mm-91 === Also, can we have things such as e^(quat) or ln(quat) where quat is a>>quaternion? Has anybody ever ?ured 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 incomplex Banach algebras, while the usual quaternions are an algebraover 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 coef?ients.Robert Israel israel@math.ubc.caDepartment 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 thesupression 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 isomorphordinary 2D-vectors (with a certain multiplication added) -You calculatethem 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.deThere 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 CasparWessel 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^nwhere n is any integer. This would start from 1 in the case where n =0 to inde?ately high. Let's call this set Q.There are an in?ite amount of such numbers since the log of in?ityis in?ity. So the cardinality is aleph-x, where x is an as yetundetermined variable.Now, by de?ition, power sets are larger than the sets associatedwith them. So the power set of Q would be equal in size to the set ofall positive integers. Or all non negative integers, if you includedthe null set.So while Q is not ?ite, its power set, which must be larger than it,is supposedly the smallest in?ite set, accordding to a certaintheory 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 inde?ately 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 allintegers that resemble an integer 2^n. What does it mean for a numberto resemble an integer? I'll suppose for the moment that you reallymeant 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 in?ite amount of such numbers since the log of in?ity> is in?ity. So the cardinality is aleph-x, where x is an as yet> undetermined variable.> In?ity is not a number, and log(in?ity) is unde?ed. Surely, thelimit lim_{x -->in?ity} log(x) fails to exist, because the functiongrows without bound, and that fact is typically written lim_{x -->in?ity} log(x) = in?ity.That does not mean that one can substitute in?ity for x in theexpression log(x), and out pops the value in?ity. The aboveexpression for the limit is a *shorthand* expression for what Ibound as x gets arbitrarily large.> Now, by de?ition, 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 de?ition* that power sets are larger than the setsassociated 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. Theoremsare hardly ever de?itions, and this theorem is surely not ade?ition.> So while Q is not ?ite, its power set, which must be larger than it,> is supposedly the smallest in?ite 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 ?ite: by itsconstruction, K is clearly of the same cardinality as N_o, the set ofnatural numbers with zero.Next, you claim that its power set 2^K (which must be of greatercardinality, according to the theorem I mentioned relating X and 2^X), is supposedly the smallest in?ite 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 de?ed as the successive cardinals, it is *this*that is a de?ition (if I understand the de?itions correctly).However, 2^K is surely not the smallest in?ite set. The existenceof the alephs has nothing to do with 2^K, and it is certain that theydo *not* show that 2^K is the smallest in?ite 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 asthe power set, appearances notwithstanding. The notation 2^X for thepower set of X is (as with the limit expression I discussed earlier)simply a shorthand for a particular construction. In this case, itrefers 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 amap 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 distinctfunctions, every subset yields a function, and every function comes froma subset.Simply taking 2^n for every element n of some set of naturals does notyield the power set. Note that the method fails already for ?ite sets.I'll take a small set X, and form your construction (I'll call it Kagain): 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 areintegers (except {} for folks who use the von Neumann constructionof 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 UTMseries).Dale.CAVEAT: I'm no set theorist, nor do I play one on TV. There are plentyof regular contributors to sci.math who are more adept at thisparticular game than I am, and they may well have more to say on thistopic. === > 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 inde?ately 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 in?ite amount of such numbers since the log of in?ity> is in?ity. So the cardinality is aleph-x, where x is an as yet> undetermined variable.> In?ity is not a number, and log(in?ity) is unde?ed. Surely, the> limit lim_{x -->in?ity} log(x) fails to exist, because the function> grows without bound, and that fact is typically written lim_{x -->in?ity} log(x) = in?ity.That does not mean that one can substitute in?ity for x in the> expression log(x), and out pops the value in?ity. The above> expression for the limit is a *shorthand* expression for what I> bound as x gets arbitrarily large.> Now, by de?ition, 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 de?ition* 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 de?itions, and this theorem is surely not a> de?ition.> So while Q is not ?ite, its power set, which must be larger than it,> is supposedly the smallest in?ite 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 ?ite: 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 in?ite 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 de?ed as the successive cardinals, it is *this*> that is a de?ition (if I understand the de?itions correctly).> However, 2^K is surely not the smallest in?ite 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 in?ite set.> However, 2^K is the set of all natural numbers. And the set of allnatural numbers is of cardinality N_0. However, K is also ofcardinality N_0. If the power set of any set is strictly larger (IEhas a greater cardinality) then how is it possible? This would requirethat 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 powerof 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 ofnonnegative integers from 0 to 2^(X+1) if you count the empty set aspart of its power set, and is bijective with the set of positiveintegers from 1 to 2^(X+1) if you don't count the empty set as part o?s power set.Then take X as X -> inf.So K is of cardinality N_0, from what you said, but so is its powerset, since its power set is bijective with the set of natural numbers.This is the contradiction I'm talking about. The power set is alwaysof a larger cardinality than the set that generates it. But this wouldmean 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 ?ite reasoning to in?itesets. In?ite sets have different properties, for instancebeing in one-to-one-correspondence with proper subsets.Let's see where reasoning about the ?ite 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 ?ite) means that you havea bijection with 1, 2, ..., 2^(X+1), not with0, 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 thepower set, and is included in the count of 8, i.e.,a one-to-one correspondence with the integers from1 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. Thisis one of the subtleties of dealing with in?ity. Limitprocesses don't go to in?ity. That's not what thesymbol -> inf means. You don't ever reach in?ity.What you're doing is noting that as X grows without bound,taking values in the natural numbers, then 2^X also growswithout 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 inde?ately 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 de?ed as the successive cardinals, it is *this*>>that is a de?ition (if I understand the de?itions correctly).>>However, 2^K is surely not the smallest in?ite 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 in?ite 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 notthe case that 2^K is the set of all natural numbers. Since you're theone making this assertion, I'll ask you to tell me how one associatesa natural number with an arbitrary subset of K, so that every naturalnumber is matched with a subset, and so that no two different subsetsare matched to the same natural number. According to Cantor's theoremit cannot be done. Given any set X, its collection of all subsets hasmore 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 arti?e of the powers of 2 isirrelevant to your argument. You could just as well take the set ofnatural 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, andthe other has 2^N elements. You're trying to state that since theindividual steps are all ?ite, somehow one can force a 1:1 mappingin 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 attemptto force this procedure to the limit as N --> in?ity, then you needto 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 tothe positive integers, one needs the sum to converge, but note that itdoesn't converge if X is in?ite: it has in?itely many terms, each ofwhich is a positive integer (in fact, the if the sum has in?itely manyterms, then it has terms that grow without bound). The series cannotconverge in the limit.This particular form of limiting argument therefore cannot succeed. Ifyou have another limiting argument (by the way, the phrase as N --> in?ity, the limit worksis 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 noargument at all, since it is strewn with implicit steps that have noclear 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 inde?ately 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 de?ed as the successive cardinals, it is *this*>>that is a de?ition (if I understand the de?itions correctly).>>However, 2^K is surely not the smallest in?ite 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 in?ite 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 greaterthan K. I said N was equal in size to the power set of K. Cantor'stheorum 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 numbersNum = (0, 1, 2, 3, 4, 5, 6, 7) easily. Now K is smaller than the powerset of K. However, the power set of K, as the size of the power setapproaches in?ity, also approaches the set of the natural numbers(Even if 0 is not a natural number, you can just add 1 to each memberof the set so that it goes from 1 to 8 rather than 0 to 7.) and istherefore 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 SeamanJudge 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 takingthe image of N via an invertible function. Just because I stepped fromN to a set of the same cardinality, that does not invalidate theapplication of Cantor's theorem.Your statement that N is equal in size to the power set of K is nottrue. Recall the mapping I provided (I've corrected a typo: originallymapping): 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 ofall elements of K? Every other element of K?You should realize that the only subsets you can map to N via thismapping are the FINITE ones. It is true that the collection of FINITEsubsets of N has the same cardinality as N itself. However, unless youcontest the existence of in?ite sets of integers (such as the set ofeven integers, or the set of all integers greater than 5, for instance),you must admit that those sums do NOT all converge, and thus youralleged mapping fails to be de?ed 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 in?ity, 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 *in?ite* sequences of 1s and 0s. For each subset K' of K, associate the in?ite 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 *?ite* 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 in?ite sequences of 1s and 0s.The proof of 3 is exactly Cantor's theorem: assume that there isa bijection f between N and the in?ite sequences of 1s and 0s,and show that leads to a contradiction.--Daryl McCulloughIthaca, 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 *in?ite* sequences of 1s and 0s. For each subset> K' of K, associate the in?ite 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> *?ite* 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> in?ite 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 in?ite 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 thatK and N both be of cardinality N_0.However, I don't see why there can't be a non-N_0 cardinality of Ksuch that when its power set is taken, its power set is of cardinalityN_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 betweenN and K, namely f(i) = 2^{i}. By the de?ition of having the samecardinality, N and K have the same cardinality.--Daryl McCulloughIthaca, 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 aset, not a number, and therefore 2^K is not a set of numbers andcertainly is not equal to N_0.-- Dave SeamanJudge 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 wasBIJECTIVE 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 SeamanJudge 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.htmlNeedless 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 ascompletely 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 inde?ately high. Let's call this set Q.>> There are an in?ite amount of such numbers since the log of in?ity> is in?ity. So the cardinality is aleph-x, where x is an as yet> undetermined variable.>> Now, by de?ition, 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 toa natural number. The obvious mapping would be take the sum of the elementsof the subset. You are probably thinking that this is the same as thenatural numbers in base-2, but that is wrong. A natural number must be*?ite*, but a subset of Q can have an in?ite amount of elements. Forexample, the subset {q in Q: lg(q) is odd} has a sum that is not a naturalnumber (lg x means log base 2 of x). In fact there are uncountably manysubsets of Q that are in?itely large. You are thinking of the Set of?ite subsets of Q, which is indeed countable and has a very clearbijection with the naturals.~Steven>> So while Q is not ?ite, its power set, which must be larger than it,> is supposedly the smallest in?ite 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 inde?ately high. Let's call this set Q.Do you mean ordinary ?ite integers? This resembled relation isunclear to me. I assume you mean:Q = {x in N: x = 2^n for some n in N}> There are an in?ite amount of such numbers since the log of in?ity> is in?ity. So the cardinality is aleph-x, where x is an as yet> undetermined variable.The log of in?ity is a meaningless phrase. But Q is indeedin?ite. x = 0. The construction of Q gives an injection from N->Q, and theidentity mapping is an injection from Q->N. So N and Q have the samecardinality, aleph-0.> Now, by de?ition, 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 powerset.The power set of Q has cardinality of the continuum.> So while Q is not ?ite, its power set, which must be larger than it,> is supposedly the smallest in?ite 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 de?ition 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 inde?ately 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 in?ite amount of such numbers since the log of in?ity> is in?ity.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 de?ition, 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 thesame cardinality of N. (Why did you begin your 2nd sentence in thatparagraph with so?)> So while Q is not ?ite, its power set, which must be larger than it,> is supposedly the smallest in?ite 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.htmlNeedless 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 intoR^5, but not into R^3. Before you ask whether R^4 is enough, youhave to decide on things like the degree of smoothness, embeddingversus immersion, and local versus global. Here are some threads fromfour years ago: http://www.math-atlas.org/99/embed_hyperdave === > 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 ofself-isometries should act transitively, i.e. that it contain a 2-parameterfamily of isometries. The torus does not have that. If you interpret theOP's request merely to be for the isometry group to be non-discrete, thenyes, clearly a (regular) torus is an example.dave === I am struggleing with ?ding a way to measure the linearity of a 3D ?ld.I have a set of data with coordinates (x,y,z) and the ?ld value on each given point f(x,y,z). And we expect the ?ld to be linear so that the theoretical relationship would bef(x,y,z)=a*x+b*y+c*z+dWe could do a least square ?ting of the given data to get the a,b,c,d parameters. But the problem now is how linear the ?ld 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 ?ting case, there is a correlation coef?ient, which is given by r^2=a1*a2. where two curve ?tings are done.y=a1*x+b1x=a2*y+b2The 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 de?ition of similiar coef?ient for higher dimensions.Could anybody help me out here?-- Shi Jinsj88@cornell.eduhttp://jinshi.dhs.orgFollowing the introduction, a particularly nice (and perhaps new)approximation of the inverse sine is presented. That approximationcan be used to give corresponding approximations of the sixtrigonometric functions and the ?e other inverse functions.------------------------------------ IntroductionNikolaus Cusanus (1401-1464, a.k.a. Nicholas of Cusa) gave a remarkableformula: 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 CusanusArcsin(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 relatedapproximation is 3*x/(2 + Sqrt(1-x^2)). To approximate Arcsin(x) for1/Sqrt(2) < x <= 1, using the fact that sin^2(t) + cos^2(t) = 1, we mayreplace 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 <= 1as an approximation of Arcsin(x), with |rel. error| < 0.0023 .Of course, due to simple interrelations between the inverse trigonometricfunctions, one could easily obtain corresponding approximations forthe other ?e inverse functions from (1).Furthermore, due to the algebraic simplicity of form in (1), it can beinverted 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/4as an approximation for sin(x), with |rel. error| < 0.0019 . Due to simpleinterrelations between the trigonometric functions, one could easilyobtain corresponding approximations for the other ?e functions from (2).