mm-1009 === Subject: transfinite sequences Hi all, given a transfinite sequence {S_a} of reals, for a c implies |S_a + S_b| < epsilon), then does {S_a} converge to a real number ? The question is the generalisation of the usual construction of R from Q : could R be analogously completed with respect to omega_1 ? We could obtain a bounded transfinite sequence for example by well-ordering (0,1). This only provides an existence of such a sequence : is there a way to specifically construct one - i.e. define inductively S_a for all non-limit / limit countable ordinals ? Assuming you mean |S_a - S_b|, the answer is that S_a is constant from some countable point on. Yes of course You should seek info about filters (its a way to better understand limits). If E is complete, every Cauchy filter is convergent; that apply to any type of convergence you can imagine. You can also seek info about the old fashion moore-Smith sequences, a generalisation of sequence you seek, but a little bit obsolete due to the use of filters, more easy to manipulate. http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Single Precision Montgomery Multiplication Optimization > The single precision Montgomery multiplication algorithm can be stated as > follows: > $$eqalign { > t &= x y,cr > u &= (t m^prime) bmod b,cr > v &= u m,cr > A &= (t + v / b) bmod m.cr > }$$ > where $m^prime = - m^{-1} pmod b$ and $b$ is the radix. > It would seem that the term > $$x y + (x y -m^{-1} bmod b) m$$ > can be reduced in some way so that only two or even one multiplication is I did some work in Mathematica and I discovered that $$x y + (x y (-m^{-1} bmod b) m$$ Is equivalent to $$- a x y + m lceil (a b + 1) x y / b rceil$$ where $$a = (m m^{-1} - 1) / b,$$ and $a < m$. I still need some way to perform the operation with one or two multiplications. Can anyone help me? === Subject: Re: JSH: My fear, consider this Discussion, linux) > admits he was wrong] of the 19th with one barely 2 days later with a > revisionist version, in which you were being stymied by people > winning at convincing sci.math that you were wrong? JSH is part of sci.math. Those bad guys convinced JSH he was wrong. I see no contradictions. -- Jesse Hughes Basically there are two angry groups. I am a harsh force of one. Against me is a society of mathematicians. So far its been a draw. -- JSH gives another display of keen insight. === Subject: Re: JSH: My fear, consider this >> NOBODY FUCKING CARES ABOUT THE >> TRUTH!!!!!!!!!!!!!!!!!!!!!!!!![...] > Giggle. You dont seem to be keeping count, but There were 239 exclamation marks. (Newsgroups trimmed.) === Subject: Do Prime Algebraic Numbers even exist? This is probably a slightly stupid question, but Eric W. Weissteins definition of Prime Algebraic Number: http://mathworld.wolfram.com/PrimeAlgebraicNumber.html gives no examples of such a number at all, and therefore leaves a lot to be desired. I am wondering what such a number might look like, if one exists at all. An obvious attempt would be sqrt(2), or, for that matter, the nth root of 2. However, if one considers the product sqrt(2) = 2^(1/4) * 2^(1/4) it is clear that sqrt(2) divides neither factor. Therefore, I fail to see how sqrt(2) could be prime. There are other difficulties. The number 2 + sqrt(3) is in fact a unit in the algebraic integers. Its reciprocal 2 - sqrt(3) is also an algebraic integer. (That these are indeed units is easily proven by multiplying them together; one gets 4 - 3 = 1. It is also clear that both are algebraic integers.) The only way I personally can resolve this is to suggest that primes have an order designation; this would mean that the regular ordinary primes are order 1, sqrt(2) is a prime of order 2, cuberoot 2 is a prime of order 3, etc. This basically means that, in any product of two integers of order <= n, if an algebraic prime of order n divides the product, it must divide one of the integers. Or something like that. What an ugly mess. Can anyone else shed some light on this problem? -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Do Prime Algebraic Numbers even exist? In sci.math, Arturo Magidin : >>This is probably a slightly stupid question, but Eric W. Weissteins >>definition of Prime Algebraic Number: >>http://mathworld.wolfram.com/PrimeAlgebraicNumber.html >>gives no examples of such a number at all, and therefore >>leaves a lot to be desired. I am wondering what such a >>number might look like, if one exists at all. > The definition does indeed leave much to be desired; for starters, it > forgets to specify the ring! > Taken at face value, assuming it means in the ring of algebraic > integers, then the definition is correct but vacuous, since no > algebraic integer is irreducible. Perhaps you can correspond with Mr. Weisstein on this. (Im not sure I have the technical expertise beyond pointing out what Ive already discovered -- if one can call it a discovery. But its certainly a problem.) [rest snipped but saved for future analysis] Your stuff is rather interesting, and in a completely different direction from my orders proposal; yours makes more sense. :-) To be sure, its been too long since Ive played with all this. -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Do Prime Algebraic Numbers even exist? Adjunct Assistant Professor at the University of Montana. >This is probably a slightly stupid question, but Eric W. Weissteins >definition of Prime Algebraic Number: >http://mathworld.wolfram.com/PrimeAlgebraicNumber.html >gives no examples of such a number at all, and therefore >leaves a lot to be desired. I am wondering what such a >number might look like, if one exists at all. The definition does indeed leave much to be desired; for starters, it forgets to specify the ring! Taken at face value, assuming it means in the ring of algebraic integers, then the definition is correct but vacuous, since no algebraic integer is irreducible. However, the definition becomes interesting when used in the context which usually occurs in algebraic number theory: in a number field. A number field in this case is a finite extension of Q. Call it K. Then one can consider the ring of integers of K, which is the collection of all algebraic integers which are in K. In this ring, there ->are<- irreducible elements, and there ->are<- prime elements, though in general they need not be equivalent notions (they are equivalent if and only if this ring of integers is a UFD). For example, consider the classical example K=Q(sqrt(-5)), with ring of integers equal to Z[sqrt(-5)]. There are some irreducibles that are not primes: for example, it is easy to verify that 2 is irreducible; and 2 divides (1+sqrt(-5))^2 = -4+2sqrt(-5) = 2(-2+sqrt(-5)) but 2 does not divide (1+sqrt(-5)). So 2 is irreducible, but not a prime. On the other hand, sqrt(-5) itself is irreducible, and is also a prime (any prime must be irreducible): for if sqrt(-5) divides (a+b*sqrt(-5))*(x+y*sqrt(-5)) = (ax-5by) + (ay+bx)sqrt(-5) then there exists an element r+s*sqrt(-5), r,s integers, such that sqrt(-5)*(r+s*sqrt(-5)) = -5s + r(sqrt(-5) = (ax-5by) +(ay+bx)sqrt(-5). in particular, -5 = ax-5by, so 5 must divide ax in the integers. That means that either 5|a (in which case, sqrt(-5) divides a+b*sqrt(-5)), or else 5 divides x (in which case, sqrt(-5) divides x+y*sqrt(-5)). So sqrt(-5) is a prime in this ring. The following result describes the primes of sqrt(-5) (taken from Dedekinds exposition, but a nice exercise to try): 1. The positive rational primes which are congruent to 11, 13, 17, or 19 (mod 20) are primes in Z[sqrt(-5)]. 2. sqrt(-5) is a prime in Z[sqrt(-5)]. 3. The rational prime 2 behaves like the square of a prime; that is, 2 satisfies the following two conditions: (I) If 2|a^2*b^2, with a,b in Z[sqrt(-5)], then 2|a^2 or 2|b^2; (II) There exist a in Z[sqrt(-5)] such that 2|a^2 but 2 does not divide a. 4. Each positive rational prime congruent to 1 or to 9 (mod 20) factors as the product of two different factors in Z[sqrt(-5)], each of them primes (they will be conjugate, that is, p=(a+b*sqrt(-5))*(a-b*sqrt(-5)) for suitably chosen integers a,b). 5. Each positive rational prime congruent to 3 or 7 (mod 20) behaves like the product of two different primes; that is, it satisfies the following three conditions: (I) If p divides a product x*y*z of elements of Z[sqrt(-5)], then p divides either x*z, x*y, or y*z. (II) There exist elements a,b in Z[sqrt(-5)] such that p divides a*b but p does not divide a nor b. (III) If p divides a^2, then p divides a, for all a in Z[sqrt(-5)]. It is not a good definition in mathworld, certainly... -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Do Prime Algebraic Numbers even exist? >This is probably a slightly stupid question, but Eric W. Weissteins >definition of Prime Algebraic Number: >http://mathworld.wolfram.com/PrimeAlgebraicNumber.html >gives no examples of such a number at all, and therefore >leaves a lot to be desired. I am wondering what such a >number might look like, if one exists at all. I could be wrong, but unless Im misunderstanding something its clear that there is nothing satisfying the definition on that page: A prime algebraic integer is defined to be an irreducible algebraic integer satisfying a certain condition, and its clear that there are no irreducible algebraic integers, because its very easy to see that if a is an algebraic integer then so is sqrt(a). >An obvious attempt would be sqrt(2), or, for that matter, >the nth root of 2. However, if one considers the product >sqrt(2) = 2^(1/4) * 2^(1/4) >it is clear that sqrt(2) divides neither factor. Therefore, >I fail to see how sqrt(2) could be prime. >There are other difficulties. The number 2 + sqrt(3) is in >fact a unit in the algebraic integers. Its reciprocal >2 - sqrt(3) is also an algebraic integer. (That these are >indeed units is easily proven by multiplying them together; >one gets 4 - 3 = 1. It is also clear that both are algebraic >integers.) >The only way I personally can resolve this is to suggest that >primes have an order designation; this would mean that the >regular ordinary primes are order 1, sqrt(2) is a prime of >order 2, cuberoot 2 is a prime of order 3, etc. This basically >means that, in any product of two integers of order <= n, if an >algebraic prime of order n divides the product, it must divide >one of the integers. >Or something like that. What an ugly mess. >Can anyone else shed some light on this problem? ************************ David C. Ullrich === Subject: Re: Do Prime Algebraic Numbers even exist? >Eric W. Weissteins >definition of Prime Algebraic Number: >http://mathworld.wolfram.com/PrimeAlgebraicNumber.html >gives no examples of such a number at all, and therefore >leaves a lot to be desired. I am wondering what such a >number might look like, if one exists at all. There are no primes in the ring of all algebraic integers because, as you basically observe, if a is an algebraic integer, then a^(1/2) is also an algebraic integer. But, there can be algebraic integer primes in the ring of algebraic integers of a number field. So, 2^(1/2) is a prime in the ring of integers of Q(sqrt(2)). Maybe this is what they are thinking of. John Robertson === Subject: Re: Do Prime Algebraic Numbers even exist? In sci.math, Jpr2718 Eric W. Weissteins >>definition of Prime Algebraic Number: >>http://mathworld.wolfram.com/PrimeAlgebraicNumber.html >>gives no examples of such a number at all, and therefore >>leaves a lot to be desired. I am wondering what such a >>number might look like, if one exists at all. > There are no primes in the ring of all algebraic integers because, as you > basically observe, if a is an algebraic integer, then a^(1/2) is also an > algebraic integer. > But, there can be algebraic integer primes in the ring of > algebraic integers of a number field. So, 2^(1/2) is a > prime in the ring of integers of Q(sqrt(2)). > Maybe this is what they are thinking of. Maybe, but thats not the ring of algebraic integers; thats the ring of numbers generated by the combination of any integer and sqrt(2), which is equivalent to all numbers a + b * sqrt(2), as it turns out. (The units for that ring appear to contain 1, -1, 1 + sqrt(2), -1 + sqrt(2), 1 - sqrt(2), and -1 - sqrt(2). Since (1 + sqrt(2))^2 = 3 + 2*sqrt(2), there are apparently a few additional units as well -- in fact, a + b * sqrt(2) is a unit if 2*b^2 - a^2 = 1 or -1.) > John Robertson -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Do Prime Algebraic Numbers even exist? > Computational complexity not only involves the number of operations, but > also the complexity of operations. Operations on large numbers are more > complex than operations on small numbers. When you look at it from a pure > mathematical viewpoint it is a bit different. But when you are actually > implementing stuff the largeness of numbers gets to play a big role. > Well, heres the whole computation. Complexity is in the eye of the > beholder, but this doesnt look too complex to me. Formulas for the > iterations are below. Use fixed-width font to view. Well, I have said that I did not want to use a bignum package, and indeed, the first line that goes wrong because I do not use such a package is noted below: > 18 17 5 7 11752497748 581123613 18 17 5 7 -1132404140 581123613 The second line is what I get (going to unsigned long does not help either). Going to double precision will give me some space, but not much. So to get the proper value I need a bignum package which adds computational complexity to the stuff. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Do Prime Algebraic Numbers even exist? > Dik T. Winter Dik.Winter@cwi.nl > However, when m > 0 it is not easy to always find a unit. > >What I meant here is that it may take a lot of work. There are >deterministic methods to get a unit (or the fundamental unit), >but the computational complexity becomes pretty big. I have done >extensive calculations with discriminant < 10000. (The discriminant >is equal to m when m = 1 mod 4, otherwise it is 4m.) The first one >I did not complete was D = 409 (m = 409) where in the fundamental >unit x + y.sqrt(m), x exceeds 10^11. And there are 19 more in the >range D < 1000. (Yes, I could have gone to some bignum package, but >did not feel inclined to do so...) See my last remark above... > It is very easy to find the fundamental unit in Q(sqrt(D)) for D = 409. Yes, I know it is easy to find, consider however the constraints I give... > It is very easy to find the fundamental unit in Q(sqrt(D)) for D = 409. The > length of the period of the continued fraction expansion of sqrt(409) is only > 21. So 21 iterations of the PQa algorithm (see my webpage if youre not > familiar with this well-known simple algorithm) suffice to find the minimal > positive solution of x^2 - 409y^2 = +-1. Actually I *did* use it. > Microsoft Excel does integer arithmetic to an accuracy of about 15 digits. > So even in Excel it takes a fraction of a second to find > x = 111,921,796,968, y = > 5,534,176,685 as the minimal positive solution. As I have no access to Microsoft Excel... As I said, I did not feel inclined to go to a bignum package, so the results are limited to (in my case) what a double precision number could handle. And I remained a bit on the safe side, limiting the numbers to 10^12. > The length of the period of the continued fraction expansion of sqrt(D) is > less than 4*sqrt(D)*ln(D) (at least for squarefree D). For D <= 10,000, > this is less than 3,700. Maybe a lot by hand, but easy on any computer. Computational complexity not only involves the number of operations, but also the complexity of operations. Operations on large numbers are more complex than operations on small numbers. When you look at it from a pure mathematical viewpoint it is a bit different. But when you are actually implementing stuff the largeness of numbers gets to play a big role. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Do Prime Algebraic Numbers even exist? Dik T. Winter Dik.Winter@cwi.nl > However, when m > 0 it is not easy to always find a unit. >What I meant here is that it may take a lot of work. There are >deterministic methods to get a unit (or the fundamental unit), >but the computational complexity becomes pretty big. I have done >extensive calculations with discriminant < 10000. (The discriminant >is equal to m when m = 1 mod 4, otherwise it is 4m.) The first one >I did not complete was D = 409 (m = 409) where in the fundamental >unit x + y.sqrt(m), x exceeds 10^11. And there are 19 more in the >range D < 1000. (Yes, I could have gone to some bignum package, but >did not feel inclined to do so...) This seems to be a reply to my post. Apparently you did not look at the webpage I suggested. It is very easy to find the fundamental unit in Q(sqrt(D)) for D = 409. The length of the period of the continued fraction expansion of sqrt(409) is only 21. So 21 iterations of the PQa algorithm (see my webpage if youre not familiar with this well-known simple algorithm) suffice to find the minimal positive solution of x^2 - 409y^2 = +-1. There are only 5 numbers that need to be computed at each iteration, three of the numbers never get bigger than 24, and each computation is pretty simple. Microsoft Excel does integer arithmetic to an accuracy of about 15 digits. So even in Excel it takes a fraction of a second to find x = 111,921,796,968, y = 5,534,176,685 as the minimal positive solution. There are 19 more what? The length of the period of the continued fraction expansion of sqrt(D) is less than 4*sqrt(D)*ln(D) (at least for squarefree D). For D <= 10,000, this is less than 3,700. Maybe a lot by hand, but easy on any computer. One also frequently sees the comment that solving x^2 - Dy^2 = N is very complicated. This is hardly more difficult than solving x^2 - Dy^2 = +-1. Again, see my webpage at http://hometown.aol.com/jpr2718/ John Robertson === Subject: Re: Do Prime Algebraic Numbers even exist? I am reminded of one error and one deficiency in my original post, so some additions: > In sci.math, Dik T. Winter > > : > Indeed, but there are *no* primes in the algebraic integers. Now you > could define class 1 primes to be primes in the integers, but I fail > to see how you could define class 2 primes or anything else. Not > all primes in number fields are of the form n-th root of p with p > a prime. For instance, one of the primes in Q(sqrt(2)) is 1 + sqrt(2) > (I think). I should have thought further, as I was reminded, 1 + sqrt(2) is a unit, so no prime. To get a prime we have to know something about norms in quadratic fields. In Q(sqrt(m)) the norm of a number a + b.sqrt(m) is a^2 - m.b^2. So the norm of 1 + sqrt(2) = -1, and hence it is a unit. You may verify that the norm so defined is multiplicative, also that the norm is integer if and only if a and b are both integers, or are both half of an odd integer when m = 1 mod 4. And these are precisely the algebraic integers in the field. To get a prime you look for a number with (integer) prime norm. In this case 3 + sqrt(2), with norm 7. > The number of units in quadratic fields (Q(sqrt(m)) ) is as follows: > m > 0: infinitely many > m = -1: 4 > m = -3: 6 > m < 0: 2 in all other cases. > (again, assuming m to be square free.) > However, when m > 0 it is not easy to always find a unit. What I meant here is that it may take a lot of work. There are deterministic methods to get a unit (or the fundamental unit), but the computational complexity becomes pretty big. I have done extensive calculations with discriminant < 10000. (The discriminant is equal to m when m = 1 mod 4, otherwise it is 4m.) The first one I did not complete was D = 409 (m = 409) where in the fundamental unit x + y.sqrt(m), x exceeds 10^11. And there are 19 more in the range D < 1000. (Yes, I could have gone to some bignum package, but did not feel inclined to do so...) This is similar to the case of showing the gcd of two algebraic integers, even if they are both from a quadratic field. You may have reasons to know that there is a non-unit common factor, but showing its value is something else. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Do Prime Algebraic Numbers even exist? In sci.math, Dik T. Winter : > In sci.math, Jpr2718 > algebraic integers of a number field. So, 2^(1/2) is a > prime in the ring of integers of Q(sqrt(2)). > Maybe this is what they are thinking of. > > Maybe, but thats not the ring of algebraic integers; thats > the ring of numbers generated by the combination of any > integer and sqrt(2), which is equivalent to all numbers > a + b * sqrt(2), as it turns out. > Indeed, but there are *no* primes in the algebraic integers. Now you > could define class 1 primes to be primes in the integers, but I fail > to see how you could define class 2 primes or anything else. Not > all primes in number fields are of the form n-th root of p with p > a prime. For instance, one of the primes in Q(sqrt(2)) is 1 + sqrt(2) > (I think). > Also you have to be careful with what you call the subring of integers. > Not in all quadratic fields (i.e. fields of the form Q(sqrt(m)) ) are > they just combinations of the form a + b*sqrt(m). Assume m square-free, > then the integers are of the form (a + b*sqrt(m))/2 with a and b both > even or both odd, when m = 1 mod 4. So (1 + sqrt(5))/2 is an integer > in Q(sqrt(5)). > (The units for that ring appear to contain 1, -1, > 1 + sqrt(2), -1 + sqrt(2), 1 - sqrt(2), and -1 - sqrt(2). > Since (1 + sqrt(2))^2 = 3 + 2*sqrt(2), there are apparently > a few additional units as well -- in fact, > a + b * sqrt(2) is a unit if 2*b^2 - a^2 = 1 or -1.) > The number of units in quadratic fields (Q(sqrt(m)) ) is as follows: > m > 0: infinitely many > m = -1: 4 > m = -3: 6 > m < 0: 2 in all other cases. > (again, assuming m to be square free.) > However, when m > 0 it is not easy to always find a unit. I have done > some quite extensive calculations and found that a fundamental unit > in Q(sqrt(241)) is: > 71011068 + 4574225.sqrt(241). > However, anything beyond quadratic fields has been barely studied. > the algebraic integers. Ah, good. However, theres an issue that what James has is not a quadratic field, but a field Z[(n +/- sqrt(n^2 - 32))/16], for any integer n (n != 6, -6, 9, -9, as these lead to either plain old Z, Z[1/2], Z[1/4], or Z[1/8]). I would assume absent further proof that each of these is a superset of some quadratic field, which basically means it has at least the units of that quadratic field. -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Do Prime Algebraic Numbers even exist? > However, when m > 0 it is not easy to always find a unit. For small m there are perfectly easy methods using continued fractions. See my webpage on this. For large m the solution gets so large that the challenge is to find a fast way to write it down. See either of the papers below for more info. The John Robertson H. W. Lenstra Jr., Solving the Pell equation, textit{Notices of the American Mathematical Society}, {bf 49} No. 2 (February 2002), pp. 182--192. The equation $x^2-Dy^2=pm 1$ for large $D$. H. C. Williams, Solving the Pell equation, in Bruce Berndt et al., textit{Surveys in Number Theory: Papers from the Millennial Conference on Number Theory}, A. K. Peters, 2002. Also included in textit{Number Theory for the Millennium}, Volumes 1, 2, 3, M. A. Bennett et al. editors, A. K. Peters, 2002. Williams web page gives this last reference as H. C. Williams, Solving the Pell equation, textit{Proc. Millennial Conference on Number Theory}, A. K. Peters, Natick MA, 2002, pp. 397-435. Discusses the equation $x^2-Dy^2=pm 1$. Terrific overview, including discussion when $D$ is large. === Subject: Re: Do Prime Algebraic Numbers even exist? > In sci.math, Jpr2718 > > But, there can be algebraic integer primes in the ring of > algebraic integers of a number field. So, 2^(1/2) is a > prime in the ring of integers of Q(sqrt(2)). > Maybe this is what they are thinking of. > Maybe, but thats not the ring of algebraic integers; thats > the ring of numbers generated by the combination of any > integer and sqrt(2), which is equivalent to all numbers > a + b * sqrt(2), as it turns out. Indeed, but there are *no* primes in the algebraic integers. Now you could define class 1 primes to be primes in the integers, but I fail to see how you could define class 2 primes or anything else. Not all primes in number fields are of the form n-th root of p with p a prime. For instance, one of the primes in Q(sqrt(2)) is 1 + sqrt(2) (I think). Also you have to be careful with what you call the subring of integers. Not in all quadratic fields (i.e. fields of the form Q(sqrt(m)) ) are they just combinations of the form a + b*sqrt(m). Assume m square-free, then the integers are of the form (a + b*sqrt(m))/2 with a and b both even or both odd, when m = 1 mod 4. So (1 + sqrt(5))/2 is an integer in Q(sqrt(5)). > (The units for that ring appear to contain 1, -1, > 1 + sqrt(2), -1 + sqrt(2), 1 - sqrt(2), and -1 - sqrt(2). > Since (1 + sqrt(2))^2 = 3 + 2*sqrt(2), there are apparently > a few additional units as well -- in fact, > a + b * sqrt(2) is a unit if 2*b^2 - a^2 = 1 or -1.) The number of units in quadratic fields (Q(sqrt(m)) ) is as follows: m > 0: infinitely many m = -1: 4 m = -3: 6 m < 0: 2 in all other cases. (again, assuming m to be square free.) However, when m > 0 it is not easy to always find a unit. I have done some quite extensive calculations and found that a fundamental unit in Q(sqrt(241)) is: 71011068 + 4574225.sqrt(241). However, anything beyond quadratic fields has been barely studied. the algebraic integers. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Do Prime Algebraic Numbers even exist? >(The units for that ring appear to contain 1, -1, >1 + sqrt(2), -1 + sqrt(2), 1 - sqrt(2), and -1 - sqrt(2). >Since (1 + sqrt(2))^2 = 3 + 2*sqrt(2), there are apparently >a few additional units as well -- in fact, >a + b * sqrt(2) is a unit if 2*b^2 - a^2 = 1 or -1.) Yep. All the units are (+-1)(1 + sqrt(2))^n for some integer n (positive, negative, or zero). The equation 2*b^2 - a^2 = 1 or -1 is a Pell equation. Solving the generalized Pell equation at http://hometown.aol.com/jpr2718/ John Robertson === Subject: Re: there is no such thing as infinity Darryl L. Pierce,,, wibbled: >> Ah, I remember when I first started programming. All I had was a really >> hot needle and had to write my code directly to the SIMMs. It was a major >> leap forward when I upgraded and could just type copy con file.exe... > > When I started, SIMMs hadnt been invented. We had people who had seen > so many punch cards and papertapes they could read the holes by eye. > I set the bootstrap instructions using an array of toggle switches on > one mainframe we had. > Paper cards? Ah, you were lucky! When I first started investigating > programming, we had to chisel instruction sets into very large rocks and > carry them ourselves to the CPU (Chiseling Processors Union) to get the > work done. And we were thankful... I had this void, and I thought, let there be light, and suddenly Id invented binary. Sorry. === Subject: Re: there is no such thing as infinity Imam Tashdid ul Alam wibbled: > 640K ought to be enough for anybody. > -BILL GATES, 1981 [I was born the following year :o] Well, since I recall a time when Comp Sci students did an entire undergraduate course with 5Mb of diskspace and all their programs had to fit into 120kb, maybe he had a point. Mind you that was on a machine with 60 bit words, and the compilers all used overlays extensively. === Subject: Re: there is no such thing as infinity >> 640K ought to be enough for anybody. >> -BILL GATES, 1981 [I was born the following year :o] > Well, since I recall a time when Comp Sci students did an entire > undergraduate course with 5Mb of diskspace and all their programs had to > fit into 120kb, maybe he had a point. Mind you that was on a machine > with 60 bit words, and the compilers all used overlays extensively. Ah, hell, the kids today are so spoiled! I write software for cell phones for a living and *have* to be cheap with the code since we have to live in recent desktop machines and wants to throw extra threads and verbose processes at the simplest little problems that we face... -- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people think, Mr. Feynman? === Subject: Re: there is no such thing as infinity >> 640K ought to be enough for anybody. >> -BILL GATES, 1981 [I was born the following year :o] > Well, since I recall a time when Comp Sci students did an entire > undergraduate course with 5Mb of diskspace and all their programs had to > fit into 120kb, maybe he had a point. Mind you that was on a machine > with 60 bit words, and the compilers all used overlays extensively. > Ah, hell, the kids today are so spoiled! I write software for cell phones > for a living and *have* to be cheap with the code since we have to live in more > recent desktop machines and wants to throw extra threads and verbose > processes at the simplest little problems that we face... In college, I crashed one of the largest mainframes in Ohio. The university was charging businesses $700 / CPU minute to use this computer. It took hours to reboot it. The OS required nearly a megabyte. I think the mainframe had less than 5 MB of memory. In the early 1980s, IBM was selling 300 MB disk drives for $100,000. dropped to $1 / kilobyte. Imagine being able to buy a megabyte of memory for under $1,000. The average desktop computer has more processing power than all of the computers at Houston Control when Apollo landed on the moon. Russell - yes, Im a fossil === Subject: Re: there is no such thing as infinity > Imam Tashdid ul Alam wibbled: > > 640K ought to be enough for anybody. > -BILL GATES, 1981 [I was born the following year :o] > > Well, since I recall a time when Comp Sci students did an entire > undergraduate course with 5Mb of diskspace and all their programs had to > fit into 120kb, maybe he had a point. Mind you that was on a machine > with 60 bit words, and the compilers all used overlays extensively. I know. Thats the point, actually. The richest guy of the planet aint an idiot, contrary to what geeks might think. === Subject: Re: there is no such thing as infinity Imam Tashdid ul Alam wibbled: > Imam Tashdid ul Alam wibbled: > > 640K ought to be enough for anybody. > -BILL GATES, 1981 [I was born the following year :o] > > Well, since I recall a time when Comp Sci students did an entire > undergraduate course with 5Mb of diskspace and all their programs had to > fit into 120kb, maybe he had a point. Mind you that was on a machine > with 60 bit words, and the compilers all used overlays extensively. > I know. Thats the point, actually. The richest guy of the planet > aint an idiot, contrary to what geeks might think. Some of the guys he employs arent so bright, though. === Subject: Re: there is no such thing as infinity >> Imam Tashdid ul Alam wibbled: >> >> 640K ought to be enough for anybody. >> -BILL GATES, 1981 [I was born the following year :o] >> >> Well, since I recall a time when Comp Sci students did an entire >> undergraduate course with 5Mb of diskspace and all their programs had to >> fit into 120kb, maybe he had a point. Mind you that was on a machine >> with 60 bit words, and the compilers all used overlays extensively. > I know. Thats the point, actually. The richest guy of the planet > aint an idiot, contrary to what geeks might think. But besides all that, the evidence is that Bill Gates never said any such thing. I suggest you Google for Bill Gates 640K enough. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: there is no such thing as infinity > Imam Tashdid ul Alam wibbled: > > 640K ought to be enough for anybody. > -BILL GATES, 1981 [I was born the following year :o] > > Well, since I recall a time when Comp Sci students did an entire > undergraduate course with 5Mb of diskspace and all their programs had to > fit into 120kb, maybe he had a point. Mind you that was on a machine > with 60 bit words, and the compilers all used overlays extensively. >> I know. Thats the point, actually. The richest guy of the planet >> aint an idiot, contrary to what geeks might think. > But besides all that, the evidence is that Bill Gates never said any such > thing. I suggest you Google for Bill Gates 640K enough. But he did write about factoring large primes. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: there is no such thing as infinity >> Imam Tashdid ul Alam wibbled: >> >> 640K ought to be enough for anybody. >> -BILL GATES, 1981 [I was born the following year :o] >> >> Well, since I recall a time when Comp Sci students did an entire >> undergraduate course with 5Mb of diskspace and all their programs had to >> fit into 120kb, maybe he had a point. Mind you that was on a machine >> with 60 bit words, and the compilers all used overlays extensively. > I know. Thats the point, actually. The richest guy of the planet > aint an idiot, contrary to what geeks might think. >> But besides all that, the evidence is that Bill Gates never said any such >> thing. I suggest you Google for Bill Gates 640K enough. > But he did write about factoring large primes. So it appears. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Request for ideas on tricky little problem Could this not be formulated as a mathematical programming problem: Let y(j) in {0,1} mean bus stops at stop j Let x(i,j) in {0, 1} mean person i gets off at stop j Let p(i)=k mean that person i needs to go to stop k Then: no more than z stops: sum(j, y(j)) <= z passengers only get off when bus stops: x(i,j) <= y(j) for all i,j passenger has to get off somewhere: sum(j, x(i,j)) = 1 objective: min sum(i, [abs(p(i) -sum(j, j*x(i,j))]^3) This would be an MINLP. Probably youll want to work on the objective (e.g. getting rid of the abs as this introduces a kink i.e. discontinuous derivative). ------------------------------------------------------------- -- Erwin Kalvelagen erwin@gams.com, http://www.gams.com/~erwin ------------------------------------------------------------- -- > Consider a bus route with k stops. When the bus starts its journey, p > persons are on the bus. Person j wants to get off at stop s_j<=k. Also > s_j>=1. However the bus cant stop at more than z < k stops. So a person > who wants to get off at a stop s_j which is not among the z stops has to get > off at the nearest one and then walk. Let f(x_j) = (x_j)^3 be the distance > person j has to walk to get to his destination, where x is the number of > stops he has to walk to get to his destination. Let T be the sum f(x_1) + > f(x_2) + ... + f(x_p) > The problem is now to construct a method(algorithm) where the bus stops at > stops, so that the T is minimized. Any ideas on a > decent strategy here ? === Subject: Re: Math notation question: Ô(Ô > The statement was: > There exist x, y, and z that are nonzero. Such that 1 ( y>x and xy=z. The period after nonzero looks wrong. Is it a comma? > Yeah, it shouldnt be there. Sorry. See whole whole transcription > below. > Does 1 ( y>x mean that both y & x are less than 1? Looking at the > answer that is the only thing that makes sense to me. I just dont > understand the notation. Neither do I. Have you accurately transcribed the _complete_ sentence? > If x, y, and z are nonzero numbers such that 1(y>x and xy=z, which of > the following CANNOT be true? > 1) y>z > 2) y=z > 3) z=x > 4) x>z > 5) z>0 > I think its 4) x>z, but the answer in the review says its 2) y=z. > However, when x=1, y=2, and z=2 it meets all of the criteria and y=z, > so #2 cannot be the answer. When looking at their explanation for why > #2 is the answer they say because all of the other choises CAN be > true #2 must be the correct answer. Which, to me, is not only a BS > cop-out, but wrong. > Also, in the #4 explanation it says According to the question, Y must > be greater to or equal to one. Thus *my* question on the notation. > Whatever happened to <=? I should also state that nowhere else in > the review does it mention that the Ô(Ô to <= notation substitution > exists. I think its just a shoddily put-together review. The question clearly contains a typo. All books have typos, expect them and work around them. I conjecture that the question was meant to be: If x, y, and z are nonzero numbers such that 1>=y>x and xy=z, which of the following CANNOT be true? 1) y>z 2) y=z 3) z=x 4) x>z 5) z>0 The correct answer to this question is (2). For suppose y=z. Then from xy=z (and the fact that x,y,z are nonzero), we get x = 1. But then from 1 >= y > x we would get 1 > 1, a contradiction. All the others are possible. For example for (1) let x = -1, y=0.5, and z=-0.5 For (3), let x = 0.5, y=1, and z=0.5. For (4) and (5), let x=1/4, y=1/2, and z=1/8. -Leonard Blackburn === Subject: Re: Math notation question: Ô(Ô Well its not 4 ... x=-1, y=-z z < -1 then xy = z and x > z Im guessing that the type-o was very significant to the problem and Id > The statement was: > There exist x, y, and z that are nonzero. Such that 1 ( y>x and xy=z. The period after nonzero looks wrong. Is it a comma? > Yeah, it shouldnt be there. Sorry. See whole whole transcription > below. > Does 1 ( y>x mean that both y & x are less than 1? Looking at the > answer that is the only thing that makes sense to me. I just dont > understand the notation. Neither do I. Have you accurately transcribed the _complete_ sentence? > If x, y, and z are nonzero numbers such that 1(y>x and xy=z, which of > the following CANNOT be true? > 1) y>z > 2) y=z > 3) z=x > 4) x>z > 5) z>0 > I think its 4) x>z, but the answer in the review says its 2) y=z. > However, when x=1, y=2, and z=2 it meets all of the criteria and y=z, > so #2 cannot be the answer. When looking at their explanation for why > #2 is the answer they say because all of the other choises CAN be > true #2 must be the correct answer. Which, to me, is not only a BS > cop-out, but wrong. > Also, in the #4 explanation it says According to the question, Y must > be greater to or equal to one. Thus *my* question on the notation. > Whatever happened to <=? I should also state that nowhere else in > the review does it mention that the Ô(Ô to <= notation substitution > exists. I think its just a shoddily put-together review. === Subject: Re: DIAL M FOR SLOPE by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1LLaNO23840; I have watched the discussion diverge aymptotically (or is it exponentially?)from my original question : where did the m as a notation for slope (or gradient)come from? What ever happened to my original question? I gather nobody knows the answer. Paul Bruckman === Subject: Re: DIAL M FOR SLOPE > I have watched the discussion diverge aymptotically (or is it > exponentially?)from my original question : where did the m as a notation > for slope (or gradient)come from? What ever happened to my original question? > I gather nobody knows the answer. Paul Bruckman It has been a year or two since this was asked here. It used to be a regular question. Anyway, to see some info go down to slope at... http://members.aol.com/jeff570/geometry.html -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: (A<-->B) as Set Derivative by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1LLaN023804; >It also gives other interesting and useful differentiation formulas: >(A u B) = AB = A<-->B AB = A<-->B (A U B) >which I suppose tells us that: >(f + g) = fg - (f + g) >-- >Im not interested in mathematics that might have anything >to do with reality. -- Russell Easterly, in sci.math I was expecting some reply in this direction, and Toni Lassila fulfils expectations, but notice that I pointed out that A<-->B is the analog of fg With Respect to Ô. In other words, I only indicated that (fg) goes over to (A<-->B). Before you begin dissecting whether fg should go over to something by this (and we can argue that forever in Heaven, where I expect to be soon at the age of 65), lets take a look at: 1) (A + iB)(A + iB)* = (A + iB)(A - iB) = A^2 + B^2 where I use the notation A, B in (1) instead of f, g respectively (differentiable functions) to make the point that we can generalize complex numbers to sets (had MacLane and Lawveres Category Theory accepted Generalized Sets instead of Set-trivializing Objects?). Does anybody recognize Borns Probability in Mathematical Physics in (1)? It might be more familiar in: 2) ww* = (x + iy)(x + iy)* = (x + iy)(x - iy) = x^2 + y^2 where w is the wave function. Of course, we have to define A^2 and A^2 + B^2, and we may end up having to replace + on the right-hand- side of (1) by U, but heck, nobody said it would be easy. Ah yes, what has been accomplished? Well, A + iB is one sided or Non-Symmetric, like A-->B = A U B. (A + iB)* = A - iB is also one-sided or Non-Symmetric, but in an opposite direction (for functions, f - ig is down versus the up of f + ig, although the directions only reverse from the tip of f). Notice that A-->B = A U B, while B-->A = B U A, which are reverses of a rather clear type. So we only have an explanation and a relationship between the Quantum Theory probability in Mathematical Physics and Classical Set and Classical Probability Theory. However, you might try following Max Born or Niels Bohr or Werner Heisenberg, who not only could not explain their choice of x^2 + y^2 grounded in Classical Set and Classical Probability Theory but stopped looking - perhaps they were too old to look? Osher Doctorow === Subject: Re: (A<-->B) as Set Derivative >>It also gives other interesting and useful differentiation formulas: >>(A u B) = AB = A<-->B AB = A<-->B (A U B) >>which I suppose tells us that: >>(f + g) = fg - (f + g) >>-- >>Im not interested in mathematics that might have anything >>to do with reality. -- Russell Easterly, in sci.math > I was expecting some reply in this direction, and Toni Lassila > fulfils expectations, but notice that I pointed out that A<-->B > is the analog of fg With Respect to Ô. Aha! This makes more sense, So the analogue of (fg) = fg + fg is (A<-->B) = (A<-->B) + (A<-->B). Slight problem .... it isnt true. > In other words, I only > indicated that (fg) goes over to (A<-->B). Before you begin > dissecting whether fg should go over to something by this (and > we can argue that forever in Heaven, where I expect to be soon at > the age of 65), Have a nice trip. > Ah yes, what has been accomplished? Nothing? > Well, A + iB is one sided or > Non-Symmetric, like A-->B = A U B. (A + iB)* = A - iB is also > one-sided or Non-Symmetric, but in an opposite direction (for > functions, f - ig is down versus the up of f + ig, although the > directions only reverse from the tip of f). Notice that A-->B > = A U B, while B-->A = B U A, which are reverses of a rather clear > type. Yup, nothing. Instead of justifying your bogus analogy, you start prattling on about yet another bogus one: A --> B is like A + iB. Where the **** does this come from? > So we only have an explanation No explanation of anything. >and a relationship between the Quantum > Theory probability in Mathematical Physics and Classical Set and > Classical Probability Theory. However, you might try following > Max Born or Niels Bohr or Werner Heisenberg, All conveniently dead. > who not only could not > explain their choice of x^2 + y^2 grounded in Classical Set and > Classical Probability Theory but stopped looking - perhaps they were > too old to look? And they had nothing to do with your bogus analogy between differeniation and complementation. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Great mistakes of the physicists > http://hamidvansari.topcities.com/ You see yourself this way, http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees you this way, http://www.mazepath.com/uncleal/effete3.png http://b5.sdvc.uwyo.edu/bab5/snds/argcstpd.wav http://w0rli.home.att.net/youare.swf http://www.mazepath.com/uncleal/sunshine.jpg -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: needing help from a kind soul can you please help me achieve the answer? Al and Bob are at opposite ends of a diameter of a silo in the shape of a tall right circular cylinder with radius 150 ft. al is due west of Bob. Al begins walking along the edge of the silo at 6 ft. per second at the same moment that Bob begins to walk due east at the same speed. The value closest to the time in seconds when Al first can see Bob is what? answer: 48 Choogu === Subject: Re: needing help from a kind soul >Al and Bob are at opposite ends of a diameter of a silo in the >shape of a tall right circular cylinder with radius 150 ft. al is due >west of Bob. Al begins walking along the edge of the silo at 6 ft. per >second at the same moment that Bob begins to walk due east at the same >speed. The value closest to the time in seconds when Al first can see >Bob is what? answer: 48 First, let us compute the distance they travel. When Al has walked s silo radii around the silo, he has moved s radians as measured from the center of the silo. At this time, Bob is s+1 silo radii from the center of the silo. When Al can first see Bob, the triangle formed by Bob, Al, and the center of the silo is a right triangle, with Al at the right angle. Since Al has walked s radians, the angle at the center of the silo is pi-s radians. Since the side between the center and Al is 1 silo radius and the side between the center and Bob is s+1 silo radii, we get the equation 1/(s+1) = cos(pi-s) = -cos(s) Solving this equation tells us that s is about 1.92029909426 silo radii. Multiply by the silo radius of 150 ft to get 288.044864139 ft and divide by 6 ft/sec to get 48.0074773565 sec. Rob Johnson take out the trash before replying === Subject: Re: needing help from a kind soul Choogu === Subject: Re: needing help from a kind soul 1. Al and Bob are at opposite ends of a diameter of a silo in the shape of a tall right circular cylinder with radius 150 ft. al is due west of Bob. Al begins walking along the edge of the silo at 6 ft. per second at the same moment that Bob begins to walk due east at the same speed. The value closest to the time in seconds when Al first can see Bob is what? answer: 48 2. if a, b, c, and d are nonzero numbers such that c and d are solutions of x^2+ax+b=0 and a and b are solutions of x^2+cx+d, find a+b+c+d. ans: -2 3. A boat with an ill passenger is 7.5 mi north of a straight coastline which runs east and west. A hospital on the coast is 60 miles from the point on shore south of the boat. If the boat starts toward shore at 15 mph at the same time an ambulance leaves the hospital at 60 mph and meets the ambulance, what is the total distance (to the nearest 0.5 mile) traveled by the boatand the ambulance? ans:62.5 ********************************************************* P.S.I posted this lot last time you asked Let x = 6t feet be the distance travelled by Al (A) and Bob (B) in t seconds. Sketch a tangent to the circle, (representing the cylinder), from the centre O, draw in the radius which is perpendicular to the tangent contact at C. pi - x/150 is the acute angle BOC The cosine of this angle is OC/OB =150/(150+x) = C, say Use a graphing calculator to graph y = cos (pi - x/150) and y =150/(150+x) With x from - 1 to 300, y=-1.1 to+1.1 intersection gives x=28 8.0448 hence t is close to 48 [No graphical calculator allowed? As a very rough method, a careful sketch 150 (1 - .342)/.342 =approx 288 ] ************************************** Quadratics can be written x^2 - (sum of roots)+(product of roots) = 0 c+d = -a cd = b a+b = -c ab = d b-c-d=-c b=d c=b/d=1 a=d/b=1 a+b+c+d=-(a+c)=-2 **************************** Take meets to mean arrives at the same time, and assume that boat intelligently steered towards optimum angle x west of due south. Sketch this for yourself. Let cosine x=c, sine x=s, tangent x=t, Distance travelled by boat = 7.5/c Miles Time taken by boat = 30/c Miles Miles = minutes for ambulance = 60 - 7.5 t Equating x and dividing by 7.5 4/c = 8 - t 4 = 8 c - s s = 4(2c - 1) s^2 = 16(4c^2 - 4c + 1) = 1 - c^2 65c^2 - 64c + 15 = 0 = (5c - 3)(13c - 5) Shortest time taken by boat corresponds to largest c, so select c = 3/5 which also gives t = 4/3 Distance travelled by boat = (7.5 x 5)/3 = 12.5 miles Minutes taken by boat= 4 x 12.5 miles = 50 but this is same as miles travelled by ambulance, so Total distance travelled by boat and ambulance = 62.5 [ By the way, the other solution to the quadratic also allows boat and ambulance to arrive at same time, but c = 5/13 with t = 12/5 lead to a boat distance of 19.5 miles and an ambulance distance of 78 miles. This is interpreted as the less useful solution where the boat is steered the east of south ] afterwards Ian Hutcheson === Subject: Martingales Could someone show me in detail; Check the following process X_t are martingales wrt {F_t} 1) X_t = B_t + 4t; B_t is Brownian motion 2) X_t = B^2_t === Subject: Re: Martingales > Could someone show me in detail; > Check the following process X_t are martingales wrt {F_t} > 1) X_t = B_t + 4t; B_t is Brownian motion > 2) X_t = B^2_t I assume B_t is Brownian motion wrt {F_t}. E[X_t] = E[B_t+4t] = E[B_t]+4t = 0+4t = 4t is not constant, so X_t is not a martingale... Maybe X_t is supposed to be a submartingale? Or what? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Martingales >> Could someone show me in detail; >> Check the following process X_t are martingales wrt {F_t} >> 1) X_t = B_t + 4t; B_t is Brownian motion >> 2) X_t = B^2_t >I assume B_t is Brownian motion wrt {F_t}. >E[X_t] = E[B_t+4t] = E[B_t]+4t = 0+4t = 4t is not constant, so >X_t is not a martingale... >Maybe X_t is supposed to be a submartingale? Or what? My guess would be its a homeword problem asking the reader to determine whether or not these are martingales. (The second example isnt either.) ************************ David C. Ullrich === Subject: 1st order Diff Eqns Hi there, Ive been studying some kinetics and have come up with this differential eqn from my experiment. dy/dx = 2y-xy^2 I cant for the life of me see how you can solve it. As far as I see it using the substitution y=vx doesnt help. Can anyone lend me a hand? Sarah -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: 1st order Diff Eqns > http://www.sosmath.com/diffeq/first/bernouilli/bernouilli.html Usually spelled Bernoulli, not in a pseudo-French manner Bernouilli === Subject: Re: 1st order Diff Eqns > Hi there, > Ive been studying some kinetics and have come up with this > differential eqn from my experiment. > dy/dx = 2y-xy^2 > I cant for the life of me see how you can solve it. As far as I see > it using the substitution y=vx doesnt help. Can anyone lend me a > hand? but the substitution v = 1/y will v = -y/(yy) dy/dx [-1/(yy)] = 2y [-1/(yy)] - xy^2 [-1/(yy)] -y/(yy) = -2/y + x v = -2v + x v + 2v = x your equation belongs to a family of differential equations of called Bernoulli equations: http://www.sosmath.com/diffeq/first/bernouilli/bernouilli.html > Sarah === Subject: Re: 1st order Diff Eqns > Hi there, > Ive been studying some kinetics and have come up with this > differential eqn from my experiment. > dy/dx = 2y-xy^2 > I cant for the life of me see how you can solve it. As far as I see > it using the substitution y=vx doesnt help. Can anyone lend me a > hand? > Sarah Dont you look at the replies you get? This is what I gave you on sci.physics 20 minutes after you asked: | > take | > z = 1/y | > so | > dz/dx = -1/y^2 dy/dx | > Then | > -1/y^2 dy/dx = -1/y^2 (2y - xy^2) | > = -2/y + x | > so | > dz/dx = -2z + x | > | > hth | | By the way, my taking z = 1/y was a bit of a lucky guess, induced | by your complaint about having y^2 screwing it up :-) | But in fact your equation is a simple example of the Bernoulli equation. | See for instance | http://www.sosmath.com/diffeq/first/bernouilli/bernouilli.html | In this case | p(x) = -2 | q(x) = -x | n = 2 | | Dirk Vdm Dirk Vdm -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: 1st order Diff Eqns === Subject: 1st order Diff Eqns >Ive been studying some kinetics and have come up with this >differential eqn from my experiment. >dy/dx = 2y-xy^2 Doesnt look dimensionally balanced. >I cant for the life of me see how you can solve it. >using the substitution y=vx doesnt help. 1/y dy/dx = 2 - xy u = log y du/dx = 2 - xe^u u = 2x - v 2 - dv/dx = 2 - xe^2x e^-v e^v dv/dx = xe^2x w = e^v dw/dx = xe^2x w = (1/4)(2x - 1)e^2x + c v = log w u = 2x - log w y = e^2x / w = 4 / (2x - 1 + 4ce^-2x) = 4 / (2x - 1 + ce^-2x) let c = 0 y = 4/(2x-1) y = -8/(2x - 1)^2 2y - xy^2 = 8/(2x-1) - 16x/(2x-1)^2 c /= 0 asymptotically approaches c = 0. ---- -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: 1st order Diff Eqns > Hi there, > Ive been studying some kinetics and have come up with this > differential eqn from my experiment. > dy/dx = 2y-xy^2 > I cant for the life of me see how you can solve it. As far as I see > it using the substitution y=vx doesnt help. Can anyone lend me a > hand? > Sarah Mathematica gives sol=DSolve[y[x]==2*y[x]-x*y[x]^2,y[x],x]; Print[sol]; {{y[x] -> (4*E^(2*x))/(-E^(2*x) + 2*E^(2*x)*x + 4*C[1])}} -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: 1st order Diff Eqns >Ive been studying some kinetics and have come up with this >differential eqn from my experiment. >dy/dx = 2y-xy^2 >I cant for the life of me see how you can solve it. As far as I see >it using the substitution y=vx doesnt help. Can anyone lend me a >hand? On both sides, subtract 2y and divide by y^2 to get 1 dy 1 --- -- - 2 - = -x [1] y^2 dx y Using u = 1/y and negating both sides gives du -- + 2u = x [2] dx This is of the form (D+2)u = x, where D = d/dx. Equations of this form can be handled using an integrating factor. We want to find a function m, the integrating factor, which satisfies (D+2)m = 0, because then, (D+2)(mz) = z(D+2)m + mDz = mDz [3] Substituting u/m for z, [3] becomes (D+2)u = mD(u/m) [4] Noting that (D+2)e^{-2x) = 0, combine [2] and [4] to get -2x 2x x = (D+2)u = e D(e u) [5] Multiplying both sides of [5] by e^{2x} yields 2x 2x e x = D(e u) [6] Equation [6] can be solved simply by integration; dont forget the constant of integration. To verify that [6] really is the same as [2], use the product rule: 2x 2x 2x e x = 2 e u + e Du [7] divide [7] by e^{2x}: x = 2u + Du [8] and [8] is the same as [2]. Rob Johnson take out the trash before replying -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: 1st order Diff Eqns >Hi there, >Ive been studying some kinetics and have come up with this >differential eqn from my experiment. >dy/dx = 2y-xy^2 >I cant for the life of me see how you can solve it. As far as I see >it using the substitution y=vx doesnt help. Can anyone lend me a >hand? >Sarah This is a Bernoulli equation. Multiply by -y^(-2): -y^(-2)y + 2y^(-1) = x and let u = y^(-1) Then u = -y^(-2)y giving: u + 2u = x. This has complementary solution Ce^(-2x) and particular solution (1/2)x - 1/4 so its general solution is u = Ce^(-2x) + (1/2)x - 1/4 and y = 1/u. Note this method precludes the y == 0 solution. --Lynn -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: 1st order Diff Eqns Well, Ive come up with the solution 1/y = x/2 - 1/4 + C*e^(-2x), for any constant C. -Michael. > Hi there, > Ive been studying some kinetics and have come up with this > differential eqn from my experiment. > dy/dx = 2y-xy^2 > I cant for the life of me see how you can solve it. As far as I see > it using the substitution y=vx doesnt help. Can anyone lend me a > hand? > Sarah -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Ellipses & reducing to General Equation thereof Hi - Ive read about the General Quadratic Equation (Ax^2+By^2+Cx+Dy+E=0) & the equation for an ellipse (x-h)^2+(y-k)^2=0; Ive seen a number of web pages dealing with the above. But -- Say you have an ellipse with center at the origin, foci at (-2,0) and (2,0). Now translate the ellipse 3 units in the positive x direction, and 2 units positve y. Figuring out the quadratic equation is still pretty straightforward, right? Now rotate the ellipse counterclockwise 30 degrees. How do I figure out the quadratic equation now? Craig === Subject: Re: Ellipses & reducing to General Equation thereof >Hi - >Ive read about the General Quadratic Equation (Ax^2+By^2+Cx+Dy+E=0) & the >equation for an ellipse (x-h)^2+(y-k)^2=0. Neither of the above equations are correct! A general quadratic equation has a term in x y. Your second equation specifes only the point (h,k). The general equation of an ellipse *with axes parallel to the coordinate axes* is a (x-h)^2 + b (y-k)^2 = 1, where both a and b are positive. >Ive seen a number of web pages >dealing with the above. >But -- >Say you have an ellipse with center at the origin, foci at (-2,0) and >(2,0). and, I assume, one more piece of information to specify the ellipse. >Now translate the ellipse 3 units in the positive x direction, and 2 units >positve y. Figuring out the quadratic equation is still pretty >straightforward, right? >Now rotate the ellipse counterclockwise 30 degrees. How do I figure out >the quadratic equation now? The equation for the rotated ellipse will contain a term in x y; it is no longer of the form a (x-h)^2 + b (y-k)^2 = 1. The general equation for an ellipse is a (x-h)^2 + b (x-h) (y-k) + c (y-k)^2 = 1, where a and c are positive and b^2 < 4 a c. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Ellipses & reducing to General Equation thereof Craig Reed > Ive read about the General Quadratic Equation (Ax^2+By^2+Cx+Dy+E=0) & the > equation for an ellipse (x-h)^2+(y-k)^2=0; Ive seen a number of web pages > dealing with the above. ... > Now rotate the ellipse counterclockwise 30 degrees. How do I figure out > the quadratic equation now? Its just a linear change of variables. For a rotation around the origin, replace x and y by x cos G + y sin G and x sin G - y cos G respectively, where G is the angle. LH === Subject: Re: Ellipses & reducing to General Equation thereof > Hi - > Ive read about the General Quadratic Equation (Ax^2+By^2+Cx+Dy+E=0) & the > equation for an ellipse (x-h)^2+(y-k)^2=0; Ive seen a number of web pages > dealing with the above. > But -- > Say you have an ellipse with center at the origin, foci at (-2,0) and > (2,0). > Now translate the ellipse 3 units in the positive x direction, and 2 units > positve y. Figuring out the quadratic equation is still pretty > straightforward, right? > Now rotate the ellipse counterclockwise 30 degrees. How do I figure out > the quadratic equation now? > Craig A translation of coordinates (or, equivalently, an opposite translation of the ellipse) can be achieved by replacing x by x+a and y by y+b in the equation of the ellipse. A rotation of coordinates (or, equivalently, an opposite rotation of an ellipse around origin) can be achieved by relacing x by x*a-y*b and replacing y by x*b+y*a in the equation of the ellipse, where a^2 + b^2 = 1. === Subject: Re: My fear, consider this Anonymous > I finally decided to take a look at something that youve said, rather than > just reading the commands from others and assuming that youre wrong. ... > So far, all is well. But then came this: > and notice you STILL have that f on the front. > what is the Ôfront? do you mean the fact that there is an f that appears > on the left hand side of the equation? > I dont want to hear that it isnt applicable because f isnt an > integer, as if you will have to get a polynomial reducible over Q if > you pick the right f, as thats just bogus. > What? Why isnt f an integer? What are you talking about here? Im not at all sure that Harris himself knows. Unfortunately JSH is quite dishonest, and he uses several M.O.s to evade even the clearest refutations of his claims: turning a constant into a variable, talking about a factorization x=yz without specifying a ring in which y and z are to be elements, and so on. In the sci.math archives http://mathquest.com/discuss/sci.math/a/ you can find dozens of specimens just like this latest one of his. LH === Subject: How to study for math? Im trying to see if my study habits are part of the reason why I can be hot/cold with math. So how exactly do you study for proof-based math classes? Here is what I usually do: Read the chapter laying in bed (hey, I was originally a history major). Sometimes I wont understand a part of the proof of the theorem and Ill be tempted to just skip the theorem entirely. After Im done with this, Ill go to the homework (this may not be the same day). If I dont remember one of the definitions, Ill have to go back and look at what it says. I also have a problem with waiting to look at the answer. If I cant figure out the problem in about a minute or so, Ill just look at the back of the book. Im not sure how good/bad this is considered but its a habit Ive developed nevertheless. Any comments or suggestions? === Subject: Re: How to study for math? > Im trying to see if my study habits are part of the reason why I can be > hot/cold with math. > So how exactly do you study for proof-based math classes? Im assuming you are trying to pass a test where you will be asked to prove things. 1. Re-read and *understand* all proofs you have seen in course: during class, text book, homework, etc. You should be able to repeat the proofs without looking at the solution. But not repeat them from pattern recognition, but because you understand the logical argument as to how what you have just written is proof. 2. Now re-read and memorize (unless they are accessable during the test--IMO they always should be) all axioms, theorems, corollaries, etc. from course (presented in class, text book, homework, etc.) Understand these as well. One way I learn to understand theorems/corollaries is to study the proof of the theorem/corollary. You should be able to write down a proof of the theorem/corollary without looking at the one provided. Again do not memorize and regurgitate the steps, understand the proof. 3. The best thing to do at this point is to find related proofs to practice on. The best is unassigned homework problems. You need the solutions available to compare your results to. If solutions are not available you can always as the TA/Prof to look at a few you have done. Other texts or internet perhaps if you cant find sample problems. Now you are as ready for the exam as you can be. You will be asked to provide really one of two types of proofs. A proof similar to what you have already seen, or a new proof that lends itself to be proved with one or more of the the axioms/theorems/corollaries you have seen. In either case you have the knowledge to complete the problem. Now sometimes you have all the knowledge but dont see the how. I always start with a sketch. And then work with the given. If something like prove every X that has a Y also has a Z, then start with an X that has a Y. Now condiser an X with a Y with and without a Z. Why is without a Z not possible? As you play with this kind of thinking, you will usually see a connection to some of the stuff you have been studying, i.e. you have seen this before in theroem 12.4.5. === Subject: Re: How to study for math? >Read the chapter laying in bed (hey, I was originally a history major). >> >> Hens lay, people lie. (Some more than others.) >Lying in bed refers to being supported by the bed, but laying in bed would >be OK for referring to being spread over the surface of the bed. >>Nope. Saying youre laying in bed is not right, regardless of whether >>youre spread all over the surface. (There _is_ a sense of the word >>lay under which it might be regarded as correct, but thats not >>what were talking about here.) Lay is transitive, lie is >>intransitive. >Those garlands of ßowers they use in Hawaii are called leis. I think >its correct form to say to lei someone when you mean to put the >garland on someone. When some of the other students came back from >Hawaii, they told me theyd gotten leid. It took me the longest >time to figure this out. I suspect they were just playing with your head. Kids these days, got no respect... >Still not sure what the bed has to do with it. >dave ************************ David C. Ullrich === Subject: Re: How to study for math? >>Read the chapter laying in bed (hey, I was originally a history major). > > Hens lay, people lie. (Some more than others.) >>Lying in bed refers to being supported by the bed, but laying in bed would >>be OK for referring to being spread over the surface of the bed. >Nope. Saying youre laying in bed is not right, regardless of whether >youre spread all over the surface. (There _is_ a sense of the word >lay under which it might be regarded as correct, but thats not >what were talking about here.) Lay is transitive, lie is >intransitive. Those garlands of ßowers they use in Hawaii are called leis. I think its correct form to say to lei someone when you mean to put the garland on someone. When some of the other students came back from Hawaii, they told me theyd gotten leid. It took me the longest time to figure this out. Still not sure what the bed has to do with it. dave === Subject: Re: How to study for math? >Read the chapter laying in bed (hey, I was originally a history major). >> Hens lay, people lie. (Some more than others.) >Lying in bed refers to being supported by the bed, but laying in bed would >be OK for referring to being spread over the surface of the bed. Nope. Saying youre laying in bed is not right, regardless of whether youre spread all over the surface. (There _is_ a sense of the word lay under which it might be regarded as correct, but thats not what were talking about here.) Lay is transitive, lie is intransitive. >Id guess >he was probably doing both. ************************ David C. Ullrich === Subject: Re: How to study for math? >>Read the chapter laying in bed (hey, I was originally a history major). > Hens lay, people lie. (Some more than others.) Lying in bed refers to being supported by the bed, but laying in bed would be OK for referring to being spread over the surface of the bed. Id guess he was probably doing both. -- --Tim Smith === Subject: Re: How to study for math? >Here is what I usually do: >Read the chapter laying in bed (hey, I was originally a history major). Hens lay, people lie. (Some more than others.) === Subject: Re: How to study for math? >Im trying to see if my study habits are part of the reason why I can be >hot/cold with math. >So how exactly do you study for proof-based math classes? >Here is what I usually do: >Read the chapter laying in bed (hey, I was originally a history major). >Sometimes I wont understand a part of the proof of the theorem and Ill be >tempted to just skip the theorem entirely. After Im done with this, Ill go >to the homework (this may not be the same day). If I dont remember one of the >definitions, Ill have to go back and look at what it says. I also have a >problem with waiting to look at the answer. If I cant figure out the problem >in about a minute or so, Ill just look at the back of the book. Im not sure >how good/bad this is considered but its a habit Ive developed nevertheless. Well, if you seriously want to learn mathematics you need to break that habit. Its not that its considered bad, its simply impossible to learn math without doing a lot of work, including solving problems that take a long time to figure out. If you look in the back of the book after a minute youre never going to learn to do it yourself. >Any comments or suggestions? ************************ David C. Ullrich === Subject: Re: How to study for math? >>definitions, Ill have to go back and look at what it says. I also have a >>problem with waiting to look at the answer. If I cant figure out the >>problem in about a minute or so, Ill just look at the back of the book. >>Im not sure how good/bad this is considered but its a habit Ive >>developed nevertheless. > Well, if you seriously want to learn mathematics you need to break that > habit. Its not that its considered bad, its simply impossible to > learn math without doing a lot of work, including solving problems that > take a long time to figure out. If you look in the back of the book after > a minute youre never going to learn to do it yourself. It might be a good idea for the original poster to pick up a one of the volumes of Knuths The Art of Computer Programming. They all have a lot of mathematical content, and in particular, a *lot* of the exercises are basically math problems. The nice thing about Knuth is that all the exercises include an estimate of about how long they should take. That might give the OP something to shoot for before wimping out and looking in the back. (Theres a lot of very interesting math in those exercises, too. TAoCP is an excellent set of books for mathematics students even if they have no intention of ever going near a computer. Id say volume 2 is the best, from a mathematical point of view, if you are only going to get one volume). -- --Tim Smith === Subject: Re: How to study for math? > Im trying to see if my study habits are part of the reason why I can be > hot/cold with math. > So how exactly do you study for proof-based math classes? > Here is what I usually do: > Read the chapter laying in bed (hey, I was originally a history major). Fine, if that means no distractions. Have paper and pencil with you. > Sometimes I wont understand a part of the proof of the theorem and Ill be > tempted to just skip the theorem entirely. After Im done with this, Ill go Skip the proof the first time, but at least try to understand what its saying. Eventually youll have to master the proofs, but if you dont like the ones in the book try another book. > to the homework (this may not be the same day). If I dont remember one of the > definitions, Ill have to go back and look at what it says. I also have a Fine--if you forget something theres not much else you can do, but if none of it sticks in your mind that may be telling you something (like: go back to history!). > problem with waiting to look at the answer. If I cant figure out the problem > in about a minute or so, Ill just look at the back of the book. Im not sure A minute or so is not enough. > how good/bad this is considered but its a habit Ive developed nevertheless. Try the exercises that dont have solutions in the back, try another book if necessary. > Any comments or suggestions? -- G.C. === Subject: Re: How to study for math? > definitions, Ill have to go back and look at what it says. I also have a > problem with waiting to look at the answer. If I cant figure out the problem > in about a minute or so, Ill just look at the back of the book. Im not sure > how good/bad this is considered but its a habit Ive developed nevertheless. There are problems that you must sweat to solve, and -should- sweat to solve. Dont be so quick to look up the answer. If you have trouble with a problem then find a similar problem that you can solve and carry over the skill you gain. Your one minute approach is way to superficial for you to learn anything. Bob Kolker. P.S. Do both your reading and your problem solving with pencil and paper at the ready. This means you have -sit up- to do the work. It is better posture, easier on your eyes and better for your circulation or breathing. If you have to take a break, do something physical like excercise or jog in place. That will shake the cobwebs out. === Subject: Re: integral > ...integrating from 0 to infinity e^(-(x^2))dx? > Let I be the number we want. > I^2 > = (integral from 0 to infinity e^(-(x^2))dx) > times (integral from 0 to infinity e^(-(y^2))dy) > = double integral (0 to infinity in x and y)[e^-(x^2+y^2)]dy*dx > (change to polar coordinates) > = double integral(0 <= theta <= pi/2)(0 <= r < infinity) > of the function: e^(-r^2) times r*dr*d(theta) > infinity > = (-1/2)e^(-r^2)| times pi/2 > 0 > = pi/4. > Since I^2 = pi/4, > I = (1/2)sqrt(pi). > Holy smoke! Thats neat! > I have a question ... very silly ... regarding this business. How > exactly this substitution thing work? > SPECIFICALLY: What are the necessary and sufficient conditions for a > successful substitution? > Is there anything about multiple valued functions (probably inverse > functions), open or closed domain and ranges, discontinuities, poles, > whatever topological ... that I should be aware of while doing a > substitution? I know for sure whatever can go wrong normally does not > go wrong, or else I would have done very bad in the exams! > Take this one for example ... coordinating (x,y) plane with (r,theta) > has something special about it that scares me. x runs from -infinity > to infinity, so does y, but r goes just half of that range, and > theta is BOUNDED! I know 0 <= theta <= pi sounds fine, and that > omitting the = in <= would do know harm ... et cetera ... but I just > wanted a brief review of the whole story. The reason that it goes from 0 to pi2, is if you graph where the function goes while in dxdy you will see that its only in the first quadrant. The y goes from 0 to positive infinity and the x goes from 0 to positive infinity. The whole first quadrant goes from 0 to pi/2 and thats why the d_theta has those limits. > Anyone care to go through the pain? Just to encourage you: remember > how it felt when it all came together for you the first time? -Greg === Subject: Re: integral ...integrating from 0 to infinity e^(-(x^2))dx? Let I be the number we want. I^2 > = (integral from 0 to infinity e^(-(x^2))dx) > times (integral from 0 to infinity e^(-(y^2))dy) = double integral (0 to infinity in x and y)[e^-(x^2+y^2)]dy*dx (change to polar coordinates) = double integral(0 <= theta <= pi/2)(0 <= r < infinity) of the function: e^(-r^2) times r*dr*d(theta) infinity > = (-1/2)e^(-r^2)| times pi/2 > 0 > = pi/4. Since I^2 = pi/4, I = (1/2)sqrt(pi). > Holy smoke! Thats neat! > I have a question ... very silly ... regarding this business. How > exactly this substitution thing work? > SPECIFICALLY: What are the necessary and sufficient conditions for a > successful substitution? > Is there anything about multiple valued functions (probably inverse > functions), open or closed domain and ranges, discontinuities, poles, > whatever topological ... that I should be aware of while doing a > substitution? I know for sure whatever can go wrong normally does not > go wrong, or else I would have done very bad in the exams! > Take this one for example ... coordinating (x,y) plane with (r,theta) > has something special about it that scares me. x runs from -infinity > to infinity, so does y, but r goes just half of that range, and > theta is BOUNDED! I know 0 <= theta <= pi sounds fine, and that > omitting the = in <= would do know harm ... et cetera ... but I just > wanted a brief review of the whole story. > The reason that it goes from 0 to pi2, is if you graph where the function > goes while in dxdy you will see that its only in the first quadrant. The y > goes from 0 to positive infinity and the x goes from 0 to positive infinity. > The whole first quadrant goes from 0 to pi/2 and thats why the d_theta has > those limits. > Anyone care to go through the pain? Just to encourage you: remember > how it felt when it all came together for you the first time? > -Greg God! Thats wasnt my question! === Subject: Re: Sequences Involving Binary & Prime-Factorization > First, we have the sequence formed by simply letting > a(m) = 