as N itself. However, unless you contest the existence of infinite sets of integers (such as the set of even integers, or the set of all integers greater than 5, for instance), you must admit that those sums do NOT all converge, and thus your alleged mapping fails to be defined everywhere on the power set of K. For instance, take K = (1, 2, 4). Take the power set of K = (Null, 1, > 2, 1&2, 4, 1&4, 2&4, 1&2&4). Now this relates to the natural numbers > Num = (0, 1, 2, 3, 4, 5, 6, 7) easily. Now K is smaller than the power > set of K. However, the power set of K, as the size of the power set > approaches infinity, also approaches the set of the natural numbers > (Even if 0 is not a natural number, you can just add 1 to each member > of the set so that it goes from 1 to 8 rather than 0 to 7.) and is > therefore bijective. (...Starblade Riven Darksquall...) Ignoring the bulk of subsets of 2^N doesn't buy you any credibility, BTW. Dale. ==== Starblade Darksquall says... >What I'm saying is that K is the set of all numbers which are a power >of 2... >So K is of cardinality N_0, from what you said, but so is its power >set, since its power set is bijective with the set of natural numbers. No, the power set of K is not bijective with the naturals. What is true is the following: 1. There is a bijection between the power set of K and the set of *infinite* sequences of 1s and 0s. For each subset K' of K, associate the infinite sequence s as follows: s[i] = 1 if 2^i is in K', and 0 otherwise. 2. There is a bijection between the naturals and the set of *finite* sequences of 1s and 0s (namely, you can write every natural number in binary notation). 3. But there is no bijection between the naturals and the set of infinite sequences of 1s and 0s. The proof of 3 is exactly Cantor's theorem: assume that there is a bijection f between N and the infinite sequences of 1s and 0s, and show that leads to a contradiction. -- Daryl McCullough Ithaca, NY ==== > Starblade Darksquall says... >What I'm saying is that K is the set of all numbers which are a power >of 2... So K is of cardinality N_0, from what you said, but so is its power >set, since its power set is bijective with the set of natural numbers. No, the power set of K is not bijective with the naturals. What is true is the following: 1. There is a bijection between the power set of K and the > set of *infinite* sequences of 1s and 0s. For each subset > K' of K, associate the infinite sequence s as follows: > s[i] = 1 if 2^i is in K', and 0 otherwise. 2. There is a bijection between the naturals and the set of > *finite* sequences of 1s and 0s (namely, you can write every > natural number in binary notation). 3. But there is no bijection between the naturals and the set of > infinite sequences of 1s and 0s. The proof of 3 is exactly Cantor's theorem: assume that there is > a bijection f between N and the infinite sequences of 1s and 0s, > and show that leads to a contradiction. The only way I see there being a contradiction is if you require that K and N both be of cardinality N_0. However, I don't see why there can't be a non-N_0 cardinality of K such that when its power set is taken, its power set is of cardinality N_0. That will also have to be explained to me. (...Starblade Riven Darksquall...) ==== Starblade Darksquall says... >The only way I see there being a contradiction is if you require that >K and N both be of cardinality N_0. It's not a requirement, it is a theorem. There is a bijection between N and K, namely f(i) = 2^{i}. By the definition of having the same cardinality, N and K have the same cardinality. -- Daryl McCullough Ithaca, NY ==== > However, 2^K is the set of all natural numbers. And the set of all > natural numbers is of cardinality N_0. However, K is also of > cardinality N_0. If the power set of any set is strictly larger (IE > has a greater cardinality) then how is it possible? This would require > that N_0 > N_0, which is the contradiction I'm talking about. No. 2^K is the power set of K, which is the set of all subsets of K. For example, one such subset is K_p = { 2^p : p is prime }. This is a set, not a number, and therefore 2^K is not a set of numbers and certainly is not equal to N_0. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== > However, 2^K is the set of all natural numbers. And the set of all > natural numbers is of cardinality N_0. However, K is also of > cardinality N_0. If the power set of any set is strictly larger (IE > has a greater cardinality) then how is it possible? This would require > that N_0 > N_0, which is the contradiction I'm talking about. No. 2^K is the power set of K, which is the set of all subsets of K. > For example, one such subset is K_p = { 2^p : p is prime }. This is a > set, not a number, and therefore 2^K is not a set of numbers and > certainly is not equal to N_0. I never said 2^K WAS the set of all natural numbers. I said 2^K was BIJECTIVE with the set of all natural numbers. (...Starblade Riven Darksquall...) ==== >> However, 2^K is the set of all natural numbers. And the set of all >> natural numbers is of cardinality N_0. However, K is also of >> cardinality N_0. If the power set of any set is strictly larger (IE >> has a greater cardinality) then how is it possible? This would require >> that N_0 > N_0, which is the contradiction I'm talking about. > No. 2^K is the power set of K, which is the set of all subsets of K. >> For example, one such subset is K_p = { 2^p : p is prime }. This is a >> set, not a number, and therefore 2^K is not a set of numbers and >> certainly is not equal to N_0. > I never said 2^K WAS the set of all natural numbers. I said 2^K was > BIJECTIVE with the set of all natural numbers. Wrong on two counts. 1. You did say 2^K is the set of all natural numbers. The quote is plainly visible above. 2. It is not true that 2^K is bijective with the set of all natural numbers. Cantor's theorem shows that. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== What I'm saying is that K is the set of all numbers which are a power > of 2. The power set of K is uncountable. > If it terminates at, say, 2^X, It doesn't terminate. So K is of cardinality N_0, from what you said, but so is its power > set, No. > since its power set is bijective with the set of natural numbers. No. > This is the contradiction I'm talking about. The power set is always > of a larger cardinality than the set that generates it. But this would > mean N_0 > N_0, No. > and thus the contradiction. No. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== (*Snip*) You are very unhelpful. Are you sure you're not just here to be as completely useless as possible? (...Starblade Riven Darksquall...) ==== > Suppose we had the set of all integers which resembled an integer 2^n > where n is any integer. This would start from 1 in the case where n = > 0 to indefinately high. Let's call this set Q. There are an infinite amount of such numbers since the log of infinity > is infinity. So the cardinality is aleph-x, where x is an as yet > undetermined variable. Now, by definition, power sets are larger than the sets associated > with them. So the power set of Q would be equal in size to the set of > all positive integers. Or all non negative integers, if you included > the null set. No. You are arguing that there is a function that maps each subset of Q to a natural number. The obvious mapping would be take the sum of the elements of the subset. You are probably thinking that this is the same as the natural numbers in base-2, but that is wrong. A natural number must be *finite*, but a subset of Q can have an infinite amount of elements. For example, the subset {q in Q: lg(q) is odd} has a sum that is not a natural number (lg x means log base 2 of x). In fact there are uncountably many subsets of Q that are infinitely large. You are thinking of the Set of finite subsets of Q, which is indeed countable and has a very clear bijection with the naturals. ~Steven So while Q is not finite, its power set, which must be larger than it, > is supposedly the smallest infinite set, accordding to a certain > theory which guarantees that aleph notation is exact and there are no > 'middle numbers' in between two consecutive alephs. Isn't this a contradiction? (...Starblade Riven Darksquall...) ==== > Suppose we had the set of all integers which resembled an integer 2^n > where n is any integer. This would start from 1 in the case where n = > 0 to indefinately high. Let's call this set Q. Do you mean ordinary finite integers? This resembled relation is unclear to me. I assume you mean: Q = {x in N: x = 2^n for some n in N} > There are an infinite amount of such numbers since the log of infinity > is infinity. So the cardinality is aleph-x, where x is an as yet > undetermined variable. The log of infinity is a meaningless phrase. But Q is indeed infinite. x = 0. The construction of Q gives an injection from N->Q, and the identity mapping is an injection from Q->N. So N and Q have the same cardinality, aleph-0. > Now, by definition, power sets are larger than the sets associated > with them. So the power set of Q would be equal in size to the set of > all positive integers. Or all non negative integers, if you included > the null set. You are confusing compute 2^n with compute power set. 2^n is the *size of the power set*, not the size of the *members* of the power set. The power set of Q has cardinality of the continuum. > So while Q is not finite, its power set, which must be larger than it, > is supposedly the smallest infinite set, accordding to a certain > theory which guarantees that aleph notation is exact and there are no > 'middle numbers' in between two consecutive alephs. That certain theory is the definition of the alephs. Thomas ==== > Suppose we had the set of all integers which resembled an integer 2^n > where n is any integer. resembled? > This would start from 1 in the case where n = > 0 to indefinately high. Let's call this set Q. Hmmm. Q is usually reserved to mean the set of rational numbers. Do you mean that Q = {2^n: n is a nonnegative integer? > There are an infinite amount of such numbers since the log of infinity > is infinity. Groan!!! the log of what?! :-( > So the cardinality is aleph-x, where x is an as yet > undetermined variable. The cardinaltity of my Q is aleph-zero. > Now, by definition, power sets are larger than the sets associated > with them. So the power set of Q would be equal in size to the set of > all positive integers. Or all non negative integers, if you included > the null set. size = cardinality? If so the the power-set of Q (whatever it be) could not have the same cardinality as N. As no power set has the same cardinality of N. (Why did you begin your 2nd sentence in that paragraph with so?) > So while Q is not finite, its power set, which must be larger than it, > is supposedly the smallest infinite set, No. > accordding to a certain > theory which guarantees that aleph notation is exact and there are no > 'middle numbers' in between two consecutive alephs. Isn't this a contradiction? No. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== > Since I'm on the topic, does anyone know how many dimensions you need > for a Euclidean space in which you can hyperbolic space? That's a very good question with a very complicated and, I think, incomplete, answer. The hyperbolic plane embeds isometrically into R^5, but not into R^3. Before you ask whether R^4 is enough, you have to decide on things like the degree of smoothness, embedding versus immersion, and local versus global. Here are some threads from four years ago: http://www.math-atlas.org/99/embed_hyper dave ==== > at 03:52 PM, Russell Blackadar said: >Indeed, if we're talking only about embeddings in 3-space, the torus >>fails to be an example at all. Not an example of what? The OP didn't specify rotational symmetry, and >it is certainly an example of a surface with a one parameter family of >isometries. The OP wanted any other surface S such that a piece of S can be moved around adlibitum. Some of us interpreted adlibitum to mean that the group of self-isometries should act transitively, i.e. that it contain a 2-parameter family of isometries. The torus does not have that. If you interpret the OP's request merely to be for the isometry group to be non-discrete, then yes, clearly a (regular) torus is an example. dave ==== I am struggleing with finding a way to measure the linearity of a 3D field. I have a set of data with coordinates (x,y,z) and the field value on each given point f(x,y,z). And we expect the field to be linear so that the theoretical relationship would be f(x,y,z)=a*x+b*y+c*z+d We could do a least square fitting of the given data to get the a,b,c,d parameters. But the problem now is how linear the field is? One obvious way is to measure the sum of residual squared, but suppose the data become orders of magnitude larger, the residuals are orders of magnitude larger. But obvously the linearity doesn't change in this case. I know in curve fitting case, there is a correlation coefficient, which is given by r^2=a1*a2. where two curve fittings are done. y=a1*x+b1 x=a2*y+b2 The a1*a2 is between 0 and 1 and the closer it is to 1, the more linear the data are. This is a very good measurement of linearity of one-dimensional data. But I am having some trouble getting the definition of similiar coefficient for higher dimensions. Could anybody help me out here? -- Shi Jin sj88@cornell.edu http://jinshi.dhs.org Following the introduction, a particularly nice (and perhaps new) approximation of the inverse sine is presented. That approximation can be used to give corresponding approximations of the six trigonometric functions and the five other inverse functions. ------------------------------------ Introduction Nikolaus Cusanus (1401-1464, a.k.a. Nicholas of Cusa) gave a remarkable formula: In a right triangle with sides a < b < c, the smallest angle is a/(b + 2*c)*172 degrees. Of course, this is just an approximation, although it seems that Cusanus Arcsin(x) by 43/45*Pi*x/(2 + Sqrt(1-x^2)) for 0 <= x <= 1/Sqrt(2). Noting that 43/45*Pi is only slightly larger than 3, a closely related approximation is 3*x/(2 + Sqrt(1-x^2)). To approximate Arcsin(x) for 1/Sqrt(2) < x <= 1, using the fact that sin^2(t) + cos^2(t) = 1, we may replace x by Sqrt(1-x^2), and vice versa, and then take the complement (that is, subtract that result from Pi/2). Altogether, we obtain ( 3*x/(2 + Sqrt(1-x^2)) for 0 <= x <= 1/Sqrt(2), (1) ( ( Pi/2 - 3*Sqrt(1-x^2)/(2 + x) for 1/Sqrt(2) < x <= 1 as an approximation of Arcsin(x), with |rel. error| < 0.0023 . Of course, due to simple interrelations between the inverse trigonometric functions, one could easily obtain corresponding approximations for the other five inverse functions from (1). Furthermore, due to the algebraic simplicity of form in (1), it can be inverted easily, giving ( x*(6 + Sqrt(9-3*x^2))/(9 + x^2) for -Pi/4 <= x <= Pi/4 (2) ( ( (18 + 3*Sqrt(9-3*(Pi/2-x)^2))/(9 + (Pi/2-x)^2) - 2 for Pi/4 < x <= 3*Pi/4 as an approximation for sin(x), with |rel. error| < 0.0019 . Due to simple interrelations between the trigonometric functions, one could easily obtain corresponding approximations for the other five functions from (2). (This ends the introduction, in which I presume that there is really nothing new.) ------------------------------------ We now primarily consider approximating Arcsin(x) for 0 <= x <= 1/Sqrt(2), knowing that, as indicated above, corresponding approximations can be obtained easily for Arcsin(x) for 1/Sqrt(2) < x <= 1, and ultimately for all of the trigonometric functions and their inverses. The first part of (1) may be thought of as being in the form (3) x/(1 - h*(1 - Sqrt(1-x^2))) with h = 1/3. The question then arises: Can the approximation given by (3) be improved by choosing a different value of h? Answer: Yes. My favorite choice is h = (Sqrt(2) + 1)*(Sqrt(2) - 4/Pi) = 0.340341385... Using it, we obtain an expression giving Arcsin(x) precisely at both 0 and 1/Sqrt(2) and overestimating it for 0 < x < 1/Sqrt(2), with rel. error < 0.00057 . Unfortunately, the improvement in accuracy is not dramatic. It might also be noted that this approximation is differentiable at x = 1/Sqrt(2); by contrast, (1) has a jump discontinuity there. But could the form of (3) be altered slightly -- say, by introducing another parameter -- so as to get much better accuracy? Yes! Choosing the form (4) x/(1 - h*(1 - Sqrt(1-(k*x)^2))) allows relative error to be reduced substantially, to approximately 1/50 of that given by (1). The parameters h and k were determined so that (4) would give Arcsin(x) precisely at 1/2 and 1/Sqrt(2). [Slightly different values of h and k would surely reduce error even a bit more, but I think it's neat to use the two criteria already mentioned.] Although the exact expressions for h and k are indeed messy, namely (3+2*Sqrt(2))*(2*(3-Sqrt(2))-Pi/2-5/Pi)*(6*Sqrt(2)+Pi*(3-2*Sqrt(2))-9) h = ---------------------------------------------------------------------- (Pi-3)*(6-Pi-2*Sqrt(2)) and Sqrt((Pi-3)*(2*(3*Pi-6*Sqrt(2)+4)-Pi^2)) k = ----------------------------------------- , (1+Sqrt(2))*(Pi*(3-Sqrt(2))-(Pi/2)^2-5/2) we may of course just use approximations of these parameters: h = 0.30777349... and k = 1.0419890... As previously noted, we correspondingly approximate Arcsin(x) for 1/Sqrt(2) < x <= 1 by taking the complement of (4) once x has been replaced by Sqrt(1-x^2). Taken together then, we get an approximation of Arcsin(x) for 0 <= x <= 1 with |rel. error| < 0.000046 . Using exact values for h and k, the approximation gives the exact values of Arcsin(x) for x = 0, 1/2, 1/Sqrt(2), Sqrt(3)/2, and 1, and is differentiable on [0,1). Similar comments could be made concerning the corresponding approximations for the trigonometric functions and for the five other inverse trigonometric functions. ------------------------------------- How does this approximation of Arcsin(x) compare with other approximations? Of course there are far more accurate approximations. Some of them are also related to Cusanus's formula; see Lou Talman's Some (Almost) Rational Thoughts , for example. However, AFAIK, they are of more complicated forms, thereby making their algebraic inversion more difficult -- in fact, impossible in most cases. Abramowitz and Stegun mention a four-parameter approximation, their item 4.4.45 (see ) which provides accuracy almost identical to our two-parameter approximation. That polynomial approximation has the form Pi/2 - Sqrt(1-x)*(a_0 + a_1*x + a_2*x^2 + a_3*x^3) in which the values of the coefficients seem to have been determined numerically. ------------------------------------ Your thoughtful comments are welcome. David W. Cantrell ==== How do you prove that the circumference of a circle is proportional to it's radius? What's the modern version and how did the greeks prove it? /david ==== > How do you prove that the circumference of a circle is proportional to > it's radius? > What's the modern version and how did the greeks prove it? /david > The approximations to that ratio are based on inscribing/circumscribing regular polygons, each of which may be partitioned into congruent isosceles triangles. For any such polygon, the odd(circuferential) side of any of those isosceles triangles is proportional to the other (radial) sides by similar triangles. Thus each approximate circumference is proportional to its radius. ==== > How do you prove that the circumference of a circle is proportional to > it's radius? > What's the modern version and how did the greeks prove it? The approximations to that ratio are based on > inscribing/circumscribing regular polygons, each of which may be > partitioned into congruent isosceles triangles. For any such polygon, the odd(circuferential) side of any of those > isosceles triangles is proportional to the other (radial) sides by > similar triangles. Thus each approximate circumference is proportional to its radius. Now if you know/accept that perimeter of inscribed polygon < circumference of circle < perimeter of circumscribed polygon, then it's easy to complete the proof. Now the first inequality, perimeter of inscribed polygon < circumference of circle, is clear; a straight line is the shortest distance between two points. The other inequality, circumference of circle < perimeter of circumscribed polygon, is certainly plausible, but is it actually possible to prove it using only tools available to the ancients? -- ==== How do you prove that the circumference of a circle is proportional to >> it's radius? >> What's the modern version and how did the greeks prove it? > The approximations to that ratio are based on >> inscribing/circumscribing regular polygons, each of which may be >> partitioned into congruent isosceles triangles. > For any such polygon, the odd(circuferential) side of any of those >> isosceles triangles is proportional to the other (radial) sides by >> similar triangles. > Thus each approximate circumference is proportional to its radius. Now if you know/accept that perimeter of inscribed polygon > < circumference of circle > < perimeter of circumscribed polygon, then it's easy to complete the proof. Now the first inequality, perimeter of inscribed polygon < circumference of circle, is clear; a straight line is the shortest distance between two points. The other inequality, circumference of circle < perimeter of circumscribed polygon, is certainly plausible, but is it actually possible to prove it >using only tools available to the ancients? Well, I don't know what Euclid's _definition_ of the circumference of a circle was, but if he had one I can't imagine that it was not something essentially equivalent to what we would call the least upper bound of the perimeter of an inscribed polygon. Assuming that, then first we need to show that (*) perimeter of inscribed polygon < perimeter of circumscribed polygon; that shows that the least upper bound mentioned above _exists_, and also makes the other inequlalities clear. (Regardless of how much of this they were explicit about, I do tend to suspect that they would have said the question was the same as (*), for some reason.) The circumcribed polygon is the union of triangles with height equal to the radius of the circle and base one of the sides of the polygon, so the _area_ of the circumcribed polygon is equal to r times its circumference. Otoh, let r' < r. Note that adding points to the inscribed polygon increases the circumference, by the triangle inequality. So starting with an inscribed polygon we can find another with larger circumference, and with area > r' * perimeter. Since the relation between the areas is clear, it follows that r' * inscribed perimeter < r * circumscribed perimeter for all r' < r, hence (*). David C. Ullrich ==== Of some relevance: once we know/accept that the area of a circle is proportional to the square of the radius (A = K*r^2) and that the circumference of a circle is proportional to the radius (C = l*r), how do we establish l = 2*k (especially in case the ancients are listening)? Here is a 'heuristic' argument of debatable merit (that I recently came up with, assuming nonetheless that it must have been known for quite a while): Partition a disk of radius r into a disk of radius r/2 and a ring of width r/2 ... and express the ring's area as both the difference between the areas of the two disks and the area of a trapezoid (that is stretched-out ring) of bases l*r, l*r/2 and height r/2: l = 2*k follows easily from K*r^2 - k*(r/2)^2 = (1/2)*(l*r + l*r/2)*(r/2). [In a similar manner one may compute the volume of the torus -- I recall that on my first Calculus final I asked the students to do that using Calculus, using Pappus' theorem, and ... using just their imagination :-) ] baloglouAToswego.edu ==== >Of some relevance: once we know/accept that the area of a circle is >proportional to the square of the radius (A = K*r^2) and that the >circumference of a circle is proportional to the radius (C = l*r), how do >we establish l = 2*k (especially in case the ancients are listening)? If P is a circumscribed polygon then as noted above the area of P is (r/2) times the perimeter of P. (If we don't define the circumference of a circle as the sup of the perimeter of inscribed polygons then I don't know how we _do_ define it. If we do define it that way then it follows easily that it's the inf of the perimeter of circumscribed polygons, qed.) >Here is a 'heuristic' argument of debatable merit (that I recently came up >with, assuming nonetheless that it must have been known for quite a while): Partition a disk of radius r into a disk of radius r/2 and a ring of width >r/2 ... and express the ring's area as both the difference between the >areas of the two disks and the area of a trapezoid (that is stretched-out >ring) of bases l*r, l*r/2 and height r/2: l = 2*k follows easily from >K*r^2 - k*(r/2)^2 = (1/2)*(l*r + l*r/2)*(r/2). [In a similar manner one may compute the volume of the torus -- I recall >that on my first Calculus final I asked the students to do that using >Calculus, using Pappus' theorem, and ... using just their imagination :-) ] baloglouAToswego.edu David C. Ullrich ==== A problem: * Find all primes of the form 4^n + n^Û I just can't do it! 4^1 + 1^4 = 5, is prime * If 2 divides n, 2 divides 4^n + n^4 so it can't be prime. * If 2 does noes not divide n and 5 doesn't either 5 divides 4^n + n^4 so it isn't prime. (4^n always ends in 4 for odd n and n^4 always ends in 1 for odd n that aren't divisible by 5. Thus 4^n + n^4 always ends in 5 and is divisible by 5. If anyone has a more elegant way of proving this, please let me know.) * This leaves us odd n that are multipiles of 5. I suspect these are all compsite too. 4^5 + 5^4 = 1649 = 17*97, at least. But how do you prove it? /david ==== >* Find all primes of the form 4^n + n^Û [This line ends with a non-ASCII character after a ^] >* If 2 divides n... >* If 2 does noes not divide n... This is one of those funny cases where a problem gets easier when you make it harder: I claim at that point you're actually looking at the broader question, Find all primes of the form n^4 + 4 m^4. Just for the record, I seem to recall that the polynomial n^4 + k^2 was recently proven to yield infinitely many prime values (as opposed to the polynomial j^2 + k^2, which is easy, and the polynomial 1 + k^2, which is still unknown). dave ==== > A problem: > * Find all primes of the form 4^n + n^Û Borrowing one of Einstein's ideas, we have 4^N+N^4 = (2^N+N^2)^2 - 2^(N+1)N^2, and the rest is relatively easy. ==== david escribi.97 en el mensaje > A problem: > * Find all primes of the form 4^n + n^Û I just can't do it! 4^1 + 1^4 = 5, is prime > * If 2 divides n, 2 divides 4^n + n^4 so it can't be prime. * If 2 does noes not divide n and 5 doesn't either 5 divides 4^n + n^4 > so it isn't prime. (4^n always ends in 4 for odd n and n^4 always ends > in 1 for odd n that aren't divisible by 5. Thus 4^n + n^4 always ends in > 5 and is divisible by 5. If anyone has a more elegant way of proving > this, please let me know.) * This leaves us odd n that are multipiles of 5. I suspect these are all > compsite too. 4^5 + 5^4 = 1649 = 17*97, at least. But how do you prove it? > Fon even n is obvious. For odd n, replace n = 2k+1. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== Fon even n is obvious. For odd n, replace n = 2k+1. > For odd n such that 5 does not divide n, this is obvious too ... but what about the case n = 5*i, where i is odd? ==== Julien Santini escribi.97 en el mensaje Fon even n is obvious. For odd n, replace n = 2k+1. > For odd n such that 5 does not divide n, this is obvious too ... but what > about the case n = 5*i, where i is odd? For n = 2k + 1, n^4 + 4^n = (n^2 + 2^(k + 1)n + 2^n)(n^2 - 2^(k + 1)n + 2^n) -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== Julien Santini escribi.97 en el mensaje Fon even n is obvious. For odd n, replace n = 2k+1. > For odd n such that 5 does not divide n, this is obvious too ... but what > about the case n = 5*i, where i is odd? For n = 2k + 1, n^4 + 4^n = (n^2 + 2^(k + 1)n + 2^n)(n^2 - 2^(k + 1)n + 2^n) -- Ignacio Larrosa Ca.96estro > A Coru.96a (Espa.96a) > ilarrosaQUITARMAYUSCULAS@mundo-r.com I searched for n^4+4^n = semiprime, i.e. (n^2 + 2^(k + 1)n + 2^n) and (n^2 - 2^(k + 1)n + 2^n) both prime and found only n=3,5,15,35,55 e.g. 3^4+4^3=145=5*29, 5^4+4^5=1649=17*97, ... can we find more terms or is there a fundamental reason that no semiprime representations exist beyond n=55? Hugo Pfoertner ==== >I searched for n^4+4^n = semiprime, i.e. >(n^2 + 2^(k + 1)n + 2^n) and (n^2 - 2^(k + 1)n + 2^n) both prime >and found only n=3,5,15,35,55 >e.g. 3^4+4^3=145=5*29, 5^4+4^5=1649=17*97, ... >can we find more terms or is there a fundamental reason that no >semiprime representations exist beyond n=55? It's quite reasonable that there should be only finitely many. Heuristically the probability of a number the size of these being prime is on the order of 1/n. The probability of both being prime would be on the order of 1/n^2. Since sum_n 1/n^2 < infinity, we should expect only a finite number of examples. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >I searched for n^4+4^n = semiprime, i.e. >(n^2 + 2^(k + 1)n + 2^n) and (n^2 - 2^(k + 1)n + 2^n) both prime >and found only n=3,5,15,35,55 >e.g. 3^4+4^3=145=5*29, 5^4+4^5=1649=17*97, ... >can we find more terms or is there a fundamental reason that no >semiprime representations exist beyond n=55? It's quite reasonable that there should be only finitely many. > Heuristically the probability of a number the size of these being > prime is on the order of 1/n. The probability of both being prime > would be on the order of 1/n^2. Since sum_n 1/n^2 < infinity, we > should expect only a finite number of examples. Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 There are no further solutions up to k=5000 -> n=5001. So the practical chances of finding another solution (if one exists) are rather limited. Hugo Pfoertner ==== There are no further solutions up to k=5000 -> n=5001. So the practical n=10001 sorry > chances of finding another solution (if one exists) are rather limited. Hugo Pfoertner ==== > For n = 2k + 1, n^4 + 4^n = (n^2 + 2^(k + 1)n + 2^n)(n^2 - 2^(k + 1)n + 2^n) > Magical !! Julien Santini ==== >>Fon even n is obvious. For odd n, replace n = 2k+1. >> > For odd n such that 5 does not divide n, this is obvious too ... but what > about the case n = 5*i, where i is odd? Forget the divisibility by 5, treat all the odd n the same. (you might benefit from writing one of the 4s differently) ==== > A problem: > * Find all primes of the form 4^n + n^åÛ I just can't do it! 4^1 + 1^4 = 5, is prime > * If 2 divides n, 2 divides 4^n + n^4 so it can't be prime. * If 2 does noes not divide n and 5 doesn't either 5 divides 4^n + n^4 > so it isn't prime. (4^n always ends in 4 for odd n and n^4 always ends > in 1 for odd n that aren't divisible by 5. Thus 4^n + n^4 always ends in > 5 and is divisible by 5. If anyone has a more elegant way of proving > this, please let me know.) * This leaves us odd n that are multipiles of 5. I suspect these are all > compsite too. 4^5 + 5^4 = 1649 = 17*97, at least. But how do you prove it? I set this as a challenge problem in my number theory course: http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html . -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== I set this as a challenge problem in my number theory course: > http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html . > But you didn't provide a proof in the answer sheet! =( ==== >> I set this as a challenge problem in my number theory course: >> http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html . >> But you didn't provide a proof in the answer sheet! =( > A short hint. Complete something. ==== I set this as a challenge problem in my number theory course: > http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html . > But you didn't provide a proof in the answer sheet! =( >> A short hint. Complete something. > The proof? ==== >> I set this as a challenge problem in my number theory course: >> http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html . > > But you didn't provide a proof in the answer sheet! =( > A short hint. Complete something. > The proof? > Something else first. ==== >> I set this as a challenge problem in my number theory course: >> http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html . But you didn't provide a proof in the answer sheet! =( Heh, heh, heh! -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== > 11/4/03 Dear teaching colleague, > I could sure use your help. I was just asked to speak next January at > the National Institute for the Teaching of Psychology and would like > to ask you for a few minutes of your time. My presentation at NITOP is > titled, What's difficult to teach in introductory statistics and how > to do it. Can you please take ten minutes and help me prepare for > this presentation by answering the following questions? 