(This ends the introduction, in which I presume that there is reallynothing new.)------------------------------------We now primarily consider approximating Arcsin(x) for 0 <= x <= 1/Sqrt(2),knowing that, as indicated above, corresponding approximations can beobtained easily for Arcsin(x) for 1/Sqrt(2) < x <= 1, and ultimately forall of the trigonometric functions and their inverses.The ?st 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 favoritechoice is h = (Sqrt(2) + 1)*(Sqrt(2) - 4/Pi) = 0.340341385...Using it, we obtain an expression giving Arcsin(x) precisely at both 0 and1/Sqrt(2) and overestimating it for 0 < x < 1/Sqrt(2), withrel. error < 0.00057 . Unfortunately, the improvement in accuracy is notdramatic. It might also be noted that this approximation is differentiableat x = 1/Sqrt(2); by contrast, (1) has a jump discontinuity there.But could the form of (3) be altered slightly -- say, by introducinganother parameter -- so as to get much better accuracy? Yes! Choosing theform(4) x/(1 - h*(1 - Sqrt(1-(k*x)^2)))allows relative error to be reduced substantially, to approximately 1/50of 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 differentvalues of h and k would surely reduce error even a bit more, but I thinkit's neat to use the two criteria already mentioned.] Although the exactexpressions 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) for1/Sqrt(2) < x <= 1 by taking the complement of (4) once x has been replacedby 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 andk, 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 approximationsfor the trigonometric functions and for the ?e other inverse trigonometricfunctions.------------------------------------- How does this approximation of Arcsin(x) compare with other approximations?Of course there are far more accurate approximations. Some of them arealso related to Cusanus's formula; see Lou Talman's Some (Almost) RationalThoughts , forexample. However, AFAIK, they are of more complicated forms, thereby makingtheir algebraic inversion more dif?ult -- in fact, impossible in mostcases.Abramowitz and Stegun mention a four-parameter approximation, their item4.4.45 (see ) whichprovides 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 coef?ients seem to have been determinednumerically.------------------------------------ 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 ?st 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 ?st 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 _de?ition_ of the circumferenceof a circle was, but if he had one I can't imagine that it was notsomething essentially equivalent to what we would call the leastupper bound of the perimeter of an inscribed polygon. Assumingthat, then ?st 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 questionwas the same as (*), for some reason.)The circumcribed polygon is the union of triangles with heightequal to the radius of the circle and base one of the sides ofthe polygon, so the _area_ of the circumcribed polygon isequal to r times its circumference.Otoh, let r' < r. Note that adding points to the inscribedpolygon increases the circumference, by the triangle inequality.So starting with an inscribed polygon we can ?d anotherwith larger circumference, and with area > r' * perimeter.Since the relation between the areas is clear, it followsthat r' * inscribed perimeter < r * circumscribed perimeterfor 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 upwith, 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 fromK*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 recallthat on my ?st Calculus ?al I asked the students to do that usingCalculus, 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 de?e the circumference of a circle as the sup of theperimeter of inscribed polygons then I don't know how we _do_ de?eit. If we do de?e it that way then it follows easily that it's theinf 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 ?st Calculus ?al I asked the students to do that using>Calculus, using Pappus' theorem, and ... using just their imagination :-) ]>> baloglouAToswego.eduDavid C. Ullrich === A problem:* Find all primes of the form 4^n + n^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^[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 lookingat 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 in?itely many prime values (as opposedto the polynomial j^2 + k^2, which is easy, and the polynomial1 + k^2, which is still unknown).dave === > A problem:> * Find all primes of the form 4^n + n^Borrowing one of Einstein's ideas, we have4^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^>> 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.96estroA 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 whatabout 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.96estroA 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.comI 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 primeand found only n=3,5,15,35,55e.g. 3^4+4^3=145=5*29, 5^4+4^5=1649=17*97, ...can we ?d more terms or is there a fundamental reason that nosemiprime 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 ?d 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 ?itely many.Heuristically the probability of a number the size of these beingprime is on the order of 1/n. The probability of both being primewould be on the order of 1/n^2. Since sum_n 1/n^2 < in?ity, weshould expect only a ?ite number of examples.Robert Israel israel@math.ubc.caDepartment 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 ?d 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 ?itely 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 < in?ity, we> should expect only a ?ite 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 1Z2There are no further solutions up to k=5000 -> n=5001. So the practicalchances of ?ding 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 ?ding 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 bene? from writing one of the 4s differently) === > A problem:> * Find all primes of the form 4^n + n^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.htmlNeedless 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 ?st. === >> 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.htmlNeedless 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 dif?ult 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 ?d most> dif?ult 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 realizethe power of taking a sample to ?d a population. I pass out to the class abunch of mini-bags of M&Ms, and ask them to predict how many red M&Ms are inthe room. I write their prediction on the board.Then, I have ONE student open their bag. If they have, say 5 red M&Ms, andthe class has 20 kids, all the predictions range close to 100 total, with abit of variance (which we can calculate) because some students astutelypredict that there may be bags out there loaded with red ones. Then anotherstudent opens their bag, and the predictions become less variant.In a class of 20 students, they all start feeling quite certain of theirpredictions after only 4 or 5 bags being opened. It gives them an excellentsense of how the number of samples is related to the degree of certainty ofthe predictions. We graph the range of predictions vs. the number of openbags, and they see how their variance and predictions became very stableafter very few samples. Also, they realize how little an outlier bag affectsthe sum total. And they get a good understanding of how unusual it would befor that outlier bag to be one of the ?st bags sampled, but how it ispossible.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 heardthatBasic Partial Differential equations by David Bleecker, George Csordas, andDarko Grundler book is nice, but I haven't had the chance to take a look atit. Any suggestions would be appreciated!TIALurch === 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!TIALurchThis is not a direct answer but can't you search onPartial differential equations lecture notesand then make some kind of judgment about the book, derived from whatthe 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, thatis (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 haveheard> 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 lookat> 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!TIALurch1. Partial Differential Equations: An Introduction by Walter A. Strauss2. Partial Differential Equations 4e by Fritz John === >> Could someone recommend a good Introduction to P.D.E.'s book ? I haveheard> 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 lookat> 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, GeorgeCsordas,> and> Darko Grundler book is nice, but I haven't had the chance to take alook> 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 andone on integral equations. They are in a book with the strange titleMathematics 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 S12 5 23 1 14 19 = 74 Audrey sat at the MS Society table turning the street fair downtown on2nd Ave. She was the ?st of the three nubile sweeties seated at this tableto provide stats. And Audrey was the second person to provide family statstable, it was the next table over. Audrey liked the work I did on Shadia'sstats and then told me about her family. Audrey and I meet on the 1288th dayof the century, there are 1288 verses in Numbers.74+ Dad 5 4 40 96/270 +616274+ Mom 21 3 45 80/285 +435174+ Sib 21 11 /4074+ Sib 22 10 /7074+ Sib 1 3 /30574+ Sib 21 1 21/194 Carmen 9 9 76 253/113 7144Carmen 54 Frances 66 Lewans 7474+ Sib 19 9 /103213 Audrey 9 5 80 130/236 8482Audrey 74 Lynn 65 Lewans 7474+ Sib 31 12 /0 And I met sister Carmen on January 17th 1998 (2001 days earlier), Carmenalso gave out birthdays for all 10 family members and claimed that dad wasborn on April 7th 1940, claimed that mom was born on March 21st 1944, andthe kids are a sister born on November 26th 1969, then a sister on October25th 1971, followed by a brother on March 1st 1973, then another brother onJanuary 20th 1975, followed by Carmen, followed by a sister on September19th 1978, followed by another sister (Audrey) on May 9th 1980, and thenfollowed by the last sister on December 12th 1982. Between Carmen andAudrey, they gave out different birthdays for both parents and differentbirthdays 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), bothCarmen 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 21210 64 116 164 213 <-166th12 65 117 165 21414 66 118 166 21515 68 119 168 21616 69 120 169 21718 70 121 170 21820 72 122 171 21921 74 123 172 22022 75 124 174 22124 76 125 175 22225 77 126 176 22426 78 128 177 22527 80 129 178 22628 81 130 180 22830 82 132 182 23032 84 133 183 23133 85 134 184 23234 86 135 185 23435 87 136 186 23536 88 138 187 23638 90 140 188 23739 91 141 189 23840 92 142 190 24042 93 143 192 24244 94 144 194 24345 95 145 195 24446 96 146 196 24548 98 147 198 24649 99 148 200 24750 100 150 201 24851 102 152 202 24952 104 153 203 25054 105 154 204 25255 106 155 205 25356 108 156 206 254 Audrey has 6 lettered ?st and last names, 66.666...% of her names are 6lettered. Her 6 lettered names both add to 74 (a factor of 666). Her ?sttwo 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 onday 130 (Numbers 13) while her name adds to 213, prettier as 21 is the 13thnon-prime. Her given names add together for 139, her middle name adds to amultiple of 13. Her name adds to 213 (166th non-prime). Her ?st 13 lettersadd 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. Theparents were born on days of the month averaging 13 (6th prime). Carmen'sand Audrey's birthdays are likely correct, these two sisters are separatedby 3.66 years. And note that Carmen (6 letters) has a middle name adding to66.1-50 - Genesis51-90 - Exodus91-117 - Leviticus118-153 - Numbers154-187 - Deuteronomy188-211 - Joshua930-957 - Matthew958-973 - Mark974-997 - Luke998-1018 - John1019-1046 - Acts1047-1062 - 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 661062 <-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 61070 <-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 herparents were together born 485 days closer to the beginning of their yearsthan to the end of their years (perhaps). Audrey's unrepeated letters add to85.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 ?st 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 differin 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'sbirthdays. Perhaps the parents were an average of 37.61 years old whenAudrey was born. Audrey's vowels add to 37.41% of her consonants. Her ?stand 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 in80. 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 1211days remaining in their years or at least generally together have theirbirthdays when there are 1211 days remaining in their years, it's exactly173 weeks (the 40th prime). Or according to Carmen, the last 4 kids wereborn on days of the year adding to 1010, corresponding to the 81st chapterof the New Testament, pretty as Carmen claims that mom was born on the 81stday of the year.Primes 2 61 149 3 67 151 5 71 157 7 73 16311 79 16713 83 173 <-40th17 89 17919 97 18123 101 19129 103 19331 107 19737 109 19941 113 21143 127 22347 131 22753 137 22959 139 233 Dad was born in 40, mom in 45, while Audrey's 10 different letters add to142 (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 170thnon-prime, while chapter 170 is Deuteronomy 17. And Numbers 7 (124) is the7x7th prime (227) short of the numbers up to the 17th non-prime. Numbersopens at chapter 118 (twice the 17th prime) and terminates at 153, it's 1through 17 and is the 117th non-prime, it's the number of ?h in the net inJohn 21 (7+7+7), while there are 21 (7+7+7) chapters in Book 7. There are 45chapters 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+17verses. 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 numbersadding to 17 (1+3+13=17). The ?st 17 versed chapters in the Bible are atchapters 91 and 93, together for 184, or the 167 verses of Book 17 plus 17more. Leviticus contains 859 verses, it ends in 59 (the 17th prime). The?st 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 9100 10101 11102 12103 13 59 <-17th prime104 14105 15106 16107 17108 18109 19110 20111 21112 22113 23114 24115 25116 26117 27 34 <-17+17 Mom was born on the 21st and managed to give birth twice on the 21st (ifthe stats Audrey provided are correct). Audrey and mom were born on days ofthe year adding to 210. Audrey's name adds to 213. Perhaps the second of thekids was born 215 days after mom's birthday. Audrey's ?st 7 and last 7letters 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 that21, the 21st prime (73) and the 21st non-prime (32) average 42 (21+21). Idon't want to spend a lot of time on this family for Carmen and Audrey haveprovided 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 2310 2911 31 <- 3112 3713 41 <- 4114 4315 4716 5317 59 <- 59 --- 167 Esther Book 17 The parents were together 27477 days old when Audrey was born. Audrey wasborn exactly 7 weeks after mom's birthday. Audrey was born 34 (17+17) daysafter 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 repeatingletters add to 128 (2 to 7th). Audrey's 130th day of birth adds with her 213valued 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 missingfrom her given names add to 251 (the ?st 17 non-primes). In her givennames, her repeating letters exceed her unrepeated letters by 17. In herfull name, her unrepresented letters exceed her represented letters by 67(Exodus 17). Audrey is 23.17 years old. Audrey might marry Marcia and me andtake our 117 valued last names, and then become a Queen like Esther in Book17. And note that just 14 (7+7) days ago I met Emily Karen Lewans at BurgerKing, her name also presently adds to 187.187 Dar 17 2 57 48/317 00Daryl 60 Shawn 65 Kabatoff 62187 Marcia 6 8 80 219/147 8571Marcia 45 Veronica 87 Acevedo 55187 Emily 31 10 81 304/61 9022Emily 64 Karen 49 Lewans 74256 Audrey 9 5 80 130/236 8482Audrey 74 Lynn 65 Acevedo-Kabatoff 55-62 Anyway, if you people think that you have the right to use my abusiveparents as tools and arrest and torture me, then I think that I should havethe right to ask women to marry me, or to marry Marcia and me, our lastnames add together for the 117 verses of Song of Solomon, it's the Bible'sBook of Love. The nubile sweety was born on the 6th and has a 6 lettered?st 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, greedor lust, or who limit their love for traditions and for people who abide bytheir traditions. You people have Egyptian penises on the roofs of yourchurches and lined city streets with representations of penises, and had metortured for years for saying so, others just sat back in silence while theywere doink this to me, and similarly you remain silent and compassionlessnow that the arrests and torture have ceased. You people spent millions ofdollars having me tortured, and then annually you spend billions on yourdecorated trees, I begged and begged for assistance to ?e country(they tortured me for years at the U of S) and you people are so cheap thatyou can't even offer to buy me a cookie when I bust my ass to show youevidence that your very name is a gift from God, you cheap, ?thy,compassionless assholes won't even spend 48 cents on a stamp so that youlife to show you evidence that your very name