1, if the sum of all exponents of the prime-factorization of m > has no carries when summed in in base-2. > a(m) = 0, if there are any carries in the summing of the exponents of > the prime-factorization of m. > So, for example, > a(12) = 1 because 12 = 2^2 *3^1, > and, in base-2, 2 = Ô10, 1 = Ô1, > and Ô10 and Ô1 have their ones in different positions. > But, > a(24) = 0, because 24 = 2^3 *3^1, > and, in base-2, 3 = Ô11, 1 = Ô1, > which both share a rightmost one. > Unless I made an error, this sequence is not yet in the Encyclopedia > Of Integer Sequences. > a(k) -> 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1,... > (Defining a(1) = 1 makes b()-recursion below work.) > Now, let b(m) = > sum{k=1 to m} a(k). > b(m) gives the number of positive integers <= m and with an a() of 1, > obviously. > My main math result related to these sequences: > (a somewhat interesting method of calculating {b()} using recursion): > b(1) = 1; > b(m) = > b(ßoor(sqrt(m)) + sum{p=primes<=m} b(ßoor(sqrt(m/p))), > (or, if we choose to define 1 as a prime: > b(m) = > sum{p=primes<=m} b(ßoor(sqrt(m/p))) ) > And, unless I made an error, this sequence is not yet in the > Encyclopedia Of Integer Sequences either. > b(k) -> 1, 2, 3, 4, 5, 5, 6, 7, 8, 8, 9, 10, 11, 11, 11, 12, 13,... I wondered about the asymptotics of {a(k)} and {b(k)}. We can rewrite the b-recursion (which the original version I give below in copied message), so as to get: b(m) = b(ßoor(sqrt(m))) + sum{k=1 to ßoor(sqrt(m))} a(k) pi(m/k^2), where pi(x) is the number of primes <= x, for x = positive real. So, very roughly (and shunning rigor), I assume, using this new recursion, that b(m) = O(pi(m)) = O(m/ln(m)), using the prime-number theorem (of course). More precisely (and still unrigorously), I get: limit{m -> oo} (b(m) - b(ßoor(sqrt(m)))) *ln(m)/(m - sqrt(m)) = sum{k=1 to oo} a(k)/k^2 = C, which of course converges, since the as are each either 0 or 1. So, on average, each a(k) has a C/ln(k) probability of being 1. Am I right in each of my assumptions? And, what is the constant C = sum{k=1 to oo} a(k)/k^2 numerically? Leroy Quet === Subject: Re: Squares Of Certain Areas: puzzle I must admit that my formula for part-2 is simply a partial sum, although an interesting partial sum, interesting since it is an unexpected alternate representation of the straight-forward partial-sum which gives the area as the sums of squares areas (taken over the ks from part-1). If anyone can find a more simple representation of part-2s answer, this would be interesting. But I will wait a couple days to give answer to this puzzle. Leroy Quet > You have a FIXED sequence of distinct positive integers, {k}, which is > to be determined. > For all ks <= EACH positive integer m, > we can take a number n (n <= m) of (geometric) squares, > each of *integer* side-length, > where the j_th square is of the largest square > with an integer area that is <= m/(k_j). > But {k} is such that, if we lined the squares up vertically so that > their horizontally-running edges coincide and they do not overlap, > the length from the upper side of the top square to the lower side of > the bottom square will ALWAYS be m. > 2 questions: > 1) Which integers make up {k_j}? > 2) Give a formula for the total area of the (n number of) squares. > Example: > For m = 4, > We have 1, 2, 3 in {k}, but not 4. > So, we have the squares, as lined up: > ------ > ! ! ! > ------ k=1, area = 4 <= 4/1 > ! ! ! > ==---- > ! ! k=2, area = 1 <= 4/2 > == > ! ! k=3, area = 1 <= 4/3 > -- > Height of the stack is m = 4. > Area is 6. > Leroy Quet === Subject: Re: Squares Of Certain Areas: puzzle Answer below: > I must admit that my formula for part-2 is simply a partial sum, > although an interesting partial sum, interesting since it is an > unexpected alternate representation of the straight-forward > partial-sum which gives the area as the sums of squares areas (taken > over the ks from part-1). > If anyone can find a more simple representation of part-2s answer, > this would be interesting. > But I will wait a couple days to give answer to this puzzle. > Leroy Quet > You have a FIXED sequence of distinct positive integers, {k}, which is > to be determined. > > For all ks <= EACH positive integer m, > > we can take a number n (n <= m) of (geometric) squares, > each of *integer* side-length, > > where the j_th square is of the largest square > with an integer area that is <= m/(k_j). > > > But {k} is such that, if we lined the squares up vertically so that > their horizontally-running edges coincide and they do not overlap, > the length from the upper side of the top square to the lower side of > the bottom square will ALWAYS be m. > > 2 questions: > > 1) Which integers make up {k_j}? > > 2) Give a formula for the total area of the (n number of) squares. > > Example: > For m = 4, > > We have 1, 2, 3 in {k}, but not 4. > > So, we have the squares, as lined up: > > ------ > ! ! ! > ------ k=1, area = 4 <= 4/1 > ! ! ! > ==---- > ! ! k=2, area = 1 <= 4/2 > == > ! ! k=3, area = 1 <= 4/3 > -- > > Height of the stack is m = 4. > > Area is 6. > > Leroy Quet My solution: | | V | | V | | V | | V | | V part 1: {k} is the sequence of squarefree integers (including 1). In other words, m --- | ----- | / | / m/ | --- |_V / k _| k=1 k=squarefree always equals m. (In linear-mode: sum{1<=k<=m,k=squarefree} ßoor(sqrt(m/k)) = m, for all positive integers m.) Part 2: Straightforwardly, we have: area = m --- 2 | ----- | / | / m/ | --- |_V / k _| k=1 k=squarefree = sum{1<=k<=m,k=squarefree} ßoor(sqrt(m/k))^2, which also is: -m + 2 *sum{k=1 to m} a(k), where a(k) = squareroot of largest square number dividing k. So, even though this is not a closed-form in the strictest sense, S(m) - S(m-1) = -1 +2*a(m), is, where S(m) = the sum of the squares areas for m. Leroy Quet === Subject: Re: Few problems about number. > Here they are: > 1. Draw 3 diagrams to show the relationship among numbers. > 2. What are the different between number, numeral & digits? > 3. What are the 3 functions & numbers? If you tell us what steps youve already taken to do your homework, we can give you some more hints. === Subject: Learning about cryptography What are the basic prerequisites to learning about cryptography and actually learning how to do it? About me: No computer science background but a pretty good (abstract) math background. I ask because neither the math nor the computer science departments offers a course on cryptography and Im not really sure what I need to learn. Is it something you learn in graduate school? If so, what do I need to learn? If it is something that is strictly computer science, can you get into a CS grad school with just a math background to learn about cryptography? I would not mind learning how to program if I need to know that. Any book recommendations would be appreciated. === Subject: Re: Learning about cryptography > What are the basic prerequisites to learning about cryptography and actually > learning how to do it? About me: No computer science background but a pretty > good (abstract) math background. I ask because neither the math nor the > computer science departments offers a course on cryptography and Im not really > sure what I need to learn. Is it something you learn in graduate school? If > so, what do I need to learn? If it is something that is strictly computer > science, can you get into a CS grad school with just a math background to learn > about cryptography? I would not mind learning how to program if I need to know > that. Any book recommendations would be appreciated. 1. See Chapter 2 (IIRC) of the Handbook for Appliced Cryptography (HAC) available online. It is a summary of most of the math that is needed. The math area depends on where you want to be and can include number theory, abstract algebra, probability, coding theory, information theory, elliptic curves and others. See counterpane.com for a writeup as to why. 2. http://www.ciphersbyritter.com/LEARNING.HTM (see writeup plus list of references) 3. Some references: [Bressoud89] D. Bressoud. Factorization and primality testing. Undergraduate Texts in Mathematics. Springer-Verlag, 1989. [Cohen93] Henri Cohen. A course in algorithmic algebraic number theory, volume 138 of Graduate Texts in Mathematics. Springer--Verlag, 1993. [Knuth81] D. E. Knuth. The Art of Computer Programming: Seminumerical Algorithms. Addison-Wesley, 2nd edition, 1981. [Koblitz87b] Neal Koblitz. A course in number theory and cryptography, volume 114 of Graduate Texts in Mathematics. Springer--Verlag, 1987. [Kranakis1986] Evangelos Kranakis. Primality and cryptography. Wiley-Teubner Series in Computer Science. John Wiley & Sons, 1986. [Riesel85] Hans Riesel. Prime numbers and computer methods for factorization, volume 57 of Progress in Mathematics. Birkh.8auser, 2nd edition, 1985. === Subject: Re: Learning about cryptography My main problem is my total lack of computer science background. I dont know where to start there in terms of what to learn -- or if this is even reasonable on my part. I know some abstract algebra and number theory just for the record. === Subject: Re: Learning about cryptography > My main problem is my total lack of computer science background. I dont know > where to start there in terms of what to learn -- or if this is even > reasonable on my part. > I know some abstract algebra and number theory just for the record. Well, for software packages, you can look at: 1. Crypto++ (C++) 2. libtomcrypt (C) 3. Bouncy Castle (Java) things like large integer math, symmetric ciphers, assymetric ciphers and the like. Of course, each item above assumes you know something of programming. However, one can learn about crypto without programming. Learn the terminology and rules and then just get a HW or SW library. HTH, Flip === Subject: Re: e is transcendental (was: classes of transcendental numbers ? >>Why do I even try? Are you even TRYING to learn? Or are you >>so closed-minded that you cant be bothered? >>So? The real part of exp(i pi) is cos(pi), and its imaginary part >>is sin(pi), so all you are saying is that cos(pi) = -1 and >>sin(pi) = 0, and we were already aware of these facts. There is >>NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect. > Panagiotis Stefanides >>Yes, but you DONT tell us what your reason is. You cant expect us to >>accept your claims without giving support for those claims. So what >>possible reason could you have for expecting us to agree with your claim >>that exp(i pi) = 0? >The reason is simply that exp(ipi0=-1 should be accompanied >by the statement that this is the real part solution. >Is it fair? >>No, that is not a fair comment. exp(i pi), the complex number, is >>equal to -1, the complex number. There is no need to appeal to the >>real part. Also, what does exp(i pi) = -1 is the real part solution >>mean? You look like you are using terminology in a manner not >>recognized in mathematics. >>David McAnally >>-------------- >e^[i*pi] ,as accepted ,is a phasor. >>In most of the relevant fields of mathematics, e^[i pi] is a complex >>number. >It is only fair to state that its >polar representation is : >e^[i*pi] = MOD 1 , ARG 180 . >>That is Arg 180 degrees, not just Arg 180. And so what? That does >>not lead to your assertion that exp[i pi] = 0, a result for which >>you have given absolutely no support. Why dont you just give up? >>David McAnally >> At the moment, they (the Time Lords) are far from being all-powerful. >> Thats why its been left up to me and me and me. >> quote by: Patrick Troughton in The Three Doctors >>------- >I, have made myself very explicit. >My original question of the implication >of the imaginary component: e^[ipi]=j*0 (to the related proof) >Complex Notation, chapter 12 ,Electrical Technology 3RD ED. >Edward Hughes Longmans ,page 338: >Stares: > OA*=OB+jOC=OA(COStheta+jSINtheta) >* Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,.. >Here is very clear the difference between PHASOR and MAGNITUDES >>I know the difference between a complex number and its modulus >>(or magnitude). You dont have to explain the difference to >>me. >[which(MAGNITUDES) I, referred to as REAL PART or IMMAGINAR PART, >>The magnitude of a complex number is generally not equal to either >>its real or imaginary part. How could you claim that it is? >I, gave a reference: >Complex Notation, chapter 12 ,Electrical Technology 3RD ED. >Edward Hughes Longmans ,page 338: >States: > OA*=OB+jOC=OA(COStheta+jSINtheta) >* Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,.. > What are these magnitudes then ? OB (which can obviously be either positive, zero or negative) is the real part of the complex number. OC (which can obviously be either positive, zero or negative) is the imaginary part of the complex number. OA (which is presumably positive or zero) is the magnitude of the complex number. theta is the argument of the complex number. Note that neither OB nor OC is the magnitude of the complex number. > I should have stated COMPONENTS ]. >>Perhaps you mean that (the imaginary part of e^[i pi]) = 0, in which >>case you are correct, but you have had a lot of problem expressing >>yourself, especially in view of the way that you initially made >>the claim by stating that e^[i pi] = 0, which you described as the >>imaginary part solution, using a terminology that nobody but you >>knows. > e[i*theta]=COStheta+iSINtheta >>I know that it is true that exp[i theta] = cos(theta)+i sin(theta). >>It follows that exp[i pi] = -1. Incidentally, e[i theta] = i e theta, >>unlike what you have written. >thetas could be given and calculations could be performed >for numerical evaluations. >>And for other results as well. >In books is stated that it is FORMULA >and also terms such as evaluate:(-1+i*sqrt[3])^10 >Are these not solutions to problems? >>No. Evaluate (-1+i sqrt(3))^10 is a problem for which you can >>get a solution using exp(i theta) = cos(theta)+i sin(theta). This >>does not mean that exp(i theta) = cos(theta)+i sin(theta) is itself >>a solution. You need a problem before you can describe anything as >>a solution. >Of course ,this crops up when theta is substituted by a given angle. This has nothing to do with your unorthodox usage of the word solution. >>Nobody knows what you mean when you make a statement like exp[i pi] = -1 >>is the real part solution of exp[i pi] = -1, or that exp[i pi] = 0 >>is the imaginary part solution of exp[i pi] = -1. I asked you to >>explain your terminology but you havent bothered. >I, doubt it but ,still I, exlpained it anyway. >>I did ask you. And you did not explain your bizarre terminology. > There >>was no problem, hence there is no solution, whether real part or >>imaginary part. >How else would do You call them,I, said: >I should have stated COMPONENTS I call them the real part of the complex number (or the equation) and the imaginary part of the complex number (or the equation). I do not call them solutions. David McAnally At the moment, they (the Time Lords) are far from being all-powerful. Thats why its been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Re: e is transcendental > Since e^[iPi]=cosPi+isinPi > or , e^[iPi]=-1+i[0] > then there are two solutions here, to the given equatio: > > A) e^[ipi]=-1 the real part solution and > > B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio. >>No, the conclusion from e^[iPi]=-1+i[0] is >>Re(e^[iPi]) = Re(-1+i[0]) = -1 AND >>Im(e^[iPi]) = Im(-1+i[0]) = 0 >>-- >>Daniel W. Johnson >>panoptes@iquest.net >>http:// members.iquest.net/~pano ptes/039 53 36 N / 086 11 55 W >Daniel, >Just saw Your message. >I, thank You. >This is it. >Your help is great, and resolves my question completely. I pointed out exactly the same thing, but you never accepeted it when I made the point. I presume that it means that you wont make any more silly comments along the lines of exp(i pi) = 0. David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. === Subject: Re: Integration by parts help needed ETAtAhRVpGHiuxBQYt8NzeGNgX1i6q/r8AIVAKDhSeoSmE1/uDnNxUU/ Z6pLhHv6 You need a different choice of u and v to sart. If you set u = x and dv = cos 2x dx, things should work out nicely. We are motivated to this substitution, instead of the one you tried, by knowing that dv = x dx will lead to a higher power of x and that does not look good for simplifying the integral. --OL === Subject: Re: Integration by parts help needed > You need a different choice of u and v to sart. If you set u = x and dv > = cos 2x dx, things should work out nicely. > We are motivated to this substitution, instead of the one you tried, by > knowing that dv = x dx will lead to a higher power of x and that does > not look good for simplifying the integral. comfortable with integration by parts now. However I now have another question: How do I integrate cos^3(5x)sin(5x)? I know how to integrate integrals in the form sin(mx)cos(nx) dx and the form sin^mx cos^nx dx, but not a combination of both. === Subject: Re: Integration by parts help needed >> You need a different choice of u and v to sart. If you set u = x >> and dv = cos 2x dx, things should work out nicely. >> We are motivated to this substitution, instead of the one you tried, >> by knowing that dv = x dx will lead to a higher power of x and that >> does not look good for simplifying the integral. > fairly comfortable with integration by parts now. > However I now have another question: > How do I integrate cos^3(5x)sin(5x)? > I know how to integrate integrals in the form sin(mx)cos(nx) dx and > the form sin^mx cos^nx dx, but not a combination of both. I = Int(cos^3(5x)sin(5x), x) = (1/5)Int(cos^3(5x)sin(5x)5, x) I = -(1/5)Int(t^3, t) Don forget reverse the change of variable after integration ... -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Integration by parts help needed > Im having trouble integrating by parts. For example: > I = Integral(x cox(2x) dx) > dv = x, v = x^2/2 > u = cos(2x), du = (1/2)sin(2x) > I = ((x^2/2)cos(2x)) - ((1/4)Integral((x^2)sin(2x) dx)) > Then Im not sure what to do because the integral on the right seems just as > complex as the one I started with...? > BTW what do we use on here for the integration symbol? Try it the other way round with u = x and dv = cos(2 x)dx. A bit of practice is needed before one gets good at choosing what should be the u and what the dv in applying integration by parts. === Subject: Re: Integration by parts help needed In this case if you swap u and v you should find it easier. (See Virgils post) To avoid getting a more complicated result, select the term in x, (or powers of x), to be u, the one that gets differentiated, and select the trig function to be v, which will be the one integrated. Eventually, (perhaps by repeated application), the term in powers of x will reduce to 1. [However, your effort is not wasted, because the integral and other terms that you got by doing it the wrong way, are equal to the results you get when doing it as above; thus you get the result for a further new integral.] Ian Hutcheson === Subject: Re: Simple vector spaces theorem >I am trying to prove the following theorem regarding vector spaces: >If u_1, u_2, ..., u_n span the vector space V, then some subset of {u_1, >u_2, ..., u_n} is a basis of V. [...] Someone else has verified your proof, but here is a much simpler one: Let A be a maximal linearly independent subset of {u1,..., u_n}. It is farily straightforward to show A spans V. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: how important is r^2 in statistics? ETAsAhQpDZkGgB83RrVcOUwH9NR9ZCODVwIUfwAZfvf/ tPkkbZkdfzckmej0VDU= It depends on what you want to do with your regression. If you want to determine which factors affect your response then your t or F value is more important. But if you want to quantitatively model your independent variable you need a high r^2, and 0.14 wont cut it. --OL === Subject: Re: JSH: Non-uniqueness of factorization > I cant imagine what kludge would prevent one from reading headers to > determine whether the message is a mime message. I think what > happened is this. Anyone may correct my mistakes, of course. > > Prior to the widespread use of mime, folks sent messages with > uuencoded files in them. > Good call. Heres MSs page on the error. > http://support.microsoft.com/?kbid=265230 Indeed. Especially the workaround is illuminating. There is simply *nothing* the receiver (who experiences the bug) can do. So this is not a workaround. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Confused . . . expectation value of a random variable > Definition: > A *probability distribution* for a discrete random variable X with > values > { x_k | k in Z } > where Z is the set of integers, is a set of numbers (I think they mean > real) > { p_k | k in Z } > such that > P(X = x_k) = p_k >= 0 > and > sum_{k = -infty}^infty p_k = 1 > > Definition: > If X is a discrete random variable with values > { x_k | k in Z } > and probability distribution > { p_k | l in Z } > then the *mean* or *expected value* of X is > E(X) = sum_{k = -infty}^infty x_k p_k > > Then comes the confusing bit: > > Proposition: > Let Y = g(X) be a function on X. Then then *mean* or *expected value* > of Y is > E(Y) = sum{k = -infty}^infty g(x_k) p_k > > My question is, is that a definition? Because, in case of continuous > random variable, they are clean in saying that it is. However, here, > they comment: > > It is possible to give a general proof of this proposition, but we > shall not do so here. > It is not a definition. If X is as described, then Y = g(X) is another > discrete random variable, and it has value g(x_k) in the event that > X has value x_k. The reason the proposition is not immediately obvious > is that you could have g(x_k)=g(x_j) for different x_k, x_j. So > the sets {Y = y_k} could be unions of several of the sets {X = x_k}, > possibly even infinitely many such sets. Alright. I get it. Subtle. :) === Subject: Re: Confused . . . expectation value of a random variable >> >Definition: >A *probability distribution* for a discrete random variable X with >values >{ x_k | k in Z } >where Z is the set of integers, is a set of numbers (I think they mean >real) >{ p_k | k in Z } >such that >P(X = x_k) = p_k >= 0 >and >sum_{k = -infty}^infty p_k = 1 >Definition: >If X is a discrete random variable with values >{ x_k | k in Z } >and probability distribution >{ p_k | l in Z } >then the *mean* or *expected value* of X is >E(X) = sum_{k = -infty}^infty x_k p_k >Then comes the confusing bit: >Proposition: >Let Y = g(X) be a function on X. Then then *mean* or *expected value* >of Y is >E(Y) = sum{k = -infty}^infty g(x_k) p_k >My question is, is that a definition? Because, in case of continuous >random variable, they are clean in saying that it is. However, here, >they comment: >It is possible to give a general proof of this proposition, but we >shall not do so here. > >>It is not a definition. If X is as described, then Y = g(X) is another >>discrete random variable, and it has value g(x_k) in the event that >>X has value x_k. The reason the proposition is not immediately obvious >>is that you could have g(x_k)=g(x_j) for different x_k, x_j. So >>the sets {Y = y_k} could be unions of several of the sets {X = x_k}, >>possibly even infinitely many such sets. >> >Alright. I get it. Subtle. :) Indeed, this is sometimes called the Law of the Unconscious Statistician, since so many statisticians think of it as the definition. By the way, this is *not* the definition for the continuous case. That is, if X has density f and Y = g(X), the fact that EY = integral( g(x) f(x) dx) is provable from the definition of expectation. More generally, let X have distribution function F and Y = g(x). Y has its own distribution function, call it G: G(y) = P{Y <= y}= P{g(X) <= y}. By the definition presented in most introductory applied probability texts, EY = integral( y dG(y)). (*) By the Law of the Unconscious Statistician, EY = integral(g(x) dF(X)). In both cases, the integrals are Stieltjes. I put definition in quotes, since the real definition of expectation involves integration with respect to the probability measure on the application of the Law of the Unconscious Statistician. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: complex problem ETAtAhUAqC33of97FNSkmxUDfnUmkHcp1zgCFDmXvmtNCGZ54vCFV0qH1oeXN0 VA You can combine these equations to get a quadratic equation in either unknown. This is soluble via the quadratic formula, but you end up taking the square root of a complex number. So how do you do that? Let (x+yi)^2 = a+bi with a and b known. Then (x^2-y^2) + (2xy)i = a+bi, so x^2-y^2 = a. Also |x+yi|^2 = |a+bi| so x^2+y^2 = sqrt(a^2+b^2). Combining these results gives: x^2 = (sqrt(a^2+b^2)+a)/2 y^2 = (sqrt(a^2+b^2)-a)/2 Taking square roots gives x and y. Unless b = 0, which will force x^2 or y^2 as computed above to be 0, there are two square roots in each case so we end up with four possible values for x+yi! Whats wrong? We forgot to fully reckon with the imaginary-part equation from above: 2xy = b. That gives us the sign information we need to distinguish roots. If b is positive, x and y have the same sign; if b is negative, they have opposite sign. For b = 0 x^2 or y^2 above will collapse to 0 and the ambiguity vanishes. Armed with this information, you should be able to crunch through the quadratic formula. Before you do, however, know what to expect: TWO solutions! Your contradtctory results could be just the two different roots. --OL === Subject: Re: Core error proof, simpler, shorter James Harris > It turns out you can prove that theres an error in core with rather > basic math, using a quadratic: Its on JSHs blog, which has the unlikely URL http://mathforprofit.blogspot.com Some highlights: The thing is that an algebraic integer cant be the root of a non-monic polynomial that doesnt have rational roots! [Back in reality, 2x^4 - 5x^2 +2 is non-monic, has sqrt(2) for a root, and has no rational roots.] ... The trouble is, nothing mathematically says that theres only one possibility--no axioms nor rules of logic nor anything other than mathematicians fixated on this definition of algebraic integers with the notion that it covers every possibility. [and in fact this blog starts with one defn of algebraic integer and ends up with another] ... Oddly enough thats new to mathematicians as algebraic integers have been around for over a hundred years, and they decided a while back that every number thats more like 8 than 1/8 is included in that category. Well I found out that theyre wrong, and its a BIG issue, as hey, theyre currently teaching the ßawed mathematics. LOL === Subject: Re: Core error proof, simpler, shorter > James Harris > It turns out you can prove that theres an error in core with rather > basic math, using a quadratic: > Its on JSHs blog, which has the unlikely URL > http://mathforprofit.blogspot.com > Some highlights: > The thing is that an algebraic integer cant be the root of a non-monic > polynomial that doesnt have rational roots! [Back in reality, 2x^4 - 5x^2 > +2 is non-monic, has sqrt(2) for a root, and has no rational roots.] > ... > The trouble is, nothing mathematically says that theres only one > possibility--no axioms nor rules of logic nor anything other than > mathematicians fixated on this definition of algebraic integers with the > notion that it covers every possibility. [and in fact this blog starts with > one defn of algebraic integer and ends up with another] > ... > Oddly enough thats new to mathematicians as algebraic integers have been > around for over a hundred years, and they decided a while back that every > number thats more like 8 than 1/8 is included in that category. Well I > found out that theyre wrong, and its a BIG issue, as hey, theyre > currently teaching the ßawed mathematics. LOL The only ßaw in the mathematics that Jimmy Harris objects to is that it wont make him rich and famous. === Subject: Re: Core error proof, simpler, shorter > It turns out you can prove that theres an error in core with rather > basic math, using a quadratic: > Let > P(x) = (x+8a)(x+b), and ab = 1, so P(x) = x^2 + (8a + b)x + 8. > Then > a = 1/b and b = 1/a > and to highlight the simple case let (8a + b) = 9, and substituting > with > a = 1/b gives, > 8(1/b) + b = 9, so 8 + b^2 = 9b, so > b^2 - 9b + 8 = (b-8)(b-1) = 0. > But why two solutions? Obviously, you can just have a=1/8, and get > (x + 8(1/8))(x + 8) = (x + 8)(x + 1) > so its trivially easy whats going on with those two solutions. Okay. You have (8a + b) = k, where k is some integer and ab = 1. The solutions to these two equations satisfy 8a^2 - ka + 1 = 0 and b^2 - kb + 8 = 0 When k = 9, these two equations become (8a - 1)(a - 1) = 0 = (b - 8)(b - 1) so we have two sets of solutions: (a, b) = (1/8, 8) and (1, 1). The other special case is when k = 6, where we have (4a - 1)(2a - 1) = 0 = (b - 4)(b - 2) which has solutions (a, b) = (1/4, 4) and (1/2, 2) For any other integer value of k we have two irreducible (over Q) polynomials 8a^2 - ka + 1 and b^2 - kb + 8 So for these values, the roots b will be algebraic integers and the corresponding roots a wont be. I fail to see anything remotely surprising about this. Notice, for later, that while a isnt an algebraic integer, 8a is, since 8a^2 - ka + 1 = 0 so 8(8a^2 - ka + 1) = (8a)^2 - k(8a) + 8 = 0 a = (17 +/- sqrt(257))/16 > but its clear that *only* one of those roots can be like 1/8 before, > while the other is like 8 from before, but *neither* is an algebraic > integer! What does it mean for one of those roots to be like 1/8 or 8? Are you attempting to generalize some property from the special, reducible, case k = 9? > So whats the core error? > The assumptions of some mathematicians would mean that > P(x) = (x+8a)(x+b), > is impossible in an ring where Ôa and Ôb have properties like > integers, because its impossible in the ring of algebraic integers, > if ab = 1 and (8a + b) is an integer, when the result is a polynomial > irreducible over Q that has Ôa as a root! Are you suggesting that under the assumptions 1. P(x) = (x + 8a)(x + b) 2. 8a + b is a rational integer 3. a, b are algebraic integers with ab = 1 4. P(x) is irreducible over Q (we dont actually need this) That P(x) = 0 does not have both solutions (x = -8a, x = -b) in the algebraic integers? That, of course, is wrong since weve already established that both b and 8a are algebraic integers (and, in fact, are conjugates when P is irreducible). Rick === Subject: Re: Core error proof, simpler, shorter The solutions to these two equations satisfy > 8a^2 - ka + 1 = 0 > and > b^2 - kb + 8 = 0 > When k = 9, these two equations become > (8a - 1)(a - 1) = 0 = (b - 8)(b - 1) > so we have two sets of solutions: (a, b) = (1/8, 8) and (1, 1). > The other special case is when k = 6, where we have > (4a - 1)(2a - 1) = 0 = (b - 4)(b - 2) > which has solutions (a, b) = (1/4, 4) and (1/2, 2) I missed k = -9 and k = -6 and the solutions (a, b) = (-1/8, 8), (-1, -1); (-1/4, -4), (-1/2, 2) respectively. Everything else goes through as before. Rick === Subject: Re: Fundamental Theorems of Calculus > I am a high school Calculus student and weve just covered the > Fundamental Theorems in class. While I understand how to use the > equations and dont have a problem with doing the assigned problems, > Im still having trouble logically working out exactly how these > Theorems work. I have a few specific questions, and Id appreciate it > if someone could tell me where my thought process is wrong or advise > how I might better understand how the Theorems work. In this post, FT1 > means The Fundamental Theorem, Part I, and FT2 means The > Fundamental Theorem, Part II or The Integral Evalulation Theorem. > 1. In FT1, can Ôa (the lower limit of integration) be chosen to be > any value at all (even if it winds up being greater than or equal to > this question is yes, but that seems to cause problems. For example, > what if a=4 and we were evaluating the equation when x=4 as well. > Wouldnt the answer then be zero, no matter what f(t) or f(x) is? > Consider the same situation again, when we are evaluating when x=4. > Then if Ôa was chosen to be less than 4, lets assume the answer was > positive. Then if Ôa was changed to 4, it would be zero and if Ôa > was changed to greater than four, the answer would be negative. If I > continue this train of thought, I realize that if Ôa is changed at > all the answer would be different, since youre measuring a different > area under the curve. It seems that the f(x) produced by FT1 would > completely depend on that Ôa value, and any change in a would result > in a different function f(x). I dont understand how Ôa can be any > value at all and FT1 still yields the same result: f(x). well, isnt it like an integral from a to b = - the same integral from b to a? no, choose a different a, and you end up with a different function. however, the good news is that the new function is your old function + a constant. > 2. When studying FT2, the book defined F(x) as the signed area > function: > F(x) = INT[from a to x](f(t)dt) > When evaluating definite integrals, the formula F(b) - F(a) is used, negatives > whatever the Ôa value was chosen to be in the original definition. > However, I was wondering if F(x) can be used by itself, for example > would a question ever give you a function f(x) and ask you to find > F(2). Since theres no subtraction, just as in my first question, it > seems here that the Ôa value would affect the answer. obviously. yes. the thing is, that this a must be provided somehow. consider an initial value problem, like: initially the tub is empty. provided, the question itself is in a sense incomplete. you provide the guy who asked with a generic answer, and say, see that constant? put any value here and it is still an answer to your question. > 3. Ive gone through the derivation in my textbook of FT1 and I > basically understand how they find that theorem, but Im still having > trouble with actually grasping how that equation works. What is wrong > with my thought process in the following: Lets say were trying to > find f(2), and we know from FT1 that f(x) = d/dx INT[from a to > x](f(t)dt). Therefore, lets start by evaluating the integral part > when x=2. In this case, the definite integral from a (whatever that > is) to x (which is 2) is simply a number, since its just giving you > the signed area under the curve between those two points. You then > differentiate that number (the d/dx part), and, since the derivative > of a constant is zero, the answer for f(2) must be zero. Under this > reasoning, any function comes out to be y=0 at all points. I dont see > how you can take a definite integral (which yields a constant), then > differentiate and get anything but zero. > However, maybe my reasoning is wrong because were dealing with x so > the answer is really a function, not a constant (but I cant > understand how this can be the case). Even if we do get a function, it > still doesnt make sense to me. First, youre integrating f(t)dt, so > you get some function in terms of t. You then differentiate with > respect to x, so youd have to use implicit differentiation and youd > wind up with both a t and an x in the answer. If the result of that > integral is a function in terms of the variable t, where does the t go > in the answer? see, your limit has an x in it. you *keep* that x a variable, otherwise the differentiation gets meaningless, i mean, look at it, it *says* that it is differentiating with respect to x. no, you dont use implicit differentiation. this t here is the dummy variable...the end result does not depend on t. 1 + 2 + .... + n = n(n-1)/2 write it as sum i [1..n] = n(n-1)/2 where did i go? really means: int[from some unspecified constant to x] {t dt} = x^2/2 + another unspecified constant which actually is a function of the unspecified constant before. it is confusing that we use the same symbol, but the thing is, the integral is a function of the limits only. > 4. I was trying to visualize what the signed area function F(x) > actually looks like, so I considered the two functions 2x and x^2. If > I understand the inverse nature of differentiation and integration > correctly, x^2 should be an integral of 2x or the F(x) that were > dealing with when 2x is our f(x). In this situation, I see how the > right half of the graph makes sense and the quadratic line shows how > the area under the straight line is increasing exponentially, but it > doesnt seem to work on the left side of the graph. Here, 2x goes > negative and the x*2 line, it seems, should also be negative since > were dealing with signed area. The only way I could see this working > is if the Ôa value is zero. Then the area left of the graph would be > taken going from right to left, and this going backwards > out the other negative value finishing with a net positive value. > However, this makes me wonder about that Ôa even more. What if a=1? > Then the area would be correct for everything to the right of one but > if you tried to find the area between zero and one youd be going > backwards and the answer would be negative even though that portion of > the graph is above the x-axis. The value chosen for Ôa seems to be > causing problems in my reasoning no matter how I look at it. this is perhaps the first time your reasoning is unclear (or i misunderstood you). the signed area thing works in many ways, different people think of it differently, and as far as i am concerned, its a mnemonic. you are not *multiplying* 2x and x^2, are you? the signed area is negative on the left side of the graph. the a, just forget about the a. draw two lines, x = -2 and x = 2. what is the area bounded? clearly you can see you have two congruent triangles. do you add them, or substract them? the signed area thing says, substract them. why, 2^2 [i.e., x^2 for x=2] - (-2)^2 [i.e., x^2 for x=-2] = 0 ! isnt that what it is about? > I can proceed with my work in the class, but I dont like to go ahead > on assumptions without really understanding the theorems that Im > using, and Im pretty confused as to how FT1 and FT2 work with an Ôa > that can be chosen arbitrarily. Id appreciate any input that would > help lead me in the right direction. > Johnathan right direction? no ... its *you* who has to decide whether its right or not. === Subject: Re: Fundamental Theorems of Calculus > I did some investigating with my graphing calculator last night while > considering what the two previous responses had said, and I think Im > starting to understand. Just to make sure, would it be safe to say > that the following is correct: > 1. F(x) = INT[from a to x] (f(t)dt) = f(x) + C what went wrong? int_a^x f(t)dt is certainly not f(x) + C e.g., int_0^x cos(t)dt = sin(x), which is not cos(x) i believe you meant to say, int_a^x f(t)dt = F(x) - F(a) > 2. d/dx F(x) = f(x) and this. the two boils down to d/dx [int_a^x f(t)dt] = F(x), making it clear that the choice of a (and, ahem, F(a)), is arbitrary and doesnt really concern us. > Johnathan youre welcome. :) === Subject: Re: Fundamental Theorems of Calculus I did some investigating with my graphing calculator last night while considering what the two previous responses had said, and I think Im starting to understand. Just to make sure, would it be safe to say that the following is correct: 1. F(x) = INT[from a to x] (f(t)dt) = f(x) + C 2. d/dx F(x) = f(x) Johnathan === Subject: Re: Fundamental Theorems of Calculus > I am a high school Calculus student and weve just covered the > Fundamental Theorems in class. While I understand how to use the > equations and dont have a problem with doing the assigned problems, > Im still having trouble logically working out exactly how these > Theorems work. I have a few specific questions, and Id appreciate it > if someone could tell me where my thought process is wrong or advise > how I might better understand how the Theorems work. In this post, FT1 > means The Fundamental Theorem, Part I, and FT2 means The > Fundamental Theorem, Part II or The Integral Evalulation Theorem. > 1. In FT1, can Ôa (the lower limit of integration) be chosen to be > any value at all (even if it winds up being greater than or equal to > this question is yes, but that seems to cause problems. For example, > what if a=4 and we were evaluating the equation when x=4 as well. > Wouldnt the answer then be zero, no matter what f(t) or f(x) is? > Consider the same situation again, when we are evaluating when x=4. > Then if Ôa was chosen to be less than 4, lets assume the answer was > positive. Then if Ôa was changed to 4, it would be zero and if Ôa > was changed to greater than four, the answer would be negative. If I > continue this train of thought, I realize that if Ôa is changed at > all the answer would be different, since youre measuring a different > area under the curve. It seems that the f(x) produced by FT1 would > completely depend on that Ôa value, and any change in a would result > in a different function f(x). I dont understand how Ôa can be any > value at all and FT1 still yields the same result: f(x). > 2. When studying FT2, the book defined F(x) as the signed area > function: > F(x) = INT[from a to x](f(t)dt) > When evaluating definite integrals, the formula F(b) - F(a) is used, negatives > whatever the Ôa value was chosen to be in the original definition. > However, I was wondering if F(x) can be used by itself, for example > would a question ever give you a function f(x) and ask you to find > F(2). Since theres no subtraction, just as in my first question, it > seems here that the Ôa value would affect the answer. Yes it does, but some functions are only defined when a is not negative, so its not really a problem. If f(x) = sqrt(x) or log(x), then both a and b have to non-negatve for the problem to even to make sense for an integral of a function over the real numbers. > 3. Ive gone through the derivation in my textbook of FT1 and I > basically understand how they find that theorem, but Im still having > trouble with actually grasping how that equation works. What is wrong > with my thought process in the following: Lets say were trying to > find f(2), and we know from FT1 that f(x) = d/dx INT[from a to > x](f(t)dt). Therefore, lets start by evaluating the integral part > when x=2. In this case, the definite integral from a (whatever that > is) to x (which is 2) is simply a number, since its just giving you > the signed area under the curve between those two points. You then > differentiate that number (the d/dx part), and, since the derivative > of a constant is zero, the answer for f(2) must be zero. Under this > reasoning, any function comes out to be y=0 at all points. I dont see > how you can take a definite integral (which yields a constant), then > differentiate and get anything but zero. > However, maybe my reasoning is wrong because were dealing with x so > the answer is really a function, not a constant (but I cant > understand how this can be the case). Even if we do get a function, it > still doesnt make sense to me. First, youre integrating f(t)dt, so > you get some function in terms of t. You then differentiate with > respect to x, so youd have to use implicit differentiation and youd > wind up with both a t and an x in the answer. If the result of that > integral is a function in terms of the variable t, where does the t go > in the answer? > 4. I was trying to visualize what the signed area function F(x) > actually looks like, so I considered the two functions 2x and x^2. If > I understand the inverse nature of differentiation and integration > correctly, x^2 should be an integral of 2x or the F(x) that were > dealing with when 2x is our f(x). In this situation, I see how the > right half of the graph makes sense and the quadratic line shows how > the area under the straight line is increasing exponentially, but it > doesnt seem to work on the left side of the graph. Here, 2x goes > negative and the x*2 line, it seems, should also be negative since > were dealing with signed area. The only way I could see this working > is if the Ôa value is zero. Then the area left of the graph would be > taken going from right to left, and this going backwards > out the other negative value finishing with a net positive value. > However, this makes me wonder about that Ôa even more. What if a=1? > Then the area would be correct for everything to the right of one but > if you tried to find the area between zero and one youd be going > backwards and the answer would be negative even though that portion of > the graph is above the x-axis. The value chosen for Ôa seems to be > causing problems in my reasoning no matter how I look at it. > I can proceed with my work in the class, but I dont like to go ahead > on assumptions without really understanding the theorems that Im > using, and Im pretty confused as to how FT1 and FT2 work with an Ôa > that can be chosen arbitrarily. Id appreciate any input that would > help lead me in the right direction. Youre right. the formula doesnt work all the time. It only works as the book says it does for absolute functions, where d_n/dx_n[ f(x) ] actually exists for every n = 0,1,2, ... infinity. And for most functions it doesnt. > Johnathan === Subject: Re: Letter to Prof Ullrich and others >I am just curious: why are people like you >interested in even acknowleding James Harris and his ilk ? >This guy has been oopsing on sci.math for ages (at least several >years) I see. >Is it not worthwhile for everyone to just ignore him at this point ? >There is real danger of course that someone would take him seriously, >but I feel thats remote. >It seems that he has taken to calling universities to complain. >Lets not giving him any more attention. Without attention, he will >shrivel up and disappear. >Is there a history or an obvious reason that I am missing ? Because mathematicians are losers and have no lives. Well, at least the ones that respond to JSH fall into this category. Mathematics is esoteric and boring. No socially well adjusted person would ever choose to pursue mathematics as a profession. There isnt even the excuse of financial rewards as there are in other technological areas such as computing or engineering. Those that pursue mathematics are simply forced to due to their lack of social prowess. Otherwise theyd be out doing fun things with people rather than investing the time and effort required to become ßuent in an unrewarding and stigmatised area of study. Mathematicians usually spend their childhood and their teen years being shunned and rejected by other people. Or they are simply used at exam times then promptly discarded. They gradually form a deep seated bitterness that comes to the fore when they perceive an opportunity to prove someone wrong. Proving someone wrong is a display of intellectual strength and the mathematicians intellect is his only solace within an isolated existence. Hence, they tend to jump at these opportunities with a great deal of fervour and gusto. Even a target such as James Harris, a person who appears to be emotionally retarded, is still fair game for the self-esteem challenged mathematicians that haunt sci.math. Also, perhaps James Harris is a mirror to the darkest corners of many mathematicians souls. James Harris is what they may have become had they not had the dedication and intellect to pursue mathematics. Then, like James Harris, they would simply be social outcasts with no redeeming characteristics at all. Hence they are prepared to battle hard to remove this unsettling spectre from disturbing their precariously sedate existences. === Subject: Re: Letter to Prof Ullrich and others Discussion, linux) > Mathematics is esoteric and boring. No socially well adjusted person > would ever choose to pursue mathematics as a profession. What about Usenet trolling? Evidence of an endlessly fascinating character, is it? -- Yup, you guessed it. If worse comes to worse, I *will* turn to the Army to help me with mathematicians. And then mathematicians dont think the NSA or CIA can save your asses, as generals LIKE me. -- James Harriss latest foray into mathematical logic. === Subject: Re: Letter to Prof Ullrich and others >>I am just curious: why are people like you >>interested in even acknowleding James Harris and his ilk ? >>This guy has been oopsing on sci.math for ages (at least several >>years) I see. >>Is it not worthwhile for everyone to just ignore him at this point ? >>There is real danger of course that someone would take him seriously, >>but I feel thats remote. >>It seems that he has taken to calling universities to complain. >>Lets not giving him any more attention. Without attention, he will >>shrivel up and disappear. >>Is there a history or an obvious reason that I am missing ? >Because mathematicians are losers and have no lives. Well, at least >the ones that respond to JSH fall into this category. >Mathematics is esoteric and boring. No socially well adjusted person >would ever choose to pursue mathematics as a profession. There isnt >even the excuse of financial rewards as there are in other >technological areas such as computing or engineering. Those that >pursue mathematics are simply forced to due to their lack of social >prowess. Otherwise theyd be out doing fun things with people rather >than investing the time and effort required to become ßuent in an >unrewarding and stigmatised area of study. Not only that, its well-known that mathematicians are ugly. They lack the minimal social skills to even get decent sex so they have to pay for it. I bet that many mathematicians are deviants... But Im not saying more. They also have bad breath. >Mathematicians usually spend their childhood and their teen years >being shunned and rejected by other people. Or they are simply used >at exam times then promptly discarded. They gradually form a deep >seated bitterness that comes to the fore when they perceive an >opportunity to prove someone wrong. Proving someone wrong is a >display of intellectual strength and the mathematicians intellect is >his only solace within an isolated existence. Hence, they tend to >jump at these opportunities with a great deal of fervour and gusto. >Even a target such as James Harris, a person who appears to be >emotionally retarded, is still fair game for the self-esteem >challenged mathematicians that haunt sci.math. >Also, perhaps James Harris is a mirror to the darkest corners of many >mathematicians souls. James Harris is what they may have become had >they not had the dedication and intellect to pursue mathematics. >Then, like James Harris, they would simply be social outcasts with no >redeeming characteristics at all. Hence they are prepared to battle >hard to remove this unsettling spectre from disturbing their >precariously sedate existences. G. Rodrigues === Subject: Re: Letter to Prof Ullrich and others >I am just curious: why are people like you >interested in even acknowleding James Harris and his ilk ? >This guy has been oopsing on sci.math for ages (at least several >years) I see. >Is it not worthwhile for everyone to just ignore him at this point ? >There is real danger of course that someone would take him seriously, >but I feel thats remote. >It seems that he has taken to calling universities to complain. >Lets not giving him any more attention. Without attention, he will >shrivel up and disappear. >Is there a history or an obvious reason that I am missing ? > Because mathematicians are losers and have no lives. Well, at least > the ones that respond to JSH fall into this category. > Mathematics is esoteric and boring. No socially well adjusted person > would ever choose to pursue mathematics as a profession. There isnt > even the excuse of financial rewards as there are in other > technological areas such as computing or engineering. Those that > pursue mathematics are simply forced to due to their lack of social > prowess. Otherwise theyd be out doing fun things with people rather > than investing the time and effort required to become ßuent in an > unrewarding and stigmatised area of study. > Mathematicians usually spend their childhood and their teen years > being shunned and rejected by other people. Or they are simply used > at exam times then promptly discarded. They gradually form a deep > seated bitterness that comes to the fore when they perceive an > opportunity to prove someone wrong. Proving someone wrong is a > display of intellectual strength and the mathematicians intellect is > his only solace within an isolated existence. Hence, they tend to > jump at these opportunities with a great deal of fervour and gusto. > Even a target such as James Harris, a person who appears to be > emotionally retarded, is still fair game for the self-esteem > challenged mathematicians that haunt sci.math. > Also, perhaps James Harris is a mirror to the darkest corners of many > mathematicians souls. James Harris is what they may have become had > they not had the dedication and intellect to pursue mathematics. > Then, like James Harris, they would simply be social outcasts with no > redeeming characteristics at all. Hence they are prepared to battle > hard to remove this unsettling spectre from disturbing their > precariously sedate existences. Its not that. James Harris attempts at mathematics are as grating to the nerves of mathematicians the attempts of a tone deaf person trying to sing are to musicians. But just as musical tone deafness is incurable, James mathematical version of tone-deafness seems to be incurable. That he has serious psychological problems doesnt help. === Subject: Alice in Wonderland and Machine-arithmetic Please, Could anyone explain to me what this excerpt from Alice in Wonderland has to do with Machine-arithmetic except the fact that the computation is done in base 18 + 3n? Mike Let me see: four times five is twelve, and four times six is thirteen, and four times seven is-oh dear! I shall never reach twenty at that rate!... === Subject: Re: Alice in Wonderland and Machine-arithmetic > ... > I myself prefer the first, but whatever is the case, there is no relation > to machine arithmetic at all. No, I dont think there is, so why does the op suggest that such a link been claimed? -- G.C. === Subject: Re: Alice in Wonderland and Machine-arithmetic > Could anyone explain to me what this excerpt from Alice in Wonderland > has to do with Machine-arithmetic except the fact that the computation > is done in base 18 + 3n? ... > Let me see: four times five is twelve, and four times six is > thirteen, and four times seven is-oh dear! I shall never reach twenty > at that rate!... > I dont know, but why do you think it ought to? I was hoping to find to > find something in Martin Gardners Annotated Alice, but it has no > index so I havent looked much. Which chapter is it in? Chapter 2, footnote 3. Quote: The simplest explanation of why Alice will never get to 20 is this: the multiplication table traditionally stops with the twelves, so if you continue this nonsense progression - 4 times 5 is 12, 4 times 6 is 13, 4 times 7 is 14, and so on - you end with 4 times 12 (the highest she can go) is 19 - just one short of 20. A. L. Taylor, in his book The White Knight, advances an interesting but more complicated theory. Four times 5 actually is 12 in a number system using a base of 18. Four times 6 is 13 in a system with a base of 21. If we continue this progression, always increasing the base by 3, our products keep increasing by one until we reach 20, where for the first time the scheme breaks down. Four times 13 is not 20 (in a number system with a base of 42), but ñ1î followed by whatever symbol is adopted for ñ10.î I myself prefer the first, but whatever is the case, there is no relation to machine arithmetic at all. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Alice in Wonderland and Machine-arithmetic > Please, > Could anyone explain to me what this excerpt from Alice in Wonderland > has to do with Machine-arithmetic except the fact that the computation > is done in base 18 + 3n? > Mike > Let me see: four times five is twelve, and four times six is > thirteen, and four times seven is-oh dear! I shall never reach twenty > at that rate!... > I dont know, but why do you think it ought to? I was hoping to find to > find something in Martin Gardners Annotated Alice, but it has no > index so I havent looked much. Which chapter is it in? Chapter II The Pool of Tears http://www-2.cs.cmu.edu/People/rgs/alice-II.html `Im sure Im not Ada, she said, `for her hair goes in such long ringlets, and mine doesnt go in ringlets at all; and Im sure I cant be Mabel, for I know all sorts of things, and she, oh! she knows such a very little! Besides, SHES she, and Im I, and--oh dear, how puzzling it all is! Ill try if I know all the things I used to know. Let me see: four times five is twelve, and four times six is thirteen, and four times seven is--oh dear! I shall never get to twenty at that rate! However, the Multiplication Table doesnt signify: lets try Geography. London is the capital of Paris, and Paris is the capital of Rome, and Rome--no, THATS all wrong, Im certain! I must have been changed for Mabel! Ill try and say How doth the little-- and she crossed her hands on her lap as if she were saying lessons, and began to repeat it, but her voice sounded hoarse and strange, and the words did not come the same as they used to do:-- > -- > G.C. Hugo === Subject: Re: Alice in Wonderland and Machine-arithmetic > Please, > Could anyone explain to me what this excerpt from Alice in Wonderland > has to do with Machine-arithmetic except the fact that the computation > is done in base 18 + 3n? > Mike > Let me see: four times five is twelve, and four times six is > thirteen, and four times seven is-oh dear! I shall never reach twenty > at that rate!... I dont know, but why do you think it ought to? I was hoping to find to find something in Martin Gardners Annotated Alice, but it has no index so I havent looked much. Which chapter is it in? -- G.C. === Subject: Question on notation I am reading a book on set theory and I have become unsure about the usage of function notation (ie f(x)). Specifically, the book I am reading says that f(x) is allowed to be used only when f is a function and x in dom f. (*) Now lets say I was assuming the following formula in a proof: f is a function and x in dom f => f(x) = 1. To check that the notation f(x) was used correctly, I would reason that on the left of the =>, I said f is a function and x in dom f. But this doesnt seem to be very precise reasoning. Is there a better way to say why the notation is justified above (without stating the reason (*) verbatim). I guess I am confused because if I were to instead say to assume (the equivalent formula) f(x) != 1 => f isnt a function or x notin dom f it would seem that I shouldnt be able to use the notation f(x) even though this is the same thing as the first case. Leon === Subject: Re: Question on notation > I am reading a book on set theory and I have become unsure about the usage > of function notation (ie f(x)). I assume that the book defines a relation F to be a subset of the cartesian product AxB, where A is the domain and B is the range. I assume that the book defines a function F to be a relation that has exactly one image in the range for each element in the domain. > Specifically, the book I am reading says > that f(x) is allowed to be used only when f is a function and x in dom f. The book is trying to make the point that the function notation f(x) makes sense only when f is a function because, for each x in the domain, there is exactly one y in the range, which y we denote by f(x). The function notation f(x) will not work for the more general concept of relation because there may be multiple images y in the range for an x in the domain. Consider the following two examples. Let A = {1, 2, 3} and B = {a, b, c, d}. Let the function f = { (1,a), (2,d), (3, b)}. Since f is a function, I can use the function notation and write f(1) = a, f(2) = d, and f(3) = b. Let the relation f = { (1, a), (2, d), (3, b), (1, c)}. I cannot use the function notation because f(1) has the multiple possibilities of either a or c. That is, there is not a unique value for 1 in the domain, so that f(1) is not determined. -- Bill Hale === Subject: Re: Question on notation > I assume that the book defines a relation F to be a subset of the > cartesian product AxB, where A is the domain and B is the range. > I assume that the book defines a function F to be a relation > that has exactly one image in the range for each element in the domain. > Specifically, the book I am reading says > that f(x) is allowed to be used only when f is a function and x in dom f. > The book is trying to make the point that the function notation f(x) > makes sense only when f is a function because, for each x in the domain, > there is exactly one y in the range, which y we denote by f(x). The problem I am having is not understanding why such a requirement is in place, but how to check that it is in place. In your examples below youve previously mentioned that f satisfies the criteria to use f(x) for x=1,2,3, so it is clear to me that the notation is justified in its usage. But what about examples where the justification is in the same statement as the usage as I gave in the original post - these are the ones I am unsure about. Ie f is a function w/ 1 in dom f => f(1) in A Now in the latter part of the implication, f(1)=2 only makes sense if the former part is true. So it doesnt matter if you convert f(1)=2 to something like forall z, (1,z) in f => z in A, eliminating the use of f(1) since you wont obtain contradictions. So then something like forall x, x in {} => f(x) = 1 should be okay since xin {} is never true? But this seems artificial. Leon > The function notation f(x) will not work for the more general concept > of relation because there may be multiple images y in the range > for an x in the domain. > Consider the following two examples. Let A = {1, 2, 3} and B = {a, b, c, d}. > Let the function f = { (1,a), (2,d), (3, b)}. Since f is a function, > I can use the function notation and write f(1) = a, f(2) = d, and f(3) = b. > Let the relation f = { (1, a), (2, d), (3, b), (1, c)}. > I cannot use the function notation because f(1) has the multiple > possibilities of either a or c. That is, there is not a unique value > for 1 in the domain, so that f(1) is not determined. > -- Bill Hale === Subject: Re: Question on notation > I am reading a book on set theory and I have become unsure about the usage > of function notation (ie f(x)). Specifically, the book I am reading says > that f(x) is allowed to be used only when f is a function and x in dom f. > (*) > Now lets say I was assuming the following formula in a proof: > f is a function and x in dom f => f(x) = 1. > To check that the notation f(x) was used correctly, I would reason that on > the left of the =>, I said f is a function and x in dom f. But this > doesnt seem to be very precise reasoning. Is there a better way to say why > the notation is justified above (without stating the reason (*) verbatim). > I guess I am confused because if I were to instead say to assume (the > equivalent formula) > f(x) != 1 => f isnt a function or x notin dom f > it would seem that I shouldnt be able to use the notation f(x) even > though this is the same thing as the first case. > Leon One can mention unicorns in stating that they do not exist. === Subject: Re: Question on notation > One can mention unicorns in stating that they do not exist. The above is okay because unicorns are well-defined despite the fact that they dont exist. They are horses with horns and magical powers and they can ßy I think. But function notation is not well-defined if f isnt a function with x in dom f. So you cannot talk about f(x) if you do not have existence. That is why I am curious about checking to see if a formula uses notation properly. Leon === Subject: Re: Question on notation > One can mention unicorns in stating that they do not exist. > The above is okay because unicorns are well-defined despite the fact that > they dont exist. They are horses with horns and magical powers and they > can ßy I think. But function notation is not well-defined if f isnt a > function with x in dom f. So you cannot talk about f(x) if you do not > have existence. That is why I am curious about checking to see if a > formula uses notation properly. > Leon For the function defined for non-zero reals by f(x) = sin(x)/x, it would be quite acceptable to say that this f(x) does not exist at x = 0. It would be a little less acceptable to say that f(0) did not exist, but it would still probably be understood correctly. === Subject: Re: Question on notation > One can mention unicorns in stating that they do not exist. > The above is okay because unicorns are well-defined despite the fact > that they dont exist. They are horses with horns and magical powers > and they can ßy I think. Hmm. they can ßy I think doesnt make it sound as if unicorn is well defined! (I think youre confusing Pegasus with a unicorn.) A unicorn resembles a horse, but has one horn. Unicorns were, most unfortunately, hunted to extinction for their horns, which were highly valued, especially for their supposed aphrodisiacal properties. [The sentence above is, of course, utter fabricaton.] > But function notation is not well-defined if > f isnt a function with x in dom f. So you cannot talk about f(x) > if you do not have existence. That is why I am curious about > checking to see if a formula uses notation properly. > For the function defined for non-zero reals by f(x) = sin(x)/x, it would > be quite acceptable to say that this f(x) does not exist at x = 0. > It would be a little less acceptable to say that f(0) did not exist, but > it would still probably be understood correctly. Agreed. BTW, some people (not including me!) would, in absence of an explicitly stated domain, automatically consider the implied domain of f(x) = sin(x)/x to be R, taking f(0) to be 1. David === Subject: Re: The Lost Proof of Fermat These Fermat characters are HILARIOUS!!! http://www.fermatproof.com/ Is the proof valid? He appears to be using the rule of Pythagorean triples, trying to extend it out to all values of n > 2. Then using reducto ad absurdum to establish the absurdity of n > 2 Interesting diagrams with the right triangles ......2 .....22 ....222 ...1111 = 4^2 2+2+2+2+2+2+1+1+1+1 = 4^2 6+4 = 10 1+2+3+4 10 + 5 = 15 10*2 + 5*1 = 5^2 15 + 6 = 21 15*2 + 6*1 = 6^2 N(N+1)/2 = 1+2+3+...+ N Interesting... Z = [X^p + Y^p]/[Z^(p-1)] For p > 2, Z cannot be an integer... === Subject: moddiff inequality & 1 question on moivres formula how one would go to prove that (for complex zs), | |z_1| - |z_2| | <= |z_1 + z_2| ? i was trying using x + iy but it didnt work very well. another question (not related to the above) is: studying de moivres formula, i find that cos(t_1) * cos(t_2) - sin(t_1) * sin(t_2) + i [ sin(t_1) * cos(t_2) + cos(t_1) * sin(t_2) ] = cos(t_1 + t_2) + i sin(t_1 + t_2). i understand that cos(t_1) * cos(t_2) = cos(t_1 + t_2), but i dont get the equation above. === Subject: Re: moddiff inequality & 1 question on moivres formula > how one would go to prove that (for complex zs), > | |z_1| - |z_2| | <= |z_1 + z_2| ? Lemma: |z_1 + z_2| will be minimal, for fixed |z_1| > 0 and fixed |z_2| > 0, when z_1/z_2 is a negative real number === Subject: Re: moddiff inequality & 1 question on moivres formula Daniel C Bastos > how one would go to prove that (for complex zs), > | |z_1| - |z_2| | <= |z_1 + z_2| ? Let me switch to ascii-friendly notation | |A| - |B| | <= |A + B|. This is the same as this pair of inequalities: 1) |A| - |B| <= |A + B| 2) |B| - |A| <= |A + B| By the triangle inequality |A| <= |A+B| + |-B| = |A+B| + |B| and (1) follows. (2) follows from (1) by swapping A and B. > studying de moivres formula, i find that > cos(t_1) * cos(t_2) - sin(t_1) * sin(t_2) + > i [ sin(t_1) * cos(t_2) + cos(t_1) * sin(t_2) ] > = cos(t_1 + t_2) + i sin(t_1 + t_2). > i understand that cos(t_1) * cos(t_2) = cos(t_1 + t_2), but > i dont get the equation above. The equation means that the real parts on the two sides are equal, and likewise the imaginary parts. So it is merely this pair of formulas from trigonometry: 3) cos (A+B) = cos A cos B - sin A sin B 4) sin (A+B) = sin A cos B + cos A sin B (4) is nicely proved here: http://www.cut-the-knot.org/proofs/sine_cosine.shtml Actually (3) and (4) are practically the same. In (4), replace A by -A, and B by pi/2 - B, and use sin x = cos (pi/2 - x). When the smoke clears, you have (3). LH === Subject: Re: moddiff inequality & 1 question on moivres formula >how one would go to prove that (for complex zs), >| |z_1| - |z_2| | <= |z_1 + z_2| ? >i was trying using x + iy but it didnt work very well. Do you know the triangle inequality? >another question (not related to the above) is: >studying de moivres formula, i find that >cos(t_1) * cos(t_2) - sin(t_1) * sin(t_2) + >i [ sin(t_1) * cos(t_2) + cos(t_1) * sin(t_2) ] >= cos(t_1 + t_2) + i sin(t_1 + t_2). >i understand that cos(t_1) * cos(t_2) = cos(t_1 + t_2), but >i dont get the equation above. If thats what you understand, you wont get very much. Look again at the addition formulas for cos and sin. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Differential equation with implicit solution > Maybe that Ive misunderstood your question, but it seems to me that > all that you are supposed to do is to solve the equation > -2x^2*y + y^2 = 1 > in y. Youll get two solutions: y = x^2 + sqrt(x^4 + 1) and > y = x^2 - sqrt(x^4 + 1). > Jose Carlos Santos Sorry about my badly phrased question, but what I was really getting at was how do you solve -2x^2y + y^2 = 1 for y? How did you arrive at your solution above? R === Subject: Re: Differential equation with implicit solution >>Maybe that Ive misunderstood your question, but it seems to me that >>all that you are supposed to do is to solve the equation >>-2x^2*y + y^2 = 1 >>in y. Youll get two solutions: y = x^2 + sqrt(x^4 + 1) and >>y = x^2 - sqrt(x^4 + 1). > Sorry about my badly phrased question, but what I was really getting > at was how do you solve -2x^2y + y^2 = 1 for y? How did you arrive at > your solution above? This is a second degree polynomial equation! If you cant solve this, then I suggest that you study polynomial equations before dealing with differential equations. Jose Carlos Santos === Subject: Re: Differential equation with implicit solution > Maybe that Ive misunderstood your question, but it seems to me that > all that you are supposed to do is to solve the equation > > -2x^2*y + y^2 = 1 > > in y. Youll get two solutions: y = x^2 + sqrt(x^4 + 1) and > y = x^2 - sqrt(x^4 + 1). > > > Jose Carlos Santos > Sorry about my badly phrased question, but what I was really getting > at was how do you solve -2x^2y + y^2 = 1 for y? How did you arrive at > your solution above? Quadratic Formula! === Subject: Re: Differential equation with implicit solution >> My differential equations book asks the following: >> >> Show that -2x^2y + y^2 = 1 is a solution of the differential equation >> (in differential form) 2xy dx + (x^2 - y) dy = 0. Ok, thats trivial, >> one just implicitly differentiates the first equation, etc. >both? solve the quadratic for y, i get y = x^2 (+/-) sqrt(x^4 + 1). >check which branch is your solution (if any). >i have found an even nicer ones -- y = 0, 2x^2! > This is hardly surprising given they are trivial linear combinations > of the two general solutions. God! I know! === Subject: RSA-576 factored by NOT James Harris Oh if I only had another two or three lifetimes of the universe to complete my factoring by trial division technique I too could have acheived such fame, wealth , and glory. http://www.rsasecurity.com/rsalabs/challenges/factoring/ numbers.html Kudos to J. Franke et al. I guess that Unlce Als challenge to James is gonna have to go up a notch. We will, of course, be praying for the day when James accepts the challenge and locks himself in his garage without contact from the outside world for a few years to crack an RSA challenge. === Subject: Re: RSA-576 factored by NOT James Harris Wait, there is the RSA-640. I am heading that way. -suresh > Oh if I only had another two or three lifetimes of the universe to > complete my factoring by trial division technique I too could have > acheived such fame, wealth , and glory. > http://www.rsasecurity.com/rsalabs/challenges/factoring/ numbers.html > Kudos to J. Franke et al. > I guess that Unlce Als challenge to James is gonna have to go up a > notch. We will, of course, be praying for the day when James accepts > the challenge and locks himself in his garage without contact from the > outside world for a few years to crack an RSA challenge. === Subject: Re: RSA-576 factored by NOT James Harris > Oh if I only had another two or three lifetimes of the universe to > complete my factoring by trial division technique I too could have > acheived such fame, wealth , and glory. > http://www.rsasecurity.com/rsalabs/challenges/factoring/ numbers.html > Kudos to J. Franke et al. > I guess that Unlce Als challenge to James is gonna have to go up a > notch. We will, of course, be praying for the day when James accepts > the challenge and locks himself in his garage without contact from the > outside world for a few years to crack an RSA challenge. Dont be silly. The poster Uncle Al is a critic troll and simply posts in attempts to elicit endless replies so that he can post some more! And it turns out that I *have* worked on RSA fairly steadily for a while, without success. Remember, my posting strategy is to post MORE not LESS when Im working on a problem!!! If you see me posting a lot then theres probably something Im working on. That means that you have it backwards: the more I work on various problems the more LIKELY I am to post as I HATE THE IDEA of locking ones self away to work on a problem. To me its just a waste as I say post your ideas so that other people can critique them! Its worked for me!!! Now it does bother me that any of you actually take Uncle Al seriously, as it requires I think a tremendous effort to not understand how that poster operates. So you should have known that Uncle Al is trying to get me to post more--as long as I reply to him in anger--so that he can reply more. James Harris === Subject: Re: RSA-576 factored by NOT James Harris >[...] >To me its just a waste as I say post your ideas so that other people >can critique them! You mean so you can curse people who explain why your ideas are wrong. >Its worked for me!!! In what sense has it worked? Youve succeeded in not proving FLT, in not demonstrating that any of various well-known facts are wrong, in coming up with a program to count primes that involves no ideas less than 200 years old and is slower than molasses. Most people could accomplish all that in much less time. >Now it does bother me that any of you actually take Uncle Al >seriously, as it requires I think a tremendous effort to not >understand how that poster operates. >So you should have known that Uncle Al is trying to get me to post >more--as long as I reply to him in anger--so that he can reply more. >James Harris ************************ David C. Ullrich === Subject: Re: RSA-576 factored by NOT James Harris > That means that you have it backwards: the more I work on various > problems the more LIKELY I am to post as I HATE THE IDEA of locking > ones self away to work on a problem. > To me its just a waste as I say post your ideas so that other people > can critique them! > Its worked for me!!! By what standard to you claim it has worked for you? You have accomplished exactly *nothing* in the years you have been posting -- except to reveal to all that you are an ignorant CRANK. > James Often in error, but never in doubt. Harris James, if you dont want to elevate yourself to the intellectual status of a crash dummy, or rise to the demands of working as a thrift shop mannequin, why not rent yourself out as packing material? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: RSA-576 factored by NOT James Harris > Oh if I only had another two or three lifetimes of the universe to > complete my factoring by trial division technique I too could have > acheived such fame, wealth , and glory. > http://www.rsasecurity.com/rsalabs/challenges/factoring/ numbers.html > Kudos to J. Franke et al. > I guess that Unlce Als challenge to James is gonna have to go up a > notch. We will, of course, be praying for the day when James accepts > the challenge and locks himself in his garage without contact from the > outside world for a few years to crack an RSA challenge. What will we all do for laughs then? === Subject: Re: Probability Instinct - Sufficiently Discrete to Survive in the Jungle > Discrete enough to outrun a predator or the slowest member of your own > species. > THE PROBABILITY INSTINCT > It looks as if Kant, who thought our minds structure our perceptions, was > right. Probability was built into our minds. Our minds, the electrochemical > symphony that our narrowly evolved neural ganglia play, impose an > infrastructure on our thinking. The mind imposes a background of time and > space and causal connectedness. Scientists have never seen a causality in > the wild. They have seen, and they predict, only space-time events that > follow space-time events. Apples on the tree, then apples in the air, then > apples on the ground. Equations and correlations have replaced causes, just > as science has largely replaced philosophy and religion as a theory of > things. No causal germ in one event unfolds into another event. But the > mind, as eighteenth-century philosopher David Hume observed, makes it seem > so and inserts the causal links in the event chain. > Probability seems to be part of the same mental infrastructure. It forms > part of our mental background or viewing screen along with time and space > and causality and similarity and the topological notions of continuity and > connectedness. We see probability everywhere because it lies in our glasses. I am strongly going to suggest that you look at orthomodular logics and quantum logics for this in addition to Bayesian decision models and whatever else you find. I had been looking a Paul Halmos paper on Hilbert spaces and discovered one of the inadequacies of quantum logic for quantum mechanics--the Heisenberg uncertainty principle cannot be represented. On the other hand, the fact that logic on a Hilbert space is not fixed means that it is sufficiently free for the situation-theoretic analysis that does not presume predication and object awareness outside of series of perspectives. Thus, the Hilbert space formalism can be viewed as evolving with a stateful, defeasible belief system against which secondary judgeable content is oriented. Just a thought. :-) mitch === Subject: Re: anybody can read out eigenvalues/vectors by eye-inspection? a-s b c d-s ad-bc - (a+d)s + s^2 = det - s*trace + s^2 = 0 s = [trace +- sqr(trace^2 - 4*det)]/2 = trace/2 +- sqr((trace/2)^2 - det)) trace/2 = average value of diagonal entry s1 + s2 = trace; s1 s2 = det ---- === Subject: Re: anybody can read out eigenvalues/vectors by eye-inspection? > For simple matrices, such as [...] > [a b; > b c] > [a b; > -b c] > [3 2; > 2 3] The eigenvalues are extremely simple: >> A := matrix([[a,b],[b,a]]) +- -+ | a, b | | | | b, a | +- -+ >> linalg::charpoly(A,s) 2 2 2 s - (2 a) s + (a - b ) >> linalg::eigenvalues(A) {a + b, a - b} >> A := matrix([[a,b],[-b,a]]) +- -+ | a, b | | | | -b, a | +- -+ >> linalg::charpoly(A,s) 2 2 2 s - (2 a) s + (a + b ) >> linalg::eigenvalues(A) {a - I b, a + I b} Z.H. === Subject: projective geometry axioms Why not combine the first two axioms of projective geometry? It seems like any proof that involved either two would be upheld with an =1 . Ill list them all here from mathworld: 1. If A and B are distinct points on a plane, there is at least one line containing both A and B. 2. If A and B are distinct points on a plane, there is not more than one line containing both A and B. 3. Any two lines in a plane have at least one point of the plane (which may be the point at infinity) in common. 4. There is at least one line on a plane. 5. Every line contains at least three points of the plane. 6. All the points of the plane do not belong to the same line === Subject: Re: projective geometry axioms > Why not combine the first two axioms of projective geometry? It seems > like any proof that involved either two would be upheld with an =1 . > Ill list them all here from mathworld: > 1. If A and B are distinct points on a plane, there is at least one > line containing both A and B. > 2. If A and B are distinct points on a plane, there is not more than > one line containing both A and B. > 3. Any two lines in a plane have at least one point of the plane > (which may be the point at infinity) in common. > 4. There is at least one line on a plane. > 5. Every line contains at least three points of the plane. > 6. All the points of the plane do not belong to the same line That sort of mess is a long way out of date. In the early 20th century various such axioms were proposed, usually something like: (1) Any two points both lie on a unique line, (2) Any two lines both lie on a unique point, (3), (4), ... Non-triviality conditions such as your 4, 5, 6 above. But in 1943 Marshall Hall produced a much simpler non-triviality condition: (3) There exist four points of which no three are collinear, i.e. there exists a quadrangle. This is a purely existential statement about finitely many points, as against the older universal-existential things like your 5. (1), (2), (3) above are now widely regarded as the basic axioms for plane projective geometry. An important special case satisfies also the Desargues condition. An important special case of _that_ satisfies also the Pappus condition (which implies Desargues, by Hessenbergs Theorem). But you may not be interested in those refinements just yet. Ken Pledger. === Subject: Re: projective geometry axioms Todd Smith > Why not combine the first two axioms of projective geometry? It seems > like any proof that involved either two would be upheld with an =1 . > Ill list them all here from mathworld: ... http://mathworld.wolfram.com/ProjectiveGeometry.html That entry definitely needs work. For a real set of axioms for a projective _plane_ see Coxeter Projective Geometry or Artin Geometric Algebra or Hilbert Foundations of Geometry or ... LH === Subject: Re: projective geometry axioms > Why not combine the first two axioms of projective geometry? It seems > like any proof that involved either two would be upheld with an =1 . Huh? What you mean an=1? > Ill list them all here from mathworld: > 1. If A and B are distinct points on a plane, there is at least one > line containing both A and B. > 2. If A and B are distinct points on a plane, there is not more than > one line containing both A and B. Two distinct points determine a unique line. > 3. Any two lines in a plane have at least one point of the plane > (which may be the point at infinity) in common. > 4. There is at least one line on a plane. > 5. Every line contains at least three points of the plane. > 6. All the points of the plane do not belong to the same line === Subject: Density of Pythagorean triples Let a be positive, T(a) = {(x,y) in R^2: x and y nonnegative, x^a + y^a = 1}, and B(a) be the intersection of Q^2 and T(a). Is B(2) dense in T(2) (under the usual topology)? Nowhere dense? For n a positive integer, B(1/n) is dense in T(1/n). By FLT, B(n) is finite (in fact, has only two points) for integer n > 2. What can be said along these lines (density, cardinality) for general real positive a? [I could swear I asked this question some time ago, but search of the group with Google doesnt turn up anything.] -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Density of Pythagorean triples > Stephen J. Herschkorn > Let a be positive, T(a) = {(x,y) in R^2: x and y nonnegative, x^a > + y^a = 1}, and B(a) be the intersection of Q^2 and T(a). Is > B(2) dense in T(2) (under the usual topology)? > Even a fraction of the Pythagorean triples is dense. Consider (3+4i)^n for > positive n. Their imaginary parts are all 4 mod 5. So none of them is real, > and it follows that the arguments of these complex numbers are dense in > (0,2pi). That means that one of these numbers projects onto any given > non-empty open sector of the unit circle. > LH Thats a much nicer answer. The complex numbers x+yi of norm 1, i.e., with x^2+y^2=1, form a group under multiplication. Primitive Pythagorean triples (a,b,c), when written as (a/c)+(b/c)i, form a subgroup. As you point out, the powers of any nontrivial Pythagorean triple will be dense in the circle. More generally, this is true of the powers of any x+yi = exp(2*pi*i*t), unless t is a rational number. Even more is true. The powers of x+yi are not simply dense, they are uniformly distributed on the unit circle, in the sense that if you take any open arc of the circle, then the proportion of (x+yi)^n with n < N will approach the length of that arc divided by 2*pi. Joe Silverman === Subject: Re: Density of Pythagorean triples Stephen J. Herschkorn > Let a be positive, T(a) = {(x,y) in R^2: x and y nonnegative, x^a > + y^a = 1}, and B(a) be the intersection of Q^2 and T(a). Is > B(2) dense in T(2) (under the usual topology)? Even a fraction of the Pythagorean triples is dense. Consider (3+4i)^n for positive n. Their imaginary parts are all 4 mod 5. So none of them is real, and it follows that the arguments of these complex numbers are dense in (0,2pi). That means that one of these numbers projects onto any given non-empty open sector of the unit circle. LH === Subject: Re: Density of Pythagorean triples > Let a be positive, T(a) = {(x,y) in R^2: x and y nonnegative, x^a > + y^a = 1}, and B(a) be the intersection of Q^2 and T(a). Is > B(2) dense in T(2) (under the usual topology)? Nowhere dense? > For n a positive integer, B(1/n) is dense in T(1/n). By FLT, > B(n) is finite (in fact, has only two points) for integer n > 2. > What can be said along these lines (density, cardinality) for general > real positive a? > [I could swear I asked this question some time ago, but search of the > group with Google doesnt turn up anything.] The set of points (x/z,y/z) for integers x,y and z such that x^2+y^2=z^2 and z <> 0 are dense on the unit circle. There is a way, which I do not recall offhand, of finding such a point between any two such given points. === Subject: Re: Density of Pythagorean triples [...] > The set of points (x/z,y/z) for integers x,y and z such that x^2+y^2=z^2 > and z <> 0 are dense on the unit circle. > There is a way, which I do not recall offhand, of finding such a point > between any two such given points. I derived formulae I use for rational trigonometric values. For any side angle A where you can find a rational value f = tan (A/2) cos(A) = (1-f^2) / (1 + f^2) sin(A) = 2f / (1 + f^2) Since it should be possible to find tan(A/2) as a rational value arbitrarily close to side angle A (0 < A < 90), you can also find a Pythagorean triple arbitrarily close to that arbitrary angle and trigonometric values for a non-distorting rational rotational transformation. http://homework.jhax.net/math/Pythagorean_Triples.jsp (Also: I also derived similar formula for 120-degree triples, where you are looking for integer triangles with a 120-degree angle (such as 3-5-7) which satisfy x^2 + y^2 + xy = z^2 from an arbitrary side A (0 < A < 60) demonstrated at http://homework.jhax.net/math/Natural_Triples.jsp). Apologies in advance if it is not a proper answer, as I am clearly only a hobbiest at Math as it is applicable to my computer work, but I thought I had discovered simple useful formulae I had seen no one else mention, that seem to relate to density of rational trigonometric values on a unit circle. Ray Whitmer ray@xmission.com === Subject: Re: Density of Pythagorean triples > Let a be positive, T(a) = {(x,y) in R^2: x and y nonnegative, x^a > + y^a = 1}, and B(a) be the intersection of Q^2 and T(a). Is > B(2) dense in T(2) (under the usual topology)? Nowhere dense? Yes, its dense. If (x,y) belong to Q^2 are such that x^2 + y^2 = 1, then there is a rational number q such that x = (1 - q^2)/(1 + q^2) and that y = = 2 q/(1 + q^2). Now, take (x,y) in T(2) and take the number z = = y/(1 + x). The number z is limit of some sequence (q_n)_n of rational numbers. Take x_n = (1 - q_n^2)/(1 + q_n^2) and take y = = 2 q_n/(1 + q_n^2). Then lim_n (x_n,y_n) = (x,y). Jose Carlos Santos === Subject: Re: Density of Pythagorean triples >Let a be positive, T(a) = {(x,y) in R^2: x and y nonnegative, x^a >+ y^a = 1}, and B(a) be the intersection of Q^2 and T(a). Is >B(2) dense in T(2) (under the usual topology)? Nowhere dense? Dense, because the unit circle has a rational parametrization x = (t^2-1)/(t^2+1), y = 2t/(t^2+1) >For n a positive integer, B(1/n) is dense in T(1/n). By FLT, >B(n) is finite (in fact, has only two points) for integer n > 2. >What can be said along these lines (density, cardinality) for general >real positive a? Any given (x,y) is in B(a) for at most one a. So B(a) is empty for all but countably many a. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Density of Pythagorean triples >>Let a be positive, T(a) = {(x,y) in R^2: x and y nonnegative, x^a >>+ y^a = 1}, and B(a) be the intersection of Q^2 and T(a). Is >>B(2) dense in T(2) (under the usual topology)? Nowhere dense? >> >Dense, because the unit circle has a rational parametrization >x = (t^2-1)/(t^2+1), y = 2t/(t^2+1) >>For n a positive integer, B(1/n) is dense in T(1/n). By FLT, >>B(n) is finite (in fact, has only two points) for integer n > 2. >>What can be said along these lines (density, cardinality) for general >>real positive a? >> >Any given (x,y) is in B(a) for at most one a. So B(a) is empty for all >but countably many a. Slight correction: B(a) has at least two points for all a. Conjecture: B(a) is dense in T(a) iff 2/a is an integer. Otherwise, B(a) is finite. What do you think? As hard as (or equivalent to?) FLT? More: Let T(n,a) be the boundary of the unit ball in R^n under the l_a norm; let D(n,a) be the intetersection of Q^n and T(n,a). For what values of a is D(n,a) dense in T(n,a)? Is D(n,a) finite if it is not dense? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Pascal and Fermat Is it true that in 1640, there werent (compared to today) very many people doing math, and the math wasnt as complicated and deep as it is today, so any reasonably bright math nerd could have made the same discoveries as people like Pascal. And later Fermat was brilliant, but maybe not so much more than any other of todays leading researchers on our newer, complicated mathematics. Which doesnt put those famous ones on a level of intellectual might that others cant reach, but just makes them maths forefathers? So whose to say how many people we have working in their fields that arent at the same level as Ramanujans calculus? === Subject: Re: Pascal and Fermat > Is it true that in 1640, there werent (compared to today) very many > people doing math, and the math wasnt as complicated and deep as it > is today, so any reasonably bright math nerd could have made the same > discoveries as people like Pascal. And later Fermat was brilliant, but > maybe not so much more than any other of todays leading researchers > on our newer, complicated mathematics. Which doesnt put those famous > ones on a level of intellectual might that others cant reach, but > just makes them maths forefathers? So whose to say how many people we > have working in their fields that arent at the same level as > Ramanujans calculus? Well, there are probably a couple of people as good as Newton or Euler alive today, but theyre not famous. In a hundred years or so, with hindsight, well pick some heroes. Similarly, Bach is famous today, and will be in a few hundred years, but I doubt my children will have ever heard of, say, Britney Spears. === Subject: how to finding eigenvalues and eigen vectors hi its hasber i belong to the field of computer science. i am working on face recognition systems. my problem is that i have a matrix of size 4096*4096 and i have to find its eigenvalues and eigen vetors. anyone describle me how to calculate it, i want to implement the routine for finding both in c/c++. in case any stuff already present plz inform me at hasber57@hotmail.com hasber uraan soft pvt ltd === Subject: Re: how to finding eigenvalues and eigen vectors >hi >its hasber >i belong to the field of computer science. >i am working on face recognition systems. >my problem is that i have a matrix of size >4096*4096 and i have to find its eigenvalues and eigen vetors. [...] As others have said, thats a fairly big calculation. But I very much doubt you need to do it. Probably, you have n greyscale images, each 64 by 64 pixels, and n is a lot less than 4096. Probably, you only need to find the eigenvalues and eigenvectors of a n by n matrix which is real, symmetric, and has all eigenvalues real and non-negative. Probably, you are only interested in the largest 10 to 100 eigenvalues and their associated eigenvectors. All this makes your real problem much easier than the one you asked about, but it is still non-trivial. I suggest you Google on eigenfaces and pca, or post to sci.stat.math. -- Graham Jones http://www.visiv.co.uk Please add a j just before the @ to ensure delivery === Subject: Re: how to finding eigenvalues and eigen vectors > hi > its hasber > i belong to the field of computer science. > i am working on face recognition systems. > my problem is that i have a matrix of size > 4096*4096 and i have to find its eigenvalues and eigen vetors. Do you remember where you last saw them? Try to retrace your steps since the last time you had them. Also, look on the dining room table - thats often where I find things Ive lost. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: how to finding eigenvalues and eigen vectors >i belong to the field of computer science. >i am working on face recognition systems. >my problem is that i have a matrix of size >4096*4096 and i have to find its eigenvalues and eigen vetors. >anyone describle me how to calculate it, i want to implement the >routine for finding both in c/c++. Implementing it yourself is not recommended. There are some very capable existing packages for linear algebra that should do it quite well (assuming you have sufficient computing power available - this is a nontrivial problem). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: how to finding eigenvalues and eigen vectors > hi > its hasber > i belong to the field of computer science. > i am working on face recognition systems. > my problem is that i have a matrix of size > 4096*4096 and i have to find its eigenvalues and eigen vetors. > anyone describle me how to calculate it, i want to implement the > routine for finding both in c/c++. > in case any stuff already present plz inform me at > hasber57@hotmail.com > hasber > uraan soft pvt ltd As far as I know, theres no general way to do it, unless its a special matrix which is diagonalizable. In this case, just diagonalize it and the diagonal entires are the eigenvalues and the images of (1, 0, 0, ...), (0, 1, 0, ..), etc. in the original basis are the eigenvectors in that basis. There are some numerical routines which find approximations to eigenvalues, but Im not exactly sure how they work. === Subject: Re: how to finding eigenvalues and eigen vectors Well i just wanted ti help, and i thought that you can find eigenvalues only if you have square matrix. === Subject: Re: how to finding eigenvalues and eigen vectors > hi > its hasber > i belong to the field of computer science. > i am working on face recognition systems. > my problem is that i have a matrix of size > 4096*4096 and i have to find its eigenvalues and eigen vetors. > anyone describle me how to calculate it, i want to implement the > routine for finding both in c/c++. > in case any stuff already present plz inform me at > hasber57@hotmail.com > hasber > uraan soft pvt ltd Is this a matrix with nonnegative entries? That will help, perhaps. H. Minc, NONNEGATIVE MATRICES, Wiley, 1988. Chapter II, as I recall, is on computation of the largest eigenvalue. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: how to finding eigenvalues and eigen vectors In order to find eigenvalues you must find determinant of your matrix. Calculating determinant of THAT BIG matrix is a problem. Anyway ill tell you how to do it (mathematically). If you have matrix A you must find such vector x and scalar ¤ for which equation Ax=¤x is correct. If you rewrite this in matrix form (A-¤I)x=0, where I is identity matrix. First you must find det(A-¤I). When you calculate it you will have some polynomial of ¤ (P(¤)). All values of ¤ for which equation P(¤)=0 is satisfied are eigenvalues. Then you return theese values in equation (A-¤I)x=0. After multiplication you will have system of n linear equations with n values (theese are coordinates of your eigenvector)to find. n is dimension of your matrix. When you find theese you will have your eigenvectors. If matrix is not normal it is impossible to solve this problem. === Subject: Re: how to finding eigenvalues and eigen vectors >In order to find eigenvalues you must find determinant of your matrix. >Calculating determinant of THAT BIG matrix is a problem. It is, and the algorithm you give is O(N!) for an NxN matrix so not very useful. Not that the matrix itself is very big. Maybe it would have been more useful to point him to the group sci.math.num-analysis instead where they actually know something about these things? -- Im not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: JSH: Understand now? Frustration? > Using P(x) = (x+8a)(x+b), where ab=1, and considering when (8a + b) is > an integer can give you a perspective on what Ive been saying, I > hope. > Since b = 1/a, by your requirement, then (8a + b) = (8a + 1/a). Are you > requiring that this expression be an integer? To be fair, he said consider those cases where (8a+b) _is_ an integer. In a previous post he chose (8a+b) = 9, ab=1 which has the two solutions [b=8, a=1/8] and [b=1, a=1] so both (8a + b) and (8a + 1/a) are indeed integer even though a=1/8 is not. KeithK > Basically the mathematical position of the mainstream--why its a > core error--is that you can never have a factor like 8 in an arena > where -1 and 1 are the only integer units because you cant in the > ring of algebraic integers, except for simple radical factors, if you > dont have integer Ôa and Ôb. > Let a = b = 1, or a = b = -1. Now what? > Wrong again, stupid. The core error is obviously between your ears -- or > in the Ôsplitting field between your rear cheeks! > James Often in error, but never in doubt. Harris > How many times do you intend to keep placing your unparalleled ignorance > on display? > -- > There are two things you must never attempt to prove: the unprovable -- > and the obvious. > -- > Democracy: The triumph of popularity over principle. > -- > http://www.crbond.com === Subject: Re: JSH: Understand now? Frustration? > Using P(x) = (x+8a)(x+b), where ab=1, and considering when (8a + b) is > an integer can give you a perspective on what Ive been saying, I > hope. > Since b = 1/a, by your requirement, then (8a + b) = (8a + 1/a). Are you > requiring that this expression be an integer? > To be fair, he said consider those cases where (8a+b) _is_ an integer. > In a previous post he chose (8a+b) = 9, ab=1 which has the two solutions > [b=8, a=1/8] and [b=1, a=1] > so both (8a + b) and (8a + 1/a) are indeed integer even though a=1/8 is not. > KeithK I dont believe I was being unfair. I was simply asking for clarification. James has a history of posting contradictory requirements and leaving the burden of choosing what to work with to the reader. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Understand now? Frustration? > Thats has to do with why at times Ive just gotten totally pissed and > called Arturo Magidin evil, as its just so frustrating to deal with > people so adept at obscuring the truth. > Is this why you accepted Prof. Magidins offer to stop responding to you, > so that you could continue this cowardly sniping at him without fear > of getting a response? Youre just demonstrating that you arent man > enough to confront him directly -- something that comes as no surprise > to the rest of us, of course. > -- > Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise > fwbrown@bellsouth.net | if youre good enough. Otherwise you give > | your pelt to the trapper. > e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock Wayne, James isnt a man, period. David Moran === Subject: Re: JSH: Understand now? Frustration? > Thats has to do with why at times Ive just gotten totally pissed and > called Arturo Magidin evil, as its just so frustrating to deal with > people so adept at obscuring the truth. Is this why you accepted Prof. Magidins offer to stop responding to you, so that you could continue this cowardly sniping at him without fear of getting a response? Youre just demonstrating that you arent man enough to confront him directly -- something that comes as no surprise to the rest of us, of course. -- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Understand now? Frustration? Using P(x) = (x+8a)(x+b), where ab=1, and considering when (8a + b) is > an integer can give you a perspective on what Ive been saying, I > hope. Since b = 1/a, by your requirement, then (8a + b) = (8a + 1/a). Are you > requiring that this expression be an integer? To be fair, he said consider those cases where (8a+b) _is_ an integer. > In a previous post he chose (8a+b) = 9, ab=1 which has the two solutions [b=8, a=1/8] and [b=1, a=1] so both (8a + b) and (8a + 1/a) are indeed integer even though a=1/8 is not. KeithK > I dont believe I was being unfair. I was simply asking for clarification. James > has a history of posting contradictory requirements and leaving the burden of > choosing what to work with to the reader. > Ive been arguing with posters like Arturo Magidin for YEARS, and > throughout that period, Ive found my attempts to clarify were > hampered by attempts by other posters like here C. Bond to confuse, > and I thank KeithK for catching one as he did. I was not implying that C. Bond was attempting to confuse. I simply provided a clarification. KeithK > The problem isnt with C. Bond using 8a + 1/a as that is correct, as > long as you consider the possibility of Ôa being a factor of 1 in > some more inclusive ring than the ring of algebraic integers. > Circular arguments have gone on and on and on for some years here, as > people like Arturo Magidin, have successfully dodged being pinned down > on this issue as they repeatedly relied on certain numbers not being > algebraic integers to support their claims. > Notice, Im NOT saying that Ôa is an algebraic integer when its > irrational! > If you say that 1/a is not an algebraic integer, when irrational, yes, > you are correct. > But my point is that the label algebraic integer does not contain > all numbers such that 1/a is contained and -1 and 1 are the only > integer units. Theres a more inclusive ring, where -1 and 1 are the > only integer units, where 1/a is a unit, and Ôa is not an algebraic > integer. > Here it takes effort if some poster comes in to confuse you, relying > on the *assumption* that if 1/a is not an algebraic integer then its > more like a fraction like 1/2 than 1/1. > But think of x=(1+sqrt(-3))/2, and someone giving you 1/x, if you were > mostly experienced with integers. > Think of how easy it might be for someone to confuse you with > 2/(1+sqrt(-3)) if you werent careful and it was new to you. > Thats has to do with why at times Ive just gotten totally pissed and > called Arturo Magidin evil, as its just so frustrating to deal with > people so adept at obscuring the truth. > My hope is that at least some of you will play with > x^2 + (8a + b)x + 8, with ab = 1, and (8a + b) an integer > so that you can see why *logically* a possibility must still be there > when Ôa and Ôb are irrational, even if Ôa is not an algebraic > integer. > If you can get a handle on that possibility, then maybe discussions > can be more fruitful about those numbers lost in the shufße that > irrational Ôa here represents. > You know, they are *most* numbers given their cardinality. > James Harris === Subject: Re: JSH: Understand now? Frustration? > Notice, Im NOT saying that Ôa is an algebraic integer when its > irrational! > If you say that 1/a is not an algebraic integer, when irrational, yes, > you are correct. > But my point is that the label algebraic integer does not contain > all numbers such that 1/a is contained and -1 and 1 are the only > integer units. Theres a more inclusive ring, where -1 and 1 are the > only integer units, where 1/a is a unit, and Ôa is not an algebraic > integer. Right, the algebraic integers are not *that* ring. But you can adjoin 1/a to the ring of algebraic integers and get a more inclusive ring, such that 1/a is in that ring. On the other hand, when you do adjoin 1/a to the ring of algebraic integers, it is almost certain that -1 and 1 are no longer the only integer units. It is to you to prove that a ring as you intend does exist. > Here it takes effort if some poster comes in to confuse you, relying > on the *assumption* that if 1/a is not an algebraic integer then its > more like a fraction like 1/2 than 1/1. But it *is* more like 1/2, 3/2, etc. Note that when p is an algebraic number that is not an algebraic integers, it can be written as r/s, where r is an algebraic integer and s is a normal integer. That is all pretty standard. > But think of x=(1+sqrt(-3))/2, and someone giving you 1/x, if you were > mostly experienced with integers. > Think of how easy it might be for someone to confuse you with > 2/(1+sqrt(-3)) if you werent careful and it was new to you. I do not understand what you mean here, but indeed, x is a divisor of 1. So, the algebraic integers have more divisors of 1 than the normal integers. So have the Gaussian integers, the Eisenstein integers (which contain this x), and what you have. And it is easy to understand when you see that [(1 + sqrt(-3))/2][(1 - sqrt(-3))/2] = 1. So I do not see what the problem is. > Thats has to do with why at times Ive just gotten totally pissed and > called Arturo Magidin evil, as its just so frustrating to deal with > people so adept at obscuring the truth. *What* obscuring? > My hope is that at least some of you will play with > x^2 + (8a + b)x + 8, with ab = 1, and (8a + b) an integer > so that you can see why *logically* a possibility must still be there > when Ôa and Ôb are irrational, even if Ôa is not an algebraic > integer. Well, obviously the polynomial is: x^2 + (8a + b)x + 8ab which is reducible. So there is no sense looking at the quadratic. Note: again you fail to distinguish reducible polynomials from irreducible polynomials. The theory goes about irreducible polynomials because you can say much more about the roots in that case (see Galois theory). But I can guide you. Let b be an algebraic integer that divides 8, in that case 8a = 8/b is also an algebraic integer. Now, can it be an integer? Or can we split 8 in two factors p and q such that p + q is an integer? The answer is, yes, we can. Let r be an arbitrary integer and consider the polynomial x^2 + rx + 8, the roots multiply to give 8 and add to give an integer, so there you are. The roots are: (-r + sqrt(r^2 - 32))/2 and (-r - sqrt(r^2 - 32))/2 Set b equal to the first and a to 1/8 times the second and you are done. Fill in any algebraic integer number r you can think off. > If you can get a handle on that possibility, then maybe discussions > can be more fruitful about those numbers lost in the shufße that > irrational Ôa here represents. Eh? > You know, they are *most* numbers given their cardinality. Eh? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Understand now? Frustration? > Ive been arguing with posters like Arturo Magidin for YEARS, and > throughout that period, Ive found my attempts to clarify were > hampered by attempts by other posters like here C. Bond to confuse, The greatest impediment to JSHs attempts to clarify have been JSHs refusal to learn anything about mathematics. > and I thank KeithK for catching one as he did. > The problem isnt with C. Bond using 8a + 1/a as that is correct, as > long as you consider the possibility of Ôa being a factor of 1 in > some more inclusive ring than the ring of algebraic integers. In the field of complex numbers, everything except zero is a factor of everything. > Circular arguments have gone on and on and on for some years here, as > people like Arturo Magidin, have successfully dodged being pinned down > on this issue as they repeatedly relied on certain numbers not being > algebraic integers to support their claims. JSH is an expert at circular arguments, he makes so many of them. And JSH only badmouths Arturo as a person because Arturos mathematics have so often shot down JSHs maunderings. > Notice, Im NOT saying that Ôa is an algebraic integer when its > irrational! > If you say that 1/a is not an algebraic integer, when irrational, yes, > you are correct. You, on the other hand, are not correct even when you are rational. > But my point is that the label algebraic integer does not contain > all numbers such that 1/a is contained and -1 and 1 are the only > integer units. Is this supposed to mean something? Theres a more inclusive ring, where -1 and 1 are the > only integer units, where 1/a is a unit, and Ôa is not an algebraic > integer. Show us such a ring. > Here it takes effort if some poster comes in to confuse you, relying > on the *assumption* that if 1/a is not an algebraic integer then its > more like a fraction like 1/2 than 1/1. The poster who is trying to confuse us is James S Harris. > But think of x=(1+sqrt(-3))/2, and someone giving you 1/x, if you were > mostly experienced with integers. > Think of how easy it might be for someone to confuse you with > 2/(1+sqrt(-3)) if you werent careful and it was new to you. > Thats has to do with why at times Ive just gotten totally pissed and > called Arturo Magidin evil, as its just so frustrating to deal with > people so adept at obscuring the truth. It does not take an adept to conceal the truths from someone as reluctant to accept them as JSH. What JSH does not want to accept, however limpidly clear and true it may be, he will find some way to reject. He has done this so often now that it is a habit that he cannot break. And even after he is forced to accept a truth, as he did recently, he may turn around and later reject it again, as he did recently. As this post of his demonstrates. He seems to have errors hardwired into his brain so deeply that truth can only gain a temporary foothold in his thoughts. > My hope is that at least some of you will play with > x^2 + (8a + b)x + 8, with ab = 1, and (8a + b) an integer > so that you can see why *logically* a possibility must still be there > when Ôa and Ôb are irrational, even if Ôa is not an algebraic > integer. What possibility is that? If a*b = 1 and (8*a+b) is an integer, then 8*a and b are roots of x^2 - (8a + b)x + 8*a*b = x^2 - (8a + b)x + 8 = 0, so 8*a and b are algebraic integers. This does not require that a, or 2*a, or 4*a be algebraic integers. SO? > If you can get a handle on that possibility, then maybe discussions > can be more fruitful about those numbers lost in the shufße that > irrational Ôa here represents. > You know, they are *most* numbers given their cardinality. Who are they? As both a and b must be algebraic numbers, and the cardinality of the algebraic numbers is countable, and the cardinality of the trascendental(non-algebraic) numbers is uncountable, there are more numbers not solutions to equation x^2 - (8a + b)x + 8*a*b = 0,where a*b = 1 abd (8*a+b) is an integer, than numbers which are solutions, but why should that be a surprise to anyone? That is the case with most conditional equations. > James Harris Off in his dreamworld again. === Subject: Re: JSH: Understand now? Frustration? Using P(x) = (x+8a)(x+b), where ab=1, and considering when (8a + b) is > an integer can give you a perspective on what Ive been saying, I > hope. Since b = 1/a, by your requirement, then (8a + b) = (8a + 1/a). Are you > requiring that this expression be an integer? > To be fair, he said consider those cases where (8a+b) _is_ an integer. > In a previous post he chose (8a+b) = 9, ab=1 which has the two solutions > [b=8, a=1/8] and [b=1, a=1] > so both (8a + b) and (8a + 1/a) are indeed integer even though a=1/8 is not. > KeithK > I dont believe I was being unfair. I was simply asking for clarification. James > has a history of posting contradictory requirements and leaving the burden of > choosing what to work with to the reader. Ive been arguing with posters like Arturo Magidin for YEARS, and throughout that period, Ive found my attempts to clarify were hampered by attempts by other posters like here C. Bond to confuse, and I thank KeithK for catching one as he did. The problem isnt with C. Bond using 8a + 1/a as that is correct, as long as you consider the possibility of Ôa being a factor of 1 in some more inclusive ring than the ring of algebraic integers. Circular arguments have gone on and on and on for some years here, as people like Arturo Magidin, have successfully dodged being pinned down on this issue as they repeatedly relied on certain numbers not being algebraic integers to support their claims. Notice, Im NOT saying that Ôa is an algebraic integer when its irrational! If you say that 1/a is not an algebraic integer, when irrational, yes, you are correct. But my point is that the label algebraic integer does not contain all numbers such that 1/a is contained and -1 and 1 are the only integer units. Theres a more inclusive ring, where -1 and 1 are the only integer units, where 1/a is a unit, and Ôa is not an algebraic integer. Here it takes effort if some poster comes in to confuse you, relying on the *assumption* that if 1/a is not an algebraic integer then its more like a fraction like 1/2 than 1/1. But think of x=(1+sqrt(-3))/2, and someone giving you 1/x, if you were mostly experienced with integers. Think of how easy it might be for someone to confuse you with 2/(1+sqrt(-3)) if you werent careful and it was new to you. Thats has to do with why at times Ive just gotten totally pissed and called Arturo Magidin evil, as its just so frustrating to deal with people so adept at obscuring the truth. My hope is that at least some of you will play with x^2 + (8a + b)x + 8, with ab = 1, and (8a + b) an integer so that you can see why *logically* a possibility must still be there when Ôa and Ôb are irrational, even if Ôa is not an algebraic integer. If you can get a handle on that possibility, then maybe discussions can be more fruitful about those numbers lost in the shufße that irrational Ôa here represents. You know, they are *most* numbers given their cardinality. James Harris === Subject: Re: JSH: Understand now? Frustration? > Using P(x) = (x+8a)(x+b), where ab=1, and considering when (8a + b) is > an integer can give you a perspective on what Ive been saying, I > hope. Yup. It is still wrong. And for much the same reasom, that you, Jimmy, have no clue as to what is really going on in the relevant mathematics. Your abysmal ignorance keeps you from getting anything more complicated than high school algebra correct. Your overweening arrogance keeps you trying. The combination leads to your frequent mathematical pratfalls and the resulting temper tantrums. We all get another good laugh. Keep it up. === Subject: Re: Can something be undeterminable? >Has it been (or can it be) proved whether there exist questions >that are *absolutely* unanswerable in the sense of there being >no sufficiently powerful mathematical system to answer them? > Suppose S is your current system, and Q is a question that it > cant answer (i.e. there is no proof in S that the answer is yes, > and there is no proof in S that the answer is no). Then you can > produce two more powerful systems: > S1, by adding to S the axiom the answer to Q is yes; > S2, by adding to S the axiom the answer to Q is no. > Both of these systems can answer the question. And, if its > a meaningful yes-or-no question, one of these answers will be correct. Cant there be meaningful yes / no answers where either answer can be true? For example, do two parallel lines meet at infinity? Russell - 2 many 2 count === Subject: Re: Can something be undeterminable? [typo correction] >Has it been (or can it be) proved whether there exist questions >that are *absolutely* unanswerable in the sense of there being >no sufficiently powerful mathematical system to answer them? > Suppose S is your current system, and Q is a question that it > cant answer (i.e. there is no proof in S that the answer is yes, > and there is no proof in S that the answer is no). Then you can > produce two more powerful systems: > S1, by adding to S the axiom the answer to Q is yes; > S2, by adding to S the axiom the answer to Q is no. > Both of these systems can answer the question. And, if its > a meaningful yes-or-no question, one of these answers will be correct. > Cant there be meaningful yes / no answers where either answer > can be true? For example, do two parallel lines meet at infinity? Thats a good point, imo; however, Im interested in certain well-defined questions such as the following, to which (it seems to me) there must be exactly one *correct* answer, and that answer is either Yes or No: Are there infinitely many primes in Rados Sigma sequence (1,4,6,13,...)? At least I *think* this is an example of what I have in mind. What I want is some property, say P, concerning a non-computable sequence, such that it is either True or False that the sequence has the property, *but* is such that non-computability makes it *certain* that it cannot under any circumstances be known which is the correct answer. Im not sure whether containing infinitely many primes is such a property. --r.e.s. === Subject: Re: Can something be undeterminable? >Has it been (or can it be) proved whether there exist questions >that are *absolutely* unanswerable in the sense of there being >no sufficiently powerful mathematical system to answer them? > Suppose S is your current system, and Q is a question that it > cant answer (i.e. there is no proof in S that the answer is yes, > and there is no proof in S that the answer is no). Then you can > produce two more powerful systems: > S1, by adding to S the axiom the answer to Q is yes; > S2, by adding to S the axiom the answer to Q is no. > Both of these systems can answer the question. And, if its > a meaningful yes-or-no question, one of these answers will be correct. > Cant there be meaningful yes / no answers where either answer > can be true? For example, do two parallel lines meet at infinity? Thats a good point, imo; however, Im interested in certain well-defined questions such as the following, to which (it seems to me) there must be exactly one *correct* answer, and that answer is either True or False: Are there infinitely many primes in Rados Sigma sequence (1,4,6,13,...)? At least I *think* this is an example of what I have in mind. What I want is some property, say P, concerning a non-computable sequence, such that it is either True or False that the sequence has the property, *but* is such that non-computability makes it *certain* that it cannot under any circumstances be known which is the correct answer. Im not sure whether containing infinitely many primes is such a property. --r.e.s. === Subject: Re: Can something be undeterminable? > Cant there be meaningful yes / no answers where either answer > can be true? For example, do two parallel lines meet at infinity? Not if there is one true ideal geometry... === Subject: Re: Can something be undeterminable? <29KdndPF5sOq8aXdRVn-jg@comcast.com Suppose S is your current system, and Q is a question that it > cant answer (i.e. there is no proof in S that the answer is yes, > and there is no proof in S that the answer is no). Then you can > produce two more powerful systems: > S1, by adding to S the axiom the answer to Q is yes; > S2, by adding to S the axiom the answer to Q is no. > Both of these systems can answer the question. And, if its > a meaningful yes-or-no question, one of these answers will be correct. > Cant there be meaningful yes / no answers where either answer > can be true? For example, do two parallel lines meet at infinity? That is essentially what he is saying... you can build a system S1 for which (in your example) two parallel lines meet at infinity. Or you can build an equally-consistent system S2 in which parallel lines never meet. S2 in this case is the standard euclidean geometry while S1 is known as projective geometry. A bit more about projective geometry in the plane: all (parallel) lines with slope m meet at some point P(m). There is a separate P(m) for every slope m, and all the points P(m) (over all m) are colinear. J === Subject: OT observation Sometime it seems that the exact mening of a statement is not clear and that it is easier to rephrase the general meaning in ones own terms with a negative attached. Its nice to see that John is happy? Well, he certaintly isnt unhappy. (I dont know what you mean by happiness but I agree that he seemed to be in better shape than the last time we saw him.) === Subject: Re: OT observation > Yeah, there seems to be some kind of internal order relationship implied in > a lot of the use of the negation of contraries (I think my use of the term > double-negative was likely improper, and accept the analyses of several > others on this point). > So when we say John is happy, it establishes a point happiness as being a > part of the state of John. When we say that John is unhappy, it establishes > a different point unhappiness as not being a part of the state of John. > I asume you meant When we say that John is not unhappy because that > makes a lot more sense for the argument with which you follow, but it > also works this way. > Something like > happiness ------------------ unhappiness > and we seem to internally identify gradations between the states. The > difference between the statements then seems to be that the negation of > contraries does not assert the positive condition, it merely asserts that > the contrary position is not the case and that anywhere elsewhere on the > continuum is possible. > Ok it sounds very reasonable BUT it has the major disadvantage of > being worthless, because it works in this particular case but cannot > be generalized. > I can identify at least to levels at which this fails. The first is > that for certain adjectives there can be no gradation. Take for > example dead. The second is, that it also fails for any adjective > depending on the context of the assertion. > say: > 1.- I am not unaware of the difficulties related with my > argumentation. > 2.- I am NOT unaware of my duties. > For 1.- your hypothesis holds clearly. For 2.- it clearly doesnt, > because not unaware in this case clearly means aware. > Not being certain of the affirmative position for > whatever reason is one exemplar of this usage. > Natural languages are simply too slippery to cage with this kind of > analysis. You keep trying to use this multivalue approach and oversee > that though the problems that you attack display such structures, the > approach itself doesnt benefit from them. Hilberts programm We can > know and we will know was shattered by G.9adel. Much of science has > since has been realising that it will never know everything. But there > are some as yourself, who think its just a matter of sharpening the > tools of science. I personally believe that we will always be limited > in our ability to know by the limitations of our instruments. For > mathematics at least, G.9adel showed that our instruments will always be > blunt in the end. > This kind of negation can be modelled. One particularly nice thing I like > in Heyting models is that one already gets relationships such as > ~(~(a)) -> a > as an order relationship. Sorry for my carelessness. Yes not unhappy should be in place of unhappy. === Subject: Re: OT observation > Yeah, there seems to be some kind of internal order relationship implied in > a lot of the use of the negation of contraries (I think my use of the term > double-negative was likely improper, and accept the analyses of several > others on this point). > So when we say John is happy, it establishes a point happiness as being a > part of the state of John. When we say that John is unhappy, it establishes > a different point unhappiness as not being a part of the state of John. I asume you meant When we say that John is not unhappy because that makes a lot more sense for the argument with which you follow, but it also works this way. > Something like > happiness ------------------ unhappiness > and we seem to internally identify gradations between the states. The > difference between the statements then seems to be that the negation of > contraries does not assert the positive condition, it merely asserts that > the contrary position is not the case and that anywhere elsewhere on the > continuum is possible. Ok it sounds very reasonable BUT it has the major disadvantage of being worthless, because it works in this particular case but cannot be generalized. I can identify at least to levels at which this fails. The first is that for certain adjectives there can be no gradation. Take for example dead. The second is, that it also fails for any adjective depending on the context of the assertion. say: 1.- I am not unaware of the difficulties related with my argumentation. 2.- I am NOT unaware of my duties. For 1.- your hypothesis holds clearly. For 2.- it clearly doesnt, because not unaware in this case clearly means aware. > Not being certain of the affirmative position for > whatever reason is one exemplar of this usage. Natural languages are simply too slippery to cage with this kind of analysis. You keep trying to use this multivalue approach and oversee that though the problems that you attack display such structures, the approach itself doesnt benefit from them. Hilberts programm We can know and we will know was shattered by G.9adel. Much of science has since has been realising that it will never know everything. But there are some as yourself, who think its just a matter of sharpening the tools of science. I personally believe that we will always be limited in our ability to know by the limitations of our instruments. For mathematics at least, G.9adel showed that our instruments will always be blunt in the end. > This kind of negation can be modelled. One particularly nice thing I like > in Heyting models is that one already gets relationships such as > ~(~(a)) -> a > as an order relationship. === Subject: Re: OT observation > ~(~(a)) -> a This says it very clearly and gives me a chance to explain what I meant. The above is not true in all cases in the real world. And that is because the meaning of a can change and that change is what makes a kind of double negative useful. Im belaboring this poing because a guy who lead a Transactional Analysis group picked on a guy who did this and I thought the guy was right. This makes me think that since the leader of the group is a smart guy, that maybe this is a fine point. Anyway, maybe people have a different idea of what happy means. It might mean sexually satisfied. It might mean contented. It might mean a higher happiness as explained by Aristotle... So if A says that B is happy to C, then maybe C would say, well he isnt unhappy, which can be true of As statement, which might be assumed because of ones trust in As statement. C can let A knows that he (C) doesnt know what A means by Ôhappy, and can be agreeable in general without pretending to know what A means exactly. Sorry this is so convoluted and its really not worth thinking about too much because if its important you probably would have noticed it already, if you havent. === Subject: Re: OT observation : Sometime it seems that the exact mening of a statement is : not clear and that it is easier to rephrase the general meaning : in ones own terms with a negative attached. : : Its nice to see that John is happy? : Well, he certaintly isnt unhappy. : : (I dont know what you mean by happiness but : I agree that he seemed to be in better shape than : the last time we saw him.) Yeah, there seems to be some kind of internal order relationship implied in a lot of the use of the negation of contraries (I think my use of the term double-negative was likely improper, and accept the analyses of several others on this point). So when we say John is happy, it establishes a point happiness as being a part of the state of John. When we say that John is unhappy, it establishes a different point unhappiness as not being a part of the state of John. Something like happiness ------------------ unhappiness and we seem to internally identify gradations between the states. The difference between the statements then seems to be that the negation of contraries does not assert the positive condition, it merely asserts that the contrary position is not the case and that anywhere elsewhere on the continuum is possible. Not being certain of the affirmative position for whatever reason is one exemplar of this usage. This kind of negation can be modelled. One particularly nice thing I like in Heyting models is that one already gets relationships such as ~(~(a)) -> a as an order relationship. -- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: OT observation >Sometimes it seems that the exact meaning of a statement is >not clear and that it is easier to rephrase the general meaning >in ones own terms with a negative attached. > When the exact meaning of a statement is unclear, just rephrasing the > general meaning without a negative attached is NOT the easiest way to > clarify. > If a statement is clear, it is NOT necessary to restate it in any > form. It happens a lot and makes sense to me. So maybe I didnt explain it well enough. Oh well. === Subject: Re: OT observation >Sometimes it seems that the exact meaning of a statement is >not clear and that it is easier to rephrase the general meaning >in ones own terms with a negative attached. When the exact meaning of a statement is unclear, just rephrasing the general meaning without a negative attached is NOT the easiest way to clarify. If a statement is clear, it is NOT necessary to restate it in any form. webster2000@my-deja.com (web2k) === Subject: Re: OT observation > Sometime it seems that the exact mening of a statement is > not clear and that it is easier to rephrase the general meaning > in ones own terms with a negative attached. > Its nice to see that John is happy? > Well, he certaintly isnt unhappy. > (I dont know what you mean by happiness but > I agree that he seemed to be in better shape than > the last time we saw him.) The opposite of a statement is the conTrapositive, not the negative. Even Aristotle knew that. There are no unique antonyms in 1:1 correspondence with adjectives. What is the antonym of tepid? -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: OT observation Is your post related to mine? I said it was OT. Didnt it make >> sense to you? >OT^2: Are you sure this should be squared and not negated? > ..When the great coyote covered the ßowing spring waters and used >to beget the net, wouldnt it have been nice if the nettles in their >first naming frenzy could have chosen a name for off topic which was >not also an abbreviation of on topic? No, no, no. OT is for overtime. Re is on topic. /BAH Subtract a hundred and four for e-mail. === Subject: Re: OT observation Sometime it seems that the exact mening of a statement is > not clear and that it is easier to rephrase the general meaning > in ones own terms with a negative attached. Its nice to see that John is happy? > Well, he certaintly isnt unhappy. (I dont know what you mean by happiness but > I agree that he seemed to be in better shape than > the last time we saw him.) > The opposite of a statement is the conTrapositive, not the negative. > Even Aristotle knew that. The contrapositive is logically equivalent to the original statement: so in what sense is it the opposite? > There are no unique antonyms in 1:1 > correspondence with adjectives. What is the antonym of tepid? Right. So therefore we must use the contrapositive instead: This cup of tea is tepid. The non-tepid does not include this cup of tea. Ah ... what was your point again? > Is your post related to mine? I said it was OT. Didnt it make > sense to you? OT^2: When the great coyote covered the ßowing spring waters and used to beget the net, wouldnt it have been nice if the nettles in their first naming frenzy could have chosen a name for off topic which was not also an abbreviation of on topic? === Subject: Re: OT observation > Sometime it seems that the exact mening of a statement is > not clear and that it is easier to rephrase the general meaning > in ones own terms with a negative attached. > Its nice to see that John is happy? > Well, he certaintly isnt unhappy. > (I dont know what you mean by happiness but > I agree that he seemed to be in better shape than > the last time we saw him.) > The opposite of a statement is the conTrapositive, not the negative. > Even Aristotle knew that. There are no unique antonyms in 1:1 > correspondence with adjectives. What is the antonym of tepid? Is your post related to mine? I said it was OT. Didnt it make sense to you? === Subject: Math causes brain cancer Because math is too difficult for some brains, they overload and cancer will be produced. So better donÇt do math if you are not 100% fit and healthy or your head may burst. === Subject: Re: Math causes brain cancer In sci.math, Ralf Dieholt cancer will be produced. > So better donÇt do math if you are not 100% fit and healthy or your > head may burst. Well, a more likely result is being shot by an opponent in a duel over a woman. :-/ (A pity, for Galois had lots of potential.) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Math causes brain cancer > Because math is too difficult for some brains, they overload and > cancer will be produced. > So better donÇt do math if you are not 100% fit and healthy or your > head may burst. Sounds like an excuse 4th-graders use when they hate math... -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Math causes brain cancer G. A. Edgar <9syjyb402@sneakemail.com> scribbled the following: >> Because math is too difficult for some brains, they overload and >> cancer will be produced. >> So better donÇt do math if you are not 100% fit and healthy or your >> head may burst. > Sounds like an excuse 4th-graders use when they hate math... Or when they *try* to hate math. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ Its time, its time, its time to dump the slime! - Dr. Dante === Subject: Re: Math causes brain cancer >So better donÇt do math if you are not 100% fit and healthy or your >head may burst. Unfortunately, long-time clinical tests on sci.math cranks have failed to produce this effect. -- Im not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: Math causes brain cancer > Because math is too difficult for some brains, they overload and > cancer will be produced. > So better donÇt do math if you are not 100% fit and healthy or your > head may burst. You mean, like on that ßick Scanners, but with maths instead of telepathy? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Candy Inspiration (in the news) |It would still be interesting to know, though, whether theres any shape of |ellipsoid whose *best* packing density exceeds the *best* density for |spheres. |http://mathworld.wolfram.com/EllipsoidPacking.html I was expecting the answer either to be no or for the density to exceed it by a bit more! Theyve found how to pack identical ellipsoids with a density of between 0.75 and 0.76, whereas spheres (presumably) pack at a density of between 0.74 and 0.75. Im also a little surprised to read packings... of densities arbitrarily close to.... Is there no compactness argument that shows theres one of density equal to the limit? Say we have packings of density D1>R1, sufficient to make the density inside that ball exceed (D1+D2)/2. Then switch to the next one out to a radius of R3>>R3, sufficient to make the density inside that ball exceed (D2+D3)/2. Sure, there are awkward gaps around the spheres of radii R1, R2, R3,... but we can space them out distantly. Perhaps the packings in question are all periodic or something? Keith Ramsay === Subject: Re: SymbMath.com: web-based computer algebra system > Would you please show these results? Hey I tried some of the functions and they gave results inconsistent > with matlab. Do you think I should tell them to fix their program? > www.SymbMath.com === Subject: bounds for x in Incomplete Gamma function I need to find a superior bound for x, knowing that G(a,x) / G(a) > 0.001 where G(a,x) is the Incomplete Gamma function and G(a) is the usual Gamma function Any clues? Javier === Subject: Re: bounds for x in Incomplete Gamma function > G(a,x) / G(a) > 0.001 > where G(a,x) is the Incomplete Gamma function > and G(a) is the usual Gamma function > Any clues? For a> 0, x>0 let G(a,x)= Integral_{t=x to t=infty}t^{a-1}e^{-t}dt , G(a)=G(a,0) , g(a,x)=G(a)-G(a,x). Denote P(a,x)= a*x^{-a}e^{-x} , Q(a,x)= x^{1-a}e^x . Try to use inequalities ((a+1)(a+2)-x)/((a+1)*(a+2+x)) < P(a,x)*g(a,x) < (a+1)/(a+1+x) x/(x+1-a) < Q(a,x)*G(a,x) < (x+1)/(x+2-a) . Perhaps help, Alex= === Subject: Re: Explanation with math, error in math world (sci.math added) > It turns out that it might actually be easier to explain the problem > that mathematicians have by using mathematics, and rather simple math > at that as consider Hmm, so youve been proved wrong so many times in sci.math that you are just doing to post elsewhere, and hope noone notices? === Subject: Re: Explanation with math, error in math world In sci.logic, Paul Murray : > (sci.math added) >> It turns out that it might actually be easier to explain the problem >> that mathematicians have by using mathematics, and rather simple math >> at that as consider > Hmm, so youve been proved wrong so many times in sci.math that you are > just doing to post elsewhere, and hope noone notices? Well, look at it this way: he has such a momentous discovery that he just has to share it. Erm...well, the word starts with an Ôm, anyway. But hes a stubborn one, Ill grant him that. (Not that it helps his math any.) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Category Theorys De-Emphasis of Sets is a Serious Mistake by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1MEDvT04990; Rather than deal with personalities (of whom there are sufficiently many charismatic personalities involved in this topic to make a Hollywood blockbuster), Ill list six directions of inquiry in mathematics which were seriously downplayed by Category Theorys mistaken (in my opinion) De-Emphasis of Sets. A. Logic-Based Probability (LBP) and Rare Event Theory (RET), which although Non-Mainstream involve set causal probabilities of form P(A-->B), P(A<-->B), and relate to a Mainstream area that arose after LBP and RET, namely Large Deviations (Fat-Tailed Distribution Theory) in Probability-Statistics. Actually, RET has much more powerful results that Large Deviations, since the latter mainly uses one tool: approximation. B. Nonsmooth Analysis (e.g., F. H. Clarke, Y. S. Ledyaev, R. J. Stern, P. R. Wolenski (1998) of Russia, Quebec, etc.) which almost entirely use sets and is of considerable importance in control theory and elsewhere. C. Set-Valued and Set-Domain Forces (e.g., Morton E. Gurtin, Configurational Forces as Basic Concepts of Continuum Physics, Springer: N.Y. 2000; Set-Valued Force Laws Dynamics of Non-Smooth Systems by Christoph Glocker (Zurich) Springer: Berlin 2001. These are largely applied to continuum mechanics, but have much potential elsewhere. D. Geometric Nonlinear Functional Analysis (in fact, a monograph volume of that title by Benyamini and Lindenstrauss of Israel was published in 2000 by the American Mathematical Society). Sets Play a very important role, e.g., Gauss Null Sets, Aronszajn Null Sets, Cube Null Sets, etc. E. Resolution of Paradoxes, Anomalies, Infinities, in Mathematical Physics, especially relating to Quantum Gravity, Superstring Theory, Topological Quantum Field Theory, Effective Gauge Quantum Field Theory, and also to the explanation of the enormous gap between the microscopic Quantum concepts and the macroscopic concepts which would be considerably helped by comparing and even defining what sets are involved even if a generalization of sets (other than objects which is a turn-off rather than a generalization word) were required. F. Jacobson and other Radicals in Commutative and Noncommutative Rings, Modules, and even Topology, which share a number of similar properties with sets of Geometric Nonlinear Functional Analysis and Rare Event Theory. These were pioneered by Nathan Jacobson. Osher Doctorow === Subject: Re: Category Theorys De-Emphasis of Sets is a Serious Mistake > Here is a list of some of the implicit or explicit errors of the > Category Theory approach regardless of its direct negative effects > on the Set-related Theories that I mentioned in the last posting. > A. The assumption that an Object is more general than a General- > ized Set. The claim is often made that there is a paradox in > defining a Set of Sets, which is allegedly absent from an > Object. Never heard that before. > B. The assumption that Functions are more useful than Sets. I Are you German? > had an argument with a Belgian Computer-Oriented Logician about > this some years ago (Professor Smets who to my recollection founded > the famous IRIDIA in Belgium and France), and we didnt conclude > that one was more useful than the other although each of us > remained convinced personally that our favorite was probably more > useful. However, the assumption that Functions are more useful > than Sets in Real Analysis would involve the assumption for example > that the Lebesgue Integral is more useful than the Lebesgue Measure > (Measures being on sigma-algebras, which are types of Sets), which > is silly. Which is true. The Lebesgue integral subsumes Lebesgue measure --- which can be recovered as the Lebesgue integrals of the indicator functions. > E. The Mis-direction of Composition. The Category Theory claims > priority for Composites of Morphisms, or in Functions Composites > of Functions. The composition f o g (x) = f(g(x)) is not the > same as fg(x) = f(x)g(x) or (f + g)(x) = f(x) + g(x) and so on, > although one could always claim that the latter two are somehow > special cases of the former with a slight generalization. Well done! f o g is not the same as fg or f+g ... youre doing better than most of my algebra students here :-) Of course fg anf f+g only make sense if the codomain of f and g admit addition/multiplication operations. > In > Rare Event Theory, it is shown that fg and f + g are far more > important than f o g. What the **** is Rare Event Theory? Some pet non-concept of yours? > F. The failure of Category Theory to emphasize discovery of new > ideas and new concepts in its preoccupation with organizing old > ones. It does nothing with Anomalies, Paradoxes, Infinities, > Causation, Correlation, Motion, and so on in this respect. This > includes Category Theorys failure to match Rare Event Theorys > discovery of 1 + y - x as a form across geometry, But that (as evidenced by Doctor Osherows repeated postings to geometry.remedial) is a non-discovery. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Category Theorys De-Emphasis of Sets is a Serious Mistake Reply to title alone. Set theory is still the dominant mahematical paradigm. Of course the categorical approach to set theory emphasises functions between sets rather than the sets themselves. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Category Theorys De-Emphasis of Sets is a Serious Mistake > Rather than deal with personalities (of whom there are sufficiently > many charismatic personalities involved in this topic to make a > Hollywood blockbuster), Ill list six directions of inquiry in > mathematics which were seriously downplayed by Category Theorys > mistaken (in my opinion) De-Emphasis of Sets. > A. Logic-Based Probability (LBP) and Rare Event Theory (RET), which > although Non-Mainstream involve set causal probabilities of form > P(A-->B), P(A<-->B), and relate to a Mainstream area that arose > after LBP and RET, namely Large Deviations (Fat-Tailed Distribution > Theory) in Probability-Statistics. Actually, RET has much more > powerful results that Large Deviations, since the latter mainly > uses one tool: approximation. So is this logic-based probability all about misinterpeting complementation of events? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: e is transcendental (was: classes of transcendental numbers ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1MEDvE05014; >Why do I even try? Are you even TRYING to learn? Or are you >so closed-minded that you cant be bothered? So? The real part of exp(i pi) is cos(pi), and its imaginary part >is sin(pi), so all you are saying is that cos(pi) = -1 and >sin(pi) = 0, and we were already aware of these facts. There is >NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect. >> Panagiotis Stefanides Yes, but you DONT tell us what your reason is. You cant expect us to >accept your claims without giving support for those claims. So what >possible reason could you have for expecting us to agree with your claim >that exp(i pi) = 0? >The reason is simply that exp(ipi0=-1 should be accompanied >>by the statement that this is the real part solution. >>Is it fair? No, that is not a fair comment. exp(i pi), the complex number, is >equal to -1, the complex number. There is no need to appeal to the >real part. Also, what does exp(i pi) = -1 is the real part solution >mean? You look like you are using terminology in a manner not >recognized in mathematics. David McAnally -------------- >>e^[i*pi] ,as accepted ,is a phasor. In most of the relevant fields of mathematics, e^[i pi] is a complex >number. >It is only fair to state that its >>polar representation is : >e^[i*pi] = MOD 1 , ARG 180 . That is Arg 180 degrees, not just Arg 180. And so what? That does >not lead to your assertion that exp[i pi] = 0, a result for which >you have given absolutely no support. Why dont you just give up? David McAnally At the moment, they (the Time Lords) are far from being all-powerful. > Thats why its been left up to me and me and me. > quote by: Patrick Troughton in The Three Doctors ------- >I, have made myself very explicit. >My original question of the implication >>of the imaginary component: e^[ipi]=j*0 (to the related proof) >>Complex Notation, chapter 12 ,Electrical Technology 3RD ED. >>Edward Hughes Longmans ,page 338: >>Stares: >> OA*=OB+jOC=OA(COStheta+jSINtheta) >>* Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,.. >>Here is very clear the difference between PHASOR and MAGNITUDES >I know the difference between a complex number and its modulus >(or magnitude). You dont have to explain the difference to >me. >>[which(MAGNITUDES) I, referred to as REAL PART or IMMAGINAR PART, >The magnitude of a complex number is generally not equal to either >its real or imaginary part. How could you claim that it is? >>I, gave a reference: >>Complex Notation, chapter 12 ,Electrical Technology 3RD ED. >>Edward Hughes Longmans ,page 338: >>States: >> OA*=OB+jOC=OA(COStheta+jSINtheta) >>* Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,.. >> What are these magnitudes then ? >OB (which can obviously be either positive, zero or negative) is the real >part of the complex number. OC (which can obviously be either positive, >zero or negative) is the imaginary part of the complex number. OA (which >is presumably positive or zero) is the magnitude of the complex number. >theta is the argument of the complex number. >Note that neither OB nor OC is the magnitude of the complex number. >> I should have stated COMPONENTS ]. >Perhaps you mean that (the imaginary part of e^[i pi]) = 0, in which >case you are correct, but you have had a lot of problem expressing >yourself, especially in view of the way that you initially made >the claim by stating that e^[i pi] = 0, which you described as the >imaginary part solution, using a terminology that nobody but you >knows. >> e[i*theta]=COStheta+iSINtheta >I know that it is true that exp[i theta] = cos(theta)+i sin(theta). >It follows that exp[i pi] = -1. Incidentally, e[i theta] = i e theta, >unlike what you have written. >>thetas could be given and calculations could be performed >>for numerical evaluations. >And for other results as well. >>In books is stated that it is FORMULA >>and also terms such as evaluate:(-1+i*sqrt[3])^10 >>Are these not solutions to problems? >No. Evaluate (-1+i sqrt(3))^10 is a problem for which you can >get a solution using exp(i theta) = cos(theta)+i sin(theta). This >does not mean that exp(i theta) = cos(theta)+i sin(theta) is itself >a solution. You need a problem before you can describe anything as >a solution. >>Of course ,this crops up when theta is substituted by a given angle. >This has nothing to do with your unorthodox usage of the word solution. >Nobody knows what you mean when you make a statement like exp[i pi] = -1 >is the real part solution of exp[i pi] = -1, or that exp[i pi] = 0 >is the imaginary part solution of exp[i pi] = -1. I asked you to >explain your terminology but you havent bothered. >>I, doubt it but ,still I, exlpained it anyway. >I did ask you. And you did not explain your bizarre terminology. >> There >was no problem, hence there is no solution, whether real part or >imaginary part. >>How else would do You call them,I, said: >>I should have stated COMPONENTS >I call them the real part of the complex number (or the equation) and the >imaginary part of the complex number (or the equation). I do not call >them solutions. Please ref: All Discussions : sci.math : Topic : Message || next > Message: Re: e is transcendental === Subject: Re: e is transcendental Author: Daniel W. Johnson > Since e^[iPi]=cosPi+isinPi > or , e^[iPi]=-1+i[0] > then there are two solutions here, to the given equatio: > A) e^[ipi]=-1 the real part solution and > B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio. No, the conclusion from e^[iPi]=-1+i[0] is Re(e^[iPi]) = Re(-1+i[0]) = -1 AND Im(e^[iPi]) = Im(-1+i[0]) = 0 -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W However ,I, thank You too. Panagiotis Stefanides >David McAnally === Subject: Re: e is transcendental (was: classes of transcendental numbers ? >Of course, all the recent discussion has obscured my original comment on >the generalized Lindemann theorem being stated and proven in Chapter 1 >of Ivan Nivens Irrational Numbers, a topic of greater intrinsic >interest than subsequent discussions. As a reminder, the theorem states >that if a_1, ..., a_n are distinct algebraic numbers, then exp(a_1), ..., >exp(a_n) are linearly independent over the algebraic numbers. The >transcendence of e follows by taking a_1 = 0, a_2 = 1. The transcendence >of pi follows from taking a_1 = 0, a_2 = i pi. It is also possible to >use the theorem to prove that: exp(b), sinh(b), cosh(b), tanh(b), coth(b), >sech(b), csch(b), sin(b), cos(b), tan(b), cot(b), sec(b), and cosec(b), >are all transcendental if b is a nonzero algebraic number, and also that >all nonzero values of log(b), ar sinh(b), ar cosh(b), ar tanh(b), >ar coth(b), ar sech(b), ar csch(b), arc sin(b), arc cos(b), arc tan(b), >arc cot(b), arc sec(b), and arc cosec(b), for algebraic numbers b, are >transcendental. Where b is nonzero in the cases of log(b), ar sech(b), ar csch(b), arc sec(b) and arc cosec(b), b is distinct from 1 and -1 in the case of ar tanh(b) and ar coth)b), and b is distinct from i and -i in the case of arc tan(b) or arc cot(b). >Other chapters of the book are also of interest. The second chapter >discusses the (generalized?) Gelfond-Schneider Theorem: if a_1, a_2, ..., >a_n, are nonzero algebraic numbers, and if values of log(a_1), ..., >log(a_n), are such that log(a_1), ..., log(a_n), are linearly independent >over Q, then 1, log(a_1), ..., log(a_n), are linearly independent over the >algebraic numbers. Taking a_1 = 1 and log(a_1) = 2 i pi, it immediately >follows that pi is transcendental. The famous result that if a is an >algebraic number distinct from 0 and 1, and b is an irrational algebraic >number, then all values of a^b are transcendental, also follows from the >theorem. Specifically, taking a = i and b = i, it follows that exp(pi) >is transcendental. The transendence of e is proven by taking a_1 = e and log(a_1) = 1. >Later chapters discuss the constants associated with elliptic and related >functions, giving further proofs that some minimum number of such >constants from a given set are transcendental. David McAnally At the moment, they (the Time Lords) are far from being all-powerful. Thats why its been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Re: e is transcendental (was: classes of transcendental numbers ? Of course, all the recent discussion has obscured my original comment on the generalized Lindemann theorem being stated and proven in Chapter 1 of Ivan Nivens Irrational Numbers, a topic of greater intrinsic interest than subsequent discussions. As a reminder, the theorem states that if a_1, ..., a_n are distinct algebraic numbers, then exp(a_1), ..., exp(a_n) are linearly independent over the algebraic numbers. The transcendence of e follows by taking a_1 = 0, a_2 = 1. The transcendence of pi follows from taking a_1 = 0, a_2 = i pi. It is also possible to use the theorem to prove that: exp(b), sinh(b), cosh(b), tanh(b), coth(b), sech(b), csch(b), sin(b), cos(b), tan(b), cot(b), sec(b), and cosec(b), are all transcendental if b is a nonzero algebraic number, and also that all nonzero values of log(b), ar sinh(b), ar cosh(b), ar tanh(b), ar coth(b), ar sech(b), ar csch(b), arc sin(b), arc cos(b), arc tan(b), arc cot(b), arc sec(b), and arc cosec(b), for algebraic numbers b, are transcendental. Other chapters of the book are also of interest. The second chapter discusses the (generalized?) Gelfond-Schneider Theorem: if a_1, a_2, ..., a_n, are nonzero algebraic numbers, and if values of log(a_1), ..., log(a_n), are such that log(a_1), ..., log(a_n), are linearly independent over Q, then 1, log(a_1), ..., log(a_n), are linearly independent over the algebraic numbers. Taking a_1 = 1 and log(a_1) = 2 i pi, it immediately follows that pi is transcendental. The famous result that if a is an algebraic number distinct from 0 and 1, and b is an irrational algebraic number, then all values of a^b are transcendental, also follows from the theorem. Specifically, taking a = i and b = i, it follows that exp(pi) is transcendental. Later chapters discuss the constants associated with elliptic and related functions, giving further proofs that some minimum number of such constants from a given set are transcendental. David McAnally At the moment, they (the Time Lords) are far from being all-powerful. Thats why its been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Re: e is transcendental (was: classes of transcendental numbers ? >I, gave a reference: >Complex Notation, chapter 12 ,Electrical Technology 3RD ED. >Edward Hughes Longmans ,page 338: >States: > OA*=OB+jOC=OA(COStheta+jSINtheta) >* Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,.. > What are these magnitudes then ? >>OB (which can obviously be either positive, zero or negative) is the real >>part of the complex number. OC (which can obviously be either positive, >>zero or negative) is the imaginary part of the complex number. OA (which >>is presumably positive or zero) is the magnitude of the complex number. >>theta is the argument of the complex number. >>Note that neither OB nor OC is the magnitude of the complex number. I would have thought that since you asked about OB and OC, you might have actually been interested in an answer to that question. Or has being told that ::Re(e^[iPi]) = Re(-1+i[0]) = -1 AND ::Im(e^[iPi]) = Im(-1+i[0]) = 0 made you lose interest in all other questions that you asked? Have you lost all interest in complex numbers as a consequence of finding out that ::I,have a reason , with my due respect. ::Panagiotis Stefanides when I saId that you had no reason to conclude that exp(i pi) = 0, and ::I, have made myself very explicit. ::My original question of the implication ::of the imaginary component: e^[ipi]=j*0 (to the related proof) where you treated e^[i pi] as the imaginary part of itself? David McAnally At the moment, they (the Time Lords) are far from being all-powerful. Thats why its been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Re: e is transcendental by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1MEDvk05006; >> Since e^[iPi]=cosPi+isinPi >> or , e^[iPi]=-1+i[0] >> then there are two solutions here, to the given equatio: >> >> A) e^[ipi]=-1 the real part solution and >> >> B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio. >No, the conclusion from e^[iPi]=-1+i[0] is >Re(e^[iPi]) = Re(-1+i[0]) = -1 AND >Im(e^[iPi]) = Im(-1+i[0]) = 0 >-- >Daniel W. Johnson >panoptes@iquest.net >http:// members.iquest.net/~pano ptes/>Daniel, >>Just saw Your message. >>I, thank You. >>This is it. >>Your help is great, and resolves my question completely. >I pointed out exactly the same thing, but you never accepeted it when I >made the point. I presume that it means that you wont make any more >silly comments along the lines of exp(i pi) = 0. >David McAnally > Despite anything you may have heard to the contrary, > the rain in Spain stays almost invariably in the hills. My,question has been resolved neatly and Laconicaly, within two lines: Re(e^[iPi]) = Re(-1+i[0]) = -1 AND Im(e^[iPi]) = Im(-1+i[0]) = 0 Since there is acceptance to this, any more saying has no value. Panagiotis Stefanides http://www.stefanides.gr === Subject: Re: e is transcendental >My,question has been resolved neatly and Laconicaly, >within two lines: >Re(e^[iPi]) = Re(-1+i[0]) = -1 AND >Im(e^[iPi]) = Im(-1+i[0]) = 0 >Since there is acceptance to this, >any more saying has no value. Except to point out that it means that your original conclusion that exp(i pi) = 0 was not validly drawn from what preceded it. David McAnally At the moment, they (the Time Lords) are far from being all-powerful. Thats why its been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Im guessing this is an easy problem - but I cant solve it! Working through calculus... Stuck on a problem of factorisation: Please help! question is: find lim u--> 1 f(u) where f(u) = (u^4 - 1) / (u^3 -1) Obviously - I cannot just replace u with 1 as this will give me a division by zero. I tried replacing (u^2 - 1) (u^2 +1) on top as this seemed most logical - but I cant figure out what to do with the bottom now. Im probably going to kick myself for not figuring this one out - but Ive spent hours on this and Im still lost. Help! === Subject: Re: Im guessing this is an easy problem - but I cant solve it! Finally found the solution myself... using (a^3 - b^3) = (a-b)(a^2 + ab + b^2) > Working through calculus... Stuck on a problem of factorisation: Please > help! > question is: > find lim u--> 1 f(u) > where f(u) = (u^4 - 1) / > (u^3 -1) > Obviously - I cannot just replace u with 1 as this will give me a division > by zero. I tried replacing (u^2 - 1) (u^2 +1) on top as this seemed most > logical - but I cant figure out what to do with the bottom now. > Im probably going to kick myself for not figuring this one out - but Ive > spent hours on this and Im still lost. > Help! === Subject: Re: Im guessing this is an easy problem - but I cant solve it! >Working through calculus... Seeing this phrase makes it clear that the use of LHopitals rule (suggested by other posters) is inappropriate here. >Stuck on a problem of factorisation: Please help! Yes, you almost certainly need to factor(ise). Others posted the formula for the factorization of u^n - 1 ; you can then factor of u-1 from numerator and denominator. >question is: >find lim u--> 1 f(u) > where f(u) = (u^4 - 1) / > (u^3 -1) >Obviously - I cannot just replace u with 1 as this will give me a division >by zero. Its actually a bit worse than this, and it is because no one else was pointing this out that I am following up. You question is about computing a LIMIT as u APPROACHES 1. On the face of it, that makes it completely inappropriate to find the VALUE of the the function when u is EQUAL to 1. Those are, in general, two different things to compute about a function. If I saw students doing willy-nilly what you are proposing, I would certainly set an exam question in which setting u=1 does _not_ lead to a division-by-zero error (or any other error) but which does not give the limit either. Of course, in practice, people evaluate limits simply by substitution, as you wanted to do, but it is important to know that this only gives the correct answer when the function is CONTINUOUS (here, at u=1). Whenever you decide to compute a limit by simple substitution, you need to (be ready to) defend your step by explaining why you know the function is contiuous there. This particular example is typical: after function whose denominator does not vanish at u=1; theres a theorem (surely in your textbook) that says such a function is continuous at u=1. (You also need to know that when two functions agree except at one point, then their limits at that point agree. That is, you dont get a The typical calculus student wants to learn to do the problems by manipulation as if calculus were simply a kind of algebra. Thats really kind of pointless; I mean, its expected that you will learn to do the manipulations, but thats not really the most important thing about the calculus. Learning what the manipulations tell you about the underlying phenomena is much more important. This problem is a good example to show the difference between manipulation and understanding. (I would _almost_ rather have the student plug in u=1.00001 and get 1.333366667 than go through the algebra and plug in u=1 without commenting about the limit process.) dave === Subject: Re: Im guessing this is an easy problem - but I cant solve it! > Working through calculus... Stuck on a problem of factorisation: Please > help! > question is: > find lim u--> 1 f(u) > where f(u) = (u^4 - 1) / > (u^3 -1) > Obviously - I cannot just replace u with 1 as this will give me a division > by zero. I tried replacing (u^2 - 1) (u^2 +1) on top as this seemed most > logical - but I cant figure out what to do with the bottom now. Your statement above about cant do division by 0 is FLAWED. When it comes to limits and by direct substitution gives a non zero number over 0 then you are done!--the limit is undefined (or maybe +/- oo). However if you directly substitute and get 0/0 then you are not done. If the number x=a make a polynomial 0 then (x-a) must be a factor!! Since u=1 made both the numerator and denominator 0 then both the numerator and denominator will have u-1 as one of its factors. Just use long division to factor the denominator or realize u^3 - 1 as (u)^ - (1)^3 and factor using the difference of cubes formula. > Im probably going to kick myself for not figuring this one out - but Ive > spent hours on this and Im still lost. > Help! === Subject: Re: Im guessing this is an easy problem - but I cant solve it! >Working through calculus... Stuck on a problem of factorisation: Please >help! >question is: >find lim u--> 1 f(u) > where f(u) = (u^4 - 1) / > (u^3 -1) >Obviously - I cannot just replace u with 1 as this will give me a division >by zero. I tried replacing (u^2 - 1) (u^2 +1) on top as this seemed most >logical - but I cant figure out what to do with the bottom now. Three ways to approach this problem come to mind: 1. LHospital: Since u^4-1 and u^3-1 both tend to 0 as u tends to 1, differentiate both numerator and denominator u^4-1 4u^3 4 lim ----- = lim ---- = - u->1 u^3-1 u->1 3u^2 3 2. Expand near 1: Let u = 1+v, then use the Binomial Theorem to get u^4-1 4v+6v^2+4v^3+v^4 4+6v+4v^2+v^3 ----- = ---------------- = ------------- u^3-1 3v+3v^2+v^3 3+3v+v^2 As u->1, v->0 and you can easily see that the RHS above tends to 4/3. 3. Factor: u^4-1 (u-1)(u^3+u^2+u+1) u^3+u^2+u+1 ----- = ------------------ = ----------- u^3-1 (u-1)(u^2+u+1) u^2+u+1 and as u->1, the RHS tends to 4/3. Rob Johnson take out the trash before replying === Subject: Re: Im guessing this is an easy problem - but I cant solve it! > Three ways to approach this problem come to mind: > 1. LHospital: > Since u^4-1 and u^3-1 both tend to 0 as u tends to 1, differentiate > both numerator and denominator > u^4-1 4u^3 4 > lim ----- = lim ---- = - > u->1 u^3-1 u->1 3u^2 3 > 2. Expand near 1: > Let u = 1+v, then use the Binomial Theorem to get > > u^4-1 4v+6v^2+4v^3+v^4 4+6v+4v^2+v^3 > ----- = ---------------- = ------------- > u^3-1 3v+3v^2+v^3 3+3v+v^2 > As u->1, v->0 and you can easily see that the RHS above tends to 4/3. > 3. Factor: > u^4-1 (u-1)(u^3+u^2+u+1) u^3+u^2+u+1 > ----- = ------------------ = ----------- > u^3-1 (u-1)(u^2+u+1) u^2+u+1 > and as u->1, the RHS tends to 4/3. Heres another one: Let v = u^3. As u -> 1, v^3 -> 1. So the limit in question is the same as lim_{v->1) (v^(4/3) - 1)/(v - 1), which by definition is the derivative of the function v^(4/3) at v = 1, ie 4/3. === Subject: Another way > Working through calculus... Stuck on a problem of factorisation: Please > help! > question is: > find lim u--> 1 f(u) > where f(u) = (u^4 - 1) / > (u^3 -1) Hi! LHopitals Rule works well. One iteration, and the 4/3 just falls out. Skip === Subject: Re: Another way >>Working through calculus... Stuck on a problem of factorisation: Please >>help! >>question is: >>find lim u--> 1 f(u) >> where f(u) = (u^4 - 1) / >> (u^3 -1) > Hi! > LHopitals Rule works well. One iteration, and > the 4/3 just falls out. Another other way is to note that u^4 - 1 = (u - 1)(u + 1)(u^2 + 1) and u^3 - 1 = (u - 1)(u^2 + u + 1). (u + 1)(u^2 + 1)/(u^2 + u + 1). Setting u = 1, we get 4/3. Look Ma! No limits! Bob Kolker > Skip === Subject: Re: Im guessing this is an easy problem - but I cant solve it! > Working through calculus... Stuck on a problem of factorisation: > Please help! > question is: > find lim u--> 1 f(u) > where f(u) = (u^4 - 1) / > (u^3 -1) > Obviously - I cannot just replace u with 1 as this will give me a > division by zero. I tried replacing (u^2 - 1) (u^2 +1) on top as this > seemed most logical - but I cant figure out what to do with the > bottom now. > Im probably going to kick myself for not figuring this one out - but > Ive spent hours on this and Im still lost. > Help! u^4-1=(u-1)(u^3+u^2+u+1) u^3-1=(u-1)(u^2+u+1) Lim -->4/3 -- If its Monday then I am a fool but not ignorant. === Subject: Re: Im guessing this is an easy problem - but I cant solve it! > Working through calculus... Stuck on a problem of factorisation: Please > help! > question is: > find lim u--> 1 f(u) > where f(u) = (u^4 - 1) / > (u^3 -1) > Obviously - I cannot just replace u with 1 as this will give me a division > by zero. I tried replacing (u^2 - 1) (u^2 +1) on top as this seemed most > logical - but I cant figure out what to do with the bottom now. > Im probably going to kick myself for not figuring this one out - but Ive > spent hours on this and Im still lost. > Help! This should help: u^n - 1 = ( u - 1 ) * ( u^(n-1) + u^(n-2) + ... + u^2 + u + 1 ) Dirk Vdm === Subject: Are these equations solvable? Good Morning Yall, I am not a math major. I doing some research in which I ended up with the following four equations. My gut and distant memories of my college math classes tells me that they are unsolvable but I cannot be sure. Is there a way to find out the value of even one variable? a + b + c = 165 b + c + d = 168 c + d + e = 168 d + e + f = 162 a, b, c, d, e, f are all positive integers. To give you a background, the totals represent three day rolling totals and I am trying to find out the total for each day. Much Appreciated! === Subject: Re: Are these equations solvable? > Good Evening Yall, > but I assumed was implied, that I was looking for a unique solution. > As all of you have pointed out, there is no one unique solution. > Let me ask you a follow up question (and I think the answer to this is > negative too). > If I have an infinite number of daily rolling totals: > a+b+c = 165 > b+c+d = 168 > c+d+e = 168 > d+e+f = 162 > e+f+g = 167 > and so on where every day a new variable is added and the oldest taken > off, I am assuming that even given an infinite number of such > equations this will never have a unique solution? This set gives { a = g - 8 { b = 167 - (f+g) { c = f + 6 { d = g - 5 { e = 167 - (f+g) so, supposing that we need strictly positive integers, we have to choose f and g such that { g - 8 > 0 { f + 6 > 0 { g - 5 > 0 { f+g < 167 i.o.w. { g > 8 { f > 0 { f+g < 167 So we have f+g 9+1 9+2 ... 9+157 --> 157 10+1 10+2 ... 10+156 --> 156 ... 165+1 165+2 --> 2 166+1 --> 1 giving 1 + 2 + ... + 156 + 157 = 157*158/2 = 12403 solutions. Hm, weve seen that number before, didnt we? Coincidence :-) Dirk Vdm === Subject: Re: Are these equations solvable? > Good Evening Yall, > and so on where every day a new variable is added and the oldest taken > off, I am assuming that even given an infinite number of such > equations this will never have a unique solution? You are always short of two bits of information But over the long run the average value --> the missing value Carl -- If its Monday then I am a fool but not ignorant. If its Tuesday then I am lost === Subject: Re: Are these equations solvable? > Good Morning Yall, > I am not a math major. I doing some research in which I ended up with > the following four equations. My gut and distant memories of my > college math classes tells me that they are unsolvable but I cannot be > sure. Is there a way to find out the value of even one variable? no > a + b + c = 165 > b + c + d = 168 > c + d + e = 168 > d + e + f = 162 > a, b, c, d, e, f are all positive integers. > To give you a background, the totals represent three day rolling > totals and I am trying to find out the total for each day. > Much Appreciated! Liner reduction 6 vairables you need two more bits of information |a,b, c, 0, 0, 0,:165| |0,b, c, d, 0, 0,:168| |0,0, c, d, e, 0,:168| |0,0, 0, d, e, f,:162| |a,0, 0,-d, 0 0,:-03| |0,b, 0, 0,-e, 0,:000| |0,0, c, 0, 0,-f,:006| |a,b, 0,-d,-e, 0,:-03| |0,b, c, 0,-e,-f,:006| |a,0,-c,-d, 0, f,:-09| b = e :the same number traveled on day b as day e a+3 = d :3 more traveled on day d as on day a c = f+6 :6 more traveled on day c as on day f a,b,c,d,e,f>0 d>=4 c>=6 Max a = 165-(1+6)=158 Min a = 1 Max b = 158 Min b = 1 ~~ estimate ~~a=55 Ie 165/3 ~~b=55 ~~c=55 ~~d=58 ~~e=55 ~~c=168-58-55=55 ~~f=162-58-55=49 - - If its Monday then I am a fool but not ignorant. === Subject: Re: Are these equations solvable? I see that lots of folks gave you solutions. In their zeal to give you the solution(s), they forgot to tell you the important thing: This is one of many problems which are called Diophantine problems. I believe that they were first posed by Archimedes. The basic idea is that theyre like any linear algebra problem, except that there are not enough equations to truly solve the problem by normal methods. The catch is that theres an unspoken constraint, that all solutions must be positive integers. Often, as many have pointed out, non-zero integers. I once got my tail singed, but good, in a little barbershop in a small town in Florida. The barber asked me if I went to college. I said proudly that I was studying to be an engineer. He said, Then Ill bet you can solve this problem for me, and posed a typical Diophantine problem involving sheep, cows, goats, etc. You can guess that this problem is Diophantine because one cant have 3 2/3 goats. I couldnt solve it. I kept proving to myself that 0 was equal to 0, but that was about it. In truth, I wasnt much of a mathematician in those days, so the only method I knew to solve it was direct substitution (as opposed to a matrix inversion). If I had known the latter, I would surely have seen immediately that my matrix wasnt square. My haircut was long since over. I had elected to stay, and borrowed a pencil and paper, and stayed about a half-hour proving that 0 was equal to 0. _FINALLY_, I got around to counting unknowns and equations, and realized that there werent enough of the former. I announced belligerantly, This problem cant be solved! You didnt give me enough information. The barber asserted that he had. He said, So youre saying that the problem cant be solved? I said, Thats right. He said, So how about this answer, and proceeded to give me a solution that worked. I said (rather proud of myself for getting around to counting equations after nearly an hour of work), No, you dont understand. The problem is underdetermined. Theres an infinity of solutions. He said, Fine. Give me one more. I couldnt of course. I ended up slinking out of the shop in disgrace. It took me years to realize that Id been had; that the barber had set me up. Your problem is rather unique in the sense that you have two free variables; the usual number is one. Jack > Good Morning Yall, > I am not a math major. I doing some research in which I ended up with > the following four equations. My gut and distant memories of my > college math classes tells me that they are unsolvable but I cannot be > sure. Is there a way to find out the value of even one variable? > a + b + c = 165 > b + c + d = 168 > c + d + e = 168 > d + e + f = 162 > a, b, c, d, e, f are all positive integers. > To give you a background, the totals represent three day rolling > totals and I am trying to find out the total for each day. > Much Appreciated! === Subject: Re: Are these equations solvable? >but I assumed was implied, that I was looking for a unique solution. >As all of you have pointed out, there is no one unique solution. >Let me ask you a follow up question (and I think the answer to this is >negative too). >If I have an infinite number of daily rolling totals: >a+b+c = 165 >b+c+d = 168 >c+d+e = 168 >d+e+f = 162 >e+f+g = 167 >and so on where every day a new variable is added and the oldest taken >off, I am assuming that even given an infinite number of such >equations this will never have a unique solution? No, it might have a unique solution in positive integers (even with finitely many equations). Lets write your equation #j as x_j + x_{j+1} + x_{j+2} = b_j, for j=1,2,3,... Now given x_1 and x_2 you have x_3 = b_1 - x_1 - x_2 x_4 = b_2 - x_2 - x_3 = b_2 - b_1 + x_1 x_5 = b_3 - x_3 - x_4 = b_3 - b_2 + x_2 x_6 = b_4 - x_4 - x_5 = b_1 + b_4 - b_3 - x_1 - x_2 etc. So the requirements x_j >= 1 for j = 3, 6, 9, ... give you upper bounds on x_1 + x_2, while x_j >= 1 for j = 4, 7, 10, ... give lower bounds on x_1 and x_j >= 1 for j = 5, 8, 11, ... give lower bounds on x_2. Its possible for the lower bounds and upper bounds to uniquely determine x_1 and x_2. For example, if b_1 = 3 we immediately have x_1 = x_2 = 1. Or if b_1 to b_6 are 350,300,150,100,300,203, equations 2, 4 and 6 tell us x_4 = -50 + x_1 >= 1 so x_1 >= 51, x_6 = 300 - x_1 - x_2 >= 1 so x_1 + x_2 <= 299, x_8 = -247 + x_2 >= 1 so x_2 >= 248, and therefore x_1 = 51, x_2 = 248. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Are these equations solvable? Good Evening Yall, but I assumed was implied, that I was looking for a unique solution. As all of you have pointed out, there is no one unique solution. Let me ask you a follow up question (and I think the answer to this is negative too). If I have an infinite number of daily rolling totals: a+b+c = 165 b+c+d = 168 c+d+e = 168 d+e+f = 162 e+f+g = 167 and so on where every day a new variable is added and the oldest taken off, I am assuming that even given an infinite number of such equations this will never have a unique solution? === Subject: Re: Are these equations solvable? > Good Morning Yall, > I am not a math major. I doing some research in which I ended up with > the following four equations. My gut and distant memories of my > college math classes tells me that they are unsolvable but I cannot be > sure. Is there a way to find out the value of even one variable? > a + b + c = 165 > b + c + d = 168 > c + d + e = 168 > d + e + f = 162 > a, b, c, d, e, f are all positive integers. > To give you a background, the totals represent three day rolling > totals and I am trying to find out the total for each day. > Much Appreciated! When you solve for (for instance) a, b, c, d, you get a = 159 - (e+f) b = e c = 6 + f d = 162 - (e+f) If everything must be positive and integer, then your e and f can take any value, provided their sum is less than 159 (to make sure a is positive). How many ways are there to find two positive integers with a sum less than 159? 1+1 1+2 ... 1+158 --> 158 2+1 2+2 ... 2+157 --> 157 3+1 3+2 ... 3+156 --> 156 ... 157+1 157+2 --> 2 158+1 --> 1 So, you have 1 + 2 + 3 + ... + 157 + 158 = 12561 solutions. Dirk Vdm === Subject: Re: Are these equations solvable? |> I am not a math major. I doing some research in which I ended up with |> the following four equations. My gut and distant memories of my |> college math classes tells me that they are unsolvable but I cannot be |> sure. Is there a way to find out the value of even one variable? |> a + b + c = 165 |> b + c + d = 168 |> c + d + e = 168 |> d + e + f = 162 |> a, b, c, d, e, f are all positive integers. |> To give you a background, the totals represent three day rolling |> totals and I am trying to find out the total for each day. | When you solve for (for instance) a, b, c, d, you get | a = 159 - (e+f) | b = e | c = 6 + f | d = 162 - (e+f) Also, solving for D, d = a + 3 _____________________________Gerard S. | If everything must be positive and integer, then your e and f can | take any value, provided their sum is less than 159 (to make sure | a is positive). | How many ways are there to find two positive integers with a sum | less than 159? | 1+1 1+2 ... 1+158 --> 158 | 2+1 2+2 ... 2+157 --> 157 | 3+1 3+2 ... 3+156 --> 156 | ... | 157+1 157+2 --> 2 | 158+1 --> 1 | So, you have 1 + 2 + 3 + ... + 157 + 158 = 12561 solutions. === Subject: Re: Are these equations solvable? |> Good Morning Yall, |> I am not a math major. I doing some research in which I ended up with |> the following four equations. My gut and distant memories of my |> college math classes tells me that they are unsolvable but I cannot be |> sure. Is there a way to find out the value of even one variable? |> a + b + c = 165 |> b + c + d = 168 |> c + d + e = 168 |> d + e + f = 162 |> a, b, c, d, e, f are all positive integers. |> To give you a background, the totals represent three day rolling |> totals and I am trying to find out the total for each day. |> Much Appreciated! | When you solve for (for instance) a, b, c, d, you get | a = 159 - (e+f) | b = e | c = 6 + f | d = 162 - (e+f) | If everything must be positive and integer, then your e and f can | take any value, provided their sum is less than 159 (to make sure | a is positive). | How many ways are there to find two positive integers with a sum | less than 159? | 1+1 1+2 ... 1+158 --> 158 | 2+1 2+2 ... 2+157 --> 157 | 3+1 3+2 ... 3+156 --> 156 | ... | 157+1 157+2 --> 2 | 158+1 --> 1 | So, you have 1 + 2 + 3 + ... + 157 + 158 = 12561 solutions. Yes, there are 12,561 solutions for A, E, F. But, C and D can still vary, so there are 805 more solutions for the four equations, for a total of 13,366. _______________________________________Gerard S. === Subject: Re: Are these equations solvable? > |> Good Morning Yall, > |> I am not a math major. I doing some research in which I ended up with > |> the following four equations. My gut and distant memories of my > |> college math classes tells me that they are unsolvable but I cannot be > |> sure. Is there a way to find out the value of even one variable? > |> a + b + c = 165 > |> b + c + d = 168 > |> c + d + e = 168 > |> d + e + f = 162 > |> a, b, c, d, e, f are all positive integers. > |> To give you a background, the totals represent three day rolling > |> totals and I am trying to find out the total for each day. > |> Much Appreciated! > | When you solve for (for instance) a, b, c, d, you get > | a = 159 - (e+f) > | b = e > | c = 6 + f > | d = 162 - (e+f) > | If everything must be positive and integer, then your e and f can > | take any value, provided their sum is less than 159 (to make sure > | a is positive). > | How many ways are there to find two positive integers with a sum > | less than 159? > | 1+1 1+2 ... 1+158 --> 158 > | 2+1 2+2 ... 2+157 --> 157 > | 3+1 3+2 ... 3+156 --> 156 > | ... > | 157+1 157+2 --> 2 > | 158+1 --> 1 > | So, you have 1 + 2 + 3 + ... + 157 + 158 = 12561 solutions. > Yes, there are 12,561 solutions for A, E, F. But, C and D can > still vary, so there are 805 more solutions for the four equations, > for a total of 13,366. > _______________________________________Gerard S. hm, Im not with you here. Once e and f are chosen, clearly the values of a, b, c, d are fixed, so I dont see how there could be more solutions than the number I gave, provided (!) of course we take my correction into account, namely depending on whether we count 0 as a positive integer or not, there are either 2880 or 12403 solutions. Dirk Vdm === Subject: Re: Are these equations solvable? |>|> Good Morning Yall, |>|> I am not a math major. I doing some research in which I ended up with |>|> the following four equations. My gut and distant memories of my |>|> college math classes tells me that they are unsolvable but I cannot be |>|> sure. Is there a way to find out the value of even one variable? |>|> a + b + c = 165 |>|> b + c + d = 168 |>|> c + d + e = 168 |>|> d + e + f = 162 |>|> |> a, b, c, d, e, f are all positive integers. |>|> To give you a background, the totals represent three day rolling |>|> totals and I am trying to find out the total for each day. |>|> Much Appreciated! |>|> | When you solve for (for instance) a, b, c, d, you get |>| a = 159 - (e+f) |>| b = e |>| c = 6 + f |>| d = 162 - (e+f) |>| If everything must be positive and integer, then your e and f can |>| take any value, provided their sum is less than 159 (to make sure |>| a is positive). |>| How many ways are there to find two positive integers with a sum |>| less than 159? |>| 1+1 1+2 ... 1+158 --> 158 |>| 2+1 2+2 ... 2+157 --> 157 |>| 3+1 3+2 ... 3+156 --> 156 |>| ... |>| 157+1 157+2 --> 2 |>| 158+1 --> 1 |>| So, you have 1 + 2 + 3 + ... + 157 + 158 = 12561 solutions. |> Yes, there are 12,561 solutions for A, E, F. But, C and D can |> still vary, so there are 805 more solutions for the four equations, |> for a total of 13,366. | hm, Im not with you here. | Once e and f are chosen, clearly the values of a, b, c, d are fixed, | so I dont see how there could be more solutions than the number | I gave, provided (!) of course we take my correction into account, | namely depending on whether we count 0 as a positive integer or | not, there are either 2880 or 12403 solutions. My bad. I let F go negative in: f=c-6 (for c=6+f, above) With that error corrected, I now get 12,404 (which counts zero NOT as a positive integer). _____________________________________Gerard S. === Subject: Re: Are these equations solvable? > | hm, Im not with you here. > | Once e and f are chosen, clearly the values of a, b, c, d are fixed, > | so I dont see how there could be more solutions than the number > | I gave, provided (!) of course we take my correction into account, > | namely depending on whether we count 0 as a positive integer or > | not, there are either 2880 or 12403 solutions. > My bad. I let F go negative in: f=c-6 (for c=6+f, above) > With that error corrected, I now get 12,404 (which counts zero NOT > as a positive integer). _____________________________________Gerard S. I count 1+1 1+2 ... 1+157 --> 157 2+1 2+2 ... 2+156 --> 156 ... 155+1 155+2 155+3 --> 3 156+1 156+2 --> 2 157+1 --> 1 Sum( i=1 to 157; i ) = 157*158/2 = 12403. Im curious about your one extra solution :-) Dirk Vdm === Subject: Re: Are these equations solvable? |>| hm, Im not with you here. |>| Once e and f are chosen, clearly the values of a, b, c, d are fixed, |>| so I dont see how there could be more solutions than the number |>| I gave, provided (!) of course we take my correction into account, |>| namely depending on whether we count 0 as a positive integer or |>| not, there are either 2880 or 12403 solutions. |> My bad. I let F go negative in: f=c-6 (for c=6+f, above) |> With that error corrected, I now get 12,404 (which counts zero NOT |> as a positive integer). _____________________________________Gerard S. | I count | 1+1 1+2 ... 1+157 --> 157 | 2+1 2+2 ... 2+156 --> 156 | ... | 155+1 155+2 155+3 --> 3 | 156+1 156+2 --> 2 | 157+1 --> 1 | Sum( i=1 to 157; i ) = 157*158/2 = 12403. | | Im curious about your one extra solution :-) | | Dirk Vdm My bad again. That was a typo. I should have said, I now get 12,403. __________________________Gerard S. === Subject: Re: Are these equations solvable? > My bad again. That was a typo. > I should have said, I now get 12,403. __________________________Gerard S. Ha okay.... are you sure its not Gerard47? Sorry, couldnt resist ;-) Dirk Vdm === Subject: Re: Are these equations solvable? > Good Morning Yall, > I am not a math major. I doing some research in which I ended up with > the following four equations. My gut and distant memories of my > college math classes tells me that they are unsolvable but I cannot be > sure. Is there a way to find out the value of even one variable? > a + b + c = 165 > b + c + d = 168 > c + d + e = 168 > d + e + f = 162 > a, b, c, d, e, f are all positive integers. > To give you a background, the totals represent three day rolling > totals and I am trying to find out the total for each day. > Much Appreciated! > When you solve for (for instance) a, b, c, d, you get > a = 159 - (e+f) > b = e > c = 6 + f > d = 162 - (e+f) > If everything must be positive and integer, then your e and f can > take any value, provided their sum is less than 159 (to make sure > a is positive). > How many ways are there to find two positive integers with a sum > less than 159? > 1+1 1+2 ... 1+158 --> 158 > 2+1 2+2 ... 2+157 --> 157 > 3+1 3+2 ... 3+156 --> 156 > ... > 157+1 157+2 --> 2 > 158+1 --> 1 > So, you have 1 + 2 + 3 + ... + 157 + 158 = 12561 solutions. Oops, little mistake here. If the positive integers do not include zero, then we have 1+1 1+2 ... 1+157 --> 157 2+1 2+2 ... 2+156 --> 156 etc... resulting in 1 + 2 + ... + 157 = 12403 solutions. If o.t.o.h. the positive integers *do* include zero, then we have 0+0 0+1 ... 0+159 --> 160 1+0 1+1 ... 