1. In your intro stat class, what three topics do you find most > difficult to teach your students? > 2. How do you teach these topics? What techniques, strategies, ideas, > and tools help you? > 3. What's the one coolest, most fun thing you do in your intro stat > class to help students learn? Let me partially answer your questions with an answer to the last one, Neil. When I teach (HS) introductory statistics, I want the students to realize the power of taking a sample to find a population. I pass out to the class a bunch of mini-bags of M&Ms, and ask them to predict how many red M&Ms are in the room. I write their prediction on the board. Then, I have ONE student open their bag. If they have, say 5 red M&Ms, and the class has 20 kids, all the predictions range close to 100 total, with a bit of variance (which we can calculate) because some students astutely predict that there may be bags out there loaded with red ones. Then another student opens their bag, and the predictions become less variant. In a class of 20 students, they all start feeling quite certain of their predictions after only 4 or 5 bags being opened. It gives them an excellent sense of how the number of samples is related to the degree of certainty of the predictions. We graph the range of predictions vs. the number of open bags, and they see how their variance and predictions became very stable after very few samples. Also, they realize how little an outlier bag affects the sum total. And they get a good understanding of how unusual it would be for that outlier bag to be one of the first bags sampled, but how it is possible. Overall, an excellent intro to sampling. And then we eat the M&Ms. --riverman ==== Could someone recommend a good Introduction to P.D.E.'s book ? I have heard that Basic Partial Differential equations by David Bleecker, George Csordas, and Darko Grundler book is nice, but I haven't had the chance to take a look at it. Any suggestions would be appreciated! TIA Lurch ==== Could someone recommend a good Introduction to P.D.E.'s book ? I have heard > that > Basic Partial Differential equations by David Bleecker, George Csordas, and > Darko Grundler book is nice, but I haven't had the chance to take a look at > it. Any suggestions would be appreciated! TIA Lurch This is not a direct answer but can't you search on Partial differential equations lecture notes and then make some kind of judgment about the book, derived from what the professor does in his course. Just a thought, of course. David Ames ==== Fritz has an original set of lectures notes from when he taught in NY, that is (in my opinion), better than his book. It's the best. Rogers and Renardy is not too bad either. MB Could someone recommend a good Introduction to P.D.E.'s book ? I have heard > that > Basic Partial Differential equations by David Bleecker, George Csordas, and > Darko Grundler book is nice, but I haven't had the chance to take a look at > it. Any suggestions would be appreciated! TIA Lurch This is not a direct answer but can't you search on > Partial differential equations lecture notes > and then make some kind of judgment about the book, derived from what > the professor does in his course. Just a thought, of course. David Ames ==== Could someone recommend a good Introduction to P.D.E.'s book ? I have heard > that > Basic Partial Differential equations by David Bleecker, George Csordas, and > Darko Grundler book is nice, but I haven't had the chance to take a look at > it. Any suggestions would be appreciated! TIA Lurch 1. Partial Differential Equations: An Introduction by Walter A. Strauss 2. Partial Differential Equations 4e by Fritz John ==== Could someone recommend a good Introduction to P.D.E.'s book ? I have heard > that > Basic Partial Differential equations by David Bleecker, George Csordas, and > Darko Grundler book is nice, but I haven't had the chance to take a look at > it. Any suggestions would be appreciated! TIA Lurch 1. Partial Differential Equations: An Introduction by Walter A. Strauss > 2. Partial Differential Equations 4e by Fritz John ==== Charlie Johnson maky m. > Charlie Johnson Could someone recommend a good Introduction to P.D.E.'s book ? I have > heard > that > Basic Partial Differential equations by David Bleecker, George Csordas, > and > Darko Grundler book is nice, but I haven't had the chance to take a look > at > it. Any suggestions would be appreciated! TIA Lurch 1. Partial Differential Equations: An Introduction by Walter A. Strauss > 2. Partial Differential Equations 4e by Fritz John > Also by Fritz John are two good overviews for physicists, one on PDEs and one on integral equations. They are in a book with the strange title Mathematics Applied to Physics published in the 70's by Springer-UNESCO. It would be hard to buy, but a university library might have it. Larry ==== L E W A N S 12 5 23 1 14 19 = 74 Audrey sat at the MS Society table turning the street fair downtown on 2nd Ave. She was the first of the three nubile sweeties seated at this table to provide stats. And Audrey was the second person to provide family stats table, it was the next table over. Audrey liked the work I did on Shadia's stats and then told me about her family. Audrey and I meet on the 1288th day of the century, there are 1288 verses in Numbers. 74+ Dad 5 4 40 96/270 +6162 74+ Mom 21 3 45 80/285 +4351 74+ Sib 21 11 /40 74+ Sib 22 10 /70 74+ Sib 1 3 /305 74+ Sib 21 1 21/ 194 Carmen 9 9 76 253/113 7144 Carmen 54 Frances 66 Lewans 74 74+ Sib 19 9 /103 213 Audrey 9 5 80 130/236 8482 Audrey 74 Lynn 65 Lewans 74 74+ Sib 31 12 /0 And I met sister Carmen on January 17th 1998 (2001 days earlier), Carmen also gave out birthdays for all 10 family members and claimed that dad was born on April 7th 1940, claimed that mom was born on March 21st 1944, and the kids are a sister born on November 26th 1969, then a sister on October 25th 1971, followed by a brother on March 1st 1973, then another brother on January 20th 1975, followed by Carmen, followed by a sister on September 19th 1978, followed by another sister (Audrey) on May 9th 1980, and then followed by the last sister on December 12th 1982. Between Carmen and Audrey, they gave out different birthdays for both parents and different birthdays for kids 1, 2 and 4. Shown above are the stats as Audrey provided (except that she never provided Carmen's name nor year of birth), both Carmen and Audrey may have provided incorrect birthdays. Non-Primes 1 57 110 158 207 4 58 111 159 208 6 60 112 160 209 8 62 114 161 210 9 63 115 162 212 10 64 116 164 213 <-166th 12 65 117 165 214 14 66 118 166 215 15 68 119 168 216 16 69 120 169 217 18 70 121 170 218 20 72 122 171 219 21 74 123 172 220 22 75 124 174 221 24 76 125 175 222 25 77 126 176 224 26 78 128 177 225 27 80 129 178 226 28 81 130 180 228 30 82 132 182 230 32 84 133 183 231 33 85 134 184 232 34 86 135 185 234 35 87 136 186 235 36 88 138 187 236 38 90 140 188 237 39 91 141 189 238 40 92 142 190 240 42 93 143 192 242 44 94 144 194 243 45 95 145 195 244 46 96 146 196 245 48 98 147 198 246 49 99 148 200 247 50 100 150 201 248 51 102 152 202 249 52 104 153 203 250 54 105 154 204 252 55 106 155 205 253 56 108 156 206 254 Audrey has 6 lettered first and last names, 66.666...% of her names are 6 lettered. Her 6 lettered names both add to 74 (a factor of 666). Her first two initials add to 13 (6th prime), her last two initials add to the 24 (6+6+6+6) chapters of Bible Books 6 and 10 (6th non-prime). She was born on day 130 (Numbers 13) while her name adds to 213, prettier as 21 is the 13th non-prime. Her given names add together for 139, her middle name adds to a multiple of 13. Her name adds to 213 (166th non-prime). Her first 13 letters add to 179 (the 13th prime in prime position). She has 10 different letters (6th non-prime). She has 6 vowels and 10 (6tgh non-prime) consonants. The parents were born on days of the month averaging 13 (6th prime). Carmen's and Audrey's birthdays are likely correct, these two sisters are separated by 3.66 years. And note that Carmen (6 letters) has a middle name adding to 66. 1-50 - Genesis 51-90 - Exodus 91-117 - Leviticus 118-153 - Numbers 154-187 - Deuteronomy 188-211 - Joshua 930-957 - Matthew 958-973 - Mark 974-997 - Luke 998-1018 - John 1019-1046 - Acts 1047-1062 - Romans 123 <-Numbers 6, it is three times the 13th prime (41+41+41), keeping in mind that 13 is the 6th prime 188 <-the opening chapter of Book 6 is 6x6x6 short of the 404 verses of Bible Book 66, it is the 6th prime squared (13x13) short of the 357 verses of Daniel (also in part about 666) 193 <-Book 6 chapter 6 is the 44th prime, while 44 is in turn 66.666...% of 66 211 <-the terminating chapter of Book 6 is approximately 66.6% of the 66th prime (317) 357 <-the opening chapter of Book 6 plus the 6th prime squared is the 357 verses of Daniel (in part about 666) 404 <-the 6th prime squared (13x13) plus the 6th prime squared (13x13) plus 66 adds to the 404 verses of Bible Book 66 1062 <-666 plus 6x66 is a combination of the 658 verses of Bible Book 6 plus the 404 verses of Bible Book 66, and is the terminating chapter of New Testament Book 6 1070 <-666 plus the 404 verses of Book 66 is the 1070 verses of Job (Book 6+6+6) 1213 <-Exodus terminates at chapter 90 (66th non- prime) with 1213 verses (the 198th or the 66+66+66th prime) 1292 <-the 658 verses of Book 6 plus twice the 66th prime (317) is the 1292 verses of Isaiah (the Book contains 66 chapters) The parents were born in years adding to 85 (perhaps). Audrey's and her parents were together born 485 days closer to the beginning of their years than to the end of their years (perhaps). Audrey's unrepeated letters add to 85. Primes Non-Primes 2 1 3 4 5 6 7 <-4th-> 8 <-Bible Book 8 contains 11 4 chapters, pretty as 13 8 is the 4th non-prime 17 19 -- 77 <-the first 8 primes plus 8 more adds to the 85 verses of Bible Book 8 Perhaps the parents were born on days 5 and 21, these Bible Books differ in length by 737 verses. Perhaps the family was born on days 5, 21, 21, 22, 1, 21, 9, 19, 9 and 31, together these Bible Books contain 7376 verses. Perhaps the kids were born an average of 1737 days after their parent's birthdays. Perhaps the parents were an average of 37.61 years old when Audrey was born. Audrey's vowels add to 37.41% of her consonants. Her first and last names both add to 74 (37+37). Perhaps dad was born on the 5th, corresponding to Deuteronomy with 959 (7x137) verses. Perhaps mom was born on the 80th day of the year while Audrey was born in 80. The last 4 kids may have been born on days of the year adding to 1009 (the 80th chapter of The New Testament). Dad was born in 40, mom on day 80 (40+40). Audrey was born in 80 (40+40). The kids were either born with 1211 days remaining in their years or at least generally together have their birthdays when there are 1211 days remaining in their years, it's exactly 173 weeks (the 40th prime). Or according to Carmen, the last 4 kids were born on days of the year adding to 1010, corresponding to the 81st chapter of the New Testament, pretty as Carmen claims that mom was born on the 81st day of the year. Primes 2 61 149 3 67 151 5 71 157 7 73 163 11 79 167 13 83 173 <-40th 17 89 179 19 97 181 23 101 191 29 103 193 31 107 197 37 109 199 41 113 211 43 127 223 47 131 227 53 137 229 59 139 233 Dad was born in 40, mom in 45, while Audrey's 10 different letters add to 142 (40.45% of the possible total). That possible total is 351 (117+117+117), it is 1 through to the 17th non-prime (26). Numbers 17 (134) is 217 short of the numbers up to the 17th non-prime while 217 is the 170th non-prime, while chapter 170 is Deuteronomy 17. And Numbers 7 (124) is the 7x7th prime (227) short of the numbers up to the 17th non-prime. Numbers opens at chapter 118 (twice the 17th prime) and terminates at 153, it's 1 through 17 and is the 117th non-prime, it's the number of fish in the net in John 21 (7+7+7), while there are 21 (7+7+7) chapters in Book 7. There are 45 chapters in the Bible that contain the length of 17 verses (198-153=45, where 198 is the 153rd non-prime). Leviticus begins with 17 verses and terminates at chapter 117 with 17+17 verses. There are 17 verses at chapters 1 and 3, and 59 (the 17 prime) verses at chapter 13, so the 17's and the 17th prime are at chapter numbers adding to 17 (1+3+13=17). The first 17 versed chapters in the Bible are at chapters 91 and 93, together for 184, or the 167 verses of Book 17 plus 17 more. Leviticus contains 859 verses, it ends in 59 (the 17th prime). The first 17's in the Bible surround chapter 92 (the 4x17th non-prime): Leviticus --------- 91 1 17 92 2 <-68th (4x17th) non-prime 93 3 17 94 4 95 5 96 6 97 7 98 8 99 9 100 10 101 11 102 12 103 13 59 <-17th prime 104 14 105 15 106 16 107 17 108 18 109 19 110 20 111 21 112 22 113 23 114 24 115 25 116 26 117 27 34 <-17+17 Mom was born on the 21st and managed to give birth twice on the 21st (if the stats Audrey provided are correct). Audrey and mom were born on days of the year adding to 210. Audrey's name adds to 213. Perhaps the second of the kids was born 215 days after mom's birthday. Audrey's first 7 and last 7 letters add together for 174, corresponding to Deuteronomy 21 (7+7+7), keeping in mind that Bible Book 7 contains 21 (7+7+7) chapters. Note that 21, the 21st prime (73) and the 21st non-prime (32) average 42 (21+21). I don't want to spend a lot of time on this family for Carmen and Audrey have provided different birthdays for 5 of the 10 family members. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 --- 167 Esther Book 17 The parents were together 27477 days old when Audrey was born. Audrey was born exactly 7 weeks after mom's birthday. Audrey was born 34 (17+17) days after dad's birthday. Audrey's vowels add to 58 (the 7 primes up to 17) while her consonants add to the 155 verses of Bible Book 7x7. Her repeating letters add to 128 (2 to 7th). Audrey's 130th day of birth adds with her 213 valued name for 343 (7x7x7). Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 Audrey's given names add to 139 (17+17th prime). The 17 letters missing from her given names add to 251 (the first 17 non-primes). In her given names, her repeating letters exceed her unrepeated letters by 17. In her full name, her unrepresented letters exceed her represented letters by 67 (Exodus 17). Audrey is 23.17 years old. Audrey might marry Marcia and me and take our 117 valued last names, and then become a Queen like Esther in Book 17. And note that just 14 (7+7) days ago I met Emily Karen Lewans at Burger King, her name also presently adds to 187. 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 187 Emily 31 10 81 304/61 9022 Emily 64 Karen 49 Lewans 74 256 Audrey 9 5 80 130/236 8482 Audrey 74 Lynn 65 Acevedo-Kabatoff 55-62 Anyway, if you people think that you have the right to use my abusive parents as tools and arrest and torture me, then I think that I should have the right to ask women to marry me, or to marry Marcia and me, our last names add together for the 117 verses of Song of Solomon, it's the Bible's Book of Love. The nubile sweety was born on the 6th and has a 6 lettered first name). Isaiah is the Book with 66 chapters, pretty as it is Book 23, or the 6th prime plus the 6th non-prime (13+10=23). Isaiah 4, 12 and 20 (adds to 6x6) each contains 6 verses. Isaiah 4:1 is about Marcia (and me), and 6 other women who are capable of feeling shame rather than pride, greed or lust, or who limit their love for traditions and for people who abide by their traditions. You people have Egyptian penises on the roofs of your churches and lined city streets with representations of penises, and had me tortured for years for saying so, others just sat back in silence while they were doink this to me, and similarly you remain silent and compassionless now that the arrests and torture have ceased. You people spent millions of dollars having me tortured, and then annually you spend billions on your decorated trees, I begged and begged for assistance to flee the country (they tortured me for years at the U of S) and you people are so cheap that you can't even offer to buy me a cookie when I bust my ass to show you evidence that your very name is a gift from God, you cheap, filthy, compassionless assholes won't even spend 48 cents on a stamp so that you life to show you evidence that your very name is a gift from God!!! All you are really good for is to have your stats posted on the usenet and be used as an example to others, and look, here you are!!! Should Audrey marry me, great, but if Marcia marries me and then Audrey marries Marcia and me, then any of Audrey's siblinks that are not Catholic (necromancers) are goink to win themselves a shiny new Cadillac!!! Good luck and may God bless you!!! Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! ==== L E W A N S >12 5 23 1 14 19 = 74 Audrey sat at the MS Society table turning the street fair downtown on >2nd Ave. She was the first of the three nubile sweeties seated at this table >to provide stats. And Audrey was the second person to provide family stats << The following (courtesy of Waxy.org) is sort of an unofficial FAQ explaining the psychotic nonsense posted to Usenet by Shawn Daryl Kabatoff AKA Dar, AKA Probababbilities. And now AKA marcia and me. WARNING: Read below before even thinking about responding to this twit. http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml#000643 Usenet has the tendency to provide a public forum for those who would normally be scribbling in a closet. For example, take Daryl Shawn Kabatoff. For the last few years, he's methodically gathered statistics from various sources, ranging from local newspaper obituary pages to the food court of the Saskatoon Midtown Plaza mall. With all the raw data he's collected, he's attempting to prove daily that our full names are in mathematical harmony with our birthdays. about, starting with calculations related to their birthdate and full names, blending in whatever other personal information about their family members, spouses, birthplace, and career he's been able to zealotry, and personal torment. I've never seen anything like it. With all the prime numbers, Fibonacci sequences and biblical references, it's like reading the notebooks of Maximillian Cohen and John Nash combined. Unsurprisingly, several posts unfold to reveal a history of painful mental illness. If you have some time, take a look. I've detailed his posting history and a several sample posts below. January 27, 1999 to July 5, 2000 as Catsco@home.com December 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.ca Oct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.ca posts have been removed from Google Groups archive) Selected Posts: Tessa Lynne Smith Dastageer Sakhizai and Helen Smith Brett David Maki Andrew Meredith Cotton Amanda Dawn Newton Mona Marie Etcheverry Tony Peter Nuspl Lisa Charlene McMillan Grant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my current residence. and every single place he's mentioned in his posts (most notably nervous harold's and the roastary) were either places i've been (as it's a small city of 200K) or hangouts, ie. the two places out if they know of him, they (my friends that is) being of the broadway-centred slacker ilk. myself, too, until i got out of there. eh, anyways. thought it odd to see all this. midtown mall. i ate my meals there, whilst waiting several days in line for star wars episode one, at the theatre across the street. posted by andy raad on May 22, 2002 06:20 PM Fascinating. It's like he's trying to take chaos and bind it into whatever rules he can find, religious, logical and otherwise. Numbers and math have a reliable pattern, something that can always be proven to true or false. People and religion do not. It reminds me of Darren Aronofsky's movie Pi. It's the story of an paraniod genius who is trying to find a pattern in Pi. A group that takes interest in his work is convinced that the existence of Pi, a number whose existence can be proven but no quantified, is proof of the existence of God. Kabatoff's hunt for patterns in something as random as name selection is a way to reconcile his deeply logical thought process with his conflicting religious views. posted by matt on May 23, 2002 11:19 AM asking him if he'd be willing to create a numerological analysis for me. I also asked him if he had seen either Pi or A Beautiful Mind, and what he thought of them. If he replies, I'll be sure to post it. posted by Andy Baio on May 23, 2002 11:24 AM I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbed pumpkin pie all over my breasts for him, and my breasts turned orange. I am a pumpkin for Shawn. posted by Trisha Blondie on July 24, 2002 10:41 PM Um, that's swell. So, you're in love with him? posted by Andy Baio on July 25, 2002 07:10 AM Shawn once went to a funeral for a Jehovah Witness that shot himself and the lemon tarts were very bad, they were not only sour but were rubbery as well. Shawn said that the guy was some kind of Jehovah Witness prophet, he saw in advance that the lemon tarts at his funeral were to be very very bad, and so he shot himself. Shawn said that he never ate pumpkin pie at a funeral but would like to some day. Shawn likes pumpkin pie and so I have been practicing to make very good pumpkin pies. posted by Trisha Blondie on July 25, 2002 02:49 PM Shawn said that the lemon tarts were sour, bitter and rubbery. posted by Trisha Blondie on July 30, 2002 12:32 AM I don't think this guy takes notes. I think he has Total Recall, and it has driven him insane... posted by Todd Smith on December 26, 2002 11:00 AM Oh... I almost forgot... I didnt spend thousands of dollars a day tormenting Daryl... We got a deal on tormenting that fiscal year, it only came to about 37cents a day.... posted by Dr Claw on December 30, 2002 01:56 AM Mr. Kabatoff attempts to portray himself as a victim, but in fact he is a violent predatory pedophile who is well known to his local law enforcement. In his post to multiple newsgroups with the subject Collecting Mail For The Coming Anti-Christ, he encourages mothers to send him photos of their naked daughters. Mr Kabatoff explains, I Ant-Christ) that were of underage children unless the parent was signing consent. He is banned from virtually all the shopping malls in his community because he stalks young people and sexually harasses them. He has an extensive arrest record which includes sexual molestation charges. He's been hospitalized in mental institutions about his contact with young girls in many posts. Search newsgroup archives for posts by him containing the word nubile. As part of his harrassment, he provides personal details in a public forum, such as the real names of real children, in these and other posts. About one wanted her and her sister dead. http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+dead+or+in+my+bed&hl= en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%40ID-136124.news.dfncis.de&rnu He not only curses children and prays for their death in his posts, he also enjoys attending the funerals of young people: And so, since nubile sweeties are found in greatest abundance at the funerals of high school students, then it is the funerals of high school students that make the very very best funerals, especially if there is food... I stuff my face (and my pockets) with all the good food and look at all the pretty nubile sweeties and have the time of my life.. .http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+nubile+sex&hl=en&l r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%40twister.socal.rr. com&rnum=1 Many of his posts are sent to alt.teens.advice. However, he liberally spams, floods and crossposts his off-topic threatening and offensive missives to countless newsgroups. Some people HAVE problems and some folks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. When he sends these posts to any newgroup, please help by reporting him to I knew of him when I was attending the University of Saskatchewan. He'd hang out in the Arts computer lab and all you'd see is screens of numbers racing by on his laptop. I have an original copy of his Collecting Mail for the Coming Anti-Christ pamphlet, and have seen him be hauled away by campus security on more than one occasion. My friends and I refer to him as Crazy Number Man. I've been posting to (and about) Shawn for over two years with big gaps in between. He has seen Pi and didn't like it and didn't think it resembled him at all. (Wrong, it fits him to a tee) He doesn't have total recall and has stated that he travels with a lap top to notate items. Also, he uses cut n' paste a lot if you read all the way through his ramblings. He is anti-social as shown by his angry statements towards those who, by his own admission, have been kind (but not kind enough) to him. Still, he's intelligent and seems to be able to take a joke on occassion. That's where I came in. ALOHA if it comes from anyone not already in my addressbook. I'll never even see it) ==== >L E W A N S >12 5 23 1 14 19 = 74 Audrey sat at the MS Society table turning the street fair downtown on >2nd Ave. She was the first of the three nubile sweeties seated at this table >to provide stats. And Audrey was the second person to provide family stats > << explaining the psychotic nonsense posted to Usenet by Shawn Daryl > Kabatoff AKA Dar, AKA Probababbilities. And now AKA marcia and > me. WARNING: Read below before even thinking about responding to this > twit. http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml#000643 > Usenet has the tendency to provide a public forum for those who would > normally be scribbling in a closet. For example, take Daryl Shawn > Kabatoff. For the last few years, he's methodically gathered > statistics from various sources, ranging from local newspaper > obituary pages to the food court of the Saskatoon Midtown Plaza mall. > With all the raw data he's collected, he's attempting to prove daily > that our full names are in mathematical harmony with our birthdays. > about, starting with calculations related to their birthdate and full > names, blending in whatever other personal information about their > family members, spouses, birthplace, and career he's been able to > zealotry, and personal torment. I've never seen anything like it. > With all the prime numbers, Fibonacci sequences and biblical > references, it's like reading the notebooks of Maximillian Cohen and > John Nash combined. Unsurprisingly, several posts unfold to reveal a > history of painful mental illness. If you have some time, take a look. > I've detailed his posting history and a several sample posts below. > January 27, 1999 to July 5, 2000 as Catsco@home.com December 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.ca Oct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.ca posts have been removed from Google Groups archive) > Selected Posts: > Tessa Lynne Smith > Dastageer Sakhizai and Helen Smith > Brett David Maki > Andrew Meredith Cotton > Amanda Dawn Newton > Mona Marie Etcheverry > Tony Peter Nuspl > Lisa Charlene McMillan > Grant Allyn Wood Comments > scarier still is that saskatoon is my hometown, though not my current > residence. and every single place he's mentioned in his posts (most > notably nervous harold's and the roastary) were either places i've > been (as it's a small city of 200K) or hangouts, ie. the two places > out if they know of him, they (my friends that is) being of the > broadway-centred slacker ilk. myself, too, until i got out of there. > eh, anyways. thought it odd to see all this. midtown mall. i ate my > meals there, whilst waiting several days in line for star wars episode > one, at the theatre across the street. > posted by andy raad on May 22, 2002 06:20 PM Fascinating. It's like he's trying to take chaos and bind it into > whatever rules he can find, religious, logical and otherwise. Numbers > and math have a reliable pattern, something that can always be proven > to true or false. People and religion do not. It reminds me of Darren > Aronofsky's movie Pi. It's the story of an paraniod genius who is > trying to find a pattern in Pi. A group that takes interest in his > work is convinced that the existence of Pi, a number whose existence > can be proven but no quantified, is proof of the existence of God. > Kabatoff's hunt for patterns in something as random as name selection > is a way to reconcile his deeply logical thought process with his > conflicting religious views. > posted by matt on May 23, 2002 11:19 AM asking him if he'd be willing to create a numerological analysis for > me. I also asked him if he had seen either Pi or A Beautiful Mind, and > what he thought of them. If he replies, I'll be sure to post it. > posted by Andy Baio on May 23, 2002 11:24 AM I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbed > pumpkin pie all over my breasts for him, and my breasts turned orange. > I am a pumpkin for Shawn. > posted by Trisha Blondie on July 24, 2002 10:41 PM Um, that's swell. So, you're in love with him? > posted by Andy Baio on July 25, 2002 07:10 AM Shawn once went to a funeral for a Jehovah Witness that shot himself > and the lemon tarts were very bad, they were not only sour but were > rubbery as well. Shawn said that the guy was some kind of Jehovah > Witness prophet, he saw in advance that the lemon tarts at his funeral > were to be very very bad, and so he shot himself. Shawn said that he > never ate pumpkin pie at a funeral but would like to some day. Shawn > likes pumpkin pie and so I have been practicing to make very good > pumpkin pies. > posted by Trisha Blondie on July 25, 2002 02:49 PM Shawn said that the lemon tarts were sour, bitter and rubbery. > posted by Trisha Blondie on July 30, 2002 12:32 AM I don't think this guy takes notes. I think he has Total Recall, and > it has driven him insane... > posted by Todd Smith on December 26, 2002 11:00 AM Oh... I almost forgot... I didnt spend thousands of dollars a day > tormenting Daryl... We got a deal on tormenting that fiscal year, it > only came to about 37cents a day.... > posted by Dr Claw on December 30, 2002 01:56 AM Mr. Kabatoff attempts to portray himself as a victim, but in fact he > is a violent predatory pedophile who is well known to his local law > enforcement. In his post to multiple newsgroups with the subject > Collecting Mail For The Coming Anti-Christ, he encourages mothers to > send him photos of their naked daughters. Mr Kabatoff explains, I > Ant-Christ) that were of underage children unless the parent was > signing consent. He is banned from virtually all the shopping malls > in his community because he stalks young people and sexually harasses > them. He has an extensive arrest record which includes sexual > molestation charges. He's been hospitalized in mental institutions > about his contact with young girls in many posts. Search newsgroup > archives for posts by him containing the word nubile. As part of his > harrassment, he provides personal details in a public forum, such as > the real names of real children, in these and other posts. About one > wanted her and her sister dead. > http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+dead+or+in+my+bed&hl= en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%40ID-136124.news.dfncis.de&rnu > He not only curses children and prays for their death in his posts, he > also enjoys attending the funerals of young people: And so, since > nubile sweeties are found in greatest abundance at the funerals of > high school students, then it is the funerals of high school students > that make the very very best funerals, especially if there is food... > I stuff my face (and my pockets) with all the good food and look at > all the pretty nubile sweeties and have the time of my life.. > .http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+nubile+sex&hl=en&l > r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%40twister.socal.rr. > com&rnum=1 > Many of his posts are sent to alt.teens.advice. However, he liberally > spams, floods and crossposts his off-topic threatening and offensive > missives to countless newsgroups. Some people HAVE problems and some > folks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. When > he sends these posts to any newgroup, please help by reporting him to I knew of him when I was attending the University of Saskatchewan. > He'd hang out in the Arts computer lab and all you'd see is screens of > numbers racing by on his laptop. I have an original copy of his > Collecting Mail for the Coming Anti-Christ pamphlet, and have seen > him be hauled away by campus security on more than one occasion. My > friends and I refer to him as Crazy Number Man. I've been posting to (and about) Shawn for over two years with big > gaps in between. He has seen Pi and didn't like it and didn't think it > resembled him at all. (Wrong, it fits him to a tee) He doesn't have > total recall and has stated that he travels with a lap top to notate > items. Also, he uses cut n' paste a lot if you read all the way > through his ramblings. He is anti-social as shown by his angry > statements towards those who, by his own admission, have been kind > (but not kind enough) to him. Still, he's intelligent and seems to be > able to take a joke on occassion. That's where I came in. ALOHA if it comes from anyone not already in my addressbook. > I'll never even see it) I hope somebody *did* report this to his ISP, because I don't know how to, and are not going to lose time in trying to find out how to, I just set him on my Blocked Senders List. H. ==== >> I hope somebody *did* report this to his ISP, because I don't know how to, > and are not going to lose time in trying to find out how to, I just set him > on my Blocked Senders List. H. What about reporting Thomas's abuse of the NG? Anyone up for that? Myself I've just killfiled the pair of 'em. Dave. ==== Someone asked a way to calculate integral from 0 to infinity of sin(x) / x without using residues. Probably the most elegant way (imho) is to use so called Feynman's trick: Consider an integral (containing a parameter p) from 0 to infinity of e^(-px)*sin(x) wich we denote by I(p). It is easy to evaluate integrating by parts: I(p)=1/(p^2+1) Now the trick is to integrate I(p) from 0 to infinity with respect to p under the integral sign : int(0->oo) 1/(p^2+1) =pi/2 ==== > Someone asked a way to calculate integral from 0 to infinity of sin(x) / x > without using residues. Probably the most elegant way (imho) is to use so called Feynman's trick: Consider an integral (containing a parameter p) from 0 to infinity of > e^(-px)*sin(x) wich we denote by I(p). It is easy to evaluate integrating by > parts: I(p)=1/(p^2+1) Now the trick is to integrate I(p) from 0 to infinity with respect to p > under the integral sign : int(0->oo) 1/(p^2+1) =pi/2 Wonderful! That was just the kind of evaluation I was thinking of. Quite a standard trick, actually: Introduce an additional unknown parameter and solve the more general problem. It's like one of them factoring algorithm where - starting from some large number n - the first step involves multiplying the number by some small fixed constant. I think it is the Number Field Sieve, but please correct me if I'm wrong. -Michael. ==== >> Someone asked a way to calculate integral from 0 to infinity of sin(x) / x >> without using residues. Here's yet another derivation: 1. integral from 0 to infinity of sin(x) / x = limit as t->0 of sum from n=1 to infinity of sin(nt)/n 2. sum from n=1 to infinity of sin(nt)/n = imaginary part of sum from n=1 to infinity of exp(int)/n 3. sum from n=1 to infinity of exp(int)/n = log(1/(1 - exp(it))) (valid for 0 < t < 2pi) 4. 