is a gift from God!!! All youare really good for is to have your stats posted on the usenet and be usedas 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, thenany of Audrey's siblinks that are not Catholic (necromancers) are goink towin themselves a shiny new Cadillac!!! Good luck and may God bless you!!!Daryl Shawn KabatoffBox 7134Saskatoon SaskatchewanCanadaS7K 4J1Isaiah 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 ?st 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 unof?ial FAQexplaining the psychotic nonsense posted to Usenet by Shawn DarylKabatoff AKA Dar, AKA Probababbilities. And now AKA marcia andme.WARNING: Read below before even thinking about responding to thistwit.http://www.waxy.org/archive/2002/05/21/dar_ kaba.shtml#000643Usenet has the tendency to provide a public forum for those who wouldnormally be scribbling in a closet. For example, take Daryl ShawnKabatoff. For the last few years, he's methodically gatheredstatistics from various sources, ranging from local newspaperobituary pages to the food court of the Saskatoon Midtown Plaza mall.With all the raw data he's collected, he's attempting to prove dailythat our full names are in mathematical harmony with our birthdays.about, starting with calculations related to their birthdate and fullnames, blending in whatever other personal information about theirfamily members, spouses, birthplace, and career he's been able tozealotry, and personal torment. I've never seen anything like it.With all the prime numbers, Fibonacci sequences and biblicalreferences, it's like reading the notebooks of Maximillian Cohen andJohn Nash combined. Unsurprisingly, several posts unfold to reveal ahistory 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.comDecember 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.caOct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.caposts have been removed from Google Groups archive)Selected Posts:Tessa Lynne SmithDastageer Sakhizai and Helen SmithBrett David MakiAndrew Meredith CottonAmanda Dawn NewtonMona Marie EtcheverryTony Peter NusplLisa Charlene McMillanGrant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my currentresidence. and every single place he's mentioned in his posts (mostnotably nervous harold's and the roastary) were either places i'vebeen (as it's a small city of 200K) or hangouts, ie. the two placesout if they know of him, they (my friends that is) being of thebroadway-centred slacker ilk. myself, too, until i got out of there.eh, anyways. thought it odd to see all this. midtown mall. i ate mymeals there, whilst waiting several days in line for star wars episodeone, at the theatre across the street.posted by andy raad on May 22, 2002 06:20 PMFascinating. It's like he's trying to take chaos and bind it intowhatever rules he can ?d, religious, logical and otherwise. Numbersand math have a reliable pattern, something that can always be provento true or false. People and religion do not. It reminds me of DarrenAronofsky's movie Pi. It's the story of an paraniod genius who istrying to ?d a pattern in Pi. A group that takes interest in hiswork is convinced that the existence of Pi, a number whose existencecan be proven but no quanti?d, is proof of the existence of God.Kabatoff's hunt for patterns in something as random as name selectionis a way to reconcile his deeply logical thought process with hiscon?g religious views.posted by matt on May 23, 2002 11:19 AMasking him if he'd be willing to create a numerological analysis forme. I also asked him if he had seen either Pi or A Beautiful Mind, andwhat he thought of them. If he replies, I'll be sure to post it.posted by Andy Baio on May 23, 2002 11:24 AMI baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbedpumpkin 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 PMUm, that's swell. So, you're in love with him?posted by Andy Baio on July 25, 2002 07:10 AMShawn once went to a funeral for a Jehovah Witness that shot himselfand the lemon tarts were very bad, they were not only sour but wererubbery as well. Shawn said that the guy was some kind of JehovahWitness prophet, he saw in advance that the lemon tarts at his funeralwere to be very very bad, and so he shot himself. Shawn said that henever ate pumpkin pie at a funeral but would like to some day. Shawnlikes pumpkin pie and so I have been practicing to make very goodpumpkin pies.posted by Trisha Blondie on July 25, 2002 02:49 PMShawn said that the lemon tarts were sour, bitter and rubbery.posted by Trisha Blondie on July 30, 2002 12:32 AMI don't think this guy takes notes. I think he has Total Recall, andit has driven him insane...posted by Todd Smith on December 26, 2002 11:00 AMOh... I almost forgot... I didnt spend thousands of dollars a daytormenting Daryl... We got a deal on tormenting that ?cal year, itonly came to about 37cents a day....posted by Dr Claw on December 30, 2002 01:56 AMMr. Kabatoff attempts to portray himself as a victim, but in fact heis a violent predatory pedophile who is well known to his local lawenforcement. In his post to multiple newsgroups with the subjectCollecting Mail For The Coming Anti-Christ, he encourages mothers tosend him photos of their naked daughters. Mr Kabatoff explains, IAnt-Christ) that were of underage children unless the parent wassigning consent. He is banned from virtually all the shopping mallsin his community because he stalks young people and sexually harassesthem. He has an extensive arrest record which includes sexualmolestation charges. He's been hospitalized in mental institutionsabout his contact with young girls in many posts. Search newsgrouparchives for posts by him containing the word nubile. As part of hisharrassment, he provides personal details in a public forum, such asthe real names of real children, in these and other posts. About onewanted 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&rnuHe not only curses children and prays for their death in his posts, healso enjoys attending the funerals of young people: And so, sincenubile sweeties are found in greatest abundance at the funerals ofhigh school students, then it is the funerals of high school studentsthat 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 atall the pretty nubile sweeties and have the time of my life...http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff +nubile+sex&hl=en&lr=&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 liberallyspams, ?and crossposts his off-topic threatening and offensivemissives to countless newsgroups. Some people HAVE problems and somefolks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. Whenhe sends these posts to any newgroup, please help by reporting him toI 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 ofnumbers racing by on his laptop. I have an original copy of hisCollecting Mail for the Coming Anti-Christ pamphlet, and have seenhim be hauled away by campus security on more than one occasion. Myfriends and I refer to him as Crazy Number Man.I've been posting to (and about) Shawn for over two years with biggaps in between. He has seen Pi and didn't like it and didn't think itresembled him at all. (Wrong, it ?s him to a tee) He doesn't havetotal recall and has stated that he travels with a lap top to notateitems. Also, he uses cut n' paste a lot if you read all the waythrough his ramblings. He is anti-social as shown by his angrystatements towards those who, by his own admission, have been kind(but not kind enough) to him. Still, he's intelligent and seems to beable 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 ?st of the three nubile sweeties seated at thistable>to provide stats. And Audrey was the second person to provide familystats> <<> The following (courtesy of Waxy.org) is sort of an unof?ial 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 ?d, 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 ?d 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 quanti?d, 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> con?g 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 ?cal 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, ?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 ?s 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 ?d out how to, I just set himon 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 ?d 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 kill?ed the pair of ?em.Dave. === Someone asked a way to calculate integral from 0 to in?ity of sin(x) / xwithout 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 in?ity ofe^(-px)*sin(x) wich we denote by I(p). It is easy to evaluate integrating byparts:I(p)=1/(p^2+1)Now the trick is to integrate I(p) from 0 to in?ity with respect to punder the integral sign :int(0->oo) 1/(p^2+1) =pi/2 === > Someone asked a way to calculate integral from 0 to in?ity of sin(x) / x> without using residues.>> Probably the most elegant way (imho) is to use so called Feynman'strick:>> Consider an integral (containing a parameter p) from 0 to in?ity of> e^(-px)*sin(x) wich we denote by I(p). It is easy to evaluate integratingby> parts:>> I(p)=1/(p^2+1)>> Now the trick is to integrate I(p) from 0 to in?ity with respect to p> under the integral sign :>> int(0->oo) 1/(p^2+1) =pi/2Wonderful! That was just the kind of evaluation I was thinking of. Quite astandard trick, actually: Introduce an additional unknown parameter andsolve the more general problem.It's like one of them factoring algorithm where - starting from some largenumber n - the ?st step involves multiplying the number by some small?ed constant. I think it is the Number Field Sieve, but please correct meif I'm wrong.-Michael. === >> Someone asked a way to calculate integral from 0 to in?ity of sin(x) / x>> without using residues.Here's yet another derivation:1. integral from 0 to in?ity of sin(x) / x = limit as t->0 of sum from n=1 to in?ity of sin(nt)/n2. sum from n=1 to in?ity of sin(nt)/n = imaginary part of sum from n=1 to in?ity of exp(int)/n3. sum from n=1 to in?ity 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/2Taking the limit as t->0 gives the result:integral from 0 to in?ity 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 in?ite differentials with respect to t?Or any other formal de?ition on this?FredReading all these responses is very interesting, as I had no idea thatsmooth could mean different things. When I was in grad school inReal Analysis, I recall dealing with smooth functions which werede?ed to be those which had a continuous ?st derivative. Eachtime I ran into smooth thereafter, I just assumed it meant the samething.Jonathan HoyleGene 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>>de?ition 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 de?ition of smoothness for an implicitly>de?ed curve is the maximal degree of smoothness of a *non-singular*>parametrization of the curve [non-singular means the ?st 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>> de?ition 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 de?ition of smoothness for an implicitly>de?ed curve is the maximal degree of smoothness of a *non-singular*>parametrization of the curve [non-singular means the ?st derivative>is nowhere zero).As I pointed out in the paragraph you omitted. >John MitchellDavid 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 theeigenvector of C from A and B, respectively?I remember in Quantum Computation, if C is the tensor product (ordensity product, sorry, I am not sure) of two matrices A and B, theeigenvectors of Ccan be computed by the eigenvectors of A and eigenvectors of B.But how about the regular matrix product?Any ideas? thanks a lotroy === >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 bothA and B, I can sort of rebuild A and B, compute C, and then computeeigenvalues and eigenvectors. But without full information aboutthe eigenstructure of A and B, I don't see how you can get muchinformation about C. Think geometrically. Suppose you have a rubber sheet nailed at one pointto the table top. One person comes in and stretches the sheet, ?st in one direction (pulling both sides away from the nail) and then stretchesit in the perpendicular direction by a different amount. A second personthen comes in and does likewise, but with a whole new pair of directionsand a new pair of stretching factors. With complete informationabout the four factors and two directions, you can express how thecombination of their moves can be accomplished with just a singlepair of stretches. With incomplete information, there's not much youcan say in general. (Try a simple case: each person simply doubleslengths in a single direction and then stops. That's not enough informationto 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 matricestogether.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 donot know one of the three variables. Let's suppose we know C and wewant to ?d A and B. What would help is some information regardingthe distance between A and Bor A towards C or B towards C.If this information is not available the only thing we do know is thatA<= SQR(C) and B>=SQR(C)David === : > : > Random thoughts on creating a theory of sets prior to a theory of : > propositions and quanti?rs: : > : > Let's start with the empty set, 0, and logical identity, =, then we can : > de?e T, for true, by : > : > T =def 0 = 0 : > : > Let's de?e ordered pair a la Kuratowski, then we can de?e : > conjunction by : > Kuratowksy de?es 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 satis?s the ordered pair axiom because of : > the Axiom of Extensionality. Those axioms both seem to require the : > apparatus of ?st order logic to be formulated. I already said that. Why don't *I* rate a reply? : Well, we cannot de?e 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 de?e : {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 isanybody 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 claimthat you've presumed some set theory, if you STILL have NO wayof deciding whether z is or isn't *in* {x,y} ? === >> Kuratowksy de?es as {a,{a,b}}.>Just for the record:http://www-gap.dcs.st-and.ac.uk/~history/ Mathematicians/Kuratowski.htmlF. === This question arises from economics, but I'll give you the question?st, the context later:Suppose we have a set of n objects, and a binary relation S that is alinear ordering, ie irre?, asymmetric, complete, transitive. Isthere any way to use this relation to induce some soft of ordering onthe power set of our set of objects?This arises from an examination of equilibrium in the jungle: we haveN agents, and S is strength(1 is stronger than 2, etc.) To see ifthis equilibirum is coalition-proof we need some way to ordercoalitions 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 irre?, 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 preferencesover subsets. A good reference is Roth and Sotomayor'sbook on two-sided matching.Federico Echenique === > This question arises from economics, but I'll give you the question> ?st, the context later:> Suppose we have a set of n objects, and a binary relation S that is a> linear ordering, ie irre?, 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 (?ite) set with a sorted array containingexactly those elements in that set, and use the lexocographic(dictionary) order of those arrays. It will also be a linear orderwith a minimum element (the empty array) but no maximum element. === > This question arises from economics, but I'll give you the question> ?st, the context later:> Suppose we have a set of n objects, and a binary relation S that is a> linear ordering, ie irre?, 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 upof the symbols 0 and 1... 1 means this object is in the set 0 meansthis object is not in the set. But strings of 0s and 1s of length ncorrespond 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>> ?st, the context later:>> Suppose we have a set of n objects, and a binary relation S that is a>> linear ordering, ie irre?, 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> ?st, the context later:> Suppose we have a set of n objects, and a binary relation S that is a> linear ordering, ie irre?, 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 (?ite) 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 ( ?d 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>> ?st, the context later:>> Suppose we have a set of n objects, and a binary relation S that is a>> linear ordering, ie irre?, 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 (?ite) 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 ( ?d 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>> ?st, the context later:>> Suppose we have a set of n objects, and a binary relation S that is a>> linear ordering, ie irre?, 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 onthe bottom dogs. Even in US democracy, we see how well the rich top havecoaluded 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 linearlyindependent so that no sum of weights of one subset equals the sum ofweights of another subset. For example square roots of different primes.Then order subsets by sum of weights of members. The order of the subsetswill 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> ?st, the context later:> Suppose we have a set of n objects, and a binary relation S that is a> linear ordering, ie irre?, 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 de?ed 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>?st, the context later:>Suppose we have a set of n objects, and a binary relation S that is a>linear ordering, ie irre?, 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 bthe smallest element of B. Then we say that A is smaller than B if andonly 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 thestrength. Does a coalition involving strength 2 and 3 beat oneinvolving 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 youcan then order the coalitions; you don't ?st order the coalitions insome way and then use that order to ?ure out their relativestrengths. === === =========================== === === ===============It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === === =================================== === === ======Arturo Magidinmagidin@math.berkeley.edu===>> This question arises from economics, but I'll give you the question>> ?st, the context later:>> Suppose we have a set of n objects, and a binary relation S that is a>> linear ordering, ie irre?, 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}> This sets up a recursion process.