1+158 --> 159 etc... resulting in 1 + 2 + ... + 160 = 12880 solutions. Sorry, Dirk Vdm === Subject: Re: Are these equations solvable? >Good Morning Yall, >I am not a math major. I doing some research in which I ended up with >the following four equations. My gut and distant memories of my >college math classes tells me that they are unsolvable but I cannot be >sure. Is there a way to find out the value of even one variable? >a + b + c = 165 >b + c + d = 168 >c + d + e = 168 >d + e + f = 162 >a, b, c, d, e, f are all positive integers. Standard linear equation system. [ 1 1 1 0 0 0 ] [ a ] [ 165 ] [ 0 1 1 1 0 0 ] [ b ] [ 168 ] [ 0 0 1 1 1 0 ] [ c ] = [ 168 ] [ 0 0 0 1 1 1 ] [ d ] [ 162 ] [ e ] [ f ] Octave says one solution is: a = 53, b = 54.5, c = 57.5, d = 56, e = 54.5, f = 51.5. -- Im not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: Are these equations solvable? >I am not a math major. I doing some research in which I ended up with >the following four equations. My gut and distant memories of my >college math classes tells me that they are unsolvable but I cannot be >sure. Is there a way to find out the value of even one variable? >a + b + c = 165 >b + c + d = 168 >c + d + e = 168 >d + e + f = 162 >a, b, c, d, e, f are all positive integers. > Standard linear equation system. > [ 1 1 1 0 0 0 ] [ a ] [ 165 ] > [ 0 1 1 1 0 0 ] [ b ] [ 168 ] > [ 0 0 1 1 1 0 ] [ c ] = [ 168 ] > [ 0 0 0 1 1 1 ] [ d ] [ 162 ] > [ e ] > [ f ] > Octave says one solution is: a = 53, b = 54.5, c = 57.5, d = 56, e = > 54.5, f = 51.5. Not whats required (positive integers), but by reverse engineering I identified it as the solution with the minimal variance. A small adjustment gives an integer solution [a b c d e f]=[54 54 57 57 54 51] Adjustments in the directions of [2 -1 -1 2 -1 -1] and [0 1 -1 0 1 -1] will not step out of the family of solutions, so thats what I did, taking half of the former. Taking half or (-half) of the latter, I would get [a b c d e f]=[53 55 57 56 55 51] or [53 54 58 56 54 51], respectively. Yet another wrong solution: MATLAB goes for a solution with the smallest Euclidean norm, and that solution is [0 150 6 3 159 0]. > -- > Im not interested in mathematics that might have anything > to do with reality. -- Russell Easterly, in sci.math === Subject: 5 circles Find a configuration of five circles, where each pair intersects under the same angle @ (cos @12=(r1^2+r2^2-d12^2)/(2r1r2) - complex @ are allowed, r=0 or oo (points and lines) too.) Which @ are possible? -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de als man ankam wollte man werden, die geschichte schreiben, die doofen sollen sterben, der plan als man damals nach hamburg kam (Kettcar) === Subject: fiber product of map may you explain me the notion of fiber product of two maps? I know the notion of fiber product of two sets, the notion of pull back of two fiber bundles... Tern === Subject: Adjacency matrix algorithm re graph theory I am trying to prove that for every graph G a balanced orientation can be found. In other words, for a digraph, the absolute value of the difference between the sum of incoming edges and outgoing edges for every vertex is less than or equal to 1. I have re-written the problem slightly. For a square matrix, comprised with 0s and 1s, I am trying to show that an algorithm can be found to change about half of the 1s on each row and column to -1s so that the absolute value of the total sum of each row and column is always less than or equal to 1. For any square matrix, there might be an arbitrary number of 1s on each row and column to start. Any ideas? Diana -- God made the integers, all else is the work of man. L. Kronecker, Jahresber. DMV 2, S. 19. === Subject: Re: Adjacency matrix algorithm re graph theory > I am trying to prove that for every graph G a balanced orientation can be > found. In other words, for a digraph, the absolute value of the difference > between the sum of incoming edges and outgoing edges for every vertex is > less than or equal to 1. > Any ideas? Note that every edge contributes 1 to the total in-degree (ID) and 1 to the total out-degree (OD), so ID = OD. If the graph is not balanced, then some vertex has more in-degree and some vertex has more out-degree. Is there an edge between then that will decrease these? If there isnt an edge, is there a sequence of two edges (going through another vertex) for this to work? One might think that if the two vertices are in the same connected component there will be path of alternations... I think you might need a little stronger condition than connectedness. J === Subject: Re: Adjacency matrix algorithm re graph theory I am trying to prove that an algorithm can be devised to guarantee that > absolute value of the difference between the sum of incoming edges and > outgoing edges for each vertex is less than or equal to 1. The in and out > edges for each vertex have to be played with. Yes... as I said, the total indegree over all vertices must equal the total outdegree over all vertices. Let D(v) = difference of indegree-outdegree. So, say that there is vertex u with D(u) = 2 and then some vertex w with D(w) = -2. If they have an edge between them such that ßipping its orientation would modify D(u) to 1 and D(w) to -1, then ßip it. If such an edge does not exist, then maybe there is a path of edges through an intermediate vertex z such that ßipping uz and zw would improve those D() values... Again, if such edges dont exist, maybe there are two intermediate vertices, and so on... Is there always a path between u and w? Note that total indegree must equal total outdegree for any connected component, and not just the whole graph. When I said before that Ôconnectedness might not be enough, what I meant specifically is that perhaps you need to look at connectedness in the sense of directed edges. I hope this helps, J === Subject: Re: Adjacency matrix algorithm re graph theory Jim, Perhaps I am not understanding your reply. I am trying to prove that an algorithm can be devised to guarantee that absolute value of the difference between the sum of incoming edges and outgoing edges for each vertex is less than or equal to 1. The in and out edges for each vertex have to be played with. The idea is to keep the absolute value of the difference less than two. One is OK. I have been playing with adjacency matrices, but have not yet come up with an algorithm. Diana > I am trying to prove that for every graph G a balanced orientation can be > found. In other words, for a digraph, the absolute value of the difference > between the sum of incoming edges and outgoing edges for every vertex is > less than or equal to 1. > Any ideas? > Note that every edge contributes 1 to the total in-degree (ID) and 1 to > the total out-degree (OD), so ID = OD. If the graph is not balanced, then > some vertex has more in-degree and some vertex has more out-degree. Is > there an edge between then that will decrease these? If there isnt an > edge, is there a sequence of two edges (going through another vertex) for > this to work? > One might think that if the two vertices are in the same connected > component there will be path of alternations... I think you might need a > little stronger condition than connectedness. === Subject: Need help with Rudin proof Im afraid this constitutes a pain-in-the-behind question, since it means looking up something in a book, in this case Rudins Principles (3rd ed). In the Lebesque Theory section p. 304-305, he proves u*(A)=u(A) for all A in E, where E is the family of all elementary sets of R^p. My problem is with the second part, proving u(A)<= u*(A), after proving u*(A)<= u(A). Specifically, for the inequality near the top of p. 305, I cant see the reason for Sum(n=1 to N) u(A_n) <= Sum (n=1 to oo) u(A_n)+ epsilon. Any and all help explaining this is much appreciated. === Subject: Re: Axioms defining a finite field > [...] > On the other hand, Im suspicious of axiom 5: a * 1 = a. Maybe Im missing > something obvious, but can there possibly be a finite ring without identity > that has no zero divisors? You are right. First observe, that axiom 9 together with distributivity gives, that for any fixed nonzero element a in A both maps A ni x --> ax in A and A ni x --> xa in A are injective, hence bijective. Moreover both maps when restricted to B := A - {0} yield maps B ---> B. Now the usual proof for suitable (finite) semigroups applies: (1) for each b in B there is a unique b in B such that bb=b (2) for each x,b in B one has xb=x [ because xbb=xb ] (3) for each b,c in B one has b=c [ compare bb = b = bc ] So in fact all the b (b in B) coincide and act as a left and right unit. Marc === Subject: adjoining elements to a ring I am having a very hard time understanding a section of my algebra book about adjoining elements to a ring. my book (artin, algebra) discusses it breißy, and I check out herstein but that didnt help much either. In particular Im trying to how that Z[x]/(2x-1) is isomorphic to Q_2 (the dyadic rations i.e. rations with a power of 2 for its denominator). I have made the identifications j:Z-->Z[x], pi:Z[x]-->Z[x]/(2x-1), and phi:Z-->Q_2, so that if I could show ker(pi o j)=ker(phi) I would have the isomorphism. This hasnt worked so far. However, I feel that my real problem is that I have only seen very elementary examples of this construction (for example R[x]/(x^2+1)=C). Are there books or online references that discuss this more clearly? === Subject: Re: adjoining elements to a ring > I am having a very hard time understanding a section of my algebra > book about adjoining elements to a ring. my book (artin, algebra) > discusses it breißy, and I check out herstein but that didnt help > much either. In particular Im trying to how that Z[x]/(2x-1) is > isomorphic to Q_2 (the dyadic rations i.e. rations with a power of 2 > for its denominator). > I have made the identifications j:Z-->Z[x], pi:Z[x]-->Z[x]/(2x-1), and > phi:Z-->Q_2, so that if I could show ker(pi o j)=ker(phi) I would have > the isomorphism. This hasnt worked so far. Not the way to do this at all ..... To show that Z[x]/(2x-1) is isomorphic to Q_2, all you need (by 1st isomorphism theorem) is to find a ring homomorphism phi:Z[x] -> Q with image Q_2 and kernel (2x - 1). Now there is exactly one phi with phi(x) = 1/2. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: adjoining elements to a ring >I am having a very hard time understanding a section of my algebra >book about adjoining elements to a ring. my book (artin, algebra) >discusses it breißy, and I check out herstein but that didnt help >much either. In particular Im trying to how that Z[x]/(2x-1) is >isomorphic to Q_2 (the dyadic rations i.e. rations with a power of 2 >for its denominator). >I have made the identifications j:Z-->Z[x], pi:Z[x]-->Z[x]/(2x-1), and >phi:Z-->Q_2, so that if I could show ker(pi o j)=ker(phi) I would have >the isomorphism. This hasnt worked so far. Im not sure why you want to do that. The natural map from Z to Q_2 would be an injection in my book, and have trivial kernel. (So does the other one, I dont see how that gives the isomorphism you want. You would just have two injections from Z to two other rings.) Consider instead the map from Z[x]/(2x-1) sending x to 1/2, or perhaps the map from Z[x] sending x to 1/2 and finding its kernel. >However, I feel that my real problem is that I have only seen very >elementary examples of this construction (for example R[x]/(x^2+1)=C). >Are there books or online references that discuss this more clearly? Try Stewarts Galois Theory for some nice stuff about field extensions. === Subject: Re: adjoining elements to a ring >>I am having a very hard time understanding a section of my algebra >>book about adjoining elements to a ring. my book (artin, algebra) >>discusses it breißy, and I check out herstein but that didnt help >>much either. In particular Im trying to how that Z[x]/(2x-1) is >>isomorphic to Q_2 (the dyadic rations i.e. rations with a power of 2 >>for its denominator). >>I have made the identifications j:Z-->Z[x], pi:Z[x]-->Z[x]/(2x-1), and >>phi:Z-->Q_2, so that if I could show ker(pi o j)=ker(phi) I would have >>the isomorphism. This hasnt worked so far. >> Im not sure why you want to do that. The natural map from Z to Q_2 >> would be an injection in my book, and have trivial kernel. (So does >> the other one, I dont see how that gives the isomorphism you want. >> You would just have two injections from Z to two other rings.) >> Consider instead the map from Z[x]/(2x-1) sending x to 1/2, or perhaps >> the map from Z[x] sending x to 1/2 and finding its kernel. >>However, I feel that my real problem is that I have only seen very >>elementary examples of this construction (for example R[x]/(x^2+1)=C). >>Are there books or online references that discuss this more clearly? >> Try Stewarts Galois Theory for some nice stuff about field >> extensions. >Is that what this subject is called? Field extensions? I tried looking >up information about this myself, but ring adjunction element didnt >turn up anything useful. Strictly speaking no, because youre adjoining things to ring, no necessarily a field, but if you study field extensions of Q say youll have a good idea of whats going on. C is a degree two extension of R for instance. === Subject: Re: adjoining elements to a ring >I am having a very hard time understanding a section of my algebra >book about adjoining elements to a ring. my book (artin, algebra) >discusses it breißy, and I check out herstein but that didnt help >much either. In particular Im trying to how that Z[x]/(2x-1) is >isomorphic to Q_2 (the dyadic rations i.e. rations with a power of 2 >for its denominator). >I have made the identifications j:Z-->Z[x], pi:Z[x]-->Z[x]/(2x-1), and >phi:Z-->Q_2, so that if I could show ker(pi o j)=ker(phi) I would have >the isomorphism. This hasnt worked so far. > Im not sure why you want to do that. The natural map from Z to Q_2 > would be an injection in my book, and have trivial kernel. (So does > the other one, I dont see how that gives the isomorphism you want. > You would just have two injections from Z to two other rings.) > Consider instead the map from Z[x]/(2x-1) sending x to 1/2, or perhaps > the map from Z[x] sending x to 1/2 and finding its kernel. >However, I feel that my real problem is that I have only seen very >elementary examples of this construction (for example R[x]/(x^2+1)=C). >Are there books or online references that discuss this more clearly? > Try Stewarts Galois Theory for some nice stuff about field > extensions. Is that what this subject is called? Field extensions? I tried looking up information about this myself, but ring adjunction element didnt turn up anything useful. === Subject: differential equation lost in finding the necessary differential equation. Any help would be appreciated. Marine Biologists have determined that when a shark detects the presence of blood in the water, it will swim in the direction in which the concentration of the blood increases most rapidly. Based on certain tests in seawater, the concentration of blood (in parts per million) at a point P(x,y) on the surface is approximated by C(x,y)= e^ {(-x^2+2y^2)/10^4}where x and y are measured in meters in a rectangular coordinate system with the blood source at the origin. Suppose a shark is at the point (Xo,Yo) when it first detects the presence of blood in the water. Find an equation of the sharks path by setting up and solving a differential equation. === Subject: Re: differential equation >lost in finding the necessary differential equation. Any help would be >appreciated. Marine Biologists have determined that when a shark detects the >presence of blood in the water, it will swim in the direction in which the >concentration of the blood increases most rapidly. Based on certain tests >in seawater, the concentration of blood (in parts per million) at a point >P(x,y) on the surface is approximated by C(x,y)= e^ {(-x^2+2y^2)/10^4}where >x and y are measured in meters in a rectangular coordinate system with the >blood source at the origin. Suppose a shark is at the point (Xo,Yo) when it >first detects the presence of blood in the water. Find an equation of the >sharks path by setting up and solving a differential equation. Hint: the gradient of a function points in the direction of most rapid increase. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: [HELP] Martingales Can someone explain me following in detail; Check whether the following process are martingales wrt {F_t} 1) X_t = B_t + 4t (B_t: brownian motion) 2) X_t = B^2_t 3) X_t = t^2B_t - 2tB_t+2int_0^t B_s ds 4) X_t = B_1(t)B_2(t) (B_1(t), B_2(t)) is 2-dimentional Brownian motion === Subject: zero-one law I read the following corollary to Komogorovs Zero-One law: Corollary: Let X_1, X_2, ... be a sequence of independent random variables, and f a measurable tail function. Then the random variable f(X_1, X_2, ...) is constant almost surely. Now, consider the sequence X, 0, 0, ... of independent random variables. Now let f = X. (This is a tail function since given any tail 0, 0, ..., we still know f = X.) Then the conclusion is: X is constant almost surely. But this cant be right; it says that all random variables are constant almost surely. What did I do wrong? === Subject: Re: zero-one law melnick says... >I read the following corollary to Komogorovs Zero-One law: >Corollary: Let X_1, X_2, ... be a sequence of independent random variables, >and f a measurable tail function. Then the random variable f(X_1, X_2, ...) >is constant almost surely. >Now, consider the sequence X, 0, 0, ... of independent random variables. Now >let f = X. (This is a tail function since given any tail 0, 0, ..., we still >know f = X.) I think that maybe you have the wrong definition of tail function. The idea that I have is that f(X_1, X_2, ...) is a tail function if it is unaffected by any change to the *head*. That is, if Y_i and X_i are two infinite sequences that only differ in finitely many places, then f(X_1, X_2, ...) = f(Y_1, Y_2, ...) Some examples are: 1. f(X_1, X_2, ...) = limit as n->infinity of 1/n * sum from i=0 to n of X_i 2. f(X_1, X_2, ...) = limit as n->infinity of X_n 3. f(X_1, X_2, ...) = smallest real r such that there are infinitely many i such that X_i > r Obviously, f(X_1, X_2, ...) = X_1 is not a tail function, since its value depends on the head. -- Daryl McCullough Ithaca, NY === Subject: Re: zero-one law > I read the following corollary to Komogorovs Zero-One law: > Corollary: Let X_1, X_2, ... be a sequence of independent random variables, > and f a measurable tail function. Then the random variable f(X_1, X_2, ...) > is constant almost surely. > Now, consider the sequence X, 0, 0, ... of independent random variables. Now > let f = X. (This is a tail function since given any tail 0, 0, ..., we still > know f = X.) Then the conclusion is: > X is constant almost surely. > But this cant be right; it says that all random variables are constant > almost surely. What did I do wrong? No it doesnt. Since f(X) is not even defined if X is a random variable. Let E be the expectation operator. Then it is obvious, that f(E[X]) is a constant for all f and X even to probabilty theorists. So Kolmorgovs Corollary concerns E[ f(X) ], for a pre-conditioned f, and X any sequence of random variables. If the conditions on f meet the constraints on E, then it is again obvious that E[ f(X) ] is a constant for all f and X. Hence Kolmorov claims, *do not* let X = X_1, 0, 0, 0, 0. Since then X is not a tail function, but a tail. let X = X_1, X_2, X_3, X_4, .... where X_1, X_2, X_3, X_4 are *independent* random variables. So then we know that both E( X_i * X_j) and E( X_i + X_j ) are identically 0 for all i and j = 1,2,3,.... So we now that f(X) is a constant for any linear f and all X. And since we are sure that given a chance to pick an arbitrary function f, a mathematician will almost definitely do the stupid thing and pick f linear, we also know that f(X) is a constant almost surely. === Subject: Re: zero-one law >I read the following corollary to Komogorovs Zero-One law: >Corollary: Let X_1, X_2, ... be a sequence of independent random variables, >and f a measurable tail function. Then the random variable f(X_1, X_2, ...) >is constant almost surely. >Now, consider the sequence X, 0, 0, ... of independent random variables. Now >let f = X. (This is a tail function No its not. >since given any tail 0, 0, ..., we still >know f = X.) Im not sure exactly what you mean by that, but its not the definition of tail function. If youre actually talking about a _function_ f, that function is defined by f(X1, X2, ...) = X1. Now f(1, 0, 0, ...) is not equal to f(2, 0, 0, ...), which shows that f is _not_ a tail function, because the two sequence (1, 0, 0, ...) and (2, 0, 0, ...) are eventually equal but f takes different values. > Then the conclusion is: >X is constant almost surely. >But this cant be right; it says that all random variables are constant >almost surely. What did I do wrong? ************************ David C. Ullrich === Subject: Re: zero-one law > I read the following corollary to Komogorovs Zero-One law: > Corollary: Let X_1, X_2, ... be a sequence of independent random variables, > and f a measurable tail function. Then the random variable f(X_1, X_2, ...) > is constant almost surely. > Now, consider the sequence X, 0, 0, ... of independent random variables. Now > let f = X. (This is a tail function since given any tail 0, 0, ..., we still > know f = X.) Then the conclusion is: > X is constant almost surely. > But this cant be right; it says that all random variables are constant > almost surely. What did I do wrong? It is because in the corollary, f is a function, not a random variable. (I know my comment is kind of cryptic, but if you think about it long enough you will get it.) === Subject: Re: zero-one law >>It is because in the corollary, f is a function, not a random variable. >>(I know my comment is kind of cryptic, > Since a random variable is in fact a function, one might regard > it as a little more than kind of cryptic... > The point is that the function in the theorem is supposed to have > sequences of reals for its domain, not points in the sample space. >>but if you think about it long enough you >>will get it.) Yes, but I think that he got it. === Subject: Re: zero-one law >> I read the following corollary to Komogorovs Zero-One law: >> Corollary: Let X_1, X_2, ... be a sequence of independent random variables, >> and f a measurable tail function. Then the random variable f(X_1, X_2, ...) >> is constant almost surely. >> Now, consider the sequence X, 0, 0, ... of independent random variables. Now >> let f = X. (This is a tail function since given any tail 0, 0, ..., we still >> know f = X.) Then the conclusion is: >> X is constant almost surely. >> But this cant be right; it says that all random variables are constant >> almost surely. What did I do wrong? >It is because in the corollary, f is a function, not a random variable. >(I know my comment is kind of cryptic, Since a random variable is in fact a function, one might regard it as a little more than kind of cryptic... The point is that the function in the theorem is supposed to have sequences of reals for its domain, not points in the sample space. >but if you think about it long enough you >will get it.) ************************ David C. Ullrich === Subject: Re: zero-one law Hmmm...OK, yeah that comment makes sense. One more question about this zero-one type of stuff. I know at, if X_1, X_2, ... is a sequence of i.r.v.s, then, given a constant r, {limsup (X_1 + ... + X_n)/n > r} is a tail event. Does it follow, then, that {limsup X_1 + ... + X_n > r} is also a tail event? It seems like it would be...I dont really see how not dividing by n -- considering the sum, not the mean average -- untails it. > I read the following corollary to Komogorovs Zero-One law: > Corollary: Let X_1, X_2, ... be a sequence of independent random variables, > and f a measurable tail function. Then the random variable f(X_1, X_2, ...) > is constant almost surely. > Now, consider the sequence X, 0, 0, ... of independent random variables. Now > let f = X. (This is a tail function since given any tail 0, 0, ..., we still > know f = X.) Then the conclusion is: > X is constant almost surely. > But this cant be right; it says that all random variables are constant > almost surely. What did I do wrong? > It is because in the corollary, f is a function, not a random variable. > (I know my comment is kind of cryptic, but if you think about it long enough you > will get it.) === Subject: Re: zero-one law > Hmmm...OK, yeah that comment makes sense. > One more question about this zero-one type of stuff. I know at, if X_1, X_2, > ... is a sequence of i.r.v.s, then, given a constant r, > {limsup (X_1 + ... + X_n)/n > r} > is a tail event. Does it follow, then, that > {limsup X_1 + ... + X_n > r} > is also a tail event? It seems like it would be...I dont really see how not > dividing by n -- considering the sum, not the mean average -- untails it. Well in the first case, X_1/n -> 0 in the lim sup, so the value of X_1 does not effect the set (and similarly for changing any finite number of the Xs). In the second case, changing X_1 will clearly change the set. So it is not a tail event. === Subject: Re: zero-one law >Hmmm...OK, yeah that comment makes sense. >One more question about this zero-one type of stuff. I know at, if X_1, X_2, >... is a sequence of i.r.v.s, then, given a constant r, >{limsup (X_1 + ... + X_n)/n > r} >is a tail event. Does it follow, then, that >{limsup X_1 + ... + X_n > r} >is also a tail event? Certainly not. Say X_1 = 1, X_j = 0 for j > 1. Say all the Y_j = 0. Then the sequences (X_j) and (Y_j) are eventually equal, but limsup X_1 + ... + X_n > 0 while limsup Y1 + ... + Y_n = 0. >It seems like it would be...I dont really see how not >dividing by n -- considering the sum, not the mean average -- untails it. Well, do you know how to _prove_ that {limsup (X_1 + ... + X_n)/n > r} is a tail event? If you know that proof you can look and see where the division by n is important. (And if you dont know how to prove it then you dont know _why_ its a tail event, so you shouldnt expect to see why some detail makes a difference.) >> I read the following corollary to Komogorovs Zero-One law: >> Corollary: Let X_1, X_2, ... be a sequence of independent random >variables, >> and f a measurable tail function. Then the random variable f(X_1, X_2, >...) >> is constant almost surely. >> Now, consider the sequence X, 0, 0, ... of independent random variables. >Now >> let f = X. (This is a tail function since given any tail 0, 0, ..., we >still >> know f = X.) Then the conclusion is: >> X is constant almost surely. >> But this cant be right; it says that all random variables are constant >> almost surely. What did I do wrong? >> It is because in the corollary, f is a function, not a random variable. >> (I know my comment is kind of cryptic, but if you think about it long >enough you >> will get it.) ************************ David C. Ullrich === Subject: Action of Super Cosmos II. However, Kiehns note is more general based on a kind of pre-metrical action principle. One can start with the conjecture that the Action of Super Cosmos has Pfaff topological dimension 4 i.e. turbulent system (chaotic inßation?) But sooner or later one has to add Heisenberg uncertainty and equivalence principle to ßesh out that skeleton. re: http://qedcorp.com/destiny/ All classical considerations of pressure from the kinetic theory of the issue of dark energy and dark matter in cosmology, which are purely quantum effects together with the equivalence principle beyond the scope of classical nonrelativistic physics. w = pressure/energy density Heisenbergs uncertainty principle + equivalence principle imply w = -1 for all zero point vacuum ßuctuations for all dynamical fields. Proof is on p. 26 of John Peacocks Cosmological Physics. In Peter Milonnis book Quantum Vacuum is proof that QED virtual electron-positron pairs have negative energy density hence equal and opposite positive pressure (attractive gravity) and virtual photons have positive energy density hence equal and opposite negative pressure (repulsive gravity - not possible in classical physics) Einsteins theory demands for any stuff Guv ~ (energy density)(1 + 3w) (weak field limit) In particular for zero point energies w = -1. The only issue then is what is the value of the total zero point energy density in each space-time region and on what scale? I have a theory of vacuum coherence on how that works - another topic. We do not seem to need extra dimensions and supersymmetry to understand the observations? More with less. The problem appears to be much simpler than the Pundits all imagine? === Subject: little problem Let xi and eta be two differentiable vector bundles over a same differentiable manifold M. Suppose that each fiber xi_x of xi is isomorphic to the fiber eta_x of eta. Then xi and eta are isomorphic. May you give me some ideas to proof the above proposition? Tern === Subject: Re: little problem > Let xi and eta be two differentiable vector bundles over a same > differentiable manifold M. > Suppose that each fiber xi_x of xi is isomorphic to the fiber eta_x of > eta. > Then xi and eta are isomorphic. > May you give me some ideas to proof the above proposition? With difficulty, Ôcos it aintt true. The tangent bundle of a sphere T(S^2) has the same fibres as the trivial bundle S^2 x R^2 yet these undles are not isomorphic (S^2 is not parallelizable). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: little problem > With difficulty, Ôcos it aintt true. > The tangent bundle of a sphere T(S^2) has the same fibres > as the trivial bundle S^2 x R^2 yet these undles are not > isomorphic (S^2 is not parallelizable). > -- You are true. My proposition isnt true in general. One moment, consider the fiber bundle P^k(xi,eta) whose fiber over x is the space P^k(xi_x,eta_x) of homogeneous polynomial maps of degree k of xi_x to eta_x ; and the fiber bundle L^k_s(xi,eta) whose fiber over x is the space L^k_s(xi_x,eta_x) of k-linear maps of (xi_x) ^k into eta. P^k(xi,eta) and L^k_s(xi,eta) are fiber bundle over the same manifold, say M. After having shown that L^k_s(xi_x,eta_x) and P^k(xi_x,eta_x) are isomorphic, Palais says that it follows that L^k_s(xi,eta) and P^k(xi,eta) are naturally equivalent. May you help me please? Tern === Subject: differential equations lost in finding the necessary differential equation. Any help would be appreciated. Marine Biologists have determined that when a shark detects the presence of blood in the water, it will swim in the direction in which the concentration of the blood increases most rapidly. Based on certain tests in seawater, the concentration of blood (in parts per million) at a point P(x,y) on the surface is approximated by C(x,y)= e^ {(-x^2+2y^2)/10^4}where x and y are measured in meters in a rectangular coordinate system with the blood source at the origin. Suppose a shark is at the point (Xo,Yo) when it first detects the presence of blood in the water. Find an equation of the sharks path by setting up and solving a differential equation. === Subject: Re: differential equations at a > lost in finding the necessary differential equation. Any help would be > appreciated. Marine Biologists have determined that when a shark detects the > presence of blood in the water, it will swim in the direction in which the > concentration of the blood increases most rapidly. Based on certain tests > in seawater, the concentration of blood (in parts per million) at a point > P(x,y) on the surface is approximated by C(x,y)= e^ {(-x^2+2y^2)/10^4}where > x and y are measured in meters in a rectangular coordinate system with the > blood source at the origin. Suppose a shark is at the point (Xo,Yo) when it > first detects the presence of blood in the water. Find an equation of the > sharks path by setting up and solving a differential equation. > The shark should head in the direction of the gradient vector, which at any > point (x,y) is a positive multiple of (-2x,4y). This looks pretty bogus. > For example, if the shark is on the y-axis he should swim directly away > from the blood source? Are you sure it isnt e^{-(x^2+2y^2)/10^4}? Yes,yes, I know all that and I showed my fiend this but the instruction to the problem requested solving this with a differential equation. This is all your help and I would appreciate a reply. === Subject: Re: differential equations > lost in finding the necessary differential equation. Any help would be > appreciated. Marine Biologists have determined that when a shark detects the > presence of blood in the water, it will swim in the direction in which the > concentration of the blood increases most rapidly. Based on certain tests > in seawater, the concentration of blood (in parts per million) at a point > P(x,y) on the surface is approximated by C(x,y)= e^ {(-x^2+2y^2)/10^4}where > x and y are measured in meters in a rectangular coordinate system with the > blood source at the origin. Suppose a shark is at the point (Xo,Yo) when it > first detects the presence of blood in the water. Find an equation of the > sharks path by setting up and solving a differential equation. The shark should head in the direction of the gradient vector, which at any point (x,y) is a positive multiple of (-2x,4y). This looks pretty bogus. For example, if the shark is on the y-axis he should swim directly away from the blood source? Are you sure it isnt e^{-(x^2+2y^2)/10^4}? === Subject: Re: differential equations >lost in finding the necessary differential equation. Any help would be >appreciated. Marine Biologists have determined that when a shark detects the >presence of blood in the water, it will swim in the direction in which the >concentration of the blood increases most rapidly. Based on certain tests >in seawater, the concentration of blood (in parts per million) at a point >P(x,y) on the surface is approximated by C(x,y)= e^ {(-x^2+2y^2)/10^4}where >x and y are measured in meters in a rectangular coordinate system with the >blood source at the origin. Suppose a shark is at the point (Xo,Yo) when it >first detects the presence of blood in the water. Find an equation of the >sharks path by setting up and solving a differential equation. Hint: Grad(C) is proportional to <-2x, 4y> so you need dy/dx = -2y/x for the orthogonal trajectories to the level curves of C... --Lynn === Subject: Re: differential equations > lost in finding the necessary differential equation. Any help would be > appreciated. Marine Biologists have determined that when a shark detects the > presence of blood in the water, it will swim in the direction in which the > concentration of the blood increases most rapidly. Based on certain tests > in seawater, the concentration of blood (in parts per million) at a point > P(x,y) on the surface is approximated by C(x,y)= e^ {(-x^2+2y^2)/10^4}where > x and y are measured in meters in a rectangular coordinate system with the > blood source at the origin. Suppose a shark is at the point (Xo,Yo) when it > first detects the presence of blood in the water. Find an equation of the > sharks path by setting up and solving a differential equation. direction of most rapid increase is related to gradient. Look it up in a multivariable calculus text. === Subject: Re: Who can tell me the recent develepoment on the GCH? >W. Huge Woodin said he gave a solution that yield CH is false. What >was it gonging on? Hugh Woodin has a program with proposed axioms under which the power of the continuum comes out to be Aleph_2. --hbe === Subject: Re: No Set Contains Every Computable Natural :> An outright admission that a TM cant decide if a string is infinite. :> Duh. : But a human can, right? Can you? Tell me if the following string is infinite or not. Stephen === Subject: Re: No Set Contains Every Computable Natural > So there is no way to determine if a string is infinite? > Not according to the scope in this thread. Of course if you have a human > holding both ends of the tape.... > Actually theres another question, if you have a theoretical peice of > string, and you have an end in each hand but the middle of it extends to > an infinite distance away from you before coming back is it of finte > length? Does such a string even exist? > Why do you believe some strings are finite and other > strings are infinite if there is no possible way to tell > the difference? > We can proove the exitence of an infinte set. We cant (by it > definition) give you a peice of paper with every member listed on it as > it is impossible to list the second to last member. You cant prove that an infinite set exists. You have to assume it exists. This is why ZFC has the Axiom of Infinity. You can prove that a set, if it exists, can not be finite. Lets look at some of these type of proofs. We are defining an algorithm as anything that can be computed by a TM in a finite number of steps. Since we assume a TM requires a step to read or write a symbol, this requires the TMs input and output to be finite. Consider Euclids proof of the infinitude of primes. The input is a finite number of primes. The algorithm consists of multiplying a finite number of primes together. This can certainly be done in a finite number of steps. The output is the product which is finite. Euclids proof fits our definition of algorithm. Now consider Cantors diagonal proof. The input is an infinite list of real numbers. Obviously, this list is not finite. The algorithm requires that we compute an infinite number of digits. Again, not finite. The output is a real number. This number probably doesnt have a finite number of digits. Cantors proof completely fails to be an algorithm. There is no way that Cantors proof that the reals are uncountable can be computed by a TM in a finite number of steps. If we redefine algorithm to include TMs that can perform an infinite number of operations, we have to deal with the many paradoxes that result. For example, Cantors first proof of uncountability can be easily modified to find the smallest positive real number. There is my proof that the natural numbers are uncountable. The list is endless. If we accept Cantors proof then we must also accept these other proofs as well. Russell - 2 many 2 count