1/(1 - exp(it)) = exp(-it/2 + i pi/2)/(2 sin(t/2)) 5. imaginary part of (log(1/(1 - exp(it)))) = -t/2 + pi/2 Taking the limit as t->0 gives the result: integral from 0 to infinity of sin(x) / x = pi/2 -- Daryl McCullough ==== alternative ending to matrix revolutions (no spoilers) oracle: everything that has a beginning has an end neo: (thinks long and hard) ...erm, what about the positive integers 1,2,3,4,5... ? oracle: er...(thinks long and hard, self destructs, causes system failure) neo: whoa! i did it. ==== > alternative ending to matrix revolutions (no spoilers) oracle: everything that has a beginning has an end neo: (thinks long and hard) ...erm, what about the positive integers > 1,2,3,4,5... ? oracle: er...(thinks long and hard, self destructs, causes system > failure) neo: whoa! i did it. > ted: excellent adventure, dude. ==== > When we say a function f(t) is smooth, does this mean that > f has infinite differentials with respect to t? Or any other formal definition on this? Fred Reading all these responses is very interesting, as I had no idea that smooth could mean different things. When I was in grad school in Real Analysis, I recall dealing with smooth functions which were defined to be those which had a continuous first derivative. Each time I ran into smooth thereafter, I just assumed it meant the same thing. Jonathan Hoyle Gene Codes Corporation ==== > > > >... >> >I have seen smooth used for once continuously differentiable, twice >continuously differentiable, C^infty and presumably anything in >between, ... > >Indeed, I have seen it used for something like the following function: >> f(x) = sqrt(x) when x >= 0 >> = -sqrt(-x) when x <= 0. >>Has certainly a pretty smooth appearance. There is no standard >>definition in analysis. > >The graph of this f is certainly a smooth curve. Hmm, can we come up >with an analytic map g: (0,1) -> R^2 whose image is this curve, > >Don't think so, haven't thought about it. Probably more important >>is to point out that the smoothness of a curve is typically not >>measured by how smooth a map has the curve as its _image_. >>... >>David C. Ullrich >> > >True, but a reasonable definition of smoothness for an implicitly >defined curve is the maximal degree of smoothness of a *non-singular* >parametrization of the curve [non-singular means the first derivative >is nowhere zero). > So here is another interesting example: The function f: R -> R, x |-> cubrt(x) is not even C1, but the curve g: (-pi/2, pi/2) -> R^2, t |-> (tan^3 t, tan t) is analytic and non-singular. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== >> ... >> I have seen smooth used for once continuously differentiable, twice >> continuously differentiable, C^infty and presumably anything in >> between, ... Indeed, I have seen it used for something like the following function: >> f(x) = sqrt(x) when x >= 0 >> = -sqrt(-x) when x <= 0. >> Has certainly a pretty smooth appearance. There is no standard >> definition in analysis. >The graph of this f is certainly a smooth curve. Hmm, can we come up >with an analytic map g: (0,1) -> R^2 whose image is this curve, Don't think so, haven't thought about it. Probably more important >>is to point out that the smoothness of a curve is typically not >>measured by how smooth a map has the curve as its _image_. >> ... >>David C. Ullrich True, but a reasonable definition of smoothness for an implicitly >defined curve is the maximal degree of smoothness of a *non-singular* >parametrization of the curve [non-singular means the first derivative >is nowhere zero). As I pointed out in the paragraph you omitted. >John Mitchell David C. Ullrich ==== In sci.math and sci.math.num-analysis alex asked about > Integrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2 + 2*Cos[x]]],{x,0,Pi}] I responded in the latter newsgroup but basically quit simplifying when > it became clear the answer would involve elliptic integrals. him understand modular forms and I thought I could take this opportunity > to add some details to my prior post. [Follow-ups set to sci.math. ] < --- very nice explanations snipped --- > Is there something readable about that 'geometric' view (not restricted to modular forms) in a book? I liked it. --- ==== Given the product of two matrices A and B, denoted as C=AB; suppose A is on site 1, and B is on site 2; if it is not possible to move A and B together to compute C, is it possible to compute the eigenvector of C from A and B, respectively? I remember in Quantum Computation, if C is the tensor product (or density product, sorry, I am not sure) of two matrices A and B, the eigenvectors of C can be computed by the eigenvectors of A and eigenvectors of B. But how about the regular matrix product? Any ideas? thanks a lot roy ==== >Given the product of two matrices A and B, denoted as C=AB; >suppose A is on site 1, and B is on site 2; if it is not possible to >move A and B together to compute C, is it possible to compute the >eigenvector of C from A and B, respectively? Well, if you give me the (complex) eigenvectors and the eigenvalues of both A and B, I can sort of rebuild A and B, compute C, and then compute eigenvalues and eigenvectors. But without full information about the eigenstructure of A and B, I don't see how you can get much information about C. Think geometrically. Suppose you have a rubber sheet nailed at one point to the table top. One person comes in and stretches the sheet, first in one direction (pulling both sides away from the nail) and then stretches it in the perpendicular direction by a different amount. A second person then comes in and does likewise, but with a whole new pair of directions and a new pair of stretching factors. With complete information about the four factors and two directions, you can express how the combination of their moves can be accomplished with just a single pair of stretches. With incomplete information, there's not much you can say in general. (Try a simple case: each person simply doubles lengths in a single direction and then stops. That's not enough information to know whether the sheet is stretched by a factor of 4 somewhere, or by a factor of 2 in every direction, or by some intermediate amounts.) BTW, there is not really any mathematical meaning to moving two matrices together. dave ==== I am very interested in how handle this problem. If I understand correctly then C=A*B. I can only think practical, and I see that we have a problem if we do not know one of the three variables. Let's suppose we know C and we want to find A and B. What would help is some information regarding the distance between A and B or A towards C or B towards C. If this information is not available the only thing we do know is that A<= SQR(C) and B>=SQR(C) David ==== : > : > Random thoughts on creating a theory of sets prior to a theory of : > propositions and quantifiers: : > : > Let's start with the empty set, 0, and logical identity, =, then we can : > define T, for true, by : > : > T =def 0 = 0 : > : > Let's define ordered pair a la Kuratowski, then we can define : > conjunction by : > Kuratowksy defines as {a,{a,b}}. But how do you make sense of : > that latter notation at this stage of the presentation? Note that in : > ZF, you can only make sense of it because of the Axiom of Pairing; you : > can only verify that it satisfies the ordered pair axiom because of : > the Axiom of Extensionality. Those axioms both seem to require the : > apparatus of first order logic to be formulated. I already said that. Why don't *I* rate a reply? : Well, we cannot define un-ordered pair in context by : : u in {x,y) iff u = x or u = y : : since we don't have or. So let's take {x,y} as primitive and define : {x} =def {x,x} : =def {{x},{x,y}} Nobody is impressed. You'd be much better off just taking as primitive. Your mission, now, which will be much harder, is, given that you've taken {x,y} as primitive, how on earth is anybody supposed to parse {x,y,z} ? More to the point, you said before that we don't have 'or', but you have a much bigger problem: you don't have *in*, EITHER. What good does it do you to take {x,y} as primitive, and to claim that you've presumed some set theory, if you STILL have NO way of deciding whether z is or isn't *in* {x,y} ? ==== Kuratowksy defines as {a,{a,b}}. > Just for the record: http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Kuratowski.html F. ==== This question arises from economics, but I'll give you the question first, the context later: Suppose we have a set of n objects, and a binary relation S that is a linear ordering, ie irreflexive, asymmetric, complete, transitive. Is there any way to use this relation to induce some soft of ordering on the power set of our set of objects? This arises from an examination of equilibrium in the jungle: we have N agents, and S is strength(1 is stronger than 2, etc.) To see if this equilibirum is coalition-proof we need some way to order coalitions of agents. Which I don't know how to do. ==== > Suppose we have a set of n objects, and a binary relation S that is a > linear ordering, ie irreflexive, asymmetric, complete, transitive. Is > there any way to use this relation to induce some soft of ordering on > the power set of our set of objects? Yes, there are several ways. For example a max-min criterion is: Let < be the linear order. Any subset of A has a smallest element, i(A) in <. Say that A is smaller than B if i(A) =< i(B). You can also do min-max. More generally, you want responsive preferences over subsets. A good reference is Roth and Sotomayor's book on two-sided matching. Federico Echenique ==== > This question arises from economics, but I'll give you the question > first, the context later: > Suppose we have a set of n objects, and a binary relation S that is a > linear ordering, ie irreflexive, asymmetric, complete, transitive. Is > there any way to use this relation to induce some soft of ordering on > the power set of our set of objects? Yes: Associate each (finite) set with a sorted array containing exactly those elements in that set, and use the lexocographic (dictionary) order of those arrays. It will also be a linear order with a minimum element (the empty array) but no maximum element. ==== > This question arises from economics, but I'll give you the question > first, the context later: > Suppose we have a set of n objects, and a binary relation S that is a > linear ordering, ie irreflexive, asymmetric, complete, transitive. Is > there any way to use this relation to induce some soft of ordering on > the power set of our set of objects? This arises from an examination of equilibrium in the jungle: we have > N agents, and S is strength(1 is stronger than 2, etc.) To see if > this equilibirum is coalition-proof we need some way to order > coalitions of agents. Which I don't know how to do. Subsets of an n-element set correspond to strings of length n made up of the symbols 0 and 1... 1 means this object is in the set 0 means this object is not in the set. But strings of 0s and 1s of length n correspond to binary representations of the integers from 0 to 2^n-1. There is a natural way to order THIS set, right? ==== > This question arises from economics, but I'll give you the question >> first, the context later: >> Suppose we have a set of n objects, and a binary relation S that is a >> linear ordering, ie irreflexive, asymmetric, complete, transitive. Is >> there any way to use this relation to induce some soft of ordering on >> the power set of our set of objects? This arises from an examination of equilibrium in the jungle: we have >> N agents, and S is strength(1 is stronger than 2, etc.) To see if >> this equilibirum is coalition-proof we need some way to order >> coalitions of agents. Which I don't know how to do. Subsets of an n-element set correspond to strings of length n made up > of the symbols 0 and 1... 1 means this object is in the set 0 means > this object is not in the set. But strings of 0s and 1s of length n > correspond to binary representations of the integers from 0 to 2^n-1. > There is a natural way to order THIS set, right? but not one that in any way uses S. The idea is to extend the idea of strength of individuals to strength of coalitions. -- ==== > This question arises from economics, but I'll give you the question > first, the context later: > Suppose we have a set of n objects, and a binary relation S that is a > linear ordering, ie irreflexive, asymmetric, complete, transitive. Is > there any way to use this relation to induce some soft of ordering on > the power set of our set of objects? This arises from an examination of equilibrium in the jungle: we have > N agents, and S is strength(1 is stronger than 2, etc.) To see if > this equilibirum is coalition-proof we need some way to order > coalitions of agents. Which I don't know how to do. In order to transfer unambiguously the order on the underlying ordered set to an order on (finite) subsets or coalitions, the underlying order must do something like allowing comparisons of intervals, or differences in order, so that, say, given w, x, y and z in the underlying set, one may compare w-x to y-z ( find whether w as much greater than x as y is greater than z). Since this requires more than simple ordering, there cannot be any unique extention to coalitions based only on simple ordering. The question is just how much more than simple ordering is required on the underlying set in order to make such comparisons of coalitions meaningful or useful. ==== Right. So strength will be a function of some sort, that representive relative power, and lets us say how much stronger 1 is than 2, etc. The properties of the equilibirum will then depend pretty strongly on how we specify this function, won't it? > This question arises from economics, but I'll give you the question >> first, the context later: >> Suppose we have a set of n objects, and a binary relation S that is a >> linear ordering, ie irreflexive, asymmetric, complete, transitive. Is >> there any way to use this relation to induce some soft of ordering on >> the power set of our set of objects? This arises from an examination of equilibrium in the jungle: we have >> N agents, and S is strength(1 is stronger than 2, etc.) To see if >> this equilibirum is coalition-proof we need some way to order >> coalitions of agents. Which I don't know how to do. In order to transfer unambiguously the order on the underlying > ordered set to an order on (finite) subsets or coalitions, the > underlying order must do something like allowing comparisons of > intervals, or differences in order, so that, say, given w, x, y and > z in the underlying set, one may compare w-x to y-z ( find whether w > as much greater than x as y is greater than z). Since this requires more than simple ordering, there cannot be any > unique extention to coalitions based only on simple ordering. The question is just how much more than simple ordering is required > on the underlying set in order to make such comparisons of > coalitions meaningful or useful. -- ==== >> This question arises from economics, but I'll give you the question >> first, the context later: >> Suppose we have a set of n objects, and a binary relation S that is a >> linear ordering, ie irreflexive, asymmetric, complete, transitive. Is >> there any way to use this relation to induce some soft of ordering on >> the power set of our set of objects? This arises from an examination of equilibrium in the jungle: we have >> N agents, and S is strength(1 is stronger than 2, etc.) To see if >> this equilibirum is coalition-proof we need some way to order >> coalitions of agents. Which I don't know how to do. > What do you mean coalition proof? Can't the top dogs always beat up on the bottom dogs. Even in US democracy, we see how well the rich top have coaluded to succefully control, own and extort the poorer bottom. > Subsets of an n-element set correspond to strings of length n made up > of the symbols 0 and 1... 1 means this object is in the set 0 means > this object is not in the set. But strings of 0s and 1s of length n > correspond to binary representations of the integers from 0 to 2^n-1. > There is a natural way to order THIS set, right? but not one that in any way uses S. The idea is to extend the idea of > strength of individuals to strength of coalitions. > Use weights. Give each a weight or strength. Make the weights linearly independent so that no sum of weights of one subset equals the sum of weights of another subset. For example square roots of different primes. Then order subsets by sum of weights of members. The order of the subsets will depend upon the weights assigned each individual. Why insist upon a complete order? ==== This question arises from economics, but I'll give you the question > first, the context later: > Suppose we have a set of n objects, and a binary relation S that is a > linear ordering, ie irreflexive, asymmetric, complete, transitive. >> Is > there any way to use this relation to induce some soft of ordering on > the power set of our set of objects? >> This arises from an examination of equilibrium in the jungle: we >> have > N agents, and S is strength(1 is stronger than 2, etc.) To see if > this equilibirum is coalition-proof we need some way to order > coalitions of agents. Which I don't know how to do. > What do you mean coalition proof? Can't the top dogs always beat up on > the bottom dogs. Even in US democracy, we see how well the rich top have > coaluded to succefully control, own and extort the poorer bottom. > Well, it's a well defined concept in general equilibrium theory. >> Subsets of an n-element set correspond to strings of length n made up >> of the symbols 0 and 1... 1 means this object is in the set 0 means >> this object is not in the set. But strings of 0s and 1s of length n >> correspond to binary representations of the integers from 0 to 2^n-1. >> There is a natural way to order THIS set, right? but not one that in any way uses S. The idea is to extend the idea of >> strength of individuals to strength of coalitions. > Use weights. Give each a weight or strength. Make the weights linearly > independent so that no sum of weights of one subset equals the sum of > weights of another subset. For example square roots of different primes. > Then order subsets by sum of weights of members. The order of the > subsets > will depend upon the weights assigned each individual. Why insist upon a complete order? Because if I don't have one I can say whether or not a particular coalition would be able to appropriate some allocation from another person. -- ==== >This question arises from economics, but I'll give you the question >first, the context later: >Suppose we have a set of n objects, and a binary relation S that is a >linear ordering, ie irreflexive, asymmetric, complete, transitive. Is >there any way to use this relation to induce some soft of ordering on >the power set of our set of objects? There are many ways to do it. One way would be to use a variant of the lexicographic ordering: (a) The empty set is the smallest set. (b) Let A and B be nonempty. Let a be the smallest element of A, and b the smallest element of B. Then we say that A is smaller than B if and only if (i) a1, order the subsets of k elements as follows: if A and B both have k elements, let a be the smallest element of A and b the smallest element of B. Then AThis arises from an examination of equilibrium in the jungle: we have >N agents, and S is strength(1 is stronger than 2, etc.) To see if >this equilibirum is coalition-proof we need some way to order >coalitions of agents. Which I don't know how to do. How you order the coalitions will depend on the properties of the strength. Does a coalition involving strength 2 and 3 beat one involving 1 and 4? You seem to be putting the horse before the cart: you want to understand how the strength of a coalition works, so you can then order the coalitions; you don't first order the coalitions in some way and then use that order to figure out their relative strengths. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >> This question arises from economics, but I'll give you the question >> first, the context later: >> Suppose we have a set of n objects, and a binary relation S that is a >> linear ordering, ie irreflexive, asymmetric, complete, transitive. Is >> there any way to use this relation to induce some soft of ordering on >> the power set of our set of objects? There are many ways to do it. One way would be to use a variant of the lexicographic ordering: (a) The empty set is the smallest set. > (b) Let A and B be nonempty. Let a be the smallest element of A, and b > the smallest element of B. Then we say that A is smaller than B if and > only if (i) a (ii) a=b and A-{a} well-ordering, so it makes sense to talk about smallest element. However, this ordering has a number of unnatural features. A > slightly more interesting and natural way is to extend the natural > partial ordering that exists in P(S) under inclusion. Order P(S) as follows: (1) Two subsets of different size are ordered by their size; that is, > if A and B are subsets, and A has fewer elements than B, then A k>1, order the subsets of k elements as follows: if A and B both > have k elements, let a be the smallest element of A and b the > smallest element of B. Then A A-{a} ordering of S, you need to consider your problem. > This arises from an examination of equilibrium in the jungle: we have >> N agents, and S is strength(1 is stronger than 2, etc.) To see if >> this equilibirum is coalition-proof we need some way to order >> coalitions of agents. Which I don't know how to do. How you order the coalitions will depend on the properties of the > strength. Does a coalition involving strength 2 and 3 beat one > involving 1 and 4? You seem to be putting the horse before the cart: > you want to understand how the strength of a coalition works, so you > can then order the coalitions; you don't first order the coalitions in > some way and then use that order to figure out their relative > strengths. > Well, i was hoping there was some way to do it without getting any more specific on the properties of S. If there was some natural way that S could in some sense extend to the power set, then that would have been nice. It seems that I'm going to have to move away from S to some cardinal representation of it, some function from the set of agents to the reals, so say exactly how strong each agent is in relation to each other, and then have some sort of shapely value of strength for agent. This is a much weaker result, though, since we'll get different allocations depending onthe function I use for S. ====================================================================== > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > ====================================================================== Arturo Magidin > magidin@math.berkeley.edu > -- ==== > Well, i was hoping there was some way to do it without getting any more > specific on the properties of S. If there was some natural way that S > could in some sense extend to the power set, then that would have been > nice. It seems that I'm going to have to move away from S to some > cardinal representation of it, some function from the set of agents to the > reals, so say exactly how strong each agent is in relation to each other, > and then have some sort of shapely value of strength for agent. This is a > much weaker result, though, since we'll get different allocations > depending onthe function I use for S. Since there are many natural ways, you can work with an arbitrary order that is somehow consistent with the order on the individual elements of S: Look at how people in two-sided matching theory do it (see my previous post). ==== I S M A I L 9 19 13 1 9 12 = 63 Shadia works for the CBC and was seated at the CBC booth during today's downtown street fair on 2nd Ave. in Saskatoon, she was the first of 10 people to provide stats today. 63+ Dad 12 8 36 225/141 +7494 63+ Mom 1 5 46 121/244 +3945 63+ Bro 15 1 71 15/350 5080 63+ Bro 16 2 72 47/319 5477 63+ Bro 16 2 72 47/319 5477 105 Shadia 10 8 76 223/143 7114 the year than to the beginning of the year (the first 13 primes minus the first 13 non-primes). Mom was born 123 days closer to the beginning of the year than to the end of the year, or 3 times the 13th prime (41). The parents were together born 39 (13+13+13) days closer to the beginning of their years than to the end of their years. The parents were born on days of the month adding to 13 and in months adding to 13. The parents were born in years averging 41 (13th prime). The kids were born in months adding to 13. The kids were born on days of the month adding to 57, it's the 41st non-prime (Genesis 41 contains 57 verses) while 41 in turn is the 13th prime. Twins arrived on the 16th (Nehemiah with 13 chapters). The family was born on days of the year adding to 678 (6x113). The first and last kids were born on days of the year adding to 238 (the first 13 primes). The brothers were born on days 15 and 47, together for 62 (the 13th prime plus the 13th non-prime). Mom gave birth a total of 942 days after her birthdays (the number of verses in Bible Book 13). Mom was 9422 days old when she gave birth to twins (942 verses in Bible Book 13). Shadia was at the CBC booth, set up in front of the CBC Radio Canada studio at 144 2nd Ave. South (the 13th Fibonacci). Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 18 55 199 37 20 89 322 41 <-13th-> 21 <-13th-> 144 <-13th-> 521 --- --- 238 154 <-Lamentations Shadia was born 101 days after mom's birthday (26th prime) and 364 days after dad's birthday (First Chronicles 26). She was born in 76 (Exodus 26). Born in 76 while her brothers were born on days of the century adding to 78635, the brothers were born an average of 71.76 years into the century. Primes Non-Primes Lucas 2 1 1 3 4 3 5 6 4 7 8 7 11 9 11 13 10 18 17 12 29 19 14 47 23 15 76 29 16 123 31 18 199 37 20 322 41 21 521 43 22 843 47 <-15th-> 24 <-15th-> 1364 <-Jeremiah is Book 24 with 1364 verses Dad was born on day 225 (15x15), dad was born with 141 days remaining in the year, corresponding to Numbers 24 (15th non-prime), pretty as 141 is 47+47+47 (3 times the 15th prime). The first of the kids was born on the 15th. The twins arrived on the 47th day of the year (15th prime). The brothers were born in years adding to 215. Shadia's day, month and year of birth adds to 94 (twice the 15th prime). Her name adds to 105 (7x15 and is Leviticus 15). Her vowels add to 30 (2x15), consonants add to 75 (5x15). The kids are together 1515 days closer in age than the parents. The parents were on average born 47% into their years, their first kid arrived on the 25947th day of the century and then they got twins on the 47th day of the year. The family was born on days of the century adding to 136911, they were born an average of 62.47 years into the century. Primes Non-Primes Numbers 2 1 1 3 4 2 5 6 3 7 8 4 11 9 5 13 10 6 17 12 7 19 14 8 23 15 9 29 16 10 31 18 11 37 20 12 41 21 13 43 <-14th-> 22 <-14th-> 14 --- --- --- 281 176 105 Dad was born on day 225 (Judges 14). Dad was born in 36, or 14 plus the 14th non-prime (22), it's the number of chapters in Bible Book 14. The brothers were together born 879 days closer to the beginning of their years than to the end of their years, it's the number of verses in Gospel John, Bible Book 43 (14th prime). The brothers were born in 71 and 72, together for 143, while Shadia was born with 143 days remaining in the year (43 is the 14th prime). Shadia's name adds to 105 (1 through 14 and is the first 14 primes minus the first 14 non-primes). Her initials average 14. Her initials add to the 28 (14+14) chapters of Bible Book 14. The first 4 primes plus the first 4 non-primes add together for the 36 chapters of Bible Book 4 Numbers: Primes Non-Primes 2 1 3 4 5 6 7 <-4th-> 8 -- -- 17 19 The first 4 primes in prime positions add together for the 36 chapters of Bible Book 4: Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 -- 36 Seven plus the 7th prime plus the 7th non-prime adds together for the 36 chapters of Bible Book 4, pretty as 7 is the 4th prime: Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 <-7th-> 12 -- -- 58 50 We are meeting on the 1288th day of the century (the number of verses in Numbers). Dad was born in 36, there are 36 chapters in Numbers. Dad was born in 36 (6x6), his average age when the kids were born amounts to 36.36 years. 1-50 - Genesis 51-90 - Exodus 91-117 - Leviticus 118-153 - Numbers 154-187 - Deuteronomy 188-211 - Joshua 930-957 - Matthew 958-973 - Mark 974-997 - Luke 998-1018 - John 1019-1046 - Acts 1047-1062 - Romans 123 <-Numbers 6, it is three times the 13th prime (41+41+41), keeping in mind that 13 is the 6th prime 188 <-the opening chapter of Book 6 is 6x6x6 short of the 404 verses of Bible Book 66, it is the 6th prime squared (13x13) short of the 357 verses of Daniel (also in part about 666) 193 <-Book 6 chapter 6 is the 44th prime, while 44 is in turn 66.666...% of 66 211 <-the terminating chapter of Book 6 is approximately 66.6% of the 66th prime (317) 357 <-the opening chapter of Book 6 plus the 6th prime squared is the 357 verses of Daniel (in part about 666) 404 <-the 6th prime squared (13x13) plus the 6th prime squared (13x13) plus 66 adds to the 404 verses of Bible Book 66 1062 <-666 plus 6x66 is a combination of the 658 verses of Bible Book 6 plus the 404 verses of Bible Book 66, and is the terminating chapter of New Testament Book 6 1070 <-666 plus the 404 verses of Book 66 is the 1070 verses of Job (Book 6+6+6) 1213 <-Exodus terminates at chapter 90 (66th non- prime) with 1213 verses (the 198th or the 66+66+66th prime) 1292 <-the 658 verses of Book 6 plus twice the 66th prime (317) is the 1292 verses of Isaiah (the Book contains 66 chapters) Shadia's names are 6 and 6 lettered, her first name adds to 66.666...% of her last name, she is the 6th of 6 family members. She was born with 143 days remaining in the year, pretty as chapter 666 brings Ecclesiastes up to 143 verses. Bible chapter 666 contains 29 verses and brings Ecclesiastes up to 143 verses, this 143 is the 109th non-prime while 109 in turn is the 29th prime, while 29 is 6 plus the 6th prime (13) plus the 6th non-prime (10). She is the 6th of 6 family members and was born on the 10th (6th non-prime). Her brothers were born on days of the year adding to 109 (29th prime). 389 <-77th prime 104 <-77th non-prime 77 <-77 --- 570 The Four 57's Genesis 41 -> 41 Leviticus 14 -> 104 Judges 9 -> 220 <-I dreamt of 220 roofs blown John 11 -> 1008 off homes in the Dakotas ---- 1373 <-220th prime Chapter 57 is Exodus 7 with 25 verses Book 57 is Philemon with 25 verses -- -- 41st non-prime 16th non-prime <-together for 57-> Major Books of End-Times Prophecy (Daniel and Revelation are in part about 666 while Isaiah contains 66 chapters): Daniel - 357 verses Revelation - 404 verses <-57 plus the 57th prime plus the 57th non-prime Isaiah - 1292 verses <-an average of 19.575757... verses per chapter The kids were born on days of the month adding to 57. Mom gave birth a total of 2.57 years after her birthdays. The brothers were together born 1373 days after their parent's birthdays, the 57's are at chapters 41, 104, 220 and 1008, together for 1373 (the 220th prime). The brothers were born on days of the year adding to 109 while Shadia was born on day 223 (57+57 difference). Kids were born on days 15, 47 and 223, together for 285 (5x57). Shadia's odd valued letters exceed her even valued letters by 57. Shadia's consonants add to 75, pretty as the 57's are at chapter numbers adding to 75. Her name adds to 105, it's the 78th non-prime while 78 in turn is the 57th non-prime (105 is the 57th non-prime in non-prime position). Mom was born on the 1st, corresponding to Genesis (there is number play involving 57 in Genesis in dozens of ways). Mom and Shadia were born on days 1 and 10, together these Bible Books contain 2228 (4x557) verses. Dad was 12574 days old when the first of the kids was born. Mom's combined ages when she gave birth amounts to 106.57 years. Shadia might marry Marcia and me, pretty as Marcia and Shadia are separated by 1457 days. 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 159 Shadia 10 8 76 223/143 7114 Shadia 42 Acevedo-Kabatoff 55-62 Anyway, if you people think that you have the right to use my abusive parents as tools and arrest and torture me, then I think that I should have the right to ask women to marry me, or to marry Marcia and me, our last names add together for the 117 verses of Song of Solomon, it's the Bible's Book of Love. The nubile sweety was born on the 6th and has a 6 lettered first name). Isaiah is the Book with 66 chapters, pretty as it is Book 23, or the 6th prime plus the 6th non-prime (13+10=23). Isaiah 4, 12 and 20 (adds to 6x6) each contains 6 verses. Isaiah 4:1 is about Marcia (and me), and 6 other women who are capable of feeling shame rather than pride, greed or lust, or who limit their love for traditions and for people who abide by their traditions. You people have Egyptian penises on the roofs of your churches and lined city streets with representations of penises, and had me tortured for years for saying so, others just sat back in silence while they were doink this to me, and similarly you remain silent and compassionless now that the arrests and torture have ceased. You people spent millions of dollars having me tortured, and then annually you spend billions on your decorated trees, I begged and begged for assistance to flee the country (they tortured me for years at the U of S) and you people are so cheap that you can't even offer to buy me a cookie when I bust my ass to show you evidence that your very name is a gift from God, you cheap, filthy, compassionless assholes won't even spend 48 cents on a stamp so that you life to show you evidence that your very name is a gift from God!!! All you are really good for is to have your stats posted on the usenet and be used as an example to others, and look, here you are!!! Should Shadia marry me, great, but if Marcia marries me and then Shadia marries Marcia and me, then Shadia's brothers are each goink to win themselves a shiny new Cadillac!!! Good luck and may God bless you!!! Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! What a wonderful weddink there will be What a wonderful day for you and me Church bells will chime You will be mine In apple blossom time... ==== I S M A I L >9 19 13 1 9 12 = 63 Shadia works for the CBC and was seated at the CBC booth during today's >downtown street fair on 2nd Ave. in Saskatoon, she was the first of 10 >people to provide stats today. << The following (courtesy of Waxy.org) is sort of an unofficial FAQ explaining the psychotic nonsense posted to Usenet by Shawn Daryl Kabatoff AKA Dar, AKA Probababbilities. And now AKA marcia and me. WARNING: Read below before even thinking about responding to this twit. http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml#000643 Usenet has the tendency to provide a public forum for those who would normally be scribbling in a closet. For example, take Daryl Shawn Kabatoff. For the last few years, he's methodically gathered statistics from various sources, ranging from local newspaper obituary pages to the food court of the Saskatoon Midtown Plaza mall. With all the raw data he's collected, he's attempting to prove daily that our full names are in mathematical harmony with our birthdays. about, starting with calculations related to their birthdate and full names, blending in whatever other personal information about their family members, spouses, birthplace, and career he's been able to zealotry, and personal torment. I've never seen anything like it. With all the prime numbers, Fibonacci sequences and biblical references, it's like reading the notebooks of Maximillian Cohen and John Nash combined. Unsurprisingly, several posts unfold to reveal a history of painful mental illness. If you have some time, take a look. I've detailed his posting history and a several sample posts below. January 27, 1999 to July 5, 2000 as Catsco@home.com December 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.ca Oct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.ca posts have been removed from Google Groups archive) Selected Posts: Tessa Lynne Smith Dastageer Sakhizai and Helen Smith Brett David Maki Andrew Meredith Cotton Amanda Dawn Newton Mona Marie Etcheverry Tony Peter Nuspl Lisa Charlene McMillan Grant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my current residence. and every single place he's mentioned in his posts (most notably nervous harold's and the roastary) were either places i've been (as it's a small city of 200K) or hangouts, ie. the two places out if they know of him, they (my friends that is) being of the broadway-centred slacker ilk. myself, too, until i got out of there. eh, anyways. thought it odd to see all this. midtown mall. i ate my meals there, whilst waiting several days in line for star wars episode one, at the theatre across the street. posted by andy raad on May 22, 2002 06:20 PM Fascinating. It's like he's trying to take chaos and bind it into whatever rules he can find, religious, logical and otherwise. Numbers and math have a reliable pattern, something that can always be proven to true or false. People and religion do not. It reminds me of Darren Aronofsky's movie Pi. It's the story of an paraniod genius who is trying to find a pattern in Pi. A group that takes interest in his work is convinced that the existence of Pi, a number whose existence can be proven but no quantified, is proof of the existence of God. Kabatoff's hunt for patterns in something as random as name selection is a way to reconcile his deeply logical thought process with his conflicting religious views. posted by matt on May 23, 2002 11:19 AM asking him if he'd be willing to create a numerological analysis for me. I also asked him if he had seen either Pi or A Beautiful Mind, and what he thought of them. If he replies, I'll be sure to post it. posted by Andy Baio on May 23, 2002 11:24 AM I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbed pumpkin pie all over my breasts for him, and my breasts turned orange. I am a pumpkin for Shawn. posted by Trisha Blondie on July 24, 2002 10:41 PM Um, that's swell. So, you're in love with him? posted by Andy Baio on July 25, 2002 07:10 AM Shawn once went to a funeral for a Jehovah Witness that shot himself and the lemon tarts were very bad, they were not only sour but were rubbery as well. Shawn said that the guy was some kind of Jehovah Witness prophet, he saw in advance that the lemon tarts at his funeral were to be very very bad, and so he shot himself. Shawn said that he never ate pumpkin pie at a funeral but would like to some day. Shawn likes pumpkin pie and so I have been practicing to make very good pumpkin pies. posted by Trisha Blondie on July 25, 2002 02:49 PM Shawn said that the lemon tarts were sour, bitter and rubbery. posted by Trisha Blondie on July 30, 2002 12:32 AM I don't think this guy takes notes. I think he has Total Recall, and it has driven him insane... posted by Todd Smith on December 26, 2002 11:00 AM Oh... I almost forgot... I didnt spend thousands of dollars a day tormenting Daryl... We got a deal on tormenting that fiscal year, it only came to about 37cents a day.... posted by Dr Claw on December 30, 2002 01:56 AM Mr. Kabatoff attempts to portray himself as a victim, but in fact he is a violent predatory pedophile who is well known to his local law enforcement. In his post to multiple newsgroups with the subject Collecting Mail For The Coming Anti-Christ, he encourages mothers to send him photos of their naked daughters. Mr Kabatoff explains, I Ant-Christ) that were of underage children unless the parent was signing consent. He is banned from virtually all the shopping malls in his community because he stalks young people and sexually harasses them. He has an extensive arrest record which includes sexual molestation charges. He's been hospitalized in mental institutions about his contact with young girls in many posts. Search newsgroup archives for posts by him containing the word nubile. As part of his harrassment, he provides personal details in a public forum, such as the real names of real children, in these and other posts. About one wanted her and her sister dead. http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+dead+or+in+my+bed&hl= en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%40ID-136124.news.dfncis.de&rnu He not only curses children and prays for their death in his posts, he also enjoys attending the funerals of young people: And so, since nubile sweeties are found in greatest abundance at the funerals of high school students, then it is the funerals of high school students that make the very very best funerals, especially if there is food... I stuff my face (and my pockets) with all the good food and look at all the pretty nubile sweeties and have the time of my life.. .http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+nubile+sex&hl=en&l r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%40twister.socal.rr. com&rnum=1 Many of his posts are sent to alt.teens.advice. However, he liberally spams, floods and crossposts his off-topic threatening and offensive missives to countless newsgroups. Some people HAVE problems and some folks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. When he sends these posts to any newgroup, please help by reporting him to I knew of him when I was attending the University of Saskatchewan. He'd hang out in the Arts computer lab and all you'd see is screens of numbers racing by on his laptop. I have an original copy of his Collecting Mail for the Coming Anti-Christ pamphlet, and have seen him be hauled away by campus security on more than one occasion. My friends and I refer to him as Crazy Number Man. I've been posting to (and about) Shawn for over two years with big gaps in between. He has seen Pi and didn't like it and didn't think it resembled him at all. (Wrong, it fits him to a tee) He doesn't have total recall and has stated that he travels with a lap top to notate items. Also, he uses cut n' paste a lot if you read all the way through his ramblings. He is anti-social as shown by his angry statements towards those who, by his own admission, have been kind (but not kind enough) to him. Still, he's intelligent and seems to be able to take a joke on occassion. That's where I came in. ALOHA if it comes from anyone not already in my addressbook. I'll never even see it) ==== Friday April 26th 2002 116/249 16504 R I Z Z U T O 18 9 26 26 21 20 15 = 135 I am posting Vito Rizzuto's stats again because Shriner/Freemason Don Ocean and/or his friend James Takayama is repeatedly calling me a pedophile. I have never sexually assaulted anybody, I have never been charged with sexual assault, I have never been arrested for sexual assault, and I am not sexually attracted to children. In 1988 I criticized the phallic worship in the Protestant and Catholic churches and was repeatedly arrested and tortured at the U of S for years. Ruby would have me arrested for failing to kiss her God-damned ass and wear the clothes she was always trying to force upon me (and because Protestants and Catholics were upset with my words and lobbied her to shut me up), Ruby would have me arrested and chemically lobotomized, and then she would come to the psychiatric ward and force her choice of clothes upon me there. Eventually I found myself internet access at the U of S and posted Collecting Mail For The Coming Anti-Christ, in the essay I spoke in defense of people to wear their own choice of clothes, and this included defending the rights of women and girls to go topless if they wanted. Now Don Ocean calls me a pedophile, and as a result of Don Ocean, James Takayama in Hawaii has begun doing the same. First the fellow libels me in the WaxyOrg website using the name of Nospam, then using the name of Thomas (aka MauiCop), James Takayama quotes his own material posted as Nospam. Now another website has begun to quote the libel as being truthful, saying that I have been repeatedly arrested for sexual assault and that I was a pedophile (and it is possible that James Takayama is responsible for this website and may start others where he will claim I have been repeatedly arrested for sexual assault). Now as a result, people in Saskatoon are calling me a pedophile and are threatening to beat me up and kill me. The Saskatoon police are not in the habit of charging people for assault when they give me beatings (in fact they arrested me after James De Witt brutally assaulted me at the Seventh Day Adventist Church, I was brutally assaulted and then the police arrested me and took me to the U of S relievers when these people crack and break my ribs. The situation is quite unfair, and now Vito Rizzuto will have his stats posted daily until the situation is resolved. I get tortured year after year and begged people in futility for assistance to flee the country, now watch Vito spend huge piles of cash again this year turning trees into decorated idols, for his compassion is limited to traditions. 203 Nicolo Nicolo 68 Rizzuto 135 225 Libertina Libertina 90 Rizzuto 135 201 Vito 21 2 46 52/313 +4014 Vito 66 Rizzuto 135 177 Maria Maria 42 Rizzuto 135 Mom's first name adds to 90 (66th non-prime), Exodus contains 1213 verses (66+66+66th prime) and terminates at chapter 90 (66th non-prime). Vito adds to 66, mom and the little sister have first names averaging 66. The kids have first names adding together for 108 (the first 6 primes in prime positions). The kids have first names differing in value by the 24 chapters of Bible Books 6 and 10 (6th non-prime). Dad has a 6 lettered first name, there are 24 letters in all the first names, the number of chapters in Books 6 and 10 (6th non-prime). All first names add together for 266. Dad and Vito have first names adding together for 134, corresponding to Numbers 17 with 13 verses (the 6th prime). Mom and Vito have first names averaging 78 (6 times the 6th prime). Dad and Vito have full names adding together for the 404 verses of Bible Book 66, Revelation. Vito was born on the 21st, corresponding to Ecclesiastes (chapters 660 to 671). Vito has consonants adding to 132 (66+66). Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 --- 108 Vito Rizzuto (135) was hot for Cammalleni (83), the names differ in value by 52, pretty as Vito was born on the 52nd day of the year. Giovanna (83) Cammalleni (83) was born with names adding to the 23rd prime and to the 23rd prime, together for 46, and she gets a husband that was born in 46. 166 Giovanna 48 Giovanna 83 Cammalleni 83 Vito married Giovanna (83) Cammalleni (83). She was born with 18 (6+6+6) letters adding to 166. Both of her names added to the 83verses of Second Timothy (the 16th Book of the New Testament). Her names added to 83 and 83, corresponding to Exodus 33 and Exodus 33, together for 66. At birth, Vito and Giovanna had 29 letters in their names, or 6 plus the 6th prime (13) plus the 6th non-prime (13), there are 29 (6+6p+6np) chapters in Bible Book 13 (the 6th prime), there are 29 (6+6p+6np) verses in chapter 666 (Ecclesiastes 7). At birth Vito and Giovanna had 29 letters adding together for 367 (First Chronicles 29). Vito's sister's first name adds to 42 (the 29th non-prime), Bible Book 29 is Joel (42). Now Giovanna's name adds to 218 (twice the 29th prime). Nicolo's name adds to 203 (7x29). Leonardo's name adds to 219, or 3 times the 73 verses of Bible Book 29. Libertina was born in 73 (the length of Book 29 and is the Lucas numbers up to 29). Leonardo and Libertina have first names adding together for 174 (6x29). Dad and the first two kids have first names adding together for 218 (twice the 29th prime). The males were born in years adding to 182 (Deuteronomy 29 with 29 verses). 201 Vito 21 2 46 52/313 +4014 Vito 66 Rizzuto 135 218 Giovanna 48 Giovanna 83 Rizzuto 135 203 Nicolo 67 Nicolo 68 Rizzuto 135 219 Leonardo 69 Leonardo 84 Rizzuto 135 225 Libertina 73 Libertina 90 Rizzuto 135 Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime The have three kids, the first and last of the kids bear Vito's parent's names, so the daughter's first name adds to 90 (66th non-prime). The kids have first names adding together for 242 (First Samuel 6), all names in the family add together for 1066. The kids were born in 67, 69 and 73, these are the 19th prime, 50th non-prime and the 21st prime, together for 90 (66th non-prime and the value of the daughter's first name). Mom and her sons have first names adding together for 235... the 184th prime (1097) and the 184th non-prime (235) averages 666. The brothers have names averaging 211, it is approximately 66.6% of the 66th prime (317) and is the terminating chapter of Bible Book 6. Book 6 chapter 6 (193) plus the terminating chapter of Book 6 (211) adds together for the 404 verses of Bible Book 66. Daughter's first name not only adds to 90 (66th non-prime), but her first name adds to 66.666...% of her last name. This is a family of 5, the 4014 days dad is older than me is the 959 verses of Bible Book 5 short of the 666th prime (4973). Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 26 61 27 67 28 71 30 73 32 79 33 83 34 89 35 97 36 101 38 103 39 107 40 109 42 113 44 127 45 131 46 137 48 139 49 149 50 151 51 157 52 163 54 167 55 173 56 179 57 181 58 191 60 193 62 197 63 199 64 211 65 223 <-48th-> 66 227 68 229 69 233 70 239 72 241 74 251 75 257 76 263 77 269 78 271 80 277 81 281 82 283 84 293 85 307 86 311 87 313 88 317 <-66th-> 90 Vito adds to 66 (48th non-prime). Vito's first name adds to 48.888...% of his last name, God gives him a wife that was born in 48. Vito and his wife have first names adding together for the 149 verses of Bible Book 48, Galatians. Vito's 201 valued name exceeds his 52nd day of birth by the 149 verses of Bible Book 48. Rizzuto (135) was hot for Cammalleni (83), the 83 value of mom's maiden name is 61.48% of the 135 value of Rizzuto. The males were born in years adding to 182, the females in years adding to 121 (66.48%). The kids are missing 15 letters from their first names, these missing letters add to 248. The 4014 days dad is older than me is 18 times the 48th prime (223), prettier as I was born on the 48th day of the year and with 317 days remaining in the year (66th prime). 1-50 - Genesis 51-90 - Exodus 91-117 - Leviticus 118-153 - Numbers 154-187 - Deuteronomy 188-211 - Joshua 930-957 - Matthew 958-973 - Mark 974-997 - Luke 998-1018 - John 1019-1046 - Acts 1047-1062 - Romans 123 <-Numbers 6, it is three times the 13th prime (41+41+41), keeping in mind that 13 is the 6th prime 188 <-the opening chapter of Book 6 is 6x6x6 short of the 404 verses of Bible Book 66, it is the 6th prime squared (13x13) short of the 357 verses of Daniel (also in part about 666) 193 <-Book 6 chapter 6 is the 44th prime, while 44 is in turn 66.666...% of 66 211 <-the terminating chapter of Book 6 is approximately 66.6% of the 66th prime (317) 357 <-the opening chapter of Book 6 plus the 6th prime squared is the 357 verses of Daniel (in part about 666) 404 <-the 6th prime squared (13x13) plus the 6th prime squared (13x13) plus 66 adds to the 404 verses of Bible Book 66 1062 <-666 plus 6x66 is a combination of the 658 verses of Bible Book 6 plus the 404 verses of Bible Book 66, and is the terminating chapter of New Testament Book 6 1070 <-666 plus the 404 verses of Book 66 is the 1070 verses of Job (Book 6+6+6) 1213 <-Exodus terminates at chapter 90 (66th non- prime) with 1213 verses (the 198th or the 66+66+66th prime) 1292 <-the 658 verses of Book 6 plus twice the 66th prime (317) is the 1292 verses of Isaiah (the Book contains 66 chapters) Vito is 4014 days older than me, his name adds to 201 (Joshua 14), his mom and daughter share the same 225 valued name (Judges 14). The kids have odd valued letters in their first names adding together for 140. The letters that are neither prime nor square in the kids' given names add to 196 (14x14). Vito and his daughter have names differing in value by 14. All first names add to 391 (314th non-prime). Dad's name adds to 67+67+67. The first of the kids was born in 67 and has names differing in value by 67. These first three family members have first names adding to 66, 83 and 68, corresponding to Exodus 16, 33 and 18, together for 67. This 67 is the 19th prime, the second kid was born in 69 (Exodus 19) and has a name adding to 219. The brothers were born in years adding to 136 (Numbers 19). Vito and the last of the kids were born in years adding to 119. Perhaps mom was 19 years old when she gave birth in 67 (the 19th prime). Mom's name adds to twice 109 (Leviticus 19) while dad's name adds to 201 (3 times the 19th prime). The parents have names adding together for 419 (Nehemiah 6 with 19 verses). Vito and his kids were born in years adding to 255 (First Samuel 19). Vito and his kids have first names averaging 77 (the primes up to 19). The kids were born in years adding to 5909 (19x311). Libertina's last name adds to 150% of her first name (150 chapters in Bible Book 19). Vito was born a multiple of 19 days into the century (16853=19x887). Vito's vowels add to 69 (Exodus 19). The 2460 verses of Bible Book 19 is 19x19+19x19+19x19+19x19+19x19+19x19+19x19 minus 67 (the 19th prime). Vito's names differ in value by 69 and his second kid was born in 69. Vito was born on the 21st and his third kid was born in 73 (21st prime). Vito was born on the 21st and his name adds to 201, and he was likely 21 years old when the first of the kids was born. Vito was born in 46, his 201 valued name exceeds it's 155th non-prime position by 46. The kids have prime and square valued letters in their first names adding together for 46. Vito was born on day 52, it's the number of chapters in Bible Book 24, while his daughter gets a name that exceeds his by 24. Vito and I were together born 530 days closer to the beginning of our years than to the end of our years (Psalm 52). This 530 is a combination of the first three perfect numbers (6, 28 and 496), they have factors that add to form themselves: Perfects 6 - 1, 2, 3 28 - 1, 2, 4, 7, 14 496 - 1, 2, 4, 8, 16, 31, 62, 124, 248 The males have first names adding together for 218, and mom's name adds to 218. Mom was born with 18 letters and took a last name that begins with the 18th letter of the alphabet, it is a last name that adds to 135 (Numbers 18). Vito and his sister have first names adding together for 108 (Leviticus 18). Rizzuto adds to 135 (the 103rd non-prime), the 11 different letters utilized in the construction of the first names for the kids add together for 103. 3x3x3x3x3x3 3x3x3 103 <- the 3x3x3rd prime ----------- 859 <- the number of verses in Bible Book 3 The parents were born in 46 and 48, Bible Books 46 and 48 contain 437 and 149 verses. These Books contain an average of 293 verses while the 437 is 293.28% of the 149. The 437 and 149 are the 353rd non-prime and 35th prime, and so it is pretty that there would be 35 letters in the family first names, and pretty that the last name would add to 135. Mom and her sons have first names adding together for 235. The brothers have first names adding together for 152 (Numbers 35). Libertina Kabatoff adds to 152 (Numbers 35), prettier as 293 is the 62nd prime while Kabatoff adds to 62. Rizzuto adds to 135, or 57 plus the 57th non-prime (78). The main Books of end-times prophecy are Daniel with 357 verses and Revelation with 404 verses, or 57 plus the 57th prime (269) plus the 57th non-prime (78). Primes Non-Primes 2 1 3 4 5 6 7 8 11 <-5th-> 9 -- -- 28 28 Vito and his sister Maria have first names adding to 66 and 42, together these Bible Books contain 1555 verses. Vito has 11 letters and a first name adding to a multiple of 11, his kids have first names adding together for 11x11+11x11 (11 is the 5th prime). Then Vito marries Giovanna (83) Cammalleni (83), there are 83 verses in Bible Book 55, and she soon finds herself in a family of 5. The 5th of 5 family members has a name adding to 225, prettier as 25 is not only 5x5 but is 5 plus the 5th prime (11) plus the 5th non-prime (9), keeping in mind that Bible Book 5 contains 959 verses. Perhaps mom was 25 (5x5 or 5+5p+5np) years old when she gave birth to the 5th of 5 family members. Libertina's first name adds to a multiple of 5 and also a multiple of 9 (5th non-prime). Libertina's (the 5th of 5 family members) first 5 letters add together for dad's 46th year of birth. My name adds to 187 (the terminating chapter of Bible Book 5), it is 5x5x5 plus twice the 5th prime in prime position (31). If Libertina took my 62 (twice the 5th prime in prime position) valued last name, then her names would have an average value of 76 (55th non-prime). Lamentations is Bible Book 25 with 154 verses, the 154th prime is 887 while Vito was born on the 19x887th day of the century, pretty because if Libertina married me then he would be lamenting. Vito and his kids already have first names adding together for 308, or twice the 154 verses of Lamentations. Note that Bible Books 5 and 5x5 differ in length by 805 verses (the 666th non-prime). And Old Testament Book 9 (the 5th non-prime) and New Testament Book 9 (the 5th non-prime) together contain 959 verses (the number of verses of Book 5). Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 --- 167 Esther Book 17 Vito (66) was born on day 52, it is an average of 59 (the 17th prime). have names averaging 189 (the first 17 primes minus the first 17 non-primes). Vito and Giovanna have names differing in value by 17 (7th prime, the primes up to 7 add to 17). Giovanna's name adds to 218 (Book 7 chapter 7). Vito and Giovanna were born with last names adding together for 218 (Book 7 chapter 7). The first of the kids gets a first name adding to a multiple of 17 and he was born in 67 (Exodus 17). The second gets a first name adding to 7 times the 7th non-prime (12). The brothers have first names adding to 68 and 84 (a span of 17). The daughter gets a name adding to 225 (Book 7 chapter 7+7). The kids were born in 67, 69 and 73, corresponding to Exodus 17, 19 and 23, together for 59 (the 17th prime). Libertina's names differ in value by 45 (45 chapters contain the length of 17 verses), she might take my name and end up with 17 letters. Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 I wanted 17 French fry girls at my weddink, they would throw French fries high up into the air, the french fries would fall to the ground and get dirty, and nobody would ever eat french fries again. And I wanted 59 big nosed Greeks with their big noses throwing hamburglers at us from across the street. And I wanted American Noel Nibblett blowing his bugle and leading a marching regiment of cadets up and down the street while American Don Ocean and his Shriner friends ride circles around them on their little motorscooters. Scientists have recently discovered that tomatoes contain properties that help to prevent prostate cancer, in men, and since A&W allows one to take as much ketchup as they wanted, the Great A&W Rootbear was on the fast track to becoming an international symbol of health and fertility... so of course I wanted the Great A&W Rootbear to be the best man at my weddink. My weddink was soon approaching and it was goink to be a glorious affair, probababbly. 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 207 Libertina 73 Libertina 90 Acevedo-Kabatoff 55-62 Anyway, if you people think that you have the right to use my abusive parents as tools and arrest and torture me, then I think that I should have the right to ask women to marry me, or to marry Marcia and me, our last names add together for the 117 verses of Song of Solomon, it's the Bible's Book of Love. The nubile sweety was born on the 6th and has a 6 lettered first name). Isaiah is the Book with 66 chapters, pretty as it is Book 23, or the 6th prime plus the 6th non-prime (13+10=23). Isaiah 4, 12 and 20 (adds to 6x6) each contains 6 verses. Isaiah 4:1 is about Marcia (and me), and 6 other women who are capable of feeling shame rather than pride, greed or lust, or who limit their love for traditions and for people who abide by their traditions. You people have Egyptian penises on the roofs of your churches and lined city streets with representations of penises, and had me tortured for years for saying so, others just sat back in silence while they were doing this to me, and similarly you remain silent and compassionless now that the arrests and torture have ceased. You people spent millions of dollars having me tortured, and then annually you spend billions on your decorated trees, I begged and begged for assistance to flee the country (they tortured me for years at the U of S) and you people are so cheap that you can't even offer to buy me a cookie when I bust my ass to show you evidence that your very name is a gift from God!!! Should Libertina marry me, great, but if Marcia marries me and then Libertina marries Marcia and me, then Nicolo and Leonardo are both goink to win themselves a shiny new Cadillac!!! And if Libertina turns out to be some sort of sexual acrobat then Vito and Giovanna are both goink to win themselves a shiny new Cadillac too. Good luck and may God bless you!!! Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! What a wonderful weddink there will be What a wonderful day for you and me Church bells will chime You will be mine In apple blossom time... Note that every day Freemason Don Ocean libels me and calls me a pedophile (or such) on the usenet, then I will repost the stats for the following people: Daniel Bruner, Angel Cadwell, Brittanie Cecil, Carl Koopang, Jay Larson, Brian Lott, Adam Millikan, Cody Milliken, Christopher Ridsdale, Vito Rizzuto, Melissa Schultz, Erin Sorenson, Ann Wigdahl, Michael Winkler, or maybe more (more or less). ==== > Friday April 26th 2002 116/249 16504 R I Z Z U T O > 18 9 26 26 21 20 15 = 135 > Philip Francis (Scooter) Rizzuto was the shortstop of the New York Yankees from 1941 to 1956. In his lifetime he hit .273 and was a 5 time all-star. After his playing career he became a Yankee broadcaster for 40 years. He was elected to the Baseball Hall of Fame in 1994. A trivia item is that he was the very first mystery guest on the tv show What's My Line on February 2, 1950. I hope this clears up any concerns you may have. ==== taken this course, but I am learning it by myself, I am trying to solve all the problems of Issaac's book) A group is called metabelian if there exists some abelian normal subgroup A of G such that G/A is also abelian. a) show that G is metabelian iff G''=1 (G' is the commutator subgroup) b) show that homomorphic images of metabelian groups are metabelian c) show that subgroups of metabelian groups are metabelian I don't have clue for it can any one give me some hints? thanks a lot! ==== > taken this course, but I am learning it by myself, I am trying to solve all > the problems of Issaac's book) A group is called metabelian if there exists some abelian normal subgroup > A of G such that G/A is also abelian. a) show that G is metabelian iff G''=1 (G' is the commutator subgroup) [cut] I don't have clue for it > can any one give me some hints? > thanks a lot! It would be nice to give us an idea how far you got on your own. I will assume that you got no where and thus start at the beginning. But, you are working on a pretty advanced problem here and the things I will say should have already been mastered by this time. Anyway, here are some suggestions. I assume that you know that to prove an iff theorem of the form p iff q you must prove 1) if p then q; and 2) if q then p. I will concentrate on just the first statement: if p then q. That is, I will try to prove: Claim: If G is metabelian then G''= 1 The skeleton of the standard proof would look like: Proof: Suppose G is metabelian. ... Then, G'' = 1. Hence, by modus ponens, if G is metabelian then G''= 1. QED (Please note that I am not suggesting or recommending that you or I figure out the proof in the form I am giving here. I know that a direct proof will have the above structure, and I just scribble down the steps that I know will be needed. I think the explanation will be easier to follow if you see how the things I am recommending to do will fit into the final proof that you will write up.) However, that last line is usually left implicit. Thus, you would have: Claim: If G is metabelian then G''= 1 Proof: Suppose G is metabelian. ... Hence, G'' = 1. QED As someone previously mentioned, you should write down the facts that are given in more detail by unwinding the definitions. Thus, you would have: Claim: If G is metabelian then G''= 1 Proof: Suppose G is metabelian. Thus, I can find a subgroup A of G such that: [1] A is abelian; [2] A is normal in G; [3] G/A is abelian (where quotient group makes sense since A is normal in G). ... Hence, G'' = 1. QED Although it is not necessary, I would like to introduce simple names for the two commutator subgroups G' and G'', where I let H = G' and K = H'. Thus, I have: Claim: If G is metabelian then G''= 1 Proof: Suppose G is metabelian. Thus, I can find a subgroup A of G such that: [1] A is abelian; [2] A is normal in G; [3] G/A is abelian (where quotient group makes sense since A is normal in G). Let H = G'. Let K = H'. Note G'' = K. I wish to show that K = 1. ... Hence, G'' = 1. QED Up to now, all this is pretty mechanical. Now, you have to do some thinking. But, at least you have the available information spelled out and you have the goal K = 1 that you want to arrive at. One technique is to work backwards. You want to show that K = 1. But, K = H'. Thus, you want to show that H' = 1. But, H' is the commutator subgroup of H which is the group generated by all the commutators formed from elements of H. Note that you don't have a simple description of K or H', since all you have all its generators. But, you want to prove that H' = 1. What does that say about what the generators of H' look like? Etc. -- Bill Hale ==== >taken this course, but I am learning it by myself, I am trying to solve all >the problems of Issaac's book) A group is called metabelian if there exists some abelian normal subgroup >A of G such that G/A is also abelian. a) show that G is metabelian iff G''=1 (G' is the commutator subgroup) >b) show that homomorphic images of metabelian groups are metabelian >c) show that subgroups of metabelian groups are metabelian I don't have clue for it >can any one give me some hints? The commutator subgroup of G is defined to be the subgroup generated by all elements of the form [x,y] = x^{-1}y^{-1}xy, with x and y in G. (a) Prove that G/G' is abelian, and if N is any normal subgroup of G such that G/N is abelian, then G' is contained in N. Deduce that if G is metabelian, with A the witness, then G'very<- high-powered approach (akin to swatting flies with a nuclear device; my suggestion is to just skip it, but I'm including it just because I can...): Show that a group G is metabelian if and only if it satisfies the identity [ [a,b],[x,y] ] = e for all a,b,x,y in G. Conclude that the class of all metabelian groups is a variety (in the sense of general algebra), and therefore is closed under both subgroups and quotients. Then show that the verbal subgroup corresponding to [[a,b],[x,y]] is G'', to conclude (a). ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >taken this course, but I am learning it by myself, I am trying to solve all >the problems of Issaac's book) A group is called metabelian if there exists some abelian normal subgroup >A of G such that G/A is also abelian. a) show that G is metabelian iff G''=1 (G' is the commutator subgroup) And G'' = (G')' is the commutator subgroup of G'. And the commutator subgroup of a group is a trivial group if and only if the original group is an abelian group. So, G'' = 1 if and only if ... ? >b) show that homomorphic images of metabelian groups are metabelian >c) show that subgroups of metabelian groups are metabelian I don't have clue for it Name everything in sight (G is a metabelian group, A is an abelian normal subgroup of G, h: G->H is a homomorphism onto a group H, ...) and then consider that you are trying to find a certain thing which, after all, if it exists must depend on the things you have named--and if it can be shown to exist presumably depends on those things in a more-or-less natural way. What could it be?... This sounds like Herman Rubin's standard advice for word problems (well, except for the part after --, where natural usually should be replaced by chosen by the teacher, or maybe chosen by Nature but by no means `natural'). Maybe it is. Lee Rudolph ==== >Two questions whose answers I am seeking are: >1) What exactly might we mean by representation? >2) How do we define a mathematical object if we do not want to >identify it with some representation of it? [...] >A mathematical object is uniquely determined if we know what counts as >a model of it, so instead of working with a mathematical object we >could work with the property of being a model of it, but I think that >would be inellegant. [Analogy: in geometry, instead of talking about >points, we could talk about the property a line may have of passing >through that point, but this is inellegant.] This is the extensional view. But the same numbers can > be used as both finite ordinals and finite cardinals, and > these are not the same intensional concept. That the same > numbers can be used for both is useful. I am not sure I have understood the point of your remark. When I said mathematical object above, I did not mean something that includes intension. Nor did I mean to include intension when I mentioned the property of being a model of [the natural numbers]. When I wanted to distinguish between a mathematical object and the property of being a model of it, the distinction I wanted to draw had nothing to do with the extension-intension distinction. As you state, finite ordinals and finite cardinals are the same in extension (I am not sure if one can say without reservation that they are different in intension). To the list above of questions whose answers I am seeking may be added: 3. What might we mean by intensional object? However, I do not see any reason why knowing the answer of this question should be a prerequsite for knowing the answers of the other two. Mattias ==== > I will take the proof with me to my grave. A noblest state of the mind. The attainment of the nobility should not require any 'elite status'. Besides, entire theories should be just as weightless as a single proof to carry to one's grave effortlessly. So what is at stake here? BTP? Seriously, please let me know if there is any interest out there about my giving it a try to settle BTP. (I can assure you that this is not something worth my taking to the grave with me.) ==== Ich spreche Deutsche nicht so schlecht. Wie angegeben in anderen Post, Z steht vor Ze Set which bla bla bla. Haben Sie etwas mehr Fragen ? > I am a novice to the redneck notations accepted in the US math world, and >> intend to stay this way. To cater whims of philistines, card(N):=aleph_0 and >> card(Z):=2**aleph0 Then what is your set Z? If it is the set of integers, you are wrong > about its cardinality, and I am not aware of any common > interpretation as a set other than as the set of integers ( Z for > Zermelo, IIRC). More likely, it's Z for Zahlen. > -- > Dave Seaman > Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. > ==== > Ich spreche Deutsche nicht so schlecht. Wie angegeben in anderen Post, Z > steht vor Ze Set which bla bla bla. Haben Sie etwas mehr Fragen ? >> Then what is your set Z? If it is the set of integers, you are wrong >> about its cardinality, and I am not aware of any common >> interpretation as a set other than as the set of integers ( Z for >> Zermelo, IIRC). More likely, it's Z for Zahlen. Ich habe nichts gefragt. Was heisst Troll auf Deutsch? -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== Lerne dem Killfile zu nuetzen, als weil versuchen dem Mumia konservieren. Troll, in some cases, is what some people who'd like to run usenet but fortunately don't call those whose opinions they cannot comment on, but make themselves look really smart by saying I'm so above it. Troll heisst ein Zwerg, in diesem Fall - ein Giftzwerg, ob Sie hatte nichts gewuesst bis heute. > Ich spreche Deutsche nicht so schlecht. Wie angegeben in anderen Post, Z > steht vor Ze Set which bla bla bla. Haben Sie etwas mehr Fragen ? > Then what is your set Z? If it is the set of integers, you are wrong >> about its cardinality, and I am not aware of any common >> interpretation as a set other than as the set of integers ( Z for >> Zermelo, IIRC). More likely, it's Z for Zahlen. Ich habe nichts gefragt. Was heisst Troll auf Deutsch? > -- > Dave Seaman > Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. > ==== > Lerne dem Killfile zu nuetzen, als weil versuchen dem Mumia konservieren. > Troll, in some cases, is what some people who'd like to run usenet but > fortunately don't call those whose opinions they cannot comment on, but make > themselves look really smart by saying I'm so above it. Troll heisst ein > Zwerg, in diesem Fall - ein Giftzwerg, ob Sie hatte nichts gewuesst bis > heute. The only opinion I have expressed in this thread is that the usage of Z for the integers derives from Zahlen. *Plonk* -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== Besides, to comment on you > awareness, what languages do you fluently read in except english ? You presume that I am fluent in English? ==== Besides, to comment on you > awareness, what languages do you fluently read in except english ? You presume that I am fluent in English? ==== One knows that E_8 is the smallest example of an even unimodular lattice, and the Leech lattice (dim=24) is the smallest example of a unimodular lattice with minimum norm = 4. Are there interesting unimodular lattices with minimum norm = 6 (or larger)? What about odd minimum norms (other than 1)? ==== One knows that E_8 is the smallest example of an even unimodular > lattice, and the Leech lattice (dim=24) is the smallest example of a > unimodular lattice with minimum norm = 4. Are there interesting unimodular lattices with minimum norm = 6 (or > larger)? Yes. There are 3 known examples in rank 48. > What about odd minimum norms (other than 1)? For a start there's the odd Leech lattice of rank 24 and minimum norm 3. This is a current topic of research: see for instance M. Harada, Extremal odd unimodular lattices in dimensions C. Bachoc, G. Nebe & B. Venkov, Odd unimodular lattices of minimum 4. Acta Arith. 101 (2002), no. 2, 151--158. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== Suppose p is an odd prime congruent to 3 (mod 4). If r is any positive odd number, and 2pr+1 is prime, then is 2pr+1 | (p^r) + 1 Ever true? Any help or ideas would be greatly appreciated. GREG ==== , > Suppose p is an odd prime congruent to 3 (mod 4). If r is any positive odd > number, and 2pr+1 is prime, then is 2pr+1 | (p^r) + 1 Ever true? p = 3, r = 11. -- ==== How'd you get that? Brute force method? GREG > , Suppose p is an odd prime congruent to 3 (mod 4). If r is any positive odd > number, and 2pr+1 is prime, then is 2pr+1 | (p^r) + 1 Ever true? p = 3, r = 11. -- ==== I am going through a proof that needs the existance of a real-valued function F(x) F : R^n --> R that is C infinity, i.e., its n-th derivative exists for every n=0,1,2,... The condition on F is that it has to satisfy F(x) = 0 for norm(x) <= r F(x) = 1 for norm(x) => R where 0 < r < R My problem is that I remember (probably I'm very wrong), from complex analysis, that there is no holomorphic function that satisfies these conditions. However, this is in the Reals, so I can't use those results. Is it possible to construct this function or prove its existance? Any help is appreciated, Fernando G. del Cueto ==== > I am going through a proof that needs the existance of a real-valued > function F(x) F : R^n --> R that is C infinity, i.e., its n-th derivative exists for every n=0,1,2,... The condition on F is that it has to satisfy F(x) = 0 for norm(x) <= r > F(x) = 1 for norm(x) => R where 0 < r < R My problem is that I remember (probably I'm very wrong), from complex > analysis, that there is no holomorphic function that satisfies these > conditions. However, this is in the Reals, so I can't use those results. Is it possible to construct this function or prove its existance? Any help is appreciated, > Fernando G. del Cueto > Umm... sure. Infinitely differentiable isn't *nearly* as strong as holomorphic - it's not even as strong as analytic. In particular local behaviour doesn't tell you a lot about global behaviour, and the zeroes of your function can be (almost) arbitrarily badly behaved. The set of zeroes has to be a closed set, but I suspect that given any closed set you can probably construct an infinitely differentiable function that is zero only on that set. Or not. I don't really know, but it looks highly plausible and a brief sketch proof with some major details in need of filling in suggests that it probably works. All of which is really just a long-winded way of saying that you mustn't let your intuition about complex analysis colour your intuition about real analysis. It simply doesn't work the same way. In this particular case you should be able to express your function F as a function of ||x||. ||x|| is differentiable everywhere except at 0, and what do you know about F near 0...? David ==== Is there any characterization for analytic functions in R^n? Fernando G. del Cueto ==== Is there any characterization for analytic functions in R^n? > Fernando G. del Cueto > Hmm. Actually, now that I come to think of it, I'm not sure. They certainly aren't required to have isolated zeroes (e.g. (x, y) -> xy). I *think* they can't have compact support (i.e. be non-zero only on a bounded set) unless they're identically zero, but don't quote me on that. There isn't however, as far as I know, any nice characterisation of them. I'll think about it, but I don't expect to come up with anything particularily nice. It's probably worth noting that if I'm right about the compact support, then the function you request can't be analytic. Because if it was then 1 - F would have compact support and be analytic. Certainly at any rate all the examples I can think of are non-analytic. (I probably would have used something like W Dale Hall's example, just with a slightly different choice of bump function. Similar construction, starting from e^{-1/x^2}, but I would tend to take a slightly different route. No good reason, I just do.) David ==== > Is there any characterization for analytic functions in R^n? > >> Fernando G. del Cueto >> Hmm. Actually, now that I come to think of it, I'm not sure. They certainly >aren't required to have isolated zeroes (e.g. (x, y) -> xy). I *think* >they can't have compact support (i.e. be non-zero only on a bounded set) >unless they're identically zero, but don't quote me on that. That's certainly true. The same proof as in complex analysis works: Let S be the set where all the derivatives vanish. The fact that f is analytic, ie locally equal to a power series, shows that S is open, while it's clear that f is closed. So assuming the domain is connected, which of course we're assuming or the result is obviously false, S must be empty or the entire domain. If f vanishes on an open set then S is non-empty, so S is the whole domain. >There isn't >however, as far as I know, any nice characterisation of them. I'll think >about it, but I don't expect to come up with anything particularily nice. I can't figure out what sort of characterization we're looking for. I mean the definition seems like a pretty nice characterization... >It's probably worth noting that if I'm right about the compact support, >then the function you request can't be analytic. Because if it was then >1 - F would have compact support and be analytic. Certainly at any rate >all the examples I can think of are non-analytic. (I probably would have >used something like W Dale Hall's example, just with a slightly >different choice of bump function. Similar construction, starting from >e^{-1/x^2}, but I would tend to take a slightly different route. No good >reason, I just do.) David David C. Ullrich ==== There isn't >>however, as far as I know, any nice characterisation of them. I'll think >>about it, but I don't expect to come up with anything particularily nice. > I can't figure out what sort of characterization we're looking for. > I mean the definition seems like a pretty nice characterization... > My guess was that he wanted something analagous to 'every complex differentiable function is analytic'. Some simple statement that's easier to check than the power series definition. As I said, the chance of finding one doesn't seem too likely - real analysis isn't as friendly as complex - but I figured it can't hurt to have a bit of a think about it. David ==== > >There isn't >however, as far as I know, any nice characterisation of them. I'll think >about it, but I don't expect to come up with anything particularily nice. > >> I can't figure out what sort of characterization we're looking for. >> I mean the definition seems like a pretty nice characterization... >> My guess was that he wanted something analagous to 'every complex >differentiable function is analytic'. Some simple statement Like, every real function (on an open set U in R^n) which can be extended to be complex-differentiable on an open neighborhood of U in C^n=R^n+iR^n is real-analytic? >that's >easier to check than the power series definition. Ooops. Scratch that, then. Lee Rudolph ==== > Actually, now that I come to think of it, I'm not sure. They certainly > aren't required to have isolated zeroes (e.g. (x, y) -> xy). I *think* > they can't have compact support (i.e. be non-zero only on a bounded set) > unless they're identically zero, but don't quote me on that. Of course! One of the basic properties of analytic functions is that, if their domain is connected, then two distinct analytic functions cannot have the same restrictions to an open subset of their domain. Jose Carlos Santos ==== Actually, now that I come to think of it, I'm not sure. They certainly >> aren't required to have isolated zeroes (e.g. (x, y) -> xy). I *think* >> they can't have compact support (i.e. be non-zero only on a bounded >> set) unless they're identically zero, but don't quote me on that. > Of course! One of the basic properties of analytic functions is that, if > their domain is connected, then two distinct analytic functions cannot > have the same restrictions to an open subset of their domain. Yes, I expected as much. I just didn't feel like working through the proof to check all the details, as the proof I would use in the complex case obviously doesn't generalise at all (which probably shows it's a bad proof, but never mind. :), and didn't want to commit to something I wasn't certain of. David ==== > I am going through a proof that needs the existance of a real-valued > function F(x) F : R^n --> R that is C infinity, i.e., its n-th derivative exists for every n=0,1,2,... The condition on F is that it has to satisfy F(x) = 0 for norm(x) <= r > F(x) = 1 for norm(x) => R where 0 < r < R My problem is that I remember (probably I'm very wrong), from complex > analysis, that there is no holomorphic function that satisfies these > conditions. However, this is in the Reals, so I can't use those results. Is it possible to construct this function or prove its existance? Any help is appreciated, > Fernando G. del Cueto > What you need is what I've heard called a bump function, that is, a C-infinity function B(x) for which B(x) = 0 for x <= a B(x) > 0 for a < x < b B(x) = 0 for x >= b Note that this function q(x) (the standard C-infinity function that has all zero derivatives at x = 0): q(x) = exp(-1/x^2) for |x|>0, q(0) = 0 can be used to fashion B(x) as follows: Let q1(x) = 0 for x <= a q(x-a) for x > a and q2(x) = q(x-b) for x <= b 0 for x >= b Then q1 and q2 are C-infinity functions with support(q1) = {x | x >= a} support(q2) = {x | x <= b} Let B(x) = q1(x)q2(x), and notice that B(x) has support in {x | a <= x <= b}. Note that B is positive on the open interval (a,b). Next, a smooth step s(x) is formed by integrating B(t) from -infinity to x. Note that s(x) is identically 0 for x <= a, and takes a positive constant value for x >= b. That constant can be adjusted to your favorite value by multiplying s(x) by the appropriate factor. Your function F can be produced by the appropriate selection of constant values for a and b, and then taking F(x) = B(|x|) where |x| is the Euclidean norm |x| = sqrt(x'x). Pretty standard stuff, I think. Dale. ==== I guess I shouldn't trust in my complex intuition anymore... Fernando G. del Cueto I am going through a proof that needs the existance of a real-valued >> function F(x) F : R^n --> R that is C infinity, i.e., its n-th derivative exists for every >> n=0,1,2,... The condition on F is that it has to satisfy F(x) = 0 for norm(x) <= r >> F(x) = 1 for norm(x) => R where 0 < r < R My problem is that I remember (probably I'm very wrong), from complex >> analysis, that there is no holomorphic function that satisfies these >> conditions. However, this is in the Reals, so I can't use those results. Is it possible to construct this function or prove its existance? Any help is appreciated, >> Fernando G. del Cueto >> What you need is what I've heard called a bump function, that is, > a C-infinity function B(x) for which B(x) = 0 for x <= a > B(x) > 0 for a < x < b > B(x) = 0 for x >= b Note that this function q(x) (the standard C-infinity function > that has all zero derivatives at x = 0): q(x) = exp(-1/x^2) for |x|>0, > q(0) = 0 can be used to fashion B(x) as follows: Let q1(x) = 0 for x <= a > q(x-a) for x > a and q2(x) = q(x-b) for x <= b > 0 for x >= b Then q1 and q2 are C-infinity functions with support(q1) = {x | x >= a} > support(q2) = {x | x <= b} Let B(x) = q1(x)q2(x), and notice that B(x) has > support in {x | a <= x <= b}. Note that B is positive on the open interval (a,b). Next, a smooth step s(x) is formed by integrating B(t) > from -infinity to x. Note that s(x) is identically 0 > for x <= a, and takes a positive constant value for x >= b. > That constant can be adjusted to your favorite value by > multiplying s(x) by the appropriate factor. Your function F can be produced by the appropriate selection > of constant values for a and b, and then taking F(x) = B(|x|) where |x| is the Euclidean norm |x| = sqrt(x'x). Pretty standard stuff, I think. Dale. > ==== Most of us know that x! = Gamma(x+1) ~ x^(x+1/2) exp(-x) sqrt(2 pi) as x -> oo, by Stirling's asymptotical formula. But what about the asymptotic behavior of Gamma(x+1) as x appoaches other limits other than real positive infinity? For example: In the vicinity of x = 0, Gamma(x+1) = exp(-c*x +x^2 pi^2/12 - x^3 zeta(3) +O(x^4)), where c is Euler's constant. - And, as x -> -oo, Gamma(x+1) ~ -(-x)^(x+1/2) exp(-x) csc(pi x) sqrt(pi/2). But I am not absolutely certain about this asymptotic formula, and would not use it to investigate the behavior of Gamma(x+1) at x very near negative integers. - So, what, for example, about the asymptotics if x = 1/2 + i y, where y approaches infinity? Or how about if x approaches any other interesting finite/infinite complex/real constants? thanks, Leroy Quet ==== >Most of us know that >x! = Gamma(x+1) ~ x^(x+1/2) exp(-x) sqrt(2 pi) >as x -> oo, by Stirling's asymptotical formula. >But what about the asymptotic behavior of Gamma(x+1) as x appoaches >other limits other than real positive infinity? >For example: >In the vicinity of x = 0, >Gamma(x+1) = exp(-c*x +x^2 pi^2/12 - x^3 zeta(3) +O(x^4)), >where c is Euler's constant. Change that to zeta(3)/3; the series is convergent for |x| < 1, and the general term is (-x)^n*zeta(n)/n. >- >And, as x -> -oo, >Gamma(x+1) ~ -(-x)^(x+1/2) exp(-x) csc(pi x) sqrt(pi/2). >But I am not absolutely certain about this asymptotic formula, and >would not use it to investigate the behavior of Gamma(x+1) at x very >near negative integers. Gamma(x)*Gamma(1-x) = pi/sin(pi x). No problems here. >- >So, what, for example, about the asymptotics if x = 1/2 + i y, where y >approaches infinity? Sterling's formula with remainder is valid; I am unsure as to how good it is for the imaginary part of the logarithm. The real part of the logarithm follows from the multiplication identity above. >Or how about if x approaches any other interesting finite/infinite >complex/real constants? A very useful formula is log(x!) = (x+1/2)log(x+1/2) -x + log(2 pi)/2 + R, where R ~ (1/2 -1)*1/(12x) - (1/2^3 - 1)*(1/360x^3) ... -- 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 ==== > Most of us know that x! = Gamma(x+1) ~ x^(x+1/2) exp(-x) sqrt(2 pi) as x -> oo, by Stirling's asymptotical formula. > But what about the asymptotic behavior of Gamma(x+1) as x appoaches > other limits other than real positive infinity? For example: In the vicinity of x = 0, Gamma(x+1) = exp(-c*x +x^2 pi^2/12 - x^3 zeta(3) +O(x^4)), where c is Euler's constant. - And, as x -> -oo, > Gamma(x+1) ~ -(-x)^(x+1/2) exp(-x) csc(pi x) sqrt(pi/2). > But I am not absolutely certain about this asymptotic formula, and > would not use it to investigate the behavior of Gamma(x+1) at x very > near negative integers. - So, what, for example, about the asymptotics if x = 1/2 + i y, where y > approaches infinity? Or how about if x approaches any other interesting finite/infinite > complex/real constants? You know of course the classic log(Gamma(z)) ~ (z-1/2) log(z) - z + log(2 pi)/2 + sum_{m>0} B_m/(2 m (2 m-1) z^(2 m -1)) where B_m are the Bernoulli numbers. AFAIK, this is true for |arg(z)| < pi. HTH, Michael. -- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&& Dr. Michael Ulm FB Mathematik, Universitaet Rostock michael.ulm@mathematik.uni-rostock.de ==== > Most of us know that x! = Gamma(x+1) ~ x^(x+1/2) exp(-x) sqrt(2 pi) as x -> oo, by Stirling's asymptotical formula. But even nicer (as I have repeatedly mentioned in sci.math) is Burnside's formula: x! = Gamma(x+1) ~ sqrt(2 pi) ((x+1/2)/e)^(x+1/2) > But what about the asymptotic behavior of Gamma(x+1) as x appoaches > other limits other than real positive infinity? For example: In the vicinity of x = 0, Gamma(x+1) = exp(-c*x +x^2 pi^2/12 - x^3 zeta(3)/3 +O(x^4)), where c is Euler's constant. Here are some others: Near x = -1, Gamma(x+1) = 1/((1 + x) + c*(1 + x)^2 + (c^2/2 - Pi^2/12)*(1 + x)^3 + O(x^4)) Near x = -2, Gamma(x+1) = 1/(-(2 + x) + (1 - c)*(2 + x)^2 + (c - c^2/2 + Pi^2/12)*(2 + x)^3 + O(x^4)) Near x = -3, Gamma(x+1) = 1/(2*(3 + x) + (-3 + 2*c)*(3 + x)^2 + (1 + (-3 + c)*c - Pi^2/6)*(3 + x)^3 + O(x^4)) Near x = -4, Gamma(x+1) = 1/(-6*(4 + x) + (11 - 6*c)*(4 + x)^2 + (-6 + (11 - 3*c)*c + Pi^2/2)*(4 + x)^3 + O(x^4)) > And, as x -> -oo, Gamma(x+1) ~ -(-x)^(x+1/2) exp(-x) csc(pi x) sqrt(pi/2). Somewhat nicer IMO would be Gamma(x+1) ~ sqrt(pi/2) x csc(pi x)/((-x+1/2)/e)^(-x+1/2). > But I am not absolutely certain about this asymptotic formula, and > would not use it to investigate the behavior of Gamma(x+1) at x very > near negative integers. Right. Instead, use results similar to those in the group of four which I gave above. David Cantrell ==== > Most of us know that x! = Gamma(x+1) ~ x^(x+1/2) exp(-x) sqrt(2 pi) as x -> oo, by Stirling's asymptotical formula. > But what about the asymptotic behavior of Gamma(x+1) as x appoaches > other limits other than real positive infinity? For example: In the vicinity of x = 0, Gamma(x+1) = exp(-c*x +x^2 pi^2/12 - x^3 zeta(3) +O(x^4)), where c is Euler's constant. Whoops! I forgot to divide zeta(3) by 3. So... Gamma(x+1) = exp(-c*x +x^2 pi^2/12 - x^3 zeta(3)/3 +O(x^4)) Leroy - And, as x -> -oo, > Gamma(x+1) ~ -(-x)^(x+1/2) exp(-x) csc(pi x) sqrt(pi/2). > But I am not absolutely certain about this asymptotic formula, and > would not use it to investigate the behavior of Gamma(x+1) at x very > near negative integers. - So, what, for example, about the asymptotics if x = 1/2 + i y, where y > approaches infinity? Or how about if x approaches any other interesting finite/infinite > complex/real constants? > thanks, > Leroy Quet ==== Background: I am a 48-year old adjunct partial-load professor employed in a local college and very happy teaching on contract. My qualifications are from the techy/industry side, i.e. engineer/technologist association exams, college diploma, 28 years of technical and teaching experience. I have aways enjoyed an interest and ability in applied mathematics. Both kids have moved away and home, finances, and marriage, are now secure, hence, its time for *me*! It has always been my dream to publish papers, write text books, supervise courses, and generally influence education in applied math areas. My area of expertise is instrumentation and control, requiring Laplace, differential equations, complex numbers, matrices, etc. In order to achieve these goals and to enhance my knowledge, qualifications, and employability, I am looking for an online, correspondence, or distance Bachelor's degree which can lead to a Master's Degree or Ph.D. Q1. Has anybody out there done this? Q2. Will a math degree of this type be considered acceptable to teach in university? Q3. Will I be qualified to teach engineering students? Q4. Will publishers publish? George ==== ..is newly available for free download at http://www.tinaja.com/glib/msquant.pdf Sourcecode is provided as http://www.tinaja.com/glib/msquant.psl It is on Magic Sinewave Quantization optimizations. Also includes some DFT Fourier Transform fundamental routines. Magic sinewaves are a brand new way to synthesize high power digital sinewaves with arbitrarily low distortion at the highest possible energy efficiency. Other GuruGrams at http://www.tinaja.com/gurgrm01.asp -- Many thanks, Don Lancaster Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552 Please visit my GURU's LAIR web site at http://www.tinaja.com ==== Seems as if no one has any clue... (snicker) So I will give a couple clues below... > We have a triangular array of integers {a(m,n)}. a(1,1) = 0; And, for 1 <= n <= m, m >= 2, recursively: a(m,n) = > > m-1 > --- --- > > 1 + > ( > mu(j) ) a(m-1,k) > / / > --- --- > k=1 j|k > j>= kn/m Ascii-mode: a(m,n) = 1 + sum{k=1 to m-1} (sum{j|k, j >= kn/m} mu(j)) a(m-1,k) The inner-sum is over the divisors, j, of k, > where each j is >= kn/m. And mu() is the Mobius (Moebius) function. Now, the array's elements can be easily described with a closed-form > (ie. non-recursive) definition. What is this closed-form for {a(m,n)}? I will first give a clue in a few days, then the answer a few days > after that, if no one posts an answer before I do this. thanks, > Leroy Quet First, every a(m,n) is a positive integer for m >=2, 1 <= n <= m. Second, a(m,1) DOES always = m-1, and a(m,m) does always = 1 for m >= 2. thanks, Leroy Quet ==== sorry; I shouldn't make an implication that an inductive proof is just the reverse of a deductive proof. > there is also a proof of the isomorphism > of deductive proofs with inductive ones, > which may perhaps be amenable to combining the two forms > into a tautology, or necklace. --ils duces d'Enron! http://larouchepub.com/ ==== sci.physics snipped. >>If y is not 0, you can't use the constant term tricks that depend on y >>being 0. That's stupid. The constant term once found is distinguished by being >constant. All the trick is doing is finding it. So let me give you the example that I've used elsewhere which is >a_1(x) + 7, which has a constant term that is 7. Do you understand >what it means for it to be constant? Here's a test. If I have x=11, then I necessarily have a_1(11) + 7, right? Notice the constant term is STILL there, do you understand? Now then, if I now divide P(11) by 49, what should the constant term >be? If you answer honestly I'll be shocked. Let me try again. (a_1(y) + 7)/x is not the same as 7/x, unless a_1(y) = 0. > Ok, I can go from there Richard Henry, and note that necessarily > a_1(y) has *some* factor in common with 7, right? But for some values of y, that factor might be 1. This is the possibility you have not been dealing with. So let's call that factor f, now dividing 49 from P(y) will divide f > off from a_1(y) + 7, understand? Again, f might be 1 for some values of y. Now then, you have a_1(y)/f + 7/f, and if f does not equal 7, what > does that tell you about the *constant* term Richard Henry? Nothing, because the constant term is determined by a_1(0), and that does not deal with a_1(y). Necessarily, you have 7/f left as the constant term for a factor of > P(11)/49, and if 7/f does not equal 1, you have a contradiction. This is assuming f is not a function of y. The whole point has been that most of us believe that f *is* a function of y. Understand? The trick is to FIND the constant term, as you know that it's not > affected by the value of x, and it sits there like a rock, unaffected > by the value of x, and none of your protestations against mathematical > reality will change that fact. This trick is not as useful as you believe it is. -- Will Twentyman ==== In Lie Groups, Lie Algebras, and their Representations, V. S. Varadarajan defines a C^infinity manifold to be a second countable Hausdorff space M with a C^infinity differentiable structure, that being an assignment D:U|->D(U), for all open U contained in M, such that (i) for each open U contained in M, D(U) is an algebra of complex-valued functions on U containing 1 (ii) if V, U are open, V is contained in U, and f is in D(U), then the restriction of f to V is in D(V); if V_i (i in J) are open, V is the union of the V_i, and f is a complex-valued function defined on V such that the restriction of f to V_i is in D(V_i) for all i in J, then f is in D(V) (iii) there exists an integer m>0 with the following property; for any x in M, one can find an open set U containing x, and m real functions x_1, ..., x_m from D(U) such that (a) the map xi:y|->(x_1(y),...,x_m(y)) is a homoeomorphism of U onto an open subset of R^m, and (b) for any open set V contained in U and any complex-valued function f defined on V, f is in D(V) if and only if f circle xi^-1 is a C^infinity function on xi[V]. He then remarks that M is obviously locally connected and metrizable. It sure is obviously locally connected, but metrizable I can't see. Can anyone shed some light? ==== > In Lie Groups, Lie Algebras, and their Representations, V. S. > Varadarajan defines a C^infinity manifold to be a second countable > Hausdorff space M with a C^infinity differentiable structure, that > being an assignment D:U|->D(U), for all open U contained in M, such > that (i) for each open U contained in M, D(U) is an algebra of > complex-valued functions on U containing 1 > (ii) if V, U are open, V is contained in U, and f is in D(U), then the > restriction of f to V is in D(V); if V_i (i in J) are open, V is the > union of the V_i, and f is a complex-valued function defined on V such > that the restriction of f to V_i is in D(V_i) for all i in J, then f > is in D(V) > (iii) there exists an integer m>0 with the following property; for any > x in M, one can find an open set U containing x, and m real functions > x_1, ..., x_m from D(U) such that (a) the map > xi:y|->(x_1(y),...,x_m(y)) is a homoeomorphism of U onto an open > subset of R^m, and (b) for any open set V contained in U and any > complex-valued function f defined on V, f is in D(V) if and only if f > circle xi^-1 is a C^infinity function on xi[V]. He then remarks that M is obviously locally connected and > metrizable. It sure is obviously locally connected, but metrizable I can't see. > Can anyone shed some light? Since it's explicitly assumed that the space is 2nd-countable, this follows pretty easily from a couple of well-known results: - every locally compact Hausdorff space is completely regular - (Urysohn's Metrization Theorem) A 2nd-countable Hausdorff space is metrizable iff it is regular. Oh ... and the fact that M above is as obviously locally compact as it is locally connected (both due to its locally Euclidean nature) ... Does that help ?? ==== > In Lie Groups, Lie Algebras, and their Representations, V. S. > Varadarajan defines a C^infinity manifold to be a second countable > Hausdorff space M with a C^infinity differentiable structure, that > being an assignment D:U|->D(U), for all open U contained in M, such > that (i) for each open U contained in M, D(U) is an algebra of > complex-valued functions on U containing 1 > (ii) if V, U are open, V is contained in U, and f is in D(U), then the > restriction of f to V is in D(V); if V_i (i in J) are open, V is the > union of the V_i, and f is a complex-valued function defined on V such > that the restriction of f to V_i is in D(V_i) for all i in J, then f > is in D(V) > (iii) there exists an integer m>0 with the following property; for any > x in M, one can find an open set U containing x, and m real functions > x_1, ..., x_m from D(U) such that (a) the map > xi:y|->(x_1(y),...,x_m(y)) is a homoeomorphism of U onto an open > subset of R^m, and (b) for any open set V contained in U and any > complex-valued function f defined on V, f is in D(V) if and only if f > circle xi^-1 is a C^infinity function on xi[V]. He then remarks that M is obviously locally connected and > metrizable. It sure is obviously locally connected, but metrizable I can't see. > Can anyone shed some light? ==== I want to determine the probability that if you pick B,C randomly from [0,1], and A randomly from (0,1], that the polynomial Ax^2 + Bx + C has two distinct roots. Obviously this boils down to B^2-4AC >= 0. So I try doing a double integral of the function C=B^2/(4A), but I can't figure out what to put as my limits of integration. Anyone have a suggestion? ==== > I want to determine the probability that if you pick B,C randomly from > [0,1], and A randomly from (0,1], that the polynomial Ax^2 + Bx + C > has two distinct roots. Obviously this boils down to B^2-4AC >= 0. > So I try doing a double integral of the function C=B^2/(4A), but I > can't figure out what to put as my limits of integration. Anyone have > a suggestion? You need a triple integral, integrating over all three variables A, B, and C. The domain of integration is determined by the condition B^2 >= 4 A C. The integrand is just a unit function, and hence the integral can be interpreted geometrically as the volume of the domain of integration. The latter is some section of a hyperboloid, but of course bounded by the additional constraints that all three variables are between 0 and 1. Forgetting about the geometrics and just doing the integration is pretty straightforward, and I get the final value (5+6*ln 2) / 36 = 0.254413... -Michael. ==== > I want to determine the probability that if you pick B,C randomly from > [0,1], and A randomly from (0,1], that the polynomial Ax^2 + Bx + C > has two distinct roots. Obviously this boils down to B^2-4AC >= 0. > So I try doing a double integral of the function C=B^2/(4A), but I > can't figure out what to put as my limits of integration. Anyone have > a suggestion? You need a triple integral, integrating over all three variables A, B, and > C. The domain of integration is determined by the condition B^2 >= 4 A C. > The integrand is just a unit function, and hence the integral can be > interpreted geometrically as the volume of the domain of integration. The > latter is some section of a hyperboloid, but of course bounded by the > additional constraints that all three variables are between 0 and 1. Forgetting about the geometrics and just doing the integration is pretty > straightforward, and I get the final value (5+6*ln 2) / 36 = 0.254413... -Michael. Well, back to the geometry, what did you use for your limits of integration on each of the variables A, B, and C? Also, does anyone have handy the discriminant of a cubic? Maybe we can apply the same idea to this? ==== > You need a triple integral, integrating over all three variables A, B, and > C. The domain of integration is determined by the condition B^2 >= 4 A C. > The integrand is just a unit function, and hence the integral can be > interpreted geometrically as the volume of the domain of integration. The > latter is some section of a hyperboloid, but of course bounded by the > additional constraints that all three variables are between 0 and 1. Forgetting about the geometrics and just doing the integration is pretty > straightforward, and I get the final value (5+6*ln 2) / 36 = 0.254413... Well, back to the geometry, what did you use for your limits of > integration on each of the variables A, B, and C? Well, I threw away my calculations, but I'll try to reconstruct. The integration limits are 04*A*C, i.e. B > 2 * sqrt(A*C). I first did the integral over B. Holding A and C fixed there are two cases to consider: A*C < 1/4 and A*C > 1/4. In the former case, the integral is over 2 * sqrt(A*C) < B < 1, and hence has the value 1 - 2 * sqrt(A*C). In the latter case, the integral is over the interval from 0 to 1, and hence has the value 1. Next I eliminated the C integral, and - holding A fixed - once again had to consider two cases: A < 1/4 and A > 1/4. Are you able to do the rest from here? > Also, does anyone have handy the discriminant of a cubic? Maybe we > can apply the same idea to this? Er, what for? Do you have some application in mind? I mean, why uniform in the interval [0,1]? Why not standard deviation, say, or some other distribution? -Michael. ==== > You need a triple integral, integrating over all three variables A, B, > and > C. The domain of integration is determined by the condition B^2 >= 4 A > C. > The integrand is just a unit function, and hence the integral can be > interpreted geometrically as the volume of the domain of integration. > The > latter is some section of a hyperboloid, but of course bounded by the > additional constraints that all three variables are between 0 and 1. Forgetting about the geometrics and just doing the integration is pretty > straightforward, and I get the final value (5+6*ln 2) / 36 = 0.254413... Well, back to the geometry, what did you use for your limits of > integration on each of the variables A, B, and C? Well, I threw away my calculations, but I'll try to reconstruct. The > integration limits are 0 restriction that B^2>4*A*C, i.e. B > 2 * sqrt(A*C). I first did the integral > over B. Holding A and C fixed there are two cases to consider: A*C < 1/4 and > A*C > 1/4. In the former case, the integral is over 2 * sqrt(A*C) < B < 1, > and hence has the value 1 - 2 * sqrt(A*C). In the latter case, the integral > is over the interval from 0 to 1, and hence has the value 1. Next I > eliminated the C integral, and - holding A fixed - once again had to > consider two cases: A < 1/4 and A > 1/4. Are you able to do the rest from > here? > Also, does anyone have handy the discriminant of a cubic? Maybe we > can apply the same idea to this? Er, what for? Do you have some application in mind? I mean, why uniform in > the interval [0,1]? Why not standard deviation, say, or some other > distribution? -Michael. No particular reason, I just thought it would be interesting to see if there was some noticeable pattern in the probabilities. I suppose as n tends to infinity the probability of an arbitrary n-th degree polynomial having n real roots tends to 0, but it would still require proof. Anyway it basically boils down to just being curious. ==== >Forgetting about the geometrics and just doing the integration is pretty >straightforward, and I get the final value (5+6*ln 2) / 36 = 0.254413... That coincides with the numerical results I found. Doug ==== >I want to determine the probability that if you pick B,C randomly from >[0,1], and A randomly from (0,1], that the polynomial Ax^2 + Bx + C >has two distinct roots. Obviously this boils down to B^2-4AC >= 0. >So I try doing a double integral of the function C=B^2/(4A), but I >can't figure out what to put as my limits of integration. Anyone have >a suggestion? > Why do you assume the roots must be real? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== >I want to determine the probability that if you pick B,C randomly from >>[0,1], and A randomly from (0,1], that the polynomial Ax^2 + Bx + C >>has two distinct roots. Obviously this boils down to B^2-4AC >= 0. >>So I try doing a double integral of the function C=B^2/(4A), but I >>can't figure out what to put as my limits of integration. Anyone have >>a suggestion? >Why do you assume the roots must be real? Wouldn't make a very interesting problem otherwise. But what if we ask what is the probability that the roots have at most epsilon difference? ==== >I want to determine the probability that if you pick B,C randomly from >[0,1], and A randomly from (0,1], that the polynomial Ax^2 + Bx + C >has two distinct roots. Obviously this boils down to B^2-4AC >= 0. >So I try doing a double integral of the function C=B^2/(4A), but I >can't figure out what to put as my limits of integration. Anyone have >a suggestion? I was bored, so I ran an empirical analysis on Mathematica. With 4,000,000 trials, there were 1,018,340 successes for a p of 0.254585. With 99% confidence, the true probability of success is between 0.254024 and 0.255146. Doug ==== >I want to determine the probability that if you pick B,C randomly from >[0,1], and A randomly from (0,1], that the polynomial Ax^2 + Bx + C >has two distinct roots. Obviously this boils down to B^2-4AC >= 0. >So I try doing a double integral of the function C=B^2/(4A), but I >can't figure out what to put as my limits of integration. Anyone have >a suggestion? I was bored, so I ran an empirical analysis on Mathematica. With 4,000,000 > trials, there were 1,018,340 successes for a p of 0.254585. With 99% confidence, the true probability of success is between > 0.254024 and 0.255146. Doug Well, I should restate the problem and say that I want it to have two REAL roots, distinct or not. Sorry I was not being clear originally. Anyway, I think I've solved it. Let me see if this works: We need B^2>=4AC B>=2Sqrt[AC] So we calculate 2*Integral[{0,1},Sqrt[A]*Integral[{0,1},Sqrt[C],dC],dA]. This turns out to be 8/9. So this is the probability that the roots will be COMPLEX, so the roots that they are real is 1-8/9 = 1/9. I'm not sure if this is the test you ran, but it would be interesting to run mathematica again and see if it is close to 1/9 with the restated problem. Using a similar method except with integration in 1 variable, I find that when A is fixed to be 1, the probability is 1/12, which would make sense since we don't allow that possibility that A reduces the 4AC term. ==== >I want to determine the probability that if you pick B,C randomly from >>[0,1], and A randomly from (0,1], that the polynomial Ax^2 + Bx + C >>has two distinct roots. Obviously this boils down to B^2-4AC >= 0. >>So I try doing a double integral of the function C=B^2/(4A), but I >>can't figure out what to put as my limits of integration. Anyone have >>a suggestion? > I was bored, so I ran an empirical analysis on Mathematica. With 4,000,000 >> trials, there were 1,018,340 successes for a p of 0.254585. > With 99% confidence, the true probability of success is between >> 0.254024 and 0.255146. > Doug >Well, I should restate the problem and say that I want it to have two >REAL roots, distinct or not. Sorry I was not being clear originally. I used the same assumptions that you did in my empirical work, so I'm still under the impression that you should get a value within the above confidence interval. In fact, all I checked for success was whether the discriminant was positive. Doug ==== > I want to determine the probability that if you pick B,C randomly from > [0,1], and A randomly from (0,1], that the polynomial Ax^2 + Bx + C > has two distinct roots. Obviously this boils down to B^2-4AC >= 0. > So I try doing a double integral of the function C=B^2/(4A), but I > can't figure out what to put as my limits of integration. Anyone have > a suggestion? I'm not sure I understand the problem completely. It seems like you want to have the roots be real ( a condition not explicitly stated), and distinct. However, it seems that letting the discriminant equal zero will give you 2 repeated roots. So let me paraphrase your question and see if this is what you are saying: We let a,b,c be independent random variables distributed as unif(0,1), and we wish to find the probability that an a,b,c, are chosen such that the polynomial ax^2+bx+c has 2 distinct real roots , i.e. that b^2-4ac > 0. Is this a corect statement of your problem? MB ==== I'd like to know if there's any connection between what the algebraic integers *should* include, and the ever-shrinking Short Proof of Fermat's Last, altough I realize that monsiuer Harris doesn't have the time to make those connections in his pioneering work, what wiht less than 2 years in the programme, left. so, this falls to us, his students. again I note, FLT is the same sort of problem as the Perfect Box one, if it can be proven that there is no solution in integers for the 3 sides, 3 diagonals & one interior diagonal of a rectangular box. whether this simpler exclusionary problem throws any light on the old Fermat one, is the question. for correctness, I'd just like to refer to it, as Fermat's Old Problem -- How'd He Do It? > routine technique in analysis, will see how desperate certain people --ils duces d'Enron! http://www.movisol.org/ http://members.tripod.com/~american_almanac/ ==== > Last I remembered setting a variable to 0 to clear it out, knowing > that pulled out terms independent of it was routine in analysis Care to diagram that sentence like back in grade school? It makes not > the slightest bit of sense. The sentence slightly deobfuscated: As I remember, the techniqe of pulling out independent terms by setting a variable to 0 (clearing it out so to speak) was routine in analysis. If you want to correspond with James Harris you are going to have to deal with much worse than this. - William Hughes ==== > Last I remembered setting a variable to 0 to clear it out, knowing > that pulled out terms independent of it was routine in analysis, which > is why a lot of this is funny, ironic, and very, very sad. Yup, may be common in analysis, you might even apply it to algebra, but you should do it the correct way. Neither in analysis, nor in algebra, is it immediately clear what the remainder is. But is is blatantly nonsense when you write that in a1(9)/sqrt(7) + sqrt(7) the constant term is sqrt(7). I set the variable at 0 (I do not see any variable), and get the same value, which is a1(9)/sqrt(7) + sqrt(7). > What is happening (I hope) is that people who do analysis, if they > ever bother to notice the discussions that are going back and forth, > where certain posters are spending a lot of effort to challenge a > routine technique in analysis, will see how desperate certain people > in math society have become. Well, if you did apply your technique the correct way, yes, there would be a point. But when you state that the constant term of a1(9)/sqrt(7) + sqrt(7) is sqrt(7) you do not use the technique in the way it should be done. > They're challenging simple techniques that are very important to real > research. But why do you not use them the correct way? What *is* the constant term of a1(9)/sqrt(7) + sqrt(7) ? Use either analytical methods or algebraic, I do not mind. Sheesh. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > Last I remembered setting a variable to 0 to clear it out, knowing > that pulled out terms independent of it was routine in analysis, which > is why a lot of this is funny, ironic, and very, very sad. I'll try to keep up the tradition of cutting your posts off after the first mistake to avoid unnecessary work: You remember wrongly. And yes, you are indeed very, very sad. ==== > >Background: I am a 48-year old adjunct partial-load professor employed in a > local college and very happy teaching on contract. wow, that is weird, a person happy at being abused. > My qualifications > are from the techy/industry side, i.e. engineer/technologist > association exams, college diploma, 28 years of technical and > teaching experience. > I have aways enjoyed an interest and ability in applied mathematics. > Both kids have moved away and home, finances, and marriage, are now > secure, hence, its time for *me*! It has always been my dream to publish papers, write text books, > supervise courses, and generally influence education in applied math > areas. My area of expertise is instrumentation and control, requiring > Laplace, differential equations, complex numbers, matrices, etc. In order to achieve these goals and to enhance my knowledge, > qualifications, and employability, I am looking for an online, > correspondence, or distance Bachelor's degree which can lead to a > Master's Degree or Ph.D. i seem to recall there being a master degree by correspondence offered at florida institute of technology in applied math or operations research. now, if your real goal is to attain a full time position as a college math teacher, you will have better chances if you get a math-ed degree - reason being that, with few exceptions, colleges care less about academic expertise than they do about con-artistry abilities to please/ appease students and administrators. math education programmes will do an excellent job at preparing you for that. > Q1. Has anybody out there done this? > Q2. Will a math degree of this type be considered acceptable to teach > in university? > Q3. Will I be qualified to teach engineering students? > Q4. Will publishers publish? George ==== > God isn't responsible for the bad things that happen, Satan is. Read > your damn Bible, specifically, the Book of Job. Lol and god created Satan knowing *FULL* well what would happen > (unless hes not omniscient). > So then the question is, if God knew what would happen, then perhaps Satan isn't such a bad guy after all? Consider: he obeyed God's commands concerning how much he was allowed to afflict Job. Sounds like an obedient son of God to me... ==== There... are... four... lights! -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ To doo bee doo bee doo. - Frank Sinatra ==== >>Given reals x and y, suppose n0(k) is the least integer such that n0*x and >n0*y >>are within 10^(-k) of being integer. Is n0(k)<10^(2*k+1)? Do there exist x >>and y s.t. n0(k)<10^(2*k) for all k? > > Review the pigeonhole principle. y=sqrt(3), n0(16) is the 26 digit number 915...804? Rich Burge ==== connectivity of a complete graph is equal to n-1. I need to write a reference for that proof. Was it yours so I write your name as a reference or there was another reference as a book for example. Please let me know as soon as you can. ==== on the wayside, John, this is exactly the same sort of Diophontine problem as the FLT that James is bent on filling usenet with -- with his Simple Short Proof: the exclusion of classes of congruence. > of a simple, unsolved problem, the Perfect Box: > take a rectangular box (or parallelipiped) > with 3 edges, 3 facial digonals and 1 interior diagonal, and > find a solution in integers, or prove that there is none; > taht's 7 different lengths, although they're *dependent* > on only 3 of them, say the rectanular edges of the box. > (nothing more than the pythagorean theorem is required, > as far as algebra goes, but it's a nicety to *prove* it, > before using it. for the sake of clarity, > call the rectangular edges x, y, z, and the face diagonals a, b, c, > where A=Y+Z, B=Z+X, C=X+Y (X=xx=x^2 etc., and > D=dd is the skware of the interior diagonal) and > D = X+A = Y+B = Z+C = D.) --ils duces d'Enron! http://www.benfranklinbooks.com/20th-century.htm http://www.wlym.com/PDF-SpReps/SPRP24.pdf <2qtiqv4uegckp1j5ssg4ntcs2hsbr0goib@4ax.com> ==== >> Lonely, are you? Not at all. But don't fail to read Camaraderie of the Experts at > It explains why you Boyz go to such great lengths to fend off > assaults on one or another's 'expertise'. But you'll find nothing there about bum-fuzzling. Sorry. Yes, of course, all of the experts here are anxious over the stunning and powerful attacks of John Correy and James Harris, brilliant displays of logic and insight that threaten to bring down our charade. I don't know about the others, but I studied logic for the great respect it demands. Free valet parking, starlets swooning at the feet, things like that. Of course we must band together to defend our ill-gotten gains from your attempts to expose us. I mean, duh. (Of course, I'm not a mathematician, so I don't *really* have to defend against JSH. I only offer minor moral support on that front.) -- So, at this time, I'd like to assure you that I am not interested in making sure mathematicians worldwide get fired. I've rethought my desire to go to Congress and try to get funding for mathematicians cut. -- James Harris is a reasonable man. Whew! ==== >> Lonely, are you? Not at all. But don't fail to read Camaraderie of the Experts at > It explains why you Boyz go to such great lengths to fend off > assaults on one or another's 'expertise'. But you'll find nothing there about bum-fuzzling. Sorry. Yes, of course, all of the experts here are anxious over the > stunning and powerful attacks of John Correy and James Harris, > brilliant displays of logic and insight that threaten to bring down > our charade. I don't know about the others, but I studied logic for > the great respect it demands. Free valet parking, starlets swooning > at the feet, things like that. Of course we must band together to > defend our ill-gotten gains from your attempts to expose us. I mean, > duh. (Of course, I'm not a mathematician, so I don't *really* have to > defend against JSH. I only offer minor moral support on that front.) Whatever your level of incompetence may otherwise be, you're a mathie in spirit--and that's what counts! --John ======================================================================= You concentrated on mathematics because it is predictable, because there is always a right answer you can check in the back of the book, because you like following very precise rules, because it allows you to escape from everyday life into a world that has nothing to do with everyday life, and because mathematics does not require the creativity that you completely lack. Keith Devlin, Commencement address to the mathematics graduating class of UC Berkeley, May 23, 1997 A performance system is designed to work in a defined task domain, accepting particular goals and seeking to reach them by some kind of highly selective search. The system must be told what goal is to be reached and must be given a description of the structure and characteristics of the task domain in which it is to operate: its problem space. [...] In contrast, a learning system, is capable of acquiring a problem space, in whole or part, by interacting with the external environment and without being instructed about it directly. A performance system is designed to work in a defined task domain, accepting particular goals and seeking to reach them by some kind of highly selective search. The system must be told what goal is to be reached and must be given a description of the structure and >characteristics of the task domain in which it is to operate: its problem space. [...] In contrast, a learning system, is capable of acquiring a problem space, in whole or part, by interacting with the external environment and without being instructed about it directly. Herbert Simon ==== Consider the covering map p x p : R x R ---> S^1 x S^1 where p is the standard covering map p : R --> S^1 by p(x) = (cos (2pi x),sin (2pi x)) Consider the path f(t) = (cos (2pi t), sin(2pi t)) x (cos (4pi t), sin(4pi t)) in S^1 x S^1. Sketch what f looks like when S^1 x S^1 is identified as a doughnut. Find a lifting f ' of f to R x R and sketch it. I am completely lost here, can anyone help? Mike ==== > Consider the covering map p x p : R x R ---> S^1 x S^1 > where p is the standard covering map p : R --> S^1 by p(x) = (cos (2pi > x),sin (2pi x)) Consider the path f(t) = (cos (2pi t), sin(2pi t)) x (cos (4pi t), sin(4pi > t)) in S^1 x S^1. Sketch what f looks like when S^1 x S^1 is identified as a > doughnut. Find a lifting f ' of f to R x R and sketch it. You forgot to say what's the domain of f, but I'll assume that it's [0,1]. Since this is a text-only newsgroup, I cannot send you an image of the way that f looks on a doughnut, although it is easy to draw it with Mathematica or Maple (and it is easy to imagine too). A lifting f' is defined as f'(t) = (t,2t), since (p x p) o f' = f. Jose Carlos Santos ==== Here's a nugget for the did you ever notice bin: Did you ever notice, how Mr. Harris gives refutations based on general relations, whereas those who dispute his claims give exact, concrete counter-examples? MB > Notice, (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where you see that the constant terms match as now you have 7(7)(22) = > 1078, which is the constant term of the polynomial 49(300125 x^3 - 18375 x^2 - 360 x + 22). Various people have debated me about what happens when you divide off > 49, where for some odd reason, some of them seem to believe that you > can have w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 where the w's vary as x varies, which is a rather naive notion. That's because you can multiply *everything* out, and simplify to get (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 which should be simple enough for all of you. Now those of you who usually work in the field of complex numbers may > think that it's not a big deal, as you may think it doesn't matter if > w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but > you see, as 22 is coprime to 7 in the ring of algebraic integers, if > w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring. You know, it's like how in integers 1/2 doesn't exist. It's not an > integer, so it's not in the ring. So you see, my argument is correct and simple, and mathematicians are > indeed running from a little gut check in their field. They're > pussies too scared to handle the truth. But you should also understand, some people will be able to see that, > which is part of my plan. I can let mathematicians destroy themselves > proving they can't be trusted based on what they *see*, while they > forget what they can't see: the wearing down of the mathematician > mystique. > James Harris > http://mathforprofit.blogspot.com/ ==== > Here's a nugget for the did you ever notice bin: Did you ever notice, how Mr. Harris gives refutations based on general > relations, whereas those who dispute his claims give exact, concrete > counter-examples? There are plenty of examples of the converse as well, where somebody proves that something in general need not be true, but James comes up with an example where it is true and claims that proves the general result. - Randy ==== There are plenty of examples of the converse as well, > where somebody proves that something in general need not > be true, but James comes up with an example where it is > true and claims that proves the general result. > Well, actually this is a valid proof method (in modern Harrisanism). Let x be an arbitrary number. Now consider 0 = x. Obviously we have 0 = 0. With other words, x = 0. (Note that 0 is a constant! Hence x is a constant too!) Now since x has been arbitrary, this means x = 0 for any x! This actually proves FLT. qed. ... ==== > There are plenty of examples of the converse as well, > where somebody proves that something in general need not > be true, but James comes up with an example where it is > true and claims that proves the general result. > > Well, actually this is a valid proof method (in modern Harrisanism). Let x be an arbitrary number. Now consider 0 = x. Obviously we have > 0 = 0. With other words, x = 0. (Note that 0 is a constant! Hence x is a > constant too!) Now since x has been arbitrary, this means x = 0 for any > x! This actually proves FLT. qed. ... Nice proof, but how do you know 0 is not a variable? ==== > Notice, (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) Yeah, yeah. Weren't you supposed to leave this forum? I guess the troll needs to be fed again. ==== >Notice, (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) where you see that the constant terms match as now you have 7(7)(22) = >1078, which is the constant term of the polynomial 49(300125 x^3 - 18375 x^2 - 360 x + 22). Various people have debated me about what happens when you divide off >49, where for some odd reason, some of them seem to believe that you >can have w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 where the w's vary as x varies, which is a rather naive notion. Why is it naive? w1(x) = x+7 w2(x) = x^2+7 w3(x) = 22/((x+7)(x^2+6)) >That's because you can multiply *everything* out, and simplify to get (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 As can I (understand that I'm not claiming that these particular functions will work for this problem, I'm just pointing that that it isn't naive to assume that they don't have to be constant). >which should be simple enough for all of you. Now those of you who usually work in the field of complex numbers may >think that it's not a big deal, as you may think it doesn't matter if >w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but >you see, as 22 is coprime to 7 in the ring of algebraic integers, if >w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring. Doesn't matter. What matters is whether or not (5 b_3(x) + 22)/w_3(x) exists in the ring. Let's try a simpler example: f(x) = x(x+1) over the integers. Clearly f(x) is even. So, can we assume that either x/2 is an integer or (x+1)/2 is an integer? The answer is yes, but you can't assume it is always the same one. IOW, there exist a1 and a2 such that x/a1(x) and (x+1)/a2(x) are both integers and a1(x)a2(x)=2. But you can't assume that just because a1(0) = 2 and a2(0) = 1 that a1(x)=2 and a2(x)=1. And note that 1/2 doesn't exist in the ring, but that doesn't mean that a2(x) can't equal 2 every now and then. Alan -- Defendit numerus ==== > Notice, (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) where you see that the constant terms match as now you have 7(7)(22) = > 1078, which is the constant term of the polynomial 49(300125 x^3 - 18375 x^2 - 360 x + 22). Various people have debated me about what happens when you divide off > 49, where for some odd reason, some of them seem to believe that you > can have w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 where the w's vary as x varies, which is a rather naive notion. That's because you can multiply *everything* out, and simplify to get (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 which should be simple enough for all of you. Now those of you who usually work in the field of complex numbers may > think that it's not a big deal, as you may think it doesn't matter if > w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but > you see, as 22 is coprime to 7 in the ring of algebraic integers, if > w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring. You know, it's like how in integers 1/2 doesn't exist. It's not an > integer, so it's not in the ring. > Yes, you are right about this. 22/w_3(x) is not an algebraic integer. It's obvious, because w_3(x) is a divisor of 7. But it's irrelevant. What's relevant is that (5*b_3(x) + 22)/w_3(x) is an algebraic integer. This is easier to see if you re-write b_3(x) as it was originally, b_3(x) = a_3(x) - 3. Thus we have (5*b_3(x) + 22) = (5*a_3(x) - 15 + 22) = (5*a_3(x) + 7) Thus when we divide by w_3(x), we have 5*a_3(x)/w_3(x) + 7/w_3(x), and both a_3(x)/w_3(x) and 7/w_3(x) *are* algebraic integers. Well, you are not going to buy this. You are going to continue to insist (correctly) that 22/w_3(x) cannot be an algebraic integer and rave about how important that is and how we are trying to confuse people, etc, etc.. You are totally hung up on the idea that constant terms are inviolate and sacred in some way. They are not. No, I am NOT saying that constant terms are not constant. Of course they are. I am saying that the important thing here is not divisibility of the constant term 22 by a factor of 7. The important thing is divisibility of the *entire expression* 5_b3(x) + 22 by a factor of 7. And that is exactly what occurs. Different topic. If your method of proof is actually valid, it will apply similarly to other polynomials than just your beloved favorite, P(x). Thus, consider Q(x) = 7^2*(125*x^3 - 15*x + 22). This has the essential properties of the JSH polynomial: namely, the constant term Q(0) = 7*7*22 = 1078, and it can be factored in the form (5*b_1 + 7)*(5*b_2 + 7)*(5*c_3 + 22) where b_1, b_2, and c_3 are functions of x and are algebraic integers. Note that when x = 0, b_1 = 0, b_2 = 0, and c_3 = 0. Thus when x = 0, b_1 and b_2 are multiples of 7 (i.e., 7*0 in each case). Thus also when x = 0, the product of the constant terms equals the constant term of Q(x): Q(0) = (0 + 7)*(0 + 7)*(0 + 22) = 7*7*22 = 1078. Now according to the JSH argument, b_1 and b_2 should be divisible by 7 when x <> 0 also. So let's see what happens when x = 1. First, note that Q(1) = 7^2*(125 - 15 + 22) = 7^2*132. This can be factored as Q(1) = r * s * t, where r = 5*7^{2/3}*w + 7 s = 5*7^{2/3}*w^2 + 7, and t = 5*(7^{2/3} - 3) + 22, where w is a unit in the algebraic integers, w = (-1 + sqrt(-3))/2. Therefore Q(1) is of the required form, Q(1) = (5*b_1 + 7)*(5*b_2 + 7)*(5*c_3 + 22), where b_1 = 7^{2/3}*w, b_2 = 7*{2/3}*w^2, and c_3 = 7^{2/3} - 3, all of which are algebraic numbers. Note that b_1 and b_2 are *not* divisible by 7. Neither are they *coprime* to 7; each has the factor 7^{2/3} in common with 7. Dividing 7^{2/3} out of each factor yields: Q(1)/49 = (5*w + 7^{1/3})*(5*w^2 + 7^{1/3})*(5*(1 - 3/7^{2/3}) + 22/7^{2/3}) = (5*w + 7^{1/3})*(5*w^2 + 7^{1/3})*(5 + 7^{1/3}). It is a matter of arithmetic to check that the right-hand side equals 132 = Q(1)/49, as it should. It is worth noting also that the product of the three constant terms in the expression just above is 7^{1/3} * 7^{1/3} * (22/7^{2/3}) = 22. Note again, this is the factorization when x = 1. The coefficients b_1, b_2, and c_3 are all functions of x; the values given above are actually b_1(1), b_2(1), and c_3(1). When x = 0, b_1(0) = 0, b_2(0) = 0, and c_3(0) = 0. For values of x other than 1 or 0, these functions are difficult to compute. Note finally that the above factorization of Q(x) is *not* of the form claimed by JSH, i.e., it is *not* of the form Q(1) = (5*a_1 + 7)*(5*a_2 + 7)*(5*b_3 + 22), where a_1 and a_2 are algebraic integers which are divisible by 7. Yet all the arithmetic checks out; the constant terms as required are 7, 7, and 22. So where does the JSH logic break down for this example, yet succeed for his P(x)? > So you see, my argument is correct and simple, and mathematicians are > indeed running from a little gut check in their field. They're > pussies too scared to handle the truth. > Most of your loyal audience has probably noticed by now that you have stopped trying to respond to the substantive parts of my posts. You pounce on some superficial aspect of them and delete the rest. Makes you wonder, doesn't it? Who here is acting like he is too scared to handle the truth ? Nora B. > But you should also understand, some people will be able to see that, > which is part of my plan. I can let mathematicians destroy themselves > proving they can't be trusted based on what they *see*, while they > forget what they can't see: the wearing down of the mathematician > mystique. > James Harris > http://mathforprofit.blogspot.com/ ==== > Notice, > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where you see that the constant terms match as now you have 7(7)(22) = > 1078, which is the constant term of the polynomial > 49(300125 x^3 - 18375 x^2 - 360 x + 22). > Various people have debated me about what happens when you divide off > 49, where for some odd reason, some of them seem to believe that you > can have > w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and > (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > where the w's vary as x varies, which is a rather naive notion. > That's because you can multiply *everything* out, and simplify to get > (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 > which should be simple enough for all of you. > Now those of you who usually work in the field of complex numbers may > think that it's not a big deal, as you may think it doesn't matter if > w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but > you see, as 22 is coprime to 7 in the ring of algebraic integers, if > w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring. > You know, it's like how in integers 1/2 doesn't exist. It's not an > integer, so it's not in the ring. > > Yes, you are right about this. 22/w_3(x) is not an algebraic > integer. It's obvious, because w_3(x) is a divisor of 7. But it's irrelevant. What's relevant is that (5*b_3(x) + 22)/w_3(x) is an algebraic integer. This is easier to see if you re-write That ignores the fact that (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 wouldn't be in the ring of algebraic integers, if w_3(x) shared any non-unit factors with 7. However, that result follows from simplifying (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 so you have to at least admit that your claims push you out of the ring of algebraic integers. Can you admit that Nora Baron? James Harris ==== ... > you see, as 22 is coprime to 7 in the ring of algebraic integers, if > w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring. ... > Yes, you are right about this. 22/w_3(x) is not an algebraic > integer. It's obvious, because w_3(x) is a divisor of 7. > > But it's irrelevant. What's relevant is that > > (5*b_3(x) + 22)/w_3(x) > > is an algebraic integer. This is easier to see if you re-write > > That ignores the fact that > > (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 > > wouldn't be in the ring of algebraic integers, if w_3(x) shared any > non-unit factors with 7. Why not? The left hand expression is equal to 49.22/(w1(x).w2(x).w3(x) where I have provided you with w1(x), w2(x) and w3(x) that multiply together to get 49. w3(x) is in general *not* coprime to 7, and w1(x) and w2(x) are in general *not* divisible by 7. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== [cut] > > That ignores the fact that > > (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 > > wouldn't be in the ring of algebraic integers, if w_3(x) shared any > > non-unit factors with 7. Why not? The left hand expression is equal to > 49.22/(w1(x).w2(x).w3(x) > where I have provided you with w1(x), w2(x) and w3(x) that multiply > together to get 49. w3(x) is in general *not* coprime to 7, and > w1(x) and w2(x) are in general *not* divisible by 7. I think Harris meant to say that 22/w_3(x) wouldn't be in the ring of algebraic integers. Therefore, one is not allowed to use it on the left hand side of that equality. He is not saying that 22 wouldn't be in the ring of algebraic integers. He is objecting to using 22/w_3(x), which is not an algebraic integer. Others have explained why this is not a valid objection. -- Bill Hale ==== > Notice, > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where you see that the constant terms match as now you have 7(7)(22) = > 1078, which is the constant term of the polynomial > > 49(300125 x^3 - 18375 x^2 - 360 x + 22). > > Various people have debated me about what happens when you divide off > 49, where for some odd reason, some of them seem to believe that you > can have > > w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and > > (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > where the w's vary as x varies, which is a rather naive notion. > > That's because you can multiply *everything* out, and simplify to get > (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 > which should be simple enough for all of you. > > Now those of you who usually work in the field of complex numbers may > think that it's not a big deal, as you may think it doesn't matter if > w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but > you see, as 22 is coprime to 7 in the ring of algebraic integers, if > w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring. But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x) that must exist in the ring (and which does exist in the ring. I have no idea where you got the idea that 22/w3(x) must be in the ring. Consider P(x) = (x + 1)(x + 2), in the integers. It is always even, so it can be divided by 2. Your reasoning about constant term would lead to the result P(x)/2 = (x + 1)(x/2 + 1), because now the constant terms match. My position is that there are functions w1(x) and w2(x), defined as follows: w1(x) = gcd(x + 1, 2) w2(x) = gcd(x + 2, 2) such that P(x)/2 = [(x + 1)/w1(x)] * [(x + 2)/w2(x)] is a valid factorisation. According to your reasoning above (paraphrase): ...as you may think it doesn't matter if w1(x) has some factor of 2, despite *seeing* (1/w1(x)) but you see, as 2 is coprime to 1 in the ring of integers, if w1(x) isn't coprime to 2, (1/w1(x)) does not exist in the ring. So, although both factors in the factorisation are integer, according to you it is not a valid factorisation, so the ring of integers is flawed. > So you see, my argument is correct and simple, and mathematicians are > indeed running from a little gut check in their field. They're > pussies too scared to handle the truth. So, it is your thinking that the ring of integers is flawed? With your polynomial the situation is similar. I have given pretty *explicit* definitions of the functions w1(x) to w3(x) such that (5 a1(x) + 7)/w1(x) , (5 a2(x) + 7)/w2(x) and (5 b3(x) + 22)/w3(x) are all algebraic integer for all integer x. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > > Notice, > > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > > where you see that the constant terms match as now you have 7(7)(22) = > > 1078, which is the constant term of the polynomial > > 49(300125 x^3 - 18375 x^2 - 360 x + 22). > > Various people have debated me about what happens when you divide off > > 49, where for some odd reason, some of them seem to believe that you > > can have > > w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and > > (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = > > > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > where the w's vary as x varies, which is a rather naive notion. > > That's because you can multiply *everything* out, and simplify to get > > (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 > > which should be simple enough for all of you. > > Now those of you who usually work in the field of complex numbers may > > think that it's not a big deal, as you may think it doesn't matter if > > w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but > > you see, as 22 is coprime to 7 in the ring of algebraic integers, if > > w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring. But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x) > that must exist in the ring (and which does exist in the ring. I have > no idea where you got the idea that 22/w3(x) must be in the ring. Consider P(x) = (x + 1)(x + 2), in the integers. It is always even, so > it can be divided by 2. Why? What I've done is do a basic simplification going from (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 to (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22, which is done simply enough by multiplying everything out and letting the terms with x as a factor cancel each other out. Now you wish to avoid that Dik Winter because your assertions can't be true in the ring of algebraic integers as if w_3(x) isn't coprime to 7, there's a contradiction, as 22/w_3(x) can't be in the ring then. You're a crank Dik Winter, who has been refuted quite simply but you keep talking as if convincing *others* changes mathematical truth. It does not. James Harris ==== who are these *others* that you refer to?... perhaps, you have an audience that is not known to the rest of us, the desingated Peanut Gallery of would-be critics & helpmeets. or is it just the entire, 'virtual crowd of all of the folks who are in the googolplex, that *could* lurk on your bifurcating threads -- or maybe they should? mea culpa, dood. > keep talking as if convincing *others* changes mathematical truth. --ils dcues d'Enron! http://larouchepub.com/ ==== > who are these *others* that you refer to?... perhaps, > you have an audience that is not known to the rest of us, > the desingated Peanut Gallery of would-be critics & helpmeets. > or is it just the entire, 'virtual crowd of all > of the folks who are in the googolplex, > that *could* lurk on your bifurcating threads -- or > maybe they should? mea culpa, dood. > I have this sense that you, BQH, are more artist than, say, mathematician - and sometimes incredibly funny, and highly literate. I don't always read what you post, and when I do I don't always understand it, but there is *something*. So I hope you won't completely give up this addiction - Nora B. Les ducks d'endrun! Les quacks d'bon-bon! Ils douches de Maman! keep talking as if convincing *others* changes mathematical truth. --ils dcues d'Enron! > http://larouchepub.com/ ==== > But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x) > that must exist in the ring (and which does exist in the ring. I have > no idea where you got the idea that 22/w3(x) must be in the ring. > Now you wish to avoid that Dik Winter because your assertions can't be > true in the ring of algebraic integers as if w_3(x) isn't coprime to > 7, there's a contradiction, as 22/w_3(x) can't be in the ring then. You keep saying that 22/w_3(x) need not be in the ring. No-one is disputing this. The problem is that you seem to think that the fact that 22/w_3(x) need not be in the ring is of great importance. Indeed, you seem to assume that it is obvious why 22/w_3(x) must be in the ring. It appears that you are reasoning thus: The constant term of (b_3(x) + 22) is 22 The constant term of (b_3(x)/w_3(x) + 22/w_3(x)) cannot be b_3(x)/w_3(x) so it must be 22/w_3(x) The constant term must be in the ring, so 22/w_3(x) must be in the ring. The problem is that the constant term of (b_3(x)/w_3(x) + 22/w_3(x)) is 22/w_3(0)= 22/1 = 22. Whether or not 22/w_3(x) is in the ring for values of x other than 0 does not matter. - William Hughes ==== > But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x) > that must exist in the ring (and which does exist in the ring. I have > no idea where you got the idea that 22/w3(x) must be in the ring. Now you wish to avoid that Dik Winter because your assertions can't be > true in the ring of algebraic integers as if w_3(x) isn't coprime to > 7, there's a contradiction, as 22/w_3(x) can't be in the ring then. You keep saying that 22/w_3(x) need not be in the ring. > No-one is disputing this. The problem is that you seem to think > that the fact that 22/w_3(x) need not be in the ring is of great > importance. Indeed, you seem to assume that it is obvious why > 22/w_3(x) must be in the ring. It appears that you are reasoning thus: The constant term of (b_3(x) + 22) is 22 The constant term of (b_3(x)/w_3(x) + 22/w_3(x)) > cannot be b_3(x)/w_3(x) so it must be 22/w_3(x) The constant term must be in the ring, so 22/w_3(x) > must be in the ring. You're lying William Hughes, as my exact reasoning was posted. I just multiply out the factorization, and simplify. Now readers should note that posters like William Hughes are cranks, so of course they have to ignore the actual facts, but consider what I posted before: (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 where the w's vary as x varies, which is a rather naive notion. That's because you can multiply *everything* out, and simplify to get (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 which should be simple enough for all of you. > The problem is that the constant term of (b_3(x)/w_3(x) + 22/w_3(x)) > is 22/w_3(0)= 22/1 = 22. Whether or not 22/w_3(x) is in the > ring for values of x other than 0 does not matter. - William Hughes Denial of basic algebra is such a sad thing to display to the world as William Hughes demonstrates a rather odd irrationality, to all those readers who go through my posts. After all, I'm just talking about multiplying out and simplifying, and I'll post again because these cranks have a bad habit of creative deletion what I said previously: (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 where the w's vary as x varies, which is a rather naive notion. That's because you can multiply *everything* out, and simplify to get (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 which should be simple enough for all of you. You see, the cranks can't get arouund 22/w_3(x) NOT being in the ring of algebraic integers if w_3(x) isn't coprime to 22. And for those of you who wondered, yes, when faced with a *very* basic refutation of their positions posters who argue with me tend to just ignore the truth. Later they come back with the same position. They're irrational and are the real cranks as you can't reason with them. They believe false things but want desperately to believe and be believed. James Harris http://mathforprofit.blogspot.com/ ==== ... > That's because you can multiply *everything* out, and simplify to get > > (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 > > which should be simple enough for all of you. > > You see, the cranks can't get arouund 22/w_3(x) NOT being in the ring > of algebraic integers if w_3(x) isn't coprime to 22. Well, you know, the w's as I have defined them for you get w1(x)w2(x)w3(x) = 49, so I do not understand how it is possible that (7/w1(x))(7/w2(x))(22/w3(x)) = 22. (Note that in most cases this involves a short excursion to the algebraic numbers...) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > ... > > That's because you can multiply *everything* out, and simplify to get > > > > (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 > > > > which should be simple enough for all of you. > > > > You see, the cranks can't get arouund 22/w_3(x) NOT being in the ring > > of algebraic integers if w_3(x) isn't coprime to 22. > > Well, you know, the w's as I have defined them for you get > w1(x)w2(x)w3(x) = 49, > so I do not understand how it is possible that Of course I intended impossible here. > (7/w1(x))(7/w2(x))(22/w3(x)) = 22. > (Note that in most cases this involves a short excursion to the > algebraic numbers...) > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== [.snip.] >Well, you know, the w's as I have defined them for you get > w1(x)w2(x)w3(x) = 49, >so I do not understand how it is possible that > (7/w1(x))(7/w2(x))(22/w3(x)) = 22. You mean, how is it possible that [...] does not hold? ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x) > that must exist in the ring (and which does exist in the ring. I have > no idea where you got the idea that 22/w3(x) must be in the ring. > Now you wish to avoid that Dik Winter because your assertions can't be > true in the ring of algebraic integers as if w_3(x) isn't coprime to > 7, there's a contradiction, as 22/w_3(x) can't be in the ring then. > You keep saying that 22/w_3(x) need not be in the ring. > No-one is disputing this. The problem is that you seem to think > that the fact that 22/w_3(x) need not be in the ring is of great > importance. Indeed, you seem to assume that it is obvious why > 22/w_3(x) must be in the ring. It appears that you are reasoning thus: > The constant term of (b_3(x) + 22) is 22 > The constant term of (b_3(x)/w_3(x) + 22/w_3(x)) > cannot be b_3(x)/w_3(x) so it must be 22/w_3(x) > The constant term must be in the ring, so 22/w_3(x) > must be in the ring. You're lying William Hughes, as my exact reasoning was posted. I just > multiply out the factorization, and simplify. > Your reasoning that 22/w_3(x) need not be in the ring has been posted many, many times. No-one disagrees with your conclusion. Your reasoning that 22/w_3(x) must be in the ring has never been posted. The above is my guess as to what your reasoning is. Quiz time: What is the constant term of the following three functions? U(x) = sqrt(x^2 + x)/sqrt(x+7) + 7/sqrt(x+7) V(x) = sqrt(x^2 + x)/7 + 7/7 W(x) = sqrt(x^2 +x)/(x^2 +2x +7) + 7/(x^2 +2x +7) (hint: The constant term of b(x) is b(0) ) - William Hughes ==== > But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x) > that must exist in the ring (and which does exist in the ring. I have > no idea where you got the idea that 22/w3(x) must be in the ring. > > Now you wish to avoid that Dik Winter because your assertions can't be > true in the ring of algebraic integers as if w_3(x) isn't coprime to > 7, there's a contradiction, as 22/w_3(x) can't be in the ring then. > You keep saying that 22/w_3(x) need not be in the ring. > No-one is disputing this. The problem is that you seem to think > that the fact that 22/w_3(x) need not be in the ring is of great > importance. Indeed, you seem to assume that it is obvious why > 22/w_3(x) must be in the ring. It appears that you are reasoning thus: > The constant term of (b_3(x) + 22) is 22 > The constant term of (b_3(x)/w_3(x) + 22/w_3(x)) > cannot be b_3(x)/w_3(x) so it must be 22/w_3(x) > The constant term must be in the ring, so 22/w_3(x) > must be in the ring. > You're lying William Hughes, as my exact reasoning was posted. I just > multiply out the factorization, and simplify. > > Your reasoning that 22/w_3(x) need not be in the ring has been posted > many, many times. No-one disagrees with your conclusion. Actually, it IS in the ring, which is my point William Hughes. And you deleted out the factorization and simplification which shows that fact!!! Extraordinary behavior which is indeed crank. I give you the information, explain it clearly, and you try to just ignore it. > Your reasoning that 22/w_3(x) must be in the ring has never been posted. > The above is my guess as to what your reasoning is. The assertions are over the ring of algebraic integers, so operations are to be in that ring. I show that you're pushed out of that ring in a surprising way. > Quiz time: What is the constant term of the following three functions? Irrelevant to the issue of 22/w_3(x) having to be in the ring of algebraic integers. > U(x) = sqrt(x^2 + x)/sqrt(x+7) + 7/sqrt(x+7) V(x) = sqrt(x^2 + x)/7 + 7/7 W(x) = sqrt(x^2 +x)/(x^2 +2x +7) + 7/(x^2 +2x +7) (hint: The constant term of b(x) is b(0) ) - William Hughes What I did was multiply out the factorization and simplify to show that it's impossible for w_3(x) to not be coprime to 7 as then 22/w_3(x) is NOT in the ring of algebraic integers. It's simple, direct, and basic, and it refutes attempts at claiming that w_3(x) can share non-unit factors with 7 in the ring of algebraic integers. James Harris ==== > But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x) > that must exist in the ring (and which does exist in the ring. I have > no idea where you got the idea that 22/w3(x) must be in the ring. > > Now you wish to avoid that Dik Winter because your assertions can't be > true in the ring of algebraic integers as if w_3(x) isn't coprime to > 7, there's a contradiction, as 22/w_3(x) can't be in the ring then. > You keep saying that 22/w_3(x) need not be in the ring. > No-one is disputing this. The problem is that you seem to think > that the fact that 22/w_3(x) need not be in the ring is of great > importance. Indeed, you seem to assume that it is obvious why > 22/w_3(x) must be in the ring. It appears that you are reasoning thus: > The constant term of (b_3(x) + 22) is 22 > The constant term of (b_3(x)/w_3(x) + 22/w_3(x)) > cannot be b_3(x)/w_3(x) so it must be 22/w_3(x) > The constant term must be in the ring, so 22/w_3(x) > must be in the ring. > You're lying William Hughes, as my exact reasoning was posted. I just > multiply out the factorization, and simplify. > Your reasoning that 22/w_3(x) need not be in the ring has been posted > many, many times. No-one disagrees with your conclusion. Actually, it IS in the ring, which is my point William Hughes. And > you deleted out the factorization and simplification which shows that > fact!!! > You seem to be claiming that (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 where the w's vary as x varies, which is a rather naive notion. That's because you can multiply *everything* out, and simplify to get (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 which should be simple enough for all of you. shows that (22/2_3(x)) is in the ring of algebraic integers. Actually, it doesn't show anything. Either by using the identity (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 and the fact that (w_1(x))(w_2(x))(w_3(x)) = 49 or by direct use of the fact that (w_1(x))(w_2(x))(w_3(x)) = 49 we obtain (7)(7)(22) / ((w_1(x))(w_2(x))(w_3(x))) = 22 in the ring of algebraic integers. We can rewrite this in the complex numbers as (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 but this no more shows that (22/w_3(x)) is in the ring of algebraic integers than (6)(35)(99)/((15)(18)(7)) =(6/15)(35/18)(99/7) = 11 shows that (99/7) is an integer. You didn't attempt my quiz. Please try, It isn't hard. What is the constant term of the following three functions? U(x) = sqrt(x^2 + x)/sqrt(x+7) + 7/sqrt(x+7) V(x) = sqrt(x^2 + x)/7 + 7/7 W(x) = sqrt(x^2 +x)/(x^2 +2x +7) + 7/(x^2 +2x +7) (hint: The constant term of b(x) is b(0) ) - William Hughes ==== ... > The assertions are over the ring of algebraic integers, so operations > are to be in that ring. I show that you're pushed out of that ring in > a surprising way. But so are you! Because a1(x) is *not* divisible in the algebraic integers. You say that is a flaw of he algebraic integers. The question is, which pushing out is the better push. With: h1(x) = gcd(5 a1(x) + 7, 49) h2(x) = gcd(5 a2(x) + 7, 49) h3(x) = gcd(5 b3(x) + 22, 49) k(x) = 49/( h1(x).h2(x).h3(x) ) l1(x) = gcd(k(x), h1(x)) l2(x) = gcd(k(x)/l2(x), h2(x)) l3(x) = k(x)/( l1(x).l2(x) ) { Perhaps l3(x) = 1 for all x, just be save.} w1(x) = h1(x)/l1(x) w2(x) = h2(x)/l2(x) w3(x) = h3(x)/l3(x) where all functions defined above are functions from the algebraic integers to the algebraic integers. P(x)/49 = [(5 a1(x) + 7)/w1(x)] [(5 a2(x) + 7)/w2(x)] [(5 b3(x) + 22)/w3(x)] all three factors are algebraic integers for *all* x. But when we set: w1(x) = w2(x) = 7 and w3(x) = 1 in the above factorisation, we find that *not* all three factors are algebraic integers. That 22/w3(x) is *not* necessarily an algebraic integer is irrelevant. You do not need to calculate that value. But *even if you wish to calculate that intermediate result*, it is irrelevant. In general the case is that when you say about a function P(x) that it is a function from the ring of algebraic integers to the algebraic integers, it is not necessary that all intermediate results must also be in that ring, as long as the final result is in that ring. And as functions, all three factors with the w's as I defined them *are* functions from the algebraic integers to the algebraic integers. So P(x) is factored in three functions that are from the algebraic integers to the algebraic integers. No flaw of the algebraic integers in view. ... > What I did was multiply out the factorization and simplify to show > that it's impossible for w_3(x) to not be coprime to 7 as then > 22/w_3(x) is NOT in the ring of algebraic integers. Yes, but there is *no* 22/w3(x) in the factorisation above. It is (5 b3(x) + 22)/w3(x). Or do you claim that it is invalid in the integers to say that (x + 5)/2 is an integer for odd integer x, because (x + 5)/2 = (x/2 + 5/2) and 5/2 is not an integer? A pretty strange restriction I would say. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== [.snip.] > Or do you claim that it is invalid in >the integers to say that (x + 5)/2 is an integer for odd integer x, >because (x + 5)/2 = (x/2 + 5/2) and 5/2 is not an integer? A pretty strange restriction I would say. Perhaps a better example would be to take (x^2+3x+2)/2, which is always an integer for integer value of x; you can certainly write (x^2+3x+2)/2 = (x^2/2) + (3x/2) + 1 even though it is not always true that x^2/2 and 3x/2 are integers. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > > Notice, > > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > > where you see that the constant terms match as now you have 7(7)(22) = > > 1078, which is the constant term of the polynomial > > > > 49(300125 x^3 - 18375 x^2 - 360 x + 22). > > > > Various people have debated me about what happens when you divide off > > 49, where for some odd reason, some of them seem to believe that you > > can have > > > > w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and > > > > (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = > > > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > > > where the w's vary as x varies, which is a rather naive notion. > > > > That's because you can multiply *everything* out, and simplify to get > > (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 > > which should be simple enough for all of you. > > > > Now those of you who usually work in the field of complex numbers may > > think that it's not a big deal, as you may think it doesn't matter if > > w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but > > you see, as 22 is coprime to 7 in the ring of algebraic integers, if > > w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring. > > But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x) > that must exist in the ring (and which does exist in the ring. I have > no idea where you got the idea that 22/w3(x) must be in the ring. > > Consider P(x) = (x + 1)(x + 2), in the integers. It is always even, so > it can be divided by 2. > > Why? Why? Because your reasoning would lead to the conclusion that the ring of integers is flawed because it does not contain numbers that should be in that ring. But you are afraid to answer the questions I posed in the part you deleted. > Why? What I've done is do a basic simplification going from > (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > to (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22, > > which is done simply enough by multiplying everything out and letting > the terms with x as a factor cancel each other out. > > Now you wish to avoid that Dik Winter because your assertions can't be > true in the ring of algebraic integers as if w_3(x) isn't coprime to > 7, there's a contradiction, as 22/w_3(x) can't be in the ring then. is in the algebraic integers. If you think so you should answer the questions in the part you deleted. Because if 22/w3(x) must be in the algebraic integers, for *exactly the same reasons* 1/2 must be in the integers (see P(x) = (x + 1)(x + 2), which is divisible by 2 in the integers). > You're a crank Dik Winter, who has been refuted quite simply but you > keep talking as if convincing *others* changes mathematical truth. You dodge all my questions, because you are afraid to answer them. Now, who is the crank? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== Notice, [snip] http://www.crank.net/harris.html It's not every braying jackass that gets a whole page at crank.net Stupid is as stupid does. There is no fixing stupid because it is not broken. Harris would be much happier as a priest proclaiming god Hey Harris, your village called: Its idiot is missing. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! ==== >There is no fixing stupid because it is not >broken. What an intruguing comment! I'm not sure I get it, but it had me giggling while I was tryin to figure it out. -- Let us learn to dream, gentlemen, then perhaps we shall find the truth... But let us beware of publishing our dreams before they have been put to the proof by the waking understanding. -- Friedrich August Kekul.8e ==== > In the poission distribution like this: the monthly average number of > airplane crashes is 2.2, can I imply that the average number of airplane > crashes is 4.4 in 2 months? Yes If I use that assumption, I will lead to a different solution to the > problem: the probability of 5 crashes in 2 months, from using Poisson > process for 1 month and binomial distribution. > It shouldn't. Letting lambda = 4.4, the answer should be: P(X=5) = exp(-lambda) lambda^5 / 5! It's not obvious to me from the statement of the problem how you are to try to use the bin distbtn in solving this problem. The only way I can see it is to say: Let X1 = number crashes in month 1 X2 = number crashes month 2 X = X1 + X2 Then: P(X=5) = 2 [ P(X1=5)P(X2=0) + P(X1=4)P(X2=1)+P(X1=3)P(X2=2) ] If you work this out, P(X=5) = 2 exp(-lambda) (lambda/2)^5 [ 1/5!+ 1/4! + 1/(2!3!)] = 2 exp(-lambda) (lambda/2)^5 [ 1/5!+ 5/5! + 10/5!] =exp(-lambda) (lambda/2)^5 [ 32/5!]=exp(-lambda) lambda^5 / 5! as above. Hope this helps, MB ==== > The traditional continued fraction series can be related to a > regular tiling of the hyperbolic plane with triangles which have > all three angles equal to zero. These triangles can be paired to > produce squares, or a triangle can be grouped with its 3 neighbors to > produce a regular hexagon. The groups need to be placed symmetrically > as if the joining edge is a mirror to make a square or a hexagonal > tiling pattern. The Gaussian version is related to a 3-dimensional hyperbolic tiling > with octahedra which have 90 degree angles between faces and vertex > angles equal to zero. The Eisenstein CF expansion is related to a tiling with tetrahedra > which have 60 degree angles between faces and vertex angles equal > to zero. By surrounding one of these tetrahedra with four others, > a cube is formed, thus making a cube tiling pattern. Where can I find more information about this? Mike ==== > I know that polar concentric rings on a sphere/spheroid are known as > lines of latitude or parallels, but what are equatorially concentric > rings called? These are the small circles, having latitude of a parallel circle, form of Surface theory,using Liouville's Theorem for this spherical case one gets Tan(gama) = geodesic curvature/normal curvature = d [r*Sin (si)]/dz for all small circles, no matter whether they lie parallel to the plane of equator, perpendicular to it or make some other arbitrary intermediate angle. When gama is zero as an important special case, we have the Clairaut's Law valid on great geodesic circles. Note that normal curvature = 1/R , constant for any direction. Small circles or lines of such equal latitude are also called equal slip lines in filament winding industry of composite pressure vessels. ==== > as the geodesics on a sphere are the great circles, > two geodesics always meet in two points, taht is to say, > there are no parallel geodesics. > on the other hand, > What did you say? I'm not saying geodesics travel along these rings (anymore than they travel along latitudinal rings). Hmmm....maybe a picture will put it in perspective: http://math2.org/cgi-bin/mmb/server.pl?action=image&msgid=62326&fname=ringnam e.gif > I know that polar concentric rings on a sphere/spheroid are known as > lines of latitude or parallels, but what are equatorially concentric > rings called?--e.g., the 90¡ equatorially concentric ring would be a > line of longitude or meridian, but what about the other, semi-circle > rings **parallel TO A MERIDIAN**? > In terms of arcradius/radius of curvature, the perpendicular > meridian value is known as the normal: Is it that a meridian is > the prime normal, equals the 90¡ normal; the parallel > semi-circle/ellipse 1¡ away is the 89¡ normal, 2¡ away is the 88¡ > normal, 3¡ away is the 87¡ normal, etc., in the same way that the > equator is the 0¡ latitude, 1¡ away is the 1¡ latitude, etc.? > Or, as an annulus is a band bounded by two concentric rings, could > all of the rings comprising the annulus be something like > annulobes? ~Kaimbridge~ ----- Wanted÷Kaimbridge (w/mugshot!): http://www.angelfire.com/ma2/digitology/Wanted_KMGC.html ---------- Digitology÷The Grand Theory Of The Universe: http://www.angelfire.com/ma2/digitology/index.html ***** Void Where Permitted; Limit 0 Per Customer. ***** ==== it's impossible to see the labels on your picture. I'm not saying geodesics travel along these rings (anymore than they > travel along latitudinal rings). > Hmmm....maybe a picture will put it in perspective: http://math2.org/cgi-bin/mmb/server.pl?action=image&msgid=62326&fname=ringnam e.gif --ils duces d'Enron! http://www.benfranklinbooks.com/20th-century.htm http://www.wlym.com/PDF-SpReps/SPRP24.pdf ==== > it's impossible to see the labels on your picture. > I'm not saying geodesics travel along these rings (anymore than they > travel along latitudinal rings). > Hmmm....maybe a picture will put it in perspective: > http://math2.org/cgi-bin/mmb/server.pl?action=image&msgid=62326&fname=ringnam e.gif They seem readable at this end--are your browser settings off? ~Kaimbridge~ ----- Wanted÷Kaimbridge (w/mugshot!): http://www.angelfire.com/ma2/digitology/Wanted_KMGC.html ---------- Digitology÷The Grand Theory Of The Universe: http://www.angelfire.com/ma2/digitology/index.html ***** Void Where Permitted; Limit 0 Per Customer. ***** ==== >> it's impossible to see the labels on your picture. > >> I'm not saying geodesics travel along these rings (anymore than they >> travel along latitudinal rings). >> Hmmm....maybe a picture will put it in perspective: > http://math2.org/cgi-bin/mmb/server.pl?action=image&msgid=62326&fname=ringnam e.gif They seem readable at this end--are your browser settings off? On the outside chance the OP might not have noticed, recent Mozilla and IE browsers will shrink large images to fit the screen, by default. While inside the image, Mozilla will change the cursor to a magnifying glass with a + sign. Click to enlarge to readable. While large, cursor will contain a - sign to shrink. IE will make the image small, then you have to hover your cursor inside the lower right corner until the pillow appears. Click it to enlarge, reverse to shrink. HTH. ==== I am looking for an english reference for the Campbell theorem for >> stationary filtered point processes (not only Poisson). What Campbell theorem do you have in mind? In any event, a likely place to look is An Introduction to the Theory > of Point Processes by Daley & Vere-Jones. I have only a reference in >> Koenig/Schmidt: Einfuehrung in Punktprozesse, >> to have a reference which is easily accessible everywhere. > I am looking for the version: Let tau_n be a regular stationary point process with intensity lambda, marks k_n (having identical distributions), let a filtered point process defined by: y(t)=sum_{-oo}^{+oo} h(t,tau_n,k_n), where h is a well behaving function. Then: E[y(t)]=lambdaint_{-oo}^{+oo}E[h(t,0,k_0)]dt Any idea about this? -- Karl Breitung ==== Are there any good ways to find the minimum and maximum genera (genuses) of a graph? Let me define what I am looking for. Say that we can _cellularly embed_ a graph in a compact orientable surface S if you can embed G in S (treating G as a 1-manifold), and each connected domain of S-G (S minus G) is an open 2-cell (ie homotopic to a point). Intuitively, you are just drawing G on the surface of S while disallowing edges (of G) to cross, and making sure that each face of G does not contain any holes. Then gamma is a possible genera of G iff there is some S with genus gamma so that G can be cellularly embedded in S. For example, I think that K_4 has min genus 0 and max genus 1. I think that K_5 has min genus 1 and max genus either 2 or 3 (not sure). (Here by K_n I mean the complete graph with n vertices.) -Tyler ==== > Are there any good ways to find the minimum and > maximum genera (genuses) of a graph? Let me define what I am looking for. Say that > we can _cellularly embed_ a graph in a compact > orientable surface S if you can embed G in S > (treating G as a 1-manifold), and each connected > domain of S-G (S minus G) is an open 2-cell > (ie homotopic to a point). How is it that you know enough about the problem to define it accurately and don't know that minimum genus is NP-complete or that maximum genus is solvable in polynomial time ? If you just want to compute it for small examples, it's easy enough to try all embeddings (represented combinatorially in terms of cyclic permutations of the edges at each vertex) and test the genus of each one. There are whole books on this general subject; the ones I've used recently are Gross and Tucker, _Topological Graph Theory_ (Dover paperback, good general introduction) and Bonnington and Little, _The Foundations of Topological Graph Theory_ (Springer 1995, may be out of print, more technical material on the combinatorial representation part). -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science ==== Given the matrix product U^T U, where U^T is the transpose of matrix U, is it possible to rebuild U or is it possible to approximate U? ==== > Given the matrix product U^T U, where U^T is the transpose of matrix U, > is it possible to rebuild U or is it possible to approximate U? No, since if V is orthogonal U^t U = (VU)^t UV. But given a positive definite matrix A, one can find a unique positive definite matrix B with B^2 = A. Then the solutions of U^t U = A are the matrices VB with V orthogonal. (One has to be a little more cunning if A is singular but nonnegative definite). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== > Given the matrix product U^T U, where U^T is the transpose of matrix U, > is it possible to rebuild U or is it possible to approximate U? No, since if V is orthogonal U^t U = (VU)^t UV. I think you mean U^t U = (VU)^t VU. But given a positive definite matrix A, one can find a unique > positive definite matrix B with B^2 = A. Then the solutions > of U^t U = A are the matrices VB with V orthogonal. (One has to be > a little more cunning if A is singular but nonnegative definite). ==== We know that the geometrical figure which displays the maximum number of pairs of parallel lines (segments) when joining 5 corners is a regular pentagon, with this number, n = 5 ( 5 directions having --at least-- 1 parallel). The maximum number of parallel relationships for 6 corners is an hexagon, with n = 6. But what if we try to find not just the maximum number of parallel lines, but also rectangular lines (to another line)? I have found that for 5 points we reach the maximum number of parallel and perpendicular relationships with the square, whose corners give 4 points, and the crossing point of the two diagonals when drawed on top of the square give the fifth point. It gives n = 4 and m = 2 (perpendicular directions). The other figure for n = 4 and m = 2 that I have found is a parallelogram of sides 1 and sqrt(3), with the fifth point also the intersection of the two diagonals. Someone could please provide me with some bibliographic reference on that, and check the veracity of that assertion, i.e., that n and m are the maximum for just 5 points in those 2 figures and that there doesn't exist any more figure? I used basic analytic geometry, but based on drawings, so I am not sure if I missed some configuration which were more optimized. IR ==== I am a bit confused. >>It seems as though Multivariable Mathematics is just another name >>for Advanced Calculus, and the same applies to it as well. > The Implicit Function Theorem, the Invese Function Theorem, the > Taylor's Theorem in n-dimensions, derivatives as linear maps, > 2nd derivatives as bilinear maps, differentiability as being > distinct from having derivatives, Frobenius' Theorem, maybe > some Distribution Theory. On the integral side, Measure > Theory, different definitions for integrals, etc. Can somebody recommend a mostly formal textbook on these topics? Hopefully something about two inches thick, second or third edition, with everything proven, lots of examples, tons of problems and a complete solutions manual. I loved Ellis & Gulick, Calculus and Analytic Geometry, 2nd Ed. ==== I am a bit confused. > It seems as though Multivariable Mathematics is just another name > for Advanced Calculus, and the same applies to it as well. The Implicit Function Theorem, the Invese Function Theorem, the >> Taylor's Theorem in n-dimensions, derivatives as linear maps, >> 2nd derivatives as bilinear maps, differentiability as being >> distinct from having derivatives, Frobenius' Theorem, maybe >> some Distribution Theory. On the integral side, Measure >> Theory, different definitions for integrals, etc. Can somebody recommend a mostly formal textbook on these topics? > Hopefully something about two inches thick, second or third edition, > with everything proven, lots of examples, tons of problems and a > complete solutions manual. I loved Ellis & Gulick, Calculus and > Analytic Geometry, 2nd Ed. Buck, Advanced Calculus. Spivak, Calculus on Manifolds. You will not find any with solution manuals. Others will recommend introductory analysis texts, probably Rudin, Principles of Mathematical Analysis. Personally, I find that rough reading for self-teaching, but I have no alternative to offer. (Goldberg is readable but does no multivariable analysis.) Besides, it sounds to me as if you are more interested in the calculus direction than the analysis direction (though this distinction is somewhat vague). I really think the two books I recommend are what you seek. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== Hash: SHA1 I was struggling with trying to do a program that would numerically integrate this multivariate function (The function is complicated and isn't really important, I think), but, this thought occured to me, and I was wondering what everyone thought here: Say you have a function f(X) where X is a vector of n dimensions (terminology? but, in c parlance, I mean to say an array of n doubles) anyway, you have a function f(X) which is defined in some way, and you can't find a closed form solution to Integrate[f(X),X1,X2,...Xn] Would genetic algorithms be a good way of finding FUNCTIONS, that might approximate it on certain ranges, the way that for instance, I've seen an approximation to the Normal CDF where z > 0 as CDF = 1 - 0.5 * ((1+c1 z + c2 z^2 + c3 z^3 + c4 z^4) ^ -4); where c1 = 0.196854, c2 = 0.115194, c3 = 0.000344 c4 = 0.019527 So, I could pass the GA my function and tell it that I expect to want to numerically integrate on ranges from -100,100 and it would generate a function that would _approximate_ the integration without actually doing the numerical integration. I could then look at the function, and decide what it's error characteristics were, before plugging it into the actual app, etc. I'm expecting that the GA would be of the genetic programming type where the functions would include exp and tan, and all the usual mathematical operators. Has this been done? Does that seem like a reasonable app for a GA? I'm not stuck on the idea of generating the function via GA, so if others have different means to generate an approximate function given f that computes the integral of f, I'd be open to that as well... Binesh Bannerjee - -- 'One of the cultural barriers that separates computer scientists from regular scientists and engineers is ... the practical scientist is trying to solve tomorrow's problem with yesterday's computer; the computer scientist, we think, often has it the other way around.' -- Numerical Recipes in C SSH2 Key: http://www.hex21.com/~binesh/binesh-ssh2.pub SSH1 Key: http://www.hex21.com/~binesh/binesh-ssh1.pub OpenSSH Key: http://www.hex21.com/~binesh/binesh-openssh.pub CipherKnight Seals: http://www.hex21.com/~binesh/binesh-seal.tar.bz2.cs256 http://www.hex21.com/~binesh/binesh-seal.zip.cs256 http://www.hex21.com/~binesh/binesh-certificate.gif.cs256 Decrypt with CipherSaber2 N=256, Password=WelcomeJedi! (No quotes) iD8DBQE/sKDVtC/nHH/DrZYRApfBAKDUWs+YZHxwQA41SyB2enX4PjNG4wCeI1eZ 3XiPsv2RGD3QhqkmLW/sxoY= =QvvZ ==== You could have a look at this site: http://www.gepsoft.com/gepsoft/ I am currently using one of their programs. aprx Binesh Bannerjee a .8ecrit: >Hash: SHA1 I was struggling with trying to do a program that would >numerically integrate this multivariate function (The function >is complicated and isn't really important, I think), but, this >thought occured to me, and I was wondering what everyone thought >here: Say you have a function f(X) where X is a vector of n >dimensions (terminology? but, in c parlance, I mean to say an >array of n doubles) anyway, you have a function f(X) which is >defined in some way, and you can't find a closed form solution >to Integrate[f(X),X1,X2,...Xn] Would genetic algorithms be a good way of finding >FUNCTIONS, that might approximate it on certain ranges, the >way that for instance, I've seen an approximation to the Normal >CDF where z > 0 as > CDF = 1 - 0.5 * ((1+c1 z + c2 z^2 + c3 z^3 + c4 z^4) ^ -4); >where > c1 = 0.196854, > c2 = 0.115194, > c3 = 0.000344 > c4 = 0.019527 So, I could pass the GA my function and tell it that I expect >to want to numerically integrate on ranges from -100,100 and it >would generate a function that would _approximate_ the integration >without actually doing the numerical integration. I could then >look at the function, and decide what it's error characteristics >were, before plugging it into the actual app, etc. I'm expecting that the GA would be of the genetic programming type >where the functions would include exp and tan, and all the usual >mathematical operators. Has this been done? Does that seem like a reasonable app for a >GA? I'm not stuck on the idea of generating the function via GA, >so if others have different means to generate an approximate >function given f that computes the integral of f, I'd be open to that >as well... >Binesh Bannerjee - -- >'One of the cultural barriers that separates computer scientists from > regular scientists and engineers is ... the practical scientist is > trying to solve tomorrow's problem with yesterday's computer; the > computer scientist, we think, often has it the other way around.' > -- Numerical Recipes in C SSH2 Key: http://www.hex21.com/~binesh/binesh-ssh2.pub > SSH1 Key: http://www.hex21.com/~binesh/binesh-ssh1.pub >OpenSSH Key: http://www.hex21.com/~binesh/binesh-openssh.pub >CipherKnight Seals: > http://www.hex21.com/~binesh/binesh-seal.tar.bz2.cs256 > http://www.hex21.com/~binesh/binesh-seal.zip.cs256 > http://www.hex21.com/~binesh/binesh-certificate.gif.cs256 > Decrypt with CipherSaber2 N=256, Password=WelcomeJedi! (No quotes) >iD8DBQE/sKDVtC/nHH/DrZYRApfBAKDUWs+YZHxwQA41SyB2enX4PjNG4wCeI1eZ >3XiPsv2RGD3QhqkmLW/sxoY= >=QvvZ ==== > I am trying to calculate the probability that a gambler with capital C > and who uses a Martingale betting strategy will be wiped out in m > turns at the game. This would happen with a run of n consecutive > losses and I want to calculate the probability of this. The textbooks treat this problem at an advanced level, invoking > generating functions or difference equations, but I wonder if a > solution satisfactory for the purpose could be arrived at in a simpler > way. All I need is the probability that there would be AT LEAST one > run of length AT LEAST n. > I don't know if the following is of any use given the other approaches in the thread; just saw someone note that the recurrence had a possibility of blowing up, so perhaps there is better accuracy here... In addition to the recurrence approach, you can also use a markov process approach to this problem; this has come up before on sci.math in the context of coin flipping. Assume your event occurs independently with probability p. Let there be n states, {s_i}, with 0<=i I am trying to calculate the probability that a gambler with capital C > and who uses a Martingale betting strategy will be wiped out in m > turns at the game. This would happen with a run of n consecutive > losses and I want to calculate the probability of this. The textbooks treat this problem at an advanced level, invoking > generating functions or difference equations, but I wonder if a > solution satisfactory for the purpose could be arrived at in a simpler > way. All I need is the probability that there would be AT LEAST one > run of length AT LEAST n. > > I don't know if the following is of any use given the other approaches > in the thread; just saw someone note that the recurrence had a > possibility of blowing up, so perhaps there is better accuracy > here... In addition to the recurrence approach, you can also use a markov > process approach to this problem; this has come up before on sci.math > in the context of coin flipping. Assume your event occurs independently with probability p. Let there be n states, {s_i}, with 0<=i runs of exactly i events in row (i.e., each state represents > losing exactly i times in a row). The probability that state s_i will become state s_(i+1) (i.e., that > we have a run of exactly i events, followed by the event occuring > again) is then p; and the probability that state s_i will become s_0 > is q = 1-p. All other state transitions have probability 0. The probability that NO runs of n or more in length in m trials can > then be calculated by first letting A be an n by n matrix (indexed > here for ease of notation 0..(n-1)) with: A[0,j] = q for 0<=j A[i+1,i] = p for 0<=i<(n-1) and all other entries of A = 0.0; each A[i,j] is then the probability > of going from state s_j to state s_i. Then calculate A^m (this can be done with at most 2*log_2(m) matrix > multiplies, e.g., for m = 1488522243, there are 'only' 42 matrix > multiplies - wouldn't want to do it by hand, but... !). Let V be the vector {1.0, 0, 0, ..., 0.0}; this represents the > starting state s_0. Then the entries in V*A^m = {p_0, p_1, ..., > p_(n-1)} are the probabilities that we will be in states s_0, s_1, > etc., respectively, after m trials. Sum over the p_i in this vector, > and this gives the probability P(n,m) that you will not get n or more > events in m trials. 1 - P(n,m) is the probability that there will be > at least one run of at least n events in m trials. Using this approach, I get, with p = 0.5, n = 30, and m = 1488522243, > that you have a 50/50 chance of never getting 30 consecutive events in > m trials. > I like the simplicity of the idea behind this method of calculation. There are no cancellation problems since we are always adding. It actually executes reasonably quickly but I appear to be getting a lot of rounding error problems. My first reaction was to check my results. I added an extra row and column to get a double check on my results. I got .499999999139885 from adding up the values for 0 to 29 .500000002118337 from the extra row/column and .500000000053831 via ProbnmpApprox. I'm pretty sure that .500000000053831 is accurate, but the rounding errors via the matrix multiplication method seems a bit drastic. Can you supply the values you got, if possible it would be nice if you could add the extra row to confirm whether the problem is genuine or in my code. Ian Smith ==== >I've appended some code in VBA for the original algorithm and for the >approximation I mentioned. I make no great claims for the code but it should be adequate for any >investigations you might want to make. CONGRATULATIONS IAN SMITH! With a few lines of Basic code you have cut through mountains of > mathematical horse shit and created a program that gives quick answers > to what formerly was considered a difficult problem. This is just > what I hoped would happen. I ported your code to BC7 for DOS and ran a few tests. The results > were exactly the same as what I got using the recursive program or > through evaluating the coefficients of the generating function. May I have your permission to compile it and put it on the web site of > the Rancocas Valley Journal of Applied Mathematics, with full credit > to yourself, of course? Sam Allen My standard Conditions of use are No charges, no conditions and no guarantees. I've no idea if this is legal in the US but assuming it is, then you are more than welcome. Ian Smith ==== Let q1,q2 algebraic integers such that q1^k=n(integer),q2^k=m(integer) let K1=Q(q1),K2=Q(q2),Q:rational field. Also suppose that m,n free from k powers and K1=K2. Is there any relation of m,n? thanks costas. ==== > Given an open subset A of the unit square [0,1]x[0,1] with fixed area area(A) > it is known that the first eigenvalue of the operator u |-> - u_xx - u_yy + chi_A * u > i.e. Laplacian of u + [characteristic function of A] times u > > on the space of functions u on [0,1] x [0,1] with periodic boundary > conditions is positive. But does there exist a positive lower bound for this eigenvalue > which is independent of the choice of open subset A > as long as we keep the area a=area(A) fixed? In case anyone is interested: The answer is yes. ==== I have a normal distribution of known mean and standard deviation. For a certain case, a finite number of results will be drawn from this distribution. Is there a mathematical formula for calculating the expected range of these results? ==== > I have a normal distribution of known mean and standard deviation. For a certain case, a finite number of results will be drawn from this > distribution. Is there a mathematical formula for calculating the > expected range of these results? What exactly do you mean by expected range? I could interpret the question literally: What is the expectation value of the range? Suppose you are drawing n independent samples, X_1, X_2,... ,X_n. Here's the cdf of the maximum of the X's: P[max(X_1, X_2, ..., X_n)<=x] = = P(X_1 <=x & X_2 <=x & ... & X_n <=x] = P(X_1 <= x)^n So the pdf is p(x)= dP/dx = n*P_x(x)^(n-1)*p_x(x) where P_x(x) and p_x(x) refer to the normal distribution of individual samples. The expectation value is therefore integral(-inf,inf) n*P_x(x)^(n-1)*p_x(x)*x dx So you can in principle calculate E[max(x_i)]. Similarly, you can work out a formula for E[min(x_i)] in terms of P(x_i >= x) = 1 - P_x(x). Thus, E[max - min] = E[max] - E[min]. - Randy ==== >I have a normal distribution of known mean and standard deviation. For a certain case, a finite number of results will be drawn from this >distribution. Is there a mathematical formula for calculating the >expected range of these results? Google for confidence interval. ==== > >>I have a normal distribution of known mean and standard deviation. For a certain case, a finite number of results will be drawn from this >>distribution. Is there a mathematical formula for calculating the >>expected range of these results? >> > >Google for confidence interval. That won't provide what the OP wants. Instead, look up order statistics. Sorry, I don't have the time to provide the formula right now. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== >>I have a normal distribution of known mean and standard deviation. >>For a certain case, a finite number of results will be drawn from this >>distribution. Is there a mathematical formula for calculating the >>expected range of these results? >Google for confidence interval. This is NOT a confidence interval problem. For any sample size, the ratio of the range to the standard deviation has a known distribution, and its expected value has been tabulated. There is no simple closed form, as there is for a confidence interval. -- 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 ==== ==== > What is the largest (finite) number ever written down? it's still being written down (while watching for the competitors ...) my printer You're quite welcome. ==== Try Ackermann Functions... ==== One of those people that is very fond of exclamation marks probably once 9!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! at the end of a sentence, and thread, I think. I'm not sure. -- Quaternion ==== Quaternion scribbled the following: > One of those people that is very fond of exclamation marks probably once > 9!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! at the end of a sentence, and > thread, I think. I'm not sure. You have 30 ! signs there. This means that your number is smaller than 9^(9^(9^... containing 60 9's. I haven't counted the 9's in Jeroen Boschma's posting but I'm fairly sure he has more of them than you do. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ Nothing lasts forever - so why not destroy it now? - Quake ==== >Quaternion scribbled the following: > One of those people that is very fond of exclamation marks probably once >> 9!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! at the end of a sentence, and >> thread, I think. I'm not sure. You have 30 ! signs there. This means that your number is smaller than >9^(9^(9^... containing 60 9's. I haven't counted the 9's in Jeroen >Boschma's posting but I'm fairly sure he has more of them than you do. There was an odd competition a while ago to write a C program that would print out the largest digit possible (limited number of characters or tokens, I think). I say C because, unlike C, it had big integers of unlimited size. Things like Skewes number, numbers with lots of factorials and/or exponentials were in the first class of not really all that big. Calculations of Ackermann's number fell into the kinda big catagory. The winners were very, very, very big. I was more impressed by the fact that the person running the competition was able to work out what the programs did than that people managed to write them in the first place. Alan -- Defendit numerus ==== Joona I Palaste scribbled the following: > Quaternion scribbled the following: >> One of those people that is very fond of exclamation marks probably once >> 9!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! at the end of a sentence, and >> thread, I think. I'm not sure. > You have 30 ! signs there. This means that your number is smaller than > 9^(9^(9^... containing 60 9's. I haven't counted the 9's in Jeroen > Boschma's posting but I'm fairly sure he has more of them than you do. Erm, no. I was wrong. n! is smaller than n^n, so if we replace n with m!, we get that m!! is smaller than (m^m)^(m^m), and if we replace m with l!, we get that l!!! is smaller than ((l^l)^(l^l))^((l^l)^(l^l)), and so on. The number the variable occurs in the bigger number is 2 to the power of the number of ! signs in the smaller number. If those operations were all grouped like l^(l^(l^... then there would be 1073741824 nested exponentations, which would make Quaternion's number bigger than Jeroen's, but as they are not, I don't know which is larger. Could some of you math gurus shed more light on this? -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ Make money fast! Don't feed it! - Anon ==== à Joona I Palaste Ûçòáãå óôï íÜîùíá [snip] > Could some of you math gurus shed more light on this? Two months ago I came up with a couple of non-optimal estimates for the recursive factorial function. It really grows very fast. And I mean VERY VERY fast. Here's the link: http://users.forthnet.gr/ath/jgal/math/hfseries.html for those that know what the later means. Check the appropriate lemma in the reference link. > -- > /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- > -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ > Make money fast! Don't feed it! > - Anon -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable ==== Here's something else to ponder: It is estimated that there have been no more than 10^11 people who ever lived. Each of these people lived less than 10^10 seconds (which is more than 300 years), so it's a very safe bet to assume that each of these people has written down fewer than 10^10 numbers in their lives and it's even safer to assume that, combined, they have written down fewer than 10^21 distinct numbers. Fire up your favorite random digit generator. Have it give you 30 random nonzero digits. Write them down as a thirty-digit number. Congratulations! You have a better than one-in-a-billion chance that you are the first person to write that number down. In fact, for $29.99 plus $4.99 in postage and handling, I'll send you a certificate attesting that. ==== John Tapper scribbled the following: If you allow finite numbers constructed from a series of mathematical formulae, I think Graham's number takes the cake, by far. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ To err is human. To really louse things up takes a computer. - Anon ==== > John Tapper scribbled the following: If you allow finite numbers constructed from a series of mathematical > formulae, I think Graham's number takes the cake, by far. This might be considered cheating, but I believe N=B(B(B(B(9)))) (where B denotes the busy-beaver function) beats all the suggestions so far. The trouble is that the busy-beaver function is rather hard to calculate. (The busy-beaver function B(n) is the maximum number of 1s that can be printed by an n-state 2-colour Turing machine which halts. To show that N is bigger than an of the entry M posted so far, it suffices to come up with a Turing machine with at most B(B(B(9))) states which prints more than M 1s. Note that B(6) is at least 10^865, so this shouldn't be too hard to do.) -- Simon Nickerson ==== >> John Tapper scribbled the following: > If you allow finite numbers constructed from a series of mathematical >> formulae, I think Graham's number takes the cake, by far. This might be considered cheating, but I believe N=B(B(B(B(9)))) > (where B denotes the busy-beaver function) beats all the suggestions > so far. The trouble is that the busy-beaver function is rather hard to > calculate. (The busy-beaver function B(n) is the maximum number of 1s that can be > printed by an n-state 2-colour Turing machine which halts. To show > that N is bigger than an of the entry M posted so far, it suffices to > come up with a Turing machine with at most B(B(B(9))) states which > prints more than M 1s. Note that B(6) is at least 10^865, so this > shouldn't be too hard to do.) In the same spirit, if one wants to think of B as a not-very-fast-growing function ... http://www.cs.berkeley.edu/~aaronson/bignumbers.html ==== > John Tapper scribbled the following: If you allow finite numbers constructed from a series of mathematical > formulae, I think Graham's number takes the cake, by far. Nuh-ah. G+1. > -- > /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- > -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ > To err is human. To really louse things up takes a computer. > - Anon ==== Nuh-ah. G+1. > I'd say G+2. Or 2*G. Or G^(G^G). Yeah, that's it. /David ==== > Nuh-ah. G+1. >> >I'd say G+2. Or 2*G. Or G^(G^G). Yeah, that's it. Incidentally, G^(G^G) is the number of silly posts on Usenet since it's conception. ==== > Nuh-ah. G+1. >>I'd say G+2. Or 2*G. Or G^(G^G). Yeah, that's it. Incidentally, G^(G^G) is the number of silly posts on Usenet since > it's conception. And I thought it was a twisted anime smiley. G^G^G+1 now. Phil -- Unpatched IE vulnerability: window.open search injection Description: cross-domain scripting, cookie/data/identity theft, command execution Exploit: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-MyPage.htm ==== Incidentally, G^(G^G) is the number of silly posts on Usenet since > it's conception. LOL! /David ==== > I just did (I guess): 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9 :) ==== >> I just did (I guess): The Boschma number has 9^839 * log10(9) = 3.88 * 10^800 digits, so it's of the order 10^10^800 and is much much smaller than the 2nd Skewes Number (10^10^10^10^3). >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9 ==== >I just did (I guess): The Boschma number has 9^839 * log10(9) = 3.88 * 10^800 digits, 9^9 has 9 digits 9^9^9 has 369693099 digits 9^9^9^9 has >10^300000000 digits Where on _earth_ did 9^839 * log10(9) come from? > so > it's of the order 10^10^800 and is much much smaller than the 2nd > Skewes Number (10^10^10^10^3). Are you sure? Phil -- Unpatched IE vulnerability: Timed history injection Description: cross-domain scripting, cookie/data/identity theft, command execution Exploit: http://www.safecenter.net/liudieyu/BackMyParent2/BackMyParent2-MyPage.HTM ==== >9^9 has 9 digits >9^9^9 has 369693099 digits >9^9^9^9 has >10^300000000 digits Where on _earth_ did 9^839 * log10(9) come from? You're right, that figure is for (((9^9)^9)^9)^9)... ==== a .8ecrit : >> I just did (I guess): no , you did not :-) 1 + >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^ >9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9 ==== Is it possible to have a stroboscopic effect (where wheel spokes appear to be rotating backwards) in real life, rather than in movies where it is common? I can imagine two situations where it might happen: when a car is going through a tunnel or under a street lamp at night, or when there is a picket fence between you and it. But on an open road, under a clear sun, is it possible for your eyes to create this effect? --riverman ==== > Is it possible to have a stroboscopic effect (where wheel spokes appear to > be rotating backwards) in real life, rather than in movies where it is > common? Indeed. The first time I noticed it was near the maths building at my alma mater, Queen Mary & Westfield, in London. The compound is separated from the main street by a 6 ft. tall (or so) metallic fence, with vertical bars a few inches apart. The fence was quite long at the time, and from inside the compound you could see through the bars passing cars for a few seconds. Depending on their speed the stroboscopic effect could be seen all right; with any luck, the wheel spokes would appear to be stationary, which I found fascinating to watch. ==== Is it possible to have a stroboscopic effect (where wheel spokes appear to >be rotating backwards) in real life, rather than in movies where it is >common? > I think so. Some of the older trucks I used to drive had mirrors that rattled. Unless my memory has failed, you can sometimes get stoboscopic effects from a vibrating mirror. I never have understood why. Rich Burge ==== >Is it possible to have a stroboscopic effect (where wheel spokes appear to >be rotating backwards) in real life, rather than in movies where it is >common? Yes, easily. Try spinning something under a fluorescent light. Are you too young to remember record turntables with stroboscopic speed indicators? >But on an open road, under a clear sun, is it possible for your eyes to >create this effect? You need a varying source of light; seeing one spoked wheel through another might do it. -- Richard -- FreeBSD rules! ==== Is it possible to have a stroboscopic effect (where wheel spokes appear to >be rotating backwards) in real life, rather than in movies where it is >common? Yes, easily. Try spinning something under a fluorescent light. Are you too young to remember record turntables with stroboscopic > speed indicators? > Not at all. (But I did always wonder why they even bothered put the indicator for 78rpm on them!) >But on an open road, under a clear sun, is it possible for your eyes to >create this effect? You need a varying source of light; seeing one spoked wheel through > another might do it. > But under sunlight, no interference, etc? I'm busy convincing my 9th grade students of their inability to differentiate between reality memories and fantasy (TV) memories. Several of them swear that they've seen it on cars passing them on highways, on open roads, in broad daylight. --riverman ==== à riverman Ûçòáãå óôï íÜîùíá [snip] > But under sunlight, no interference, etc? I'm busy convincing my 9th grade > students of their inability to differentiate between reality memories and > fantasy (TV) memories. Several of them swear that they've seen it on cars > passing them on highways, on open roads, in broad daylight. > They probably don't recall right. It's a common sight on highways during the sunlit day you'd need two sets of spikes or a set of spikes and a grid to see this. > --riverman -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable ==== Let aj = a_j = sqr j-th prime. Show A = { aj | j in N } is linear independent subset of the reals as a vector space over the rationals. Related problem is to show A with 1 included is linear independent. ==== > Let aj = a_j = sqr j-th prime. Show A = { aj | j in N } is linear independent subset of the reals as a > vector space over the rationals. Related problem is to show A with 1 included is linear independent. A better problem is to show that {sqrt{a}: a in Z, a is squarefree} is linearly independent over Q. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== > There's more to my work than just arguing on Usenet, so I'd like to > point out that my paper Advanced Polynomial Factorization is slated > to be published: > The Mega Foundation is an organization of high IQ people, and I'm glad > to be associated with them. To learn further about the organization > you can use Google, or see: The Mega Foundation is an oranization of organ donors given that they're all obviously Psychologists. So IQ is irrelevent. Why a group like the Mega Foundation? > http://www.ultrahiq.org/Mega/WhyMega.htm I hope at least some of you will appreciate that often the most > important ideas in history have to get past people limited by their > lack of imagination and their prejudices, who act against scientific > progress. Science had only progressed 2 inches since math was invented, so the cerebral crowd has switched over to logic. > What I want you to see is that there's more to me than Usenet, so that > you can begin to understand that the revolution I'm giving you a > chance to be a part of is bigger than the small-minded people who > continually throughout history work to halt progress. Progress was invented by born-again historians, so it's irrelevent to science. > James Harris > http://mathforprofit.blogspot.com/ ==== > What I want you to see is that there's more to me than Usenet, so that > you can begin to understand that the revolution I'm giving you a > chance to be a part of is bigger than the small-minded people who > continually throughout history work to halt progress. > Oh Boy! ====