>> This works because any total ordering on a ?ite set is 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> (2) Order the singletons by letting {a} < {b} if and only if a> (3) Assuming all subsets of fewer than k elements have been ordered,> 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}> But of the many ways to order P(S) in some way induced by the> 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 ?st order the coalitions in> some way and then use that order to ?ure out their relative> strengths.>Well, i was hoping there was some way to do it without getting any more speci? 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 > speci? 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 withan arbitrary order that is somehow consistent withthe order on the individual elements of S: Look at howpeople in two-sided matching theory do it (see my previous post). === I S M A I L9 19 13 1 9 12 = 63 Shadia works for the CBC and was seated at the CBC booth during today'sdowntown street fair on 2nd Ave. in Saskatoon, she was the ?st of 10people to provide stats today.63+ Dad 12 8 36 225/141 +749463+ Mom 1 5 46 121/244 +394563+ Bro 15 1 71 15/350 508063+ Bro 16 2 72 47/319 547763+ Bro 16 2 72 47/319 5477105 Shadia 10 8 76 223/143 7114the year than to the beginning of the year (the ?st 13 primes minus the?st 13 non-primes). Mom was born 123 days closer to the beginning of theyear than to the end of the year, or 3 times the 13th prime (41). Theparents were together born 39 (13+13+13) days closer to the beginning oftheir years than to the end of their years. The parents were born on days ofthe month adding to 13 and in months adding to 13. The parents were born inyears 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 41stnon-prime (Genesis 41 contains 57 verses) while 41 in turn is the 13thprime. Twins arrived on the 16th (Nehemiah with 13 chapters). The family wasborn on days of the year adding to 678 (6x113). The ?st and last kids wereborn on days of the year adding to 238 (the ?st 13 primes). The brotherswere born on days 15 and 47, together for 62 (the 13th prime plus the 13thnon-prime). Mom gave birth a total of 942 days after her birthdays (thenumber of verses in Bible Book 13). Mom was 9422 days old when she gavebirth 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 (the13th 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 daysafter 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 to78635, 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 inthe year, corresponding to Numbers 24 (15th non-prime), pretty as 141 is47+47+47 (3 times the 15th prime). The ?st of the kids was born on the15th. The twins arrived on the 47th day of the year (15th prime). Thebrothers were born in years adding to 215. Shadia's day, month and year ofbirth adds to 94 (twice the 15th prime). Her name adds to 105 (7x15 and isLeviticus 15). Her vowels add to 30 (2x15), consonants add to 75 (5x15). Thekids are together 1515 days closer in age than the parents. The parents wereon average born 47% into their years, their ?st kid arrived on the 25947thday of the century and then they got twins on the 47th day of the year. Thefamily was born on days of the century adding to 136911, they were born anaverage 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 the14th non-prime (22), it's the number of chapters in Bible Book 14. Thebrothers were together born 879 days closer to the beginning of their yearsthan 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, togetherfor 143, while Shadia was born with 143 days remaining in the year (43 isthe 14th prime). Shadia's name adds to 105 (1 through 14 and is the ?st 14primes minus the ?st 14 non-primes). Her initials average 14. Her initialsadd to the 28 (14+14) chapters of Bible Book 14.The ?st 4 primes plus the ?st4 non-primes add together for the36 chapters of Bible Book 4 Numbers:Primes Non-Primes 2 1 3 4 5 6 7 <-4th-> 8 -- -- 17 19The ?st 4 primes in primepositions add together for the36 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 -- 36Seven plus the 7th prime plus the7th non-prime adds together forthe 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 inNumbers). Dad was born in 36, there are 36 chapters in Numbers. Dad was bornin 36 (6x6), his average age when the kids were born amounts to 36.36 years.1-50 - Genesis51-90 - Exodus91-117 - Leviticus118-153 - Numbers154-187 - Deuteronomy188-211 - Joshua930-957 - Matthew958-973 - Mark974-997 - Luke998-1018 - John1019-1046 - Acts1047-1062 - 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 661062 <-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 61070 <-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 ?st name adds to 66.666...% ofher last name, she is the 6th of 6 family members. She was born with 143days remaining in the year, pretty as chapter 666 brings Ecclesiastes up to143 verses. Bible chapter 666 contains 29 verses and brings Ecclesiastes upto 143 verses, this 143 is the 109th non-prime while 109 in turn is the 29thprime, 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 prime104 <-77th non-prime 77 <-77---570 The Four 57'sGenesis 41 -> 41Leviticus 14 -> 104Judges 9 -> 220 <-I dreamt of 220 roofs blownJohn 11 -> 1008 off homes in the Dakotas ---- 1373 <-220th primeChapter 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 about666 while Isaiah contains 66 chapters):Daniel - 357 versesRevelation - 404 verses <-57 plus the 57th prime plus the 57th non-primeIsaiah - 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 atotal of 2.57 years after her birthdays. The brothers were together born1373 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 ondays of the year adding to 109 while Shadia was born on day 223 (57+57difference). 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'sconsonants add to 75, pretty as the 57's are at chapter numbers adding to75. Her name adds to 105, it's the 78th non-prime while 78 in turn is the57th non-prime (105 is the 57th non-prime in non-prime position). Mom wasborn on the 1st, corresponding to Genesis (there is number play involving 57in 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 daysold when the ?st of the kids was born. Mom's combined ages when she gavebirth amounts to 106.57 years. Shadia might marry Marcia and me, pretty asMarcia and Shadia are separated by 1457 days.187 Dar 17 2 57 48/317 00Daryl 60 Shawn 65 Kabatoff 62187 Marcia 6 8 80 219/147 8571Marcia 45 Veronica 87 Acevedo 55159 Shadia 10 8 76 223/143 7114Shadia 42 Acevedo-Kabatoff 55-62 Anyway, if you people think that you have the right to use my abusiveparents as tools and arrest and torture me, then I think that I should havethe right to ask women to marry me, or to marry Marcia and me, our lastnames add together for the 117 verses of Song of Solomon, it's the Bible'sBook of Love. The nubile sweety was born on the 6th and has a 6 lettered?st 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, greedor lust, or who limit their love for traditions and for people who abide bytheir traditions. You people have Egyptian penises on the roofs of yourchurches and lined city streets with representations of penises, and had metortured for years for saying so, others just sat back in silence while theywere doink this to me, and similarly you remain silent and compassionlessnow that the arrests and torture have ceased. You people spent millions ofdollars having me tortured, and then annually you spend billions on yourdecorated trees, I begged and begged for assistance to ?e country(they tortured me for years at the U of S) and you people are so cheap thatyou can't even offer to buy me a cookie when I bust my ass to show youevidence that your very name is a gift from God, you cheap, ?thy,compassionless assholes won't even spend 48 cents on a stamp so that youlife to show you evidence that your very name is a gift from God!!! All youare really good for is to have your stats posted on the usenet and be usedas 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, thenShadia's brothers are each goink to win themselves a shiny new Cadillac!!!Good luck and may God bless you!!!Daryl Shawn KabatoffBox 7134Saskatoon SaskatchewanCanadaS7K 4J1Isaiah 45:4, Ephesians 3:15 - God gives you your name!!!What a wonderful weddink there will beWhat a wonderful day for you and meChurch bells will chimeYou will be mineIn 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 ?st of 10>people to provide stats today.<<The following (courtesy of Waxy.org) is sort of an unof?ial FAQexplaining the psychotic nonsense posted to Usenet by Shawn DarylKabatoff AKA Dar, AKA Probababbilities. And now AKA marcia andme.WARNING: Read below before even thinking about responding to thistwit.http://www.waxy.org/archive/2002/05/21/dar_ kaba.shtml#000643Usenet has the tendency to provide a public forum for those who wouldnormally be scribbling in a closet. For example, take Daryl ShawnKabatoff. For the last few years, he's methodically gatheredstatistics from various sources, ranging from local newspaperobituary pages to the food court of the Saskatoon Midtown Plaza mall.With all the raw data he's collected, he's attempting to prove dailythat our full names are in mathematical harmony with our birthdays.about, starting with calculations related to their birthdate and fullnames, blending in whatever other personal information about theirfamily members, spouses, birthplace, and career he's been able tozealotry, and personal torment. I've never seen anything like it.With all the prime numbers, Fibonacci sequences and biblicalreferences, it's like reading the notebooks of Maximillian Cohen andJohn Nash combined. Unsurprisingly, several posts unfold to reveal ahistory 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.comDecember 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.caOct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.caposts have been removed from Google Groups archive)Selected Posts:Tessa Lynne SmithDastageer Sakhizai and Helen SmithBrett David MakiAndrew Meredith CottonAmanda Dawn NewtonMona Marie EtcheverryTony Peter NusplLisa Charlene McMillanGrant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my currentresidence. and every single place he's mentioned in his posts (mostnotably nervous harold's and the roastary) were either places i'vebeen (as it's a small city of 200K) or hangouts, ie. the two placesout if they know of him, they (my friends that is) being of thebroadway-centred slacker ilk. myself, too, until i got out of there.eh, anyways. thought it odd to see all this. midtown mall. i ate mymeals there, whilst waiting several days in line for star wars episodeone, at the theatre across the street.posted by andy raad on May 22, 2002 06:20 PMFascinating. It's like he's trying to take chaos and bind it intowhatever rules he can ?d, religious, logical and otherwise. Numbersand math have a reliable pattern, something that can always be provento true or false. People and religion do not. It reminds me of DarrenAronofsky's movie Pi. It's the story of an paraniod genius who istrying to ?d a pattern in Pi. A group that takes interest in hiswork is convinced that the existence of Pi, a number whose existencecan be proven but no quanti?d, is proof of the existence of God.Kabatoff's hunt for patterns in something as random as name selectionis a way to reconcile his deeply logical thought process with hiscon?g religious views.posted by matt on May 23, 2002 11:19 AMasking him if he'd be willing to create a numerological analysis forme. I also asked him if he had seen either Pi or A Beautiful Mind, andwhat he thought of them. If he replies, I'll be sure to post it.posted by Andy Baio on May 23, 2002 11:24 AMI baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbedpumpkin 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 PMUm, that's swell. So, you're in love with him?posted by Andy Baio on July 25, 2002 07:10 AMShawn once went to a funeral for a Jehovah Witness that shot himselfand the lemon tarts were very bad, they were not only sour but wererubbery as well. Shawn said that the guy was some kind of JehovahWitness prophet, he saw in advance that the lemon tarts at his funeralwere to be very very bad, and so he shot himself. Shawn said that henever ate pumpkin pie at a funeral but would like to some day. Shawnlikes pumpkin pie and so I have been practicing to make very goodpumpkin pies.posted by Trisha Blondie on July 25, 2002 02:49 PMShawn said that the lemon tarts were sour, bitter and rubbery.posted by Trisha Blondie on July 30, 2002 12:32 AMI don't think this guy takes notes. I think he has Total Recall, andit has driven him insane...posted by Todd Smith on December 26, 2002 11:00 AMOh... I almost forgot... I didnt spend thousands of dollars a daytormenting Daryl... We got a deal on tormenting that ?cal year, itonly came to about 37cents a day....posted by Dr Claw on December 30, 2002 01:56 AMMr. Kabatoff attempts to portray himself as a victim, but in fact heis a violent predatory pedophile who is well known to his local lawenforcement. In his post to multiple newsgroups with the subjectCollecting Mail For The Coming Anti-Christ, he encourages mothers tosend him photos of their naked daughters. Mr Kabatoff explains, IAnt-Christ) that were of underage children unless the parent wassigning consent. He is banned from virtually all the shopping mallsin his community because he stalks young people and sexually harassesthem. He has an extensive arrest record which includes sexualmolestation charges. He's been hospitalized in mental institutionsabout his contact with young girls in many posts. Search newsgrouparchives for posts by him containing the word nubile. As part of hisharrassment, he provides personal details in a public forum, such asthe real names of real children, in these and other posts. About onewanted 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&rnuHe not only curses children and prays for their death in his posts, healso enjoys attending the funerals of young people: And so, sincenubile sweeties are found in greatest abundance at the funerals ofhigh school students, then it is the funerals of high school studentsthat 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 atall the pretty nubile sweeties and have the time of my life...http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff +nubile+sex&hl=en&lr=&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 liberallyspams, ?and crossposts his off-topic threatening and offensivemissives to countless newsgroups. Some people HAVE problems and somefolks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. Whenhe sends these posts to any newgroup, please help by reporting him toI 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 ofnumbers racing by on his laptop. I have an original copy of hisCollecting Mail for the Coming Anti-Christ pamphlet, and have seenhim be hauled away by campus security on more than one occasion. Myfriends and I refer to him as Crazy Number Man.I've been posting to (and about) Shawn for over two years with biggaps in between. He has seen Pi and didn't like it and didn't think itresembled him at all. (Wrong, it ?s him to a tee) He doesn't havetotal recall and has stated that he travels with a lap top to notateitems. Also, he uses cut n' paste a lot if you read all the waythrough his ramblings. He is anti-social as shown by his angrystatements towards those who, by his own admission, have been kind(but not kind enough) to him. Still, he's intelligent and seems to beable 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 16504R I Z Z U T O18 9 26 26 21 20 15 = 135 I am posting Vito Rizzuto's stats again because Shriner/Freemason DonOcean and/or his friend James Takayama is repeatedly calling me a pedophile.I have never sexually assaulted anybody, I have never been charged withsexual assault, I have never been arrested for sexual assault, and I am notsexually attracted to children. In 1988 I criticized the phallic worship inthe Protestant and Catholic churches and was repeatedly arrested andtortured at the U of S for years. Ruby would have me arrested for failing tokiss her God-damned ass and wear the clothes she was always trying to forceupon me (and because Protestants and Catholics were upset with my words andlobbied her to shut me up), Ruby would have me arrested and chemicallylobotomized, and then she would come to the psychiatric ward and force herchoice of clothes upon me there. Eventually I found myself internet accessat the U of S and posted Collecting Mail For The Coming Anti-Christ, inthe 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 ifthey wanted. Now Don Ocean calls me a pedophile, and as a result of DonOcean, James Takayama in Hawaii has begun doing the same. First the fellowlibels me in the WaxyOrg website using the name of Nospam, then using thename of Thomas (aka MauiCop), James Takayama quotes his own material postedas Nospam. Now another website has begun to quote the libel as beingtruthful, saying that I have been repeatedly arrested for sexual assault andthat I was a pedophile (and it is possible that James Takayama isresponsible for this website and may start others where he will claim I havebeen repeatedly arrested for sexual assault). Now as a result, people inSaskatoon are calling me a pedophile and are threatening to beat me up andkill me. The Saskatoon police are not in the habit of charging people forassault when they give me beatings (in fact they arrested me after James DeWitt brutally assaulted me at the Seventh Day Adventist Church, I wasbrutally assaulted and then the police arrested me and took me to the U of Srelievers when these people crack and break my ribs. The situation is quiteunfair, and now Vito Rizzuto will have his stats posted daily until thesituation is resolved. I get tortured year after year and begged people infutility for assistance to ? c ountry, now watch Vito spend huge pilesof cash again this year turning trees into decorated idols, for hiscompassion is limited to traditions.203 NicoloNicolo 68 Rizzuto 135225 LibertinaLibertina 90 Rizzuto 135201 Vito 21 2 46 52/313 +4014Vito 66 Rizzuto 135177 MariaMaria 42 Rizzuto 135 Mom's ?st name adds to 90 (66th non-prime), Exodus contains 1213 verses(66+66+66th prime) and terminates at chapter 90 (66th non-prime). Vito addsto 66, mom and the little sister have ?st names averaging 66. The kidshave ?st names adding together for 108 (the ?st 6 primes in primepositions). The kids have ?st names differing in value by the 24 chaptersof Bible Books 6 and 10 (6th non-prime). Dad has a 6 lettered ?st name,there are 24 letters in all the ?st names, the number of chapters in Books6 and 10 (6th non-prime). All ?st names add together for 266. Dad and Vitohave ?st names adding together for 134, corresponding to Numbers 17 with13 verses (the 6th prime). Mom and Vito have ?st names averaging 78 (6times the 6th prime). Dad and Vito have full names adding together for the404 verses of Bible Book 66, Revelation. Vito was born on the 21st,corresponding to Ecclesiastes (chapters 660 to 671). Vito has consonantsadding 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 2310 2911 31 <- 3112 3713 41 <- 41 --- 108 Vito Rizzuto (135) was hot for Cammalleni (83), the names differ in valueby 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 23rdprime, together for 46, and she gets a husband that was born in 46.166 Giovanna 48Giovanna 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 SecondTimothy (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, Vitoand 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 Book13 (the 6th prime), there are 29 (6+6p+6np) verses in chapter 666(Ecclesiastes 7). At birth Vito and Giovanna had 29 letters adding togetherfor 367 (First Chronicles 29). Vito's sister's ?st name adds to 42 (the29th 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 nameadds to 219, or 3 times the 73 verses of Bible Book 29. Libertina was bornin 73 (the length of Book 29 and is the Lucas numbers up to 29). Leonardoand Libertina have ?st names adding together for 174 (6x29). Dad and the?st two kids have ?st names adding together for 218 (twice the 29thprime). The males were born in years adding to 182 (Deuteronomy 29 with 29verses).201 Vito 21 2 46 52/313 +4014Vito 66 Rizzuto 135218 Giovanna 48Giovanna 83 Rizzuto 135203 Nicolo 67Nicolo 68 Rizzuto 135219 Leonardo 69Leonardo 84 Rizzuto 135225 Libertina 73Libertina 90 Rizzuto 135Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29J O E L <-Bible Book 2910 15 5 12 = 42 <-29th non-primeC O P P E R <-29th element3 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 pieceC E N T <-made out of 29th element3 5 14 20 = 42 <-29th non-prime The have three kids, the ?st and last of the kids bear Vito's parent'snames, so the daughter's ?st name adds to 90 (66th non-prime). The kidshave ?st names adding together for 242 (First Samuel 6), all names in thefamily add together for 1066. The kids were born in 67, 69 and 73, these arethe 19th prime, 50th non-prime and the 21st prime, together for 90 (66thnon-prime and the value of the daughter's ?st name). Mom and her sons have?st names adding together for 235... the 184th prime (1097) and the 184thnon-prime (235) averages 666. The brothers have names averaging 211, it isapproximately 66.6% of the 66th prime (317) and is the terminating chapterof Bible Book 6. Book 6 chapter 6 (193) plus the terminating chapter of Book6 (211) adds together for the 404 verses of Bible Book 66. Daughter's ?stname not only adds to 90 (66th non-prime), but her ?st name adds to66.666...% of her last name. This is a family of 5, the 4014 days dad isolder 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 ?st name adds to 48.888...% ofhis last name, God gives him a wife that was born in 48. Vito and his wifehave ?st 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 149verses of Bible Book 48. Rizzuto (135) was hot for Cammalleni (83), the 83value of mom's maiden name is 61.48% of the 135 value of Rizzuto. The maleswere born in years adding to 182, the females in years adding to 121(66.48%). The kids are missing 15 letters from their ?st names, thesemissing letters add to 248. The 4014 days dad is older than me is 18 timesthe 48th prime (223), prettier as I was born on the 48th day of the year andwith 317 days remaining in the year (66th prime).1-50 - Genesis51-90 - Exodus91-117 - Leviticus118-153 - Numbers154-187 - Deuteronomy188-211 - Joshua930-957 - Matthew958-973 - Mark974-997 - Luke998-1018 - John1019-1046 - Acts1047-1062 - 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 661062 <-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 61070 <-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), hismom and daughter share the same 225 valued name (Judges 14). The kids haveodd valued letters in their ?st names adding together for 140. The lettersthat 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?st names add to 391 (314th non-prime). Dad's name adds to 67+67+67. The ?st of the kids was born in 67 and hasnames differing in value by 67. These ?st three family members have ?stnames 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 yearsadding to 136 (Numbers 19). Vito and the last of the kids were born in yearsadding to 119. Perhaps mom was 19 years old when she gave birth in 67 (the19th prime). Mom's name adds to twice 109 (Leviticus 19) while dad's nameadds to 201 (3 times the 19th prime). The parents have names adding togetherfor 419 (Nehemiah 6 with 19 verses). Vito and his kids were born in yearsadding to 255 (First Samuel 19). Vito and his kids have ?st namesaveraging 77 (the primes up to 19). The kids were born in years adding to5909 (19x311). Libertina's last name adds to 150% of her ?st name (150chapters in Bible Book 19). Vito was born a multiple of 19 days into thecentury (16853=19x887). Vito's vowels add to 69 (Exodus 19). The 2460 versesof Bible Book 19 is19x19+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 21years old when the ?st of the kids was born. Vito was born in 46, his 201 valued name exceeds it's 155th non-primeposition by 46. The kids have prime and square valued letters in their ?stnames 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 weretogether born 530 days closer to the beginning of our years than to the endof our years (Psalm 52). This 530 is a combination of the ?st threeperfect numbers (6, 28 and 496), they have factors that add to formthemselves:Perfects 6 - 1, 2, 3 28 - 1, 2, 4, 7, 14496 - 1, 2, 4, 8, 16, 31, 62, 124, 248 The males have ?st names adding together for 218, and mom's name addsto 218. Mom was born with 18 letters and took a last name that begins withthe 18th letter of the alphabet, it is a last name that adds to 135 (Numbers18). Vito and his sister have ?st names adding together for 108 (Leviticus18). Rizzuto adds to 135 (the 103rd non-prime), the 11 different lettersutilized in the construction of the ?st names for the kids add togetherfor 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 and149 verses. These Books contain an average of 293 verses while the 437 is293.28% 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 ?stnames, and pretty that the last name would add to 135. Mom and her sons have?st names adding together for 235. The brothers have ?st names addingtogether 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 Booksof end-times prophecy are Daniel with 357 verses and Revelation with 404verses, 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 ?st names adding to 66 and 42, togetherthese Bible Books contain 1555 verses. Vito has 11 letters and a ?st nameadding to a multiple of 11, his kids have ?st names adding together for11x11+11x11 (11 is the 5th prime). Then Vito marries Giovanna (83)Cammalleni (83), there are 83 verses in Bible Book 55, and she soon ?dsherself in a family of 5. The 5th of 5 family members has a name adding to225, prettier as 25 is not only 5x5 but is 5 plus the 5th prime (11) plusthe 5th non-prime (9), keeping in mind that Bible Book 5 contains 959verses. Perhaps mom was 25 (5x5 or 5+5p+5np) years old when she gave birthto the 5th of 5 family members. Libertina's ?st name adds to a multiple of5 and also a multiple of 9 (5th non-prime). Libertina's (the 5th of 5 familymembers) ?st 5 letters add together for dad's 46th year of birth. My nameadds to 187 (the terminating chapter of Bible Book 5), it is 5x5x5 plustwice the 5th prime in prime position (31). If Libertina took my 62 (twicethe 5th prime in prime position) valued last name, then her names would havean average value of 76 (55th non-prime). Lamentations is Bible Book 25 with154 verses, the 154th prime is 887 while Vito was born on the 19x887th dayof the century, pretty because if Libertina married me then he would belamenting. Vito and his kids already have ?st names adding together for308, or twice the 154 verses of Lamentations. Note that Bible Books 5 and5x5 differ in length by 805 verses (the 666th non-prime). And Old TestamentBook 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 2310 2911 31 <- 3112 3713 41 <- 4114 4315 4716 5317 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 ?st 17 primes minus the ?st 17non-primes). Vito and Giovanna have names differing in value by 17 (7thprime, the primes up to 7 add to 17). Giovanna's name adds to 218 (Book 7chapter 7). Vito and Giovanna were born with last names adding together for218 (Book 7 chapter 7). The ?st of the kids gets a ?st name adding to amultiple of 17 and he was born in 67 (Exodus 17). The second gets a ?stname adding to 7 times the 7th non-prime (12). The brothers have ?st namesadding 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 toExodus 17, 19 and 23, together for 59 (the 17th prime). Libertina's namesdiffer in value by 45 (45 chapters contain the length of 17 verses), shemight 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 frieshigh up into the air, the french fries would fall to the ground and getdirty, and nobody would ever eat french fries again. And I wanted 59 bignosed Greeks with their big noses throwing hamburglers at us from across thestreet. And I wanted American Noel Nibblett blowing his bugle and leading amarching regiment of cadets up and down the street while American Don Oceanand his Shriner friends ride circles around them on their littlemotorscooters. Scientists have recently discovered that tomatoes containproperties that help to prevent prostate cancer, in men, and since A&Wallows one to take as much ketchup as they wanted, the Great A&W Rootbearwas on the fast track to becoming an international symbol of health andfertility... so of course I wanted the Great A&W Rootbear to be the best manat my weddink. My weddink was soon approaching and it was goink to be aglorious affair, probababbly.187 Dar 17 2 57 48/317 00Daryl 60 Shawn 65 Kabatoff 62187 Marcia 6 8 80 219/147 8571Marcia 45 Veronica 87 Acevedo 55207 Libertina 73Libertina 90 Acevedo-Kabatoff 55-62 Anyway, if you people think that you have the right to use my abusiveparents as tools and arrest and torture me, then I think that I should havethe right to ask women to marry me, or to marry Marcia and me, our lastnames add together for the 117 verses of Song of Solomon, it's the Bible'sBook of Love. The nubile sweety was born on the 6th and has a 6 lettered?st 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, greedor lust, or who limit their love for traditions and for people who abide bytheir traditions. You people have Egyptian penises on the roofs of yourchurches and lined city streets with representations of penises, and had metortured for years for saying so, others just sat back in silence while theywere doing this to me, and similarly you remain silent and compassionlessnow that the arrests and torture have ceased. You people spent millions ofdollars having me tortured, and then annually you spend billions on yourdecorated trees, I begged and begged for assistance to ?e country(they tortured me for years at the U of S) and you people are so cheap thatyou can't even offer to buy me a cookie when I bust my ass to show youevidence that your very name is a gift from God!!! Should Libertina marryme, great, but if Marcia marries me and then Libertina marries Marcia andme, then Nicolo and Leonardo are both goink to win themselves a shiny newCadillac!!! And if Libertina turns out to be some sort of sexual acrobatthen Vito and Giovanna are both goink to win themselves a shiny new Cadillactoo. Good luck and may God bless you!!!Daryl Shawn KabatoffBox 7134Saskatoon SaskatchewanCanadaS7K 4J1Isaiah 45:4, Ephesians 3:15 - God gives you your name!!!What a wonderful weddink there will beWhat a wonderful day for you and meChurch bells will chimeYou will be mineIn 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 followingpeople: Daniel Bruner, Angel Cadwell, Brittanie Cecil, Carl Koopang, JayLarson, Brian Lott, Adam Millikan, Cody Milliken, Christopher Ridsdale, VitoRizzuto, Melissa Schultz, Erin Sorenson, Ann Wigdahl, Michael Winkler, ormaybe more (more or less). === > Friday April 26th 2002 116/249 16504R 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 ?st 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 allthe 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 metabelianc) show that subgroups of metabelian groups are metabelianI don't have clue for itcan 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 thethings 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 theform p iff q you must prove 1) if p then q; and 2) if q then p.I will concentrate on just the ?st statement: if p then q.That is, I will try to prove: Claim: If G is metabelian then G''= 1The 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?ure out the proof in the form I am giving here. I know thata direct proof will have the above structure, and I just scribbledown the steps that I know will be needed. I think the explanationwill be easier to follow if you see how the things I am recommendingto do will ? into the ?al 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. QEDAs someone previously mentioned, you should write downthe facts that are given in more detail by unwindingthe de?itions. Thus, you would have: Claim: If G is metabelian then G''= 1 Proof: Suppose G is metabelian. Thus, I can ?d 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. QEDAlthough it is not necessary, I would like to introduce simplenames for the two commutator subgroups G' and G'', where Ilet H = G' and K = H'. Thus, I have: Claim: If G is metabelian then G''= 1 Proof: Suppose G is metabelian. Thus, I can ?d 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. QEDUp to now, all this is pretty mechanical. Now, you have to dosome thinking. But, at least you have the available informationspelled 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' isthe commutator subgroup of H which is the group generated byall the commutators formed from elements of H. Note thatyou don't have a simple description of K or H', since allyou have all its generators. But, you want to prove thatH' = 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 de?ed to be the subgroup generatedby 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 ?ith anuclear device; my suggestion is to just skip it, but I'm including itjust because I can...): Show that a group G is metabelian if and onlyif it satis?s 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 thesense of general algebra), and therefore is closed under bothsubgroups and quotients. Then show that the verbal subgroupcorresponding 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 Magidinmagidin@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 commutatorsubgroup of a group is a trivial group if and only if the originalgroup 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 itName everything in sight (G is a metabelian group, A is an abeliannormal subgroup of G, h: G->H is a homomorphism onto a group H, ...)and then consider that you are trying to ?d a certain thing which, after all, if it exists must depend on the things you havenamed--and if it can be shown to exist presumably depends on thosethings 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 shouldbe replaced by chosen by the teacher, or maybe chosen by Naturebut 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 de?e 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 ?ite ordinals and ?ite 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 thatincludes intension. Nor did I mean to include intension when Imentioned the property of being a model of [the natural numbers].When I wanted to distinguish between a mathematical object and theproperty of being a model of it, the distinction I wanted to draw hadnothing to do with the extension-intension distinction.As you state, ?ite ordinals and ?ite cardinals are the same inextension (I am not sure if one can say without reservation that theyare different in intension).To the list above of questions whose answers I am seeking may beadded:3. What might we mean by intensional object?However, I do not see any reason why knowing the answer of thisquestion should be a prerequsite for knowing the answers of the othertwo.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 ?elitestatus'. Besides, entire theories should be just as weightless as asingle 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 thereabout my giving it a try to settle BTP. (I can assure you that this isnot something worth my taking to the grave with me.) === Ich spreche Deutsche nicht so schlecht. Wie angegeben in anderen Post, Zsteht 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 SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Lerne dem Kill?e zu nuetzen, als weil versuchen dem Mumia konservieren.Troll, in some cases, is what some people who'd like to run usenet butfortunately don't call those whose opinions they cannot comment on, but makethemselves look really smart by saying I'm so above it. Troll heisst einZwerg, in diesem Fall - ein Giftzwerg, ob Sie hatte nichts gewuesst bisheute.> 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 Kill?e 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 Zfor the integers derives from Zahlen.*Plonk*-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === >> Besides, to comment on you> awareness, what languages do you ?y read in except english ?>> You presume that I am ?in English? === >> Besides, to comment on you> awareness, what languages do you ?y read in except english ?>> You presume that I am ?in English? === One knows that E_8 is the smallest example of an even unimodularlattice, and the Leech lattice (dim=24) is the smallest example of aunimodular lattice with minimum norm = 4.Are there interesting unimodular lattices with minimum norm = 6 (orlarger)? 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 instanceM. Harada, Extremal odd unimodular lattices in dimensionsC. 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.htmlNeedless 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 oddnumber, and 2pr+1 is prime, then is2pr+1 | (p^r) + 1Ever 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 is2pr+1 | (p^r) + 1Ever 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 positiveodd> 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 --> Rthat is C in?ity, i.e., its n-th derivative exists for every n=0,1,2,...The condition on F is that it has to satisfyF(x) = 0 for norm(x) <= rF(x) = 1 for norm(x) => Rwhere 0 < r < RMy problem is that I remember (probably I'm very wrong), from complex analysis, that there is no holomorphic function that satis?s 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 --> Rthat is C in?ity, i.e., its n-th derivative exists for every n=0,1,2,...The condition on F is that it has to satisfyF(x) = 0 for norm(x) <= r> F(x) = 1 for norm(x) => Rwhere 0 < r < RMy problem is that I remember (probably I'm very wrong), from complex > analysis, that there is no holomorphic function that satis?s 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.In?itely 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 in?itely 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 ?ling 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 ? 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 obviouslyfalse, S must be empty or the entire domain. If f vanishes on an openset 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 ?ure out what sort of characterization we're looking for.I mean the de?ition 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.)>>DavidDavid 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 ?ure out what sort of characterization we're looking for.> I mean the de?ition 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 de?ition. As I said, the chance of ?ding one doesn't seem too likely - real analysis isn't as friendly as complex - but I ?ured 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 ?ure out what sort of characterization we're looking for.>> I mean the de?ition 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) whichcan 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 de?ition. 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, iftheir domain is connected, then two distinct analytic functions cannothave 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 --> Rthat is C in?ity, i.e., its n-th derivative exists for every n=0,1,2,...The condition on F is that it has to satisfyF(x) = 0 for norm(x) <= r> F(x) = 1 for norm(x) => Rwhere 0 < r < RMy problem is that I remember (probably I'm very wrong), from complex > analysis, that there is no holomorphic function that satis?s 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-in?ity function B(x) for which B(x) = 0 for x <= a B(x) > 0 for a < x < b B(x) = 0 for x >= bNote that this function q(x) (the standard C-in?ity functionthat has all zero derivatives at x = 0): q(x) = exp(-1/x^2) for |x|>0, q(0) = 0can be used to fashion B(x) as follows:Let q1(x) = 0 for x <= a q(x-a) for x > aand q2(x) = q(x-b) for x <= b 0 for x >= bThen q1 and q2 are C-in?ity functions with support(q1) = {x | x >= a} support(q2) = {x | x <= b}Let B(x) = q1(x)q2(x), and notice that B(x) hassupport 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 -in?ity to x. Note that s(x) is identically 0for x <= a, and takes a positive constant value for x >= b.That constant can be adjusted to your favorite value bymultiplying s(x) by the appropriate factor.Your function F can be produced by the appropriate selectionof 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 in?ity, 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 satis?s 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-in?ity function B(x) for which B(x) = 0 for x <= a> B(x) > 0 for a < x < b> B(x) = 0 for x >= bNote that this function q(x) (the standard C-in?ity function> that has all zero derivatives at x = 0): q(x) = exp(-1/x^2) for |x|>0,> q(0) = 0can be used to fashion B(x) as follows:Let q1(x) = 0 for x <= a> q(x-a) for x > aand q2(x) = q(x-b) for x <= b> 0 for x >= bThen q1 and q2 are C-in?ity 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 -in?ity 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 thatx! = 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 appoachesother limits other than real positive in?ity?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, andwould not use it to investigate the behavior of Gamma(x+1) at x verynear negative integers.-So, what, for example, about the asymptotics if x = 1/2 + i y, where yapproaches in?ity?Or how about if x approaches any other interesting ?ite/in?itecomplex/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 in?ity?>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 in?ity?Sterling's formula with remainder is valid; I am unsure as tohow good it is for the imaginary part of the logarithm. Thereal part of the logarithm follows from the multiplication identity above.>Or how about if x approaches any other interesting ?ite/in?ite>complex/real constants?A very useful formula islog(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 viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === > Most of us know thatx! = 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 in?ity?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 in?ity?Or how about if x approaches any other interesting ?ite/in?ite> complex/real constants?You know of course the classiclog(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 truefor |arg(z)| < pi.HTH,Michael.-- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&&Dr. Michael UlmFB Mathematik, Universitaet Rostockmichael.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) isBurnside'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 in?ity?>> 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 beGamma(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 whichI gave above.David Cantrell === > Most of us know thatx! = 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 in?ity?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 in?ity?Or how about if x approaches any other interesting ?ite/in?ite> complex/real constants?> thanks,> Leroy Quet === Background: I am a 48-year old adjunct partial-load professor employed in a localcollege and very happy teaching on contract. My quali?ations arefrom the techy/industry side, i.e. engineer/technologist associationexams, 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, ?ances, and marriage, are nowsecure, hence, its time for *me*!It has always been my dream to publish papers, write text books,supervise courses, and generally in? education in applied mathareas. My area of expertise is instrumentation and control, requiringLaplace, differential equations, complex numbers, matrices, etc.In order to achieve these goals and to enhance my knowledge,quali?ations, and employability, I am looking for an online,correspondence, or distance Bachelor's degree which can lead to aMaster's Degree or Ph.D.Q1. Has anybody out there done this?Q2. Will a math degree of this type be considered acceptable to teachin university?Q3. Will I be quali?d to teach engineering students?Q4. Will publishers publish?George === ..is newly available for free download athttp://www.tinaja.com/glib/msquant.pdfSourcecode is provided as http://www.tinaja.com/glib/msquant.pslIt 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 digitalsinewaves with arbitrarily low distortion at the highest possible energyef?iency.Other GuruGrams at http://www.tinaja.com/gurgrm01.asp-- Many thanks,Don LancasterSynergetics 3860 West First Street Box 809 Thatcher, AZ 85552Please 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/mAscii-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) de?ition.What is this closed-form for {a(m,n)}?I will ?st 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 QuetFirst, every a(m,n) is a positive integer form >=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 thatan inductive proof is just the reverseof 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 ?ding 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 de?es a C^in?ity manifold to be a second countableHausdorff space M with a C^in?ity differentiable structure, thatbeing an assignment D:U|->D(U), for all open U contained in M, suchthat(i) for each open U contained in M, D(U) is an algebra ofcomplex-valued functions on U containing 1(ii) if V, U are open, V is contained in U, and f is in D(U), then therestriction of f to V is in D(V); if V_i (i in J) are open, V is theunion of the V_i, and f is a complex-valued function de?ed on V suchthat the restriction of f to V_i is in D(V_i) for all i in J, then ? in D(V)(iii) there exists an integer m>0 with the following property; for anyx in M, one can ?d an open set U containing x, and m real functionsx_1, ..., x_m from D(U) such that (a) the mapxi:y|->(x_1(y),...,x_m(y)) is a homoeomorphism of U onto an opensubset of R^m, and (b) for any open set V contained in U and anycomplex-valued function f de?ed on V, f is in D(V) if and only if fcircle xi^-1 is a C^in?ity function on xi[V].He then remarks that M is obviously locally connected andmetrizable.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 de?es a C^in?ity manifold to be a second countable> Hausdorff space M with a C^in?ity 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 de?ed 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 ?d 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 de?ed on V, f is in D(V) if and only if f> circle xi^-1 is a C^in?ity 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 de?es a C^in?ity manifold to be a second countable> Hausdorff space M with a C^in?ity 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 de?ed 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 ?d 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 de?ed on V, f is in D(V) if and only if f> circle xi^-1 is a C^in?ity 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 + Chas 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 Ican't ?ure out what to put as my limits of integration. Anyone havea 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 ?ure 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, andC. 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 beinterpreted geometrically as the volume of the domain of integration. Thelatter is some section of a hyperboloid, but of course bounded by theadditional constraints that all three variables are between 0 and 1.Forgetting about the geometrics and just doing the integration is prettystraightforward, and I get the ?al 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 ?ure 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 ?al value (5+6*ln 2) / 36 = 0.254413...-Michael.Well, back to the geometry, what did you use for your limits o?tegration on each of the variables A, B, and C?Also, does anyone have handy the discriminant of a cubic? Maybe wecan 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 AC.> 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 ?al 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. Theintegration limits are 04*A*C, i.e. B > 2 * sqrt(A*C). I ?st did the integralover B. Holding A and C ?ed there are two cases to consider: A*C < 1/4 andA*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 integralis over the interval from 0 to 1, and hence has the value 1. Next Ieliminated the C integral, and - holding A ?ed - once again had toconsider two cases: A < 1/4 and A > 1/4. Are you able to do the rest fromhere?> 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 inthe interval [0,1]? Why not standard deviation, say, or some otherdistribution?-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 ?al 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 ?st did the integral> over B. Holding A and C ?ed 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 ?ed - 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 ifthere was some noticeable pattern in the probabilities. I suppose asn tends to in?ity the probability of an arbitrary n-th degreepolynomial having n real roots tends to 0, but it would still requireproof. 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 ?al 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 ?ure 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 ?ure 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 askwhat is the probability that the roots have at most epsilondifference? === >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 ?ure 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 4,000,000trials, there were 1,018,340 successes for a p of 0.254585.With 99% con?ence, the true probability of success is between0.254024 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 ?ure 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% con?ence, the true probability of success is between> 0.254024 and 0.255146.DougWell, I should restate the problem and say that I want it to have twoREAL 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>=4ACB>=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 rootswill be COMPLEX, so the roots that they are real is 1-8/9 = 1/9. I'mnot sure if this is the test you ran, but it would be interesting torun mathematica again and see if it is close to 1/9 with the restatedproblem.Using a similar method except with integration in 1 variable, I ?dthat when A is ?ed to be 1, the probability is 1/12, which wouldmake sense since we don't allow that possibility that A reduces the4AC 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 ?ure 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% con?ence, 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, soI'm still under the impression that you should get a value withinthe above con?ence interval. In fact, all I checked for success was whether the discriminant waspositive.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 ?ure 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 tohave the roots be real ( a condition not explicitly stated), and distinct.However, it seems that letting the discriminant equal zero will give you 2repeated roots. So let me paraphrase your question and see if this is whatyou are saying: We let a,b,c be independent random variables distributed asunif(0,1), and we wish to ?d the probability that an a,b,c, are chosensuch that the polynomial ax^2+bx+c has 2 distinct real roots , i.e. thatb^2-4ac > 0. Is this a corect statement of your problem?MB === I'd like to know if there's any connectionbetween what the algebraic integers *should* include, andthe ever-shrinking Short Proof of Fermat's Last, altoughI realize that monsiuer Harris doesn't have the timeto 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 integersfor the 3 sides, 3 diagonals & one interior diagonalof a rectangular box. whether this simpler exclusionary problemthrows 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 analysisCare 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 haveto 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, butyou should do it the correct way. Neither in analysis, nor in algebra,is it immediately clear what the remainder is. But is is blatantlynonsense 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 anyvariable), 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 wouldbe a point. But when you state that the constant term ofa1(9)/sqrt(7) + sqrt(7) is sqrt(7) you do not use the technique in theway 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 constantterm 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, +31205924131home: 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 ?st 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 quali?ations> 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, ?ances, 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 in? 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,> quali?ations, 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 offeredat ? institute of technology in applied math or operationsresearch. now, if your real goal is to attain a full time position as acollege math teacher, you will have better chances if you get a math-eddegree - reason being that, with few exceptions, colleges care less aboutacademic expertise than they do about con-artistry abilities to please/appease students and administrators. math education programmes will doan 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 quali?d 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, speci?ally, 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 Satanisn't such a bad guy after all? Consider: he obeyed God's commandsconcerning how much he was allowed to af?ob. Sounds like an obedientson of God to me... === There... are... four... lights!-- /-- Joona Palaste (palaste@cc.helsinki.? ------------- Finland ---------- http://www.helsinki.?~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 writeyour name as a reference or there was another reference as a book forexample.Please let me know as soon as you can. === on the wayside, John,this is exactly the same sort of Diophontine problemas the FLT that James is bent on ?ling 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> ?d 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.htmhttp: //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 ?d nothing there about bum-fuzzling. Sorry.Yes, of course, all of the experts here are anxious over thestunning and powerful attacks of John Correy and James Harris,brilliant displays of logic and insight that threaten to bring downour charade. I don't know about the others, but I studied logic forthe great respect it demands. Free valet parking, starlets swooningat the feet, things like that. Of course we must band together todefend 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 todefend 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 inmaking sure mathematicians worldwide get ?ed. I've rethought mydesire to go to Congress and try to get funding for mathematicianscut. -- 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 ?d 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 mathiein spirit--and that's what counts!--John === === ============================== === === ============You concentrated on mathematics because it is predictable, becausethere is always a right answer you can check in the back of the book,because you like following very precise rules, because it allowsyou to escape from everyday life into a world that has nothing todo with everyday life, and because mathematics does not requirethe creativity that you completely lack.Keith Devlin, Commencement address to the mathematics graduating classof UC Berkeley, May 23, 1997A performance system is designed to work in a de?ed task domain,accepting particular goals and seeking to reach them by some kind ofhighly selective search. The system must be told what goal is to bereached and must be given a description of the structure andcharacteristics of the task domain in which it is to operate: itsproblem space. [...] In contrast, a learning system, is capable ofacquiring a problem space, in whole or part, by interacting with theexternal environment and without being instructed about it directly. A performance system is designed to work in a de?ed task domain,accepting particular goals and seeking to reach them by some kind ofhighly selective search. The system must be told what goal is to bereached and must be given a description of the structure and>characteristics of the task domain in which it is to operate: itsproblem space. [...] In contrast, a learning system, is capable ofacquiring a problem space, in whole or part, by interacting with theexternal environment and without being instructed about it directly.Herbert Simon === Consider the covering map p x p : R x R ---> S^1 x S^1where p is the standard covering map p : R --> S^1 by p(x) = (cos (2pix),sin (2pi x))Consider the path f(t) = (cos (2pi t), sin(2pi t)) x (cos (4pi t), sin(4pit))in S^1 x S^1. Sketch what f looks like when S^1 x S^1 is identi?d as adoughnut. Find a liftingf ? 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 identi?d as a> doughnut. Find a liftingf ? 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 theway that f looks on a doughnut, although it is easy to draw it withMathematica or Maple (and it is easy to imagine too). A lifting f' isde?ed asf'(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 generalrelations, whereas those who dispute his claims give exact, concretecounter-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 ?ld 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 ?ld. 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://mathforpro?.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 notbe true, but James comes up with an example where it istrue 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 aconstant too!) Now since x has been arbitrary, this means x = 0 for anyx! 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+7w2(x) = x^2+7w3(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)) = 22As can I (understand that I'm not claiming that these particularfunctions will work for this problem, I'm just pointing that thatit 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 ?ld 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. Clearlyf(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 itis 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'tassume that just because a1(0) = 2 and a2(0) = 1 that a1(x)=2 anda2(x)=1.And note that 1/2 doesn't exist in the ring, but that doesn't meanthat 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 polynomial49(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 havew_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 + 22where 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)) = 22which should be simple enough for all of you.Now those of you who usually work in the ?ld 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 algebraicinteger. 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-writeb_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 continueto insist (correctly) that 22/w_3(x) cannot be an algebraic integerand 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. Theimportant 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 yourbeloved 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 factoredin 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 algebraicintegers. 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*0in each case). Thus also when x = 0, the product of theconstant 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. Neitherare they *coprime* to 7; each has the factor 7^{2/3} incommon 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 sideequals 132 = Q(1)/49, as it should. It is worth noting also that the product of the threeconstant 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 coef?ients b_1, b_2, and c_3 are all functions of x; thevalues 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 dif?ult to compute. Note ?ally 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 termsas required are 7, 7, and 22. So where does the JSH logicbreak 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 ?ld. They're> pussies too scared to handle the truth.> Most of your loyal audience has probably noticed by now that youhave stopped trying to respond to the substantive parts of my posts. You pounce on some super?ial aspect of them anddelete 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://mathforpro?.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 ?ld 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-writeThat ignores the fact that(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22wouldn't be in the ring of algebraic integers, if w_3(x) shared anynon-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 + 22so you have to at least admit that your claims push you out of thering 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 multiplytogether to get 49. w3(x) is in general *not* coprime to 7, andw1(x) and w2(x) are in general *not* divisible by 7.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: 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 thering of algebraic integers. Therefore, one is not allowedto use it on the left hand side of that equality.He is not saying that 22 wouldn't be in the ringof 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 ?ld 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 haveno 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, soit can be divided by 2. Your reasoning about constant term wouldlead to the result P(x)/2 = (x + 1)(x/2 + 1), because now theconstant terms match. My position is that there are functions w1(x)and w2(x), de?ed 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 toyou it is not a valid factorisation, so the ring of integers is ? > So you see, my argument is correct and simple, and mathematicians are > indeed running from a little gut check in their ?ld. They're > pussies too scared to handle the truth.So, it is your thinking that the ring of integers is ?With your polynomial the situation is similar. I have given pretty*explicit* de?itions 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, +31205924131home: 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 ?ld 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 simpli?ation 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 + 22to (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22, which is done simply enough by multiplying everything out and lettingthe terms with x as a factor cancel each other out.Now you wish to avoid that Dik Winter because your assertions can't betrue in the ring of algebraic integers as if w_3(x) isn't coprime to7, 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 youkeep 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 allof the folks who are in the googolplex,that *could* lurk on your bifurcating threads -- ormaybe 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, andhighly 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 thinkthat the fact that 22/w_3(x) need not be in the ring is of greatimportance. Indeed, you seem to assume that it is obvious why22/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 thering 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 justmultiply 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 Iposted 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)) = 22which 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 HughesDenial of basic algebra is such a sad thing to display to the world asWilliam Hughes demonstrates a rather odd irrationality, to all thosereaders who go through my posts.After all, I'm just talking about multiplying out and simplifying, andI'll post again because these cranks have a bad habit of creativedeletion 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)) = 22which should be simple enough for all of you.You see, the cranks can't get arouund 22/w_3(x) NOT being in the ringof algebraic integers if w_3(x) isn't coprime to 22.And for those of you who wondered, yes, when faced with a *very* basicrefutation of their positions posters who argue with me tend to justignore the truth. Later they come back with the same position. They're irrational and are the real cranks as you can't reason withthem.They believe false things but want desperately to believe and bebelieved.James Harrishttp://mathforpro?.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 de?ed 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 thealgebraic numbers...)-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: 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 de?ed them for you get > w1(x)w2(x)w3(x) = 49, > so I do not understand how it is possible thatOf 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, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === [.snip.]>Well, you know, the w's as I have de?ed 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 Magidinmagidin@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 postedmany, 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. Andyou deleted out the factorization and simpli?ation which shows thatfact!!!Extraordinary behavior which is indeed crank.I give you the information, explain it clearly, and you try to justignore 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 operationsare to be in that ring. I show that you're pushed out of that ring ina 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 ofalgebraic 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 HughesWhat I did was multiply out the factorization and simplify to showthat it's impossible for w_3(x) to not be coprime to 7 as then22/w_3(x) is NOT in the ring of algebraic integers.It's simple, direct, and basic, and it refutes attempts at claimingthat w_3(x) can share non-unit factors with 7 in the ring of algebraicintegers.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 simpli?ation 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)) = 49or by direct use of the fact that (w_1(x))(w_2(x))(w_3(x)) = 49we obtain (7)(7)(22) / ((w_1(x))(w_2(x))(w_3(x))) = 22in the ring of algebraic integers. We can rewrite thisin the complex numbers as (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22but this no more shows that (22/w_3(x)) is in the ringof algebraic integers than (6)(35)(99)/((15)(18)(7)) =(6/15)(35/18)(99/7) = 11shows that (99/7) is an integer.You didn't attempt my quiz. Please try, It isn'thard. 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 ? he algebraic integers. The question is, whichpushing 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 de?ed above are functions from the algebraicintegers 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) = 1in the above factorisation, we ?d that *not* all three factors arealgebraic 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 tocalculate that intermediate result*, it is irrelevant.In general the case is that when you say about a function P(x) that itis a function from the ring of algebraic integers to the algebraicintegers, it is not necessary that all intermediate results must alsobe in that ring, as long as the ?al result is in that ring. Andas functions, all three factors with the w's as I de?ed them *are*functions from the algebraic integers to the algebraic integers. SoP(x) is factored in three functions that are from the algebraicintegers to the algebraic integers. No ? the algebraic integersin 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. Itis (5 b3(x) + 22)/w3(x). Or do you claim that it is invalid inthe 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, +31205924131home: 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 isalways an integer for integer value of x; you can certainly write(x^2+3x+2)/2 = (x^2/2) + (3x/2) + 1even 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 Magidinmagidin@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 ?ld 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 ringof integers is ?because it does not contain numbers that shouldbe in that ring. But you are afraid to answer the questions I posed inthe part you deleted. > Why? What I've done is do a basic simpli?ation 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 thequestions in the part you deleted. Because if 22/w3(x) must be inthe algebraic integers, for *exactly the same reasons* 1/2 must bein the integers (see P(x) = (x + 1)(x + 2), which is divisible by 2in 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, +31205924131home: 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.netStupid is as stupid does. There is no ?ing stupid because it is notbroken. Harris would be much happier as a priest proclaiming godHey Harris, your village called: Its idiot is missing.-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === >There is no ?ing 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 ?ure it out.-- Let us learn to dream, gentlemen, then perhaps we shall ?d 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 tryto use the bin distbtn in solving this problem. The only way I can see it isto say:Let X1 = number crashes in month 1X2 = number crashes month 2X = X1 + X2Then: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!)] = 2exp(-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 ?d 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 sphericalcase one gets Tan(gama) = geodesic curvature/normal curvature = d[r*Sin (si)]/dz for all small circles, no matter whether they lieparallel to the plane of equator, perpendicular to it or make someother arbitrary intermediate angle. When gama is zero as an important special case, we have the Clairaut's Law valid on great geodesiccircles. Note that normal curvature = 1/R , constant for anydirection.Small circles or lines of such equal latitude are also called equalslip lines in ?ament winding industry of composite pressurevessels. === > 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 theytravel 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=ringname.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~----- WantedKaimbridge (w/mugshot!): http://www.angel?e.com/ma2/digitology/Wanted_KMGC.html ----------DigitologyThe Grand Theory Of The Universe: http://www.angel?e.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=ringname.gif--ils duces d'Enron!http://www.benfranklinbooks.com/20th-century.htmhttp: //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=ringname.gifThey seem readable at this end--are your browser settings off? ~Kaimbridge~----- WantedKaimbridge (w/mugshot!): http://www.angel?e.com/ma2/digitology/Wanted_KMGC.html ----------DigitologyThe Grand Theory Of The Universe: http://www.angel?e.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=ringname.gif>>They seem readable at this end--are your browser settings off? On the outside chance the OP might not have noticed, recentMozilla and IE browsers will shrink large images to ? the screen, bydefault. While inside the image, Mozilla will change the cursor to amagnifying glass with a + sign. Click to enlarge to readable. Whilelarge, cursor will contain a - sign to shrink. IE will make the image small, then you have to hover yourcursor inside the lower right corner until the pillow appears. Clickit to enlarge, reverse to shrink. HTH. === I am looking for an english reference for the Campbell theorem for>> stationary ?tered 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, marksk_n (having identical distributions), let a ?tered point process de?edby: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)]dtAny idea about this?-- Karl Breitung === Are there any good ways to ?d the minimum andmaximum genera (genuses) of a graph?Let me de?e what I am looking for. Say thatwe can _cellularly embed_ a graph in a compactorientable surface S if you can embed G in S(treating G as a 1-manifold), and each connecteddomain of S-G (S minus G) is an open 2-cell(ie homotopic to a point).Intuitively, you are just drawing G on thesurface of S while disallowing edges (of G) tocross, and making sure that each face of Gdoes not contain any holes.Then gamma is a possible genera of G iff thereis some S with genus gamma so that G can becellularly embedded in S.For example, I think that K_4 has min genus 0and max genus 1. I think that K_5 has mingenus 1 and max genus either 2 or 3 (not sure).(Here by K_n I mean the complete graph withn vertices.) -Tyler === > Are there any good ways to ?d the minimum and> maximum genera (genuses) of a graph?Let me de?e 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 de?e 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 de?ite matrix A, one can ?d a uniquepositive de?ite matrix B with B^2 = A. Then the solutionsof U^t U = A are the matrices VB with V orthogonal. (One has to bea little more cunning if A is singular but nonnegative de?ite).-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless 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 de?ite matrix A, one can ?d a unique> positive de?ite 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 de?ite). === We know that the geometrical ?ure which displays the maximum numberof pairs of parallel lines (segments) when joining 5 corners is aregular pentagon, with this number, n = 5 ( 5 directions having --atleast-- 1 parallel).The maximum number of parallel relationships for 6 corners is anhexagon, with n = 6.But what if we try to ?d not just the maximum number of parallellines, but also rectangular lines (to another line)?I have found that for 5 points we reach the maximum number of paralleland perpendicular relationships with the square, whose corners give 4points, and the crossing point of the two diagonals when drawed on topof the square give the ?th point. It gives n = 4 and m = 2(perpendicular directions). The other ?ure for n = 4 and m = 2 thatI have found is a parallelogram of sides 1 and sqrt(3), with the ?thpoint also the intersection of the two diagonals.Someone could please provide me with some bibliographic reference onthat, and check the veracity of that assertion, i.e., that n and m arethe maximum for just 5 points in those 2 ?ures and that theredoesn't exist any more ?ure?I used basic analytic geometry, but based on drawings, so I am notsure if I missed some con?uration 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 de?itions 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 de?itions 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 ?d any with solution manuals.Others will recommend introductory analysis texts, probablyRudin, Principles of Mathematical Analysis.Personally, I ?d 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 wouldnumerically integrate this multivariate function (The functionis complicated and isn't really important, I think), but, thisthought occured to me, and I was wondering what everyone thoughthere: Say you have a function f(X) where X is a vector of ndimensions (terminology? but, in c parlance, I mean to say anarray of n doubles) anyway, you have a function f(X) which isde?ed in some way, and you can't ?d a closed form solutionto Integrate[f(X),X1,X2,...Xn] Would genetic algorithms be a good way of ?dingFUNCTIONS, that might approximate it on certain ranges, theway that for instance, I've seen an approximation to the NormalCDF 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 = 0.019527So, I could pass the GA my function and tell it that I expectto want to numerically integrate on ranges from -100,100 and itwould generate a function that would _approximate_ the integrationwithout actually doing the numerical integration. I could thenlook at the function, and decide what it's error characteristicswere, before plugging it into the actual app, etc.I'm expecting that the GA would be of the genetic programming typewhere the functions would include exp and tan, and all the usualmathematical operators.Has this been done? Does that seem like a reasonable app for aGA? I'm not stuck on the idea of generating the function via GA,so if others have different means to generate an approximatefunction given f that computes the integral of f, I'd be open to thatas 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.pubOpenSSH Key: http://www.hex21.com/~binesh/binesh-openssh.pubCipherKnight 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-certi?ate.gif.cs256 Decrypt with CipherSaber2 N=256, Password=WelcomeJedi! (No quotes)iD8DBQE/sKDVtC/nHH/DrZYRApfBAKDUWs+ YZHxwQA41SyB2enX4PjNG4wCeI1eZ3XiPsv2RGD3QhqkmLW/sxoY==QvvZ === =You could have a look at this site:http://www.gepsoft.com/gepsoft/I am currently using one of their programs.aprxBinesh 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>de?ed in some way, and you can't ?d a closed form solution>to Integrate[f(X),X1,X2,...Xn]>> Would genetic algorithms be a good way of ?ding>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-certi?ate.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 approachesin the thread; just saw someone note that the recurrence had apossibility of blowing up, so perhaps there is better accuracyhere...In addition to the recurrence approach, you can also use a markovprocess approach to this problem; this has come up before on sci.mathin the context of coin ?g.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 ?g.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 ?st 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. Itactually executes reasonably quickly but I appear to be getting a lotof rounding error problems.My ?st reaction was to check my results. I added an extra row andcolumn 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/columnand.500000000053831 via ProbnmpApprox.I'm pretty sure that .500000000053831 is accurate, but the roundingerrors via the matrix multiplication method seems a bit drastic.Can you supply the values you got, if possible it would be nice if youcould add the extra row to con?m whether the problem is genuine orin 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 dif?ult 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 coef?ients 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 AllenMy standard Conditions of use are No charges, no conditions and noguarantees. I've no idea if this is legal in the US but assuming itis, 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 ?ld.Also suppose that m,n free from k powers and K1=K2. Is there any relation of m,n?thankscostas. === > Given an open subset A of the unit square [0,1]x[0,1] with ?ed area area(A)> it is known that the ?st 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) ?ed? 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 ?ite number of results will be drawn from thisdistribution. Is there a mathematical formula for calculating theexpected range of these results? === > I have a normal distribution of known mean and standard deviation.For a certain case, a ?ite 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 theexpectation 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)^nSo 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 distributionof individual samples.The expectation value is thereforeintegral(-inf,inf) n*P_x(x)^(n-1)*p_x(x)*x dxSo 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 ?ite number of results will be drawn from this>distribution. Is there a mathematical formula for calculating the>expected range of these results?Google for con?ence interval. === >I have a normal distribution of known mean and standard deviation.For a certain case, a ?ite number of results will be drawn from this>>distribution. Is there a mathematical formula for calculating the>>expected range of these results?>> >>Google for con?ence 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 ?ite number of results will be drawn from this>>distribution. Is there a mathematical formula for calculating the>>expected range of these results?>Google for con?ence interval.This is NOT a con?ence interval problem. For any samplesize, the ratio of the range to the standard deviation hasa known distribution, and its expected value has beentabulated. There is no simple closed form, as there isfor a con?ence interval.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === === > What is the largest (?ite) number ever written down? it's still being written down (while watching for the competitors ...)my printerYou're quite welcome. === Try Ackermann Functions... === One of those people that is very fond of exclamation marks probably once9!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! at the end of a sentence, andthread, 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 than9^(9^(9^... containing 60 9's. I haven't counted the 9's in JeroenBoschma's posting but I'm fairly sure he has more of them than you do.-- /-- Joona Palaste (palaste@cc.helsinki.? ------------- Finland ---------- http://www.helsinki.?~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 wouldprint 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 exponentialswere in the ?st class of not really all that big. Calculations of Ackermann'snumber 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 wasable to work out what the programs did than that people managed to write themin the ?st place.Alan -- Defendit numerus === Joona I Palaste 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 withm!, we get that m!! is smaller than (m^m)^(m^m), and if we replace mwith 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 tothe power of the number of ! signs in the smaller number.If those operations were all grouped like l^(l^(l^... then there wouldbe 1073741824 nested exponentations, which would make Quaternion'snumber bigger than Jeroen's, but as they are not, I don't know which islarger. Could some of you math gurus shed more light on this?-- /-- Joona Palaste (palaste@cc.helsinki.? ------------- Finland ---------- http://www.helsinki.?~palaste --------------------- rules! --------/Make money fast! Don't feed it! - Anon === Joona I Palaste 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 therecursive 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.htmlfor those that know what the later means. Check the appropriate lemma in thereference link.> -- > /-- Joona Palaste (palaste@cc.helsinki.? ------------- Finland --------> -- http://www.helsinki.?~palaste --------------------- rules! --------/> Make money fast! Don't feed it!> - Anon--Ioannis Galidakishttp://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 ?st person to write that number down.In fact, for $29.99 plus $4.99 in postage and handling, I'll send you a certi?ate attesting that. === John Tapper scribbled the following:If you allow ?ite numbers constructed from a series of mathematicalformulae, I think Graham's number takes the cake, by far.-- /-- Joona Palaste (palaste@cc.helsinki.? ------------- Finland ---------- http://www.helsinki.?~palaste --------------------- rules! --------/To err is human. To really louse things up takes a computer. - Anon === > John Tapper scribbled the following:If you allow ?ite 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 suggestionsso far. The trouble is that the busy-beaver function is rather hard tocalculate.(The busy-beaver function B(n) is the maximum number of 1s that can beprinted by an n-state 2-colour Turing machine which halts. To showthat N is bigger than an of the entry M posted so far, it suf?es tocome up with a Turing machine with at most B(B(B(9))) states whichprints more than M 1s. Note that B(6) is at least 10^865, so thisshouldn't be too hard to do.)-- Simon Nickerson === >> John Tapper scribbled the following:> If you allow ?ite 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 suf?es 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-growingfunction ... http://www.cs.berkeley.edu/~aaronson/bignumbers.html === > John Tapper scribbled the following:>> If you allow ?ite 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.? ------------- Finland --------> -- http://www.helsinki.?~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 sinceit'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 injectionDescription: cross-domain scripting, cookie/data/identity theft, command executionExploit: 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 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^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 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, soit's of the order 10^10^800 and is much much smaller than the 2ndSkewes 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^ 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9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^>9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^>9^9^9^9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^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 digits9^9^9 has 369693099 digits9^9^9^9 has >10^300000000 digitsWhere 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 injectionDescription: cross-domain scripting, cookie/data/identity theft, command executionExploit: 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 ?ure is for (((9^9)^9)^9)^9)... === a .8ecrit :>I just did (I guess):no , you did not :-) 1 + 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9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^9^9^9^9^9^>9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^>9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^>9^9^9 ^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^9^9^9^9^9^>9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^>9^9^9^9^9^9^9^9^9^9^9^9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^>9^9^9^ 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^9^9^9^9^>9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^>9^9^9^9^9^9^9^9^9^9^9^9^9^9 ^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9 === Is it possible to have a stroboscopic effect (where wheel spokes appear tobe rotating backwards) in real life, rather than in movies where it iscommon?I can imagine two situations where it might happen: when a car is goingthrough a tunnel or under a street lamp at night, or when there is a picketfence between you and it.But on an open road, under a clear sun, is it possible for your eyes tocreate 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 ?st time I noticed it was near the maths building at myalma mater, Queen Mary & West?ld, in London. The compound is separatedfrom the main street by a 6 ft. tall (or so) metallic fence, with verticalbars a few inches apart. The fence was quite long at the time, and from inside the compound youcould see through the bars passing cars for a few seconds. Depending ontheir speed the stroboscopic effect could be seen all right; with anyluck, the wheel spokes would appear to be stationary, which I foundfascinating 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 avibrating 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 ?cent light.Are you too young to remember record turntables with stroboscopicspeed 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 throughanother might do it.-- Richard-- FreeBSD rules! === >>Is it possible to have a stroboscopic effect (where wheel spokes appearto>be rotating backwards) in real life, rather than in movies where it is>common?>> Yes, easily. Try spinning something under a ?cent 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 theindicator 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 gradestudents of their inability to differentiate between reality memories andfantasy (TV) memories. Several of them swear that they've seen it on carspassing them on highways, on open roads, in broad daylight.--riverman === riverman [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 thesunlit day you'd need two sets of spikes or a set of spikes and a grid tosee this.> --riverman--Ioannis Galidakishttp://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 avector 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.htmlNeedless 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.htmI 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 scienti?> 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://mathforpro?.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! ===